霍奇排名之理論分析 - 政大學術集成

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(1)國⽴政治⼤學應⽤數學系碩⼠學位論⽂. 立. 政治大. ‧. ‧ 國. 學. 霍奇排名之理論分析. Theoretic Aspect of HodgeRank n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 碩⼠班學⽣：陳名秀撰指導教授：蔡炎⿓博⼠中華民國 105 年 6 ⽉ 30 ⽇.

(2) 國⽴政治⼤學應⽤數學系陳名秀君所撰之碩⼠學位論⽂霍奇排名之理論分析 Theoretic Aspect of HodgeRank. 政治大業經本委員會審議通過立論⽂考試委員會委員： ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 指導教授：系主任：. 中華民國 105 年 6 ⽉ 30 ⽇.

(3) 謝經過兩年的努⼒，終於將⾃⼰的論⽂完成並通過⼝試。⾮常感謝蔡炎⿓⽼師指導我時所付出的時間與⼼⼒。在這段⽇⼦裡，⾮常幸運地可以遇到蔡炎⿓⽼師讓我對於論⽂有正確的⽅向及嚴謹審慎的架構與邏輯。若是沒有⽼師的耐⼼的指導，我的論⽂是不會這麼順利完成。對於⼝試委員也⾮常的感激，能夠給予我許多寶貴的意⾒，以及對於論⽂的建議和指點出我論⽂的問題。最後也謝謝同儕們與朋友們和家⼈們給予的⿎勵、⽀持和幫助。. 立. 政治大. ‧ 國. 學. ‧. 陳名秀謹誌于國⽴政治⼤學應⽤數學系碩⼠班中華民國 105 年 6 ⽉. n. er. io. sit. y. Nat. al. Ch. engchi. ii. i n U. v.

(4) 中⽂. 要. 霍奇排名是在近幾年才運⽤在排名的⼀種⽅法。在⼤多數現在的資料庫中，資料庫很龐⼤，有些甚⾄會需要網路連結，⽽且很多會有資料不完整或是資料不平衡的狀況。我們選擇⽤霍奇排名這種排名⽅法來處理可能會遇到的這些困擾。這篇論⽂主要⽬的是想⽤運⽤基本的線性代數來研究霍奇排名和推導組合霍奇理論。關鍵字：霍奇理論, 霍奇排名, 組合霍奇理論. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. iii. i n U. v.

(5) Abstract HodgeRank is a method of ranking that is new in recent years. In most of modern datasets, the amount of data is very large, some also need the internet connection,. 政治大 method of HodgeRank立 to deal with these difficulties.. and plenty of them have the feature that incomplete or imbalanced. We use the. ‧ 國. 學. This thesis is primary using elementary linear algebra to survey HodgeRank and deduce the combinatorial Hodge Theorem.. ‧. Keywords: HodgeRank, combinatorial Hodge theorem, combinatorial Hodge. n. al. er. io. sit. y. Nat. decomposition,. Ch. engchi. iv. i n U. v.

(6) Contents ⼝試委員會審. i. 謝. 立. 要. 3. sit. al. n. 2. Introduction. er. io. 1. Pairwise Ranking. iv v. y. Nat. List of Figures. ‧. Contents. 學. Abstract. ii iii. ‧ 國. 中⽂. 政治大. Ch. engchi U. v ni. vii 1 4. 2.1. Pairwise Ranking Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.2. Introduction to HodgeRank . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.3. Possible Applications of Pairwise Rankings . . . . . . . . . . . . . . . . . . .. 6. 2.3.1. Ranking Students in a Class . . . . . . . . . . . . . . . . . . . . . . .. 6. 2.3.2. Movie Recommendation Systems . . . . . . . . . . . . . . . . . . . .. 7. 2.3.3. Ranking Sports Team . . . . . . . . . . . . . . . . . . . . . . . . . . .. 7. 2.4. Pairwise Ranking from Voting . . . . . . . . . . . . . . . . . . . . . . . . . .. 8. 2.5. Problems of Pairwise Ranking . . . . . . . . . . . . . . . . . . . . . . . . . .. 8. Background. 9. v.

(7) 3.2. The First Isomorphism Theorem . . . . . . . . . . . . . . . . . . . . . . . . .. 13. Combinatorial Hodge Theory. 16. 4.1. Cochain Complex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 16. 4.2. Combinatorial Hodge Theory . . . . . . . . . . . . . . . . . . . . . . . . . . .. 17. 4.3. Finding the Best Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 20. Combinatorial Hodge Theory on Graphs. 22. 5.1. Graphs and Functions on Graphs . . . . . . . . . . . . . . . . . . . . . . . . .. 22. 5.2. Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 32. 5.3. Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 33. 5.4. Find the Global Ranking . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 35. 5.5. Applying Combinatorial Hodge Theory . . . . . . . . . . . . . . . . . . . . .. 35. 5.6. Practical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 立. 36. y. 40. sit. ‧. io. al. n. Bibliography. 學. Conclusion. 政治大. Nat. 6. 9. er. 5. Some Simple Facts in Linear Algebra . . . . . . . . . . . . . . . . . . . . . .. ‧ 國. 4. 3.1. Ch. n U i engch. vi. iv. 41.

(8) List of Figures 1.1. Draw the result on the graph. . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.1. The graph of pairwise ranking. . . . . . . . . . . . . . . . . . . . . . . . . . .. 5.1. A simple graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 5.2. A graph with 4 vertices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 25. 5.3. An example of globally inconsistent but locally consisitent. . . . . . . . . . . .. 31. 5.4. Pairwise comparison result. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. vii. i n U. v. 2 5 23. 37.

(9) Chapter 1 Introduction 政治大 The technique of solving ranking problems has many applications. For instance, recommen立. ‧ 國. 學. dation systems of restaurants, movies, books, and so on. It is difficult to apply traditional ranking methods to many modern datasets, since these datasets usually are incomplete and imbalanced.. difficulties.. ‧. In [6], Jiang, Lim, Yao, and Ye developed a method called HodgeRank to deal with these. y. Nat. io. sit. We use a simple example to elaborate the idea of HodgeRank. Suppose that we have four. n. al. er. candidates A, B, C, D to rank, and we have the following pairwise comparisons:. Ch. • A is better than C by 1 point,. engchi. i n U. v. • B is better than C by 1 point, • C is better than D by 1 point, and • D is better than B by 2 points. We can represent the relations by a graph, as shown in the Figure 1.1.. 1.

(10) A. B 1. 1. 2 1. C. D. Figure 1.1: Draw the result on the graph. The same relations can also be presented by a matrix. 立. 政治大 A B C D . ‧ 國. 0. −1. 0. 0. −1. 1. 1. 0. 0. −2. 1. 0   2   . −1    0. Nat. sit. y. ‧. 0. 學. A  B  Y=  C   D. . io. n. al. er. Note that Y is a skew-symmetric matrix. The goal is to find a score function. i n U. s : { A, B, C, D } → R.. Ch. engchi. v. If this can be done, then we get a global ranking of A, B, C, D. Usually the task is not trivial, for we might have some kind of inconsistency in our datasets. In our example, we can see B>C>D>B which is a contradiction. The HodgeRank approach is to apply so called combinatorial Hodge decomposition to the matrix Y Y = X + X H + XT , and the matrix X will give us the score function in desire. 2.

(11) In this thesis, we survey the HodgeRank and use elementary linear algebra to deduce the combinatorial Hodge theorem. The structure of this thesis is organized as follows. In chapter 2, we explain the main application we have in mind, which is the pairwise ranking problems; In chapter 3, we prove all important theorems we need from elementary linear algebra; In chapter 4 we introduce the combinatorial Hodge theory; In chapter 5, we apply the combinatorial Hodge theory to graphs; and finally we give the conclusion in the final Chapter.. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 3. i n U. v.

