A fast algorithm on average for solving the Hamilton Cycle problem

1 Introduction

A Hamilton cycle in a graph is a cycle that visits every vertex exactly ones. We say that a graph is Hamiltonian if it contains a Hamilton cycle. The Hamilton cycle problem, which we denoted by HAM, asks to determines whether a given graph $G$ has a Hamilton cycle; it belongs to the list of Karp’s 21 NP-complete problems [13] and if $P \neq N P$ then there does not exist an algorithm that determines the Hamiltonicity of $G$ in $p o l y (n)$ time for every graph $G$ on $n$ vertices. The fastest deterministic algorithm known today that solves HAM is the Held–Karp algorithm, also called Bellman–Held–Karp algorithm [6],[12]; it is a dynamic programming algorithm with worst-case time complexity $Θ (n^{2} 2^{n})$ time and worst-case space complexity $Θ (\sqrt{n} 2^{n})$ .

In [11], Gurevich and Shelah gave an algorithm that determines the Hamiltonicity of an $n$ -vertex graph in polynomial time on average. Equivalently, as $G (n, 1 / 2)$ is the uniform measure over all graph on $n$ vertices, the expected running time of their algorithm over the input distribution $G \sim G (n, p)$ is polynomial in $n$ when $p = 1 / 2$ . The last statement raises the question for which values of $p$ there exists an algorithm that solves HAM in polynomial expected running time over the input distribution $G \sim G (n, p)$ . Gurevich and Shelah [11], Thomason [18] and Alon and Krivelevich [1] gave such an algorithm for $p \in [0, 1]$ being constant, $p \geq 12 n^{- 1 / 3}$ and $p \geq 70 p^{- 1 / 2}$ respectively.

In this paper we introduce CertifyHAM, a deterministic algorithm that solves HAM. Our main result is the following.

Theorem 1.1.

CertifyHAM solves HAM in $O (\frac{n}{p})$ expected running time over the input distribution $G \sim G (n, p)$ for $p \geq \frac{2000}{n}$ .

In the above Theorem we assume that we are given $A_{G}$ , the adjacency matrix of $G$ , and that each entry of $A_{G}$ can be accessed in $O (1)$ time. Observe that any algorithm that solves HAM has to identify at least $n$ edges of $G$ ; hence w.h.p.¹¹1We say that a sequence of events ${E_{n}}_{n \geq 1}$

holds with high probability (w.h.p. in short) if

l i m_{n \to \infty} P r (E_{n}) = 1

. it has to read at least

(1 + o (1)) n / p

entries of

A_{G}

. Thus the expected running time of CertifyHAM is optimal.

One of the most interesting figures of Theorem 1.1 is that it can also deal with $G (n, p)$ below the Hamiltonicity threshold. A cerebrated result of Bollobás [8] and of Komlós and Szemerédi [14] state that the threshold for Hamiltonicity is the same as the threshold for minimum degree 2 in the model $G (n, p)$ and it is equal to $(1 + o (1)) \frac{log n}{n}$ . For a complete survey on the Hamilton Cycles in random graphs see [10].

Theorem 1.1 falls into the area of average-case analysis of algorithms. For a complete survey on this area see [7]. It is motivated by the idea that the worst-time complexity of an algorithm is usually based on some well orchestrated instance and such instances are atypical. Thus the average time complexity may be a better measure of the performance of an algorithm. In addition using algorithms that run fast on average is one way to cope with NP-hard problems.

A distributional problem is a pair $(L, D)$ where $L$ is a decision problem and $D = {D_{n}}_{n \geq 1}$ where $D_{n}$

is a probability distribution on the inputs of

L

of size

n

. We say that a distributional problem

(L, D)

belongs to

A v g P

if there exists an algorithm

A

(with running time

t_{A} (x)

for an instance

x

L

), a constant

ϵ

and a polynomial

p ()

with the property that

E_{x \sim D_{n}} (t_{A}^{ϵ} (x)) = O (p (n)) .

Let $P_{p} = {G (n, p)}_{n \geq 1}$ . Theorem 1.1 and the upper bound on $P r (δ (G) \geq 2)$ given in Lemma 2.9, give the following.

Corollary 1.2.

$(H A M, P_{p})$ belongs to $A v g P$ for all $p \geq 0$ .

1.1 Description of CertifyHAM

CertifyHAM impliments 2 algorithms. First it implements CerHam1. CerHam1 solves $H A M$ with input $G \sim G (n, p)$ in $O (\frac{n}{p})$ time with probability $1 - o (n^{- 7})$ . In addition, if $δ (G) < 2$ then CerHam1 returns “ $G$ in not Hamiltonian”. If CerHam1 fails to solve HAM for a given instance $G$ then CertifyHAM implements CerHam2 whose expected run time of CerHam2 is $O (n^{7})$ . For an instance $x$ denote by $t_{1} (x), t_{2} (x)$ and $t (x)$ the running time with input $x$ of CerHam1, CerHam2 and CertifyHam respectively. Then, for $p \geq \frac{2000}{n}$ ,

	$E_{G \sim G (n, p)} (t (G))$	$\leq E_{G \sim G (n, p)} (t_{1} (G)) + E_{G \sim G (n, p)} (t_{2} (G) \| CerHam1 fails) P r (CerHam1 fails)$
		$\leq O (\frac{n}{p}) + n^{- 7} \cdot O (n^{7}) = O (\frac{n}{p}) .$

The description of CerHam1 is given in [4]. In this paper we introduce and analyse CerHam2.

