Concatenating bipartite graphs

TL;DR

This paper studies a function related to bipartite graph concatenation, exploring its properties, symmetries, and discontinuities, and raises open questions about its behavior.

Contribution

It introduces the function (x,y) for bipartite graph concatenation, analyzes its properties, symmetry, and points of discontinuity, and proposes new questions and conjectures.

Findings

(x,y) is symmetric in x and y

(x,y) has a discontinuity at x=y=1/k for integers k>1

The paper raises open questions about the function's properties

Abstract

Let $x,y\in(0,1]$ and let $A,B,C$ be disjoint nonempty subsets of a graph $G$, where every vertex in $A$ has at least $x|B|$ neighbours in $B$, and every vertex in $B$ has at least $y|C|$ neighbours in $C$. We denote by $\phi(x,y)$ the maximum $z$ such that, in all such graphs $G$, there is a vertex $v$ in $C$ that is joined to at least $z|A|$ vertices in $A$ by two-edge paths. The function $\phi$ is interesting, and we investigate some of its properties. For instance, we show that it is symmetric in $x$ and $y$, and that it has a discontinuity at $x=y=1/k$ for all integers $k>1$. We raise a number of questions and conjectures.

Equations

$$\varphi (x,y)\le \frac{\lceil kx\rceil +\lceil ky\rceil -1}{k}$$

$$\max (x,y)\le \varphi (x,y)\le \psi (x,y)\le \xi (x,y)\le \frac{\lceil kx\rceil +\lceil ky\rceil -1}{k}$$

$$y=\sum _{v\in B}y{w}^{\prime }(v)\ge s+\sum _{v\in B}\sum _{u\in N(v)}{w}^{\prime }(v)f(u)=s+\sum _{u\in A}\sum _{v\in N(u)}f(u)w^{\prime }(v)\ge s+\sum _{u\in A}yf(u)=s+y.$$

$$w(a)$$

$$w(v)$$

$$w(b)$$

$$w(v)$$

$$w(c)$$

$$w(v)$$

$$\varphi (x,y)\le \frac{\lceil kx\rceil +\lceil ky\rceil -1}{k}=x+y-1/k<1.$$

$$\frac{x^{\prime }}{1-(k-1)x^{\prime }}+\frac{y^{\prime }}{1-(k-1)y^{\prime }}=\frac{x}{1-kx}+\frac{y}{1-ky}\le 1,$$

$$A$$

$$B$$

$$C$$

$$krN=1/k-x.$$

$$x$$

$$y$$

$$x$$

$$1/k$$

$$x\le \frac{z-z^{\prime }+x^{\prime }(1-z)}{1-z^{\prime }},$$

$$x\le 1/(2-x^{\prime })\le 1/(2-z^{\prime })\le z.$$

$$w(a)$$

$$w(v)$$

$$w(b)$$

$$w(v)$$

$$w(c)$$

$$w(v)$$

$$\max \left(1-z,\frac{x-x^{\prime }}{1-x^{\prime }}\right)\le p$$

$$\max \left(y,\frac{x-x^{\prime }}{1-x^{\prime }}\right)\le q$$

$$w(a)$$

$$w(v)$$

$$w(b)$$

$$w(v)$$

$$w(c)$$

$$w(v)$$

$$\max \left(x,1-z\right)\le p$$

$$\max \left(x,\frac{y-y^{\prime }}{1-y^{\prime }}\right)\le q$$

$$w(a_i)$$

$$w(a_i^{\prime })$$

$$w(a^{\ast })$$

$$w(b_i)$$

$$w(b_i^{\prime })$$

Full text

• •

$\phi(x,y)=\phi(y,x)$ for all $x,y$; and

• •

for each integer $k>1$, there is a discontinuity in $\phi(x,x)$ when $x=1/k$: $\phi(x,x)\leq 1/k$ when $x\leq 1/k$, and $\phi(x,x)\geq\frac{2k-1}{2k(k-1)}$ when $x>1/k$.

We raise several questions and conjectures.

1 Introduction

All graphs in this paper are finite, and have no loops or multiple edges. We denote the semi-open interval $\{x:0<x\leq 1\}$ of real numbers by $(0,1]$. Let $x,y\in(0,1]$; and let $A,B,C$ be disjoint nonempty subsets of a graph $G$, where every vertex in $A$ has at least $x|B|$ neighbours in $B$, and every vertex in $B$ has at least $y|C|$ neighbours in $C$. If we ask for a real number $z$ such that we can guarantee that some vertex in $A$ can reach at least $z|C|$ vertices in $C$ by two-edge paths, then $z$ must be at most $y$, since perhaps all the vertices in $B$ have the same neighbours in $C$. But in the reverse direction the question becomes much more interesting; that is, we ask for $z$ such that some vertex in $C$ can reach at least $z|A|$ vertices in $A$ by two-edge paths. Then there might well be values of $z>\max(x,y)$ with this property.

Let us say this more precisely. A tripartition of a graph $G$ is a partition $(A,B,C)$ of $V(G)$ where $A,B,C$ are all nonempty stable sets. For $x,y\in(0,1]$, we say a graph $G$ is $(x,y)$-constrained, via a tripartition $(A,B,C)$, if

• •

every vertex in $A$ has at least $x|B|$ neighbours in $B$;

• •

every vertex in $B$ has at least $y|C|$ neighbours in $C$; and

• •

there are no edges between $A$ and $C$.

For $v\in V(G)$, $N(v)$ denotes its set of neighbours, and $N^{2}(v)$ is the set of vertices with distance exactly two from $v$. We write $N^{2}_{A}(v)$ for $N^{2}(v)\cap A$, and so on. A first observation:

1.1

Let $x,y\in(0,1]$, and let $Z$ be the set of all $z\in(0,1]$ such that, for every graph $G$, if $G$ is $(x,y)$-constrained via $(A,B,C)$ then $|N^{2}_{A}(v)|\geq z|A|$ for some $v\in C$. Then $\sup\{z\in Z\}$ belongs to $Z$.

Proof.  Let $z^{\prime}=\sup\{z\in Z\}$, and let $G$ be an $(x,y)$-constrained graph, via $(A,B,C)$. We must show that $|N^{2}_{A}(v)|\geq z^{\prime}|A|$ for some $v\in C$. We may assume that $z^{\prime}>0$; so there exists $z$ with $0<z<z^{\prime}$, such that $\lceil z|A|\rceil=\lceil z^{\prime}|A|\rceil$. Since $z^{\prime}=\sup\{z\in Z\}$ and $z<z^{\prime}$, and $Z$ is an initial interval of $(0,1]$, it follows that $z\in Z$, and so $|N^{2}_{A}(v)|\geq z|A|$ for some $v\in C$. Consequently $|N^{2}_{A}(v)|\geq\lceil z|A|\rceil\geq z^{\prime}|A|$, as required. This proves 1.1.

We define $\phi(x,y)$ to be $\sup\{z\in Z\}$, as defined in 1.1. The objective of this paper is to study the properties of the function $\phi$. We have a trivial lower bound:

1.2

$\phi(x,y)\geq\max(x,y)$* for all $x,y>0$.*

Proof.  Let $G$ be $(x,y)$-constrained, via $(A,B,C)$. Since every vertex in $A$ has at least $x|B|$ neighbours in $B$, and $B\neq\emptyset$, there exists $u\in B$ with at least $x|A|$ neighbours in $A$; let $v\in C$ be adjacent to $u$ (this is possible since $y>0$), and then $|N^{2}_{A}(v)|\geq x|A|$. Consequently $\phi(x,y)\geq x$. Now every vertex in $A$ can reach at least $y|C|$ vertices in $C$ by two-edge paths (since $x>0$); and so by averaging, some vertex in $C$ can reach at least $y|A|$ vertices in $A$ by two-edge paths. Hence $\phi(x,y)\geq y$. This proves 1.2.

And a trivial upper bound:

1.3

For all $x,y\in(0,1]$,

[TABLE]

for every integer $k\geq 1$.

Proof.  Let $x,y\in(0,1]$, and let $k\geq 1$ be an integer. Let $A,B,C$ be three disjoint sets each of cardinality $k$, where $A=\{a_{1},\ldots,a_{k}\}$, $B=\{b_{1},\ldots,b_{k}\}$ and $C=\{c_{1},\ldots,c_{k}\}$. Make a graph $G$ with vertex set $A\cup B\cup C$ as follows. Let $g=\lceil kx\rceil$, and for $1\leq i\leq k$ make $a_{i}$ adjacent to $b_{i},b_{i+1},\ldots,b_{i+g-1}$ (reading subscripts modulo $k$). Now let $h=\lceil ky\rceil$, and for $1\leq i\leq k$ make $b_{i}$ adjacent to $c_{i},c_{i+1},\ldots,c_{i+h-1}$ (reading subscripts modulo $k$). Then $G$ is $(x,y)$-constrained via $(A,B,C)$; and for $1\leq i\leq k$, $N^{2}_{A}(c_{i})=\{a_{i},a_{i-1},\ldots,a_{i-g-h+2}\}$ (again, reading subscripts modulo $k$). Consequently $\phi(x,y)\leq(g+h-1)/k$. This proves 1.3.

In particular, we have:

1.4

For every integer $k\geq 1$, if $x,y>0$ and $\max(x,y)=1/k$ then $\phi(x,y)=1/k$.

Proof.  From 1.2, $\phi(x,y)\geq 1/k$; and the graph consisting of $k$ disjoint three-vertex paths shows that $\phi(x,y)\leq 1/k$. (This also follows from 1.3, since $\lceil kx\rceil,\lceil ky\rceil=1$.) This proves 1.4.

What makes the function $\phi$ interesting is that for some values of $x,y$, 1.2 is far from best possible, and indeed 1.3 seems closer to the truth. We were originally motivated by the hope of extending Kneser’s theorem from additive group theory [5] to a general graph-theoretic setting, and a corresponding wild conjecture that the bound in 1.3 is always best possible, that is, that for all $x,y\in(0,1]$, there is an integer $k>0$ with $\phi(x,y)=\frac{\lceil kx\rceil+\lceil ky\rceil-1}{k}$. This turns out to be false, but perhaps not ridiculously false; maybe something like it is true.

There are two other related problems:

• •

Let us say $G$ is $(x,y)$-biconstrained (via $(A,B,C)$) if $G$ is $(x,y)$-constrained via $(A,B,C)$, and in addition

• –

every vertex in $B$ has at least $x|A|$ neighbours in $A$, and

• –

every vertex in $C$ has at least $y|B|$ neighbours in $B$.

• •

Say $G$ is $(x,y)$-exact (via $(A,B,C)$) if $G$ is $(x,y)$-constrained via $(A,B,C)$, and in addition there exist $x^{\prime}\geq x$ and $y^{\prime}\geq y$ such that

• –

every vertex in $A$ has exactly $x^{\prime}|B|$ neighbours in $B$;

• –

every vertex in $B$ has exactly $x^{\prime}|A|$ neighbours in $A$;

• –

every vertex in $B$ has exactly $y^{\prime}|C|$ neighbours in $C$; and

• –

every vertex in $C$ has exactly $y^{\prime}|B|$ neighbours in $B$.

We shall sometime use “mono-constrained” to clarify that we mean the $(x,y)$-constrained case and not the $(x,y)$-biconstrained case. Let $\psi(x,y)$ be the analogue of $\phi(x,y)$ for biconstrained graphs; that is, the maximum $z$ such that for all $G$, if $G$ is $(x,y)$-biconstrained via $(A,B,C)$, then $|N^{2}_{A}(v)|\geq z|A|$ for some $v\in C$. (As before, this maximum exists.) Similarly, let $\xi(x,y)$ be the analogue of $\phi$ and $\psi$ for the exact case. Then we have

1.5

For all $x,y\in(0,1]$,

[TABLE]

for every integer $k\geq 1$.

The proof of the non-trivial part of this is the same as the proof of 1.3. One might hope that $\psi$ (and even more $\xi$) are better-behaved than $\phi$.

Let us see an example. Start with the graph of figure 1. Each vertex has a number written next to it in the figure; replace each vertex $v$ by a set $X_{v}$ of new vertices of the specified cardinality, and for each edge $uv$ of the figure make every vertex in $X_{u}$ adjacent to every vertex in $X_{v}$. This results in a graph with $81$ vertices, divided into three sets of $27$ corresponding to the three rows of the figure; call these $A,B,C$. The graph produced is $(13/27,1/9)$-biconstrained via $(A,B,C)$, and yet $|N^{2}_{A}(v)|=13$ for every vertex $v\in C$; so this proves that $\psi(13/27,1/9)\leq 13/27$ (and therefore equality holds, by 1.2). This shows that there need not exist an integer $k$ with $\psi(x,y)=\frac{\lceil kx\rceil+\lceil ky\rceil-1}{k}$. The same graph, used from bottom to top, shows that $\psi(1/9,13/27)=13/27$.

The example is not yet $(13/27,1/9)$-exact, because some vertices in $A$ have three, four or five neighbours in $B$, and vice versa. We can make it exact as follows. For each edge $uv$ of the figure with $u$ in the second row and $v$ in the third, the two sets $X_{u},X_{v}$ have the same cardinality, one of three, four, five. Delete some edges between $X_{u}$ and $X_{v}$ such that every vertex in $X_{u}$ has exactly three neighbours in $X_{v}$ and vice versa. Then the modified graph is $(13/27,1/9)$-exact, and shows that $\xi(13/27,1/9)=13/27$. Consequently, even for the supposedly nicest function $\xi$ of our three functions, there is not always an integer $k$ with $\xi(x,y)=\frac{\lceil kx\rceil+\lceil ky\rceil-1}{k}$.

So what can we prove about the functions $\phi$ and $\psi$? For which $x,y,z$ is $\phi(x,y)\geq z$, or $\psi(x,y)\geq z$? In order to make the question a little more manageable, we focus on seven special cases, $x=y$, and $z=1/2,2/3,1/3,3/4,2/5,3/5$, but in each case the results for $\phi$ and for $\psi$ are quite different. The paper is organized as follows:

• •

We begin with a proof that $\phi(x,y)=\phi(y,x)$ for all $x,y$.

• •

Then we give some general upper bounds on $\phi(x,y)$ and $\psi(x,y)$, particularly focussing on the case when $x=y$.

• •

Next we consider when $\phi(x,y)\geq 1/2$, or $\psi(x,y)\geq 1/2$. There are several theorems that this is true for certain pairs $(x,y)$, and their union fills a good part of the $(x,y)$-square. We also give a number of constructions that shows the statement is not true for certain pairs $(x,y)$. Ideally this would fill the complementary part of the square, but there is an “undecided” band of varying width down the middle.

• •

Then we do the same for $2/3$ instead of $1/2$; and then for $1/3,3/4,2/5,3/5$.

• •

Finally, we discuss some other questions and approaches.

Some of our results appear in [3] and [4].

2 Weighted graphs and some linear programming

In this section we prove that $\phi(x,y)=\phi(y,x)$ for all $x,y$. The argument uses linear programming, and we need some preparation. We denote the set of real numbers by $\mathbb{R}$, and the non-negative reals numbers by $\mathbb{R}_{+}$. A weighted graph $(G,w)$ consists of a graph $G$ together with a function $w:V(G)\rightarrow\mathbb{R}_{+}$. If $X\subseteq V(G)$, we denote $\sum_{v\in X}w(v)$ by $w(X)$. Let $(G,w)$ be a weighted graph, and $(A,B,C)$ a tripartition of $G$. If $x,y\in(0,1]$, a weighted graph $(G,w)$ is $(x,y)$-constrained via $(A,B,C)$, if:

• •

$\sum_{v\in A}w(v)=\sum_{v\in B}w(v)=\sum_{v\in C}w(v)=1$;

• •

for each $v\in A$, $w(N(v)\cap B)\geq x$; and

• •

for each $v\in B$, $w(N(v)\cap C)\geq y$.

Similarly, we say $(G,w)$ is $(x,y)$-biconstrained via $(A,B,C)$, if in addition:

• •

for each $v\in B$, $w(N(v)\cap A)\geq x$; and

• •

for each $v\in C$, $w(N(v)\cap B)\geq y$.

To make the graph of figure 1 into an appropriate weighted graph, divide all the numbers by 27.

2.1

For $x,y,z\in(0,1]$, the following are equivalent:

• •

$\phi(x,y)\geq z$;

• •

$w(N^{2}_{A}(v))\geq z$* for some $v\in C$, for every weighted graph $(G,w)$ that is $(x,y)$-constrained via a tripartition $(A,B,C)$.*

Similarly, the following are equivalent:

• •

$\psi(x,y)\geq z$;

• •

$w(N^{2}_{A}(v))\geq z$* for some $v\in C$, for every weighted graph $(G,w)$ that is $(x,y)$-biconstrained via a tripartition $(A,B,C)$.*

Proof.  To prove the “if” direction of the first statement, let $G$ be $(x,y)$-constrained via $(A,B,C)$. Define $w(v)=1/|A|$, for each $v\in A$, and $w(v)=1/|B|$ for $v\in B$ and similarly for $v\in C$. Then $(G,w)$ is an $(x,y)$-constrained weighted graph, and the claim follows. The “if” direction of the second statement is proved similarly.

For the “only if” direction, let $(G,w)$ be a weighted graph, $(x,y)$-constrained via $(A,B,C)$, and suppose such a weighted graph can be chosen with $w(N^{2}_{A}(v))<z$ for each $v\in C$. Consequently we may choose $(G,w)$ such that in addition, $w$ is rational-valued. Choose an integer $N>0$ such that $Nw(v)$ is an integer for each $v\in G$. For each $v\in V(G)$, take a set $X_{v}$ of $Nw(v)$ new vertices; and make a graph $G^{\prime}$ with vertex set $\bigcup_{v\in V(G)}X_{v}$, by making every vertex of $X_{u}$ adjacent to every vertex of $X_{v}$ for all adjacent $u,v\in V(G)$. Let $A^{\prime}=\bigcup_{v\in A}X_{v}$, and define $B^{\prime},C^{\prime}$ similarly; then $(A^{\prime},B^{\prime},C^{\prime})$ is a tripartition of $G^{\prime}$, and $G^{\prime}$ is $(x,y)$-constrained via $(A^{\prime},B^{\prime},C^{\prime})$. Since in $G$, $w(N^{2}_{A}(v))<z$ for each $v\in C$, it follows that in $G^{\prime}$, $|N^{2}_{A^{\prime}}(v^{\prime})|<z|A^{\prime}|$ for each $v^{\prime}\in C^{\prime}$, a contradiction. The “only if” direction of the second statement is similar. This proves 2.1.

Let $G$ be a graph with a bipartition $(A,B)$, and let $w:B\rightarrow\mathbb{R}_{+}$ be some function. We define $w(A\rightarrow B)$ to mean the minimum, over all $u\in A$, of $w(N(u))$ (taking $w(A\rightarrow B)=0$ if $A=\emptyset$).

2.2

Let $G$ be a graph with a bipartition $(A,B)$, and let $w:B\rightarrow\mathbb{R}_{+}$ be some function such that $w(B)=1$. Then either

• •

there is a function $w^{\prime}:B\rightarrow\mathbb{R}_{+}$, such that $w^{\prime}(B)=1$ and $w^{\prime}(A\rightarrow B)\geq w(A\rightarrow B)$, and such that $w^{\prime}(v)=0$ for some $v\in B$; or

• •

there is a function $f:A\rightarrow\mathbb{R}_{+}$, such that $f(A)=1$ and $f(B\rightarrow A)\geq w(A\rightarrow B)$.

Proof.  We may assume that $A\neq\emptyset$. If some vertex in $A$ has no neighbour in $B$, then $w(A\rightarrow B)=0$ and the second bullet holds; so we assume that each vertex in $A$ has a neighbour in $B$.

Let $x=w(A\rightarrow B)$. The function $w^{\prime}$, defined by $w^{\prime}(v)=1/|B|$ for each $v\in B$, satisfies $w^{\prime}(A\rightarrow B)>0$, since every vertex in $A$ has a neighbour in $B$. Thus we may assume that $x>0$, replacing $w$ by $w^{\prime}$ if necessary.

Let $M$ be the $0/1$-matrix $(a_{uv}:u\in A,v\in B)$, where $a_{uv}=1$ if and only if $u,v$ are adjacent. Let ${\bf 1}_{A}\in\mathbb{R}^{A}$ be the vector of all $1$’s, and define ${\bf 1}_{B}$ similarly. Then $w\in\mathbb{R}_{+}^{B}$ satisfies:

• •

${\bf 1}_{B}^{T}w=1$; and

• •

$Mw\geq x{\bf 1}_{A}$.

Consequently $b=w/x$ satisfies $b\in\mathbb{R}_{+}^{B}$, and

• •

${\bf 1}_{B}^{T}b=1/x$; and

• •

$Mb\geq{\bf 1}_{A}$.

Choose $q\in\mathbb{R}_{+}^{B}$ with $Mq\geq{\bf 1}_{A}$, with ${\bf 1}_{B}^{T}q$ minimum. (This is possible by compactness.) Thus ${\bf 1}_{B}^{T}q\leq 1/x$. Since $Mq\geq{\bf 1}_{A}$ and $G$ has an edge, it follows that ${\bf 1}_{B}^{T}q>0$; let $1/y={\bf 1}_{B}^{T}q$, and define $w^{\prime}=yq$. Then $y\geq x$, and ${\bf 1}_{B}^{T}w^{\prime}=1$ and $Mw^{\prime}\geq y{\bf 1}_{A}$, and so we may assume that $w^{\prime}(v)>0$ for each $v\in B$, because otherwise the first bullet holds.

Now $q$ minimizes ${\bf 1}_{B}^{T}q$ subject to the linear programme $q\in\mathbb{R}_{+}^{B}$ and $Mq\geq{\bf 1}_{A}$. From the linear programming duality theorem, there exists $p\in\mathbb{R}_{+}^{A}$ such that $p^{T}M\leq{\bf 1}_{B}^{T}$, and $p^{T}{\bf 1}_{A}={\bf 1}_{B}^{T}q=1/y$. Define $f=yp$. Then $f:A\rightarrow\mathbb{R}_{+}$ satisfies $f(A)=1$, and $f(N(v))\leq y$ for each $v\in B$.

Let $v^{\prime}\in B$; we claim that $f(N(v^{\prime}))=y$. This follows from the “complementary slackness” principle, but we give the argument in full, as follows. Let $s=w^{\prime}(v^{\prime})(y-f(N(v^{\prime})))$. Thus $s\geq 0$, and we will show $s=0$. We have

[TABLE]

Consequently $s=0$, as claimed. Hence $f$ satisfies the second bullet. This proves 2.2.

From 2.2 we deduce a very useful result.

2.3

If $x,y\in(0,1]$ then $\phi(x,y)=\phi(y,x)$.

Proof.  Let $z=\phi(x,y)$, and choose a weighted graph $(G,w)$ that is $(x,y)$-constrained via $(A,B,C)$, such that $w(N^{2}_{A}(v))\leq z$ for each $z\in C$. Moreover, choose $G$ with $|V(G)|$ minimum. If there is a function $w^{\prime}:B\rightarrow\mathbb{R}_{+}$, such that $w^{\prime}(B)=1$ and $w^{\prime}(A\rightarrow B)\geq w(A\rightarrow B)$, and such that $w^{\prime}(v)=0$ for some $v\in B$, then we may replace $w$ by a new weight function, changing $w$ to $w^{\prime}$ on $B$ and otherwise keeping $w$ unchanged, and then we may delete the vertex $v\in B$ with $w^{\prime}(v)=0$, contrary to the minimality of $|V(G)|$. Thus there is no such $w^{\prime}$, and so by 2.2, there is a function $f:A\rightarrow\mathbb{R}_{+}$, such that $f(A)=1$ and $f(B\rightarrow A)\geq w(A\rightarrow B)\geq x$. Similarly, there is a function $g:B\rightarrow\mathbb{R}_{+}$, such that $g(B)=1$ and $g(C\rightarrow B)\geq y$. Let $H$ be the graph with bipartition $(A,C)$ in which $u\in A$ and $v\in C$ are adjacent if $u\notin N^{2}_{A}(v)$ in $G$. Thus, in $H$, $w(C\rightarrow A)\geq 1-z$; and so from 2.2 and the minimality of $|V(G)|$, there is a function $h:C\rightarrow\mathbb{R}_{+}$, such that $h(C)=1$ and (in $H$) $h(A\rightarrow C)\geq 1-z$. Let $w^{\prime}$ be defined by the union of $f,g$ and $h$ in the natural sense; then $(G,w^{\prime})$ is a weighted graph and is $(y,x)$-constrained via $(C,B,A)$, and $w^{\prime}(N^{2}_{C}(v))\leq z$ for each $v\in A$. This proves that $\phi(y,z)\leq z$, and so proves 2.3.

We remark that we have not been able to prove an analogue of 2.3 for the biconstrained case, or for the exact case, although we have no counterexample for either one.