(12) Chapter 2 Pairwise Ranking 政治大. 立 Problems Pairwise Ranking. 學. ‧ 國. 2.1. Let V = {v1 , v2 , . . . , vn } be a collection of alternatives (or canditates). We would like to. ‧. give the set of alternatives a global ranking. For instance, V can be a collection of sport teams, movies, books, or students from a class, and so on. Suppose that we can do some pairwise. y. Nat. io. sit. ranking. For example, team A is better than team B by 5 points; book C is 2.3 points better. n. al. er. than D. However, these datasets usually unbalanced, incomplete, even inconsistence. In this. i n U. v. chapter, we will describe the type of ranking problems we consider and discuss the difficulties. Ch. engchi. we will encounter. For more details we refer to [2, 3, 9–14].. 2.2. Introduction to HodgeRank. In this thesis, we use the method HodgeRank [6] to deal with pairwise ranking. To elaborate the idea of HodgeRank, we give an example. Support we have four sport teams V =. { A, B, C, D }. Four games was held and the results are as follows • A beats C for 1 point, • B beats C by 1 point, • C beats D by 1 point, and 4.

(13) • D beats B by 2 points. Then we can present the results on a graph, as figure 2.1 shows.. A. B 1. 1. 2 1. C. D. 政治大. Figure 2.1: The graph of pairwise ranking.. 立. One problem is that we can’t intuitively determine which team is the best for the results are. ‧ 國. 學. inconsistent. For example, C beats D and D beats B. However, B beats C by 1 points. We can also present the results by the corresponding skew-symmetric matrix,. n. al. Ch. 0. 0. −1. 0. 0. −1. 1. 1. 0. e n0 g−c2h i1. D. . y. C. sit. B. 0   2   . −1    0. er. io. A  B  Y=  C   D. A. ‧. Nat. . i n U. v. Thus a pairwise ranking can be represented by a skew-symmetric matrix. The observation gives us the following definition. Definition 2.2.1 (Pairwise Ranking). A pairwise ranking of n alternatives is an n × n skewsymmetric matrix. Many different methods of dealing pairwise ranking have been used in many fields, like psychology, management science, social choice theory, and statistics [1, 4, 5, 7, 14]. Our goal is to get a global ranking from a pairwise ranking. For our example, that means that we want to find a function s : { A, B, C, D } → R. 5.

(14) We call the function s a score function. Applying so called the combinatorial Hodge decomposition to Y, we get. Y = X + X H + XT ,. where X is exactly what we want: . .  0 −1.33 −1 −0.67   1.33 0 0.33 0.66    X= .   1 −0.33 0 0.33     0.67 −0.66 −0.33 0. 政治大. 立. ‧ 國. 學. Remember that we have an inconsistent pairwise ranking, but now X is consistent. Set a score for A then we can get the function s in desire. For example, let s( A) = 0.75. From the. Nat. n. al. s(C ) = −0.25, and. Ch. s( D ) = 0.08.. engchi. sit. s( B) = −0.58,. er. io. 2.3. y. ‧. matrix X, we get the score function. i n U. v. Possible Applications of Pairwise Rankings. Suppose we want to rank n items v1 , v2 , . . . , vn , the key of our approach is to model our problem into a pairwise ranking. We give some possible examples as follows.. 2.3.1 Ranking Students in a Class Many pairwise rankings raised from grading or voting from some voters. We denote all voters by Λ = {α1 , α2 , . . . , αm }. For example, the peer assessment in MOOCs, students will give other’s homework scores online with the criterion. For most people, rating 100 alternatives. 6.

(15) at a time is harder than rating only 2 alternatives at a time. In fact, it has been observed that most people can rank only between 5 and 9 alternatives at a time. [8] And pairwise ranking has fewer values missing. Take the Netflix problem in the [6] for an example, there is almost 99% of its values missing but only 0.22% of the pairwise comparison values are missing. Moreover, in certain things, such as a badminton tournament, only pairwise comparison is possible.. 2.3.2 Movie Recommendation Systems Suppose the movie recommendation system wants to rank the movies in 2016, candidates are. 政治大. all movies that be released in the movie theaters in 2016 and voters are all viewers that had ever seen some of these movies. The dataset is all the movies, and then the movie recommendation. 立. system wants to globally rank all the movies. Now for simplicity, suppose we only want to rank. ‧ 國. 學. these three movies, “Me Before You,” “Independence Day: Resurgence,” and “The Conjuring 2,” which we denote by M1 , M2 and M3 . Each time the voters answer that “preferring M1 or. ‧. M2 ?”, “preferring M2 or M3 ?”, or ”preferring M1 or M3 ”?” But there may be a case : if voter. Nat. sit. y. A thinks M1 is better than M2 , voter B prefers M2 to M3 and voter C votes M3 is better than. n. al. er. io. M1 . It results an inconsistent cyclic preference relation. CMh1 > M2 > M3 >UMn1.i engchi. v. (2.3.1). 2.3.3 Ranking Sports Team We can use a graph to represent the dataset: the vertices are all candidates we have to rank and the edges means that the end points of each edge have been compared. In other words, if we want to rank football teams, the vertex set is all the football teams and if two teams have a match to each other then there is an edge between these two vertices. In pairwise comparison, we can assign the edge a direction and weight, as edge flows, on a diagraph. There are many different ways to define the weight, for instance, we can let the weight to represent the scores difference, the score of winning candidate subtracts the score of the losing candidate, and use an arrow to represent the direction from the winning candidate to the losing one. 7.

(16) 2.4. Pairwise Ranking from Voting. Let V = {v1 , v2 , . . . , vn } be the set of candidates to be ranked and Λ = {α1 , α2 , . . . , αm } be the set of voters. For each α ∈ Λ, the pairwise comparisons of voter α is a skew-symmetric matrix X α ∈ Rn×n . For each ordered paired (i, j) ∈ V × V, we have Xijα = − X jiα . In matrix X α , Xijα means that how much the degree of αth voter preferring jth candidate to i-th candidate. If αth voter does not compare ith and jth candidate, Xijα is a missing value and we let Xijα be zero. The diagonal entries on the corresponding matrix X are all zero since when we compare someone and itself, there is no one of them is better than the other. There are several ways to. 政治大 of 1, it is clearly that there is another one losses for 1 point. So we can construct a corresponding 立. define the degree of preference, like the score difference. For a game, if one wins by a majority. skew-symmetric matrix and we can apply the combinatorial Hodge Theory to obtain a global. ‧ 國. ‧. Problems of Pairwise Ranking. sit. y. Nat. 2.5. 學. ranking.. al. er. io. As we have seen, many problems can be transferred into a pairwise ranking. However, there. v. n. are three major problems in raw pairwise ranking datasets. The pairwise rankings raised from real-world problems usually are. Ch. engchi. i n U. inconsistent: As we have seen in 2.3.1. incomplete: In most cases, not all voters will vote (grade) all candidates. imbalanced: As in the movie recommendation system, one movie might be scored by many voter, but another might be scored by just one or two persons. Use the method of HodgeRank, we can solve all these problems.. 8.

(17) Chapter 3 Background 政治大. Some Simple 立 Facts in Linear Algebra. 學. ‧ 國. 3.1. In this chapter, the definition of inner product space is an inner product space over the filed. ‧. R or C (Hermitian product space).. sit. y. Nat. Lemma 3.1.1. Let V be a finite dimensional inner product space over R, and let g : V → R be. er. al. n. all v ∈ V.. io. a linear transformation.Then there exists a unique vector v′ ∈ V such that g(v) = ⟨v, v′ ⟩ for. Ch. engchi. i n U. v. Theorem 3.1.2. Let V, W be two finite dimensional inner product space. Let T : V → W be a linear transformation. Then there is unique linear transformation T ∗ : W → V such that. ⟨ T (v), w⟩W = ⟨v, T ∗ (w)⟩V . We call T ∗ the adjoint linear transformation of T. Proof. Let w ∈ W, and define g : W → R by ⟨ T (v), w⟩W for all T (v) ∈ W.. 9.

