HAMILTONIAN CYCLES

Vesal HakamiOctober 30, 2010

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Hamiltonian Cycles Demystified

• Brief History• Basic Definitions, Initiatives, Sufficient

Conditions, A Backtracking Algorithm • Chiba & Nishezeki’s Linear

Implementation for Internally 4-Connected Plane (I4CP) Graphs (Journal of Algorithms, 1989 – Elsevier)

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History• Icosian game invented by Sir. William Rowan Hamilton (1805–

1865) in 1857 and sold to a London game dealer in 1859 for 25 pounds.

• Game Description: The corners of a regular dodecahedron are labeled with the names of cities; the task is to find a circular tour along the edges of the dodecahedron visiting each city exactly once.

• Solution Model: Hamiltonian cycle; i.e. look for a cycle in the corresponding dodecahedral graph which contains each vertex exactly once.

The Platonic Solid used in Icosian game;the corresponding Hamiltonian cycle is designated by darkened edges.

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Basic Definitions

• Hamiltonian Path: A Hamiltonian path is a path (and not a cycle) in G that contains each vertex. It is possible, however, for a graph to have a Hamiltonian path without having a Hamiltonian cycle.

Remarks:• There is no helpful connection between the idea of Eulerian circuits (or

trails) and Hamiltonian cycles (or paths).• There do not exist necessary and sufficient* conditions on a graph G

that guarantee the existence of a Hamiltonian cycle (path).• One may resort to trial and error for particular graphs along with a

number of helpful observations.

• Hamiltonian Cycle: If G = (V, E) is a graph or multi-graph with |V|>=3, we say that G has a Hamiltonian cycle if there is a cycle in G that contains every vertex in V.

Hamiltonian path Hamiltonian cycle

* Necessary and Sufficient Conditions for Hamiltonian based on Linear Diophantine Equation Systems with Cycle Vector, 2009 3rd Intl. Conf. on Genetic and Evolutionary Computing.

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Example I- A Connected Graph with no Cycle of Odd Length

• Does G contain a Hamiltonian path?• Starting from vertex a, alternatively label each vertex and its

adjacents with x’s and y’s. If G is to have a Hamiltonian path, there must be an alternating sequence of five x's and five y's.

• Recall: A graph G = (V, E) is bipartite iff it contains no cycle of odd length.

G (no cycle of odd length)

x-y vertex labeling

4 x’s and 5 y’s No Hamiltonian pathG is, in fact, a bipartite graph

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Helpful Hints on Finding Hamiltonian Cycles

The complete graph Kn is Hamiltonian.

If G = (V, E) has a Hamiltonian cycle, then for all vV, deg(v) 2.

If a V and deg(a) = 2, then the two edges incident with vertex a must appear in every Hamiltonian cycle for G.

If a V and deg(a) > 2, then as we try to build aHamiltonian cycle, once we pass throughvertex a, any unused edges incident with a aredeleted from further consideration.

In building a Hamiltonian cycle for G, we cannotobtain a cycle for a sub-graph of G unless itcontains all the vertices of G.

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Example II- Tournament Graphs• Theorem. A complete digraph (called a tournament)- in which for each distinct pair x, y

of vertices, exactly one of the edges (x, y) or (y, x) is in - always contains a (directed) Hamiltonian path.

• Proof: • Let m 2 with Pm a path containing the m-1 edges (v1, v2), (v2, v3), .... (vm-1, vm).• if m = n, we're finished.• If not,

– let v be a vertex that does not appear in Pm. – If (v, v1) is an edge in , we can extend Pm by adjoining this edge.– If not,

• (v1, v) must be an edge.

• Suppose that (v, v2) is in the graph.

• We have the larger path: (v1, v), (v, v2), (v2, v3), ..., (vm-1, vm).

• If (v, v2) is not an edge in , then (v2, v) must be.

– Continue this process; there are only two possibilities:• (a) For some 1 k m-1, the edges (vk, v), (v, vk+1) are in and we replace (vk, vk+1) with this pair of edges.

• (b) (vm, v) is in and we add this edge to Pm.