There is another useful application of 2.2, the following:

2.4

Let $(G,w)$ be an $(x,y)$-constrained weighted graph, via $(A,B,C)$, such that $w(N^{2}_{A}(v))\leq z$ for each $v\in C$. Suppose that there exists $X\subseteq A$ with $|X|<z^{-1}$ such that $\bigcup_{v\in X}N^{2}_{C}(v)=C$. Then there exists $u\in A$ and a weighted graph $(G^{\prime},w^{\prime})$ such that

• •

$G^{\prime}$* is obtained from $G$ by deleting $u$;*

• •

$(G^{\prime},w^{\prime})$* is $(x,y)$-constrained via $(A^{\prime},B,C)$, where $A^{\prime}=A\setminus\{u\}$;*

• •

in $G^{\prime}$, $w^{\prime}(N^{2}_{A^{\prime}}(v))\leq z$ for all $v\in C$; and

• •

$w^{\prime}(u)=w(u)$* for all $u\in B\cup C$.*

Proof.  Suppose not. Let $H$ be the graph with bipartition $(A,C)$, in which $u\in A$ and $v\in C$ are adjacent if $u\notin N^{2}_{A}(v)$ in $G$. Then by 2.2, applied to $H$, there is a function $h:C\rightarrow\mathbb{R}_{+}$, such that $h(C)=1$ and (in $H$) $h(A\rightarrow C)\geq 1-z$. Consequently, in $G$, $h(N^{2}_{C}(v))\leq z$ for each $v\in A$. In particular, $h(N^{2}_{C}(v))\leq z$ for each $v\in X$, and so $h(C)\leq z|A|<1$, a contradiction. This proves 2.4.

Let $x,y,z\in(0,1]$. We say that $(x,y,z)$ is triangular if no triangle-free graph $G$ admits a tripartition $A,B,C$ of $V(G)$ with the following properties:

• •

$A,B,C$ are nonempty stable sets;

• •

every vertex in $A$ has at least $x|B|$ neighbours in $B$;

• •

every vertex in $B$ has at least $y|C|$ neighbours in $C$; and

• •

every vertex in $C$ has at least $z|A|$ neighbours in $A$.

It is possible to reformulate results about $\phi(x,y)$ in terms of triangular triples, because we have:

2.5

For $x,y,z\in(0,1]$, $\phi(x,y)>1-z$ if and only if $(x,y,z)$ is triangular. Consequently the three statements $\phi(x,y)\leq 1-z$, $\phi(z,x)\leq 1-y$, and $\phi(y,z)\leq 1-x$ are equivalent.

Proof.  Suppose that $(x,y,z)$ is not triangular. Then there is a triangle-free graph $G$ with a tripartition $(A,B,C)$, satisfying the three bullets in the definition of “triangular”. Let $H$ be the subgraph of $G$ with $V(H)=V(G)$, obtained by deleting all edges between $A$ and $C$. If $v\in C$, then $N^{2}_{A}(v)$ (defined with respect to $H$) contains only vertices in $A$ that are nonadjacent to $v$ in $G$, since $G$ is triangle-free; and so $|N^{2}_{A}(v)|\leq|A|-z|A|$, since in $G$, $v$ has at least $z|A|$ neighbours in $A$. Consequently $\phi(x,y)\leq 1-z$.

For the reverse implication, suppose that $\phi(x,y)\leq 1-z$, and let $H$ be $(x,y)$-constrained via $(A,B,C)$, such that $|N^{2}_{A}(v)|\leq|A|-z|A|$ for each $v\in C$. Make a graph $G$ by adding certain edges to $H$, namely for each $v\in C$ and $u\in A$, add an edge $uv$ if $u\notin N^{2}_{A}(v)$. Then $G$ is triangle-free, and every vertex $v\in C$ is adjacent in $G$ to at least $|A|-(1-z)|A|=z|A|$ vertices in $A$; and so $(x,y,z)$ is not triangular.

In particular, $(x,y,z)$ is triangular if and only if $(z,x,y)$ is triangular; so it follows that $\phi(x,y)\leq 1-z$ if and only if $\phi(z,x)\leq 1-y$, and similarly if and only if $\phi(y,z)\leq 1-x$. This proves 2.5.

We call the equivalence of the second statement of 2.5 “rotating”.

3 Constructions

In this section we construct some graphs to prove upper bounds on $\phi(x,y)$ or $\psi(x,y)$ for certain values of $x,y$. We begin with:

3.1

Let $x,y\in(0,1]$, and let $z\in(0,1]$ such that $z/(1-z)=\phi(x/(1-x),y/(1-y))$; then $\phi(x,y)\leq z$.

Proof.  Let $(G^{\prime},w^{\prime})$ be a weighted graph that is $(x/(1-x),y/(1-y))$-constrained via some tripartition $(A^{\prime},B^{\prime},C^{\prime})$, such that $w^{\prime}(N^{2}_{A^{\prime}}(v))\leq z/(1-z)$ for each $v\in C^{\prime}$. Add three new vertices $a,b,c$ to $G^{\prime}$, and two edges $ab$ and $bc$, forming $G$. Define $w$ by

[TABLE]

Then $G$ is $(x,y)$-constrained via $(A^{\prime}\cup\{a\},B^{\prime}\cup\{b\},C^{\prime}\cup\{c\})$ and shows that $\phi(x,y)\leq z$. This proves 3.1.

3.2

Let $k\geq 0$ be an integer, and let $x,y\in(0,1]$ with $\frac{x}{1-kx}+\frac{y}{1-ky}\leq 1$, with strict inequality if $x$ or $y$ is irrational; then $\phi(x,y)<\frac{1}{k+1}$.

Proof.  By increasing $x$ and $y$ if necessary, we may assume that $x,y$ are rational. Suppose first that $k=0$; then we may assume that $x+y=1$. Choose an integer $k\geq 1$ such that $kx$ (and hence $ky$) is an integer. By 1.3,

[TABLE]

This completes the proof for $k=0$. For general $k$ we proceed by induction on $k$. We may assume that $k>0$; let $x,y\in(0,1]$ with $\frac{x}{1-kx}+\frac{y}{1-ky}\leq 1$, with strict inequality if $x$ or $y$ is irrational. Let $x^{\prime}=x/(1-x)$, and $y^{\prime}=y/(1-y)$. Thus $x^{\prime},y^{\prime}\in(0,1]$ with

[TABLE]

with strict inequality if $x^{\prime}$ or $y^{\prime}$ is irrational. From the inductive hypothesis, $\phi(x^{\prime},y^{\prime})<1/k$. Let $z$ satisfy $z/(1-z)=\phi(x^{\prime},y^{\prime})$; then $z/(1-z)<1/k$, and so $z<1/(k+1)$. From 3.1, $\phi(x,y)\leq z<1/(k+1)$. This proves 3.2.

3.3

Let $k\geq 0$ be an integer, and let $x,y\in(0,1]$ with $x+ky\leq 1$ and $kx+y\leq 1$, with strict inequality in both if $x$ or $y$ is irrational; then $\psi(x,y)<\frac{1}{k}$.

Proof.  Again, we may assume that $x,y$ are rational. Let $s=\max(x,y)$; thus, $s<1/k$. Choose an integer $N\geq 1$ such that $p=xN/(1-(k-1)s)$ and $q=yN/(1-(k-1)s)$ are integers. (It follows that $p+q\leq N$, from the hypothesis.) Let $G$ be a graph with vertex set partitioned into three sets $A,B,C$, with $|A|=N+k$ and $|B|=|C|=N+k-1$; let

[TABLE]

Let $G$ have the following edges:

• •

for $1\leq i\leq N$, $a_{i}$ is adjacent to $b_{i},b_{i+1},\ldots,b_{i+p-1}$ reading subscripts modulo $N$;

• •

for $1\leq i\leq N$, $b_{i}$ is adjacent to $c_{i},c_{i+1},\ldots,c_{i+q-1}$ reading subscripts modulo $N$;

• •

for $1\leq i\leq k-1$, $a_{i}^{\prime}$ is adjacent to $b_{i}^{\prime}$, and $b_{i}^{\prime}$ is adjacent to $c_{i}^{\prime}$.

• •

$a^{*}$ is adjacent to $b_{i}$ for $1\leq i\leq N$.

(Thus, this is the same as in the proof of 3.2, except for the extra vertex $a^{*}$.) Let $r$ satisfy

[TABLE]

Thus $r>0$. For each $v\in V(G)$, define $w(v)$ as follows:

• •

$w(v)=(k-1)r(N+1)/N$ for $v\in\{a_{1},\ldots,a_{N}\}$; $w(v)=1/k-r$ for $v\in\{a_{1}^{\prime},\ldots,a_{k-1}^{\prime}\}$;

• •

$w(a^{*})=1/k-N(k-1)r$;

• •

$w(v)=(1-(k-1)s)/N$ for $v\in\{b_{1},\ldots,b_{N}\}$; $w(v)=s$ for $v\in\{b_{1}^{\prime},\ldots,b_{k-1}^{\prime}\}$; and

• •

$w(v)=(1-(k-1)y)/N$ for $v\in\{c_{1},\ldots,c_{N}\}$; $w(v)=y$ for $v\in\{c_{1}^{\prime},\ldots,c_{k-1}^{\prime}\}$.

Then $(G,w)$ is a weighted graph. We claim it is $(x,y)$-biconstrained via $(A,B,C)$, and $w(N^{2}_{A}(v))<1/k$ for each $v\in C$. To see this we must verify:

[TABLE]

The first and third hold with equality from the definitions of $p,q$, and the second follows since $y\leq s$. The fourth follows from the definition of $r$. For the fifth, on substituting for $p$ and simplifying, we need to show that $r(k-1)(N-x(N+1)/(1-(k-1)s))\leq 1/k-x$, and this follows from the definition of $r$. Finally, the sixth simplifies to $(p+q-1)(N+1)/N<N$, and this is true since $p+q\leq N$. Consequently $\psi(x,y)<\frac{1}{k}$, by 2.1. This proves 3.3.

We also need the next three results:

3.4

Let $x^{\prime},y^{\prime},z^{\prime}\in(0,1)$, with $\psi(x^{\prime},y^{\prime})\leq z^{\prime}$. If $x,y,z\in(0,1]$ satisfy $x\leq 1/(2-x^{\prime})$, $y\leq y^{\prime}/(1+y^{\prime})$, $x+(1-x^{\prime})y/y^{\prime}\leq 1$, $z\geq 1/(2-z^{\prime})$, and

[TABLE]

*then $\psi(x,y)\leq z$. *

Proof.  Since $x^{\prime}\leq z^{\prime}$ (because $\psi(x^{\prime},y^{\prime})\leq z^{\prime}$) it follows that

[TABLE]

Let $G^{\prime}$ be $(x,y)$-biconstrained via $(A,B,C)$, such that $|N_{A}^{2}(w)|\leq z^{\prime}|A|$ for all $w\in C$. Add three vertices $a,b,c$ to the graph, and edges from $a$ to every vertex in $B$, edges from $b$ to every vertex in $A$, and an edge between $b$ and $c$. Let this new graph be $G$. Assign weights as follows:

[TABLE]

We will choose $p,q$ such that the weighted graph $(G,w)$ is $(x,y)$-biconstrained via $(A\cup\{a\},B\cup\{b\},C\cup\{c\})$ and $w(N_{A\cup\{a\}}^{2}(w))\leq z$ for all $w\in C\cup\{c\}$. The conditions are: $1-p\geq x$, $1-q\geq x$, $q\geq y$, $(1-q)x^{\prime}+q\geq x$, $(1-p)x^{\prime}+p\geq x$, $(1-y)y^{\prime}\geq y$, $(1-q)y^{\prime}\geq y$, $1-p\leq z$, and $(1-p)z^{\prime}+p\leq z$. These are equivalent to the following:

[TABLE]

Thus, it suffices to show that the lower bound on $p$ is at most the upper bound on $p$, and the same for $q$. We obtain eight conditions, which simplify to those given in the theorem statement. This proves 3.4.

3.5

Let $x^{\prime},y^{\prime},z^{\prime}\in(0,1)$, with $\psi(x^{\prime},y^{\prime})\leq z^{\prime}$. If $x,y\in(0,1]$ satisfy $y\leq 1/(2-y^{\prime})$, $x\leq x^{\prime}/(1+x^{\prime})$, $x\leq x^{\prime}z$, $(1-y^{\prime})x/x^{\prime}+y\leq 1$, $z\geq 1/(2-z^{\prime})$, and $x\leq(z-z^{\prime})/(1-z^{\prime})$, then $\psi(x,y)\leq z$.

Proof.  Let $G^{\prime}$ be $(x,y)$-biconstrained via $(A,B,C)$, such that $|N_{A}^{2}(w)|\leq z^{\prime}|A|$ for all $w\in C$. Add three vertices $a,b,c$ to $G^{\prime}$, with an edge from $a$ to $b$, edges from $b$ to every vertex in $C$, and edges from $c$ to every vertex in $B$. Let this new graph be $G$. Assign weights as follows:

[TABLE]

The conditions that the weighted graph $(G,w)$ is $(x,y)$-biconstrained via $(A\cup\{a\},B\cup\{b|],C\cup\{c\})$ with $w(N_{A\cup\{a\}}A^{2}(w))\leq z$ for all $w\in C\cup\{c\}$ can be written as follows:

[TABLE]

We need to check that the lower bound for $p$ is at most the upper bound for $p$, and the same for $q$. This gives eight conditions, which simplify (using that $1-y^{\prime}>x^{\prime}$, since $\psi(x^{\prime},y^{\prime})<1$) to those given in the theorem. This proves 3.5.

3.6

Let $s,t\geq 1$ be integers with $s/t\leq 1/2$. Let $x,y\in(0,1]$, satisfying $tx/s+y\leq 1$, $x+ty/s\leq 1$, and either $sy\leq x$ or $sx\leq y$. Furthermore, if either $x$ or $y$ is irrational, let strict inequality hold in all of these, that is, $tx/s+y<1$, $x+ty/s<1$, and either $sy<x$ or $sx<y$. Then $\psi(x,y)<s/t$.

Proof.  By increasing $x$ or $y$ if necessary, we may assume that $x,y$ are both rational. Let $k+1=\frac{t}{s}$. In terms of $k$, the hypotheses become $k\geq 1$, $(k+1)x+y\leq 1$, $x+(k+1)y\leq 1$, and either $sy\leq x$ or $sx\leq y$.

Suppose first that $sy\leq x$. Choose an integer $N\geq 1$ such that $p=xN/(1-kx)$ and $q=yN/(1-kx)$ are integers, and thus $p+q\leq(x+y)N/(x+y)=N$. Let $G_{1}$ be the graph with vertices $\{a_{1},...,a_{N},a^{*},b_{1},...,b_{N},c_{1},...,c_{N}\}$ where (reading subscripts modulo $N$) each $a_{i}$ is adjacent to $b_{i},...,b_{i+p-1}$, each $b_{i}$ is adjacent to $c_{i},...,c_{i+q-1}$, and $a^{*}$ is adjacent to all of the $b_{i}$.

Let $m=t-s$. Let $G_{2}$ be the graph with vertex set $\{a_{1}^{\prime},...,a_{m}^{\prime},b_{1}^{\prime},...,b_{m}^{\prime},c_{1}^{\prime},...,c_{m}^{\prime}\}$, where each $a_{i}^{\prime}$ is adjacent to $b_{i}^{\prime},...,b_{i+s-1}^{\prime}$ (reading subscripts modulo $m$), and each $b_{i}^{\prime}$ is adjacent to $c_{i}^{\prime}$. Let $G$ be the disjoint union of $G_{1}$ and $G_{2}$. Let $A=\{a_{1},...,a_{N},a^{*},a_{1}^{\prime},...,a_{m}^{\prime}\}$, and $B=\{b_{1},...,b_{N},b_{1}^{\prime},...,b_{m}^{\prime}\}$ and define $C$ similarly.

Let $r$ satisfy $(k+1)rN=\frac{1}{k+1}-x$. Assign weights as follows:

[TABLE]

This defines a weighted graph $(G,w)$, $(x,y)$-biconstrained via $(A,B,C)$, such that $w(N_{A}^{2}(v))<s/t=1/(k+1)$ for all $v\in C$, and so $\psi(x,y)<s/t$, as desired.

Now suppose $sy>x$, and consequently $sx\leq y$. Choose an integer $N\geq 1$ such that $p=xN/(1-ky)$, $q=yN/(1-ky)$ are integers, and thus $p+q\leq N$. Let $G_{1}$ be as before. Let $m=b-a$, and let $G_{2}$ be the graph with vertex set $\{a_{1}^{\prime},...,a_{m}^{\prime},b_{1}^{\prime},...,b_{m}^{\prime},c_{1}^{\prime},...,c_{m}^{\prime}\}$, where each $a_{i}^{\prime}$ is adjacent to $b_{i}^{\prime}$, and each $b_{i}^{\prime}$ is adjacent to $c_{i}^{\prime},...,c_{i+a-1}^{\prime}$, reading subscripts modulo $m$ (thus, this is the earlier graph $G_{2}$ flipped). Let $G$ be the disjoint union of $G_{1}$ and $G_{2}$, and define $A,B,C$ as before. Let $r>0$ satisfy $(k+1)rN\leq 1/(k+1)-x$ and $r\leq 1/(k+1)-y$. Assign weights as follows:

[TABLE]

Then $(G,w)$ is a weighted graph, and is $(x,y)$-biconstrained via $(A,B,C)$, and $w(N_{A}^{2}(v))<s/t$ for all $v\in C$, showing that $\psi(x,y)<s/t$. This proves 3.6.

3.7

Let $x,y,z\in(0,1]$, with $y\leq 1/2<x$. If $\phi\left(2-1/x,y/(1-y)\right)\leq 2-1/z$, then $\phi(x,y)\leq z$.

Proof.  Let $x^{\prime}=(2x-1)/x$, and $y^{\prime}=y/(1-y)$, and let $z^{\prime}=\phi\left(2-1/x,y/(1-y)\right)$. Let $G^{\prime}$ be a graph that is $(x,y)$-constrained via $(A,B,C)$, such that $|N_{A}^{2}(w)|\leq z^{\prime}|A|$ for all $w\in C$. Add three vertices $a,b,c$ to the graph, and edges from $a$ to every vertex in $B$, edges from $b$ to every vertex in $A$, and an edge between $b$ and $c$. Let this new graph be $G$.

Assign weights $w(v)\;(v\in V(G))$ as follows:

[TABLE]

Then the weighted graph $(G,w)$ is $(x,y)$-constrained via $(A\cup\{a\},B\cup\{b\},C\cup\{c\})$, since $xx^{\prime}+(1-x)=x$ and $(1-y)y^{\prime}=y$. Moreover, $w(N_{A}^{2}(w))\leq(1-z)+zz^{\prime}\leq z$ for all $w\in C$; and $w(N_{A\cup\{a\}}^{2}(c))=z$. Thus $\phi(x,y)\leq z$. This proves 3.7.

4 Biconstrained graphs

In this section we prove some lower bounds on $\psi(x,y)$. On the diagonal $x=y$, $\psi(x,y)$ behaves perfectly; it turns out that for all $x$, $\psi(x,x)=1/k$, where $k$ is the largest integer with $1/k\geq x$. That follows from:

4.1

For all integers $k\geq 1$, if $x,y\in(0,1]$ with $x+ky>1$ and $kx+\frac{x}{1-(k-1)y}\geq 1$, then $\psi(x,y)\geq 1/k$.

Proof.  By 1.2 we may assume that $x,y<1/k$. Let $G$ be $(x,y)$-biconstrained, via $(A,B,C)$. We must show that $|N^{2}_{A}(v)|\geq|A|/k$ for some $v\in C$. Suppose not. Choose $K\subseteq C$ with $|K|\leq k$, and subject to that with $|K|$ maximum such that the sets $N(v)\;(v\in K)$ are pairwise disjoint. Let $I\subseteq A$ be the union of the sets $N^{2}_{A}(v)\;(v\in K)$, and let $J\subseteq B$ be the union of the sets $N(v)\;(v\in K)$. It follows that

(1) $|A\setminus I|>(1-|K|/k)|A|$, and $|B\setminus J|\leq(1-|K|y)|B|$.

If $|K|=k$, then by (1), $|B\setminus J|\leq(1-ky)|B|<x|B|$, and since every vertex in $A$ has $x|B|$ neighbours in $B$, it follows that every vertex in $A$ has a neighbour in $J$, that is, $I=A$, contrary to (1). Thus $|K|<k$.

Since each vertex in $A\setminus I$ has at least $x|B|$ neighbours in $B$, and they all belong to $B\setminus J$, some vertex $t\in B\setminus J$ has at least

[TABLE]

neighbours in $A\setminus I$ by (1). Since $|K|\leq k-1$ and $ky<1$, and therefore $|K|y<1$, it follows that

[TABLE]

and so $t$ has at least $\frac{x|A|}{k(1-(k-1)y)}$ neighbours in $A\setminus I$. Let $u\in C$ be adjacent to $t$. From the maximality of $K$, $u$ has a neighbour $w\in N(v)$ for some $v\in K$. Since $w$ has at least $x|A|$ neighbours in $I$, it follows that

[TABLE]

a contradiction. This proves 4.1.

We deduce:

4.2

For all integers $k\geq 1$, if $x,y\in(0,1]$ with $x+ky>1$ and $kx+y\geq 1$, then $\psi(x,y)\geq 1/k$.

Proof.  If $k=1$ the result is easy (and follows from 5.2 below), so we assume that $k\geq 2$; and hence we may assume that $x,y<1/k\leq 1/2$ by 1.2. By 4.1 we may assume (for a contradiction) that $kx+\frac{x}{1-(k-1)y}<1$. Consequently $kx+\frac{x}{1-(k-1)(1-kx)}<1$. Let $t=1-kx$. Then $\frac{1-t}{1-(k-1)t}<kt$, and so $k(k-1)t^{2}-(k+1)t+1<0$. This is quadratic in $t$, with discriminant $(k+1)^{2}-4k(k-1)$, and the latter is negative if $k>2$; so we may assume that $k=2$. Then $2t^{2}-3t+1<0$, so $(2t-1)(t-1)<0$, that is, $1/2<t<1$. But $t=1-2x$, so $1/2<1-2x<1$, that is, $x<1/4$. But $2x+y\geq 1$ and $y<1/2$, a contradiction. This proves 4.2.

Consequently we have:

4.3

For all $x\geq 0$, $\psi(x,x)=1/k$, where $k$ is the largest integer with $1/k\geq x$.

Proof.  Certainly $\psi(x,x)\leq 1/k$, since by 1.5,

[TABLE]

Equality holds by 4.2. This proves 4.3.

Next we need a lemma:

4.4

Let $k\geq 1$ be an integer, let $(k-1)/k^{2}\leq y\leq 1$, and let $(A,B,C)$ be a tripartition of a graph $G$, such that:

• •

every vertex in $B$ has at least $y|C|$ neighbours in $C$; and

• •

$|N^{2}_{A}(v)|<|A|/k$* for each $v\in C$.*

Then there exist $v_{1},\ldots,v_{k}\in A$ such that $N(v_{i})\cap N(v_{j})=\emptyset$ for $1\leq i<j\leq k$.

Proof.  If some vertex $v$ in $A$ has degree zero, then we may take $v_{1}=\cdots=v_{k}=v$. So we assume that every vertex in $A$ has a neighbour in $B$. For each $v\in A$, let $c(v)=|N^{2}_{C}(v)|$, and let $A(v)\subseteq A$ be the set of vertices in $A$ that have a neighbour in $N(v)$. Let $|A(v)|=a(v)$.

(1) For each $v\in A$, $c(v)>kya(v)|C|/|A|$.

If we choose $u\in N^{2}_{C}(v)$ independently at random, then since every vertex in $A(v)$ has at least $y|C|$ second neighbours in $N^{2}_{C}(v)$, the probability that a given vertex $w\in A(v)$ belongs to $N^{2}_{A}(u)$ is at least $y|C|/c(v)$, and so the expectation of $|N^{2}_{A}(u)|$ is at least $(y|C|/c(v))a(v)$. On the other hand, the expectation of $|N^{2}_{A}(u)|$ is less than $|A|/k$. This proves (1).

Let $H$ be the graph with vertex set $A$, in which distinct $u,v$ are adjacent if (in $G$) $u,v$ have a common neighbour in $B$. Thus every vertex $v$ has degree $a(v)-1$ in $H$. So $2|E(H)|=\sum_{v\in A}(a(v)-1)$; but

[TABLE]

Consequently

[TABLE]

By Turán’s theorem, $H$ has a stable set of cardinality $k$. This proves 4.4.

4.5

Let $k\geq 1$ be an integer, and let $x,y\in(0,1]$ where $y\geq(k-1)/k^{2}$ and $kx+y>1$. Let $G$ be $(x,y)$-constrained via $(A,B,C)$, such that every vertex in $C$ has at least $y|B|$ neighbours in $B$. Then $|N^{2}_{A}(v)|\geq|A|/k$ for some $v\in C$. Consequently $\psi(x,y)\geq 1/k$.