(18) Let T (v1 ), T (v2 ) ∈ W and c ∈ R. g(cv1 + v2 ) = ⟨ T (cv1 + v2 ), w⟩W. = ⟨cT (v1 ) + T (v2 ), w⟩W = c ⟨ T ( v 1 ) , w ⟩W + ⟨ T ( v 2 ) , w ⟩W = cg(v1 ) + g(v2 ). Hence, g is linear. By Lemma 3.1.1 , there exists v′ ∈ V such that g(v) = ⟨v, v′ ⟩V , i.e ⟨ T (v), w⟩W = ⟨v, v′ ⟩V .. 政治大. Define T ∗ : W → V by T ∗ (w) = v′ , we have ⟨ T (v), w⟩W = ⟨v, T ∗ (w)⟩V .. 立. To show T ∗ is linear. Let w1 , w2 ∈ W and c ∈ R.For all v ∈ V ,. ‧ 國. 學. ⟨v, T ∗ (cw1 + w2 )⟩V = ⟨ T (v), cw1 + w2 ⟩W. ‧. = c ⟨ T ( v ) , w 1 ⟩W + ⟨ T ( v ) , w 2 ⟩W. = ⟨v, cT ∗ (w1 ) + T ∗ (w2 )⟩V .. n. al. er. io. sit. y. Nat. = c⟨v, T ∗ (w1 )⟩V + ⟨v, T ∗ (w2 )⟩V. Since v is arbitrary ,. T ∗ (cw. 1. + wC 2) =. Finally, we show T ∗ is unique.. cT ∗ (w. i ). 2n U. ) + T ∗ (w. h e n g1c h i. v. Given U : W → V is linear and ⟨ T (v), w⟩W = ⟨v, U (w)⟩V , for all v ∈ V and w ∈ W. Then ⟨v, T ∗ (w)⟩V = ⟨ T (v), w⟩W = ⟨v, U (w)⟩V , for all v ∈ V and w ∈ W. So we get T ∗ = U . Theorem 3.1.3. Let V, W be two finite dimensional inner product space. Let T : V → W be a linear transformation. If β, γ be arbitrary orthonormal bases for V, W, respectively, and γ. A = [ T ] β , then. [ T ∗ ]γβ = A∗ .. 10.

(19) Proof. Let β = {v1 , v2 , ..., vn } and γ = {w1 , w2 , ..., wm } . Then . .  ⟨ T ( v 1 ) , w 1 ⟩W ⟨ T ( v 2 ) , w 1 ⟩W · · · ⟨ T ( v n ) , w 1 ⟩W     ⟨ T ( v 1 ) , w 2 ⟩W ⟨ T ( v 2 ) , w 2 ⟩W · · · ⟨ T ( v n ) , w 2 ⟩W    A= , .. .. .. ..   . . . .     ⟨ T ( v 1 ) , w m ⟩W ⟨ T ( v 2 ) , w m ⟩W · · · ⟨ T ( v n ) , w m ⟩W and we get . .  ⟨ T ( v 1 ) , w 1 ⟩W ⟨ T ( v 1 ) , w 2 ⟩W · · · ⟨ T ( v 1 ) , w m ⟩W     ⟨ T ( v 2 ) , w 1 ⟩W ⟨ T ( v 2 ) , w 2 ⟩W · · · ⟨ T ( v 2 ) , w m ⟩W    A∗ =  . .. .. .. ..   . . . .     ⟨ T ( v n ) , w 1 ⟩W ⟨ T ( v n ) , w 2 ⟩W · · · ⟨ T ( v n ) , w m ⟩W. 立. 政治大. ‧ 國. 學. n. al. ···. y. 2 ), v1 ⟩V. ⟨ T ∗ (w. . m ), v1 ⟩V .  ⟨ T ∗ ( w2 ) , v 2 ⟩ V · · · ⟨ T ∗ ( w m ) , v 2 ⟩ V    .. .. ...  . .  v i n C h ⟨T∗ (w2 ), vn ⟩V U· · · ⟨T∗ (wm ), vn ⟩V  engchi  ∗ ∗ ⟨v1 , T (w2 )⟩V · · · ⟨v1 , T (wm )⟩V   ⟨v2 , T ∗ (w2 )⟩V · · · ⟨v2 , T ∗ (wm )⟩V    .. .. ..  . . .   ⟨vn , T ∗ (w2 )⟩V · · · ⟨vn , T ∗ (wm )⟩V  ⟨ T ( v 1 ) , w 2 ⟩W · · · ⟨ T ( v 1 ) , w m ⟩W   ⟨ T ( v 2 ) , w 2 ⟩W · · · ⟨ T ( v 2 ) , w m ⟩W    .. . ..  .. . .   ⟨ T ( v n ) , w 2 ⟩W · · · ⟨ T ( v n ) , w m ⟩W. sit. io. 1 ), v1 ⟩V    ⟨ T ∗ ( w1 ) , v 2 ⟩ V  B= ..  .   ⟨ T ∗ ( w1 ) , v n ⟩ V  ∗  ⟨v1 , T (w1 )⟩V   ⟨v2 , T ∗ (w1 )⟩V  = ..  .   ⟨vn , T ∗ (w1 )⟩V   ⟨ T ( v 1 ) , w 1 ⟩W   ⟨ T ( v 2 ) , w 1 ⟩W  = ..  .   ⟨ T ( v n ) , w 1 ⟩W. ⟨ T ∗ (w. er. ⟨ T ∗ (w. Nat. . ‧. Let B = [ T ∗ ]γβ , then. = A∗ .. 11.

(20) Theorem 3.1.4. Let V, W be two finite dimensional inner product space. Let T : V → W be a linear transformation. Then V = ker( T ) ⊕ im( T ∗ ), and W = im( T ) ⊕ ker( T ∗ ). Proof. We just prove V = ker( T ) ⊕ im( T ∗ ), for the arguments for W = im( T ) ⊕ ker( T ∗ ) are exactly the same.. 政治大. Since we have V = im( T ∗ ) ⊕ im( T ∗ )⊥ . We only need to show that ker( T ) = im( T ∗ )⊥ .. 立. Let x ∈ ker( T ). We have T ( x ) = 0. Then. ‧ 國. 學 ⟨ T ( x ), w⟩W = 0, for all w ∈ W.. ‧ y ⟨ x, T ∗ (w)⟩V = 0, for all w ∈ W.. n. al. er. io. sit. Nat. The definition of T ∗ gives us. Ch. engchi. i n U. v. We conclude that x ∈ im( T ∗ )⊥ . The proof in the reverse direction is similar. Lemma 3.1.5. Let V, W be two finite dimensional inner product space. Let T : V → W be a linear transformation. Then 1. ker( T ∗ ) = im( T )⊥ ,and 2. ker( T ∗ T ) = ker( T ). Proof.. 1. Let x ∈ ker( T ∗ ). We have T ∗ ( x ) = 0. Then. ⟨v, T ∗ ( x )⟩V = 0, for all v ∈ V.. 12.