• Either case results in a path Pm+1 that includes m+1 vertices and has m edges.

– Repeat this process until we have a path Pn on n vertices.

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Hamiltonian Path: Sufficient Condition

• Theorem. Let G= (V, E) be a loop-free graph with |V|= n 2. If deg(x) + deg(y) n - 1 for all x, y V, x y, then G has a Hamiltonian path.

• Proof. • I) Connectivity (Proof by Contradiction): G consists of two components C1 (|

C1| = n1) and C2 (|C2| = n2). Let x C1 and y C2. Obviously, deg(x) n1 - 1 and deg(y) n2 - 1; that is, deg(x) + deg(y) (n1 + n2) – 2 n - 2, contradicting the condition given in the theorem; thus, G is connected.

• II) Hamiltonian path construction: – For m 2, let Pm be the path {v1, v2}, {v2, v3}, ... , {vm-1, vm} of length m - 1.

– If v1 is adjacent to any vertex v other than v2, v3, ..., vm, we add the edge {v, v1} to Pm to get Pm+1.

– The same procedure is carried out if vm is adjacent to a vertex other than v1, v2, ..., vm-1.

– If we are able to enlarge Pm to Pn in this way, we get a Hamiltonian path.

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Hamiltonian Path: Sufficient Condition- Ctd.

• Otherwise,– Pm: {v1, v2}, ..., {vm-1, vm} has v1 and vm adjacent only to vertices in Pm and m < n.– Now, we claim that G has a cycle on these vertices:

• If v1 and vm are adjacent, then the cycle is {v1, v2}, {v2, v3}, ..., {vm-1, vm}, {vm, v1}.• Otherwise,

– v1 is adjacent to a subset S of the vertices in {v2, v3, ..., vm-1}.– If there is a vertex vt S such that vm is adjacent to vt-1, then we can get the

cycle by adding {v1, vt}, {vt-1, vm} to Pm and deleting {vt-1, vt}, as shown in Fig. (a).

– If not,» Let |S| = k < m - 1. » Then, deg(v1) = k and deg(vm) (m - 1) - k» Contradiction! deg(v1) + deg(vm) m - 1 < n - 1.

– Hence, there is a cycle connecting v1, v2, ..., vm.

Fig. (a)

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Hamiltonian Path: Sufficient Condition- End.

• Now, consider a vertex v V that is not found on this cycle.• G is connected, so there is a path from v to a first vertex vr in the cycle, as shown in Fig. (b).

• Removing the edge {vr-1, vr} (or {v1, vt} if r = t), we get the path (longer than the original Pm) show in Fig. (c).

• Repeating this whole process for the path in Fig. (c), we continue to increase the length until it includes every vertex of G.

Fig. (b)

Fig. (c)

Corollary. The loop-free graph G = (v, E) with |V| 2 has a Hamiltonian path if deg(v) (n - 1)/2 for all v V.

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Hamiltonian Cycle: Sufficient Condition- Ore (1960)

• Theorem. Let G= (V, E) be a loop-free graph with |V|= n 3. If deg(x) + deg(y) n for all x, y V, x y, then G has a Hamiltonian cycle.

• Proof. • Assume that G does not contain a Hamiltonian cycle. • We add edges to G until we arrive at a sub-graph H of Kn, where H

has no Hamiltonian cycle, but, for any edge e (of Kn) not in H, H + e does have a Hamiltonian cycle.

• Since H Kn, there are vertices a, b V, where {a, b} is not an edge of H but H + {a, b} has a Hamiltonian cycle C.

• The graph H has no such cycle, so the edge {a, b} is a part of cycle C.• The vertices of H (and G) on cycle C are as follows:

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Hamiltonian Cycle: Sufficient Condition- Ore (1960)- End.

• For each 3 i n, if the edge {b, vi} is in the graph H,– then we claim that the edge {a, vi-1} cannot be an edge of H.

