A Simple Linear Time Algorithm for the Domatic Partition Problem on Strongly Chordal Graphs

A Simple Linear Time Algorithm for the Domatic

Partition Problem on Strongly Chordal Graphs

Speaker: Chuang-Chieh Lin ( 林莊傑 )Computation Theory Laboratory

Advisor: Professor Maw-Shang Chang

Shen-Lung Peng and Maw-Shang Chang

Information Processing Letters, 43 (1992), pp. 297-300.

‧2‧

Outline

• Introduction• Strongly Chordal Graphs• Algorithm• Correctness• Complexity• Conclusions• References

‧3‧

Outline


‧4‧

Dominating set

• A graph of vertices D is a dominating set of a graph G = (V, E) if every vertex in V \ D is adjacent to a vertex in D.

v2

v3

v1

v4v5 v8

v7v6v1

v4v5 v8

v7

D = {v1, v4, v5, v7, v8}

‧5‧

Dominating set

• Another dominating set

v2

v3

v1

v4v5 v8

v7v6v1

v4v5 v8

v7v2

v3

v6v2

D = {v2, v3, v6}D = {v3, v6}

Q: Is there another smaller dominating set?

‧6‧

Dominating set

• What is the dominating set of the following graph?

v2

v3

v1

v4v5 v8

v7v6v2

v3

v1

v4v5 v8

v7v6

‧7‧

Domatic number

• The domatic number of G, denoted by d(G), is the maximum number of pairwise disjoint dominating sets in G.

v3

v1

v4 v8

v7v6

v5

v2

{v1, v4, v5, v7, v8}, {v2, v3, v6}

v1

v4 v8

v7

v5v3

v6v2

‧8‧

Domatic number

• Another family of pairwise disjoint dominating sets in G:

v3

v1

v4 v8

v7v6

v5

v2

{v1, v4, v5, v7, v8}, {v3, v6}

v1

v4 v8

v7

v5v3

v6v2

‧9‧

Domatic number

• Another family of pairwise disjoint dominating sets in G:

v3

v1

v4 v8

v7v6

v5

v2

{v3, v5, v7, v8}, {v1, v4 , v6}

v3 v8

v7

v5

v2v1

v4

v6

The domatic number of the graph is “2”. (Why?)

‧10‧

The domatic partition problem

• The domatic partition problem is to partition V into d(G) disjoint dominating sets.

{{v1, v4, v5, v7, v8}, {v2, v3, v6}} is called a domatic partition.

v3

v1

v4 v8

v7v6

v5

v2v1

v4 v8

v7

v5v3

v6v2

‧11‧

• Another example:

5

1

2

3

46

7

8

{ {1, 5}, {2, 6}, {3, 8}, {4, 7}} is a domatic partition.

‧12‧

• Question:– Does any graph has a domatic partition?

‧13‧

• The domatic number was defined and studied by E. J. Cockayne and S. T. Hedetniemi [CH77].

• They showed that d(G) ≤ + 1, where is the minimum degree of vertices in G.

• G is domatically full if d(G) = + 1.

‧14‧

• Some special graph classes are domatically full, such as :– Trees, [CH77]

– Cliques, [CH77]

– Maximum outerplanar graphs, [CH77]

– Strongly chordal graphs, [F84]

– ……

‧15‧

Chord

• An edge is called a chord of a cycle if it connects two vertices of the cycle but is not itself an edge within the cycle.

1

54

2

3

G

‧16‧

Chord

• The subgraph of G induced by {1,2,3} has a cycle.

54

2

3

1

Any chord exists?

‧17‧

Chord

• The subgraph of G induced by {1,2,3,4} has cycles.

54

2

3

Any chord exists?

[2, 4], [1, 3]

1

‧18‧

Chord

• The subgraph of G induced by {1,2,4,5} has cycles.

3

Any chord exists?

[1, 4]

12

54

‧19‧

Hole

• A chordless cycle is sometimes also called a graph hole.

3

2

45

1

‧20‧

Chordal graphs

• A chordal graph is a simple graph possessing no holes (chordless cycles).

• That is, A graph is chordal if and only if every cycle of length greater than three has a chord.

‧21‧

A chordal graph:

A non-chordal graph:

3

62

7

4

1

5

1

54

2

3

62

7

4

1

5

A hole !

‧22‧

Outline


‧23‧

Relax!