CerHam2 takes as an input a graph $G$ and starts by querying a number of edges of $G$ . By querying an edge of $G$ we refer to reading the corresponding entry of $A_{G}$ the adjacency matrix of $G$ . When $G \sim G (n, p)$ we consider that initially all of the entries of $G$ are blank. Once we query an edge of $G$ we reveal whether the edge belongs to $G$ , equivalently whether the corresponding entry of $A_{G}$ is filled with 0 or 1. As $G \sim G (n, p)$ the corresponding entry is equal to 1 independently of all the entries that we have revealed so far.

Once CerHam2 has made its initial queries it proceeds and iterates over a while loop. Each iteration starts with a full path packing $P$ of $G$ and a set $S$ . A full path packing of $G$ is a set of vertex disjoint paths in $G$ that cover $V (G)$ . Here we consider single vertices to be paths of length 0. Thus $G$ has at least 1 full path packing, namely the one that consists of the $n$ paths of length 0. Then the algorithm adds to $G$ a set of edges $R$ that join the paths in $P$ into a Hamilton path. It continues by performing sequences of restricted Pósa rotations, defined in Section 2, and identifies sets $E n d$ , ${E n d_{u}}_{u \in E n d}$ such that for every $u \in E n d$ and $w \in E n d_{u}$ there exists a Hamilton path in $G \cup R$ from $u$ to $w$ . Now either (I) all of the sets $E n d, {E n d_{u}}_{u \in E n d}$ are large or (II) at least one of them is small. In the first case it identifies and queries an edge $u w$ with $u \in E n d$ and $w \in E n d_{u}$ that has not been queried yet. If $w u \in E (G)$ then it updates $P$ to a full path packing of size 1 less or claims that $G$ is Hamiltonian; in such a case we say it improves on $| P |$ . In the unlikely event that (II) occurs and one of the sets $E n d$ , ${E n d_{u}}_{u \in E n d}$ is small then it adds this set to $S$ , along with possibly some other sets. It then attempts to pack paths in $S \cup N (S)$ such that every vertex in $S$ is in the interior of some path. If no such path packing exists then it outputs “ $G$ is not Hamiltonian”. If such a path packing $P^{'}$ does exists then it adjust $P$ such that it contains $P^{'}$ as a subset. The algorithm exits the while-loop if $S$ gets too large or too many attempts to improve $| P |$ have been taken place. In the unlikely event that the CerHam2 exits its while loop it implements the Held-Karp algorithm.

The rest of the paper is organised as follows. In Section 2 we present the tools needed for the presentation and analysis of CerHam2. We then present and analyse CerHam2 in Section 3. Its analysis is split into 3 cases corresponding to the regimes $\frac{2000 log n}{n} \leq p$ , $\frac{1000 log log n}{n} \leq p \leq \frac{1000 log n}{n}$ and $\frac{2000}{n} \leq p \leq \frac{1000 log log n}{n}$ respectively. For ease of the presentation of the paper we have moved the proofs of technical results to the Appendix.

2 Preliminaries

2.1 Restricted Pósa Rotations

Given a Hamilton path $P = v_{1}, v_{2}, . . . ., v_{n}$ and $v_{i} v_{n}$ , $1 \leq i < n$ we say that the path $P^{'} = v_{1}, v_{2}, . . ., v_{i}, v_{n}, v_{n - 1}, . . ., v_{i + 1}$ is obtained from $P$ via a Pósa rotation that fixes $v_{1}$ . We call $v_{i}$ and $v_{i + 1}$ the pivot vertex and the new endpoint respectively. We call $v_{n} v_{i}$ and $v_{i} v_{i + 1}$ the inserted and the deleted edge respectively. We will not perform all Pósa rotations possible. Instead we will only perform Pósa rotations for which the pivot vertex does not belong to some prescribed set $S$ . The exact procedure which we call $R P R$ follows shortly. It takes as an input a graph $G$ , a Hamilton path $P$ of $G$ with endpoints $v, u$ and a set $S$ not containing $u$ . It then outputs sets $E n d$ and $P$ such that for each $w \in E n d$ there exists a Hamilton path in $P$ from $v$ to $w$ .

1:Let

Q = u

P = {P}

E n d = {u}

2:while

Q \neq \emptyset

3: Let

w

be the first vertex in

Q

and

P_{w}

be the unique Hamilton path in

P

from

v

w

4: For

z \in N (w) ∖ S

perform the Pósa rotation starting from

P_{w}

that fixes

v

and the inserted edge is

w z

only if the new endpoint does not belong to

E n d

. If the Pósa rotation is performed then add the new endpoint to

E n d

and to the end of

Q

and the corresponding path to

P

5: Remove

z

from

Q

6:end while

7:Output

E n d, P

Algorithm 1

R P R (G, P, u, v, S)

As every vertex enters at most once the set $Q$ , both $E n d, P$ have size at most $n$ throughout the above algorithm which runs in $O (n^{2})$ time.