Proof.  Suppose not; then there is a weighted graph $(G^{\prime},w)$, $(x,y)$-constrained via some tripartition $(A^{\prime},B^{\prime},C^{\prime})$, such that

• •

for each $v\in C^{\prime}$, $w(N(v))\geq y|B^{\prime}|$; and

• •

for each $v\in C^{\prime}$, $w(N^{2}_{A^{\prime}}(v))<1/k$.

Choose such a weighted graph $(G^{\prime},w)$ with $|V(G^{\prime})|$ minimum, and let $z<1/k$ such that $w(N^{2}_{A^{\prime}}(v))\leq z$ for each $v\in C^{\prime}$. By 4.4, there exist $v_{1},\ldots,v_{k}\in A^{\prime}$ such that $N(v_{1}),\ldots,N(v_{k})$ are pairwise disjoint. Consequently $w(N(v_{1})\cup\cdots\cup N(v_{k}))\geq kx$; and since $w(N(u))\geq y>1-kx$ for each $u\in C^{\prime}$, it follows that $\bigcup_{v\in X}N^{2}_{C^{\prime}}(v)=C^{\prime}$ where $X=\{v_{1},\ldots,v_{k}\}$. But $|X|<z^{-1}$, contrary to 2.4 and the minimality of $|V(G^{\prime})|$. This proves the first claim, and the second follows. This proves 4.5.

It is awkward to express the biconstrained problem in the language of triangular triples, but we can do so as follows. For $x,y,z\in(0,1]$ we say that $(x^{*},y,z)$ is triangular if no triangle-free graph $G$ admits a tripartition $(A,B,C)$ that satisfies the three bullets of the previous definition, and in addition satisfies

• •

every vertex in $B$ has at least $x|A|$ neighbours in $A$.

Similarly, we say $(x^{*},y^{*},z)$ is triangular if no triangle-free graph $G$ admits a tripartition $(A,B,C)$ that satisfies the three bullets of the previous definition, and in addition satisfies

• •

every vertex in $B$ has at least $x|A|$ neighbours in $A$; and

• •

every vertex in $C$ has at least $y|B|$ neighbours in $B$;

and so on. Then we have:

4.6

For $x,y,z\in(0,1]$, $\psi(x,y)>1-z$ if and only if $(x^{*},y^{*},z)$ is triangular.

This is not the same as saying that $(x^{*},y^{*},z^{*})$ is triangular, so we need to keep track of the asterisks if we rotate; but still it can be useful, as we shall see.

5 The mono-constrained case

In this section we are mostly concerned with $\phi(x,y)$ when $x=y$. We know that $\psi$ behaves well on the diagonal $x=y$, because of 4.3, so what about $\phi$? More generally, what about an analogue of 4.1 or 4.2 with $\psi$ replaced by $\phi$?

If we replace $\psi$ by $\phi$ in 4.1, it becomes false, even with $k=2$, because $\phi(3/10,4/11)\leq 4/9$, as the graph of figure 2 shows (the sets $A,B,C$ are the rows, and the numbers on the vertices are used as in figure 1). But as far as we know, 4.2 might hold with $\psi$ replaced by $\phi$. Let us state this as a conjecture:

5.1

*Conjecture: *** For all integers $k\geq 1$, if $x,y\in(0,1]$ with $x+ky>1$ and $kx+y\geq 1$, then $\phi(x,y)\geq 1/k$.

On the other hand, we have not even been able to prove what is presumably the simplest nontrivial case of this, namely that $\phi(x,y)\geq 1/2$ for all $x,y$ with $x,y>1/3$. But we do have several results approaching 5.1. First, it is true with $k=1$; we have the trivial:

5.2

For $x,y\in(0,1]$, if $x+y>1$, or $x+y=1$ and $x$ is irrational, then $\phi(x,y)=1$.

Proof.  Let $G$ be $(x,y)$-constrained via $(A,B,C)$. Then some vertex $v\in C$ has at least $y|B|$ neighbours in $B$, and strictly more if $y$ is irrational; and so $N_{A}^{2}(v)=A$, as every vertex in $A$ has at least $x|B|$ neighbours in $B$. This proves 5.2.

This is tight, in that if $x+y=1$ and $x,y$ are rational, then $\phi(x,y)<1$. We omit the proof, which is easy.

5.1 implies that $\phi(x,x)\geq 1/2$ if $x>1/3$. We have not been able to prove this, but we can show that $\phi(x,x)>3/7$ if $x>1/3$. That is implied by the following:

5.3

Let $k\geq 2$ be an integer; then for $x,y\in(0,1]$, if $y>1/k$ then

[TABLE]

Indeed if $k=2$, then $\phi(x,y)\geq 2x-x^{2}$ (which is larger).

Proof.  Let $G$ be $(x,y)$-constrained via $(A,B,C)$. If $x$ is irrational then $G$ is $(x,y)$-constrained via $(A,B,C)$, for some rational $x^{\prime}>x$; so we may assume that $x$ is rational, by increasing $x$ if necessary. Suppose that $k=2$, and choose $v_{1},v_{2}\in B$ independently and uniformly at random. For each $u\in A$, the probability that $u$ is adjacent to at least one of $v_{1},v_{2}$ is at least $2x-x^{2}$, since $u$ has at least $x|B|$ neighbours in $B$; and so we may choose $v_{1},v_{2}$ such that at least $(2x-x^{2})|A|$ vertices in $A$ are adjacent to at least one of them. But $v_{1},v_{2}$ have a common neighbour in $C$, since $y>1/2$, and the claim follows.

Thus we may assume that $k\geq 3$. By 1.2, $\phi(x,y)\geq x$, and so we may assume that

[TABLE]

that is, $x<1/(k-1)$. Consequently $x\leq(k-2)/(k-1)$ since $k\geq 3$. Define

[TABLE]

These are all non-negative, and $p$ is rational with denominator $T$ say; and by replacing each vertex by $T$ copies, we may assume that $p|A|$ is an integer. Since $x\leq(k-2)/(k-1)$ it follows that $s\leq 1$.

For $1\leq i\leq k-1$, we define $v_{i}\in B$, and a subset $P_{i}$ of $N_{A}(v_{i})$ with $|P_{i}|=p|A|$, inductively, as follows. Let $Q=P_{1}\cup\cdots\cup P_{i-1}$.

(1) There exists $v_{i}\in B$ such that $sa+b\geq x(s|Q|+|A|-|Q|)$, where $a=|N_{A}(v_{i})\cap Q|$ and $b=|N_{A}(v_{i})\setminus Q|$.

Suppose not; then summing over all $v\in B$, we deduce that

[TABLE]

But the first sum is $s$ times the number of edges between $Q$ and $B$, and so at least $xs|B|\cdot|Q|$; and the second is similarly at least $x|B|(|A|-|Q|)$, a contradiction. This proves (1).

Let $v_{i}$ be as in (1). Thus $sa+b\geq x(s|Q|+|A|-|Q|)\geq x(1-(1-s)(k-2)p)|A|$. In particular, since

[TABLE]

there exists $P_{i}\subseteq N_{A}(v_{i})$ of cardinality $p|A|$. Also, since $a\leq(k-2)p|A|$, and so

[TABLE]

it follows that

[TABLE]

and so

[TABLE]

for $1\leq h<i$. This completes the inductive definition of $v_{1},\ldots,v_{k-1}$ and $P_{1},\ldots,P_{k-1}$.

Let $P=P_{1}\cup\cdots\cup P_{k-1}$. Then $|P|\leq(k-1)p|A|$. Since every vertex in $A\setminus P$ has at least $x|B|$ neighbours in $B$, there exists $v_{k}\in B$ with at least $x(|A|-|P|)\geq x(1-(k-1)p)|A|$ neighbours in $A\setminus P$. Let $P_{k}$ be its set of neighbours in $A\setminus P$. Then for all $i$ with $1\leq i\leq k-1$,

[TABLE]

Consequently, for all distinct $v,v^{\prime}\in\{v_{1},\ldots,v_{k}\}$, $|N_{A}(v)\cup N_{A}(v^{\prime})|\geq m|A|$. But since $y>1/k$, some two of $v_{1},\ldots,v_{k}$ have a common neighbour $u\in C$, and so $|N^{2}_{A}(u)|\geq m$. This proves 5.3.

We deduce from 5.3 a version of 4.3 for the mono-constrained case:

5.4

For $y\in(0,1]$, if $y>1/k$ where $k\geq 2$ is an integer, then $\phi(1/k,y)\geq\frac{2k-3}{2k^{2}-4k+1}$.

Consequently $\phi(1/k,y)\geq 1/k+1/(2k^{2})+\Omega(k^{-3})$.

5.4 tells us in particular that $\phi(x,x)\geq\frac{2k-3}{2k^{2}-4k+1}>1/k$ when $x>1/k$ (if $k\geq 2$ is an integer), and since $\phi(1/k,1/k)=1/k$, there is a discontinuity in $\phi(x,x)$ when $x=1/k$, and the limit of $\phi(x,x)$ as $x\rightarrow 1/k$ from above is different from $\phi(1/k,1/k)$. What happens when $x\rightarrow 1/k$ from below? The next results investigate this. We will show that if $x$ is sufficiently close to $1/k$ from below, then $\phi(x,x)=1/k$.

5.5

If $k>0$ is an integer and $x\in(0,1]$ satisfies $(1-x)^{k}<x$, then $\phi(x,x)\geq 1/k$. In particular, if $x>0.382$ then $\phi(x,x)\geq 1/2$, and if $x>0.318$ then $\phi(x,x)>1/3$.

Proof.  Let $G$ be $(x,x)$-constrained via $(A,B,C)$. If we choosing $k$ vertices from $C$ uniformly at random, the number of vertices in $B$ nonadjacent to all of them is at most $(1-x)^{k}|B|$ in expectation; and so there exist $v_{1},\ldots,v_{k}\in C$ such that at most $(1-x)^{k}|B|$ vertices in $B$ are nonadjacent to all of them. Since $(1-x)^{k}|B|<x|B|$, it follows that the sets $N^{2}_{A}(v_{i})\;(1\leq i\leq k)$ have union $A$, and so one of them has cardinality at least $|A|/k$. This proves 5.5.

The proof of 5.5 is very simple, but the result is not of any value. It is of no use when $k\geq 4$ because then $(1-x)^{k}<x$ implies $x>1/k$; and we will prove in 6.6 and 8.3 that $\phi(x,x)\geq 1/2$ when $x>0.352202$, and $\phi(x,x)\geq 1/3$ when $x\geq 0.28231$, which are stronger than 5.5 when $k=2,3$. Here is another approach to the same question, more successful for larger values of $k$.

5.6

*Let $k\geq 1$ be an integer, and let $x\geq 1/k-\varepsilon$ where $\varepsilon=1/(13k^{3})$. Then $\phi(x,x)\geq 1/k$. *

Proof.  We may assume that $x=1/k-\varepsilon$. By 5.2 we may assume that $k\geq 2$. We leave the reader to check that

• •

$1/(2k)-\varepsilon>6k^{2}\varepsilon$;

• •

$x>1/(k+1)$; and

• •

$(2kx-1)/(2k-1)>(k\varepsilon)/(x+k\varepsilon)$.

(These are inequalities we will need later.) Let $G$ be $(x,x)$-constrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(v)|<|A|/k$ for each $v\in C$. Let $P$ be the set of vertices in $B$ that have at most $(1/k-2k\varepsilon)|A|$ neighbours in $A$.

(1) $|P|\leq|B|/(2k)$.

Every vertex in $B$ has fewer than $|A|/k$ neighbours in $A$, and so the number of edges between $A$ and $B$ is at most $|P|(1/k-2k\varepsilon)|A|+(|B|-|P|)|A|/k$. On the other hand, the number of such edges is at least $(1/k-\varepsilon)|A|\cdot|B|$; and so

[TABLE]

which simplifies to $2k|P|\leq|B|.$ This proves (1).

(2) If $u,v\in B\setminus P$ have a common neighbour in $C$, then $|N_{A}(u)\setminus N_{A}(v)|\leq 2k\varepsilon|A|$.

Since $u,v\in B\setminus P$ have a common neighbour in $C$, it follows that $|N_{A}(u)\cup N_{A}(v)|\leq|A|/k$. But $|N_{A}(u)|\geq(1/k-2k\varepsilon)|A|$ since $u\in B\setminus P$, and so $|N_{A}(u)\setminus N_{A}(v)|\leq 2k\varepsilon|A|$. This proves (2).

(3) There exist $v_{1},\ldots,v_{k}\in B\setminus P$ such that for $1\leq i<j\leq k$, there are at least $(1/(2k)-\varepsilon)|A|/k$ vertices in $A$ that are adjacent to $v_{j}$ and not to $v_{i}$.

Choose $v_{1},\ldots,v_{k}\in B\setminus P$ as follows. Choose $v_{1}\in B\setminus P$ arbitrarily. Inductively, suppose we have defined $v_{1},\ldots,v_{i}$ where $i<k$. Each has at most $|A|/k$ neighbours in $A$, and so the set of vertices in $A$ adjacent to one of $v_{1},\ldots,v_{i}$ has cardinality at most $(i/k)|A|\leq(1-1/k)|A|$. Let $D$ be the set of vertices in $A$ nonadjacent to each of $v_{1},\ldots,v_{i}$; then $|D|\geq|A|/k$. Since, by (1), each vertex in $D$ has at least $x|B|-|P|\geq(1/(2k)-\varepsilon)|B|$ neighbours in $B\setminus P$, there exists $v_{i+1}\in B\setminus P$ with at least $(1/(2k)-\varepsilon)|A|/k$ neighbours in $D$. This completes the inductive definition. We see that for $1\leq i<j\leq k$, there are at least $(1/(2k)-\varepsilon)|A|/k$ vertices in $A$ that are adjacent to $v_{j}$ and not to $v_{i}$. This proves (3).

Let $H$ be the bipartite graph $G[(B\setminus P)\cup C]$.

(4) For $1\leq i<j\leq k$, $v_{i}$ and $v_{j}$ belong to distinct components of $H$.

From (2), the sets $N_{C}(v_{1}),\ldots,N_{C}(v_{k})$ are pairwise disjoint, because $(1/(2k)-\varepsilon)|A|/k>2k\varepsilon|A|$. Suppose that there is a path of $H$ joining some two of $v_{1},\ldots,v_{k}$, and take the shortest such path $Q$; between $v_{i}$ and $v_{j}$ say, where $j>i$. Let $Q$ have $m$ vertices in $B$, say $u_{1},\ldots,u_{m}$ in order where $u_{1}=v_{i}$. We claim that $m\leq 4$. For suppose that $m\geq 5$. From the minimality of the length of $Q$, $u_{3}$ has no common neighbour in $C$ with any of $v_{1},\ldots,v_{k}$, and so the sets $N_{C}(u_{3}),N_{C}(v_{1}),\ldots,N_{C}(v_{k})$ are pairwise disjoint, which is impossible since $x>1/(k+1)$. Thus $m\leq 4$. By applying (2) to each pair of consecutive members of $V(Q)\cap B$, we deduce that

[TABLE]

But $|N_{A}(v_{j})\setminus N_{A}(v_{i})|\geq(1/(2k)-\varepsilon)|A|/k$, and so $(1/(2k)-\varepsilon)|A|/k\leq 6k\varepsilon|A|$, a contradiction. This proves (4).

For $1\leq i\leq k$, let $H_{i}$ be the component of $H$ containing $v_{i}$, and let $V(H_{i})\cap B=B_{i}$ and $V(H_{i})\cap C=C_{i}$. If there exists $v\in B\setminus P$ that does not belong to any of $B_{1},\ldots,B_{k}$, then the sets $N_{C}(v),N_{C}(v_{1}),\ldots,N_{C}(v_{k})$ are pairwise disjoint, which is impossible since they all have cardinality at least $x|C|$, and $(k+1)x>1$. Consequently the sets $B_{1},\ldots,B_{k}$ and $P$ form a partition of $B$.

(5) For $1\leq i\leq k$ there exists $u_{i}\in C_{i}$ adjacent to at least $\frac{(1-k\varepsilon)}{1+k(k-1)\varepsilon}|B_{i}|$ vertices in $B_{i}$.

For $1\leq i\leq k$, since $v_{i}$ has at least $x|C|$ neighbours in $C$, it follows that $|C_{i}|\geq x|C|$. Let $1\leq i\leq k$. Since $C_{1},\ldots,C_{k}$ are pairwise disjoint, and the union of the sets $C_{j}\;(j\in\{1,\ldots,k\}\setminus\{i\})$ has cardinality at least $(k-1)x|C|$, it follows that

[TABLE]

There are at least $x|B_{i}|\cdot|C|$ edges between $B_{i}$ and $C_{i}$, and so some vertex in $C_{i}$ has at least

[TABLE]

neighbours in $B_{i}$. By substituting $x=1/k-\varepsilon$, this proves (5).

For $1\leq i\leq k$, let $A_{i}=N^{2}_{A}(u_{i})$. Since $|A_{i}|<|A|/k$ for $1\leq i\leq k$, there exists $v\in A$ that is in none of $A_{1},\ldots,A_{k}$. Now $v$ has at least $x|B|$ neighbours in $B$, and they all belong to $B_{1}\cup\cdots\cup B_{k}$ except for at most $|P|$ of them. Consequently there exists $i\in\{1,\ldots,k\}$ such that $v$ has at least $(x|B|-|P|)|B_{i}|/|B\setminus P|$ neighbours in $B_{i}$. Since $v\notin A_{i}$, it follows that

[TABLE]

Since $x|B|\leq|B|$ and $|P|\leq|B|/(2k)$ by (1), it follows that

[TABLE]

and so $(2kx-1)/(2k-1)\leq k\varepsilon/(x+k\varepsilon)$, a contradiction. This proves 5.6.

For comparison, in figure 3 we give graphs of the function $\psi(x,x)$ (which we know completely, because of 4.3), and the function $\phi(x,x)$ (which we only know partially, from 5.6, 5.4 and 1.4.)

The next result is a useful general lower bound on $\phi(x,y)$.

5.7

For $x,y,z\in(0,1]$, if $y>1/2$ and $4x^{2}y(1-z)\geq(z-x)^{2}$ then $\phi(x,y)\geq z$. If in addition $4x^{2}y(1-z)>(z-x)^{2}$ then $\phi(x,y)>z$.

Proof.  Let $G$ be $(x,y)$-constrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(w)|<z|A|$ for each $w\in C$. There are at least $xy|A|\cdot|B|\cdot|C|$ two-edge paths between $A$ and $C$, and so there is a vertex $w\in C$ that is an end of at least $xy|A|\cdot|B|$ such paths. Let $w$ be an end of exactly $xq|A|\cdot|B|$ such paths; thus $y\leq q$. Let $B_{1}=N_{B}(w)$, and let $t=|B_{1}|/|B|$. Since $|N^{2}_{A}(w)|\leq z|A|$, there exists $A_{1}\subseteq A$ including $N^{2}_{A}(w)$ with $|A_{1}|=z|A|$ (we may assume the latter is an integer.) For each $u\in A_{1}$ let $u$ have exactly $d(u)|B|$ neighbours in $B_{1}$, and therefore at least $(x-d(u))|B|$ neighbours in $B\setminus B_{1}$. It follows that

[TABLE]

Let $v_{1}\in B_{1}$ and $v_{2}\in B\setminus B_{1}$, and let $A(v_{1},v_{2})=N_{A}(v_{1})\cup N_{A}(v_{2})$. For every such choice of $v_{1},v_{2}$, since $y>1/2$, there is a vertex $w$ in $C$ adjacent to both $v_{1},v_{2}$, and since $|N^{2}_{A}(w)|<z|A|$, it follows that $|A(v_{1},v_{2})|<z|A|$. Let us choose $v_{1}\in B_{1}$ and $v_{2}\in B\setminus B_{1}$, uniformly at random. It follows that the expected value of $|A(v_{1},v_{2})|$ is less than $z|A|$. The expected value of $|A(v_{1},v_{2})\cap A_{1}|$ is at least

[TABLE]

and the expected value of $|A(v_{1},v_{2})\setminus A_{1}|$ is at least

[TABLE]

Consequently the sum of these two is less than $z|A|$, and so

[TABLE]

Since $\sum_{u\in A_{1}}d(v)=xq|A|$, this simplifies to

[TABLE]

Now since $\sum_{u\in A_{1}}d(v)=xq|A|$ and $|A_{1}|=z|A|$, it follows from the Cauchy-Schwarz inequality that $\sum_{u\in A_{1}}d(u)^{2}\geq x^{2}q^{2}|A|/z$. Consequently

[TABLE]

This can be rewritten as:

[TABLE]

Since the first term above is a square, it is nonnegative, and so, since $q\geq y$, it follows that

[TABLE]

contrary to the hypothesis. This proves the first statement of the theorem, and the second is immediate by slightly increasing $z$. This proves 5.7.

6 When is $\phi(x,y)$ or $\psi(x,y)\geq 1/2$?

Another way to approach the problem is to ask, given some value $z$, for which $x,y\in(0,1]$ is $\phi(x,y)\geq z$? Or we could ask the same question for $\psi$, or ask when $\phi(x,y)>z$. For instance:

6.1

If $k\geq 1$ is an integer, then for $x,y\in(0,1]$, $\phi(x,y)>1/k$ if and only if $\max(x,y)>1/k$.

This follows trivially from 1.3 and 1.2. And the same holds with $\phi$ replaced by $\psi$. But deciding when $\psi(x,y)\geq 1/k$ or $\phi(x,y)\geq 1/k$ seems to be much less obvious. In this section we discuss when $\psi(x,y)$ or $\phi(x,y)$ is at least $1/2$; and in later sections we look at when they are at least $2/3$, and at least $1/3$.

For $x,y\in(0,1]$, we say (temporarily) that $(x,y)$ is good if $\psi(x,y)\geq 1/2$, and bad otherwise. The “map” of good and bad points is shown in the left half of figure 4. The solid black curve borders the known bad points, and the dotted curve borders the good points; between them is undecided. The borders are complicated, and we have indicated in the figure which theorem is responsible for each stretch of border.

Let us explain some of the details. First, if $\max(x,y)\geq 1/2$, then $(x,y)$ is good; and all pairs $(x,y)$ with $x+2y,2x+y\leq 1$ are bad, by 3.3. We searched by computer to find other examples of bad pairs $(x,y)$, and found about 12 maximal such pairs of rationals, with numerator and denominator at most 100. In fact we only searched for pairs $(x,y)$ where the corresponding $(x,y)$-biconstrained graph is similar to the graph obtained from figure 1, that is, it is obtained by “blowing up” the vertices of another graph in which the graph between two of the three parts is a matching. All these examples not only show that $\psi(x,y)<1/2$, but also that $\psi(y,x)<1/2$, and $\xi(x,y)<1/2$. In particular, for every bad pair $(x,y)$ we found by computer search, $(y,x)$ is another. This is just an artifact of our method of search, and is not evidence that the set of all bad pairs is closed under switching $x$ and $y$ (though it might be; it is for $\phi$, by 2.3). Anyway, for each bad pair the computer found, all pairs it dominates are also bad, and that gave us a step function bordering the area of the known bad points. We improved on this; we were able to smooth out some of the steps of the step function, by means of 3.3 and 6.5, so the step function the computer found now only survives towards the ends of the solid black curve in the figure. (These “fills” are not invariant under switching $x$ and $y$.) We give the coordinates of some bad pairs that we find particularly interesting. The apparent asymmetry between $x$ and $y$ in the left half of the figure is just asymmetry among what we have been able to prove; we have no proof of asymmetry. The right half of figure 4 does the same for $\phi$. Here there is symmetry exchanging $x$ and $y$, by 2.3, and so we only “explain” half of the border.

A graph has radius at most $r$ if there is a vertex $u$ such that every vertex has distance at most $r$ from $u$. We will need the following theorem of Erdős, Saks and Sós [1]:

6.2

Let $G$ be a connected graph with radius at least $r$, where $r\geq 1$ is an integer. Then $G$ has an induced path with $2r-1$ vertices, and consequently has a stable set of cardinality at least $r$.

When we have more than one graph defined using the same vertices, we speak of “$H$-distance” to mean distance in the graph $H$, and so on.

6.3

Let $x,y\in(0,1]$, such that

[TABLE]

Then $\psi(x,y)\geq 1/2$.

Proof.  Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(v)|<|A|/2$ for each $v\in C$. Then 1.2 implies that $x,y<1/2$. Suppose that $y>1/4$. The given inequality implies that $56x^{2}-64x+17<0$, and so $x>.41$. Since $2x+y>1$, 4.5 implies that $y\leq 1/4$, a contradiction. Thus $y\leq 1/4$. We leave the reader to verify that the following are consequences of the given inequality:

• •

$\frac{x}{3-3y}>1-2x$; and in particular, $\frac{x}{1-y}>1-2x$, so from 4.1 it follows that $x+2y\leq 1$;

• •

$\frac{y}{3-6y}>1-2x$; and so $\frac{y}{2-2y}>1-2x$, since $y\leq 1/4$; and

• •

$x+3y>1$.