• 某天上李家同教授的科技英文課，他問：「我問你們，你們知道米開朗基羅是誰嗎？」

• 某同學回答：「忍者龜！」• 李教授正在感嘆學生人文素養不夠，正要開罵時，我補了一句：

• 「另外還有三隻忍者龜，叫多納太羅、拉斐爾和達文西。」

• 從此我變成大學生人文素養不夠的範例。

‧24‧

Odd chords

• Let (v1, v2, …, vk) be an even length cycle. Then an edge e = [vi, vj] is an odd chord of the even length cycle if e is a chord and i – j is odd.

2

34

5

61

(1, 2, 3, 4, 5, 6) is an even length cycle.

Any chord else?

‧25‧

Can you find an odd chord in the following graph?

• Is it a chordal graph?

2

34

5

61

‧26‧

Strongly chordal graphs

• A graph is strongly chordal if it is chordal and every even length cycle of length six or more has an odd chord.

• Other useful properties that characterize chordal and strongly chordal graphs are introduced in the following.

Why six?

‧27‧

Simplicial vertex

• N[v] = Adj(v){v} is called the closed neighborhood of v.

• A vertex v is called simplicial if the subgraph induced by N[v] is a clique.

v

w

x

y N[v] = {v, w, x, y}

v

w

x

y v is simplicial

‧28‧

3

62

7

4

1

5

Where are the simplicial vertices in the graph?Is vertex “1” a simplicial vertex?

{7, 6}

‧29‧

Perfect elimination ordering

• Rose [R70] showed that a graph G is chordal if and only if it is possible to order the vertices (v1, v2,…, vn) in such a way that, for each i{1, 2,…, n}, vi is a simplicial vertex of Gi = G({vi, vi+1,…, vn}), where Gi is a subgraph induced by the vertex set {vi, vi+1,…, vn}.

• Such an ordering is called a perfect elimination ordering.

‧30‧

62

7

1

6 → 7 → 2 → 1 → 3 →

3

4

4 →

5

5

(6, 7, 2, 1, 3, 4, 5) is called a perfect elimination ordering of the given graph.

‧31‧

Strong elimination ordering

• Let Ni[v] denote the closed neighborhood of v in Gi.

• The ordering of vertices (v1, v2,…, vn) is called a strong elimination ordering if it is a perfect elimination ordering and, for each i < j < k, if vj and vk belong to Ni[vi] in Gi then Ni[vj] Ni[vk].

• For example,

‧32‧

7 8

31 4

965

2 3

65

43

65

2

5

2 3

7 8

4

96

7

‧33‧

Strong elimination ordering (contd.)

• In other words, for each vertex vi we not only require that vi be simplicial in Gi but in addition insist that the ordering of the vertices in Ni[vi] be consistent with the ordering of this nested family (smallest to largest).

‧34‧

Why are we concerned for the strong elimination ordering?

• Farber [F83] showed that a graph is strongly chordal if and only if it admits a strong elimination ordering.

• To date (1992), the best-known time to recognize strongly chordal graphs is O(m log n). This algorithm additionally determines a strong elimination ordering. [PT87]

‧35‧

Outline


‧36‧

Relax!如果你只剩下 3 天壽命 ......

某日，一位教授問學生：「如果你只剩下三天的壽命，你要怎麼利用？」學生競相回答：

AA ：「我要趁這幾天好好陪陪家人。」

BB ：「我要和我女朋友大玩特玩。」

CC ：「我要寫好遺囑。」

這時教授看見一個同學沈默不語，便問他：「 DD 你呢？你要怎麼利用這三天？」

DD ：「教授，我想聽你的課。」

全班同學開始罵他狗腿，為了分數不擇手段等等，而教授聽了大喜，心想現在居然還有如此尊師重道的同學，不禁再問他：「為什麼呢？」　

他突然正氣凜然的站起來：「因為聽你的課讓我有度日如年的感覺．．．」

一學期後，那同學被當了。

‧37‧

Preliminary of the algorithm

• A greedy approach is used to design this algorithm.

• First, we assume that a strong elimination ordering has been constructed for the strongly chordal graph G = (V, E).

• It has the property that the closed neighborhoods of the members of Ni[vi] form a nested family of sets.

‧38‧

Preliminary of the algorithm (contd.)

• The algorithm maintains + 1 disjoint sets. – Initially, these sets are empty.

• The algorithm visits vertices in the reverse ordering of the strong elimination ordering. A vertex is included in a set when it is visited.

‧39‧

Preliminary of the algorithm (contd.)

• We say that a vertex v is dominated by set S if there exists a vertex u S such that u N[v].

• A vertex v is completely dominated if v is dominated by + 1 dominating sets.

