Lemma of Gessel Viennot

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1 Advanced Graph Algorithms Jan-Apr 2014 Lemma of Gessel Viennot Speaker: Diptapriyo Majumdar Date: April 22, Overview In the last few lectures, we had learnt different algorithms using matrix multiplication. For example strassen s algorithm for matrix multiplication, finding max cut problem of a weighted graph, finding witness for boolean matrix multiplication, min plus product of two matrices etc etc. Today, we will learn Gessel Viennot Lemma. It is one that establishes a connection between lattice paths and determinants. It was first proved by Lindstrom, but its combinatorial significance was first seen by Gessel and Viennot. They establishes a connection between the path systems and determinant. 2 Definitions Let G = (V, E) be a directed acyclic graph, a weight function w : E R and A = (A 1, A 2,..., A n ), B = (B 1, B 2,..., B n ) be two sets of vertices which are not necessarily disjoint. Definition 1. A path system P is given by σ S n and n paths P 1 : A 1 B σ(1), P 2 : A 2 B σ(2),..., P n : A n B σ(n). Weight of a path system P is denoted as w(p) = w(p i ) and sign(p) =. Weight of a path P is given by the product of the edges in the path. w(p ) = e P w(e) When the path P is from vertex v to vertex v, it does not contain any edge since the graph is directed acyclic graph, we consider w(p ) = 1. Definition 2. A path system P is said to be vertex disjoint if for any two paths P i, P j P, there is no vertex in common. Definition 3. Path Matrix M = [m i,j ] n i,j=1 from A to B is defined as follows. m i,j = P :A i B j w(p ) 1

2 3 Interpretation of a square matrix as a path matrix of a bipartite graph Let M be a square matrix. Construct a directed bipartite graph G = (A, B, E) with A = (A 1,..., A n ) that corresponds to the rows of M and B = (B 1, B 2,..., B n ) that corresponds to columns of M. E = {(A i, B j ) i, j [n]} and w(a i, B j ) = m i,j. In this graph, consider A i and B j, there is only one path from that goes from A i to B j which is just the edge (A i, B j ) and w(p ) = w(a i, B j ) = m i,j. Now, lets consider the permutation description of determinant. det(m) = m i,σ(i) Define ψ : AllP athsystems S n as ψ(p) = σ where P is defined as σ and n paths. Lemma 4. ψ is a bijection. Proof. Consider any path system P from A to B. For that P there is unique σ S n. Let ψ(p) = σ 1 and ψ(q) = σ 2 and σ 1 = σ 2, then i [n] : σ 1 (i) = σ 2 (i). Then, let P is given by σ 1 and n unique paths. Q also gives σ 2 and n unique paths. Now, σ 1 = σ 2. Now, fix any σ S n. Then for every i [n], there is only one path from A i to B σ(i) in G. Therefore, P = Q since σ 1 = σ 2. Therefore, ψ is injective. Take any σ S n. By the construction of the graph, clearly, for this σ, there is a path system P σ with n paths P 1 : A 1 B σ(1), P 2 : A 2 B σ(2),..., P n : A n B σ(n) where weights the paths are w(p k ) = w(a k, B σ(k) ) = m k,σ(k). Therefore, ψ is surjective. Therefore, ψ is a bijection. Similarly consider any σ S n, then there is a unique path system P from A to B. w(p) = w(a i, B σ(i) ) = n det(m) = m i,j n m i,σ(i) = P sign(p) n m i,j = P sign(p)w(p). This summation is over all path systems. In this way, we interpret a given square matrix as path matrix of a directed bipartite graph. 4 Gessel Viennot Lemma Gessel Viennot Lemma generalizes this interpretation from bipartite graph to general directed acyclic graph. Statement of the lemma is as follows. 2