• If both of these edges are in H, for some 3 i n, – we get the Hamiltonian cycle:

for the graph H (which has no Hamiltonian cycle).• Therefore, for each 3 i n, at most one of the edges {b,

vi}, {a, vi-1} is in H.• Consequently,

• Since for all v V, , so we have nonadjacent (in G) vertices a, b, where

𝒅𝒆𝒈 (𝒂)+𝒅𝒆𝒈 (𝒃 )<𝒏 CONTRADICTION!

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Hamiltonian Cycle: Ore’s Sufficient Condition Corollaries

• Corollary I. If G = (V, E) is a loop-free undirected graph with |V| = n 3, and if deg(v) n/2 for all v V, then G has a Hamiltonian cycle.

• Corollary II. If G = (V, E) is a loop-free undirected graph with |V| = n 3, and if , then G has a Hamiltonian cycle.

• Proof. • Let a, b V, where {a, b} is not in E. [Since a, b are nonadjacent,

we want to show that deg(a) + deg(b) >= n.] • Remove the vertices a and b together with all edges incident on

them from the graph G.• Let H = (V', E') denote the resulting sub-graph.• |E| = |E'| + deg(a) + deg(b) because {a, b} is not in E.

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Hamiltonian Cycle: Ore’s Sufficient Condition Corollaries- End.

• Since |V'| = n - 2, H is a sub-graph of the complete graph Kn-2, so

• Consequently,

• and we find that:

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Hamiltonian Cycle: Backtracking Algorithm

• State space tree:– The initial node in level 0– All the remaining n-1 nodes in levels 1, 2, …, n-1– The number of state space tree nodes:

• Promising nodes’ characterization:– The ith node on the path must be adjacent to (i-1)th node– (n-1)th node must be adjacent to node 1 (initial node)– The ith node cannot be one of the first i-1 nodes

• Inputs:– A positive integer n (Number of vertices)– An undirected graph G with n vertices and with adjacency matrix denoted by

the 2D array W• Output:

– The Hamiltonian cycle denoted by the linear array vindex• vindex [i] is the index of the ith node on the cycle.• The Index of the initial node is located in vindex [0].

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Hamiltonian Cycle: Backtracking Algorithm- End.

• High level call:– vindex[0] = 1;– hamiltonian(0);

void hamiltonian (index i){

index j;if (promising(i)) if (i == n – 1) cout << vindex[0] through vindex[n – 1];else for (j = 2; j <= n; j++) // Try all vertices { // as next one. vindex[i + 1] = j; hamiltonian(i+1); }

}

void promising (index i){

index j;bool switch;if ( i == n – 1 && ! W[vindex[n – 1]] [vindex[0]]) // First vertex must switch = false; // be adjacent toelse if (i > 0 && ! W[vindex[i -1]] [vindex[i]]) // last. ith vertex switch = false; // must be adjacentelse { // to (i – 1)th. switch = true; j = 1; while (j < i && switch) { // Check if vertex is if (vindex[i] == vindex[j]) // already selected. switch = false; j++; }}return switch;

}

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CHIBA & NISHIZEKI’S LINEAR IMPLEMENTATION FOR I4CP GRAPHS

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Chiba & Nishizeki’s Algorithm- Overview

• The Hamiltonian cycle problem is NP-complete even for 3-connected planar graphs.

• Tutte proved that a 4-connected planar (4CP) graph necessarily contains a Hamiltonian cycle and that It can be found in polynomial time.

• Prior art for 4CP graphs:– An O(n3) algorithm– A linear algorithm for 4-connected “maximal” planar graphs

• Objective– A linear algorithm for finding a Hamiltonian cycle in

general 4CP graphs

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Chiba & Nishizeki’s Algorithm- Overview

• Solution methodology– Divide & Conquer with inductive argument:

• Decompose a graph into several small sub-graphs, then find Hamiltonian cycles in sub-graphs by applying the inductive hypothesis to them, and finally combine the Hamiltonian cycles into a Hamiltonian cycle of the whole graph.• Since the decomposed graphs are not always edge-

disjoint, it was non-trivial to verify even the polynomial bounded-ness of the algorithm.

– Key idea: • Avoid the decomposition into non-disjoint sub-graphs.

