A bound on partitioning clusters

The purpose of this note is to establish the following theorem:

Note that $\frac{3}{p}=1.72598\dots$. Thus the inequality (1) improves upon the trivial bound

$$|\{(A_{1},A_{2},A)\in X\times X\times X:A=A_{1}\uplus A_{2}\}|\leqslant|X|^{2}$$

(2)

that arises simply because there are $|X|^{2}$ pairs $(A_{1},A_{2})$, and this pair uniquely determines $A$.

Theorem 1 has applications in analyzing the running time of several dynamic programming algorithms used in phylogenetic reconstruction literature. Several published algorithms [4, 5, 6, 7] seek to find a median tree that minimizes the total distance to an input set of trees, for various definitions of a distance between two trees. For example, a method called ASTRAL [4, 5] defines the distance between two trees as the number of quartet trees induced by one tree but not the other, and seeks to find the unrooted tree that minimizes the sum of this quartet distance to its input set of unrooted trees, a problem that turns out to be NP-Hard [8]. To make this optimization problem tractable, ASTRAL uses a dynamic programming approach where the final tree is built by successively dividing each subset of leaves (called a cluster) $A$ into smaller clusters, $A^{\prime}$ and $A\backslash A^{\prime}$ while minimizing the number of quartets in the input tree set that will have to be missing from any tree that includes $A^{\prime}|A\backslash A^{\prime}$ as a bipartition (i.e., a branch; note that a branch in an unrooted tree is just a bipartition of the leaves). If all possible subsets $A^{\prime}\subset A$ are considered when dividing $A$ to two subsets, the algorithm provably returns the optimal tree (also, the optimal solution has been shown to enjoy statistical consistency under certain models of gene and species evolution [4]). However, such a dynamic programming algorithm will have to explore the power set of the set of leaves and will thus require time exponential in the number of leaves.

To give a practical alternative, ASTRAL solves a constrained version of the problem where a set $X$ of clusters is defined in advance to constrain the search space; when dividing $A$, we only look for $A^{\prime}\subset A$ such that $A^{\prime}\in X$ and $A\backslash A^{\prime}\in X$. The set $X$ is defined heuristically, and the running time of ASTRAL should be defined asymptotically as a function of $|X|$. Throughout the dynamic programming execution, ASTRAL considers all possible pairs of clusters $x,y\in X$ exactly once iff $x\cap y=\emptyset$ and $x\cup y\in X$. Therefore, establishing the asymptotic running time of ASTRAL with regards to $|X|$ requires bounding the left-hand side of (1) ASTRAL simply used the trivial $O(|X|^{2})$ upper bound in their running time analysis. We can now improve that analysis to $O(|X|^{1.72598\dots})$.

We first demonstrate why the exponent $p$ is best possible. Let $n$ be a large multiple of $3$, and let $X$ denote the collection of all sets $A\subset\{1,\dots,n\}$ whose cardinality $|A|$ is equal to either $n/3$ or $2n/3$. Clearly

$$|X|=\binom{n}{n/3}+\binom{n}{2n/3}=2\frac{n!}{(n/3)!(2n/3)!}.$$

On the other hand, if $A\in X$ has cardinality $2n/3$, then it can be partitioned in $\binom{2n/3}{n/3}$ ways into $A_{1}\uplus A_{2}$ with $A_{1},A_{2}\in X$, and no partition is available when $A$ has cardinality $n/3$. Thus the left-hand side of (1) is equal to

$$\binom{n}{2n/3}\times\binom{2n/3}{n/3}=\frac{n!}{(n/3)!(n/3)!(n/3)!}.$$

Using the Stirling approximation $n!=n^{n}e^{-n+o(n)}$ as $n\to\infty$, we conclude that the left-hand side of (1) is equal to $\exp(n(\log 3+o(1)))$, while $|S|$ is equal to $\exp(\frac{n}{3}(\log\frac{27}{4}+o(1)))$, and on sending $n\to\infty$ we conclude that (1) can fail whenever $p>\log_{3}\frac{27}{4}$.

