Algebraic approach to exact algorithms, Part III: Polynomials over nite elds of characteristic two

Transcription

1 Algebraic approach to exact algorithms, Part III: Polynomials over nite elds of characteristic two Šukasz Kowalik University of Warsaw ADFOCS, Saarbrücken, August 2013 Šukasz Kowalik (UW) Algebraic approach... August / 38

2 The Schwartz-Zippel Lemma Lemma [DeMillo and Lipton 1978, Zippel 1979, Schwartz 1980] Let p(x 1, x 2,..., x n ) be a non-zero polynomial of degree at most d over a eld F and let S be a nite subset of F. Sample values a 1, a 2,..., a n from S uniformly at random. Then, A typical application Pr[p(a 1, a 2,..., a n )] = 0] d/ S. We can eciently evaluate a polynomial p of degree d. We want to test whether p is a non-zero polynomial. Then, we pick S so that S 2d and we evaluate p on a random vector x S n. We answer YES i we got p(x) 0. If p is the zero polynomial we always get NO, otherwise we get YES with probability at least 1 2. This is called a Monte-Carlo algorithm with one-sided error. Šukasz Kowalik (UW) Algebraic approach... August / 38

3 The Schwartz-Zippel Lemma: Example Polynomial equality testing Input: Two multivariate polynomials P, Q given as an arithmetic circuit. Question: Does P Q? Note: A polynomial described by an arithmetic circuit of size s can have 2 Ω(s) dierent monomials: (x 1 + x 2 )(x 1 x 3 )(x 2 + x 4 ). Šukasz Kowalik (UW) Algebraic approach... August / 38

4 The Schwartz-Zippel Lemma: Example Polynomial equality testing Input: Two multivariate polynomials P, Q given as an arithmetic circuit. Question: Does P Q? Note: A polynomial described by an arithmetic circuit of size s can have 2 Ω(s) dierent monomials: (x 1 + x 2 )(x 1 x 3 )(x 2 + x 4 ). Solution Test whether the polynomial P Q is non-zero using the Schwartz-Zippel Lemma. Theorem Polynomial equality testing for two polynomials represented by circuits of size at most s can be solved in O(s) time with a Monte Carlo algorithm with one-sided error probability bounded by 1/2. Šukasz Kowalik (UW) Algebraic approach... August / 38

5 Question Question What if the bound of 1/2 for the probability of success is not enough for us? Šukasz Kowalik (UW) Algebraic approach... August / 38

6 Question Question What if the bound of 1/2 for the probability of success is not enough for us? Answer Repeat the algorithm 1000 times and answer YES if there was at least one YES. Then, Note Pr[error] The probability that an earthquake destroys the computer is probably 1 higher than Šukasz Kowalik (UW) Algebraic approach... August / 38

7 Finite elds of characteristic 2 In what follows, we use nite elds of size 2 k. We need to know just three things about such elds: They exist (for every k N), We can perform arithmetic operations fast, in O(k log k log log k) time, They are of characteristic two, i.e = 0. In particular, for any element a, we have a + a = a (1 + 1) = a 0 = 0 Šukasz Kowalik (UW) Algebraic approach... August / 38

8 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) Šukasz Kowalik (UW) Algebraic approach... August / 38

9 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Šukasz Kowalik (UW) Algebraic approach... August / 38

10 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Šukasz Kowalik (UW) Algebraic approach... August / 38

11 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Alon, Yuster, Zwick 1994: O((2e) k n O(1) ) Šukasz Kowalik (UW) Algebraic approach... August / 38

12 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Alon, Yuster, Zwick 1994: O((2e) k n O(1) ) Kneis, Mölle, Richter, Rossmanith 2006: O(4 k n O(1) ) Šukasz Kowalik (UW) Algebraic approach... August / 38