(We found the easiest way to check these is to have a computer plot the various curves.) Let $H$ be the bipartite graph $G[B\cup C]$.

(1) If $v,v^{\prime}\in C$ have $H$-distance at most $2t$ where $t>0$ is an integer, then

[TABLE]

Take a path $P$ of $H$ joining $v$ and $v^{\prime}$, of length at most $2t$. Let the vertices of $P$ in $C$ be

[TABLE]

in order. For $1\leq i\leq t$ let $u_{i}\in B$ be adjacent to $v_{i-1}$ and $v_{i}$. Then for $1\leq i\leq t$, $N^{2}_{A}(v_{i-1})\cap N^{2}_{A}(v_{i})$ includes $N_{A}(u_{i})$ and hence has cardinality at least $x|A|$; and since $|N^{2}_{A}(v_{i})|<|A|/2$, it follows that $|N^{2}_{A}(v_{i})\setminus N^{2}_{A}(v_{i-1})|<(1/2-x)|A|$. But the union of the $t$ sets $N^{2}_{A}(v_{i})\setminus N^{2}_{A}(v_{i-1})$ includes $N^{2}_{A}(v^{\prime})\setminus N^{2}_{A}(v)$, and so the latter has cardinality less than $t(1/2-x)|A|$. This proves (1).

(2) There do not exist $v_{1},\ldots,v_{4}\in C$, pairwise with no common neighbour in $B$.

Then every three of $N(v_{1}),\ldots,N(v_{4})$ have union of cardinality at least $3y$; and since $3y>1-x$, every vertex in $A$ has a neighbour in at least two of $N(v_{1}),\ldots,N(v_{4})$. Consequently every vertex in $A$ belongs to at least two of $N^{2}_{A}(v_{1}),\ldots,N^{2}_{A}(v_{4})$, and so one of $N^{2}_{A}(v_{1}),\ldots,N^{2}_{A}(v_{4})$ has cardinality at least $|A|/2$, a contradiction. This proves (2).

(3) $H$* has at least two components.*

Suppose not, and let $H^{\prime}$ be the graph with vertex set $C$ in which $v,v^{\prime}$ are adjacent if they have a common neighbour in $H$. By (2), it follows that $H^{\prime}$ has no stable set of cardinality four, and so has radius at most three by 6.2. Choose $v\in C$ such that every vertex in $C$ has $H^{\prime}$-distance at most three from $v$. Let $B_{1}=N_{B}(v)$ and $A_{1}=N^{2}_{A}(v)$. Every vertex in $A\setminus A_{1}$ has at least $x|B|$ neighbours in $B\setminus B_{1}$, and so some vertex $u\in B\setminus B_{1}$ has at least $x(|B|/|B\setminus B_{1}|)|A\setminus A_{1}|$ neighbours in $A\setminus A_{1}$. Let $A_{2}$ be the set of neighbours of $u$ in $A\setminus A_{1}$. Since $|B\setminus B_{1}|\leq(1-y)|B|$ and $|A\setminus A_{1}|>|A|/2$, and $\frac{x}{3-3y}>1-2x$, it follows that

[TABLE]

Let $v^{\prime}\in C$ be adjacent to $u$. Since the $H^{\prime}$-distance from $v$ to $v^{\prime}$ is at most three, the $H$-distance from $v$ to $v^{\prime}$ is at most six. By (1), $|N^{2}_{A}(v^{\prime})\setminus N^{2}_{A}(v)|<3(1/2-x)|A|$, a contradiction. This proves (3).

(4) If $H^{\prime}$ is a component of $H$ then $|V(H^{\prime})\cap B|\leq(1-x)|B|$.

Suppose that $|V(H^{\prime})\cap B|>(1-x)|B|$; then every vertex in $A$ has a neighbour in $V(H^{\prime})$. By (2), there do not exist three vertices in $C\cap V(H^{\prime})$ pairwise with no common neighbour, and so by 6.2, it follows that there is a vertex $v\in C\cap V(H^{\prime})$ with $H^{\prime}$-distance at most four from every vertex in $C\cap V(H^{\prime})$. Let $A^{\prime}=N^{2}_{A}(v)$; then $|A^{\prime}|<|A|/2$. Since every vertex in $A\setminus A^{\prime}$ has at least $y|C|$ second neighbours in $C\cap V(H^{\prime})$, and $|C\cap V(H^{\prime})|\leq(1-y)|C|$, some vertex $v^{\prime}\in C\cap V(H^{\prime})$ has at least $(y/(1-y))|A\setminus A^{\prime}|$ second neighbours in $A\setminus A^{\prime}$. By (1), $(y/(1-y))|A\setminus A^{\prime}|<2(1/2-x)|A|$. But $|A^{\prime}|<|A|/2$, so $y/(4(1-y))\leq 1/2-x$, a contradiction. This proves (4).

(5) Some component $H^{\prime}$ of $H$ satisfies $(1-x)|B|\geq|V(H^{\prime})\cap B|\geq x|B|$.

By (2) and (3), $H$ has either two or three components. If $H$ has only two components, then they both satisfy (5), by (4); so we assume there are three. Let the components of $H$ be $H_{1},H_{2},H_{3}$, and for $1\leq i\leq 3$, let $V(H_{i})\cap B=B_{i}$ and $V(H_{i})\cap C=C_{i}$; and let $|B_{i}|/|B|=b_{i}$ and $|C_{i}|/|C|=c_{i}$. Suppose that $b_{1},b_{2},b_{3}<x$. Consequently every vertex in $A$ has neighbours in at least two of $B_{1},B_{2},B_{3}$. For $1\leq i\leq 3$, let $A_{i}$ be the set of vertices in $A$ with a neighbour in $B_{i}$. Thus every vertex in $A$ belongs to at least two of $A_{1},A_{2},A_{3}$, so from the symmetry we may assume that $|A_{1}|\geq 2|A|/3$. By (2), every two vertices in $C_{1}$ have a common neighbour in $B$. Choose $v\in C_{1}$, and let $A^{\prime}=N^{2}_{A}(v)$; then $|A^{\prime}|\leq|A|/2$. Since every vertex in $A_{1}$ has at least $y|C|$ second neighbours in $C_{1}$, some vertex $v^{\prime}$ in $C_{1}$ has at least $(y/c_{1})|A_{1}\setminus A^{\prime}|$ second neighbours in $A_{1}\setminus A^{\prime}$. By (1), $(y/c_{1})|A_{1}\setminus A^{\prime}|<(1/2-x)|A|$. But $|A^{\prime}|<|A|/2$, so $|A_{1}\setminus A^{\prime}|\geq|A|/6$; and $c_{1}\leq 1-2y$, so $y/(6(1-2y))\leq 1/2-x$, a contradiction. This proves (5).

(6) Every vertex in $C$ has at most $(1-x-y)|B|$ neighbours in $B$. Consequently

[TABLE]

for each component $H^{\prime}$ of $H$.

Suppose that $v\in C$ has more than $(1-x-y)|B|$ neighbours in $B$. Choose $v^{\prime}\in C$ in a different component of $H$; so $v,v^{\prime}$ have no common neighbour in $B$. Consequently

[TABLE]

and so every vertex in $A$ has a neighbour in $N(v)\cup N(v^{\prime})$. But then one of $|N^{2}_{A}(v)|,|N^{2}_{A}(v^{\prime})|\geq|A|/2$, a contradiction. This proves the first assertion. Let $H^{\prime}$ be a component of $H$. Then $H^{\prime}$ has at least $y|C|\cdot|V(H^{\prime})\cap B|$ edges, and at most $(1-x-y)|B|\cdot|V(H^{\prime})\cap C|$ edges, so the second claim follows. This proves (6).

Let $H^{\prime}$ be as in (5), and take the union of the other (one or two) components of $H^{\prime}$. We obtain nonnull subgraphs $H_{1},H_{2}$ of $H$, pairwise vertex-disjoint and with union $H$, such that $|V(H_{i})\cap B|\geq x|B|$ for $i=1,2$. For $i=1,2$, let $V(H_{i})\cap B=B_{i}$ and $V(H_{i})\cap C=C_{i}$; and let $|B_{i}|/|B|=b_{i}$ and $|C_{i}|/|C|=c_{i}$. Thus $b_{1},b_{2}\geq x$. From (6), $b_{i}\leq(1-x-y)c_{i}/y$ for $i=1,2$; and $c_{1},c_{2}\geq y$, since every vertex in $B_{i}$ has at least $y|C|$ neighbours in $C_{i}$. Also $b_{1}+b_{2}=c_{1}+c_{2}=1$.

For $i=1,2$ let $A_{i}$ be the set of vertices $v\in A$ that have more than $(b_{i}-y)|B|$ neighbours in $B_{i}$. Let $A_{0}=A\setminus(A_{1}\cup A_{2})$. Hence if $u\in A_{0}$, then since $u$ has at least $x|B|$ neighbours in $B$, $u$ has at least $(x+y-b_{2})|B|$ neighbours in $B_{1}$, and at least $(x+y-b_{1})|B|$ neighbours in $B_{2}$.

Since $A_{1},A_{2}$ and $A_{0}$ have union $A$, we may assume that $|A_{1}|+|A_{0}|/2\geq|A|/2$. Now $A_{1}\subseteq N^{2}_{A}(v)$ for each $v\in C_{1}$, since if $u\in A_{1}$, then $u$ has more than $(b_{1}-y)|B|$ neighbours in $B_{1}$, and $v$ has at least $y|B|$ neighbours in $B$. Consequently $|N^{2}_{A}(v)\cap A_{0}|<|A_{0}|/2$ for each $v\in C_{1}$.

Let us choose $v\in C_{1}$ uniformly at random; then the expected number of second neighbours of $v$ in $A_{0}$ is less than $|A_{0}|/2$, and so for some vertex $u\in A_{0}$, the probability that $u\in N^{2}_{A}(v)$ is less than $1/2$. Let $D$ be the set of neighbours of $u$ in $B_{1}$. Then $|D|\geq(x+y-b_{2})|B|$, and the probability that $v$ has a neighbour in $D$ is less than $1/2$. Thus more than $|C_{1}|/2$ vertices in $C_{1}$ have no neighbour in $D$. On the other hand, the expectation of the number of neighbours of $v$ in $D$ is at least $|D|y/c_{1}$; and so there exists $v\in C_{1}$ with more than $2|D|y/c_{1}$ neighbours in $D$. Also there exists $v^{\prime}\in C_{1}$ with no neighbours in $D$. It follows that

[TABLE]

Some vertex in $A_{0}$ is not a second neighbour of either of $v,v^{\prime}$, and so

[TABLE]

Consequently $y+2(x+y+b_{1}-1)y/c_{1}\leq 1-x-y$. Now $c_{1}\leq(1-x-2y+yb_{1})/(1-x-y)$ since

[TABLE]

. So

[TABLE]

that is,

[TABLE]

But $b_{1}\geq x$, contrary to the hypothesis. This proves 6.3.

6.4

Let $x,y\in(0,1]$, such that $x+2y>1$, $x\geq 1/4$ and $y\geq 1/3$. Then $\psi(x,y)\geq 1/2$.

Proof.  The only lower bound constraints on $y$ are $y\geq 1-2x$ and $y\geq 1/3$, and these are both satisfied if $y=0.38$ since $x\geq 1/4$. Hence we may assume that $y\leq 0.38$, by replacing $y$ by $\min(y,0.38)$. Consequently $y^{2}-3y+1>0$, and so

[TABLE]

Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(w)|<|A|/2$ for each $w\in C$. Choose $w_{1},w_{2},w_{3}\in C$ uniformly at random. The expected number of vertices in $B$ nonadjacent to all of $w_{1},w_{2},w_{3}$ is at most $(1-y)^{3}|B|<x|B|$; so we may choose $w_{1},w_{2},w_{3}$ such that fewer than $x|B|$ vertices in $B$ are nonadjacent to all of $w_{1},w_{2},w_{3}$. For $i=1,2,3$ let $A_{i}=N^{2}_{A}(w_{i})$. Thus $A_{1}\cup A_{2}\cup A_{3}=A$. In particular, one of $A_{2},A_{3}$, say $A_{2}$, includes at least half of $A\setminus A_{1}$; and since $|A_{1}|<|A|/2$, it follows that $|A_{2}\setminus A_{1}|>|A|/4$. Since $|A_{2}|<|A|/2$, it follows that $|A_{1}\cap A_{2}|<|A|/4<x|A|$; and so $N_{B}(w_{1}),N_{B}(w_{2})$ are disjoint (because any common neighbour would have at least $x|A|$ neighbours in $A$, all belonging to $A_{1}\cap A_{2}$). Hence

[TABLE]

and so $A_{1}\cup A_{2}=A$, contradicting that $|A_{1}|,|A_{2}|<|A|/2$. This proves 6.4.

6.5

Let $x,y\in(0,1]$, such that $x\leq 13/27$ and $y\leq 1/7$ and $3x+5y\leq 2$. Then $\psi(x,y)\leq 13/27$. If in addition $y\leq 1/8$, then $\psi(y,x)<1/2$.

Proof.  We claim that, for both statements of the theorem, we may assume that $3x+5y=2$. By increasing $x$, we may assume that either $x=13/27$ or $3x+5y=2$; and if $x=13/27$ then $y\leq 1/9$, since $3x+5y\leq 2$, and by increasing $y$ we may assume that $3x+5y=2$. This proves our claim. Since $3x+5y=2$ and $x\leq 13/27$, it follows that $y\geq 1/9$; and since $y\leq 1/7$ it follows that $x\geq 3/7$.

We return to the graph of figure 1. Let $A,B,C$ be the three rows of vertices, in order where $A$ is the top row. We need to adjust the vertex weights. Define $p=1/2-x/2-y$ and $r=(x-y)/2$. With the vertices in the same order as the figure, take vertex weights as follows:

[TABLE]

One can check (it takes some time and we omit the details) that this defines an $(x,y)$-biconstrained weighted graph showing that $\psi(x,y)\leq 13/27$. For the second statement, take the same graph and same vertex weighting, except replace the third row (of all one-sevenths) in the table above, by

[TABLE]

where $p^{\prime}=1/2-3y$, and $r^{\prime}=3y/2$. This weighted graph is $(y,x)$-biconstrained via $(C,B,A)$, and shows that $\psi(y,x)<1/2$. (Again, we leave the reader to check that this works.) This proves 6.5.

Now the mono-constrained case: for which pairs $(x,y)$ is $\phi(x,y)\geq 1/2$? Now we have symmetry between $x$ and $y$, and we found some examples of pairs $(x,y)$ with $\phi(x,y)<1/2$ on a computer searching randomly. (Conjecture 5.1 says that all points above both the lines $x+2y=1$ and $2x+y=1$ should be good, and indeed, all the maximal examples the computer found lie in the wedges between the lines.)

The next result strengthens 5.5 when $k=2$:

6.6

Let $x,y\in(0,1]$, such that $2x^{2}y\geq(1-x-y)^{2}$. Then $\phi(x,y)\geq 1/2$.

Proof.  Suppose that $\phi(x,y)=1/2-\varepsilon$ where $\varepsilon>0$. Then by 2.3 and 2.5 we have $\phi(x,1/2+\varepsilon)\leq 1-y$. Let $y^{\prime}=1/2+\varepsilon$ and $z=1-y$. Since $y^{\prime}>1/2$ and $\phi(x,y^{\prime})\leq z$, the second statement of 5.7 implies that $4x^{2}y^{\prime}(1-z)\leq(z-x)^{2}$, and so $2x^{2}y<(1-x-y)^{2}$, a contradiction. This proves 6.6.

In particular, 6.6 implies that $\phi(x,x)\geq 1/2$ if $x\geq 0.352202$, which is stronger than 5.5 when $k=2$.

6.7

Let $x,y\in(0,1]$ with $y<1/2$ and $y<(1-x)^{2}/(2-4x+6x^{2})$; then $\phi(x,y)<1/2$.

Proof.  By increasing $y$ slightly if necessary, we may assume that $s=(1/y-2)^{1/2}$ is rational. Thus $s\leq 1/y-1$, $s^{2}\leq 1/y-2$ and $1+2/s\leq 1/x$, since $y\leq 1/2$ and $y\leq(1-x)^{2}/(2-4x+6x^{2})$. Choose an integer $N\geq 2$ such that $sN$ is an integer.

Choose a graph $G_{1}$ that is $(s,1-s)$-constrained via a tripartition $(A_{1},B_{1},C_{1})$, such that $|A_{1}|=|B_{1}|=|C_{1}|=N$ and $N^{2}_{A_{i}}(v)\neq A_{1}$ for each $v\in C_{1}$. (It is easy to see that such a graph exists, for instance, one of the graphs used in 1.3.) Let $G_{2}$ be isomorphic to $G_{1}$, and let $(A_{2},B_{2},C_{3})$ be the corresponding tripartition. Take the disjoint union of $G_{1}$ and $G_{2}$, and add edges to make every vertex in $B_{1}$ adjacent to every vertex in $C_{2}$. Add three more vertices $a,b,c$, where $a$ is adjacent to $b$, $b$ is adjacent to every vertex in $C_{1}$, and $c$ is adjacent to every vertex in $B_{2}$, forming $G$. We define a weighting $w$ of $G$ as follows. Let $p=1/(2N)$ and $q=1/2-1/(4N^{2})$. Define $w$ by:

[TABLE]

Define $A=A_{1}\cup A_{2}\cup\{a\}$ and define $B,C$ similarly. Then the weighted graph $(G,w)$ is $(x,y)$-constrained via $(A,B,C)$, and proves that $\phi(x,y)<1/2$. (To see the latter, observe that, for instance, if $v\in C_{1}$ then

[TABLE]

from the choice of $G_{1}$). This proves 6.7.

7 The $2/3$ level

When is $\phi(x,y)\geq 2/3$; or the same question for $\psi$? In this section we say what we know about these.

7.1

If $x>1/2$ then $\psi(x,1/3)\geq 2/3$.

Proof.  Let $G$ be $(x,1/3)$-biconstrained via $(A,B,C)$, and suppose for a contradiction that $|N^{2}_{A}(v)|<2|A|/3$ for all $v\in C$. By averaging, there exists $v_{0}\in A$ such that $|N^{2}_{C}(v_{0})|<2|C|/3$. Let $B_{0}=N(v_{0})$ and $C_{0}=N^{2}_{C}(v_{0})$. Hence $|B_{0}|\geq x|B|$, and $|C_{0}|<2|C|/3$, and there are no edges between $B_{0}$ and $C\setminus C_{0}$, and every vertex in $C_{0}$ has a neighbour in $B_{0}$.

Choose $v_{1}\in C_{0}$. Thus $N(v_{1})\cap B_{0}\neq\emptyset$. Let $B_{1}=N(v_{1})$ and $A_{1}=N^{2}_{A}(v_{1})$. So $|A_{1}|\geq x|A|$, and $|A_{1}|<2|A|/3$. Every vertex $v\in A\setminus A_{1}$ has a neighbour in $B_{0}$, since $|B\setminus B_{0}|<|B|/2<|N(v)|$. Consequently every vertex in $A\setminus A_{1}$ has at least $|C|/3\geq|C_{0}|/2$ second neighbours in $C_{0}$, and by averaging it follows that some vertex $v_{2}\in C_{0}$ has at least $|A\setminus A_{1}|/2$ second neighbours in $A\setminus A_{1}$. Let $B_{2}=N(v_{2})$ and $A_{2}=N^{2}_{A}(v_{2})$. Then $|A_{2}\setminus A_{1}|\geq|A\setminus A_{1}|/2\geq|A|/6$. If there exists $u\in B_{1}\cap B_{2}$, then since $u$ has at least $x|A|$ neighbours in $A_{1}$, and they all belong to $A_{2}$, it follows that

[TABLE]

a contradiction. Consequently $B_{1}\cap B_{2}=\emptyset$.

In particular, $|B_{1}\cup B_{2}|\geq 2|B|/3$, and so every vertex in $A$ has a neighbour in $B_{1}\cup B_{2}$; and so $A_{1}\cup A_{2}=A$. Since $|A_{1}|,|A_{2}|<2|A|/3$, it follows that $|A_{1}\cap A_{2}|<|A|/3$. For $i=1,2$, choose $b_{i}\in B_{i}\cap B_{0}$. Then $N(b_{i})\cap A\subseteq A_{i}$ for $i=1,2$, and so $|N(b_{1})\cap N(b_{2})\cap A|<|A|/3$. Consequently $|(N(b_{1})\cup N(b_{2}))\cap A|>2|A|/3$. Since $b_{1},b_{2}\in B_{0}$ and they each have at least $|C|/3$ neighbours in $C_{0}$, and $|C_{0}|<2|C|/3$, it follows that they have a common neighbour $v\in C_{0}$. But then $N(b_{1})\cup N(b_{2})\cap A\subseteq N^{2}_{A}(v)$, and so $|N^{2}_{A}(v)|\geq 2|A|/3$, a contradiction. This proves 7.1.

7.2

If $y>1/2$ then $\psi(1/3,y)\geq 2/3$.

Proof.  Let $G$ be $(1/3,y)$-biconstrained via $(A,B,C)$. There exists $v_{1}\in C$ with at least $y|B|$ neighbours in $B$ (in fact, every vertex in $C$ has this property). Let $B_{1}=N(v_{1})$ and $A_{1}=N^{2}_{A}(v_{1})$. Thus $|B_{1}|\geq y|B|$. Since every vertex in $B\setminus B_{1}$ has at least $y|C|$ neighbours in $C$, some vertex $v_{2}\in C$ has at least $y|B\setminus B_{1}|$ neighbours in $B\setminus B_{1}$. Let $B_{2}=N(v_{2})$ and $A_{2}=N^{2}_{A}(v_{2})$. Thus $|B_{2}\setminus B_{1}|\geq y|B\setminus B_{1}|$, and so

[TABLE]

In particular, since every vertex in $A$ has at least $|B|/3$ neighbours in $B$, it follows that $A_{1}\cup A_{2}=A$. But $B_{1}\cap B_{2}\neq\emptyset$, since $|B_{1}|,|B_{2}|\geq y|B|>|B|/2$, and so there exists $b\in B_{1}\cap B_{2}$; and since $b$ has at least $|A|/3$ neighbours in $A$, and they all belong to $A_{1}\cap A_{2}$, it follows that $|A_{1}\cap A_{2}|\geq|A|/3$. Since $|A_{1}\cup A_{3}|=|A|$, it follows that $|A_{1}|+|A_{2}|\geq 4|A|/3$, and so one of $|A_{1}|,|A_{2}|$ is at least $2|A|/3$. This proves 7.2.

The last two results are closely related, via reformulation into triangular language. First, we need some shorthand for results of the form “if $x^{\prime}>x$ then $(x^{\prime},y,z)$ is triangular”; let us say “$(x^{+},y,z)$ is triangular” to mean “$(x^{\prime},y,z)$ is triangular for all $x^{\prime}>x$”, and treat the other two coordinates similarly. We will mix the two systems of notation, in expressions such as “$(x^{+*},y^{+},z)$ is triangular”, meaning “$(x^{\prime*},y^{+},z)$ is triangular for all $x^{\prime}>x$”.

Thus, in triangular language, we have the following.

• •

$(1/2^{+*},1/3^{*},1/3^{+})$ is triangular: because 7.1 says that $(x^{*},1/3^{*},1/3^{+})$ is triangular when $x>1/2$.

• •

$(1/2^{+},1/3^{+},1/3^{*})$ is triangular: because the proof of 7.2 did not use that every vertex in $C$ has at least $x|B|$ neighbours in $B$, and so it proves that $(1/3^{*},1/2^{+},1/3^{+})$ is triangular, and rotating gives that $(1/2^{+},1/3^{+},1/3^{*})$ is triangular.

• •

$(1/2^{+},1/3^{+*},1/3^{*})$ is triangular; this follows from 4.1 with $k=2$ and rotating.

• •

$(1/2^{+},1/3^{*},1/3^{+*})$ is triangular; this also follows from 4.1 with $k=2$ and rotating.

These four statements are similar, but no two are equivalent, and it would be good to find a common strengthening. Note, however, that $(1/2^{+*},1/3^{*},1/3^{*})$ is not triangular, and indeed $(2/3^{*},1/3^{*},1/3^{*})$ is not triangular. We have not been able to decide whether $(1/2^{+},1/3^{+},1/3)$ and $(1/2^{+},1/3,1/3^{+})$ are triangular, or indeed whether $(1/2^{+*},1/3^{+},1/3^{+})$ is triangular.

Pursuing this further, what about $(1/2^{+},1/3^{+},x)$ when $x<1/3$ (perhaps with some sprinkling of asterisks)? How small can $x$ be such that the triple remains triangular? We have examples that show that $(5/9,5/14,4/13)$ and $(4/7,3/8,2/7)$ are not triangular, and $(3/5^{*},2/5,1/4)$ is not triangular, but as far as we know, $(1/2^{+*},1/3^{+*},1/5)$ might be triangular.

