Tree Edit Distance Cannot be Computed in Strongly Subcubic Time (unless APSP can) Karl Bringmann, Paweł Gawrychowski, Shay Mozes, Oren Weimann

Transcription

1 Karl Bringmann, Paweł Gawrychowsi, Shay Mozes, Oren Weimann

2

3 Minimum edits to transform one tree into the other

4 Minimum edits to transform one tree into the other rooted, ordered trees with node labels x y

5 Minimum edits to transform one tree into the other relabel node x to y x y A B C D relabel node y to x A B C D

6 Minimum edits to transform one tree into the other x delete node xy A B C D insert node xy A B C D

7 Minimum edits to transform one string into the other a b c d e f a b b d e e g e e

8 Minimum edits to transform one string into the other a b c d e a b b d e e g e e f

9 Minimum edits to transform one string into the other a b c d e e a b b d e e g e e

10 Minimum edits to transform one string into the other a b c d e e a b b d e e g e e

11 Minimum edits to transform one string into the other a b c d e a b b d e e g e e f

12 Minimum edits to transform one string into the other a b c d e a b b d e e g e e

13 Minimum edits to transform one string into the other a b c d e a b b d e e g e e

14 Minimum edits to transform one string into the other a b c d e a b b d e e g e e f

15 Minimum edits to transform one string into the other a b c d e a b b d e e g e f

16 Minimum edits to transform one string into the other a b c d e a b b d e e g e f

17 Minimum edits to transform one string into the other O(n 2 ) time a b c d e a b b d e e g e e f

18 Minimum edits to transform one string into the other O(n 2 ) time a b c d e f a b b d e e g e e

19 Minimum edits to transform one string into the other O(n 2 ) time O(n 4 ) time a b c d e f a b b d e e g e e

20 String Edit Distance Cannot be Computed in Strongly Subquadratic Time (unless SETH is false) [Bacurs,Indy, STOC 5] a b c d e f a b b d e e g e e

21 x y

22 x y

23 x y

24 y

25 y

26 y prefix in postorder traversal prefix in postorder traversal

27 x y

28 x

29 x

30 x prefix in postorder traversal prefix in postorder traversal

31 x y

32 y y

33 y y

34 y y

35 y y not prefix in postorder traversal not prefix in postorder traversal

36 O(n 4 ) time [Shasha Zhang 989] y y not prefix in postorder traversal not prefix in postorder traversal

37 O(n 4 ) time [Shasha Zhang 989] O(n 3 log n) time [Klein 998] y y not prefix in postorder traversal not prefix in postorder traversal

38 O(n 4 ) time [Shasha Zhang 989] O(n 3 log n) time [Klein 998] O(n 3 ) time [Demaine, Mozes, Rossman, W. 2007] y y not prefix in postorder traversal not prefix in postorder traversal

39 O(n 4 ) time [Shasha Zhang 989] O(n 3 log n) time [Klein 998] O(n 3 ) time [Demaine, Mozes, Rossman, W. 2007] y y

40 Conjecture (APSP): For any ε > 0 there exists c > 0, such that All Pairs Shortest Paths on n node graphs with edge weights in {,..., n c } cannot be solved in O(n3-ε) time.

41 Conjecture (APSP): For any ε > 0 there exists c > 0, such that All Pairs Shortest Paths on n node graphs with edge weights in {,..., n c } cannot be solved in O(n3-ε) time. Equivalent to negative triangle detection [Vassilevsa-Williams,Williams 200] j i w(i,j)+w(j,)+w(,i) < 0

42 The Hard Instance of TED

43 The Hard Instance of TED To specify a solution to TED it is enough to say which nodes are matched to which (the rest are deleted) a b b 0 a 0 c e 0 e d d 0 c 0

44 The Hard Instance of TED To specify a solution to TED it is enough to say which nodes are matched to which (the rest are deleted) The matched pairs must have the same left-right and ancestor-descendant relations a b b 0 a 0 c e 0 e d d 0 c 0