13 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Alon, Yuster, Zwick 1994: O((2e) k n O(1) ) Kneis, Mölle, Richter, Rossmanith 2006: O(4 k n O(1) ) Koutis 2008: O(2 3/2k n O(1) ) Šukasz Kowalik (UW) Algebraic approach... August / 38

14 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Alon, Yuster, Zwick 1994: O((2e) k n O(1) ) Kneis, Mölle, Richter, Rossmanith 2006: O(4 k n O(1) ) Koutis 2008: O(2 3/2k n O(1) ) Williams 2009: O(2 k n O(1) ) Šukasz Kowalik (UW) Algebraic approach... August / 38

15 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Alon, Yuster, Zwick 1994: O((2e) k n O(1) ) Kneis, Mölle, Richter, Rossmanith 2006: O(4 k n O(1) ) Koutis 2008: O(2 3/2k n O(1) ) Williams 2009: O(2 k n O(1) ) Björklund 2010: O(1.66 n n O(1) ), undirected Hamiltonian cycle (k = n) Šukasz Kowalik (UW) Algebraic approach... August / 38

16 k-path problem Problem Input: directed/undirected graph G, integer k. Question: Does G contain a path of length k? A few facts NP-complete (why?) even O(f (k)n O(1) )-time algorithm is non-trivial, Monien 1985: O(k!n O(1) ) Alon, Yuster, Zwick 1994: O((2e) k n O(1) ) Kneis, Mölle, Richter, Rossmanith 2006: O(4 k n O(1) ) Koutis 2008: O(2 3/2k n O(1) ) Williams 2009: O(2 k n O(1) ) Björklund 2010: O(1.66 n n O(1) ), undirected Hamiltonian cycle (k = n) Björklund, Husfeldt, Kaski, Koivisto 2010: O(1.66 k n O(1) ), undirected Šukasz Kowalik (UW) Algebraic approach... August / 38

17 k-path in O (2 k )-time Šukasz Kowalik (UW) Algebraic approach... August / 38

18 Notation [k] = {1,..., k} Šukasz Kowalik (UW) Algebraic approach... August / 38

19 O (2 k )-time algorithm for k-path Rough idea Want to construct a polynomial P, P 0 i G has a k-path. Šukasz Kowalik (UW) Algebraic approach... August / 38

20 O (2 k )-time algorithm for k-path Rough idea Want to construct a polynomial P, P 0 i G has a k-path. First try: P( ) = monomial(r). k-path R in G Seems good, but how to evaluate it? Šukasz Kowalik (UW) Algebraic approach... August / 38

21 O (2 k )-time algorithm for k-path Rough idea Want to construct a polynomial P, P 0 i G has a k-path. First try: P( ) = monomial(r). k-path R in G Seems good, but how to evaluate it? Second try: P( ) = monomial(w ). k-walk W in G Now we can evaluate it but we may get false positives. Šukasz Kowalik (UW) Algebraic approach... August / 38

22 O (2 k )-time algorithm for k-path Rough idea Want to construct a polynomial P, P 0 i G has a k-path. First try: P( ) = monomial(r). k-path R in G Seems good, but how to evaluate it? Second try: P( ) = monomial(w ). k-walk W in G Now we can evaluate it but we may get false positives. Final try: P( ) = monomial(w, l). k-walk W in G l:[k] [k] l is bijective We still can evaluate it, It turns out that every monomial corresponding to a walk which is not a path appears an even number of times so it cancels-out! Šukasz Kowalik (UW) Algebraic approach... August / 38

23 Our Hero P(x, y) = k 1 walk W = v 1,..., v k l:[k] [k] i=1 l is bijective k x vi,v i+1 y vi,l(i) i=1 } {{ } mon W,l Variables: a variable x e for every e E, a variable y v,l for every v V and l [k]. Šukasz Kowalik (UW) Algebraic approach... August / 38

24 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Šukasz Kowalik (UW) Algebraic approach... August / 38

25 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. Šukasz Kowalik (UW) Algebraic approach... August / 38