(21) The definition of T ∗ gives us. ⟨ T (v), x ⟩W = 0, for all v ∈ V. We conclude that x ∈ im( T )⊥ . So we get ker ( T ∗ ) ⊆ im( T )⊥ . And it is similar to proof im( T )⊥ ⊆ ker ( T ∗ ). Beside, we can prove it using the proof in Theorem3.1.4 by ( T ∗ )∗ = T. 2. Clearly, ker( T ) ⊆ ker( T ∗ T ) .. 政治大. Given x ∈ ker( T ∗ T ) , T ∗ T ( x ) = 0 .. 立. It impies. ‧ 國. 學. T ( x ) ∈ ker( T ∗ ) = im( T )⊥ .. ‧. But we have. T ( x ) ∈ im( T ).. y. Nat. io. sit. So T ( x ) = 0. Then x ∈ ker( T ) .. n. al. er. Therefore, ker( T ∗ T ) = ker( T ) .. 3.2. Ch. engchi. i n U. v. The First Isomorphism Theorem. Theorem 3.2.1. (The First Isomorphism Theorem). Let T : V → W be a linear transformation between two vector spaces over field R. Then. V/ ker( T ) = im( T ). Proof. Define ϕ : V/ ker( T ) → im( T ); that is, ϕ(v + ker( T )) = T (v) for all v ∈ V.. 13.

(22) For all v ∈ V, there exists ϕ(v + ker( T )) = T (v) ∈ im( T ). And for all v1 + ker( T ), v2 + ker( T ) ∈ V + ker( T ) with v1 + ker( T ) = v2 + ker( T ), it implies v1 − v2 ∈ ker( T ). Then T (v1 − v2 ) = 0.. 立. 學. T ( v1 ) = T ( v2 ).. ‧ 國. So we have. 政治大. Therefore, ϕ is well-defined.. ‧. Second, we show that ϕ is linear. For all ϕ(v1 + ker( T )), ϕ(v2 + ker( T )) ∈ im( T ) and c ∈ R,. y. Nat. = T ( v1 + v2 ) = T ( v1 ) + T ( v2 ).. n. al. er. io. sit. ϕ(v1 + ker( T ) + v2 + ker( T )) = ϕ((v1 + v2 ) + ker( T )). And. Ch. engchi. i n U. v. ϕ(c(v + ker( T ))) = ϕ(cv + ker ( T )). = T (cv) = cT (v) = cϕ(v + ker ( T )). So ϕ is linear. Then, for all ϕ(v1 + ker( T )), ϕ(v2 + ker( T )) ∈ im( T ) with ϕ(v1 + ker( T )) = ϕ(v2 + ker( T )), we have T (v1 ) = T (v2 ). So T (v1 ) − T (v2 ) = 0, 14.

(23) and we get T (v1 − v2 ) = 0, which implies v1 − v2 ∈ ker( T ). Then we obtain v1 + ker( T ) = v2 + ker( T ). Hence, ϕ is one-to-one. Last, for all w ∈ T (v), there exists v ∈ V such that T (v) = w. We get. 政治大. 立ϕ(v + ker(T)) = T(v) = w.. ‧ 國. 學. Therefore, ϕ is onto.. ‧. n. er. io. sit. y. Nat. al. Ch. engchi. 15. i n U. v.

(24) Chapter 4 Combinatorial Hodge Theory 政治大 In this thesis, we only focus on combinatorial Hodge theory. More information of classical 立. ‧ 國. Cochain Complex. Nat. n. a∂ l. −1. C0. Ch. ∂0. ∂1. C1. er. io ···. sit. Let (C• , ∂• ) be a cochain complex of inner product spaces:. y. ‧. 4.1. 學. Hodge theory we refer to [15, 16].. v ni. C2. engchi U. ∂2. ···. such that ∂k ’s are linear transformations and ∂k+1 ∂k = 0. Let ∂∗k be the adjoint of the linear transformation ∂k . That is, ∂∗k is the unique linear transformation from Ck+1 to Ck such that. ⟨∂k ( x ), y⟩k+1 = ⟨ x, ∂∗k (y)⟩k , for all x ∈ Ck and y ∈ Ck+1 . We now have the following maps: ∂ k −1. ···. Ck−1. ∂k. Ck ∂∗k−1. Ck+1 ∂∗k. 16. ··· ..

(25) Definition 4.1.1. Let (C• , ∂• ) be a cochain complex of inner product spaces. Define the k-th combinatorial Laplacian ∆k : Ck → Ck by ∆k = ∂k−1 ∂∗k−1 + ∂∗k ∂k . An element u in Ck is harmonic if ∆k (u) = 0.. 4.2. Combinatorial Hodge Theory. 政治大. Lemma 4.2.1. Let (C• , ∂• ) be a cochain complex of inner product spaces.. 立. 1. Ck = im(∂k−1 ) ⊕ ker(∂∗k−1 ) = im(∂∗k ) ⊕ ker(∂k ). ‧ 國. 學. It is not hard to prove the following theorem.. ‧. Theorem 4.2.2 (Combinatorial Hodge Theorem). Let (C• , ∂• ) be a cochain complex of inner product spaces. Then we have. sit. y. Nat. n. al. 2. Ck = im(∂k−1 ) ⊕ ker(∆k ) ⊕ im(∂∗k ), and. Ch. engchi. 3. ker(∆k ) = ker(∂k ) ∩ ker(∂∗k−1 ).. er. io. 1. H k (C ) = ker(∂k )/ im(∂k−1 ) ≃ ker(∆k ),. i n U. v. We call (2) the combinatorial Hodge decomposition on Ck . Proof. We first prove (3). Clearly, ker(∂k ) ∩ ker(∂∗k−1 ) ⊆ ker(∆k ). Let x ∈ ker(∆k ) = ker(∂k−1 ∂∗k−1 + ∂∗k ∂k ) , which means ∂k−1 ∂∗k−1 ( x ) + ∂∗k ∂k ( x ) = 0. Then ∂k−1 ∂∗k−1 ( x ) = −∂∗k ∂k ( x ). Multiplying ∂k , ∂k ∂k−1 ∂∗k−1 ( x ) = −∂k ∂∗k ∂k ( x ) = 0.. 17.

(26) We get ∂∗k ∂k ( x ) ∈ ker(∂k ) , but we have ∂∗k ∂k ( x ) ∈ im(∂∗k ) = ker(∂k )⊥ . So ∂∗k ∂k ( x ) = 0. We have x ∈ ker(∂∗k ∂k ) = ker(∂k ) . Multiplying ∂k−1 , ∂∗k−1 ∂k−1 ∂∗k−1 ( x ) = −∂∗k−1 ∂∗k ∂k ( x ) = 0. We get ∂k−1 ∂∗k−1 ( x ) ∈ ker(∂∗k−1 ) , but we have ∂k−1 ∂∗k−1 ( x ) ∈ im(∂k−1 ) = ker(∂∗k−1 )⊥ . So ∂k−1 ∂∗k−1 ( x ) = 0 . We have x ∈ ker(∂k−1 ∂∗k−1 ) = ker(∂∗k−1 ). Then x ∈ ker(∂k ) ∩ ker(∂∗k−1 ). We obtaine. 治政大 = ker(∆ ) .. ker(∂k ) ∩ ker(∂∗k−1 ). 立. ∂k · ∂k−1 = 0, then (∂k · ∂k−1 )∗ = ∂∗k−1 · ∂∗k = 0 .. 學. ‧ 國. Second,we prove (2).. k. sit. Nat. ker(∂∗k−1 ) = Ck ∩ ker(∂∗k−1 ). y. ‧. We obtain im(∂∗k ) ⊆ ker(∂∗k−1 ) .. er. io. = [im(∂∗k ) ⊕ ker(∂k )] ∩ ker(∂∗k−1 ). n. =a[lim(∂k ) ∩ ker(∂∗k−1 )] ⊕ [keri(v ∂k ) ∩ ker(∂∗k−1 )]. n U i e[nkerg(∂ck )h∩ ker(∂∗k−1 )] = im(∂k ) ⊕. C∗h. Now, Ck = im(∂k−1 ) ⊕ ker(∂∗k−1 ). = im(∂k−1 ) ⊕ im(∂∗k ) ⊕ [ker(∂k ) ∩ ker(∂∗k−1 )]. And by (3), ker(∂k ) ∩ ker(∂∗k−1 ) = ker(∆k ).. 18.