45 The Hard Instance of TED To specify a solution to TED it is enough to say which nodes are matched to which (the rest are deleted) The matched pairs must have the same left-right and ancestor-descendant relations a subsequence of spine nodes b b 0 a 0 subsequence of spine nodes subsequence of leaves c e 0 one middle leaf node one final spine node e d d 0 reversed subsequence of leaves c 0

46 The Hard Instance of TED To specify a solution to TED it is enough to say which nodes are matched to which (the rest are deleted) The matched pairs must have the same left-right and ancestor-descendant relations a subsequence of spine nodes one middle leaf node subsequence of leaves b b 0 c e 0 a 0 subsequence of spine nodes one middle leaf node one final spine node e d d 0 reversed subsequence of leaves c 0 one final spine node

47 APSP TED

48 APSP TED i j w(i,j)+w(j,)+w(,i) < 0 n n n n

49 APSP TED i j j w(i,j)+w(j,)+w(,i) < 0 i n n n n

50 APSP TED i j j w(i,j)+w(j,)+w(,i) < 0 i C(i,j) n n C(j,) C(j,) C(,i) n n

51 APSP TED i j j w(i,j)+w(j,)+w(,i) < 0 i C(i,j) +w(i,j) n C(j,) +w(j,) C(j,) C(,i) +w(,i) n n n

52 APSP TED Large alphabet: Σ = Θ(n) i j j w(i,j)+w(j,)+w(,i) < 0 i C(i,j) +w(i,j) n C(j,) +w(j,) C(j,) C(,i) +w(,i) n n n

53 APSP TED Large alphabet: Σ = Θ(n) i j j w(i,j)+w(j,)+w(,i) < 0 i C(i,j) +w(i,j) n C(j,) +w(j,) C(j,) C(,i) +w(,i) n n Solvable in: O( Σ 2 n 2 logn) time n

54 APSP TED j Small alphabet: Σ = O() i w(i,j)+w(j,)+w(,i) < 0

55 Max-weight APSP -Clique TED Small alphabet: Σ = O() Conjecture (Max-weight -Clique): For any ε > 0 there exists c > 0, such that for any 3 finding a maximum weight -Clique in graphs with edge weights in {,..., n c } cannot be solved in O(n(-ε)) time.

56 Max-weight APSP -Clique TED Small alphabet: Σ = O() Conjecture (Max-weight -Clique): For any ε > 0 there exists c > 0, such that for any 3 finding a maximum weight -Clique in graphs with edge weights in {,..., n c } cannot be solved in O(n(-ε)) time.

57 Max-weight APSP -Clique TED Small alphabet: Σ = O() 2 3 4

58 Max-weight APSP -Clique TED Small alphabet: Σ = O() n/3+2 n/3+2 n/3+2 n/3+2

59 Max-weight APSP -Clique TED Small alphabet: Σ = O() n/3+2 n/3+2 n/3+2 n/3+2

60 Max-weight APSP -Clique TED Small alphabet: Σ = O() n/3+2 n/3+2 n/3+2 Subcubic TED O(n (-ε) ) Max-weight -Clique (for some =(ε)) n/3+2

61 Max-weight APSP -Clique TED Small alphabet: Σ = O() Each /3 clique is a spine node j i n/3+2 n/3+2 n/3+2 n/3+2

62 Max-weight APSP -Clique TED Small alphabet: Σ = O() Each /3 clique is a spine node j Simulate matching costs with small (n 2 size) gadgets Ti. n/3+2 i Ti Tj n/3+2 n/3+2 n/3+2

63 Max-weight APSP -Clique TED Small alphabet: Σ = O() Each /3 clique is a spine node j Simulate matching costs with small (n 2 size) gadgets Ti. n/3+2 i Ti Tj n/3+2 n/3+2 n/3+2

64 Max-weight APSP -Clique TED Small alphabet: Σ = O() Each /3 clique is a spine node j Simulate matching costs with small (n 2 size) gadgets Ti. Challenging: - Ti needs to prepare for any possible Tj - we need to control which Ti can be matched to which (in APSP by height) - constant O() size alphabet n/3+2 i Ti Tj n/3+2 n/3+2 n/3+2