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Required Terms and Definitions• Cut-vertex: a vertex whose deletion increases the number of

components.• Separation pair: a separation pair of a 2-connected graph G

is two vertices whose deletion disconnects G. • Splitting: Dividing a graph G into two split graphs G1 and G2.• Merging: the inverse of splitting• Virtual edge: The edge {x, y} in G1 or G2 whether it originally

exists in G1 and G2 or is newly added to G1 and G2.• Separation triple: a separating set of three vertices.• 4-connected graph: a 3-connected graph with no separation

triples.• Block: A block of G is a maximal 2-connected sub-graph of G.

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Thomassen’s Result from Tutte’s Theorem

Lemma. Let G be a 2-connected plane graph with the outer facial cycle Z. Let s and e = (a, b) be a vertex and an edge, both on Z, and let t be any vertex of G distinct from s. Then G has a path P (called a Tutte path) going from s to t through e such that

– (i) each component of G - P is adjacent to at most three vertices of P, and– (ii) each component of G - P is adjacent to at most two vertices of P if it contains an

edge of Z.

Remark I. The above lemma implies that a 4-connected planar graph G has a Hamiltonian cycle.

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Thomassen’s Result from Tutte’s Theorem- End.

• Remark II. Let G be a 4-connected and P be a s, t - Tutte path in G. Suppose H is a (G - P)-component then, since G is 4-connected, H must have 4 vertices of attachment in P which is a contradiction because P is a Tutte path.

• Remark III. In a 4-connected planar graph G: Let s and t be two adjacent vertices on Z and let e (s, t) be an edge on Z, then the path P joining s and t through e, assured by the Lemma, must be a Hamiltonian path of G, so P + (s, t) must be a Hamiltonian cycle of G. Thus an algorithm for finding the s-t path P immediately yields an algorithm for finding a Hamiltonian cycle.

• Decomposition strategy– The original graph G is 4-connected. – The sub-graphs into which G is decomposed are no longer 4-

connected.– They inherit the “internally 4-connected (I4C)” property.

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Internally 4-Connected Plane (I4CP) Graphs

• The I4C property• Intuitively, a graph is internally 4-connected if it contains no separation

pair or triple in the interior.• Formal definition: Let G be a 2-connected plane graph with outer facial

cycle Z. Let s and t be two distinct vertices on Z, and let e = (a, b) be an arbitrary edge on Z such that e (s, t). t a , b and vertices s, a, b and t appear clockwise on Z in this order. (Fig. (a).)

• Possibly s = a as illustrated in Fig. (b). Let r be the vertex on Z counterclockwise next to s, and let f = (r, s) be the edge joining r and s.

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Internally 4-Connected Plane (I4CP) Graphs- End.

• A plane 2-connected graph G is internally 4-connected with respect to (s, t, e) if G satisfies the following two conditions:(a) If {x, y} is a separation

pair, then(i) every component of

G - {x , y} contains at least one vertex of Z

(ii) each of the three paths Psa, Pbt and Pts contains at most one of x and y.

(b) If {x, y, z} is a separation triple, then every component of G - {x, y, z} contains at least one vertex of Z.

(c), (d) and (e) violate conditions in (a).

(f) violates condition (b).

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• Vertical separation pair: a separation pair {x, y} is called vertical if

either x Psa- s and y Pbr

or x = s and y Pbt – t.

• Type I Reduction: G is decomposed into two sub-graphs w.r.t. a vertical separation pair.

• {y’, z’} is a horizontal separation pair.

Type I Reduction Using Vertical Separation Pairs:

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• Let G be an I4CP graph having no vertical separation pair, then G is decomposed into several sub-graphs, called , , and. (If S = a, G is decomposed into and .)

Type II Reduction: , , and

𝑮𝒃

𝑮𝒈𝟐

G

𝑮𝒈𝟒

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Type II Reduction- ,

• : the block of G – Psa which contains t. is cross-hatched.)

• must entirely contain Pbr, otherwise G would contain a vertical separation pair.

• Repeat splitting of at every separation pair such that one of the two split graphs entirely contains Pbr.