26 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. We dene l : [k] [k] as follows: l(b) if x = a, l (x) = l(a) if x = b, l(x) otherwise. Šukasz Kowalik (UW) Algebraic approach... August / 38

27 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. We dene l : [k] [k] as follows: l(b) if x = a, l (x) = l(a) if x = b, l(x) otherwise. (W, l) (W, l ) since l is injective. Šukasz Kowalik (UW) Algebraic approach... August / 38

28 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. We dene l : [k] [k] as follows: l(b) if x = a, l (x) = l(a) if x = b, l(x) otherwise. (W, l) (W, l ) since l is injective. k 1 k mon W,l = y vi,l(i) = k 1 x vi,v i+1 i=1 i=1 x vi,v i+1 i=1 i [k]\{a,b} y vi,l(i) y va,l(a) }{{} y vb l (b) y vb,l(b) }{{} y val (a) = mon W,l Šukasz Kowalik (UW) Algebraic approach... August / 38

29 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. We dene l : [k] [k] as follows: l(b) if x = a, l (x) = l(a) if x = b, l(x) otherwise. (W, l) (W, l ) since l is injective. mon W,l = mon W,l Šukasz Kowalik (UW) Algebraic approach... August / 38

30 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. We dene l : [k] [k] as follows: l(b) if x = a, l (x) = l(a) if x = b, l(x) otherwise. (W, l) (W, l ) since l is injective. mon W,l = mon W,l If we start from (W, l ) and follow the same way of assignment we get (W, l) back. (This is called a xed-point free involution) Šukasz Kowalik (UW) Algebraic approach... August / 38

31 Monomials corresponding to non-path walks cancel-out Let W = v 1,..., v k be a walk, and a bijection l S k. Assume v a = v b for some a < b, if many such pairs take the lexicographically rst. We dene l : [k] [k] as follows: l(b) if x = a, l (x) = l(a) if x = b, l(x) otherwise. (W, l) (W, l ) since l is injective. mon W,l = mon W,l If we start from (W, l ) and follow the same way of assignment we get (W, l) back. (This is called a xed-point free involution) Since the eld is of characteristic 2, mon W,l and mon W,l cancel out! Šukasz Kowalik (UW) Algebraic approach... August / 38

32 Half the way... Corollary If P 0 then there is a k-path. Šukasz Kowalik (UW) Algebraic approach... August / 38

33 The second half Recall: P(x, y) = k 1 walk W = v 1,..., v k l:[k] [k] i=1 l is bijective k x vi,v i+1 y vi,l(i) i=1 } {{ } mon W,l Question Why do we need exactly mon W,l = k 1 i=1 x v i,v i+1 k i=1 y v i,l(i)? What if, say, mon W,l = k i=1 y v i,l(i)? Šukasz Kowalik (UW) Algebraic approach... August / 38

34 The second half Recall: P(x, y) = k 1 walk W = v 1,..., v k l:[k] [k] i=1 l is bijective k x vi,v i+1 y vi,l(i) i=1 } {{ } mon W,l Question Why do we need exactly mon W,l = k 1 i=1 x v i,v i+1 k i=1 y v i,l(i)? What if, say, mon W,l = k i=1 y v i,l(i)? Answer Now, every labelled walk which is a path gets a unique monomial. Šukasz Kowalik (UW) Algebraic approach... August / 38

35 The second half Recall: P(x, y) = k 1 walk W = v 1,..., v k l:[k] [k] i=1 l is bijective k x vi,v i+1 y vi,l(i) i=1 } {{ } mon W,l Question Why do we need exactly mon W,l = k 1 i=1 x v i,v i+1 k i=1 y v i,l(i)? What if, say, mon W,l = k i=1 y v i,l(i)? Answer Now, every labelled walk which is a path gets a unique monomial. Corollary If there is a k-path in G then P 0. Šukasz Kowalik (UW) Algebraic approach... August / 38