We replace the classical Pósa Lemma with the first part of the following one.

Lemma 2.1.

Let $G$ be a graph, $S \subset V (G)$ , $P$ a Hamiltonian path of $G$ with endpoints $u, v$ and $E n d, P = R P R (G, P, v, u, S)$ . Then,

| N_{G} (E n d) ∖ S | < 2 | E n d | .

Proof.

Let $u = u_{1}, u_{2}, . . . ., u_{ℓ}$ be the vertices in $E n d$ in the order that are added to $Q$ and $P_{u_{1}}, P_{u_{2}},$ $. . ., P_{u_{ℓ}}$ be the corresponding Hamiltonian paths. Let $T_{1} = {u_{1}, r_{1, 1}}$ and $T_{i} = {u_{i}, r_{i, 1}, r_{i, 2}}$ for $2 \leq i \leq ℓ$ , where $r_{i, 2} u_{i}$ is the deleted edge in the Pósa rotation that results to the Hamilton path $P_{u_{i}}$ for $2 \leq i \leq ℓ$ and $r_{i, 1}$ is the vertex preceding $u_{i}$ on $P_{u_{i}}$ for $1 \leq i \leq ℓ$ .

Now let $k \leq ℓ$ and $w \in N (u_{k}) ∖ S$ . If a Pósa rotation occurs with $u_{k} w$ being the deleted edge then $w \in T_{k^{'}}$ for some $k^{'} > k$ . Else the vertex that succeeds $w$ on $P_{u_{k}}$ is $u_{k^{'}}$ for some $k^{'} < k$ . In this case let $Q_{0} = P_{u_{k^{'}}}, Q_{1}, Q_{2}, . . ., Q_{t} = P_{u_{k}}$ be the sequence of paths in $P_{u_{1}}, P_{u_{2}}, . . ., P_{u_{ℓ}}$ that correspond to the $t$ Pósa rotations that transform $P_{u_{k^{'}}}$ to $P_{u_{k}}$ . Then at every path $P$ of this sequence an edge $e$ incident to $u_{k^{'}}$ is either equal to $u_{k^{'}} r_{k^{'}, 1}$ or is an inserting edge. In both cases, both of the endpoints of $e$ belong to $\cup_{i \leq k} T_{i}$ hence $w \in \cup_{i \leq k} T_{i}$ . Therefore $N_{G} (E n d) ∖ S \subset \cup_{i \leq ℓ} T_{ℓ}$ and

| N_{G} (E n d) ∖ S | \leq 1 + 2 (| E n d | - 1) < 2 | E n d | .

∎

The key advantage of performing restricted Pósa rotations as opposed to normal Pósa rotations is given by the following observation.

Observation 2.2.

If $S = W \cup N (W)$ and $E n d, P = R P R (G, P, v, u, S)$ then $E n d$ and $W$ are disjoint. Indeed $u \notin S$ . Thereafter for a vertex $w \in [n]$ only a Pósa rotation with a pivot vertex in $N (w) ∖ S$ may result to a path with $w$ as an endpoint. As $N (w) ∖ S = \emptyset$ for $w \in W$ we have that no vertex in $W$ ever enters $E n d$ .

Observation 2.3.

If $E n d, P = R P R (G, P, u, v, S)$ then the set $E n d \cup (N (E n d) ∖ S)$ is connected in $G$ .

We use $R P R$ as a subroutine in the algorithm $R e d u c e P a t h s$ stated below. $R e d u c e P a t h s$ takes as an input a full path packing $P$ . It then performs $R P P$ and outputs sets $E n d$ , ${E n d_{u}}_{u \in E n d}$ and a set of full path packings $U$ with the following property. For $u \in E n d$ and $w \in E n d_{u}$ there exists a full path packing $P_{u, w} \in U$ of size at most $| P |$ with the property that adding $u w$ to $P_{u, w}$ creates a Hamilton cycle if $| P_{u, w} | = 1$ or joints two paths in $P_{u, w}$ and creates a full path packing of size at most $| P | - 1$ .

1:Let

P = {P_{1}, P_{2}, . . ., P_{k}}

. Let

v_{i}, u_{i}

be the endpoints of

P_{i}

1 \leq i \leq k

2:Let

R = {u_{i} v_{i + 1} : 1 \leq i \leq k - 1}

and

P = v_{1} P_{1} u_{1} v_{2} P_{2} u_{2}, . . ., u_{k - 1} v_{k} P_{k} u_{k}

E n d, P = R P R (G \cup R, P, v_{1}, v_{k}, S)

4:for

u \in E n d

5: Let

P_{u}

be the Hamilton path from

v

u

P

E n d_{u}, P_{u} = R P R (G \cup R, P_{u}, v_{1}, v_{k}, S)

7: For

w \in E n d_{u}

let

P_{u, w} \in P_{u}

be the

u

w

Hamilton path in

G \cup R

and let

P_{u, w}

be the full path packing obtained from

P_{u, w}

by removing the edges in

R

8:end for

9:Set

U = {P_{u, w} : u \in E n d, w \in E n d_{w}}

10:Output

E n d, {E n d_{u}}_{u \in E n d}, U

Algorithm 2

R e d u c e P a t h s (G, P, S)

$R e d u c e P a t h s (G, P, S)$ calls $R P R$ O(n) times, hence its running time is $O (n^{3})$ .