Now we show why (1) holds for $p=\log_{3}\frac{27}{4}$. This will be a consequence of the following convolution estimate on the discrete unit cube $\{0,1\}^{n}$ (which we view as a subset of $\mathbb{Z}^{n}$ for the purposes of defining convolution):

Let us now see why Theorem 2 implies Theorem 1. We first observe that to prove Theorem 1, it suffices to do so under the additional assumption that all the sets in $X$ are finite. Indeed, if we let $\Omega\coloneqq\bigcup_{A\in X}A$ denote the union of the sets in $X$, then $X$ partitions $\Omega$ into at most $2^{|X|}$ cells. Some of these cells may be infinite, but we may replace any such cell with a single point without affecting either the left or right-hand side of (1). After applying this replacement, every set in $X$ is now finite.

Without loss of generality, we may now assume that all the sets $A$ in $X$ are subsets of $\{1,\dots,n\}$ for some natural number $n\geqslant 1$. We now define the functions $f,g,h:\{0,1\}^{n}\to\{0,1\}$ as follows. For any $(a_{1},\dots,a_{n})\in\{0,1\}^{n}$, we set $f(a_{1},\dots,a_{n})=g(a_{1},\dots,a_{n})=1$ when the set $\{1\leqslant i\leqslant n:a_{i}=1\}$ lies in $X$, and $f(a_{1},\dots,a_{n})=g(a_{1},\dots,a_{n})=0$ otherwise. Similarly, we set $h(a_{1},\dots,a_{n})=1$ when the set $\{1\leqslant i\leqslant n:a_{i}=0\}$ lies in $X$, and $h(a_{1},\dots,a_{n})=0$ otherwise. It is then easy to see that

$$\|f\|_{\ell^{p}(\{0,1\}^{n})}=\|g\|_{\ell^{p}(\{0,1\}^{n})}=\|h\|_{\ell^{p}(\{0,1\}^{n})}=|X|^{1/p}$$

and

$$f*g*h(1^{n})=|\{(A_{1},A_{2},A)\in X\times X\times X:A=A_{1}\uplus A_{2}\}|$$

giving the claim.

The form of Theorem 2 is very amenable to an induction on dimension:

From this proposition and induction, we see that to prove Theorem 2, it suffices to do so in the one-dimensional case $n=1$. We may normalize

$$\|f\|_{\ell^{p}(\{0,1\})}=\|g\|_{\ell^{p}(\{0,1\})}=\|h\|_{\ell^{p}(\{0,1\})}=1,$$

so that we may write

	$\displaystyle f(0)$	$\displaystyle=a^{1/p}$
	$\displaystyle f(1)$	$\displaystyle=(1-a)^{1/p}$
	$\displaystyle g(0)$	$\displaystyle=b^{1/p}$
	$\displaystyle g(1)$	$\displaystyle=(1-b)^{1/p}$
	$\displaystyle h(0)$	$\displaystyle=c^{1/p}$
	$\displaystyle h(1)$	$\displaystyle=(1-c)^{1/p}$

for some $0\leqslant a,b,c\leqslant 1$. The inequality (3) then simplifies to the elementary inequality

$$(ab(1-c))^{1/p}+(bc(1-a))^{1/p}+(ca(1-b))^{1/p}\leqslant 1.$$

(4)

Observe that equality is attained here when $(a,b,c)=(0,1,1),(1,0,1),(0,1,1),(2/3,2/3,2/3)$; the final case $(a,b,c)=(2/3,2/3,2/3)$ also reveals that the inequality (4) fails if $p$ is replaced by any quantity larger than $\log_{3}\frac{27}{4}$. This is of course consistent with the second part of Theorem 1.

The fact that equality is attained in (4) in four different locations seems to rule out any quick proof of (4) using convexity-based methods such as Jensen’s inequality. Instead, we argue as follows. First observe that when $a=0$, the left-hand side of (4) simplifies to $(bc)^{1/p}$, and it is then clear that the inequality (4) holds whenever $a=0$ and is strict unless $(a,b,c)=(0,1,1)$. Next, we analyze the left-hand side of (4) for $(a,b,c)$ close to $(0,1,1)$. Writing $(a,b,c)=(\alpha,1-\beta,1-\gamma)$ for some small $\alpha,\beta,\gamma\geqslant 0$, we can write the left-hand side of (4) as

$$(\alpha\gamma)^{1/p}+((1-\beta)(1-\gamma)(1-\alpha))^{1/p}+(\alpha\beta)^{1/p}.$$