36 Where are we? Corollary There is a k-path in G i P 0. The missing element How to evaluate P eciently? (O (2 k ) is eciently enough.) Šukasz Kowalik (UW) Algebraic approach... August / 38

37 Weighted inclusion-exclusion Let A 1,..., A n U, where U is a nite set. Let w : U F be a weight function. For any X U denote w(x ) = x X w(x). Let us also denote i (U A i) = U. Then, w = ( ) ( 1) X w (U A i ). A i i [n] X [n] i X Šukasz Kowalik (UW) Algebraic approach... August / 38

38 Weighted inclusion-exclusion Let A 1,..., A n U, where U is a nite set. Let w : U F be a weight function. For any X U denote w(x ) = x X w(x). Let us also denote i (U A i) = U. Then, w = ( ) ( 1) X w (U A i ). A i i [n] X [n] i X Counting over a eld of characteristic 2 we know that 1 = 1 so we can remove the ( 1) X : w = ( ) w (U A i ). i [n] A i X [n] Šukasz Kowalik (UW) Algebraic approach... August / 38 i X

39 Evaluating P(x, y) = walk W l:[k] [k] l is bijective mon W,l (x, y) Fix a walk W. U = {l : [k] [k]} (all functions) for l U, dene the weight w(l) = mon W,l (x, y). for i = 1,..., k let A i = {l U : l 1 (i) }. Then, mon W,l (x, y) = mon W,l (x, y) = w( l:[k] [k] l is bijective l:[k] [k] l is surjective l:[k] [k] l is surjective By weighted I-E, mon W,l (x, y) = ( ) w (U A i ) = X [k] l:[k] [k]\x mon W,l (x, y) X [k] i X k A i ). Šukasz Kowalik (UW) Algebraic approach... August / 38 i=1

40 Evaluating P(x, y) = walk W l:[k] [k] l is bijective mon W,l (x, y) Fix a walk W. U = {l : [k] [k]} (all functions) for l U, dene the weight w(l) = mon W,l (x, y). for i = 1,..., k let A i = {l U : l 1 (i) }. Then, mon W,l (x, y) = mon W,l (x, y) = w( l:[k] [k] l is bijective l:[k] [k] l is surjective l:[k] [k] l is surjective By weighted I-E, mon W,l (x, y) = ( ) w (U A i ) = X [k] l:[k] X mon W,l (x, y) X [k] i X k A i ). Šukasz Kowalik (UW) Algebraic approach... August / 38 i=1

41 Evaluating P(x, y) = walk W l:[k] [k] l is bijective mon W,l (x, y) We got Hence, l:[k] [k] l is bijective mon W,l (x, y) = X [k] l:[k] X mon W,l (x, y) P(x, y) = walk W X [k] l:[k] X = X [k] walk W l:[k] X mon W,l (x, y) mon W,l (x, y) } {{ } P X (x,y) Šukasz Kowalik (UW) Algebraic approach... August / 38

42 Evaluating P X (x, y) = mon W,l (x, y) in n O(1) walk W l:[k] X of length k We use dynamic programming. (How?) Šukasz Kowalik (UW) Algebraic approach... August / 38

43 Evaluating P X (x, y) = walk W l:[k] X of length k We use dynamic programming. (How?) Fill the 2-dimensional table T, mon W,l (x, y) in n O(1) Then, T [v, d] = walk W = v 1,..., v d v 1 = v d 1 l:[k] X i=1 d x vi,v i+1 y vi,l(i) i=1 y vl when d = 1, l X T [v, d] = y vl x vw T [w, d 1] otherwise. l X (v,w) E Hence, P X (x, y) = s V T [s, k] can be computed in O(k E ) time. Šukasz Kowalik (UW) Algebraic approach... August / 38

44 Conclusion Corollary The k-path problem can be solved by a O (2 k )-time polynomial space one-sided error Monte-Carlo algorithm. Šukasz Kowalik (UW) Algebraic approach... August / 38