2.2 Covering non-expanding sets

In this subsection we introduce $C o v e r A n d A d j u s t$ which is the subroutine that we will use to deal with “problematic sets”. These sets will be non (sufficiently) expanding and will contain small sets outputted by $R e d u c e P a t h s$ . $C o v e r A n d A d j u s t$ takes as an input a graph $G^{'}$ , a set $S$ of size less than $n / 2$ and a full path packing $P$ and tries to adjust $P$ so that every vertex in $S$ is found in the interior of some path in $P$ . It starts by implementing the Held-Karp algorithm to find a path packing $P^{'}$ that covers $S$ . It then adjusts $P$ so that it contains $P^{'}$ .

1:Let

G_{S}^{'}

be the subgraph of

G^{'}

induced by

S \cup N (S)

2:for every component of

C

G_{S}^{'}

3: Add to

C

every edge spanned by

V (C) \cap N (S)

and let

G_{C}^{'}

be the obtained graph.

4: Implement Held-Karp to find a Hamilton cycle

H

G_{C}^{''}

. If no such a cycle exists output “

G^{'}

is not Hamiltonian”.

5: Remove from

H

all the edges not incident to

S

let

P^{'}

be the set of paths left of length at least 1.

6: Remove from the paths in

P

the vertices that are covered by

P^{'}

and add

P^{'}

P

7:end for

8:Output

P

Algorithm 3

C o v e r A n d A d j u s t (G^{'}, S, P)

The correctness of line 4 of the above algorithm is given by the following Lemma.

Lemma 2.4.

Let $G^{'} \subseteq G$ and $S \subset V (G)$ be such that $| S | < | V (G) | / 2$ and each vertex in $S$ has the same neighborhood in both $G$ and $G^{'}$ . Then if CoverAndAdjust( $G, S, P$ ) outputs “ $G^{'}$ is not Hamiltonian” then $G$ is not Hamiltonian.

Proof.

Assume that $G$ has a Hamilton cycle $H$ and let $C$ be an arbitrary component of $G_{S}^{'}$ . Then, as $| S | < | V (G) | / 2$ the edges of $H$ incident to $V (C) \cap S$ induce a path packing ${P_{1}, P_{2}, . . . ., P_{k}}$ that cover $S \cap V (C)$ such that the endpoints of $P_{i}$ lie in $N (S) \cap V (C)$ for $i \in [k]$ . As each vertex in $S$ has the same neighborhood in both $G$ and $G^{'}$ these paths are present in $G_{S}^{'}$ , hence in $G_{C}^{''}$ . Finally as every pair of vertices in $N (S) \cap V (C)$ forms an edge in $G_{C}^{''}$ the paths $P_{1}, P_{2}, . . . ., P_{k}$ can be used to form a Hamilton cycle in $G_{C}^{''}$ . Thus CoverAndAdjust( $G^{'}, S, P$ ) does not outputs “ $G^{'}$ is not Hamiltonian”. ∎

Observation 2.5.

After the execution of $C o v e r A n d A d j u s t (G^{'}, S, P)$ $| P |$ may increase by at most $6 | S |$ . This is because the “new” paths cover at most $3 | S |$ vertices and for each vertex covered by a new path at most 2 edges may be deleted at line 6.

The running time of $C o v e r A n d A d j u s t (G^{'}, S, P$ ) is dominated by the application of Held-Karp algorithm and it is $O (n s^{2} 2^{s}) + O (n^{2})$ where $s$ is the size of the largest component of the subgraph of $G^{'}$ induced by $S \cup N (S)$ .

2.3 A Family of Pseudorandom Graphs

Initially CerHam2 queries the edges of $G$ that belong to some graph $H_{n}$ . $H_{n}$ is constructed by taking a $d$ -regular pseudorandom graph $H_{n}^{'}$ on at least $n$ and at most $4 n$ vertices, then taking the subgraph of $H_{n}^{'}$ induced by a subset of $V (H_{n}^{'})$ of size $n$ and finally adding a number of edges incident to vertices of small degree. The construction of ${H_{n}}_{n \geq 1}$ as well as the proof of the Lemma that follows is given in Appendix A.

Lemma 2.6.

For $n$ significantly large there exists a graph $H_{n}$ on $[n]$ that can be constructed in $O (n^{7})$ time and satisfies the following.

For every pair of disjoint sets $U, W \subseteq [n]$ the number of edges spanned by $U \times W$ is at least $\frac{0.1 | U | | W |}{n} - n^{7 / 4}$ and at most $\frac{0.101 | U | | W |}{n} + n^{7 / 4}$
$H_{n}$ has minimum degree $0.1 n$ .
At most $n^{2 / 3}$ vertices in $[n]$ have degree larger than $0.101 n$ in $H_{n}$ .