• The component containing Pbr is called Gb, while each of the others is called where g = (x, y) is a virtual edge contained in the component.

𝑮𝒃

𝑮𝒈𝟐

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Type II Reduction-• maximal sub-graph of G – which can be separated from at each cut-

vertex u.• the sub-graph of G induced by the vertices of and the vertices on Psa

which are adjacent to .• Let x (resp. y) be the vertex of which is nearest to s (resp. a) along . Add a

new edge e’ = (u, y) to if it does not exist, and let be the resulting graph.

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Type II Reduction-• : schematically, is a graph formed by merging with all the and drawn above .

• Constructively, is defined as:– := ;– while has a virtual edge g’ g, iterate to merge into ; – for each cut-vertex u (x, y) of G – Psa contained in merge to ;– Construct the sub-graph of G induced by the vertices of together with the vertices of Psa adjacent to , and redefine as the sub-graph;– - x – y.

𝑮𝒈𝟐𝟒

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Type II Reduction Maintains the I4C Property

• Lemma. Let G be an I4CP graph containing no vertical separation pair. Then, all the decomposed graphs , and are I4C w.r.t. (s’ , t’ , e’) if s’ , t’ , e’ are defined as follows:

- Case s’ = bt’ = te’ =

if t r the edge clockwise incident to r on the outer facial cycle of Gb.

If t = rthe edge counterclockwise incident to b.

𝑮𝒃

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Type II Reduction Maintains the I4C Property- Ctd.

– Case s’ = yt’ = xe’ = the edge counterclockwise incident to y on the outer facial cycle of .

– Case s’ = ut’ = xe’ = (u, y).

– Case s’ = wt’ = ve’ =

if w’ v’ an arbitrary edge on outer path Pw’v’.

If w’ v’ an arbitrary edge on Pwv incident to w’ (= v’).

𝑮𝒈𝟒

𝑮𝒈𝟐

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Type II Reduction Maintains the I4C Property- End.

• Proof. We verify only the case . Suppose is not I4C w.r.t. (s’ , t’ , e’). Then, must have one of the following three:– a cut-vertex vc.– A separation pair {vp1, vp2} such that both vp1 and vp2 are contained in path Pwa’ or

Pbv’; and

– a separation triple {vt1, vt2, vt3} for which one of the components of - {vt1, vt2, vt3} contains no vertex of the outer facial cycle Z’ of

• In either case, G would contain a separation triple {vc, x, y}, {vp1, vp2, vx}, {vp1, vp2, y}, or {vt1, vt2, vt3} such that the deletion of the triple from G produces a component containing no vertex of Z. This contradicts the I4C-ness of G.

G 𝑮𝒈𝟒

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An I4C graph w.r.t. (s, t, e) contains a Hamiltonian path connecting s and t through e

• Lemma. Let G be a plane graph having an outer facial cycle Z. Let s and t be two distinct vertices on Z and let e( (s , t) ) be an edge on Z . If G is I4C w.r.t. (s, t, e), then G has a Hamiltonian path P(G, s, t, e) which connects s and t and contains e. Moreover, if G has no vertical separation pair, then P(G, s, t, e) does not contain edge f = (s, r).

• Proof. The proof is by induction on the number n of vertices of a graph G. If n = 3, the claim is clearly true, so we assume that n > 3. There are two cases to consider.– There exists a vertical separation pair {x, y}– G has no vertical separation pairs.

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An I4C graph w.r.t. (s, t, e) contains a Hamiltonian path connecting s and t through e - Ctd.

• Case 1: There exists a vertical separation pair {x, y}. – Among all vertical separation pairs of G, we choose {x, y} such that Gr has the

smallest number of vertices. We call such a separation pair the rightmost separation pair.

– Consider the case t GI * as shown in Fig.

• Let e’ = (x, y) be a virtual edge.Clearly, Gl is I4C w.r.t. (s, t, e’), while Gr is I4C w.r.t. (x, y, e) (or w.r.t. (y, x, e) if y = b).• By the inductive hypothesis, Gl has

a Hamiltonian path P(Gl , s, t, e’) and Gr has P(Gr , x, y, e).