45 k-path in undirected bipartite graphs in O (2 k/2 ) time Šukasz Kowalik (UW) Algebraic approach... August / 38

46 k-path in undirected bipartite graphs in O (2 k/2 ) time V 2 V 1 Šukasz Kowalik (UW) Algebraic approach... August / 38

47 A new hero Idea Label vertices of V 1 only. P(x, y) = walk W = v 1,..., v k l:[k/2] [k/2] l is bijective k 1 i=1 k/2 x vi,v i+1 y v2i 1,l(i) i=1 } {{ } mon W,l Variables: a variable x e for every e E (x u v = x v u), a variable y v,l for every v V and l [k/2]. Šukasz Kowalik (UW) Algebraic approach... August / 38

48 Checking the hero P(x, y) = walk W = v 1,..., v k l:[k/2] [k/2] l is bijective k 1 i=1 k/2 x vi,v i+1 y v2i 1,l(i) i=1 } {{ } mon W,l Paths do not cancel-out If there is a k-path with an endpoint in V 1 then P 0. (Proof: We can recover (W, l) from mon W,l as before.) Šukasz Kowalik (UW) Algebraic approach... August / 38

49 Checking the hero P(x, y) = walk W = v 1,..., v k Do non-path walks cancel-out? l:[k/2] [k/2] l is bijective k 1 i=1 k/2 x vi,v i+1 y v2i 1,l(i) i=1 } {{ } mon W,l Consider a non-path labelled walk (W, l), W = v 1,..., v k. Case 1 If exist i, j, i < j s.t. v i = v j, v i V 1 : pick lexicographically rst such pair; both v i and v j have labels so we swap labels as before. Case 2 As in Case 1, but v i V 2 and Case 1 does not occur: reverse the cycle: l 2 l 1 l 2 l 1 mon W,l = mon W,l, W v i = v j W v i = v j from (W, l ) we get (W, l), Does (W, l) (W, l )? Šukasz Kowalik (UW) Algebraic approach... August / 38

50 Checking the hero P(x, y) = walk W = v 1,..., v k Do non-path walks cancel-out? l:[k/2] [k/2] l is bijective k 1 i=1 k/2 x vi,v i+1 y v2i 1,l(i) i=1 } {{ } mon W,l Consider a non-path labelled walk (W, l), W = v 1,..., v k. Case 1 If exist i, j, i < j s.t. v i = v j, v i V 1 : pick lexicographically rst such pair; both v i and v j have labels so we swap labels as before. Case 2 As in Case 1, but v i V 2 and Case 1 does not occur: reverse the cycle: W v i = v j W v i = v j mon W,l = mon W,l, from (W, l ) we get (W, l), Does (W, l) (W, l )? NO! Šukasz Kowalik (UW) Algebraic approach... August / 38

51 Fixing the hero Admissible walks Walk v 1,..., v k is admissible if: For every i = 1,..., k 2, if v i V 2 and v i+1 V 1 then v i+2 v i. P(x, y) = walk W = v 1,..., v k W is admissible l:[k/2] [k/2] l is bijective k 1 i=1 k/2 x vi,v i+1 y v2i 1,l(i) i=1 } {{ } mon W,l Šukasz Kowalik (UW) Algebraic approach... August / 38

52 Checking the xed hero P(x, y) = walk W = v 1,..., v k W is admissible Do non-path walks cancel-out? l:[k/2] [k/2] l is bijective k 1 i=1 k/2 x vi,v i+1 y v2i 1,l(i) i=1 } {{ } mon W,l Consider a non-path labelled walk (W, l), W = v 1,..., v k. Case 1 If exist i, j, i < j s.t. v i = v j, v i V 1 : pick lexicographically rst such pair; both v i and v j have labels so we swap labels as before. Case 2 As in Case 1, but v i V 2 and Case 1 does not occur: reverse the cycle: mon W,l = mon W,l l 2 l 1 l, 2 l 1 from (W, l ) we get (W, l), W W (W, l) (W, l ) because W v i = v j v i = v j admissible, W is admissible. Šukasz Kowalik (UW) Algebraic approach... August / 38