2.4 A coloring process

For a graph $G$ we let $G \cap H_{n}$ be the graph on $[n]$ with edge set $E (G) \cap E (H_{n})$ . At the regimes where $p \leq \frac{1000 log n}{n}$ we first implement the following red/blue/black coloring coloring procedure: Initially all the vertices of $G \cap H_{n}$ are coloured black. While there exists a blue or black vertex $v$ with at most $5$ black neighbors in $G \cap H_{n}$ recolor $v$ red and the black neighbors of $v$ blue.

We denote by $V_{5, r e d} (G)$ , $V_{5, b l u e} (G)$ and $V_{5, b l a c k} (G)$ the sets of vertices that their final color given by the above procedure is red, blue and black respectively. A standard argument shows that $V_{5, r e d} (G)$ , $V_{5, b l u e} (G)$ and $V_{5, b l a c k} (G)$ do not depend on the order in which the vertices are being processed (see [3]). Observe that every vertex in both $V_{5, b l u e} (G)$ and $V_{5, b l a c k} (G)$ has at least $6$ neighbors in $V_{5, b l a c k} (G)$ . The motivation of the above procedure is to separate the sparse (red) portion of the graph from the dense one (black) by a cut (blue) while ensuring that the vertices in the cut are robustly connected to the dense part. In our analysis, for the sparse regimes, we use this decomposition and take advantage of the following observation and lemma. For the proof of Lemma 2.8 see Appendix A.

Observation 2.7.

During the above coloring process every time a vertex is colored red at most 5 other vertices are colored blue. Thus, for every component $C$ of the subgraph of $G$ induced by $V_{5, r e d} (G) \cup V_{5, b l u e} (G)$ we have that $| V_{5, b l u e} (G) \cap C | \leq 5 | V_{5, r e d} (G) \cap C |$ .

Lemma 2.8.

Let $\frac{2000}{n} \leq p$ and $G \sim G (n, p)$ . For $i, j, s \in N$ define the following events,

E_{r e d, s} = {| V_{5, r e d} (G \cap H_{n}) | \geq s)}

E_{e x p} = {\exists S \subset [n] : 0.02 n \leq S \leq 0.27 n and N_{G \cap H_{n}} \leq 2 s + 0.05 n)},

	$A_{i} := {$	there exists a set $S \subset [n] ∖ V_{5, r e d}$ of size at most $0.02 n$ with the property
		$that \| N_{G \cap H_{n}} (S) ∖ V_{5, r e d} \| < 2 \| S \| has % size i},$

and

	$B_{j} := {$	there exists a set $W \subset [n]$ of size at most $0.02 n$ and a subgraph $G_{R}$ of $G$
		$with the property that \| N_{G \cap H_{n}} (W) ∖ V_{5, r e d} \| < 2 \| W \| and$
		$W \cup V_{5, r e d} (G) \cup N_{G} (V_{5, r e d} (G) \cup W) spans a % component of size j in G_{R}} .$

Then, for $1 \leq i \leq 0.02 n$ , $1 \leq j \leq 0.09 n$ and $\frac{n}{{log}^{2} n} \leq s \leq 0.01 n$ the following hold.

P r (E_{r e d, s}) \leq {(\frac{e^{7} (0.1 n p)^{5} e^{- 0.09 n p} n}{s})}^{s}, P r (E_{e x p}) \leq 2^{- n},

(1)

(\frac{n}{i}) P r (A_{i}) = O (n) and 2^{j} P r (B_{j}) = O (n) .

(2)

2.5 Minimum degree 2

De facto if $G \sim G (n, p)$ does not have minimum degree 2 then $G$ is not Hamiltonian. Such a feature can be detected in $O (n^{2})$ time. In the event that $δ (G (n, p)) \geq 2$ we can initiate CerHam2 by implementing an algorithm for prepossessing $G$ that runs in time inversely proportional to $P r (δ (G) \geq 2)$ . Thus it becomes crucial to upper bound $P r (δ (G) \geq 2)$ . This is done in the following Lemma. Its proof can be found at Appendix A.

Lemma 2.9.

Let $\frac{2000}{n} \leq p \leq \frac{1000 log log n}{n}$ and $G \sim G (n, p)$ . Then,

P r (δ (G (n, p)) \geq 2) \leq n e^{- 1.1 n^{2} p e^{- 1.1 n p}} + 2^{- n} .

(3)

3 Certifying Hamiltonicity

CerHam2 takes as input a graph $G$ on $[n]$ and two sets $K, S \subseteq [n]$ . It also utilizes the graph $H_{n}$ whose construction is given in Appendix A. Throughout its execution $H$ consists of the edges of $G$ that have been queried so far and $G_{R}$ is the subgraph of $G$ with edge set the edges in $H$ that belong to $G$ .