• Now let us see the algorithm.

‧40‧

Algorithm DP

• 1. Initially, Si = for 1≤ i ≤ + 1.

• 2. FOR i = n down to 1 DO

• 3. Find the largest k such that vk N[vi] and vk is not completely dominated;

• 4. Let Sj be a set that does not dominate vk ;

• 5. Include vi in Sj, i.e., Sj = {vi} Sj;

• 6. If no such set exists, then include vi to an arbitrary set.

• 7. ENDFOR

‧41‧

8

1

7

4

95

A strong elimination ordering: (preprocessing)

6

32

8

1

7

4

95 6

32

1 → 5 → 6 → 2 → 7 → 4 → 3 → 9 → 8

N1 [2] = {2, 3, 4, 5, 6, 7}

N1 [5] = {2, 3, 5, 6}

N1 [6] = {2, 3, 5, 6, 7}

‧42‧

1 4

5

2 3

8 96 7

1 ← 5 ← 6 ← 2 ← 7 ← 4 ← 3 ← 9 ← 8

= 2

S1 =

S2 =

S3 =

Algorithm DP :

‧43‧

1 4

5

2 3

8 96 7

S1 =

S2 =

S3 =

{8}

1 ← 5 ← 6 ← 2 ← 7 ← 4 ← 3 ← 9 ← 88

Algorithm DP :

‧44‧

1 4

5

2 3

8 96 7

S1 = {8}

S2 =

S3 =

{9}

1 ← 5 ← 6 ← 2 ← 7 ← 4 ← 3 ← 99 ← 8

Algorithm DP :

‧45‧

1 4

5

2 3

8 96 7

S1 = {8}

S2 = {9}

S3 = {3}

1 ← 5 ← 6 ← 2 ← 7 ← 4 ← 33 ← 9 ← 8

Algorithm DP :

‧46‧

1 4

5

2 3

8 96 7

S1 = {8}

S2 = {9}

S3 = {3}

{4, 8}

1 ← 5 ← 6 ← 2 ← 7 ← 44 ← 3 ← 9 ← 8

Algorithm DP :

‧47‧

1 4

5

2 3

8 96 7

S1 = {4, 8}

S2 = {9}

S3 = {3}

{7, 9}

1 ← 5 ← 6 ← 2 ← 77 ← 4 ← 3 ← 9 ← 8

Algorithm DP :

‧48‧

1 4

5

2 3

8 96 7

S1 = {4, 8}

S2 = {7, 9}

S3 = {3}

{2, 4, 8}

1 ← 5 ← 6 ← 22 ← 7 ← 4 ← 3 ← 9 ← 8

Algorithm DP :

‧49‧

1 4

5

2 3

8 96 7

S1 = {2, 4, 8}S2 = {7, 9}

S3 = {3}

{6, 7, 9}

1 ← 5 ← 66 ← 2 ← 7 ← 4 ← 3 ← 9 ← 8

Algorithm DP :

‧50‧

1 4

5

2 3

8 96 7

S1 = {2, 4, 8}S2 = {6, 7, 9}S3 = {3}

1 ← 55 ← 6 ← 2 ← 7 ← 4 ← 3 ← 9 ← 8

{3, 5}

Algorithm DP :

‧51‧

1 4

5

2 3

8 96 7

S1 = {2, 4, 8}S2 = {6, 7, 9}S3 = {3, 5}

Therefore we find a domatic partition { {2, 4, 8}, {6, 7, 9}, {1, 3, 5} }

11 ← 5 ← 6 ← 2 ← 7 ← 4 ← 3 ← 9 ← 8

S3 = {1, 3, 5}

Algorithm DP :

‧52‧

• Now let us proceed with the correctness of algorithm DP.

‧53‧

Outline


‧54‧

Relax!

• 一個很虔誠的基督徒到非洲叢林探險，不幸脫隊迷失在叢林中，

• 接著更悲慘的事發生了，一隻獅子發現他，便開始追殺他，他沒命的跑啊跑啊。

• 終於讓他逃到一棵樹上，可是那獅子也不願放棄的在樹下等。

• 天黑了，他又餓又渴，於是他開始向上帝禱告：

– 『主啊！請您將這隻噬血的獅子變成基督徒吧！』

• 話說完，樹下的獅子說話了：

– 『親愛的上帝，謝謝您賜給我豐盛的晚餐。』

‧55‧

• To prove the correctness of the algorithm, we have to figure out some lemmas first.