For $\alpha,\beta,\gamma$ small enough, we have

$$(1-\beta)(1-\gamma)(1-\alpha)\leqslant 1-\frac{1}{2}(\alpha+\beta+\gamma)$$

(say), which by the concavity of $x\mapsto x^{1/p}$ implies that

$$((1-\beta)(1-\gamma)(1-\alpha))^{1/p}\leqslant 1-\frac{1}{2p}(\alpha+\beta+\gamma).$$

On the other hand, from the arithmetic mean-geometric mean inequality we certainly have

$$(\alpha\gamma)^{1/p},(\alpha\beta)^{1/p}\leqslant(\alpha+\beta+\gamma)^{2/p}.$$

Since $p<2$, we conclude that the inequality (4) holds whenever $\alpha+\beta+\gamma$ is sufficiently small, or equivalently when $(a,b,c)$ is sufficiently close to $(0,1,1)$. Since both sides of (4) depend continuously on $a,b,c$, we now see that (4) holds whenever $a$ is sufficiently small, and similarly for $b$ and $c$. Thus we may assume $a,b,c\geqslant\varepsilon$ for some small absolute constant $\varepsilon>0$.

We next consider the boundary case $a=1$, $b,c>\varepsilon$. Here, we claim strict inequality:

$$(b(1-c))^{1/p}+(c(1-b))^{1/p}<1.$$

(5)

Indeed, from the Cauchy-Schwarz inequality one has

$$(b(1-c))^{1/2}+(c(1-b))^{1/2}\leqslant\left((b^{1/2})^{2}+((1-b)^{1/2})^{2}\right)^{1/2}\left(((1-c)^{1/2})^{2}+(c^{1/2})^{2}\right)^{1/2}=1$$

and the claim follows since $\frac{1}{p}>\frac{1}{2}$.

For similar reasons, we obtain strict inequality in (4) when $b=1$ or $c=1$. By continuity, this establishes (4) in all regions except the region

$$\varepsilon\leqslant a,b,c\leqslant 1-\varepsilon$$

for some small absolute constant $\varepsilon>0$. We now work in this region.

We can rewrite (4) as

$$\left(\frac{1-c}{c}\right)^{1/p}+\left(\frac{1-a}{a}\right)^{1/p}+\left(\frac{1-b}{b}\right)^{1/p}\leqslant\frac{1}{(abc)^{1/p}};$$

writing $x\coloneqq(\frac{1-a}{a})^{1/p}$, $y\coloneqq(\frac{1-b}{b})^{1/p}$, $z\coloneqq(\frac{1-c}{c})^{1/p}$, $x,y,z$ lies in the region

$$\left(\frac{\varepsilon}{1-\varepsilon}\right)^{p}\leqslant x,y,z\leqslant\left(\frac{1-\varepsilon}{\varepsilon}\right)^{p}$$

(6)

and the above inequality transforms to

$$x+y+z\leqslant\exp\left(\frac{f(x)+f(y)+f(z)}{p}\right)$$

or equivalently

$$f(x)+f(y)+f(z)-p\log(x+y+z)\geqslant 0$$

(7)

where $f:(0,+\infty)\to(0,+\infty)$ is the function $f(z)\coloneqq\log(1+z^{p})$.

Since the region (6) is compact, and the inequality (7) is already known on the boundary of this region, it suffices to verify (7) when $(x,y,z)$ is a critical point of the left-hand side, that is to say that

$$f^{\prime}(x)=f^{\prime}(y)=f^{\prime}(z)=\frac{p}{x+y+z}.$$

Since $f^{\prime}(z)=\frac{p}{z+z^{1-p}}$, we can rewrite this condition as

$$x+x^{1-p}=y+y^{1-p}=z+z^{1-p}=x+y+z.$$

(8)

The function $x\mapsto x+x^{1-p}$ is increasing for $x<(p-1)^{1/p}$ and decreasing for $x>(p-1)^{1/p}$, so it can only attain any given value at most twice. From (8) and the pigeonhole principle, we conclude that at least two of $x,y,z$ are equal. Without loss of generality we may assume $x=y$, then from (8) we have

$$x^{1-p}=x+z$$

and

$$z^{1-p}=2x$$

and hence

$$x^{1-p}=x+(2x)^{\frac{1}{1-p}}.$$

Dividing by $x$ we obtain

$$x^{-p}=1+2^{\frac{1}{1-p}}x^{\frac{p}{1-p}}$$

and then setting $u\coloneqq x^{-p}$ we conclude that

$$u=1+(u/2)^{\frac{1}{p-1}}.$$

(9)