1:Set

c o u n t = 0

i = 1

S_{0} = K \cup S

P = [n]

and

H = E (H_{n})

2:Query the edges in

H

and set

G_{R} = ([n], E (G) \cap H)

3:while

| S_{i - 1} | < 0.25 n

and

c o u n t < 0.01 n^{2}

E n d, {E n d_{u}}_{u \in E n d}, U = R e d u c e P a t h s (G, P, S_{i} \cup N (S_{i}))

5: if there exists

S_{i}^{'} \in {E n d_{u} : u \in E n d} \cup {E n d}

such that

| S_{i}^{'} | < 0.25 n

then

6: Let

S_{i} = S_{i - 1} \cup S_{i}

7: While there exists

W \subset [n] ∖ S_{i}

s.t.

| W | \leq | S_{i} ∖ S_{0} |

and

| N_{G_{R}} (W) ∖ S_{i} | < 2 | W |

and

W \cup (N_{G_{R}} (W) ∖ S_{i})

is connected in

G_{R}

, add

W

S_{i}

8: Add to

H

all the edges incident to

S_{i}

and query the edges of

G

that have just been added to

H

. Add any edges of

G

that have just been revealed to

G_{R}

9: Execute

C o v e r A n d A d j u s t (G_{R}, S_{i}, P)

. If it outputs “

G_{R}

is not Hamiltonian” then output “

G

is not Hamiltonian”. Else let

P = C o v e r A n d A d j u s t (G_{R}, S_{i}, P)

10: Set

i = i + 1

11: else

12: Let

u \in E n d

and

w \in E n d_{u}

be such that

u w \notin H

13: Add

u w

H

and query whether

u w \in E (G)

14: Set

c o u n t = c o u n t + 1

15: if

u w \in E (G)

then let

P_{u, w}

be a full path packing in

U

such that none of

u, w

lies in the interior of a path in

P_{u, w}

. Add to the graph spanned by the paths in

P_{u, w}

the edge

u w

. If it spans a Hamilton cycle output “

G

is Hamiltonian”. Else let

P

be the induced full path-packing. Add

u w

G_{R}

16: end if

17: end if

18:end while

19:Set

t = i - 1

20:Reveal

G

21:Execute the Held-Harp algorithm with input the graph

G

Algorithm 4 CerHam2(G,K,S)

For the correctness of CerHam2 one has to verify the correctness of lines 9 and 12. If CerHam2 outputs “ $G$ is not Hamiltonian” at line 9 then this statement is indeed true due to Lemma 2.4; here we are using that we reveal all the edges in $G$ incident to $S_{i}$ at line 8 and hence the edge sets incident to $S_{i}$ in $G_{R}$ and $G$ are identical. For line 12 recall that Observation 2.2 implies that at every iteration of the while-loop the sets $E n d, {E n d}_{u \in E n d}$ and disjoint from $S_{i}$ . In addition observe that $c o u n t$ is an upper bound on the number of edges in $H ∖ E (H_{n})$ not incident to $S_{i}$ . Furthermore Lemma 2.6 states that at most $n^{2 / 3}$ vertices have degree larger than $0.101$ in $H_{n}$ . Therefore if the algorithm proceeds to line 12 then the set ${u w : u \in E n d, w \in E n d_{u} and u w \notin E (H)}$ contains at least $[(0.25 - 0.101) n]^{2} / 2 - 0.01 n^{2} - o (n^{2}) \geq 1$ edges and the corresponding choice of $u$ and $w$ is possible.

Let $s_{i} = | S_{i} |$ for $0 \leq i \leq t$ . The following lemma upper bounds the number of edges in $G$ we need to identify at line 13. We later use Lemma 3.1 to upper bound the probability that $c o u n t \geq 0.01 n^{2}$ at the end of the while-loop. For $i \leq 0.01 n^{2}$ let $X_{i} = 1$ if the $i^{t h}$ time line 13 is executed an edge $w u$ in $G$ is identified.

Lemma 3.1.

For $i \in [t]$ we have,

| N_{G \cap H_{n}} (S_{i} ∖ S_{0}) | < 2 | S_{i} ∖ S_{0} | .

(4)

Furthermore,

t \leq {log}_{2} (\frac{s_{t} - s_{0}}{s_{1} - s_{0}}) .

(5)

In addition, if $| N_{G \cap H_{n}} (S ∖ K) | < 2 | S ∖ K |$ then for $i \in [t]$ ,

| N_{G \cap H_{n}} (S_{i} ∖ K) | < 2 | S_{i} ∖ K | .

(6)

Finally, if CerHam2 exits the while-loop and $| S | < 0.25 n$ then,

0.01 n^{2} \sum i = 1 X_{i} < 6 s_{0} t + 12 s_{t} + n .

(7)

Proof.

Let $1 \leq i \leq t$ and let $Q_{1}, Q_{2}, . . ., Q_{r}$ be the sets added to $S_{0}$ , due to lines 6 and 7, in the order that they were added, to form $S_{i}$ . As throughout the algorithm $G_{R}$ is a supergraph of $G \cap H_{n}$ , with $W_{j} = S_{0} \cup (\cup_{ℓ = 1}^{j - 1} Q_{ℓ})$ , we have,

| N_{G \cap H_{n}} (S_{i} ∖ S_{0}) |

= r \sum j = 1 | N_{G \cap H_{n}} (Q_{j} ∖ W_{j}) | < r \sum j = 1 2 | Q_{j} | = 2 | S_{i} ∖ S_{0} | .