(27) Therefore, Ck = im(∂k−1 ) ⊕ ker(∂∗k−1 ). = im(∂k−1 ) ⊕ im(∂∗k ) ⊕ [ker(∂k ) ∩ ker(∂∗k−1 )] = im(∂k−1 ) ⊕ im(∂∗k ) ⊕ ker(∆k ). Last, we prove (1) . Since ∂k · ∂k−1 = 0, we get im(∂k−1 ) ⊆ ker(∂k ). Let ϕ : C → im(∂k−1 )⊥ be the projection onto the subspace im(∂k−1 )⊥ .. 政治大. C = im(∂k−1 )⊥ ⊕ im(∂k−1 ).. 立. 學. And. ‧ 國. So x = ϕ( x ) + (1 − ϕ)( x ) , for all x ∈ C .. ‧. ker(∂k ) = C ∩ ker(∂k ). y. Nat. = [im(∂k−1 )⊥ ⊕ im(∂k−1 )] ∩ ker(∂k ). er. io. sit. = [im(∂k−1 )⊥ ∩ ker(∂k )] ⊕ [im(∂k−1 ) ∩ ker(∂k )]. = [im(∂k−1 )⊥ ∩ ker(∂k )] ⊕ im(∂k−1 ).. n. al. Ch. engchi. i n U. v. We can restrict ϕ to the subspace ker(∂k ) , denoted ϕk . Then. x = ϕk ( x ) + (1 − ϕk )( x ), for all x ∈ ker(∂k ) . ϕ is surjective , so is ϕk . Then im(ϕk ) = ker(∂k ) ∩ im(∂k−1 )⊥ . By the first isomorphism theorem, ker(∂k )/ ker(ϕk ) ≃ im(ϕk ) = ker(∂k ) ∩ im(∂k−1 )⊥ .. 19.

(28) And. ∀ x ∈ ker(ϕk ), ϕk ( x ) = 0. ⇒ x ∈ im(∂k−1 ), also same in the reverse. So we get ker(ϕk ) = im(∂k−1 ). Then ker(∂k )/ im(∂k−1 ) ≃ ker(∂k ) ∩ im(∂k−1 )⊥ . By (3), ker(∆k ) = ker(∂k ) ∩ ker(∂∗k−1 ), and (1) in Lemma 3.1.5 ,im(∂k−1 )⊥ = ker(∂∗k−1 ) , we finally get ker(∂k )/ im(∂k−1 ) ≃ ker(∂k ) ∩ ker(∂∗k−1 ) = ker(∆k ).. 學. 4.3. Finding the Best Solution. n. ∂0∗. Ch. C2 . ∂1∗. engchi. sit. io. al. C1. er. Nat. ∂1. ∂0. C0. y. Let (C• , ∂• ) be a cochain complex of inner product spaces:. ‧. ‧ 國. 立. 政治大. i n U∗. v. such that for k = 0, 1, ∂k are linear transformations and ∂k be the adjoint of the linear transformation ∂k . That is, ∂0∗ is the unique linear transformation from C1 to C0 such that. ⟨∂0 ( x ), y⟩1 = ⟨ x, ∂0∗ (y)⟩0 , for all x ∈ C0 and y ∈ C1 . And ∂1∗ is the unique linear transformation from C2 to C1 such that. ⟨∂1 ( x ), y⟩2 = ⟨ x, ∂1∗ (y)⟩1 , for all x ∈ C1 and y ∈ C2 . Fix y in C1 , we want to find min ∥ ∂0 x − y ∥ . x ∈C0. 20.

(29) That is, let V = {∂0 x | x ∈ C0 } and given y ∈ C1 , we want to find the unique ∂0 x ∈ V, where x is in C0 , which is closest to y. We know that ∂0 x − y ∈ V ⊥ . So for all x ∈ C0 ,. ⟨∂0 x − y, ∂0 x ⟩1 = 0. We get. ⟨∂0∗ (∂0 x − y), x ⟩0 = 0, for all x ∈ C0 . It implies. 立. ∗ 0. 0. 學. ∂0∗ ∂0 x − ∂0∗ y = 0.. ‧. io. y. sit. Nat. x = (∂0∗ ∂0 )−1 ∂0 y.. n. al. er. Therefore,. ‧ 國. Then. 治政 ∂ (∂ x − y) = 0.大. Ch. engchi. 21. i n U. v.

(30) Chapter 5 Combinatorial Hodge Theory on Graphs 政治大. 立 Graphs and Functions on Graphs 學. ‧ 國. 5.1. In this section, we introduce some terminologies in graph theory. The basic definitions are. ‧. referred to [17]. For representing our data, we introduce special functions called flows.. sit. y. Nat. Definition 5.1.1. A graph G is a triple consisting of a vertex set V, an edge set E, and a relation. io. er. that associates with each edge two vertices (not necessarily distinct) called its endpoints. We. al. write G = (V, E) to present a graph G with vertex set V and edge set E.. n. v i n C set Note that the vertex set and the edge in general. The vertex set V having n elements U h earenfinite i h gc means that G have n vertices, denoted by | G | = n; that is V = {v1 , v2 , . . . , vn } and writing vi v j for an edge with its endpoints vi , v j ∈ V. Definition 5.1.2. A loop is an edge whose endpoints are equal and multiple edges are edges having the same pair of endpoints. A simple graph is a graph having no loops or multiple edges. A graph G = (V, E) is called a subgraph of G = (V, E) if V ⊆ V and E ⊆ E. Moreover, if the vertices of a simple graph are pairwise adjacent, it is a complete graph. And a clique is a set of pairwise adjacent vertices in a graph. That is, a complete graph is each distinct pair of vertices (vi , v j ) in V, the edge vi v j is in E and a clique of G is a nonempty complete subgraph of G. A k-clique of G is a clique of G that containing k vertices. 22.

(31) For an example, figure 5.1 is a simple graph because it does not have loop and multiple edges. The vertex set is { A, B, C, D } and the edge set is { AD, BC, CD, BD }. There is no edge between vertex A and vertex B, so it isn’t a complete graph.. 立. A. B. C. D. 政治大 Figure 5.1: A simple graph.. ‧ 國. 學. Definition 5.1.3. A path is a simple graph whose vertices can be ordered so that two vertices. ‧. are adjacent if they are consecutive in the list. A vi , v j -path, vi , v j ∈ V, means that we can travel from vi to v j and the vertices in the path are not repeated.. y. Nat. sit. A cycle is a graph with an equal number of vertices and edges whose vertices can be placed. n. al. er. io. around a circle so that two vertices are adjacent if and only if they appear consecutively along. i n U. v. the circle. We can think that a cycle is a path that the start vertex and the end vertex are the same.. Ch. engchi. On figure 5.1, there is a cycle : B − C − D − B; there is an A, B-path : A − D − C − B. There is another A, B-path, A − D − B. Definition 5.1.4. A directed graph G (or diagraph) is a triple consisting of a vertex set V, an edge set E, and a function assigning each edge an ordered pair of vertices. In the ordered pair. (vi , v j ) in V, the first vertex is called the tail of the edge and the second is the head. And when drawing a diagraph, we draw an arrow from its tail to its head. The vertex set is written by V = v1 , v2 , . . . , vn and an edge in E is written as − vi⇀ v j by its tail vi and head v j . But in our pairwise ranking problems, all is diagraph. So we only simply write an edge as vi v j by its tail vi and head v j .. 23.