The function $u\mapsto 1+(u/2)^{\frac{1}{p-1}}$ is convex, equals $2$ when $u=2$, $0$ when $u=0$, and is larger than $u$ for sufficiently large $u$. As a consequence, the equation (9) has exactly two solutions, one at $u=2$ and one with $u>2$; see Figure 2. The second solution can be computed numerically as $u=10.70297\dots$. Thus, there are two critical points $(x,y,z)$ with $x=y$, the first of which is

$$(2^{-1/p},2^{-1/p},2^{-1/p})=(0.67113\dots,0.67113\dots,0.67113\dots),$$

and the second of which can be computed numerically as

$$(x,y,z)=(0.25568\dots,0.25568\dots,2.48086\dots).$$

At the first critical point, we have $f(x)=f(y)=f(z)=\log(3/2)$, and one easily verifies that the left-hand side of (7) vanishes since $p=\log(27/4)/\log(3)$. At the second critical point, one can numerically verify that

$$f(x)=f(y)=0.089321\dots;\quad f(z)=1.766695\dots$$

and hence

$$f(x)+f(y)+f(z)-p\log(x+y+z)=0.040307\dots>0$$

at this critical point, giving the claim.

Recall from [9] that the additive energy $E(A)$ of a finite subset $A$ of an additive group $G$ is defined as the number of quadruples $(a_{1},a_{2},a_{3},a_{4})\in A\times A\times A\times A$ such that $a_{1}+a_{2}=a_{3}+a_{4}$. We have the trivial bound $E(A)\leqslant|A|^{3}$, which is attained for instance when $A$ is itself a finite group. By modifying the above arguments, we have the following refinement in the discrete cube $\{0,1\}^{n}$:

The second claim is clear, since if $A=\{0,1\}^{n}$ then one easily computes that $|A|=|\{0,1\}|^{n}=2^{n}$ and $E(A)=E(\{0,1\})^{n}=6^{n}$. As in the previous section, the theorem is proven by induction on $n$ together with an elementary inequality, namely

Now we establish Theorem 7. The claim is trivial for $n=0$, so suppose that $n\geqslant 1$ and that the claim has already been proven for $n-1$. For $A\subset\{0,1\}^{n}$, we may partition

$$A=(A_{0}\times\{0\})\uplus(A_{1}\times\{1\})$$

for some $A_{0},A_{1}\subset\{0,1\}^{n}$. We can then split

$$E(A)=E(A_{0})+4|\{(a_{0},a_{1},a^{\prime}_{0},a^{\prime}_{1})\in A_{0}\times A_{1}\times A_{0}\times A_{1}:a_{0}+a_{1}=a^{\prime}_{0}+a^{\prime}_{1}\}|+E(A_{1}).$$

By the Cauchy-Schwarz inequality (and writing $a_{0}+a_{1}=a^{\prime}_{0}+a^{\prime}_{1}$ as $a_{0}-a^{\prime}_{0}=a^{\prime}_{1}-a_{1}$) we have

$$|\{(a_{0},a_{1},a^{\prime}_{0},a^{\prime}_{1})\in A_{0}\times A_{1}\times A_{0}\times A_{1}:a_{0}+a_{1}=a^{\prime}_{0}+a^{\prime}_{1}\}|\leqslant E(A_{0})^{1/2}E(A_{1})^{1/2}$$

and hence by the induction hypothesis

$$E(A)\leqslant|A_{0}|^{p}+4(|A_{0}||A_{1}|)^{p/2}+|A_{1}|^{p}.$$

Applying Lemma 8 and noting that $|A_{0}|+|A_{1}|=|A|$, we obtain $E(A)\leqslant|A|^{p}$, closing the induction.

The authors would like to thank Siavash Mirarab for bringing this problem to our attention, and David Speyer and the anonymous referee for helpful comments. DK is supported by NSF award CCF-1553288 (CAREER). TT is supported by NSF grant DMS-1266164, the James and Carol Collins Chair, and by a Simons Investigator Award.