For the proof of (6) we repeat the above argument with $Q_{0} (= S), Q_{1}, Q_{2}, . . ., Q_{r}$ being the sets added to $K$ to form $S_{i}$ .

Now observe that due to line 7 of the algorithm we have that $| S_{i + 1}^{'} | > | S_{i} ∖ S_{0} |$ . Therefore,

s_{i + 1} - s_{0} = | S_{i + 1} ∖ S_{0} | \geq | S_{i + 1}^{'} | > 2 | S_{i} ∖ S_{0} | = 2 (s_{i} - s_{0})

for $i \in {1, 2, . . ., t - 1}$ . Hence $t \leq {log}_{2} (\frac{s_{t} - s_{0}}{s_{1} - s_{0}})$ .

If CerHam2 exits the while-loop and $| S | < 0.25 n$ then the line 14 has been executed $0.01 n^{2}$ times. Say $| P | = 0$ if a Hamilton cycle has been constructed. The $i^{t h}$ time line 9 is executed the number of paths in $P$ may be increased by at most $6 | S_{i} |$ (see Observation 2.5). On the other hand $X_{j} = 1$ implies a decrease in $| P |$ the $j^{t h}$ time line 13 is executed. If CerHam2 exits the while-loop then $| P | > 0$ throughout the algorithm and as initially $| P | = n$ we have,

	$0$	$< n + t \sum i = 1 6 s_{i} - 0.01 n^{2} \sum i = 1 X_{i} = n + 6 s_{0} t + 6 t \sum i = 1 (s_{i} - s_{0}) - 0.01 n^{2} \sum i = 1 X_{i}$
		$\leq n + 6 s_{0} t + 6 t \sum i = 1 2^{- i + t} (s_{t} - s_{0}) - 0 .01 n^{2} \sum i = 1 X_{i} \leq n + 6 s_{0} t + 12 s_{t} - 0.01 n^{2} \sum i = 1 X_{i} .$

∎

3.1 The upper range

Let $p \geq \frac{1000 log n}{n}$ and $G \sim G (n, p)$ . We execute CerHam2( $G, K, S$ ) with $K = S = \emptyset$ . For the analysis of this regime we define the following events.

E_{e x p}^{'} = {\exists S \subset [n] with% size in [0.02 n, 0.27 n] such that | N_{G \cap H_{n}} (S) | < 2 | S |},

E_{c o u n t} = {c o u n t reaches the value of 0.01 n^{2}}

and

A_{i}^{'} = {\exists S \subset [n] such that % S = i and | N_{G \cap H_{n}} (S) | < 2 | S |}, i \leq 0.02 n .

Lemma 3.2.

O (n^{3} 2^{n} P r (E_{e x p} \cup E_{c o u n t}) + 0.02 n \sum j = 0 n^{2} (\frac{n}{j}) P r (A_{j} ∖ E_{e x p} \cup E_{c o u n t}) + n^{5}) = O (n^{7})

(8)

then the expected running time of CerHam2 is $O (n^{7})$ .

Proof.

Lemma 3.1 implies that $t \leq log n$ and therefore CerHam2 may execute at most $log n + 0.01 n^{2}$ times line 4. Each execution of line 4 runs in $O (n^{3})$ time; hence line 4 takes in total $O (n^{5})$ time. This is also the running time of CerHam2 in the event that it never proceeds to line 6 and it does not exit the while-loop.

In the event $E_{e x p}^{'} \cup E_{c o u n t}$ the algorithm may exit the while loop. Before exiting the while loop it may reach line 5 at most $n$ times, each time spending at most $O (n^{2} 2^{n})$ time at lines 5 to 9. After exiting the while-loop it executes Held-Karp whose complexity is $O (n^{2} 2^{n})$ . Thus, in the event $E_{e x p}^{'} \cup E_{c o u n t}$ CerHam2 runs in $O (n^{3} 2^{n})$ time.

On the other hand, if none of $E_{c o u n t}^{'}$ , $E_{e x p}$ occurs then $| S_{i} | \leq 0.02 n$ for $0 \leq i \leq t$ and $c o u n t \leq 0.01 n^{2}$ . The inequality $| S_{i} | \leq 0.02 n$ follows from the fact that while $| S_{i} | \leq 0.02 n$ the algorithm will not search for larger sets to add to $S_{i}$ at line 7 and it will never add to it a set of size larger than $0.25 n$ at line 6. Thus if a set $W$ is added by CerHam to $S_{i}$ then $| W | \leq 0.25 n$ and $| N_{G \cap H_{n}} (W) ∖ S_{i} | \leq | N_{G_{R}} (W) ∖ S_{i} | < 2 | W |$ . In addition, by Lemma 3.1, $| N_{G \cap H_{n}} (S_{i}) | < 2 | S_{i} |$ and hence $| N_{G \cap H_{n}} (S_{i} \cup W) | \leq 2 | S_{i} \cup W |$ and $| S_{i} \cup W | \leq 0.02 n + 0.25 n \leq 0.27 n$ and as $E_{e x p}^{'}$ does not occur we have that $| S_{i} \cup W | \leq 0.02 n .$