(32) Definition 5.1.5. An edge flow X on a simple graph G = (V, E) is a real valued function on the edge set E and X : V × V → R satisfies    X (vi v j ) = − X (v j vi ), for all vi v j ∈ E   X (vi v j ) = 0, for all vi v j ∈ / E. Let C1 = { all edge flows }. We define X is zero for all pairs of vertices that are not adjacent, and X (vi vi ) = 0 since we have defined edges does not self-adjacent; if there had self-adjacent edges, there are multiple. 政治大 The edge flow function 立 can be represented by a skew-symmetric matrix [ X ] by X. edges and G is not a simple graph.. ij. ij. =. ‧ 國. 學. X (vi , v j ), and we still define Xij = 0 if vi , v j ∈ / E. If there is a n × n skew-symmetric matrix and a graph G with | G | = n, we can define an edge flow of G by letting X (vi v j ) = Xij . So the. that is satisfying. ‧. set of edge flows on G is ono-to-one corresponding to the set of n × n skew-symmetric matrices,. sit. y. Nat. er. io. { X ∈ Mn×n (R)| X T = − X and Xij = 0 if vi v j ∈ / E }.. al. n. v i n C From now on, it is not importanthtoedistinguish edges n g c h i U flows and skew-symmetric matrices. The reason is we can induce one of them if we have another. As figure 5.2, there is a graph with 4 vertices whose edge flows can be corresponding to the following 4 × 4 skew-symmetric matrix, and vice versa..  A  B  Y=   C  D. A. B. C. 0. 0. −1. 0. 0. −1. 1. 1. 0. 0. −2. 1. 24. D. . 0   2   . −1    0.

(33) A. B 1. 1. 2 1. C. D. Figure 5.2: A graph with 4 vertices.   V    k. Define. 政治大. 立. ‧ 國. It is obvious that the edge set. 學. be the set of k-elements subset of V.. ‧.   V  E ⊆  . 2. y. Nat. io. sit. Definition 5.1.6. Let V = {v1 , v2 , . . . , vn } be the set of candidates to be ranked and Λ =. n. al. er. {α1 , α2 , . . . , αm } be the set of voters. Define the weight function ω : Λ × V × V by. i n U. v.    1 , if voter α made a comparison for vi , v j ∈ V α ωij = ω (α, i, j) =   0 , if voter α didn’t make a comparison for vi , v j ∈ V.. Ch. engchi. The weight matrix W is defined by. wij =. ∑. α∈Λ. ωijα .. It is clear that W is a symmetric matrix whose entries are nonnegative values. In a simple graph G = (V, E), we can find the sets of the collections of 1-clique, 2-cliques and 3-cliques respectively. Obviously, the collection of 1-clique is just the vertex set V. The. 25.

(34) collection of 2-cliques is the edge set E :   .

(35)   

(36)  

(37)

(38) V  E = vi v j

(39) {vi , v j } ∈   , ωij > 0 , 

(40)    2

(41) where ωij = ∑α ωijα . The collection of 3-cliques is the set T : 

(42)  

(43) 

(44) 

(45) V  T = vi v j vk

(46) {vi , v j , vk } ∈   , and vi v j , v j vk , vk vi ∈ E .

(47)     3

(48)   . (5.1.1). 政治大. The 3-cliques on the graph is triangles, so we denote T. The set of k-cliques, denoted Kk ( G ), is. 立. defined by. ‧. ‧ 國. We get. 學. {vi1 , . . . , vik } ∈ Kk ( G ) ⇔ vi p viq ∈ E, ∀1 ≤ p < q ≤ k.. K1 ( G ) = V, K2 ( G ) = E and K3 ( G ) = T.. y. Nat. er. io. sit. On page 23 figure 5.1, the triple (B,C,D) is a 3-cliques and the only one. Definition 5.1.7. A score function s ∈ Rn is a real-valued function on the set of all candidates;. n. al. that is,. Ch. engchi. i n U. v. s : V → R.. Let C0 = {all score functions s }. A score function assign every candidate a score : s(vi ) = si , for all vi ∈ V. Every score function induces a global ranking. Definition 5.1.8. The combinatorial gradient or simply gradient, denoted grad, is a mapping on the vertices to an edge flow. Define grad : C0 → C1 by For all s ∈ C0 , (grad s)(vi v j ) = s j − si , for all vi , v j ∈ V.. It is obvious that gradient is an edge flow. 26.

(49) Theorem 5.1.9. Gradient flow is independent with path. In other words, if X is an gradient flow, then Xij = s j − si for vi , v j ∈ V for some s is only depending on vi , v j no matter what vi , v j -path is. In other words, gradient flow is independent with path. Proof. Let v1 − v2 − · · · − vk be a path on a graph G with | G | = n, and X be a gradient flow. Then the flow of X along the path is X ( v 1 v 2 ) + X ( v 2 v 3 ) + · · · + X ( v k −2 v k −1 ) + X ( v k −1 v k ) Since a gradient flow is X (vi v j ) = s j − si for some s, we get. 立. 政治大. X ( v 1 v 2 ) + X ( v 2 v 3 ) + · · · + X ( v k −2 v k −1 ) + X ( v k −1 v k ). ‧ 國. 學. =(s2 − s1 ) + (s3 − s2 ) + · · · + (sk−1 − sk−2 ) + (sk − sk−1 ) = s k − s1. ‧ sit. y. Nat. al. er. io. If we get a globally consistent(see Definition 5.1.14) pairwise ranking X, it is easy to deter-. v. n. mine a score function s (up to adding a scalar) by solving grad(s) = X. And we can obtain a. Ch. engchi. global ranking of candidates. For example,. i n U. Theorem 5.1.10. A gradient is a linear transformation. Proof. Let grad : C0 → C1 be a gradient flow by For all s ∈ C0 , (grad s)(vi v j ) = s j − si , for all vi , v j ∈ V. Given s, t ∈ C0 , (grad s)(vi v j ) = s j − si and (grad t)(vi v j ) = t j − ti .. 27.

(50) For all vi , v j ∈ V, we have grad(s + t)(vi v j ) = (s + t) j − (s + t)i. = ( s j + t j ) − ( si + ti ) = s j − si + t j − ti = (grad s)(vi v j ) + (grad t)(vi v j ). And for all c ∈ R,. 政治大. grad(cs)(vi v j ) = (cs) j − (cs)i. 立. = cs j − csi. ‧ 國. 學. = c ( s j − si ). = c(grad s)(vi v j ).. ‧ sit. y. Nat. al. n. that satisfies. er. io. Definition 5.1.11. The triangular flow on G = (V, E) is the function Φ : V × V × V → R. Ch. engchi. i n U. v. Φijk = Φ(vi v j vk ) = Φ(v j vk vi ) = Φ(vk vi v j ). = − Φ ( v j v i v k ) = − Φ ( v i v k v j ) = − Φ ( v k v j v i ), We define C2 = {all triangle flows}. Definition 5.1.12. Let X be an edge flow and T in page 26(5.1.1). A curl is a map from edge flows to triangle flows. Define curl : C1 → C2 by    Xij + X + X , if vi v j v ∈ T jk ki k For all X ∈ C1 , (curl X )(vi v j vk ) =   0, if v v v ∈ / T. i j k. Theorem 5.1.13. A curl is a linear transformation. 28.

(51) Proof. Let Φ : C1 → C2 be a curl by    Xij + X + X , if vi v j v ∈ T jk ki k For all X ∈ C1 , (curl X )(vi v j vk ) =   0, if vi v j vk ∈ / T.   V  Given A, B ∈ C1 and for all vi , v j , vk ∈   , 3 If vi v j vk ∈ T,. curl( A + B)(vi v j vk ) = ( A + B)ij + ( A + B) jk + ( A + B)ki. 立. ij. ij. jk. jk. + Aki + Bki. = ( Aij + A jk + Aki ) + ( Bij + Bjk + Bki ). ‧ 國. 學. = curl( A)(vi v j vk ) + curl( B)(vi v j vk ).. ‧. io. n. al. sit. y. Nat. curl( A + B)(vi v j vk ) = 0,. er. If vi v j vk ∈ / T,. 治政 = A +B +A 大 +B. i n U. curl( A)(vi v j vk ) = 0, and. Ch. engchi. v. curl( B)(vi v j vk ) = 0.. So, we get curl( A + B)(vi v j vk ) = curl( A)(vi v j vk ) + curl( B)(vi v j vk ). And for all c ∈ R,. curl(cA)(vi v j vk ) = (cA)ij + (cA) jk + (cA)ki. = cAij + cA jk + cAki = c( Aij + A jk + Aki ) = c curl( A)(vi v j vk ).. 29.