‧56‧

Lemma 1

• Assume S {vi+1,…, vn}, vk N[vi], where 1≤ i ≤ n. If S does not dominate vk , then S does not dominate vj for all j < k and vj N[vi].

• To prove Lemma 1, we consider three cases as follows.

‧57‧

Case 1: i ≤ j < k

• In this case, the lemma holds by the definition of strong elimination ordering since Ni[vj] Ni[vk].

k

j

i

‧58‧

Case 2: j < i ≤ k

• If S dominates vj and does not dominate vk , then there exists a number k such that vk S and vk N [vj].

• Since Nj[vj] and Ni[vi] are cliques, we can derive that v

i is adjacent to vk and vk is adjacent to vk .

k

j

i

k

Then a contradiction occurs!

‧59‧

Case 3: j < k < i

• In this case, we also assume that S dominates vj and does not dominate vk . Then there exists a number k such that vk S and vk N [vj].

• This implies that vk is dominated by S by the property of Nj[vi] Nj[vk ].

• A contradiction then occurs. – Why?

(Notice the order of vk and S {vi+1,…, v

n})

k

j

i

k

‧60‧

• Therefore, the proof of Lemma 1 is done.

• Any question?

‧61‧

Corollary 1

• Assume S {vi+1,…, vn}, vk N[vi], where 1≤ i ≤ n. If S dominates vk, then S dominate vj for all j > k and vjN[vi].

• This corollary can be obtained easily from Lemma 1.

‧62‧

R[v] and ndom(v)

• R[v] = {x : x N[v] and x is not in any set}

– Residual neighborhood?

• ndom(v) denotes the number of sets that do not dominate v during execution of Algorithm DP.

‧63‧

1 4

5

2 3

8 96 7

S1 = {2, 4, 8}

S2 = {7, 9}

S3 = {3}

1 ← 5 ← 6 ← 22 ← 7 ← 4 ← 3 ← 9 ← 8

R[2] = {1, 5, 6}

R[1] = {1, 5, 6} , ndom[1] = 2

, ndom[2] = 0

‧64‧

Lemma 2

• Algorithm DP maintains the following invariant:

For each i {1, 2,…, n}, R[vi] ≥ ndom(vi)

• We prove this lemma by induction.

‧65‧

Proof of Lemma 2

• Initially, | R[vi] | = degree(vi) + 1 and ndom(vi) = +1.

– It is obvious that | R[vi] | ≥ ndom(vi) for all vi in V.

• Only values of | R[vj] | and ndom(vj), where vj N[vi], may be altered when vi is included in a set.

• A vertex is included in a set when it is visited by the algorithm.

‧66‧

Proof of Lemma 2 (contd.)

• The algorithm scans N[vi] to find the largest number k such that vk N[vi] and vk is not completely dominated.

• We can partition N[vi] into two sets:

– X = {vj : j > k and vj N[vi]}

– Y = {vj : j ≤ k and vj N[vi]}

‧67‧


• Let S be a set that does not dominate vk.

• By Lemma 1, the vertices in Y are not dominated by S.

• Therefore, for each vertex vjY, both values of | R[vj] | and ndom(vj) are decremented by one after vafter vii is included is included

in S.in S.– Why?

X = {vj : j > k and vj N[vi]}

Y = {vj : j ≤ k and vj N[vi]}

‧68‧


• By Corollary 1, the vertices in X are all completely dominated, i.e., ndom(vj) = 0.

• Thus the invariant is maintained.

– Notice that if no such vk exists, that is, all vertices in N[vi] are completely dominated, then ndom(vj) = 0, where vj N[vi]. Thus vi can be included in any set and the invariant still holds.

X = {vj : j > k and vj N[vi]}

Y = {vj : j ≤ k and vj N[vi]}

‧69‧

Theorem 1

• Algorithm DP correctly partitions a strongly chordal graph into + 1 disjoint dominating sets.

• Proof:– Upon termination, | R[vi] | = 0 for each i {1, 2,…, n}.

– By Lemma 2, | R[vi] | ≥ ndom(vi), thus ndom(vi) = 0 for all vi

’s in V.

– That is, all vi’s in V are dominated by all +1 dominating sets.

‧70‧

Outline


‧71‧

Theorem 2

• Algorithm DP is linear.

• The proof is listed as follows.

‧72‧

Proof of Theorem 2

• In practical implementation, each vertex vi is associated with a variable ndom(i) and an array Li of size +1.

• Initially, ndom(i) = +1 and the values of entries in Li are all zero.

• If vi is dominated by Sj, then Li( j ) = 1.