3 Theorem 5. Let G = (V, E) be a directed acyclic graph, a weight function w : E R and A = (A 1, A 2,..., A n ), B = (B 1, B 2,..., B n ) be two sets of vertices which are not necessarily disjoint. Let M be the path matrix from (A) to (B) and let V D be the set of all vertex disjoint path systems from A to B. Then det(m) = sign(p)w(p) Proof. Lets consider the permutation description of determinant. det(m) = Fix arbitrary σ S n. Consider n m i,σ(i). m i,σ(i) = [ m i,σ(i) P 1 :A 1 B σ(1) w(p 1 )]... [ P n:a n B σ(n) w(p n )] Now, for a given σ there are collection of path systems from A to B. Define P σ = {P P is a path system from A to B given by σ} Now, we have m i,σ(i) = [ w(p 1 )]... [ w(p n )] P 1 :A 1 B σ(1) P n:a n B σ(n) = sign(p)w(p) P P σ We know that P σ = AllP athsystems. Therefore, we get m i,σ(i) = sign(r)w(r) R P σ = sign(p)w(p) P AllP athsystems we get summation over all path systems P from A to B. Therefore, we have det(m) = P sign(p)w(p ) Now, we have to prove that P will prove this theorem. sign(p)w(p ) = sign(p)w(p ). Proving the following lemma 3

4 Lemma 6. sign(p)w(p ) = P sign(p)w(p ) Proof. Let ND be the set of path systems that are not vertex disjoint path systems from A to B. We break the left hand side as follows since the set of all path systems is partitioned into V D and ND. sign(p)w(p ) = P sign(p)w(p ) + P ND sign(p)w(p ) Pick any R ND. R = (R 1,..., R n ). Among the crossing paths, we define the following things: i 0 = smallest index such that R i0 crosses with some R j with j > i 0. X = first vertex at which R i0 is intersected by some other path in R. j 0 = smallest index of all the paths in R that intersect R i0 at X (or equivalently smallest index of all paths such that X R i0 R j0 ) Now, we define involution on ND as follows. Define φ : ND ND by φ(r) = T = (T 1,..., T n ) where T k = R k when k i 0, j 0 T i0 : From A i0, traverse the edges of R i0 from X to reach B σ(j0 ). T j0 : From A j0, traverse the edges of R j0 from X to reach B σ(i0 ). till it reaches X. After that it traverses the edges of R j0 till it reaches X. After that it traverses the edges of R i0 X R i0 B σ(j0) A i0 R j0 B σ(i0) A j0 edges of R i0 edges of R i0 B σ(j0) A i0 X R i 0 B σ(i0) A j0 edges of R j0 R j 0 edges of R j0 Therefore, for the path system R, we get σ S n where σ = σ (i 0 j 0 ). Therefore, sign(σ ) = sign(i 0 j 0 ) =. Therefore, sign(r) = sign(t ). Also both T and R consists of 4

5 same set of edges. Therefore, w(t ) = w(t i ) = [ k [n]\{i 0,j 0 } w(t k )].w(t i0 ).w(t j0 ) = [ k [n]\{i 0,j 0 } w(r k )].w(t i0 ).w(t j0 ) Now, set of edges that are there in T i0, T j0 are the same set of edges that are there in R i0, R j0. Therefore w(t i0 )w(t j0 ) = w(e) = w(e) = w(r i0 )w(r j0 ) e T i0 T j0 e R i0 R j0 Therefore, w(r) = w(t ). Now, if we apply φ again, then let φ(t ) = Z. Z k = T k = R k when k i 0, j 0 Z i0 : From A i0, traverse the edges of T i0 till it reaches X. After that it traverses the edges of T j0 from X to reach B σ (j 0 ) = B σ(i0 ). This yields path which is same as R i0. Z j0 : From A j0, traverse the edges of T j0 till it reaches X. After that it traverses the edges of T i0 from X to reach B σ (i 0 ) = B σ(j0 ). This yields path R j0. Therefore, Z = R. Therefore, φ(φ(r)) = R. Since φ = φ 1, therefore, φ is a bijection. Therefore, we have found an one to one correspondance to obtain pairs of path systems (R, φ(r)) in ND where for every pairs of path systems with w(r) = w(φ(r)) and sign(r) = sign(φ(r)). Therefore, we get that sign(p)w(p ) = 0. P ND It implies that sign(p)w(p ) = sign(p)w(p ). P Proving the lemma proves that det(m) = sign(p)w(p ) which proves the theorem. 4.1 Example of a graph and path matrix We give an example of a directed acyclic graph and its corresponding path matrix. Let G be the graph as drawn in figure. Notice that A and B are not disjoint. In fact, in this path matrix, there are path systems which are not vertex disjoint(common vertex can be present even at endpoints also). 1 A 2 = B 1 A B 2 5