65 Max-weight APSP -Clique TED Small alphabet: Σ = O() Each /3 clique is a spine node j Simulate matching costs with small (n 2 size) gadgets Ti. Challenging: - Ti needs to prepare for any possible Tj - we need to control which Ti can be matched to which (in APSP by height) - constant O() size alphabet n/3+2 i Ti T i T j Tj T j n/3+2 n/3+2 n/3+2

66 We assume without loss of generality that z z. Then, an optimal solution must match dz 0 to b0xz0, d0z 0 2 to b0xz0 2,..., and d0 to b0x, for some z x >... > xz 0 P, as otherwise its D( w(uwe uy )) x, obtain cost is larger than M 2 M (2N z z ) M 2(z ). Rewriting thex,ycost M 2 M (2N 2 z + z ), so recalling our assumption z z we see that in fact z = z 0 as otherwise its cost is larger than M 8 2 M 7 2(N ). u } z { z, M 2 ). { n { of Cj is the right tree nof C 0 (, j v } z M2 rig We are now ready to state properties of the remaining micro structures. Let cmatch (T, T2 ) denote ht topmost node matched to a node from I. Looing again at the same expression, we see that 4. the right-right subtree /3werequire the cost of matching two trees T and T2. Then, that: t f l e b0 /3 and N there.are (A0, D2 such nodes, namely the nodes z. zn0 < Nz 0 =,/3 z0 6 3 (N 0 ) for.n N z 0 with 0) = cmatch /3 M M i) W (i, z every i =, 2,.., and 2,.., N, z i n See Figure 7. For the construction of C(i, j, M (/3) ) and C(i, j, M ) to be co divisible by n/ Continuing in the same fashion, we obtain that there are i nodes matched to 2. cmatch (Bz, Cj ) = M 6 M 3 (N j) W (z, j) for every z =, 2,..., N and j =, 2,..., N. need that M (/3)2 W n(/3 + )3 and M 2 M n2 n(/3 + )3, respectively, whi nodes from Ii, maing the total cost M 7 (N z) as claimed. n 2 large. We z 0. Then, match d0z 0 M n 3. cassume (Ai, Cwithout Wof (j, generality i) + W (j,that i )zfor every i = 2, 3,an...optimal, N and j solution = for 2, 3,.sufficiently..,must N. j ) = M loss match to b0x4.z0 c, d(a x >,... > xz 0 Now, asweotherwise calculateitscmatch (Ai, Cj ). C(i, j, M 2 ) contributes M 2 minus the total cost 0 2 to bx 0 5,...,3 and d to bx, for some z, N z, C ) = M M (i ) W (, i) for every i =, 2,... z 2 i match connecting cost is larger than M M 7 (2N z z 0 ) M 7 2(z 0 ). Rewriting the costtwo we subsets obtain of /3 nodes in the new graph. As the weights in the new graph are 5. cmatch (A7, Cj ) = M M (j0 ) W (j, ) for every j =, 2,..., N. rig0ht t 8 f 0 2 M 2 M (2N 2 z + z ), so recalling our assumption z z we see as v), this is exactly right as that w (u,inv)fact =lezm= z 0w(u, M 2Decrease (Mgadget (/3),xj= n2 )). j, ME Figure 5: Left: built for d =W 3, n(i = 6 and 2 + n C(i, Right: 8 7 The labelsits of cost the nodes in thethan micro structures be partitioned into disjoint subsets corregadget for u = 3, v = otherwise is larger M 2 should M 2(N ). contributes M (/3) W (i, j), so c (Ai, Cj ) = M (/3) W (i, j) M (M { n u } } z 2 n 2 n n v } 2 { z 2 le ft sponding to the following micro structures: match W,(T j, T)) = W (i, j) + the Wfirst (i segment, jof the) required. left as tree can be matched with nodes from the first segment of the rig We are now state properties Let 0,A 0,.. ready E(u D( y w(7, v(i ))match 2 ) denote 2,N7, yc. {A., A0N, Dto },M ) of the remaining micro structures., D2,..., D 2 with cost c, and similarly for the second Then,subtree if u = v we of can A match(c all n = It remains to bound the cost of matching nodes. Nodes insegments. the left i j the cost of matching two trees T0 and T2. Then, we require that: both trees, so the total cost is c= n. Otherwise, we can match at most n nodes, {B, B2,..., BN, C, C2,..., CN }, P M2 3 n >5 (right). the total cost atat least c= n + M c=, see Figure that the matched0 only to nodes of C (A ) with cost least M 4Furthermore,, exceptnote that thetotalr i) for every i =, 2,..., N and z =, 2,..., N, matching the left tree of E(u, v, c= ) with any tree is at least c= n and the cost of match 5 be matched M=. 