‧73‧

Proof of Theorem 2 (contd.)

• Thus, for each vertex vi we can take O(di) time to determine vk , where di is the degree of vi and vk N[vi] is not completely dominated.

• Then we take O(+1) time to decide in which set vi should be included.

• Finally for each vj N[vi], we take O(1) time to update ndom(i) and Lj .

By checking Lk

O (1 di)

‧74‧

Proof of Theorem 2 (contd.)

• Therefore, the algorithm takes time of

).()1(1

nmOddOn

iii

‧75‧

Theorem 3 (Concluding theorem)

• Given the strong elimination ordering of a strongly chordal graph G, we can find a domatic partition of G in linear time.

‧76‧

Outline


‧77‧

Conclusions

• We have learned the following knowledge:

– Chordal graphs, strongly chordal graphs

– Perfect elimination ordering and strong elimination ordering

– Greedy algorithm DP

‧78‧

Mutual encouragements

• Encouraging sentences

• Exercise more.

• Three states of our lives

‧79‧

• Think hard, not work hard.– Professor R. C. T. Lee

• Think hard and work smartly.– Professor M. S. Chang

• 件件工作反應自我，凡經我手必為佳作。– Dr. D. T. Lee

‧80‧

• 我國幾乎可以完全不重視學生的體能訓練。

• 我國學生的體能絕對低於歐美學生的體能。

- 節錄至李家同教授之“我國教育上的重大問題” 投影片

我國學生體能太差

‧81‧

• 某高中老師：「五千公尺不能在 22 分鐘內跑完的，就不是男人。」 (10 年前 )– 從前台南一中入學測驗包括一千五百公尺測驗

• 成功大學班際接力賽時有人連跑完 200M 都很勉強。 (6 年前 )

• 中正大學某次體育課測驗一千五百公尺時，不少人 ( 男 ) 接近 10 分鐘才勉強跑完。 ( 上個月 )

‧82‧

人生三種境界王國維

• 古今之成大事業、大學問者，必經過三種之境界：

– 昨夜西風凋碧樹。獨上高樓，望盡天涯路。

– 衣帶漸寬終不悔，為伊消得人憔悴。

– 眾裡尋他千百度，驀然回首，那人卻在，燈火闌珊處。

Thank you.

References• [B88] On the domatic number of interval graphs, A. A. Bertossi, Informati

on Processing Letters, 28 (1988), pp. 275-280.• [B85] Dominating sets and domatic number of circular arc graphs, M. A.

Bonuccelli, Discrete Applied Mathematics, 12 (1985), pp. 203-213.• [BJ82] Dominating sets in chordal graphs, K. S. Booth and J. H. Johnson,

SIAM Journal on Computing, 11 (1982), pp. 191-199.• [CH77] Towards a theory of domination in graphs, E. J. Cockayne and S.

T. Hedetniemi, Networks, 7 (1977), pp. 247-161.• [F83] Applications of l.p. duality to problems involving independence and

domination, M. Farber, Ph.D. Thesis, Rutgers University, New Brunswick, NJ, 1982.

• [F84] Domination, independent domination, and duality in strongly chordal graphs, M. Farber, Discrete Applied Mathematics, 7 (1984), pp. 115-130.

• [GJ79] Computers and Intractability: A Guide to the Theory of NP-Completeness, M. R. Garey and D. S. Johnson, Freeman, San Francisco, 1979.

• [G80] Algorithm Graph Theory and Perfect Graphs (Academic Press, New York, 1980), M. C. Golumbic.

• [LHC90] The Domatic Number Problem in Interval Graphs, T. L. Lu, P. H. Ho and G. J. Chang, SIAM Journal on Discrete Mathematics, 3 (1990), pp. 531-536.

• [PT87] Three Partition Refinement Algorithms, R. Paige and R. E. Tarjan, SIAM Journal on Computing, 16 (1987), pp. 973-989.

• [PC91] A New Approach for Domatic Number Problem on Interval Graphs, S. L. Peng and M. S. Chang, Proceedings of National Computer Symposium, 1991, Taipei, Republic of China, pp. 236-241.

• [R70] Triangulated Graphs and the Elimination Process, D. J. Rose, J. Math. Anal. Appl., 32 (1970), pp. 597-609.

• [RR89] Linear Algorithm for Domatic Number Problem on Interval Graphs, Information Processing Letters, 33 (1989), pp. 29-33.

Documents

A Simple Linear Time Algorithm for the Domatic Partition Problem on Strongly Chordal Graphs