Thus, if none of $E_{c o u n t}^{'}$ , $E_{e x p}$ occurs and we let $j = | S_{t} |$ then $0 \leq j \leq 0.02 n$ and the event $A_{j}^{'} ∖ (E_{e x p}^{'} \cup E_{c o u n t})$ occurs. In the event $A_{j}^{'} ∖ (E_{e x p}^{'} \cup E_{c o u n t})$ CerHam2 may execute lines 6-10 at most $n$ times each running in $O (\sum_{ℓ = 1}^{j} (\frac{n}{ℓ})) = O (n (\frac{n}{j}))$ time yielding a total running time for lines 6-10 of $O (n^{2} (\frac{n}{j}))$ .

Equation (8) follows from the fact that at least one of the events $E_{e x p}^{'}, E_{c o u n t}, {A_{j}^{'}}_{j = 0}^{0.02 n}$ occurs. ∎

Lemma 3.3.

Equation (8) holds.

Proof.

In the event $E_{e x p}^{'}$ we can find disjoint sets $A, B \subset [n]$ of size $0.02 n$ and $(1 - 0.27 \cdot 3) n = 0.19 n$ such that no edge from $A$ to $B$ belongs to $G$ . Thus,

P r (E_{e x p}^{'}) \leq 2^{n} \cdot 2^{n} \cdot (1 - p)^{0.02 \cdot 0.19 n^{2}} \leq 4^{n} e^{- 10^{- 3} p n^{2}} < 2^{- n} .

In the event $E_{c o u n t} ∖ E_{e x p}^{'}$ (7) implies that $\sum_{i = 1}^{0.01 n^{2}} X^{i} < n + 12 \cdot 0.02 n \leq 1.3 n$ . For $1 \leq i \leq 0.01 n^{2}$ , $X_{i} = 1$ with probability $p$ independently of $X_{1}, X_{2}, . . ., X_{i - 1}$ . Thus,

	$P r (E_{c o u n t} ∖ E_{e x p}^{'})$	$\leq 1.3 n \sum j = 0 (\frac{0.01 n^{2}}{j}) p^{j} (1 - p)^{0.01 n^{2} - j} \leq n {(\frac{0.01 n^{2} p e}{1.3 n})}^{1.3 n} e^{- 0.009 p n^{2}}$
		$\leq n (n p e^{- 0.006 p n})^{1.3 n} \leq n (1000 log n e^{- 6 log n})^{1.3 n} < 2^{- n} .$

Finally, for $j \leq 0.02 n$ , in the event

A fast algorithm on average for solving the Hamilton Cycle problem

Authors

Fast algorithms for solving the Hamilton Cycle problem with high probability

Exact and Approximate Algorithms for Computing a Second Hamiltonian Cycle

A 2^O(k)n algorithm for k-cycle in minor-closed graph families

Deep reinforced learning enables solving discrete-choice life cycle models to analyze social security reforms

Facility Deployment Decisions through Warp Optimizaton of Regressed Gaussian Processes

The 1-1 algorithm for Travelling Salesman Problem

Fast recognition of some parametric graph families

1 Introduction

Theorem 1.1.

Corollary 1.2.

1.1 Description of CertifyHAM

2 Preliminaries

2.1 Restricted Pósa Rotations

Lemma 2.1.

Proof.

Observation 2.2.

Observation 2.3.

2.2 Covering non-expanding sets

Lemma 2.4.

Proof.

Observation 2.5.

2.3 A Family of Pseudorandom Graphs

Lemma 2.6.

2.4 A coloring process

Observation 2.7.

Lemma 2.8.

2.5 Minimum degree 2

Lemma 2.9.

3 Certifying Hamiltonicity

Lemma 3.1.

Proof.

3.1 The upper range

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

A fast algorithm on average for solving the Hamilton Cycle problem

Authors

Related Research

Fast algorithms for solving the Hamilton Cycle problem with high probability

Exact and Approximate Algorithms for Computing a Second Hamiltonian Cycle

A 2^O(k)n algorithm for k-cycle in minor-closed graph families

Deep reinforced learning enables solving discrete-choice life cycle models to analyze social security reforms

Facility Deployment Decisions through Warp Optimizaton of Regressed Gaussian Processes

The 1-1 algorithm for Travelling Salesman Problem

Fast recognition of some parametric graph families

This week in AI

1 Introduction

Theorem 1.1.

Corollary 1.2.

1.1 Description of CertifyHAM

2 Preliminaries

2.1 Restricted Pósa Rotations

Lemma 2.1.

Proof.

Observation 2.2.

Observation 2.3.

2.2 Covering non-expanding sets

Lemma 2.4.

Proof.

Observation 2.5.

2.3 A Family of Pseudorandom Graphs

Lemma 2.6.

2.4 A coloring process

Observation 2.7.

Lemma 2.8.

2.5 Minimum degree 2

Lemma 2.9.

3 Certifying Hamiltonicity

Lemma 3.1.

Proof.

3.1 The upper range

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.