• G has a Hamiltonian path:P(G , s, t, e) = P(Gl , s, t, e’) + P(Gr , x, y, e) – e’or P(Gl , s, t, e’) + P(Gr , y, x, e) – e’

* The case t GI is handled with a similar understanding.

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An I4C graph w.r.t. (s, t, e) contains a Hamiltonian path connecting s and t through e - End.

• Case 2: G has no vertical separation pair. – By the inductive hypothesis, all the decomposed graphs , and have Hamiltonian

paths. From them, a Hamiltonian path P(G , s, t, e) of G can be constructed as follows

(Note that edge f = (s, r) is never included in path P):

First set P to Psb + P(Gb , b, t, e’).While there is a virtual edge g = (x, y) in Gb

If g P, thenP : = P – g + P( , y, x, e’). (replace g of P by a H-path of )Merge into Gb ;

If g p, thenConstruct , and set P := P – Pvw + P( , w, v, e’),Delete g from Gb .

For each u of the cut-vertices of G – Psa contained in Gb Set P := P – Pxy + P( , u, x, (u, y)) - (u, y).

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An Abstract Example

, , and

If g1 is on P(Gb , b, t, e’), then:

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A Concrete Example

G , ,

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A Concrete Example- Ctd.

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A Concrete Example- End.

– In the previous slide,

– The path passes through the virtual edge (1, 7), so a Hamiltonian path of is determined.

– The path passes through the virtual edges (1, 3) and (5, 7) but not (3, 5), so the Hamiltonian paths of , and are determined.

S- t Hamiltonian path

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The HPATH Algorithm of O(n2)procedure HPATH(G, s, t , e);

beginif G contains exactly three vertices

then {G is a triangle}return a (trivial) Hamiltonian path P(G, s, t, e);

else if G has a vertical separation pairthen Type I reductionelse Type II reduction

end;

• Lemma. There are at most n-3 reductions during one execution of algorithm HPATH.

• Theorem. The algorithm HPATH spends at most O(n2) time.

Time Complexity Analysis

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Time Complexity Analysis- Ctd.

• Proof. • r: the number of reductions occurred during one execution

of HPATH.• d(i), 1 i r: the number of small graphs into which a graph is

decomposed at the ith reduction; i.e. d(i) is the number of the recursive calls which occurred at the reduction.

d(i) = 2 if the ith reduction is of Type I

1 if the ith reduction is of Type II and none of,is produced.

• r': the number of reductions with d = 1• t: the number of final triangles

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Time Complexity Analysis- Ctd.• Upper bound on the number of reductions in the

recursive call tree:– Internal nodes Reductions– Children of internal nodes Recursive calls at the reduction– Leaves Final triangles

• The tree has:– t leaves– r internal nodes (r' of which have out-degree 1)

• Recall:The number of internal nodes in a binary tree is one less than the number of leaves.

• The recursive tree has at most t - 1 internal nodes of out-degree 2 or more; thus, we have:

r t – 1 + r’ (1)

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Time Complexity Analysis- Ctd.

• Upper bound for the total length of paths found in triangles:– A Type I reduction increases by one the total length of paths

which will be found in the two reduced graphs.– In general, the ith reduction increases by d(i) – 1 the total

length of paths which will be found in the d(i) reduced graphs.– Conversely, if d(i) = 1, then the length of a path which will be

found in the reduced graph decreases by at least one.– HPATH initially wishes to find a Hamiltonian path of length n

– 1. Therefore, the upper bound for the total length of paths found in triangles is:

𝒏−𝟏+ ∑𝟏≤𝒊≤ 𝒓

(𝒅 ( 𝒊 )−𝟏 )−𝒓 ′=𝒏−𝟏+𝒓+𝒕−𝟏−𝒓 −𝒓 ′

(2) =

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Time Complexity Analysis- End.

• The lengths of the Hamiltonian paths found in the final triangles total 2t, we have:

• that is,

• Combining (1) with (3), we get the claimed bound on r:

(3)

r

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THE END