6 The path systems are as follows. P 1 = [({A 1 B 1 }, {A 2 B 2 }), (1)(2)] P 2 = [({A 1 B 1 B 2 }, {A 2 B 1 }), (12)] P 3 = [({A 1 B 2 }, {A 2 B 1 }), (12)] Only P 3 is vertex disjoint path system. Now the elements of the path matrix M are as follows: m 11 = w(a 1 B 1 ) = 2 m 12 = w(a 1 B 1 B 2 ) + w(a 1 B 2 ) = = 4 m 21 = w(a 2 B 1 ) = 1 m 22 = w(a 2 B 2 ) = 2 Therefore, det(m) = 2 4 = 2 and. sign(p) w(p i ) = sign(p 3 ).w(p 3 ) = ( 1).w({A 1 B 2 }).w(({a 2 B 1 }) = ( 1).2.1 = 2 Therefore, we can see that the Gessel Viennot Lemma holds. 5 Applications of Gessel Viennot Lemma to Matrix Properties It has applications to Matrix Theorems as well as some combinatorial applications. Consider these applications in matrix theorems. 5.1 Application 1: For any square matrix M, det(m) = det(m T ). Proof. For M, we consstructed a directed bipartite graph G = (A, B, E) with bipartitions A, B and E = {(A i, B j ) A i A, B j B} Now, m T ij = m ji. Now, we construct the bipartite graph H = (C, D, F ) where C = A, D = B and F = {(D i, C j ) D i D, C j C} Now, since = sign(σ 1 ), therefore we get det(m T ) = P sign(p)w(p) = m T i,σ(i) 6

7 = = sign(σ 1 ) k=1 k=1 = det(m) m k,σ 1 (k) m k,σ 1 (k) 5.2 Application 2: For any two square matrices of same size M 1, M 2, det(m 1.M 2 ) = det(m 1 )det(m 2 ). Proof. Construct these set of vertices A = (A 1,..., A n ), B = (B 1,..., B n ), C = (C 1,..., C n ). We construct the directed graph with vertex A, B, C where edges are directed from A to B and also from B to C with w(a i, B j ) = m 1 [i, j] and w(b j, C k ) = m 2 [j, k]. A corresponds to the rows of M 1. B corresponds to the columns of M 1 as well as to the rows of M 2. C corresponds to the columns of M 2. Let M = M 1.M 2. Then, m[i, j] = n k=1 m 1 [i, k].m 2 [k, j]. Now, we consider an arbitrary vertexd disjoint path systems P from A to C. P must come through vertices of B. Since P is vertex disjoint, therefore any path system which is a subpath system of P is also vertex disjoint. Now, we see that any vertex disjoint path system P from A to C is divided into two parts Q and R where Q is a vertex disjoint path system from A to B and R is a vertex disjoint path system from B to C. Let W be the set of all vertex disjoint path systems from A to B and Z be the set of all vertex disjoint path systems from B to C. Now consider det(m 1 ).det(m 2 ) = Q W = P W Z sign(q)w(q) R Z sign(r)w(r) sign(r)sign(q)w(r)w(q) Now W Z is the set of all ordered pair of vertex disjoint path systems where the for every pair (Q, R) where Q is a vertex disjoint path system from A to B and R is a vertex disjoint path system from B to C. Now, Q (σ Q, Q 1, Q 2,..., Q n ) where i [n] : Q i : A i B σr (i). R (σ R, R 1, R 2,..., R n ) where i [n] : R i : B i C σr (i). Now, consider P i : A i B σq (i) C σr (σ Q (i)). Composition of two permutations is also a permutation. Let σ Q σ R = σ. So, composition of Q and R gives a path system P. 7