53 Evaluating P(x, y) = admissible walk W l:[k/2] [k/2] l is bijective mon W,l (x, y) As before, from inclusion-exclusion principle we can get l:[k/2] [k/2] l is bijective Hence, as before: mon W,l (x, y) = X [k/2] l:[k/2] X mon W,l (x, y) P(x, y) = = admissible walk W X [k/2] l:[k/2] X X [k/2] admissible walk W l:[k/2] X mon W,l (x, y) mon W,l (x, y) } {{ } P X (x,y) Note: Only 2 k/2 polynomials P X to evaluate. Šukasz Kowalik (UW) Algebraic approach... August / 38

54 Evaluating P X (x, y) = Dynamic programming: Then, T [v, w, d] = admissiblel:[k/2] X walk W of length k admissible walk W = v 1,..., v d v 1 = v v 2 = w l:[k/2] X i=1 mon W,l in poly-time k 1 k/2 x vi,v i+1 y v2i 1,l(i) x vw l X y vl when d = 2 and v V 1, x vw l X y wl when d = 2 and v V 2, y vl x vw T [w, u, d 1] when d > 2 and v V 1, T [v, w, d] = l X (w,u) E x vw T [w, u, d 1] when d > 2 and v V 2. (w,u) E u v Šukasz Kowalik (UW) Algebraic approach... August / 38 i=1

55 Conclusion Theorem (Björklund, Husfeldt, Kaski, Koivisto 2010) The k-path problem in undirected bipartite graphs can be solved in O (2 k/2 ) = O (1.42 k ) time and polynomial space. Šukasz Kowalik (UW) Algebraic approach... August / 38

56 Conclusion Theorem (Björklund, Husfeldt, Kaski, Koivisto 2010) The k-path problem in undirected bipartite graphs can be solved in O (2 k/2 ) = O (1.42 k ) time and polynomial space. Šukasz Kowalik (UW) Algebraic approach... August / 38

57 k-path in undirected graphs in 3 O (24 k ) time Šukasz Kowalik (UW) Algebraic approach... August / 38

58 k-path in undirected graphs in O (2 3 4 k ) time Choose a random bipartition V = V 1 V 2, V 1 V 2 1. (V 1 and V 2 need not be independent now.) Where does the bipartite case algorithm fail? W W Then (W, l) = (W, l ). v i = v j v i = v j Šukasz Kowalik (UW) Algebraic approach... August / 38

59 k-path in undirected graphs in O (2 3 4 k ) time Choose a random bipartition V = V 1 V 2, V 1 V 2 1. (V 1 and V 2 need not be independent now.) Where does the bipartite case algorithm fail? W W Then (W, l) = (W, l ). v i = v j v i = v j What if we forbid also? Šukasz Kowalik (UW) Algebraic approach... August / 38

60 k-path in undirected graphs in O (2 3 4 k ) time Choose a random bipartition V = V 1 V 2, V 1 V 2 1. (V 1 and V 2 need not be independent now.) Where does the bipartite case algorithm fail? W W Then (W, l) = (W, l ). v i = v j v i = v j What if we forbid also? Then we run into another trouble: W W W contains the forbidden conguration. Šukasz Kowalik (UW) Algebraic approach... August / 38

61 The solution Forbidden conguration as before: Add more labels: label each V 2 V 2 -edge: W W l 2 l 1 l 2 l 1 Now l l. v i = v j v i = v j Šukasz Kowalik (UW) Algebraic approach... August / 38

62 How many labels do we need now? a dierent label for each i = 1,..., k s.t. v i V 1 a dierent label for each i = 1,..., k s.t. v i v i+1 V 2 Šukasz Kowalik (UW) Algebraic approach... August / 38