(52) Definition 5.1.14. Let X : V × V → R be a edge flow on a pairwise comparison graph G =. (V, E). 1. X is called consistent on vi v j vk ∈ T if it is curl-free on vi v j vk . That is,. (curl X )(vi v j vk ) = Xij + X jk + Xki = 0.. 2. X is called globally consistent if it is a gradient of a score function. That is,. 立. 政治大 X = grad s for some s : V → R.. ‧ 國. 學. 3. X is called locally consistent or triangular consistent if it is curl-free on every triangle in. ‧. T.. sit. y. Nat. 4. X is called a cyclic ranking if it contains any inconsistencies. That is,. n. al. er. io. there exists vi , v j , . . . , v p , vq ∈ V such that Xij + X jk + · · · + X pq + Xqi ̸= 0. Ch. engchi. i n U. v. Global consistency implies local consistence but the inverse may not be true. There is may be a case that is globally consistent but not locally consistent, as figure 5.3 shown. There is a closed path, A → B → C → D → A, adding all edge flows along the path is X AB + XBC + XCD + XDE + XEA = 1 + 1 + 1 + 1 + 1 = 5 ̸= 0. We get a nonzero net weight along the path, so figure 5.3 is globally inconsistent. On the other hand, we look at the only triangle in the graph, A → B → C → A. Since curl( A, B, C ) = X AB + XBC + XCA = 1 + 1 − 2 = 0, figure 5.3 is locally consistent. Lemma 5.1.15. curl ◦ grad = 0. 30.

(53) 1. C. 1. D. B 2 1. 1. E. 1. A. Figure 5.3: An example of globally inconsistent but locally consisitent.. 政治大. Proof. Let X = grad(s) for some s ∈ Rn , and vi v j vk ∈ T. Then. 立. ‧ 國. 學. curl( X )ijk = Xij + X jk + Xki. = ( s j − si ) + ( s k − s j ) + ( si − s k ). ‧. =0. er. io. sit. y. Nat. From the lemma, we know that global consistency implies local consistency. Then a curl-. n. al. Ch. free on a complete graph must be a gradient.. engchi. i n U. v. Definition 5.1.16. A divergence div ∈ Rn is a real-valued function on the set of all candidates; that is, div : V → R. Define div : C1 → Rn by for all X ∈ C1 , div X (vi ) =. ∑. v j ,( v i v j ∈ E ). Xij .. And if div X (vi ) = 0 on each vertex vi ∈ V, the graph G is called divergence-free. We also take page 25 figure 5.2 for an example: div( A) = −1, div( B) = 1, div(C ) = 1, div( D ) = −1,. and figure 5.2 is not divergence-free. 31.

(54) Definition 5.1.17. Let X be an edge flow on the graph G. The flow X is a harmonic flow if it both curl-free and divergence-free.. 5.2. Matrices. We can use skew-symmetric matrices to represent edge flows, gradient flows, harmonic flows, and curl flows. Let A be the set of all skew-symmetric matrices:. 政治大. A = { X ∈ Mn × n ( R ) | X T = − X }. 立. Define the set of gradient matrices:. ‧ 國. 學. MG = { X ∈ A | Xij = s j − si for some s ∈ Rn },. (5.2.1). ‧. io. sit. y. Nat. and the set of triangular consistent matrices :. n. al. er. MT = { X ∈ A | Xij + X jk + Xki = 0 for all vi v j vk ∈ T }.. Ch. engchi. i n U. v. By lemma 5.1.15, every X ∈ MG is in MT ; that is, MG ⊆ MT . With the usual inner product. ⟨·, ·⟩ in vector space Mn×n (R) : ⟨ A, B⟩ = tr( A∗ B) = ∑ aij bij for all A, B ∈ Mn×n (R), i,j. we let M H the orthogonal complement of MG in MT ; that is, ⊥ M H = MT ∩ MG .. It is equivalent to MT = MG ⊕ M H .. 32. (5.2.2).

(55) The elements in M H is harmonic, which are curl-free and divergence-free. Since MG and MT are subspaces of A, we can get ⊥ A = MG ⊕ MG. A = MT ⊕ MT⊥. (5.2.3). So we can obtain A = MT ⊕ MT⊥ by equation(5.2.3). 政治大. = MG ⊕ M H ⊕ MT⊥ by equation(5.2.2). 立 Inner Product. ‧ 國. 學. 5.3. (5.2.4). (5.3.1). io. er. i,j. sit. Nat. ⟨ X, Y ⟩W = ∑ ωij xij yij ,. y. ‧. We define the weighted inner product in C1 by. al. v i n Ch C1 is all the set of all n × n skew-symmetric i U the inner product in C1 defined in e n g c hmatrices, n. for all edge flows X, Y in C1 . And equation( 5.2.4) is still hold.. (5.3.1) is referred to [6]. But there is a little problem, we need to be careful of it. If we have a pairwise ranking problem, we obtained the weighted matrix and define the inner product in C1 in the function (5.3.1). If we give the weight matrix W and A ∈ C1 as following:  0  1  W= 1   1. . . 0 1 1 1 0 0   0 0 1 0 0 0   ,A =  0 −1 0 0 0 0     0 −1 −1 0 0 0. 33.  0  1  . 1   0. (5.3.2).

(56) Then. ⟨ A, A⟩W = ∑ ωij a2ij = 0 i,j. but A is not O. It is not a inner product. We try to give some constraints to let it be a inner product. First, all candidates must be voted at least once. So all our graphs are connected. But it is clear that the graph related the matrix X = O is not a connected graph. So the set of all weighted adjacency matrices induced by connected graphs is not a vector space. So we consider the set S = span{ X ∈ Mn×n (R) | X is a weighted adjacency matrix induced by a connected graph}.. 立. 政治大 . 1 0 −1 0 0    0 0 1 1 1     ,C =  .   1 1  0 −1 0     0 1 −1 −1 0. y. sit. io. n. al. Clearly, B and C are skew-symmetric matrices. . . ‧. Nat. 0 0  0   0 0 1  B=  0 −1 0   −1 −1 −1. . Ch  engchi 1 0 0 0. er. . 學. ‧ 國. We want (5.3.1) defined in S to be a inner product. But if we take. i n U. v. . . 0 0 0 −1 0 0 0          0   0 1 1 1 1  1   0 0  0 0 B+C =  + =      1  0 −1 0  0 −1 0 1 0 −1 0      −1 −1 −1 0 1 −1 −1 0 0 −1 −1.  0.   1  =A  1  0. A is in S. And it is the same as the example in (5.3.2). We still check that it is not a inner product in S. We failed to let the definition in (5.3.1) be a inner product in the subspace of the set of all skew-symmetric matrices. Therefore, we need to give more constraints to get the subspace of C1 and let our definition in (5.3.1) be a inner product .. 34.

