A Book Proof of the Middle Levels Theorem

Abstract

We give a short constructive proof for the existence of a Hamilton cycle in the subgraph of the $(2n+1)$-dimensional hypercube induced by all vertices with exactly n or $n+1$ many 1s.

The n-dimensional hypercube $Q_n$ is the graph that has as vertices all bitstrings of length n, and an edge between any two bitstrings that differ in a single bit. The weight of a vertex x of $Q_n$ is the number of 1s in x. The kth level of $Q_n$ is the set of vertices with weight k.

FormalPara Theorem 1

For all $n\ge 1$, the subgraph of $Q_{2n+1}$ induced by levels n and $n+1$ has a Hamilton cycle.

Theorem 1 solves the well-known middle levels conjecture, and it was first proved in [2] (see this paper for a history of the problem). A shorter proof was presented in [1] (12 pages). Here, we present a proof from ‘the book’.

FormalPara Proof

We write $D_n$ for all Dyck words of length 2n, i.e., bitstrings of length 2n with weight n in which every prefix contains at least as many 1s as 0s. We also define $D:=\bigcup _{n\ge 0} D_n$. Any $x\in D_n$ can be decomposed uniquely as $x=1u0v$ with $u,v\in D$. Furthermore, Dyck words of length 2n can be identified by ordered rooted trees with n edges as follows; see Fig. 1: Traverse the tree with depth-first search and write a 1-bit for every step away from the root and a 0-bit for every step towards the root. For any bitstring x, we write $\sigma ^s(x)$ for the cyclic right rotation of x by s steps. We write $A_n$ and $B_n$ for the vertices of $Q_{2n+1}$ in level n or $n+1$, respectively, and we define $M_n:=Q_{2n+1}[A_n\cup B_n]$. For any $x\in D_n$, $b\in \{0,1\}$ and $s\in \{0,\ldots ,2n\}$ we define $\langle x,b,s\rangle :=\sigma ^s(xb)$. Note that we have $A_n=\{\langle x,0,s\rangle \mid x\in D_n\wedge 0\le s\le 2n\}$ and $B_n=\{\langle x,1,s\rangle \mid x\in D_n\wedge 0\le s\le 2n\}$. Thus, we think of every vertex of $M_n$ as a triple $\langle x,b,s\rangle$, i.e., an ordered rooted tree x with n edges referred to as the nut, a bit $b\in \{0,1\}$, and an integer $s\in \{0,\ldots ,2n\}$ referred to as the shift.

The first step is to construct a cycle factor in the graph $M_n$. For this we define a mapping $f:A_n\cup B_n\rightarrow A_n\cup B_n$ as follows. Given an ordered rooted tree $x=1u0v\in D_n$ with $u,v\in D$, a tree rotation yields the tree $r(x):=u1v0\in D_n$; see Fig. 2. We define $f(\langle x,0,s\rangle ):=\langle r(x),1,s+1\rangle$ and $f(\langle x,1,s\rangle ):=\langle x,0,s\rangle$. It is easy to see that f is a bijection. Indeed, the inverse mapping is $f^{-1}(\langle x,0,s\rangle )=\langle x,1,s\rangle$ and $f^{-1}(\langle x,1,s\rangle )=\langle r^{-1}(x),0,s-1\rangle$. Furthermore, f changes only a single bit. To see this observe that for $x=1u0v$ with $u,v\in D$ the bitstrings $\langle x,0,s\rangle =\sigma ^s(1u0v0)$ and $f(\langle x,0,s\rangle )=\langle r(x),1,s+1\rangle =\sigma ^{s+1}(u1v01)=\sigma ^s(1u1v0)$ differ only in the bit between the substrings u and v. We also note that $f^2(\langle x,0,s\rangle )=\langle r(x),0,s+1\rangle \ne \langle x,0,s\rangle$. Consequently, for any vertex y of $M_n$, the sequence $C(y):=\big (y,f(y),f^2(y),\ldots \big )$ is a cycle, and $F_n:=\{C(y)\mid y\in A_n\cup B_n\}$ is a cycle factor in $M_n$.