63 L-admissible walks Walk W = v 1,..., v k is L-admissible when For every i = 1,..., k 2, if v i V 2 and v i+1 V 1 then v i+2 v i. {i : v i V 1 } + {i : v i v i+1 V 2 } = L Šukasz Kowalik (UW) Algebraic approach... August / 38

64 The ultimate hero P L (x, y) = walk W = v 1,..., v k W is L-admissible l:[l] [L] l is bijective k 1 x vi,v i+1 i=1 i=1 L y f (i),l(i), where f (i) = i-th labeled object (V 1 -vertex or V 2 V 2 -edge) in walk W. f (3) f (5) f (4) f (1) f (2) f (6) P = 3 4 k L=k/2 P L Šukasz Kowalik (UW) Algebraic approach... August / 38

65 Correctness We have checked that: P 0 exists k-path (i.e. non-path walks cancel-out) Šukasz Kowalik (UW) Algebraic approach... August / 38

66 Correctness We have checked that: P 0 exists k-path (i.e. non-path walks cancel-out) The opposite implication not always true! (why?) Šukasz Kowalik (UW) Algebraic approach... August / 38

67 Correctness We have checked that: P 0 exists k-path (i.e. non-path walks cancel-out) The opposite implication not always true! (why?) it may happen that the only (say) solution P is not L-admissible for all L 3 4 k. Šukasz Kowalik (UW) Algebraic approach... August / 38

68 Correctness We have checked that: P 0 exists k-path (i.e. non-path walks cancel-out) The opposite implication not always true! (why?) it may happen that the only (say) solution P is not L-admissible for all L 3 4 k. But... E[ {i : v i V 1 } + {i : v i v i+1 V 2 } ] = k + k 1 = 3k Šukasz Kowalik (UW) Algebraic approach... August / 38

69 Correctness We have checked that: P 0 exists k-path (i.e. non-path walks cancel-out) The opposite implication not always true! (why?) it may happen that the only (say) solution P is not L-admissible for all L 3 4 k. But... E[ {i : v i V 1 } + {i : v i v i+1 V 2 } ] = k + k 1 = 3k So, by Markov inequality Pr[P is not L-admissible for all L 3 (3k 1)/4 4k ] 3 4 k + 1 = 1 1/O(k) Šukasz Kowalik (UW) Algebraic approach... August / 38

70 Correctness We have checked that: P 0 exists k-path (i.e. non-path walks cancel-out) The opposite implication not always true! (why?) it may happen that the only (say) solution P is not L-admissible for all L 3 4 k. But... E[ {i : v i V 1 } + {i : v i v i+1 V 2 } ] = k + k 1 = 3k So, by Markov inequality Pr[P is not L-admissible for all L 3 (3k 1)/4 4k ] 3 4 k + 1 = 1 1/O(k) If we repeat the algorithm k log n times this probability drops to (1 1/O(k)) k log n = (e 1/O(k) ) k log n = e O(log n) = 1/n Ω(1) Šukasz Kowalik (UW) Algebraic approach... August / 38

71 Conclusion Theorem (Björklund, Husfeldt, Kaski, Koivisto 2010) The k-path problem in undirected graphs can be solved in O (2 3k/4 ) = O (1.682 k ) time and polynomial space. Šukasz Kowalik (UW) Algebraic approach... August / 38

72 Conclusion Theorem (Björklund, Husfeldt, Kaski, Koivisto 2010) The k-path problem in undirected graphs can be solved in O (2 3k/4 ) = O (1.682 k ) time and polynomial space. Exercises: tune the algorithm to get O (1.66 k ). Corollary (Björklund 2009) The Hamiltonian Cycle problem in undirected graphs can be solved in O (1.66 k ) time and polynomial space. Šukasz Kowalik (UW) Algebraic approach... August / 38