8. The cost matching node of of E(u, v, c= ) is c=. a node of Ai to a node of Cj, for i, j Figure 6: Schematic illustration of a connection gadget for /3with = 2cost and n cmatch ) = be M M only (N if their j) labels W (z,belong j) for toevery z =, 2,...,The N and jof =least, 2,...M, N. (/3) (for the nodes of C(i, j, M (/3)2 ) or at least M 2 (for the either so2.that two(b nodes matched the same subset. cost at z, Cj can Connection gadget C(i, j, M ). For any i, j 2 {,..., N } and sufficiently large M 2 {0,.. 2. and right tree of C(i, j, M ) is M W (i, j). The left tree d matching any node of A0i, Dz 0, Bz, Cj0 should be at least M 6. The cost of matching any node of 2 )), so for sufficiently large M }, the distance ofm the left C(i, j, M ateditleast 22. cmatch (A0i, Dz 0 ) = M 6 M 3 (N 3. {A, A2,..., AN, C, C2,..., CN }, W (i, z 0 ) (Aibe, Catj )least = M W (j, that i) + the W (j )j )for i = 2,to 3, the...,root N and A3. should M, except root, of iai (C canevery be matched of Cj = 2, 3,..., N. depend on j and the right tree does not depend on i. i, Ccjmatch 5 5 Let {u,..., u/3 } and {v,..., v/3 } be the /3-cliques corresponding to i and j, respe (A ) with cost larger than M M but at most M, and, for any non-root node of Ai (Cj ) and 5 (, ) 5 3 where u < u2 M <..., uw 4 /3 and v < v2 <... < v/3. Recall 4. any cmatch (Ai, Cnode (icost ) matching W (, i) fortotal every =, 2,.M..,N,/3 Aj) (because C that W (i, j) denotes the total w for non-root of CM thea of is larger than icost M. Finally, every A and ) = (A ),M i We can clearly construct solution with M n /3 W (i, 2 all edges connecting two nodes in the i-th clique or a node in the i-th clique with a node in t Cj should consist of O(n2 ) nodes. Now we can show that, assuming these properties, any optimal clique, where we assume that w(u, u) = 0. We construct the gadget C(i, j, M ) as follows. T 5 3 have enumerated the to i jso that 5. cmatch, Cj ) =structure. M clique M (j corresponding ) W (j, ) for every =, 2,...u,N.< u2 <... < u/3 ). the left tree has degree + /3 and the root of the right tree has P degree + n. Their rig solution has(a a specific P We claim ofthat, we correspond to the root of the left and the right trees of D( w(u, u )), respe or The appropriately chosen MMmicro and M better isinto possible. Let W = u,v w(u, v).children 2,ofno Lemma sufficiently large, the total cost an should optimal matching is EveryWe other child of the left root can be matched with every other child of the right root wi labels5.offorthe nodes in the structures be solution partitioned disjoint subsets correm (we fix M and M later). Intuitively, we would lie the x-th child left xm =8 W the (/3 enough to2 guarantee that the cost contributed by nodes in with5 the u -th child of the right root, and then contribute P ofw(uthe, the 7 following 5 sponding micro M be matched v ) to th 2to M 2(N +)). M 6This 2 structures: Mis M 3 2N + M max{w (i, z) + W (z, j) + W (j,total i)}. i,j,z M x5 =, 2,..., /3 we obtain the desired sum. To this end, we cost, so that summing up over M ll decrease at least M. The total cost contributed by nodes in all equality gadgets is 0,..., A0,is the left tree of E(u,, M ) and the right tree of D( ) to