As $f^2(\langle x,0,s\rangle )=\langle r(x),0,s+1\rangle$, moving two steps forward along a cycle of $F_n$ applies a tree rotation to the nut, and increases the shift by $+1$. As the ordered rooted tree $x\in D_n$ has n edges, we have $x=r^{2n}(x)$. Consequently, the minimum integer $t>0$ such that $x=r^t(x)$ must divide 2n. It follows that $\gcd (t,2n+1)=1$, hence all shifts of the nut x are contained in the cycle $C(\langle x,0,0\rangle )$, i.e., $\langle x,0,s\rangle \in C(\langle x,0,0\rangle )$ for all $s\in \{0,\ldots ,2n\}$. Therefore, the cycles of $F_n$ are in bijection with equivalence classes of ordered rooted trees with n edges under tree rotation, also known as plane trees. In particular, the number of cycles of $F_n$ is the number of plane trees with n edges (OEIS A002995).

The second step is to glue the cycles of the factor $F_n$ to a single Hamilton cycle. We call an ordered rooted tree $x\in D_n$ pullable if $x=110u0v$ for $u,v\in D$, and we define $p(x):=101u0v\in D_n$. We refer to p(x) as the tree obtained from x by a pull operation. In words, the leftmost leaf of x is in distance 2 from the root, and the edge leading to this leaf is removed and reattached as the new leftmost child of the root in p(x); see Fig. 3. For any pullable tree $x=110u0v\in D_n$ with $u,v\in D$, we define $y:=\langle x,0,0\rangle =x0$ and $z:=\langle p(x),0,0\rangle =p(x)0$, and we consider the 6-cycle $G(x):=(y,f(y),f^6(y),f^5(y),z,f(z))=(110u0v0,110u1v0,100u1v0,101u1v0,101u0v0,111u0v0)$, which has the edges (y, f(y)) and $(f^6(y),f^5(y))$ in common with the cycle C(y), and the edge (z, f(z)) in common with the cycle C(z); see Fig. 4. Consequently, if C(y) and C(z) are two distinct cycles, then the symmetric difference between the edge sets of C(y), C(z) and G(x) is a single cycle on the same set of vertices, i.e., G(x) glues the cycles C(y) and C(z) together.

We define $S(x):=\{f^i(y)\mid i=0,\ldots ,6\}\cup \{z,f(z)\}$, and we claim that for any two pullable trees $x\ne x'$, we have $S(x)\cap S(x')=\emptyset$, i.e., the cycles C(x) and $C(x')$ are (vertex-)disjoint. To see this, consider the shifts of the vertices in S(x) and $S(x')$, which are 0, 1, 1, 2, 2, 3, 3, 0, 1. It follows that if $S(x)\cap S(x')\ne \emptyset$, then we have $x=x'$, $p(x)=x'$, or $x=p(x')$. These cases are ruled out by the assumption $x\ne x'$, the fact that $p(x)=10\cdots$ and $x'=11\cdots$ differ in the second bit, and that $x=11\cdots$ and $p(x')=10\cdots$ differ in the second bit, respectively.

To complete the proof, it remains to show that the cycles of the factor $F_n$ can be glued to a single cycle via gluing cycles G(x) for a suitable set of pullable trees $x\in D_n$. As argued before, none of the gluing operations interfere with each other. Using the interpretation of the cycles of $F_n$ as equivalence classes of ordered rooted trees under tree rotation, it suffices to prove that every cycle can be glued to the cycle that corresponds to the star with n edges. As each gluing cycle corresponds to a pull operation, this amounts to proving that any ordered rooted tree $x\in D_n$ can be transformed to the star $(10)^n$ via a sequence of tree rotations and/or pulls.

Fig. 1

A Dyck word (left) and the corresponding ordered rooted tree (right)

Fig. 2

Tree rotation

Fig. 3

Pull operation

Fig. 4

Gluing 6-cycle G(x)

Fig. 5

Illustration of steps (i)–(iii) that make a tree more star-like

Indeed, this is achieved as follows: We fix a vertex c of x to become the center of the star (this vertex never changes), and we repeatedly perform the following three steps; see Fig. 5: (i) rotate x to a tree $x'$ such that c is root and the leftmost leaf of $x'$ is in distance $d>1$ from c; (ii) apply $d-2$ rotations to $x'$ to obtain a tree $x''$ whose leftmost leaf has distance 2 from the root; (iii) perform a pull. As step (iii) decreases the sum of distances of all vertices from c, we reach the star after finitely many steps.

This completes the proof of the theorem. $\square$