8 We have to show that W Z exhausts the set of all vertex disjoint path systems from A to C. Take any vertex disjoint path system P given by σ S n from A to C. Since it is vertex disjoint and every path from A to C comes through B and P exhausts all vertices of B (otherwise P would not be vertex disjoint). Therefore, P is decomposed into Q (vertex disjoint path system from A to B) and R (vertex disjoint path system from B to C). It easily follows that w(p) = w(q)w(r). Also, since = sign(σ Q σ R ) = sign(σ Q )sign(σ R ), therefore, sign(p) = sign(q).sign(r). Therefore, we have the followings. det(m 1 ).det(m 2 ) = P W Z sign(r)sign(q)w(r)w(q) = P sign(p)w(p) = det(m 1.M 2 ) 5.3 Application 3 This application is a generalization of application 2. It is called Cauchy Binet Formula. Theorem 7. Let M 1 be an n r matrix and M 2 be an r n matrix with n r. Then we have the following det(m 1.M 2 ) = det(m 1 [X]).det(M 2 [X]) X [r], X =n where M 1 [X] is the matrix restricted to the columns indexed by X and M 2 [X] is the matrix restricted to the rows indexed by X. Proof. Construct the directed graph G = (A B C, E) where A = (A 1,..., A n ), B = (B 1,..., B r ), C = (C 1,..., C n ). E = {(A i, B j ) i [n], j [r]} {(B j, C k ) j [r], k [n]}. Also, we assign weights to the edges as w(a i, B j ) = m 1 [i, j] and w(b j, C k ) = m 2 [k, j]. Let M = M 1.M 2. m[i, j] = r k=1 m 1 [i, k]m 2 [k, j]. Fix any arbitrary X [r]. Let P AX be the set of all vertex disjoint path systems from A to B[X ](B[X] be the subset of B restricted to the indices denoted by X) and P XB be the set of all vertex disjoint path systems from B[X ] to C. Then, consider det(m 1 [X]).det(M 2 [X]). det(m 1 [X]).det(M 2 [X]) = Q P AX sign(q)w(q). R P XB sign(r)w(r) Now, similar to the Lemma 6, we know that P AX P XB exhausts the set of all vertex disjoint path systems from A to C that passes through this B[X]. Therefore we have that Q P AX sign(q)w(q). R P XB sign(r)w(r) = P P AX P XB sign(p)w(p) 8

9 Now, in order to get over all X [r]. X [r], X =n det(m 1 [X]).det(M 2 [X]), we have to sum the expression [ P P AX P XB sign(p)w(p)] This summation will also give the summation over the set of all vertex disjoint path systems from A to C. Therefore we have, X [r], X =n det(m 1 [X]).det(M 2 [X]) = X [r], X =n [ P P AX P XB sign(p)w(p)] = P sign(p)w(p) = det(m 1.M 2 ) 6 Combinatorial Applications of Gessel Viennot Lemma Gessel Viennot Lemma also has some interesting combinatorial applications such as Binomial Determinants in terms of Lattice Paths, Rhombic Tilings are all available in [1]. References [1] Martin Aigner: A Course in Enumeration, Springer. [2] Justin Chan: The Gessel Viennot Lemma and its Applications to Combinatorics, Math 821, December 14, [3] Martin Aigner, Gunter M. Zeigler: Proofs From THE BOOK, Springer, Fourth Edition 9