This extends to weighted graphs in the natural way. For instance, the weighted graph of figure 6 (identify the vertices on the left with those on the right, in order) shows that $(4/7,2/7,3/8)$ is not triangular.

The graph of figure 6 shows that $\phi(4/7,2/7)\leq 5/8$, and so we studied $\psi(4/7,2/7)$, and proved the following.

7.3

If $x,y\in(0,1]$ such that $\max(x,y)>1/2$, $x\geq 1/3$, $x+2y>1$, and $3x+y/(1-y)>2$, then $\psi(x,y)\geq 2/3$.

Proof.  Let $G$ be $(x,y)$-biconstrained, via $(A,B,C)$, and suppose for a contradiction that $|N_{A}^{2}(v)|<2|A|/3$ for each $v\in C$. By 7.2, $y\leq 1/2$ since $x\geq 1/3$; and so $x>1/2$ since $\max(x,y)>1/2$. Hence $y<1/3$ by 7.1. Also $x<2/3$, by 1.2.

(1) For all $v_{1},v_{2}\in C$, if $|N(v_{1})\cup N(v_{2})|>(1-x)|B|$ then $N(v_{1})\cap N(v_{2})=\emptyset$, $N^{2}_{A}(v_{1})\cup N^{2}_{A}(v_{2})=A$, and $|N^{2}_{A}(v_{1})\cap N^{2}_{A}(v_{2})|<|A|/3$.

Every vertex in $A$ has a neighbour in $N(u)\cup N(v)$, and so $N_{A}^{2}(u)\cup N_{A}^{2}(v)=A$. Since $|N_{A}^{2}(u)|<2|A|/3$ and $|N_{A}^{2}(v)|<2|A|/3$ it follows that $|N_{A}^{2}(u)\cap N_{A}^{2}(v)|<|A|/3$, and so there is no vertex in $N(u)\cap N(v)$ (since any such vertex would have at least $x|A|$ neighbours in $A$, all belonging to $N_{A}^{2}(u)\cap N_{A}^{2}(v)$). This proves (1).

(2) There exist $v_{1},v_{2}\in C$ with $N(v_{1})\cap N(v_{2})=\emptyset$.

Choose $v_{1}\in C$. Since every vertex in $A\setminus N^{2}_{A}(v_{1})$ has at least $x|B|$ second neighbours in $B\setminus N(v_{1})$, some vertex $u_{2}\in B\setminus N(v_{1})$ has at least $(x/(3(1-y)))|A|$ neighbours in $A\setminus N^{2}_{A}(v_{1})$. Let $v_{2}\in C$ be adjacent to $u_{2}$. If $v_{1},v_{2}$ have a common neighbour $u_{1}$, then since $N_{A}(u_{1})\subseteq N^{2}_{A}(v_{2})$, it follows that $|N^{2}_{A}(v_{2})|\geq(x/(3(1-y))+x)|A|$, and so $x/(3(1-y))+x<2/3$, that is, $(4-3y)x<2-2y<(4-3y)/2$, and so $x<1/2$, a contradiction. This proves (2).

(3) If $v_{1},v_{2},v_{3}\in C$ and $N(v_{1})\cap N(v_{2})=\emptyset$ then $N(v_{3})$ is disjoint from exactly one of $N(v_{1}),N(v_{2})$.

If $N(v_{3})$ is disjoint from both $N(v_{1}),N(v_{2})$, then every two of $N(v_{1}),N(v_{2}),N(v_{3})$ have union of cardinality more than $(1-x)|B|$, and so every vertex in $A$ belongs to at least two of $N^{2}_{A}(v_{i})\;(i=1,2,3)$. Consequently one of $N^{2}_{A}(v_{i})\;(i=1,2,3)$ has cardinality at least $a|A|/3$, a contradiction. Now suppose that $N(u_{3})$ has nonempty intersection with both $N(v_{1}),N(v_{2})$. Thus $|N^{2}_{A}(v_{i})\cap N^{2}_{A}(v_{3})|\geq x|A|$ for $i=1,2$, and since $|N^{2}_{A}(v_{1})\cap N^{2}_{A}(v_{2})|<|A|/3$, it follows that $|N^{2}_{A}(v_{3})|\geq(2x-1/3)|A|\geq 2|A|/3$ since $x\geq 1/2$, a contradiction. This proves (3).

Let $H$ be the bipartite graph $G[B\cup C]$. From (2) and (3), $H$ has exactly two components $H_{1}$ and $H_{2}$ say. Let $C_{i}=V(H_{i})\cup C$ and $B_{i}=V(H_{i})\cap B$ for $i=1,2$. Then from (3), every two vertices in $C_{i}$ have a common neighbour in $B_{i}$, for $i=1,2$. Let $c_{i}=|C_{i}|/|C|$, for $i=1,2$. Thus $c_{1}+c_{2}=1$. We may assume that $b_{1}\geq 1/2$. Choose $v_{1}\in C_{1}$. Since $|A\setminus N^{2}_{A}(v_{1})|>|A|/3$, and every vertex in $A\setminus N^{2}_{A}(v_{1})$ has at least $y|C|$ second neighbours in $C_{1}$, some vertex $v_{2}\in C_{1}$ has at least $(y/(3c_{1}))|A|$ second neighbours in $A\setminus N^{2}_{A}(v_{1})$. But since $v_{1},v_{2}\in C_{1}$, they have a common neighbour in $B_{1}$; therefore $|N^{2}_{A}(v_{2})|\geq(y/(3c_{1})+x)|A|$, and so $y/(3c_{1})+x<2/3$. Now $c_{1}\leq 1-y$, so $3x+y/(1-y)<2$, contrary to the hypothesis. This proves 7.3.

If $A\subseteq V(G)$ and $X\subseteq V(G)\setminus A$, $N_{A}(X)$ denotes the set of vertices in $A$ with a neighbour in $X$. The next result is a useful lemma which says, roughly speaking, the larger $X$ is, the larger $N_{A}(X)$ is. In this section we only use it for $k=1$, but we will use it with $k=2$ in the section discussing when $\psi(x,y)\geq 3/4$.

7.4

Let $x,y,z\in(0,1]$, and suppose that $G$ is $(x,y)$-biconstrained via $(A,B,C)$, and $|N_{A}^{2}(w)|<z|A|$ for all $w\in C$. Then for all integers $k\geq 1$, if $B^{\prime}\subseteq B$ with

[TABLE]

then $|N_{A}(B^{\prime})|>(x+k(1-z))|A|$.

Proof.  We proceed by induction on $k$, and so we assume that either $k=1$ or the result holds for $k-1$. Let $A^{\prime}=N_{A}(B^{\prime})$. Every vertex in $B\setminus B^{\prime}$ has at least $y|C|$ neighbours in $C$, so there exists $v\in C$ with at least $y|B\setminus B^{\prime}|$ neighbours in $B\setminus B^{\prime}$. By hypothesis, $|B^{\prime}|/|B|>1-x/(1-y)$, that is, $x|B|+y|B\setminus B^{\prime}|>|B\setminus B^{\prime}|$. But every vertex in $A\setminus A^{\prime}$ has at least $x|B|$ neighbours in $B\setminus B^{\prime}$, and therefore has one such neighbour adjacent to $v$; and so $A\setminus A^{\prime}\subseteq N^{2}_{A}(v)$. Let $B^{\prime\prime}=B^{\prime}\cap N_{B}(v)$, and $A^{\prime\prime}=N_{A}(B^{\prime\prime})$.

(1) $|A^{\prime\prime}|\geq(x+(k-1)(1-z))|A|$.

Since $|N_{B}(v)|\geq y|B|$, it follows that $|B^{\prime\prime}|\geq y|B|-(|B|-|B^{\prime}|)=|B^{\prime}|-(1-y)|B|$. If $k=1$, then $|B^{\prime}|/|B|>1-y$ by hypothesis, and therefore $B^{\prime\prime}\neq\emptyset$, and so $|A^{\prime\prime}|\geq x|A|$ as claimed. If $k\geq 2$, then since $|B^{\prime}|/|B|>(k-1)(1-y)+\max(1-y,1-x/(1-y))$, it follows that

[TABLE]

and so $|A^{\prime\prime}|\geq(x+(k-1)(1-z))|A|$ from the inductive hypothesis. This proves (1).

Since $A^{\prime\prime}\subseteq A^{\prime}$, and $A^{\prime\prime}\cup(A\setminus A^{\prime})\subseteq N^{2}_{A}(v)$, it follows that

[TABLE]

and so $|A^{\prime}|\geq(x+k(1-z))|A|$. This proves 7.4.

7.5

Let $x,y\in(0,1]$, such that $y>1/2$, $x+3y>2$ and $x>2(1-y)^{2}/(2-y)$. Then $\psi(x,y)\geq 2/3$.

Proof.  Let $G$ be $(x,y)$-biconstrained, via $(A,B,C)$, and suppose for a contradiction that $|N_{A}^{2}(v)|<2|A|/3$ for each $v\in C$.

(1) If $B^{\prime}\subseteq B$ with $|B^{\prime}|/|B|>\max(1-y,1-x/(1-y))$, then there are at least $(x+1/3)|A|$ vertices in $A$ with a neighbour in $B^{\prime}$. In particular, if $|B^{\prime}|\geq(x+2y-1)|B|$ then the same conclusion holds.

The first statement follows from 7.4 with $k=1$. For the second, $x+2y-1>1-y$ since $x+3y\geq 2$; and $x+2y-1>1-x/(1-y)$ since $x>2(1-y)^{2}/(2-y)$. Consequently $x+2y-1>\max(1-y,1-x/(1-y))$. This proves the second statement and so proves (1).

Say $w_{1},w_{2}\in C$ are close if $|N_{B}(w_{1})\cup N_{B}(w_{2})|\leq(1-x)|B|$.

(2) There exists $w_{1}\in C$ such that the set of vertices in $C$ that are close to $w_{1}$ has cardinality at least $|C|/2$.

This is trivial if every two vertices in $C$ are close; so we assume there exist $w_{1},w_{2}\in C$ that are not close. Consequently every vertex in $A$ has a neighbour in $N_{B}(w_{1})\cup N_{B}(w_{2})$, and so $N_{A}^{2}(w_{1})\cup N^{2}_{A}(w_{2})=A$. If there exists $w\in C$ that is not close to either of $w_{1},w_{2}$ then similarly $N_{A}^{2}(w_{1})\cup N^{2}_{A}(w)=A$ and $N_{A}^{2}(w_{2})\cup N^{2}_{A}(w)=A$; and so every vertex in $A$ belongs to at least two of $N^{2}_{A}(w_{1}),N^{2}_{A}(w_{2}),N^{2}_{A}(w)$, and therefore one of these three sets has cardinality at least $2|A|/3$, a contradiction. Thus, exchanging $w_{1},w_{2}$ if necessary, we may assume that at least half of all vertices in $C$ are close to $w_{1}$. This proves (2).

Let $C_{1}$ be the set of vertices in $C$ that are close to $w_{1}$; thus $|C_{1}|\geq|C|/2$. Let $B_{1}=N_{B}(w_{1})$ and $A_{1}=N^{2}_{A}(w_{1})$. Since $|B_{1}|\geq y|B|$ and every vertex in $A\setminus A_{1}$ has at least $x|B|$ neighbours in $B\setminus B_{1}$, there exists $v_{2}\in B\setminus B_{1}$ with at least $\frac{x}{1-y}|A\setminus A_{1}|$ neighbours in $A\setminus A_{1}$. Since $y>1/2$ and $|C_{1}|\geq|C|/2$, $v_{2}$ has a neighbour $w_{2}\in C_{1}$. Since $w_{2}$ is close to $w_{1}$ it follows that $|N_{B}(w_{1})\cup N_{B}(w_{2})|\leq(1-x)|B|$, and so $|N_{B}(w_{1})\cap N_{B}(w_{2})|\geq(x+2y-1)|B|$. From (1), there are at least $(x+1/3)|A|$ vertices in $A$ with a neighbour in $N_{B}(w_{1})\cap N_{B}(w_{2})$. These vertices all belong to $A_{1}$, and so

[TABLE]

Consequently $\frac{x}{3(1-y)}+x+1/3<2/3$, that is, $\frac{x}{1-y}+3x<1$. By hypothesis, $x>2(1-y)^{2}/(2-y)$, and so substitution for $x$ yields

[TABLE]

which simplifies to $(3y-2)(2y-3)<0$, contrary to 1.2. This proves 7.5.

7.6

Let $x,y\in(0,1]$. If either

• •

$x\leq 11/17$* and $y\leq 1/7$ and $x+3y\leq 1$; or*

• •

$x<1/8$* and $y\leq 11/17$ and $3x+y\leq 1$*

then $\psi(x,y)<2/3$.

Proof.  Take the graph consisting of seven disjoint copies of a three-vertex path, numbered $a_{i},b_{i},c_{i}$ in order $(1\leq i\leq 7)$. For $1\leq i\leq 3$ and $4\leq j\leq 7$, make $a_{i}$ adjacent to $b_{j}$ and make $a_{j}$ adjacent to $b_{i}$, forming $G$. Let $A=\{a_{i}\;:1\leq i\leq 7\}$ and define $B,C$ similarly.

For the first statement, we may assume by increasing $x,y$ that $x+3y=1$. It follows that $x\geq 4/7$ (because $y\leq 1/7$) and similarly $y\geq 2/17$, and $4x=4-12y\geq 1+9y$. For $1\leq i\leq 3$, let $w(a_{i})=3/17$, $w(b_{i})=(4x-1)/9$, and $w(c_{i})=1/7$. For $4\leq i\leq 7$, let $w(a_{i})=2/17$, $w(b_{i})=(1-x)/3$, and $w(c_{i})=1/7$. Then this weighted graph is $(x,y)$-biconstrained via $(A,B,C)$ and shows that $\psi(x,y)<2/3$. This proves the first statement.

For the second statement, we may assume that $3x+y=1$, and so $x\geq 2/17$ and $y>5/8$, and so $3y=3-9x\leq 1+8x$. Let us take the same graph and redefine $w$, as follows. For $1\leq i\leq 3$, let $w(a_{i})=(1-y)/2$ and $w(b_{i})=w(c_{i})=(1-4x)/3$. For $4\leq i\leq 7$, let $w(a_{i})=(3y-1)/8$ and $w(b_{i})=w(c_{i})=x$. Then this weighted graph is $(x,y)$-biconstrained via $(C,B,A)$ and shows that $\psi(x,y)<2/3$. This proves the second statement, and hence proves 7.6.

7.7

Let $x^{\prime},y^{\prime},z^{\prime}\in(0,1]$ such that $\psi(x^{\prime},y^{\prime})\leq z^{\prime}<1/2$; and let $x,y\in(0,1]$ satisfy $x\leq\frac{1}{2-x^{\prime}}$, $x<1-\frac{1-x^{\prime}}{3(1-z^{\prime})}$, $y\leq\frac{y^{\prime}}{1+y^{\prime}}$ and $x+\frac{1-x^{\prime}}{y^{\prime}}y\leq 1$. Then $\psi(x,y)<2/3$. Consequently:

• •

$\psi(x,y)<2/3$* if $x\leq 3/5$ and $y\leq 1/4$ and $x+2y\leq 1$;*

• •

$\psi(x,y)<2/3$* if $x\leq 5/8$ and $y\leq 1/6$ and $x+3y\leq 1$.*

Proof.  The first statement follows from 3.4 taking $z$ slightly less than $2/3$. The two statements in bullets follow by setting $x^{\prime}=y^{\prime}=z^{\prime}=1/3$, and then $x^{\prime}=z^{\prime}=2/5$ and $y^{\prime}=1/5$. This proves 7.7.

7.8

Let $x^{\prime},y^{\prime},z^{\prime}\in(0,1]$ such that $\psi(x^{\prime},y^{\prime})\leq z^{\prime}<1/2$; and let $x,y\in(0,1]$ satisfy $x<\frac{2x^{\prime}}{3}$, $y\leq\frac{1}{2-y^{\prime}}$, and $\frac{1-y^{\prime}}{x^{\prime}}x+y\leq 1$. Then $\psi(x,y)<2/3$. Consequently:

• •

$\psi(x,y)<2/3$* if $y\leq 3/5$ and $2x+y\leq 1$; and*

• •

$\psi(x,y)<2/3$* if $y\geq 3/5$ and $x+3y\leq 2$, and $x+3y<2$ if $x$ or $y$ is irrational.*

Proof.  The first statement follows from 3.5, taking $z$ slightly less than $2/3$. To prove the first bullet, let $2x+y\leq 1$, and so $x\leq 1/2$. If also $y\leq 1/2$ then $\psi(x,y)\leq 1/2$ by 1.4; so we may assume that $y>1/2$. We claim there is an integer $k\geq 1$ with

[TABLE]

To see this, if $y>5/9$ we can take $k=1$ (because we are given that $y\leq 3/5$), and if $y\leq 5/9$, then

[TABLE]

and so again $k$ exists. Thus $y\leq\frac{2k+1}{4k+1}$, and $x\leq\frac{1-y}{2}<\frac{2k}{6k+3}.$ Let $x^{\prime}=z^{\prime}=k/(2k+1)$, and $y^{\prime}=1/(2k+1)$. Then the claim follows from the first statement.

For the second bullet, let $x+3y\leq 2$ with $y\geq 3/5$, with $x+3y<2$ if $x$ or $y$ is irrational. Consequently, we may assume that $x,y$ are rational, by increasing them slightly if necessary. Let $x^{\prime}=2/y-3$, and $y^{\prime}=2-1/y$; it follows that $x^{\prime}+2y^{\prime}\leq 1$ and $2x^{\prime}+y^{\prime}\leq 1$, and $x^{\prime},y^{\prime}$ are rational, and so $\psi(x^{\prime},y^{\prime})<1/2$ by 3.3. The result follows from the first statement. This proves 7.8.

For the mono-constrained question, we have:

7.9

For $x,y\in(0,1]$, if $y\leq 1/2$ and $x>(1-y)^{2}/(1-2y^{2})$ then $\phi(x,y)\geq 2/3$.

Proof.  Let $G$ be $(x,y)$-constrained via $(A,B,C)$. If $x+y>1$ the result follows from 5.2, so we may assume that $x+y\leq 1$. Since $x>(1-y)^{2}/(1-2y^{2})$, we may also assume that

• •

$x,y$ are rational; and

• •

every vertex in $A$ has strictly more than $x|B|$ neighbours in $B$

by reducing $x$ and $y$ a little if necessary while retaining the property that $x>(1-y)^{2}/(1-2y^{2})$.

Let $p=(1-x-y)/(1-2y)$. Thus $p$ is rational, so we may assume (by multiplying vertices) that $p|B|$ is an integer. Also $p\leq y$, since $x>(1-y)^{2}/(1-2y^{2})$. Let $s=(x-(1-y)^{2})/(y(1-y))$. It follows that $0\leq s\leq 1$, since $x>(1-y)^{2}/(1-2y^{2})$ and $x+y\leq 1$.

Choose $v_{1}\in C$ with at least $y|B|$ neighbours in $B$, and let $B_{1}\subseteq N(v_{1})$ with $|B_{1}|=y|B|$. Choose $v_{2}\in C$ such that $sb_{0}+b_{2}\geq y(sy+(1-y))$, where $b_{0}|B|=|N(v_{2})\cap B_{1}|$ and $b_{2}|B|=|N(v_{2})\setminus B_{1}|$. (Such a vertex exists by averaging.) We claim that $b_{0}+b_{2}\geq p$; for from the definition of $s$,

[TABLE]

and $p=(1-x-y)/(1-2y)\leq xy/(1-y)$ since $x>(1-y)^{2}/(1-2y^{2})$.

Also we claim that $b_{2}\geq 1-x-y$; for from the definition of $s$,

[TABLE]

Consequently $|N(v_{1})\cup N(v_{2})|\geq(1-x)|B|$, and there exist $P_{1}\subseteq N(v_{1})$ and $P_{2}\subseteq N(v_{2})$, both of cardinality $p|B|$. Choose $v_{3}\in C$ with at least $y(1-2p)|B|$ neighbours in $B\setminus(P_{1}\cup P_{2})$. Then for $i=1,2$,

[TABLE]

Since every vertex in $A$ has strictly more than $x|B|$ neighbours in $B$, it follows that every vertex in $A$ belongs to at least two of the sets $N^{2}_{A}(v_{i})\;(i=1,2,3)$; and so one of these sets has cardinality at least $2|A|/3$. This proves 7.9.

7.10

For all $x,y\in(0,1]$ with $y\leq 1/2$, if $\frac{x}{1-x}+\frac{y}{1-2y}\leq 2$, then $\phi(x,y)<2/3$.

Proof.  We may assume that $x>1/2$, or else the result is true since $\phi(1/2,1/2)=1/2$. The claim follows from 3.7 taking $z$ slightly less than $2/3$. This proves 7.10.

8 The $1/3$ level

Next we do the same for $\psi(x,y)\geq 1/3$ and $\phi(x,y)\geq 1/3$. The figure summarizes our results.

8.1

Let $x,y\in(0,1]$ with $y>\frac{1}{5}$ and $3x+\frac{y}{3(1-y)}\geq 1$. Then $\psi(x,y)\geq\frac{1}{3}$.

Proof.  We may assume that $y\leq 1/3$, by 1.2, and so $y/(3(1-y))\leq 1/6$. Consequently $x\geq 5/18$, and in particular $x>2y/3$ (we will need this later). Also, since $1/5\leq y\leq 1/3$, it follows that $3y-y/(3(1-y))>1/2$; and so

[TABLE]

and consequently $x+y>1/2$. Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(v)|<|A|/3$ for each $v\in C$. Let $H$ be the subgraph induced on $B\cup C$, and let $H_{1},\ldots,H_{k}$ be its components. Let $B_{i}=V(H_{i})\cap B$ and $C_{i}=V(H_{i})\cap C$, and $b_{i}=|B_{i}|/|B|$, $c_{i}=|C_{i}|/|C|$, for $1\leq i\leq k$. Since $y>0$, $B_{i},C_{i}$ are both nonempty and so $b_{i},c_{i}\geq y$ for $1\leq i\leq k$. For $1\leq i\leq k$, let $A_{i}$ be the set of vertices in $A$ with a neighbour in $B_{i}$, and let $A_{i}^{*}$ be the set of vertices in $A$ such that $N(v)\subseteq B_{i}$.

(1) $k\geq 2$.

Suppose that $k=1$, and let $H^{\prime}$ be the graph with vertex set $B$ in which $u,u^{\prime}$ are adjacent if $u,u^{\prime}$ have a common neighbour in $H$. Then every stable set of $H^{\prime}$ has cardinality at most $4$. By 6.2 there is a vertex $u_{1}\in B$ with $H^{\prime}$-distance at most four to every other vertex in $B$; and so the $H$-distance from $u_{1}$ to each vertex in $B$ is at most eight. Let $v_{1}\in C$ be adjacent to $u_{1}$. Let $A^{\prime}=A\setminus N^{2}_{A}(v_{1})$ and $B^{\prime}=B\setminus N(v_{1})$. Hence $|A^{\prime}|>2|A|/3$. Since every vertex in $A^{\prime}$ has at least $x|B|$ neighbours in $B^{\prime}$, and $|B^{\prime}|\leq(1-y)|B|$, some vertex $u\in B^{\prime}$ has at least

[TABLE]

neighbours in $A^{\prime}$. Choose a path of $H$ between $u_{1}$ and $u$ of length at most eight, and let its vertices be $u_{1}\hbox{-}v_{2}\hbox{-}u_{2}\hbox{-}\cdots\hbox{-}v_{t}\hbox{-}u_{t}=u$ say, in order. Thus $t\leq 5$, and so there exists $i$ with $1\leq i\leq t-1$ such that there are at least $|N_{A^{\prime}}(u)|/4$ vertices that belong to $N_{A}(u_{i+1})\setminus N_{A}(u_{i})$. Since $|N_{A}(u_{i})|\geq x|A|$, it follows that

[TABLE]

a contradiction, since $x\geq 2y/3$ and so $x+x/(6(1-y))\geq x+y/(9(1-y))\geq 1/3$ . This proves (1).