(57) 5.4. Find the Global Ranking. Our approach to ranking is to minimize a weighted sum of pairwise loss of a global ranking. Our problem is ordinary least squares: ωijα ( xij − yijα )2 , ∑ X ∈ MG min. α,i,j. ,where MG is the set that is defined in (5.2.1) and ωijα is defined in definition 5.1.6. The optimization problem can be written as. 政治 ∑大 ω (x. min ∥ X − Y ∥22,w = min. X ∈ MG. 立. X ∈ MG. ij. ij. − yij )2 ,. (5.4.1). i,j. ‧ 國. ‧. 5.5. 學. where ωij = ∑α ωijα and yij is the average scores; that is, yij = ∑α ωijα yijα / ∑α ωijα .. Applying Combinatorial Hodge Theory. y. Nat. io. sit. If we represent functions on vertices by n vectors, edge flows by n × n skew-symmetric. n. al. er. matrices, and triangular flows by n × n × n skew-symmetric matrices. That is, C0 = {s ∈ Rn } ,. Ch. engchi. C1 = { X ∈ Mn×n (R) | Xij = − Xij }, and. i n U. v. C2 = {Φ ∈ Mn×n×n (R) | Φijk = Φ jki = Φkij = −Φ jik = −Φikj = −Φkji }. Then, by the theorem 4.2.2, we have im(∂0 ) = im(grad) = MG , ker(∂1 ) = ker(curl) = MT , ⊥ ker(∂0∗ ) = ker(div) = MG , and. im(∂1∗ ) = im(curl∗ ) = MT⊥ .. 35.

(58) By definition 4.1.1: ∆k = ∂k−1 ∂∗k−1 + ∂∗k ∂k , and in particular ∆0 = ∂0∗ ∂0 = − div ◦ grad, ∆1 = ∂0 ∂0∗ + ∂1∗ ∂1 = curl∗ ◦ curl − grad ◦ div . Then the optimal solution can be written to min ∥ X − Y ∥22,w = min ∥ ∂0 s − Y ∥22,w = min ∥ grad s − Y ∥22,w. X ∈ MG. s∈C0. s∈C0. (5.5.1). 政治大. Theorem 5.5.1. The solution of 5.5.1 satisfy. 立. ∆0 s = − div Y,. ‧ 國. 學. and the minimum solution is. ‧. s∗ = −∆0† div Y,. Nat. er. io. sit. y. where the ∆0† is the Moore-Penrose inverse (pseudo-inverse), the divergence is. n. a l div Y(vi ) = ∑ ωijYiijv, j,{v ,v }∈ E n Ch engchi U i. j.    ∑ ωij , if i = j    j ∆0 = −ωij , if vi v j ∈ E      0, otherwise.. and. 5.6. Practical Example. Suppose there are five students, A, B, C, D, and E, we want to rank their homework and know which is the best. Each time teacher compares two homework, and gives his preference by scores: • A is better than B by 2 points, 36.

(59) • A is better than E by 8 points, • B is better than C by 6 points, • B is better than D by 1 point, • C is better than A by 2 points, and • D is better than E by 6 points. Then we can have the pairwise comparison graph.. 政 A治大 8 2. 立. ‧ 國. E. 學. B. 2 1. y. D. io. sit. Nat. C. ‧. 6. 6. er. Figure 5.4: Pairwise comparison result.. al. n. v i n C h matrix to represent We can also use the skew-symmetric e n g c h i U the relations.  A. A. B. C. D. E. 0. −2. 2. 0. −8.   B  2  Y = C  −2   D 0  E 8. −6 −1. 0 6. 0. 0. 1. 0. 0. 0. 0. 6. 37. .   0    0  .   −6   0.

(60) The weight matrix is aaand the . . −8      −5      div Y =   4 .      −5   14 We can get the matrix. 3 −1 −1 0 −1      −1 3 −1 −1 0       ∆0 =  −1 −1 2 0 0  .      0 −1 0 2 −1   −1 0 0 −1 2. 政治大. 立. io. n. al. Ch. 0.318 0.409 0.045 −0.227. y. 0.864 0.682 0.409 0.591 1.045 0.227. i n U. 0.682 0.591 0.955. engchi. 0.500 0.500 0.500. We can get the score function . . 7.36     8.27     †  s = −∆0 div = 5.81      6.64   0. 38. .   −0.045   −0.136 .   0.227   0.500. sit. Nat. 0.500   0.500   † ∆0 =  0.500   0.500  0.500. er. . ‧. ‧ 國. 學. Then. . . v.

(61) And we can get the consistent skew-symmetric matrix . 8.27 − 7.36 5.81 − 7.36 6.64 − 7.36 0 − 7.36.   0 5.81 − 8.27 6.64 − 8.27 0 − 8.27   8.27 − 5.81 0 6.64 − 5.81 0 − 5.81    8.27 − 6.64 5.81 − 6.64 0 0 − 6.64  8.27 − 0 5.81 − 0 6.64 − 0 0  −1.55 −0.72 −7.36   −2.46 −1.63 −8.27   0 0.83 −5.81 .   −0.83 0 −6.64  5.81 6.64 0. ‧. ‧ 國. 立. 政治大. 學. 0   7.36 − 8.27   X= 7.36 − 5.81   7.36 − 6.64  7.36 − 0  0 0.91   −0.91 0   =  1.55 2.46    0.72 1.63  7.36 8.27. n. er. io. sit. y. Nat. al. . Ch. engchi. 39. i n U. v.

(62) Chapter 6 Conclusion 政治大 In this thesis, we primarily show the elementary linear algebra that is used to survey HodgeR立. ‧ 國. 學. ank and deduce combinatorial Hodge theorem. In the original paper, Jiang, Lim, Yao and Ye, proposed that HodgeRank is a suitable method for ranking datasets which are incomplete or. ‧. imbalanced. In other words, most values are missing or some candidates are popular that they received a great quantity of scores. These difficulties usually happened in the modern ranking. y. Nat. io. sit. problems. We use HodgeRank to deal with these troubles and each time we only compare two. n. al. er. candidate. Then we can get a global ranking so that we can give every candidate a score.. i n U. v. Pairwise comparison is practical in today’s ranking problems. For example, the teacher uses. Ch. engchi. MOOCs for the course this semester, and students will hand in their homework on the internet. Teacher use Peer assessment that students will give scores to a few classmates’ homework after the homework deadline. Online courses is popular and have more people to sign up now. It is not possible for a teacher to score all homework. In the end of the course, teacher can use HodgeRank and obtain the ranking of all students’ performance. There are a lot of applications of HodgeRank still can try : combine HodgeRank with other method to ranking or think about how to rank when adding a new candidate or there is a candidate have no scores. Also in this thesis, we mention that there is some little problem when we define the inner product. So in the following research, we can try to give more constraints to find the inner product space. To let the method be more perfect. 40.

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(64) [9] Frederick Mosteller. Remarks on the method of paired comparisons: I. the least squares solution assuming equal standard deviations and equal correlations. Psychometrika, 16(1): 3–9, 1951. [10] Frederick Mosteller. Remarks on the method of paired comparisons: Ii. the effect of an aberrant standard deviation when equal standard deviations and equal correlations are assumed. Psychometrika, 16(2):203–206, 1951. [11] Frederick Mosteller. Remarks on the method of paired comparisons: Iii. a test of significance for paired comparisons when equal standard deviations and equal correlations are. 政治大. assumed. Psychometrika, 16(2):207–218, 1951.. 立. [12] Gottfried E. Noether. Remarks about a paired comparison model. Psychometrika, 25:357–. ‧ 國. 學. 367, 1960.. ‧. [13] Donald G. Saari and Vincent R. Merlin. A geometric examination of Kemeny’s rule. Soc. Choice Welf., 17(3):403–438, 2000.. sit. y. Nat. io. er. [14] Thomas L. Saaty. A scaling method for priorities in hierarchical structures. J. Mathematical Psychology, 15(3):234–281, 1977.. n. al. Ch. engchi. i n U. v. [15] Claire Voisin. Hodge theory and complex algebraic geometry. I, volume 76 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, english edition, 2007. Translated from the French by Leila Schneps. [16] Claire Voisin. Hodge theory and complex algebraic geometry. II, volume 77 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, english edition, 2007. Translated from the French by Leila Schneps. [17] Douglas B. West. Introduction to graph theory. Prentice Hall, Inc., Upper Saddle River, NJ, 1996.. 42.

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