(2) $b_{i}\leq 1-x-y<1/2$* for $1\leq i\leq k$, and so $k\geq 3$.*

Suppose that $b_{1}>1-x-y$ say. Thus, if $u\in A\setminus A_{1}$, then $u\in N^{2}_{A}(v)$ for every $v\in C\setminus C_{1}$; and so $|A\setminus A_{1}|<|A|/3$, and so $|A_{1}|>2|A|/3$. Let $H^{\prime}$ be the graph with vertex set $C_{1}$ in which $v,v^{\prime}$ are adjacent if they have a common $H_{1}$-neighbour in $B_{1}$. Thus $H^{\prime}$ has stability number at most three (by (1)) and so has radius at most three, by 6.2. Choose $v_{1}\in C_{1}$ such that every vertex in $C_{1}$ has $H_{1}$-distance at most six from $v_{1}$. Let $A^{\prime}=A_{1}\setminus N^{2}_{A}(v_{1})$; thus $|A^{\prime}|>|A|/3$. Since every vertex in $A^{\prime}$ has a neighbour in $B_{1}$ and hence has at least $y|C|$ second neighbours in $C_{1}$, there exists $v\in C_{1}$ such that

[TABLE]

since $|C_{1}|\leq(1-y)|C|$. Choose a path of $H_{1}$ between $v_{1},v$ of length at most six, with vertices $v_{1}\hbox{-}u_{1}\hbox{-}v_{2}\hbox{-}\cdots\hbox{-}u_{t-1}\hbox{-}v_{t}=v$ say where $t\leq 4$. Then for some $i$ with $1\leq i\leq t-1$,

[TABLE]

and hence

[TABLE]

since all vertices of $N_{A}(u_{i})$ belong to $N^{2}_{A}(v_{i+1})$ and do not belong to $N^{2}_{A^{\prime}}(v_{i+1})\setminus N^{2}_{A^{\prime}}(v_{i})$. But $3x+y/(3(1-y))\geq 1$, a contradiction. This proves (2).

By (2), $k\geq 3$; and $k\leq 4$ since $y>1/5$. We may assume that $|B_{1}|,|B_{2}|\geq|B_{i}|$ for $i\geq 3$; let $B_{0}=\bigcup_{3\leq i\leq k}B_{i}$, and $C_{0}=\bigcup_{3\leq i\leq k}C_{i}$. Hence $|B_{0}|\leq|B|/2$ since $k\leq 4$. Let $b_{0}=|B_{0}|/|B|$ and $c_{0}=|C_{0}|/|C|$; let $A_{0}$ be the set of vertices in $A$ with a neighbour in $B_{0}$, and let $A_{0}^{*}$ be the set of vertices in $A$ such that $N(v)\subseteq B_{0}$. For $0\leq i<j\leq 2$, choose $A_{ij}=A_{ji}\subseteq A_{i}\cap A_{j}$ such that the sets $A_{12},A_{13},A_{23},A_{0}^{*},A_{1}^{*},A_{2}^{*}$ are pairwise disjoint and have union $A$. For $0\leq i\leq 2$ let $a_{i}=|A_{i}|/|A|$ and $a_{i}^{*}=|A_{i}^{*}|/|A|$, and for $0\leq i,j\leq 2$ with $i\neq j$ let $a_{ij}=|A_{ij}|/|A|$. Since $b_{1},b_{2}\leq 1-x-y<x+y$ and $b_{0}\leq 1/2<x+y$, we have $b_{i}<x+y$ for $i=0,1,2$. Let $0\leq i\leq 2$, and choose $v\in C_{i}$ uniformly at random. Then $A_{i}^{*}\subseteq N^{2}_{A}(v)$ because $b_{i}<x+y$, and the expected value of $|N^{2}_{A}(v)\cap(A_{i}\setminus A_{i}^{*})|$ is at least $(y/c_{i})|A_{i}\setminus A_{i}^{*}|$; so the expected value of $|N^{2}_{A}(v)|$ is at least

[TABLE]

Since $|N^{2}_{A}(v)|<|A|/3$, it follows that $a^{*}_{i}+(y/c_{i})(a_{i}-a^{*}_{i})<1/3$. Now $A^{*}_{0},A_{01},A_{02}$ are pairwise disjoint subsets of $A_{0}$, so $a_{01}+a_{02}\leq a_{0}-a_{0}^{*}$; and hence

[TABLE]

Similarly we have $a^{*}_{1}+(y/c_{1})(a_{01}+a_{12})<1/3$ and $a^{*}_{2}+(y/c_{2})(a_{02}+a_{12})<1/3$; and by summing these three inequalities and using the equation

[TABLE]

we obtain

[TABLE]

Consequently there exist distinct $i,j\in\{0,1,2\}$ with $y/c_{i}+y/c_{j}-1<0$. But $1/c_{i}+1/c_{j}\geq 4/(c_{i}+c_{j})$, and $c_{i}+c_{j}\leq 1-y$, and so $4y/(1-y)<1$, a contradiction. This proves 8.1.

For $\phi$, we need the following.

8.2

Let $x,y\in(0,1]$. Let $G$ be $(x,2/3)$-constrained via $(A,B,C)$, such that every three vertices in $B$ have a common neighbour in $C$, and every vertex $w\in C$ satisfies $|N^{2}_{A}(w)|<(1-y)|A|$. If $v_{1}\in B$ has $a|A|$ neighbours in $A$ then

[TABLE]

Proof.  Let $v_{1}\in B$ have $a|A|$ neighbours in $A$. Define $A_{1}=N_{A}(v_{1})$ and choose $C_{1}\subseteq N_{C}(v_{1})$ with $|C_{1}|=2|C|/3$ (we may assume this is an integer). Let $A_{1}^{\prime}=A\setminus A_{1}$ and $C_{1}^{\prime}=C\setminus C_{1}$.

(1) If some vertex $v_{2}\in B$ has a set $A_{2}$ of $t|B|$ neighbours in $A_{1}^{\prime}$, then

• •

every $v\in B$ has at most $(1-y-a-t)|A|$ neighbours in $A\setminus(A_{1}\cup A_{2})$; and

• •

the sum over all $v\in B$ of the number of neighbours of $v$ in $A\setminus(A_{1}\cup A_{2})$ is $(1-a-t)x|A||B|$.

The first claim follows since $v_{1},v_{2},v$ have a common neighbour in $C$. The second holds since every vertex in $A\setminus(A_{1}\cup A_{2})$ has $x|B|$ neighbours. The third follows. This proves (1).

The sum over $u\in A_{1}^{\prime}$, of $|N^{2}_{C_{1}}(u)|$, is at most $\frac{2}{3}(1-y-a)|A||C|$; and for each $u$, $|N^{2}_{C_{1}}(u)|\geq\max_{v\in N_{B}(u)}|N_{C_{1}}(v)|$. But the latter is at least

[TABLE]

It follows that

[TABLE]

Consequently

[TABLE]

Moreover, each vertex in $C_{1}^{\prime}$ has at least $x|B|$ nonneighbours in $B$, and so there are at most $(1-x)/3|B||C|$ edges between $B$ and $C_{1}^{\prime}$. Hence there are at least $(1+x)/3|B||C|$ edges between $B$ and $C_{1}$.

For each $v\in B$, let $p(v)=|N_{A_{1}^{\prime}}(v)|/|A|$. Thus $\sum_{v\in B}p(v)=x(1-a)|B|$. By setting $q(v)=3|N_{C_{1}}(v)|/|C|-1$ we deduce: for each $v\in B$ there exists $q(v)$ such that

• •

for each $v\in B$, $1/3\leq q(v)/3+1/3\leq 2/3$, that is, $0\leq q(v)\leq 1$;

• •

$\sum{v\in B}(q(v)/3+1/3)\geq(1+x)/3|B|$, that is, $\sum{v\in B}q(v)\geq x|B|$;

• •

$\sum_{v\in B}p(v)(q(v)/3+1/3)\leq\frac{2}{3}x(1-y-a)|B|$, that is, $\sum_{v\in B}p(v)q(v)\leq(x-2xy-xa)|B|$.

Let $Q\subseteq B$ be the $x|B|$ vertices in $B$ (we may assume this is an integer) with $p(v)$ smallest. Then the expression in the last bullet above is minimized by setting $q(v)=1$ for $v\in Q$, and $q(v)=0$ for $v\in B\setminus Q$. Consequently $\sum_{v\in Q}p(v)\leq(x-xa-2xy)|B|$.

Choose $v_{2}\in B\setminus Q$ with $|N_{A_{1}^{\prime}}(v_{2})|$ maximum; $A_{2}$ say, where $|A_{2}|=t|A|$. By (1), every $v\in B$ has at most $(1-y-a-t)|A|$ neighbours in $A_{1}^{\prime}\setminus A_{2}$, and the sum over all $v\in B$ of the number of neighbours of $v$ in $A_{1}^{\prime}\setminus A_{2}$ is $(1-a-t)x|A||B|$. So the number of edges between $A_{1}^{\prime}\setminus A_{2}$ and $B\setminus Q$ is at most $(1-y-a-t)(1-x)|A||B|$; and the number between $A_{1}^{\prime}\setminus A_{2}$ and $Q$ is at most $(x-xa-2xy)|A||B|$, since $\sum_{v\in Q}p(v)\leq(x-xa-2xy)|B|$. Hence the number between $A_{1}^{\prime}\setminus A_{2}$ and $B$ is at most $((1-y-a-t)(1-x)+x-xa-2xy)|A||B|$, and since this number equals $(1-a-t)x|A||B|$, it follows that

[TABLE]

that is,

[TABLE]

Consequently $t\leq(1-x-y-a+xa-xy)/(1-2x)$.

Now $t(|B|-|Q|)|A|+(x-xa-2xy)|A||B|\geq x(1-a)|A||B|$, so

[TABLE]

Hence

[TABLE]

that is,

[TABLE]

Consequently

[TABLE]

This proves 8.2.

8.3

Let $x,y\in(0,1]$ with $(1-x)^{2}>(2x+1)(1+2x-5x^{2})y$ and $x\geq 1/4$ and $4x^{2}y^{2}\geq(1-y)(x-y)^{2}$. Then $\phi(x,y)\geq 1/3$. Consequently if $x>0.28231$ then $\phi(x,x)\geq 1/3$.

Proof.  Let $\phi(x,y)=z$ and suppose that $z<1/3$. Then there is a graph $G$ that is $(x,1-z)$-constrained via $A,B,C$, such that $|N^{2}_{A}(w)|\leq(1-y)|A|$ for each $w\in C$, by 2.5 and 2.3. As in 5.7, there exists $w\in C$ such that there are at least $x(1-z)|A|\cdot|B|$ edges between $N_{B}(w)$ and $N^{2}_{A}(w)$. Define $B_{1}=N_{B}(w)$ and $B_{2}=B\setminus B_{1}$; and let $B_{2}=t|B|$. For each $u\in A$, let $u$ have $d(u)|B|$ neighbours in $B_{1}$. Let $A_{1}=N^{2}_{A}(w)$ and $A_{2}=A\setminus A_{1}$; and let $|A_{2}|=s|A|$. Thus $d(u)=0$ for each $u\in A_{2}$.

(1) $t\geq x$, and $s\geq y$, and $0\leq d(u)\leq\min(x,1-t)$ for each $u\in A$. Also

[TABLE]

We may assume that every vertex in $A$ has degree exactly $x|B|$; so $0\leq d(u)\leq\min(x,1-t)$ for each $u\in A$. Since $|A_{1}|\leq(1-y)|A|$, it follows that $s\geq y$. In particular, $A_{2}\neq\emptyset$, and so some vertex in $A_{2}$ has $x|B|$ neighbours in $B_{2}$, and so $t\geq x$. Since there are at least $x(1-z)|A|\cdot|B|$ edges between $N_{B}(w)$ and $N^{2}_{A}(w)$, it follows that $\sum_{u\in A_{1}}d(v)\geq x(1-z)|A|$. Since $|A_{1}|\leq(1-y)|A|$ and every vertex in $A_{1}$ has degree exactly $x|B|$, it follows that the number of edges between $A_{1}$ and $B_{1}$ is at most $x(1-y)|A|\cdot|B|$, and so $x(1-y)|A|\geq\sum_{u\in A_{1}}d(v)$. This proves (1).

(2) $1-t\geq\frac{2x(1-x)^{2}/3}{(1-x)^{2}-(1+2x-5x^{2})y}$. Consequently $x<1-3t/2$ and so $x<1-t$.

There are at least $2x|A|\cdot|B|/3$ edges between $B_{1}$ and $A$, since $z<1/3$. But each vertex in $B_{1}$ has at most

[TABLE]

neighbours in $A$, by 8.2, and the first claim follows. To show that $x<1-3t/2$, suppose not; then

[TABLE]

and so

[TABLE]

that is,

[TABLE]

contrary to the hypothesis. This proves (2).

Let us choose $v_{1},v_{1}^{\prime}\in A_{1}$ uniformly and independently at random, and choose $v_{2}\in A_{2}$ uniformly at random. Then for $u\in A$, the probability that all of $v_{1},v_{1}^{\prime},v_{2}$ are nonadjacent to $u$ is

[TABLE]

Since $1-y>2/3$ and so $v_{1},v_{1}^{\prime},v_{2}$ have a common neighbour in $C$, say $w^{\prime}$, and $|N^{2}_{A}(w^{\prime})|\leq(1-y)|A|$, it follows that

[TABLE]

that is,

[TABLE]

This can be rewritten as:

[TABLE]

where $f(r)$ is the polynomial $(r+t-x)(r+t-1)^{2}$. We therefore need to investigate the maximum value of $\sum_{u\in A}f(d(v))$ (which we call “the objective function”) over all choices of the numbers $d(u)(u\in A)$ satisfying the various constraints, and verify that this maximum is less than $t(1-t)^{2}(1-y)|A|$.

The derivative of $f(r)$ is zero when $3r^{2}+2(3t-x-2)r+(3t-2x-1)(t-1)=0$, which has roots $r=1-t$ and $r=(2x+1)/3-t$. Let us define $r_{0}=(2x+1)/3-t$. Since $r_{0}<1-t$, the function $f(r)$ increases for $r<r_{0}$ and for $r>1-t$, and decreases for $r_{0}<r<1-t$.

The second derivative of $f(r)$ is zero when $3r+3t-x-2=0$, that is, when $r=r_{1}$ where $r_{1}=2/3+x/3-t$. By (2), $x<r_{1}$, and we are only concerned $f(r)$ for $r$ in with the range $0\leq r\leq x$; so in particular all such $r$ are less than $r_{1}$. The function $f(r)$ is concave through the range $0\leq r\leq r_{1}$, since its second derivative is at most zero.

Let us choose real numbers $d(v)(v\in A)$ satisfying the constraints

• •

$0\leq d(u)\leq x$ for each $u\in A$;

• •

$d(u)=0$ for at least $y|A|$ vertices $u\in A$;

• •

$x(1-y)|A|\geq\sum_{u\in A_{1}}d(v)\geq 2x|A|/3$

to maximize the function $\sum_{u\in A}f(d(v))$. From the concavity of $f$, it follows that there exists $r^{*}$ with $0<r^{*}\leq x$ such that $d(v)\in\{0,r^{*}\}$ for all $v$ (because if there were $u,v$ with $d(u),d(v)$ distinct and nonzero, replacing them both by $(d(u)+d(v))/2$ would still satisfy the constraints and increase the objective function). Similarly, if there were more than $y|A|$ vertices $v$ with $d(v)=0$, then choose some one of them, $v$ say, and choose some $u$ with $d(u)>0$; then again replacing them both by $(d(u)+d(v))/2$ would still satisfy the constraints and increase the objective function. We deduce that there are exactly $y|A|$ vertices $v$ with $d(v)=0$.

Now the problem breaks into three cases, depending which of the constraints $x(1-y)|A|\geq\sum_{u\in A_{1}}d(v)\geq 2x|A|/3$ hold with equality.

Suppose first that neither holds with equality. Then from the optimality of the objective function, it follows that $r^{*}=r_{0}$, and since

[TABLE]

it follows that

[TABLE]

that is,

[TABLE]

Thus, if $t$ satisfies

[TABLE]

then there is a possible optimal solution where the objective function has value

[TABLE]

Now $f(0)=(t-x)(t-1)^{2}$, and

[TABLE]

We must therefore check that for $t$ in the given range,

[TABLE]

This simplifies to:

[TABLE]

Now the function $27(1-t)^{2}(t-2ty+xy)$ has no local minimum at $t$ with $t<1$, and so is minimized at one of the ends of the range. Since $t\geq x$, we might as well replace the lower extreme of the range by $t\geq x$ (because it makes the arithmetic easier); so to check the lower extreme, we need to check that

[TABLE]

that is, $4(1-x)<27x,$ which is true by hypothesis.

For the upper extreme, $t\leq(2x+1)/3-2x/(3-3y)<1/3$; so it suffices to check that

[TABLE]

when $t=1/3$, that is, to check $(1-x)^{3}(1-y)<(1-2y+3xy)$. But $(1-x)^{3}<1/2$ by hypothesis, and $(1-2y+3xy)/(1-y)\geq(1-2y)/(1-y)\geq 1/2$ since $y<1/3$. This finished the first of the three cases.

Now let us assume that $x(1-y)|A|=\sum_{u\in A_{1}}d(v)$. It follows from the optimality of the objective function that $r^{*}\leq r_{0}$. Moreover, since $x(1-y)|A|=\sum_{u\in A_{1}}d(v)$, it follows that $x(1-y)|A|=(1-y)|A|r^{*}$, so $r^{*}=x$. This is only possible if $x\leq r_{0}$, that is, $t<(1-x)/3$; and this is impossible since $t\geq x\geq 1/4$. This finishes the second case.

Finally, we assume that $\sum_{u\in A_{1}}d(v)=2x|A|/3$. It follows from the optimality of the objective function that $r^{*}\geq r_{0}$. Moreover, since $\sum_{u\in A_{1}}d(v)=2x|A|/3$, it follows that $(1-y)|A|r^{*}=2x|A|/3$, that is, $r^{*}=2x/(3-3y)$. We must check that

[TABLE]

This is cubic in $t$, and, collecting the various powers of $t$, it becomes:

[TABLE]

[TABLE]

[TABLE]

This simplifies to:

[TABLE]

The derivative of the left side with respect to $t$ is

[TABLE]

which can be rewritten as

[TABLE]

Since by hypothesis, $-(x-y)^{2}/(3y)+4x^{2}y/(3-3y)\geq 0$, the derivative is nonnegative, at every value of $t$. Thus we only need verify the inequality for the maximum value of $t$ that lies in the range.

By (2), $t\leq 2(1-x)/3$; so it is enough to verify that

[TABLE]

holds when $t=2(1-x)/3$. Thus we need to check that

[TABLE]

[TABLE]

According to WolframAlpha, this is true for all $x,y$ with $1/4<x<1/3$ and $0<y\leq x$. This proves 8.3.

8.4

Let $x,y\in(0,1]$ with $x\leq\frac{1}{4}$ and $y<\frac{1}{3}$ and $y<\frac{(1-2x)^{2}}{3-12x+16x^{2}}$; then $\phi(x,y)<\frac{1}{3}$.

Proof.  Apply 3.1 to 6.7. This proves 8.4.

9 The 3/4 Level

In this section we investigate when $\psi(x,y)\geq 3/4$ and $\phi(x,y)\geq 3/4$. The results are shown in figure 8.

9.1

Let $x,y\in(0,1]$, such that $y>1/3$, $x\geq 1/2$, and $2y-2y^{2}>1-x$. Then $\psi(x,y)\geq 3/4$.

Proof.  Let $G$ be a graph that is $(x,y)$-biconstrained via $(A,B,C)$. We can assume that $x,y$ are rational, and by multiplying vertices if necessary that $y|B|\in\mathbb{Z}$. Let $v_{1}\in C$, and let $B_{1}\subseteq N(v_{1})$ be such that $|B_{1}|=y|B|$. Choose $v_{2}\in C$ with at least $y|B\setminus B_{1}|=y(1-y)|B|$ neighbours in $B\setminus B_{1}$, and choose $B_{2}\subseteq N(v_{2})$ with $|B_{2}|=y|B|$. Choose $v_{3}\in C$ with at least $y(1-2y)$ neighbours in $B\setminus(B_{1}\cup B_{2})$. Thus $|N(v_{1})\cup N(v_{2})|\geq y+y(1-y)>1-x$, and for $i=1,2$, $|N(v_{i})\cup N(v_{3})|\geq y+y(1-2y)>1-x$.

For $1\leq i\leq 3$, let $A_{i}=N_{A}^{2}(v_{i})$. Since $y>1/3$, it follows that there exist $i,j$ with $1\leq i<j\leq 3$ such that $N(v_{i})\cap N(v_{j})\neq\emptyset$, and so $|A_{i}\cap A_{j}|\geq x|A|\geq|A|/2$. But $A_{i}\cup A_{j}=A$ since $|N(v_{i})\cup N(v_{j})|>1-x$, and so $|A_{i}|+|A_{j}|\geq 3|A|/2$, and therefore one of $|A_{i}|,|A_{j}|\geq 3|A|/4$. This proves 9.1.

9.2

Let $x,y\in(0,1]$, such that $x>2/3$, $x+2y>1$, and either $4(1-x)(1-y)\leq 1$ or $x>1-2y+2y^{2}$. Then $\psi(x,y)\geq 3/4$.

Proof.  If $x+y>1$ the result follows from (theorem 4.1 of the paper); so we may assume that $y\leq 1-x<1/3$. Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $|N_{A}^{2}(v)|<3|A|/4$ for each $v\in C$. Let $H$ be the graph with $V(H)=V(C)$, where distinct $u,v$ are adjacent in $H$ if and only if $u$ and $v$ have distance two in $G$ (that is, in $G$ they have a common neighbour in $B$).

(1) For all $w_{1},w_{2},w_{3}\in C$, if $N(w_{1})\cap N(w_{2})\neq\emptyset$ and $N(w_{2})\cap N(w_{3})\neq\emptyset$, then $N(w_{1})\cap N(w_{3})\neq\emptyset$. Consequently each component of $H$ is a complete graph.

Suppose that $w_{1},w_{2},w_{3}\in C$, and $N(w_{1})\cap N(w_{3})=\emptyset$, and $v_{i}\in N(w_{i})\cap N(w_{2})$ for $i=1,3$. Let $A_{i}=N_{A}^{2}(w_{i})$ for $i=1,2,3$. Since $x+2y>1$ we have $A_{1}\cup A_{3}=A$. Consequently $|A_{1}\cap A_{3}|<|A|/2$, and since $N_{A}(v_{1})\cap N_{A}(v_{3})\subseteq A_{1}\cap A_{3}$, it follows that $|N_{A}(v_{1})\cap N_{A}(v_{3})|<|A|/2$. Thus

[TABLE]

a contradiction. This proves (1).

Let $\alpha$ be the size of the largest stable set in $H$, that is, the number of components of $H$. Let the vertex sets of the components of $H$ be $C_{1},\ldots,C_{\alpha}$, and for $1\leq i\leq\alpha$ let $B_{i}$ be the set of vertices in $B$ with a neighbour in $C_{i}$. The sets $B_{1},\ldots,B_{\alpha}$ have union $B$, and from the definition of $H$, they are pairwise disjoint. For $1\leq i\leq\alpha$, let $w_{i}\in C_{i}$ and let $A_{i}=N^{2}_{A}(w_{i})$.

(2) For $1\leq i<j\leq\alpha$, $A_{i}\cup A_{j}=A$, and so $|A_{i}\cap A_{j}|<|A|/2$. Consequently $\alpha\leq 3$.

Since $w_{i},w_{j}$ have no common neighbour in $B$, it follows that $|N(w_{i})\cup N(w_{j})|\geq 2y|B|>(1-x)|B|$, and so $A_{i}\cup A_{j}=A$. Since $|A_{i}|,|A_{j}|<3|A|/4$, it follows that $|A_{i}\cap A_{j}|<|A|/2$. This proves the first assertion. Suppose that $\alpha\geq 4$. By the first assertion, every vertex in $A$ belongs to at least three of $A_{1},\ldots,A_{4}$. Consequently some $A_{i}$ has cardinality at least $3|A|/4$, a contradiction. This proves (2).

(3) $\alpha\neq 1$.

Suppose that $\alpha=1$. Every vertex in $A\setminus A_{1}$ has at least $x|B|$ neighbours in $B\setminus N(w_{1})$, so we may choose $v\in B$ with at least

[TABLE]

neighbours in $A\setminus A_{1}$. Let $w\in C$ be a neighbour of $v$. Since $w_{1},w$ have a common neighbour, it follows that

[TABLE]

a contradiction. This proves (3).

(4) For $1\leq i\leq\alpha$, if $|B_{i}|>(1-x)|B|$ then $|C_{i}|>\frac{y}{3-4x}|C|$.

Suppose that $|B_{i}|>(1-x)|B|$ say, and let $|C_{i}|=c|C|$. Since $x>2/3$, every vertex $u\in A\setminus A_{i}$ has a neighbour in $B_{i}$, and so $|N^{2}_{C}(u)\cap C_{i}|\geq y|C|=(y/c)|C_{i}|$. Hence there exists $w\in C_{i}$ such that

[TABLE]

But $w,w_{i}$ have a common neighbour, and so $|N^{2}_{A}(w)\cap A_{i}|\geq x|A|$, and therefore

[TABLE]

Consequently $3/4>x+y/(4c)$, and so $c>y/(3-4x)$. This proves (4).

(5) $x>1-2y+2y^{2}$* and $\alpha=2$.*

Since $\alpha\leq 3$, we may assume without loss of generality that $|B_{1}|\geq|B|/3>(1-x)|B|$. Since each $|C_{i}|\geq y|C|$, it follows that $|C_{1}|\leq(1-(\alpha-1)y)|C|$. By (4), $1-(\alpha-1)y>y/(3-4x)$, and since $\alpha\geq 2$, it follows that $1-y>y/(3-4x)$, that is, $4(1-x)(1-y)>1$. From the hypothesis it follows that $x>1-2y+2y^{2}$. This proves the first claim. Suppose that $\alpha>2$; then

[TABLE]

which simplifies to $(1-y)(1-4y)^{2}<0$, a contradiction. This proves (5).

We may assume without loss of generality that $|C_{1}|\leq|C|/2$. By (4), $|B_{1}|\leq(1-x)|B|<|B|/2$, and so $|B_{2}|\geq|B|/2$.

Every vertex in $B_{2}\setminus N(w_{2})$ is adjacent to at least a fraction $y/(1-y)$ of the vertices of $C_{2}$, and hence there exists $w\in C_{2}$ with

[TABLE]

Thus

[TABLE]

Since $|B_{2}|\geq x|B|$ and $|N(w_{2})|\geq y|B|$, it follows that

[TABLE]

(because $x>1-2y+2y^{2}$ by (5)). Thus, $N_{A}^{2}(w)\cup N_{A}^{2}(w_{2})=A$, and since $w$ and $w_{2}$ have a common neighbour in $B_{2}$ it follows that

[TABLE]

and so one of $|N^{2}_{A}(w)|,|N^{2}_{A}(w_{2})|\geq 3|A|/4$, a contradiction. This proves 9.2.

9.3

Let $x,y\in(0,1]$, with $x>1/3$, $y>1/2$, $x+3y>2$, and either $x\geq(5-6y)/(11-12y)$ or $x\geq(3-4y)/(4-4y)$. Then $\psi(x,y)\geq 3/4$.

Proof.  Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $N^{2}_{A}(v)<3|A|/4$ for each $v\in C$. By (theorem 4.1 of the paper) it follows that $x+y\leq 1$. Let $H$ be the graph with $V(H)=V(C)$, in which distinct $u,v$ are adjacent if and only if $|N(u)\cup N(v)|\leq(1-x)|B|$.

(1) For all $u,v\in C$, if $u,v$ are nonadjacent in $H$ then $N^{2}_{A}(u)\cup N^{2}_{A}(v)=A$. If $u,v$ are adjacent in $H$ then $|N^{2}_{A}(u)\cap N^{2}_{A}(v)|>(x+1/4)|A|$.

If $u,v$ are nonadjacent in $H$, then $|N(u)\cup N(v)|>(1-x)|B|$, and so every vertex in $A$ has a neighbour in $N(u)\cup N(v)$, that is, $N^{2}_{A}(u)\cup N^{2}_{A}(v)=A$. Now we assume that $u,v$ are adjacent in $H$. Consequently

[TABLE]

by hypothesis. Moreover, $x>1/3$ and $y>1/2$ imply that

[TABLE]

Thus, 7.4 implies that $|N^{2}_{A}(u)\cap N^{2}_{A}(v)|>(x+1/4)|A|$. This proves (1).

If there exist $w_{1},\ldots,w_{4}\in C$, pairwise nonadjacent in $H$, then by (1) each pair of the sets of the sets $N^{2}_{A}(w_{i})\;(1\leq i\leq 4)$ has union $A$, and so each vertex in $A$ belongs to at least three of the four sets; and so one of the four sets has cardinality at least $3|A|/4$, a contradiction. Thus we may choose $w_{1},w_{2},w_{3}\in C$ such that every other vertex in $C$ is adjacent in $H$ to at least one of $w_{1},w_{2},w_{3}$. Choose a partition $C=C_{1}\cup C_{2}\cup C_{3}$ such that for $1\leq i\leq 3$, every vertex in $C_{i}$ is equal or adjacent in $H$ to $w_{i}$.

(2) $|C_{i}|\leq(1-y)|C|$* for $1\leq i\leq 3$.*

Suppose that $|C_{1}|>(1-y)|C|$. Define $B_{1}=N(w)$ and $A_{1}=N^{2}_{A}(w)$. Choose $v\in B\setminus N(w)$ with at least $x|A\setminus A_{1}|/(1-y)>x|A|/(4-4y)$ neighbours in $A\setminus A_{1}$. Since $|C_{1}|>(1-y)|C|$, there exists $w\in C_{1}$ adjacent to $v$. Then

[TABLE]

since $x>1/3$ and $y>1/2$, a contradiction. This proves (2).

(3) Every vertex in $B$ has neighbours in exactly two of $C_{1},C_{2},C_{3}$.

Since each $|C_{i}|\leq(1-y)|C|<y|C|$ by (2), it follows that every vertex in $B$ has neighbours in at least two of $C_{1},C_{2},C_{3}$. Suppose that $v\in B$ has a neighbour $w_{i}^{\prime}\in C_{i}$ for $i=1,2,3$. Let $A_{i}=N_{A}^{2}(w_{i}^{\prime})$ for $i=1,2,3$. For $1\leq i<j\leq 3$, $A_{i}\cup A_{j}=A$ by (1), and so every vertex of $A$ belongs to at least two of $A_{1},A_{2},A_{3}$, and $N_{A}(u)$ is a subset of all three of $A_{1},A_{2},A_{3}$. Consequently

[TABLE]

and so some $|A_{i}|\geq 3|A|/4$, a contradiction. This proves (3).

From (3) we may partition $B$ into $B_{1},B_{2},B_{3}$ such that every vertex in $B_{1}$ has neighbours in $C_{2}$ and in $C_{3}$ but not in $C_{1}$, and similarly for $B_{2},B_{3}$. Without loss of generality, we may assume that $|B_{1}|\leq 1/3$. Let $A_{1}=N^{2}_{A}(w_{1})$.

(4) $|N^{2}_{A}(w)\setminus A_{1}|<(2-4x)|A\setminus A_{1}|$* for each $w\in C_{1}$.*

By (1) $|N^{2}_{A}(w_{1})\cap N^{2}_{A}(w)|\geq(x+1/4)|A|$, and since $|N^{2}_{A}(w)|<3|A|/4$, it follows that

[TABLE]

This proves (4).

This has two consequences. The first is that $x<(3-4y)/(4-4y)$. To see this, by (4) we may choose $u\in A\setminus A_{1}$ such that $|N^{2}_{C}(u)\cap C_{1}|<(2-4x)|C_{1}|$. Since $|B_{1}|\leq|B|/3$, $u$ has a neighbour $v\in B_{2}\cup B_{3}$, and we may assume that $v\in B_{2}$ from the symmetry. So at least $y|C|$ neighbours of $v$ belong to $C_{1}\cup C_{3}$, and therefore at least $(y-(1-y))|C|$ neighbours belong to $C_{1}$, since $|C_{2}|\leq(1-y)|C|$. So $(2y-1)|C|<(2-4x)|C_{1}|\leq(2-4x)(1-y)|C|$, and hence $x<(3-4y)/(4-4y)$ as claimed.

The second consequence is that $x<(5-6y)/(11-12y)$. To see this, let $S=B\setminus(B_{1}\cup N(w_{1}))$. By (4) and since each vertex in $B_{2}\cup B_{3}$ has a neighbour in $C_{1}$, it follows that each vertex $v\in B_{2}\cup B_{3}$ has fewer than $(2-4x)|A\setminus A_{1}|$ neighbours in $A\setminus A_{1}$. Since $S\subseteq B_{2}\cup B_{3}$, it follows that some vertex $u\in A\setminus A_{1}$ has fewer than $(2-4x)|S|$ neighbours in $S$. But $u$ has no neighbours in $N(w_{1})$, and only at most $r|B|$ neighbours in $B_{1}$; and since it has at least $x|B|$ neighbours in total, we deduce that

[TABLE]

(since $B_{1}\cap N(w)=\emptyset$ and $|N(w_{1})|\geq y|B|$ and therefore $|S|\leq(1-r-y)|B|$). Consequently

[TABLE]

and so $x<(5-6y)/(11-12y)$.

We have shown then that $x<(3-4y)/(4-4y)$ and $x<(5-6y)/(11-12y)$; but this contradicts the hypothesis. This proves 9.3.

9.4

Let $x,y\in(0,1]$ with $1/2<y<2/3$ and $x>6y^{2}-8y+3$. Then $\psi(x,y)\geq 3/4$.

Proof.  We may assume that $y\in\mathbb{Q}$, by decreasing $y$ if necessary. Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(w)|<3|A|/4$ for each $w\in C$. By multiplying vertices if necessary, we may assume that $y|B|\in\mathbb{Z}$. Since $x>6y^{2}-8y+3=6(y-2/3)^{2}+1/3$, it follows that $x>1/3$. Let $H$ be the graph with $V(H)=C$ in which distinct $u,v\in C$ are adjacent if and only if $|N(u)\cup N(v)|\leq(1-x)|B|$. It follows that if $u,v$ are nonadjacent in $H$, then $N^{2}_{A}(u)\cup N^{2}_{A}(v)=A$. As in the proof of 9.3, there do not $w_{1},\ldots,w_{4}\in C$, pairwise nonadjacent in $H$; and so we may choose $w_{1},w_{2},w_{3}\in C$ and a partition $C=C_{1}\cup C_{2}\cup C_{3}$ such that for $1\leq i\leq 3$, every vertex in $C_{i}$ is equal or adjacent in $H$ to $w_{i}$. Let $|C_{i}|=c_{i}|C|$, and choose $B_{i}\subseteq N_{(}w_{i})$ with $|B_{i}|=y|B|$ for $1\leq i\leq 3$. Let $F$ be the set of all edges $vw$ of $G$ with $v\in B$ and $w\in C$, such that for $1\leq i\leq 3$, not both $v\in B_{i}$ and $w\in C_{i}$.

(1) $\frac{|F|}{|B||C|}\leq(1-x-y)<(2-3y)(2y-1)$.

Let $w\in C$, with $w\in C_{i}$ say; then since $w,w_{i}$ are adjacent in $H$, it follows that $w$ has at most $(1-x-y)|B|$ neighbours in $B\setminus B_{i}$. Thus $|F|\leq(1-x-y)|B|\cdot|C|$. But $1-x-y<(2-3y)(2y-1)$ since $x>6y^{2}-8y+3$. This proves (1).

Let $p_{1}=|B_{1}\setminus(B_{2}\cup B_{3})|/|B|$, and define $p_{2},p_{3}$ similarly. Let $q_{1}=|(B_{2}\cup B_{3})\setminus B_{1}|/|B|$, and define $q_{2},q_{3}$ similarly. Let $p_{0}=|B\setminus(B_{1}\cup B_{2}\cup B_{3})|/|B|$, and $q_{0}=|B_{1}\cup B_{2}\cup B_{3}|/|B|$. Let $q=q_{1}+q_{2}+q_{3}$. Thus

[TABLE]

By subtracting the last three of these from the first, we obtain

[TABLE]

and so $p_{0}=2q_{0}+q-3y+1$.

Every vertex in $B\setminus(B_{1}\cup B_{2}\cup B_{3})$ is incident with at least $y|C|$ edges in $F$, every vertex in $B_{1}\setminus(B_{2}\cup B_{3})$ is incident with at least $(y-c_{1})|C|$ edges in $F$, and every vertex in $(B_{2}\cup B_{3})\setminus B_{1}$ is incident with at least $\max(y-c_{2}-c_{3},0)=\max(y+c_{1}-1,0)$ edges in $F$ (and similar statements hold for $c_{2},c_{3}$). Summing, we deduce that

[TABLE]

Since $p_{i}=y-q_{0}-q+q_{i}$ for $i=1,2,3$, and $c_{1}+c_{2}+c_{3}=1$, it follows that

[TABLE]

Also, $p_{0}=2q_{0}+q-3y+1$, and so

[TABLE]

This simplifies to

[TABLE]

where $I$ is the set of $i\in\{1,2,3\}$ such that $c_{i}<1-y$. From (1) we deduce that

[TABLE]

In particular it follows that $(1-y)q_{0}<(2-3y)(2y-1)\leq(1-y)(2y-1)$, and so $q_{0}<2y-1$. Moreover, since $|B_{2}\cup B_{3}|\leq|B|$, it follows that $|B_{2}\cap B_{3}|\geq(2y-1)|B|$, and so $q_{1}\geq 2y-1-q_{0}$, and the same holds for $q_{2},q_{3}$. Consequently

[TABLE]

and so

[TABLE]

since $2y-1-q_{0}>0$. This simplifies to $(2y-1)q_{0}<0$, a contradiction. This proves 9.4.

9.5

Let $x,y\in(0,1]$ with $x\geq 1/4$ and $y>2/3$. Then $\psi(x,y)\geq 3/4$.

Proof.  Suppose that $G$ is $(x,y)$-biconstrained via $(A,B,C)$, and $|N^{2}_{A}(w)|<3|A|/4$ for each $w\in C$. Since $x\geq 1/4$ and $y>2/3$, it follows that $x>2(1-y)^{2}$, that is, $y>(1-y)+1-x/(1-y)$ ; and also $y>2-2y$. Let $w\in C$. By 7.4 with $k=2$ and $B^{\prime}=N(w)$, it follows that

[TABLE]

which is a contradiction. This proves 9.5.

9.6

Let $x,y\in(0,1]$ with $y>2/3$, $x+4y>3$ and $x>3(1-y)^{2}/(2-y)$. Then $\psi(x,y)\geq 3/4$.

Proof.  Suppose that $G$ is $(x,y)$-biconstrained via $(A,B,C)$, and $|N^{2}_{A}(w)|<3|A|/4$ for each $w\in C$. Consequently $y\leq 3/4$, and so $x\geq 3/20$ since $x>3(1-y)^{2}/(2-y)$. From the hypotheses it follows that

[TABLE]

If $w_{1},w_{2}\in C$ with $|N(w_{1})\cup N(w_{2})|\leq(1-x)|B|$ then $|N(w_{1})\cap N(w_{2})|\geq(2y+x-1)|B|$. Thus, 7.4 applied with $k=2$ tells us that, for all such $w_{1},w_{2}\in C$, more than $(x+1/2)|A|$ vertices in $A$ have a neighbour in $N(v_{1})\cap N(v_{2})$. Let $H$ be the graph with vertex set $C$, in which $w_{1},w_{2}$ are adjacent if $|N(w_{1})\cup N(w_{2})|\leq(1-x)|B|$. As in the proof of 9.3, there is no stable set of size at least four in $H$. It follows that there exist $w_{1},w_{2},w_{3}\in C$ and a partition $C=C_{1}\cup C_{2}\cup C_{3}$ such that for $1\leq i\leq 3$, every vertex in $C_{i}$ is equal to or adjacent in $H$ to $w_{i}$. We assume without loss of generality that $|C_{1}|\geq 1/3$. Since $y>2/3$, every vertex in $B$ has a neighbour in $C_{1}$. Let $B_{1}=N(w_{1})$ and $A_{1}=N_{A}^{2}(w_{1})$, and choose $v\in B\setminus B_{1}$ with more than $x|A|/(4-4y)$ neighbours in $A\setminus A_{1}$. Since $y>2/3$, there exists $w\in C_{1}$ adjacent to $v$. Then

[TABLE]

since $x\geq 3/20\geq 1/7$ and $y\geq 2/3$, a contradiction. This proves 9.6.

9.7

Let $x,y\in(0,1]$ with $y>1/2$ and $x^{2}y\geq(3/4-x)^{2}$. Then $\phi(x,y)\geq 3/4$.

Proof.  Apply 5.7 with $z=3/4$.

9.8

Let $x,y\in(0,1]$ with $y<1/3$ and $x>\frac{1-y}{1+y-3y^{2}}$. Then $\phi(x,y)\geq 3/4$.

Proof.  Suppose that $G$ is $(x,y)$-constrained via $(A,B,C)$, and $|N^{2}_{A}(w)|<3|A|/4$ for each $w\in C$. Consequently $x+y\leq 1$. By reducing $x$ or $y$ if necessary, we may assume that every vertex in $A$ has strictly more than $x|B|$ neighbours in $B$, and that $x,y$ are rational. Let

[TABLE]

By multiplying vertices, we may also assume that $y|B|$ and $p|B|$ are integers. Note that the hypotheses imply that $p<xy$.

(1) There exists $s\in[0,1]$ such that for all $b,c$, if $0\leq a\leq y$ and $sa+b\geq y(sy+1-y)$, then $a+b\geq p$ and $b\geq 1-x-y$.

We claim first that

[TABLE]

To see this, we need to check that $0\leq\frac{2y-y^{2}+x-1}{y-y^{2}}$, and $\frac{p-y(1-y)}{y^{2}}\leq 1$, and $\frac{p-y(1-y)}{y^{2}}\leq\frac{2y-y^{2}+x-1}{y-y^{2}}$. The first is true since

[TABLE]

The second is true since $p\leq xy\leq y$. The third simplifies to $p/y\leq x/(1-y)$, and this is true since $p\leq xy$. This proves the claim, and so there exists $s$ such that

[TABLE]

We will show that $s$ satisfies (1). Suppose that $0\leq a\leq y$ and $sa+b\geq y(sy+1-y)$. Then

[TABLE]

and

[TABLE]

(and therefore $b\geq 1-x-y$). This proves (1).

(2) There exists $t\in[0,1]$ such that for all $a,b$, if $0\leq a\leq 2p$ and $ta+b\geq y(1-2p(1-t))$ then $a+b\geq p$ and $p+b\geq 1-x$.

We claim first that

[TABLE]

To see this we must check that $0\leq\frac{x+y+p-2py-1}{2p(1-y)}$, and $\frac{2py+p-y}{2py}\leq 1$, and $\frac{2py+p-y}{2py}\leq\frac{x+y+p-2py-1}{2p(1-y)}.$ The first is true since

[TABLE]

The second is true since $p\leq xy\leq y$; and the third simplifies to $p\leq xy$ and so is true. This proves the claim, and so there exists $t$ with

[TABLE]

We will show that $t$ satisfies (2). Let $a,b$ satisfy $0\leq a\leq 2y$ and $ta+b\geq y(1-2p(1-t))$. Then

[TABLE]

and

[TABLE]

(and so $b\geq 1-x-p$). This proves (2).

Choose $w_{1}\in C$ with at least $y|B|$ neighbours in $B$.

(3) There exists $w_{2}\in C$ such that $|N(w_{2})|\geq p|B|$ and $|N(w_{1})\cup N(w_{2})|\geq(1-x)|B|$.

Choose $B_{1}\subseteq N(w_{1})$ with $|B_{1}|=y|B|$. Choose $s$ as in (1). Then

[TABLE]

Consequently we may choose $w_{2}\in C$ such that

[TABLE]

Since

[TABLE]

the choice of $s$ implies that $|N(w_{2})|\geq p|B|$ and $|N(w_{2})\setminus B_{1}|\geq(1-x-y)|B|$, and so $|N(w_{1})\cup N(w_{2})|\geq(1-x)|B|$. This proves (3).

(4) There exists $w_{3}\in C$ such that $|N(w_{3})|\geq p|B|$, and $|N(w_{i})\cup N(w_{3})|\geq(1-x)|B|$ for $i=1,2$.

Since $|N(w_{1})|,|N(w_{2})|\geq p|B|$ and $p|B|$ is an integer, we may choose $R\subseteq B$ with $|R|=2p|B|$ such that $|N(w_{1})\cap R|,|N(w_{2})\cap R|\geq p|B|$. Choose $t$ as in (2). As in the proof of (3), there exists $w_{3}\in C$ with

[TABLE]

From the choice of $t$, it follows that $|N(w_{3})|/|B|\geq p$, and $|N(w_{3})\setminus R|\geq 1-x-p$, and consequently $|N(w_{1})\cup N(w_{3})|,|N(w_{2})\cup N(w_{3})|\geq(1-x)|B|$. This proves (4).

For $1\leq i\leq 3$, choose $B_{i}\subseteq N(w_{i})$ with $|B_{i}|=p|B|$. Since $|B_{1}\cup B_{2}\cup B_{3}|\leq 3p$, we may choose $w_{4}\in C$ with at least $y(1-3p)|B|$ neighbours in $B\setminus(B_{1}\cup B_{2}\cup B_{3})$. Then for all $1\leq i\leq 3$ we have:

[TABLE]

by the definition of $p$. It follows that for $1\leq i<j\leq 4$, $|N(w_{i})\cup N(w_{j})|\geq(1-x)|B|$, and so (since every vertex in $A$ has strictly more than $x|B|$ neighbours in $B$) it follows that $N^{2}_{A}(w_{i})\cup N^{2}_{A}(w_{j})=A$. Thus every vertex in $A$ belongs to at least three of the four sets $N^{2}_{A}(w_{i})\;(1\leq i\leq 4)$, and so one of them has cardinality at least $3|A|/4$, a contradiction. This proves 9.8.

9.9

Let $x,y\in(0,1]$. Then $\psi(x,y)<3/4$ if either:

• •

$x\leq 1/6$* and $y\leq 5/7$ and $2x+y\leq 1$; or*

• •

$x\leq 3/20$* and $x+4y\leq 3$, and $x+4y<3$ if $x$ is irrational; or*

• •

$x\leq 17/23$* and $y\leq 1/8$ and $x+3y\leq 1$; or*

• •

$x\leq 5/7$* and $y\leq 1/6$ and $x+2y\leq 1$.*

Proof.  If $x^{\prime},y^{\prime}$ with $2x^{\prime}+y^{\prime}\leq 1$ and $y^{\prime}\leq 3/5$, then $\psi(x^{\prime},y^{\prime})<2/3$ by the first bullet of 7.8. Given $x,y$ as in the first bullet, the hypotheses imply that there is a choice of $x^{\prime},y^{\prime}$ with $2x^{\prime}+y^{\prime}\leq 1$ and $y^{\prime}\leq 3/5$, and which also satisfy the hypotheses of 3.5 with $z^{\prime}=\psi(x^{\prime},y^{\prime})$ and $z$ slightly less than $3/4$ (checking this needs some lengthy calculation, which we omit); and so the first statement follows from 3.5. The second statement follows similarly from 3.5 and the second bullet of 7.8. The third statement follows from 3.4 and the first bullet of 7.6; and the fourth follows by applying 3.4 with $z=\max(2/7,x)$, taking $x^{\prime}=3/5$, $y^{\prime}=1/5$ and $z^{\prime}=3/5$. This proves 9.9.

9.10

If $x,y\in(0,1]$, with $y\leq 1/3$ and

[TABLE]

then $\phi(x,y)<3/4$.

Proof.  Apply 3.7 with $z$ slightly less than $3/4$ to 7.10. This proves 9.10.

10 The 2/5 Level

Next, we analyze when $\psi,\phi\geq 2/5$. The results are shown in figure 9.

10.1

Let $x,y\in(0,1]$ with $x\geq 1/5$, $y>1/3$, and $3y-2y^{2}>1-x$; then $\psi(x,y)\geq 2/5$.

Proof.  Suppose that $G$ is $(x,y)$-biconstrained via $(A,B,C)$, and $|N^{2}_{A}(w)|<(2/5)|A|$ for each $w\in C$. Let $w_{1}\in C$, and let $A_{1}=N_{A}^{2}(w_{1})$. By averaging, there exists $w_{2}\in C$ such that

[TABLE]

where $A_{2}=N^{2}_{A}(w_{2})$. Since $|A_{2}|<2|A|/5$, it follows that

[TABLE]

and so $N(w_{1})\cap N(w_{2})=\emptyset$. Let $B^{\prime}=N(w_{1})\cup N(w_{2})$; thus $|B^{\prime}|\geq 2y|B|$. By averaging, there exists $w_{3}\in C$ such that $|N(w_{3})\setminus B^{\prime}|\geq y|B\setminus B^{\prime}|$, and so

[TABLE]

Hence, setting $A_{3}=N_{A}^{2}(w_{3})$, it follows that $A_{1}\cup A_{2}\cup A_{3}=A$.

Since $y>1/3$, some pair of $N(w_{1}),N(w_{2}),N(w_{3})$ have nonempty intersection, and so some pair of $A_{1},A_{2},A_{3}$ have intersection of cardinality at least $x|A|\geq|A|/5$. But then

[TABLE]

which is impossible since $|A_{i}|<(2/5)|A|$ for $1\leq i\leq 3$. This proves 10.1.

10.2

Let $x,y\in(0,1]$ with $x>1/3$, $x+3y>1$, and either $y\geq 1/4$ or $x+y/(10(1-2y))\geq 2/5$; then $\psi(x,y)\geq 2/5$.

Proof.  Suppose that $G$ is $(x,y)$-biconstrained via $(A,B,C)$, and $|N^{2}_{A}(w)|<(2/5)|A|$ for each $w\in C$. Choose $w_{1},\ldots,w_{\alpha}\in C$ with $\alpha$ maximum such that $N(w_{1}),\ldots,N(w_{\alpha})$ are pairwise disjoint.

(1) $\alpha=3$.

Suppose that $\alpha\leq 2$. Let $A^{\prime}$ be the union of the sets $N^{2}_{A}(w_{i})$ for $1\leq i\leq\alpha-1$, and let $B^{\prime}$ be the union of the sets $N(w_{i})$ for $1\leq i\leq\alpha-1$. So $|A^{\prime}|<(2\alpha/5)|A|\leq(4/5)|A|$, and $|B^{\prime}|\geq y|B|$. By averaging, there exists $v\in B\setminus B^{\prime}$ such that

[TABLE]

let $w\in C$ be adjacent to $v$. Since $N(w)$ has nonempty intersection with $N(w_{i})$ for some $i<\alpha$, it follows that $|N^{2}_{A}(w)\cap A^{\prime}|\geq x|A|$. Adding, we deduce that

[TABLE]

a contradiction.

Thus $\alpha\geq 3$; suppose that $\alpha\geq 4$. Since $x+3y>1$ it follows that every vertex in $A$ belongs to at least two of the sets $N^{2}_{A}(w_{i})\;(1\leq i\leq 4)$, and so one of these four sets has cardinality at least $|A|/2\geq(2/5)|A|$, a contradiction. This proves (1).

(2) If $w\in C$ then $N(w)\cap N(w_{i})$ is nonempty for exactly one value of $i\in\{1,2,3\}$.

Since $\alpha=3$, it follows that $N(w)\cap N(w_{i})$ is nonempty for at least one such $i$; suppose that $N(w)$ has nonempty intersection with both $N(w_{1}),N(w_{2})$ say. Let $A_{i}=N^{2}_{A}(w_{i})$ for $i=1,2,3$, and $A_{0}=N^{2}_{A}(w)$. Since $A_{1}\cup A_{2}\cup A_{3}=A$, and each $|A_{i}|<(2/5)|A|$, there are fewer than $|A|/5$ vertices in $A$ that belong to more than one of $A_{1},A_{2},A_{3}$, and in particular $|A_{1}\cap A_{2}|<|A|/5$. But $|A_{0}\cap A_{i}|\geq x|A|$ for $i=1,2$, and so $|A_{0}|\geq(2x-1/5)|A|\geq(2/5)|A|$, a contradiction. This proves (2).

From (2), we can partition $B=B_{1}\cup B_{2}\cup B_{3}$, and partition $C=C_{1}\cup C_{2}\cup C_{3}$, such that all six of these sets are nonempty, and for all distinct $i,j\in\{1,2,3\}$ there is no edge between $B_{i}$ and $C_{j}$, and for all $i\in\{1,2,3\}$ and all $w,w^{\prime}\in C_{i}$, $N(w)\cap N(w^{\prime})\neq\emptyset$. Let $|B_{i}|=b_{i}|B|$ and $C_{i}|=c_{i}|C|$ for $i=1,2,3$. Without loss of generality we may assume that $b_{3}\leq 1/3<x$, and so every vertex in $A$ has a neighbour in $B_{1}\cup B_{2}$.

(3) $x+y/(10(1-2y))<2/5$* and so $y\geq 1/4$.*

Suppose that $x+y/(10(1-2y))\geq 2/5$. Without loss of generality, we may assume that at least $|A|/2$ vertices in $A$ have a neighbour in $B_{1}$. Choose $w\in C_{1}$. Since $|N^{2}_{A}(w)|<(2/5)|A|$, there are at least $|A|/10$ vertices $u\in A\setminus N^{2}_{A}(w)$ that have a neighbour in $B_{1}$. For each such $u$, $|N^{2}_{C}(u)\cap C_{1}|\geq y|C|$, and since $|C_{1}|\leq(1-2y)|C|$ (because $|C_{2}|,|C_{3}|\geq y|C|$), it follows that $|N^{2}_{C}(u)\cap C_{1}|\geq(y/(1-2y))|C_{1}|$. Consequently there exists $w^{\prime}\in C_{1}$ such that $N^{2}_{A}(w^{\prime})$ contains at least $(y/(10(1-2y)))|A|$ vertices in $A\setminus N^{2}_{A}(w)$. Since $w,w^{\prime}$ have a common neighbour, it follows that $|N^{2}_{A}(w)\cap N^{2}_{A}(w^{\prime})|\geq x|A|$, and so

[TABLE]

a contradiction. Thus $x+y/(10(1-2y))<2/5$, and so $y\geq 1/4$ from the hypothesis. This proves (3).

Since $(b_{1}-y)+(b_{2}-y)+(b_{3}-y)=1-3y<x,$ it follows that for every vertex $u\in A$, there exists $i\in\{1,2,3\}$ such that $|N(u)\cap B_{i}|\geq(b_{i}-y)|B|$; and consequently there is a partition $A=A_{1}\cup A_{2}\cup A_{3}$ such that for $i=1,2,3$, every vertex in $A_{i}$ has more than $(b_{i}-y)|B|$ neighbours in $B_{i}$. It follows that $A_{i}\subseteq N^{2}_{A}(w)$ for each $w\in C_{i}$. Let $|A_{i}|=a_{i}|A|$ for $i=1,2,3$.

For $i=1,2$, let $D_{i}$ be the set of vertices in $A_{3}$ with a neighbour in $B_{i}$, and let $d_{i}=|D_{i}|/|A|$. For $i=1,2$, if $u\in D_{i}$ then

[TABLE]

and so there exists $w\in C_{i}$ such that

[TABLE]

and since $A_{i}\subseteq N^{2}_{A}(w)$, it follows that $(y/(1-2y))d_{i}+a_{i}<2/5$. Since $d_{1}+d_{2}\geq a_{3}$, and $a_{1}+a_{2}=1-a_{3}$, summing for $i=1,2$ yields that

[TABLE]

that is, $(1-3y)a_{3}/(1-2y)>1/5$; and since $a_{3}<2/5$, this implies that $y<1/4$, a contradiction. This proves 10.2.

10.3

If $x,y\in(0,1]$ and $12x^{2}y\geq 5(1-x-y)^{2}$, then $\phi(x,y)\geq 2/5$.

Proof.  Suppose not. Then $\phi(x,y)=1-(3/5+\epsilon)$ for some $\epsilon>0$, so by rotating we have $\phi(x,3/5+\epsilon)\leq 1-y$. But 5.7 gives $\phi(x,3/5+\epsilon)>1-y$, a contradiction. This proves 10.3.

10.4

If $x,y\in(0,1]$ and $x>1/3$ and $y\geq(5-\sqrt{3})/11$, then $\phi(x,y)\geq 2/5$.

Proof.  Suppose not. Then $\phi(x,y)=1-(3/5+\epsilon)$ for some $\epsilon>0$, so by rotating we have $\phi(3/5+\epsilon,y)\leq 1-x<2/3$. But $3/5\geq(1-y)^{2}/(1-2y^{2})$, since $y\geq(5-\sqrt{3})/11$; and so 7.9 gives that $\phi(3/5+\epsilon,y)\geq 2/3$, a contradiction. This proves 10.4.

10.5

If $x,y\in(0,1]$, and either $5x/2+y\leq 1$ and $2y\leq x$, or $x+5y/2\leq 1$ and $2x\leq y$, then $\psi(x,y)<2/5$.

Proof.  Apply 3.6 with $s/t=2/5$. This proves 10.5.

10.6

If $x,y\in(0,1]$ with $y\leq 1/3$ and $x/(1-2x)+y/(1-3y)\leq 2$, then $\phi(x,y)<2/5$.

Proof.  Apply 3.1 (with $z$ slightly less than $2/3$) to 7.10. This proves 10.6.

11 The 3/5 Level

Next, we analyze when $\psi\geq 3/5$, and similarly for $\phi$. The results are shown in figure 10.

11.1

If $x,y\in(0,1]$ with $y>1/2$, $x\geq 1/5$, and

[TABLE]

then $\psi(x,y)\geq 3/5$.

Proof.  Suppose not, and let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, such that $|N_{A}^{2}(w)|<3|A|/5$ for all $w\in C$. We can assume that $x,y$ are rational, and by multiplying vertices if necessary, we can assume that both $x|B|$ and $|C|/5$ are integers. By averaging, there exists $u\in A$ such that $|N_{C}^{2}(u)|<3|C|/5$. Choose $B^{\prime}\subseteq N(u)$ with $|B^{\prime}|=x|B|$, and choose $C^{\prime}\subseteq C$ with $N_{C}^{2}(u)\subseteq C^{\prime}$ and $|C^{\prime}|=3|C|/5$.

(1) There exist $w_{1}\in C^{\prime}$ and $w_{2}\in C\setminus C^{\prime}$ such that $|N(w_{1})\cup N(w_{2})|>(1-x)|B|$.

Choose $w_{1}\in C^{\prime}$ and $w_{2}\in C\setminus C^{\prime}$ uniformly and independently at random, and let $S=N(w_{1})\cup N(w_{2})$. We will compute the expectation of $|S|$. Let $t=2/5$. For each $v\in B_{1}$, $v$ has at least $y|C|$ neighbours in $C^{\prime}$, so the probability it is in $S$ is at least $y/(1-t)$. For each $v\in B\setminus B^{\prime}$, define $d(v)=|N(v)\cap(C\setminus C_{1})|/|C|$. Then the probability that $v$ is a neighbour of $w_{2}$ is $d(v)/t$, and the probability that $v$ is a neighbour of $w_{1}$ and not a neighbour of $w_{2}$ is at least

[TABLE]

Thus the expectation of $|S|$ is at least

[TABLE]

Choose $q$ with

[TABLE]

Each vertex $w\in C\setminus C^{\prime}$ has at least $y|B|$ neighbours in $B\setminus B^{\prime}$, so it follows that $q\geq y$. Thus, the expectation of $|S|$ is at least

[TABLE]

Since $\sum_{v\in B\setminus B^{\prime}}d(v)=qt|B|$ and $|B\setminus B^{\prime}|=(1-x)|B|$, it follows by Cauchy-Schwarz that

[TABLE]

Thus the expectation of $|S|$ is at least

[TABLE]

To prove (1), it suffices to show that the expectation of $|S|$ is more than $(1-x)|B|$, and so it suffices to show that

[TABLE]

Remembering that $t=2/5$, this is

[TABLE]

and the derivative of the left-hand side with respect to $q$ is

[TABLE]

for $q\geq y$. It follows that the left-hand side is minimized when $q=y$, so it suffices to show that

[TABLE]

which is equivalent to the hypothesis. This proves (1).

Let $w_{1},w_{2}$ be as in (1), and let $A_{i}=N_{A}^{2}(w_{i})$ for $i=1,2$. Then $A=A_{1}\cup A_{2}$, and since $y>1/2$ we have $N(v_{1})\cap N(v_{2})\neq\emptyset$, and consequently $|A_{1}\cap A_{2}|\geq x|A|\geq|A|/5$. Then

[TABLE]

and so we have $|A_{i}|\geq 3|A|/5$ for some $i$, a contradiction. This proves 11.1.

11.2

If $x,y\in(0,1]$ with $x>1/2$ and $x+2y>1$, then $\psi(x,y)\geq 3/5$.

Proof.  Let $G$ be $(x,y)$-biconstrained via $(A,B,C)$, and suppose that $|N^{2}_{A}(w)|<(3/5)|A|$ for each $w\in C$. Choose $\alpha$ maximum such that there exist $w_{1},\ldots,w_{\alpha}\in C$ where $N(w_{i})\cap N(w_{j})=\emptyset$ for $1\leq i<j\leq\alpha$.

(1) $\alpha=2$.

Since $x+2y>1$ it follows that $N^{2}_{A}(w_{i})\cup N^{2}_{A}(w_{j})=A$ for all distinct $i,j\in\{1,\ldots,\alpha$, and so if $\alpha\geq 3$ then every vertex in $A$ belongs to at least two of the sets $N^{2}_{A}(w_{1}),N^{2}_{A}(w_{2}),N^{2}_{A}(w_{3})$, which is impossible since they each have cardinality less than $(3/5)|A|$. So $\alpha\leq 2$.

Suppose that $\alpha=1$; then every $w_{2}\in C$ satisfies $N(w_{1})\cap N(w_{2})\neq\emptyset$. Let $B_{1}=N(w_{1})$ and $A_{1}=N_{A}^{2}(w_{1})$; thus $|A_{1}|<(3/5)|A|$ and $|B_{1}|\geq y|B|$. Choose $v\in B\setminus B_{1}$ with $v$ at least $x|A\setminus A_{1}|/(1-y)>2x|A|/(5(1-y))$ neighbours in $A\setminus A_{1}$, and let $w_{2}\in C$ be a neighbour of $v$. Then

[TABLE]

since $x>1/2$, a contradiction. This proves (1).

Since $\alpha=2$, every vertex $w\in C$ shares a neighbour with at least one of $w_{1},w_{2}$. Let $A_{i}=N^{2}_{A}(w_{i})$ for $i=1,2$. Since $x+2y>1$, we have $A=A_{1}\cup A_{2}$, and so $|A_{1}\cap A_{2}|<|A|/5$ because $|A_{1}|,|A_{2}|<3|A|/5$. Then, if some $w\in C$ shares a neighbour with $w_{1}$ and shares a neighbour with $w_{2}$, it follows that $|N_{A}^{2}(w)|>2x|A|-|A|/5>4|A|/5$, a contradiction.

Thus, every vertex in $C$ shares a neighbour with exactly one of $w_{1}$ and $w_{2}$. Let $H$ be the bipartite graph $G[B\cup C]$. It follows that there are exactly two components of $H$, say $H_{1},H_{2}$, where $w_{i}\in V(H_{i})$ for $i=1,2$. Let $B_{i}=B\cap H_{i}$ and $C_{i}=C\cap H_{i}$ for $i=1,2$. Without loss of generality we may assume that $|B_{1}|\geq|B|/2$. It follows that for each $u\in A$, $u$ has a neighbour in $B_{1}$ and consequently

[TABLE]

because $|C_{2}|\geq y|C|$, and thus $|C_{1}|\leq(1-y)|C|$. Since $|A\setminus A_{1}|>(2/5)|A|$, there exists $w\in C_{1}$ with more than $2|A|y/(5(1-y))$ neighbours in $A\setminus A_{1}$. But $w$ and $w_{1}$ share a common neighbour, so

[TABLE]

since the last inequality is equivalent to $5y^{2}-5y+1\geq 0$, which is true because $y\leq 1/4$ (since $x+2y>1$, and $x>1/2$). This proves 11.2.

11.3

If $x,y\in(0,1]$ with $y>1/2$ and $40x^{2}y\geq(3-5x)^{2}$, then $\phi(x,y)\geq 3/5$.

Proof.  Apply 5.7 with $z=3/5$. This proves 11.3.

11.4

If $x,y\in(0,1]$ with $x\leq 4/7$ and $y\leq 1/2$ and $x+3y\leq 1$, then $\psi(x,y)<3/5$.

Proof.  We may assume that $x>1/2$ since $\psi(1/2,1/2)<3/5$; and so $y<1/6$ since $x+3y\leq 1$. The claim follows from applying 3.4 with $z$ slightly less than $3/5$ and $x^{\prime}=y^{\prime}=z^{\prime}=1/4$. This proves 11.4.

11.5

If $x,y\in(0,1]$, such that $3x+y\leq 1$, and $x+5y\leq 3$, with strict inequality in both if $x$ or $y$ is irrational, then $\psi(x,y)<3/5$.

Proof.  By increasing $x,y$ if necessary, we may assume that $x,y$ are rational. Suppose that $\psi(x,y)\geq 3/5$. We claim first that:

(1) $x<1/6$, and $y>1/2$, and $5xy+15y<9$, and $x<(1-y)/(5y)$, and $x\leq 3(1-y)^{2}/(1+5y)$.

Since $3x+y\leq 1$, it follows that $x<1/3$, and $y>1/2$ since $\psi(1/2,1/2)=1/2<3/5$. Thus $x<1/6$, since $3x+y\leq 1$. This proves the first two statements. Since $x+5y\leq 3$, it follows that $y<3/5$, and so $5xy+15y<3x+15y\leq 9$. This proves the third statement. For the fourth, $5x<3x/y$ (since $y<3/5$), and $3x\leq 1-y$, and so $5x<(1-y)/y$. Finally, for the fifth statement, if $y\leq 4/7$, then $1+5y\leq 9-9y$, and so

[TABLE]

and if $y\geq 4/7$, then

[TABLE]

and so

[TABLE]

This proves (1).

Since $x<1/6$, it follows that $x/(1-x)<(5x)/3$ and $(1-y)/(3y)<1/3$. The hypotheses (via (1)) imply that

[TABLE]

and

[TABLE]

Consequently there exists a rational $x^{\prime}$ with $x/(1-x)<5x/3<x^{\prime}$, and

[TABLE]

Thus

[TABLE]

choose a rational $y^{\prime}$ between them. Then $x^{\prime}+3y^{\prime}\leq 1$ and $3x^{\prime}+y^{\prime}\leq 1$, and so $\psi(x^{\prime},y^{\prime})<1/3$, by (theorem 3.3 of the paper). Let $\psi(x^{\prime},y^{\prime})=z^{\prime}<1/3$, and choose $z<3/5$ with $(1-z)/z\leq 1-z^{\prime}$, and $(1-z)/(1-x)\leq 1-z^{\prime}$, and $z\geq x/x^{\prime}$. Then from 3.5, $\psi(x,y)\leq z<3/5$, a contradiction. This proves 11.5.

11.6

If $x,y\in(0,1]$ with $y<1/3$ and $\frac{2x-1}{2-3x}+\frac{y}{1-3y}\leq 1$, and with strict inequality if $x$ or $y$ is irrational, then $\phi(x,y)<3/5$.

Proof.  We may assume that $x,y$ are rational. Let $x^{\prime}=2-1/x$ and $y^{\prime}=y/(1-y)$; it follows that $x^{\prime},y^{\prime}<1/2$, and $\frac{x^{\prime}}{1-2x^{\prime}}+\frac{y^{\prime}}{1-2y^{\prime}}\leq 1$. By 3.2 for $k=2$, it follows that $\phi(x^{\prime},y^{\prime})<1/3$. Then applying 3.7 with $z$ slightly less than $3/5$ gives the result. This proves 11.6.

12 Peaceful coexistence

We have not been able to evaluate $\phi(x,y)$ in general, but here is an easier question (that we also cannot do, but it seems to be less far out of reach). It is always true that $\phi(x,y)\geq x$, by 1.2, but if $y$ is sufficiently small then equality may hold. For fixed $x$, what is the largest $y$ such that $\phi(x,y)=x$?

Let $(G,w)$ be a weighted graph. We say it is $x$-regular via a bipartition $(A,B)$ if

• •

$|A|=|B|$, and $w(v)>0$ for each $v\in V(G)$;

• •

the $0,1$-adjacent matrix between $A$ and $B$ is nonsingular;

• •

$\sum_{u\in A}w(u)=\sum_{v\in B}w(v)=1$; and

• •

for each $u\in V(G)$, $\sum_{v\in N(u)}w(v)=x$.

(Note that the fourth bullet is required to hold both for $u\in A$ and for $u\in B$.) Its order is $|A|$, and its min-weight is $\min_{v\in B}w(v)$. We will show:

12.1

For $x,y\in(0,1]$, $\phi(x,y)=x$ if and only if there is an $x$-regular bipartite weighted graph with order at most $1/y$.

Proof.  If there is a such a weighted graph $(G,w)$, via $(A,B)$, where $|A|=|B|=n$ say, let $C$ be a set of $n$ new vertices, and add a perfect matching between $B$ and $C$. Extend $w$ to $C$ by defining $w(v)=1/n$ for each $v\in C$. The weighted graph just made is $(x,1/n)$-constrained, and shows that $\phi(x,1/n)\leq x$, and consequently $\phi(x,y)\leq x$ (and so $\phi(x,y)=x$).

For the converse, suppose that $G$ is $(x,y)$-constrained via $(A,B,C)$, and $|N^{2}_{A}(v)|\leq x|A|$ for each $v\in C$.

(1) Each vertex in $A$ has exactly $x|B|$ neighbours in $B$, and each vertex in $B$ has exactly $x|A|$ neighbours in $A$.

Each vertex $u\in B$ has at most $x|A|$ neighbours in $A$, since $u$ has a neighbour $v\in C$ and $|N^{2}_{A}(v)|\leq x|A|$. Since each vertex in $A$ has at least $x|B|$ neighbours in $B$, averaging shows that equality holds throughout. That proves (1).

Say two vertices in $A$ are twins if they have the same neighbour set in $B$, and two vertices in $B$ are twins if they have the same neighbour set in $A$. This defines equivalence relations of $A$ and $B$, and we call the equivalence classes twin classes.

(2) For each vertex $v\in C$, all its neighbours in $B$ are twins, and so $N(v)$ is a subset of a twin class of $B$.

By (1) each vertex in $N(v)$ has $x|A|$ neighbours in $A$, and all these vertices belong to $N^{2}_{A}(v)$; and since $|N^{2}_{A}(v)|=x|A|$, equality holds, and in particular, all vertices in $N(v)$ are twins. This proves (2).

Let $\mathcal{T}$ be the set of all twins classes of $B$. For each $T\in\mathcal{T}$, let $C(T)$ be the set of all $v\in C$ with $N(v)\subseteq T$. Thus the sets $C(T)\;(T\in\mathcal{T})$ are nonempty, pairwise disjoint and have union $C$. There is one of cardinality at most $|C|/|\mathcal{T}|$, say $C(T)$; and then each vertex in $T$ has only at most $|C|/|\mathcal{T}|$ neighbours in $C$, and so $y\leq 1/|\mathcal{T}|$.

Choose one vertex from each twin class of $A$ and of $B$, and let $H$ be the subgraph induced on this set. For each vertex $v$ of $H$, let $w(v)=|T|/|B|$ if $v\in T$ for some twin class $T$ of $B$, and $w(v)=|T|/|A|$ if $v\in T$ for some twin class $T$ of $A$. Then we have:

• •

$(H,w)$ is a bipartite graph, with bipartition $(A_{0},B_{0})$ say;

• •

$\sum_{u\in A_{0}}w(u)=\sum_{v\in B_{0}}w(v)=1$;

• •

for each $u\in V(H)$, $\sum_{v\in N(u)}w(v)=x$; and

• •

$|B_{0}|\leq 1/y$.

Let us choose a weighted graph $(H,w)$ and bipartition with these properties, with $|V(H)|$ minimum. If there is a function $f:A\rightarrow\mathbb{R}$ such that $\sum_{u\in N(v)}f(u)=0$ for each $v\in B$, not identically zero, then by adding a suitable multiple of $f$ to the restriction of $w$ to $A$, we can arrange that $w(u)=0$ for some $u\in A$, and then $u$ can be deleted, contrary to the minimality of $|V(H)|$. Thus there is no such $f$, and similarly there is no $f:B\rightarrow\mathbb{R}$ such that $\sum_{v\in N(u)}f(v)=0$ for each $u\in A$, not identically zero. Consequently $|A_{0}|=|B_{0}|=n$ say, and the adjacency matrix between $A_{0}$ and $B_{0}$ is nonsingular. Moreover $w(v)>0$ for each $v\in V(H)$, from the minimality of $V(H)$. This proves 12.1.

By 2.3, $\phi(x,y)=x$ if and only $\phi(y,x)=x$, so this also answers the analogous question for $\phi(y,x)$. If $x$ is irrational, there is no $x$-regular bipartite weighted graph, and so $\phi(x,y)>x$ for all $y>0$. If $x\in(0,1]$ is rational, let us define the order of $x\in(0,1]$ to be the minimum order of $x$-regular bipartite weighted graphs. If $x=p/q$ say where $p,q>0$ are integers, then the order of $x$ is at most $q$, because one can construct an appropriate cyclic shift graph. But the order of $x$ can be strictly less than $q$. For instance, the top part of the graph of figure 1 is $13/27$-regular (take as vertex-weights the numbers given, divided by 27), and so the order of $13/27$ is at most seven. Figure 11 gives a smaller example, showing that the order of $2/5$ is at most four.

We can prove that the order is also bounded below by a function of $q$ that goes to infinity with $q$. More exactly, if $G$ is $p/q$-regular (in lowest terms) and has order $n$, then $q$ is at most $(n+1)^{(n+1)/2}$. This follows from a theorem of Hadamard [2], that every $n\times n$ $0,1$-matrix has determinant at most $(n+1)^{(n+1)/2}2^{-n}$. We do not know whether there are weighted bipartite graphs with order $n$ that are $p/q$-regular (in lowest terms), where $q$ is exponentially large in $n$. (Hadamard $n\times n$ $0,1$-matrices have determinant that achieve Hadamard’s bound, and they exist when $n+1$ is a power of two, but they give weighted bipartite graphs that are vertex-transitive, and which therefore are $p/q$-regular with $q=n$.)

One could ask the same question for the biconstrained problem: given $x$, for which values of $y$ is it true that $\psi(x,y)=x$? A similar analysis (we omit the details) shows:

12.2

For $x,y\in(0,1]$, the following are equivalent:

• •

$\psi(x,y)=x$;

• •

$\psi(y,x)=x$; and

• •

there is an $x$-regular bipartite weighted graph with min-weight at least $y$.