the. {A0, Agadgets D, D,..., D }, 2 2 Px-th child of the left root. Simila Proof. Consider i, j, z N maximizing W (i, z)n+ W (z, j) + W (j, i). We may assume that i j. Then, attach the right tree of E(, t, M ) and the left tree of D( w(t, v )) to the t-th child of th t least M to nchoose /3. Consequently, setting M2 = M n /3 + M guarantees that any optimal it is possible the following matching: root. Here we use to emphasise that a particular tree does not depend on the particular v {B, B2,..., BN, C, C2,..., CN }, parameter. All decrease gadgets are of the same type. See Figure 6. olution. bmust all x3 (j= ),+2,W.(j,. ))., /3 we themust 0 match M 8, children of the left root, so in fact, for every D(M 3 (i ) + W (, i)) D(M to cj with cost l ef t left {Ax-th, A2,..child., AN, C C2,..left., CNroot }, to some child of the right root. Because matching the, the match3.2.the of left tree 0 some nodes from the copy of I being the left child of cj to some spine nodes below bz with 5 total M 7 (N z),contributes so that twocost nodes can be matched only at if their to the same subset. The choice cost of of MA f any decrease gadget leastlabelswbelong to the total cost, by the an optimal i Cj 0,8D 0, B, C 0 should be at least 6. The cost of matching any node of 0 matching any node of A M 3. a to d with cost M, z z i j i olution in fact must match2 the x-th2child of the left root with the ux -th child of the right root, as Ai, C4.j some should be from at least Mof,I except that the root of Ai (Cspine be matched to the root of C j ) can nodes the copy M being the right child of a0i 5to some nodes below gadget dz with therwise we lose at least due to the corresponding equality that cannot be compensated 5 (A ) with M but at most M, and, for any non-root node of Ai (Cj ) and totalcost cost larger M 7 (Nthan z), M yp matching its node accompanying gadget. Finally, decrease gadget adds for any 0non-root of C (A ), thedecrease cost of matching is larger than the M 4. corresponding Finally, every Ai and 5. b to d0z, b02 to d0z 2,2..., b0z to d0 with cost M 7 2 each, Cj should consist of O(n ) nodes. Now we can show that, assuming as MtheseMproperties, Moptimal n /3 the total cost is 2 /3 +any y w(ux, vy ) to the total cost. Therefore,3 as long 2 6. ai has aspecific M 2 + M each, solution i to cj structure.,..., ai j+ to c with cost ndeed Mto cja,w (i, j) after choosing the cost of matching the roots to be M +M2 /3+M n / Ai 5. to DFor cost M large M (N W (i, z), z with sufficiently M, i) the the total cost of optimal matching is For Lemma any7. node in a decrease gadget, cost ofanmatching is at least W, for any node in an equality Bz tocost Cj with M M (N j) M W (z,and j),3 7 ofcost adget, finally the of the roots is M 8the 2 M 2(Nmatching ) M 6 is 2 M 5, M 2N + M 2 the maxcost {W (i,of z) matching + W (z, j) + W (j, children i)}. i,j,z t le 9. Athe to matching Cj,..., Ai j+2 costsc(i, M 2j, W W (j, W (j least, i ),M. For the correctness i to Ccost j, Ai of 2 with of M2, so anyto Cnode ) i)is+ at of the ht gh ft i r l 2 2 eft M W (j, i ) + W (j 2, i 2),..., M + 2) + W (, i j + ). rig Proof. Consider j, z maximizing +W (z, j)w +(2, Wi (j,ji). We may assume that i j. Then, onstruction it isi, enough that W M(i,isz) at least C 0 (i, j, M 4 ) C(i, j, M 3 (/3)2 ) x x<y 2 2 j + ). x y right right x it is0. possible to Cchoose the following Ai j+ to M 5 M 3 (imatching: j) W (, i with cost y y y x y

67 Open Problems

68 Open Problems TED to APSP reduction?

69 Open Problems TED to APSP reduction? Largest common subforest: unlabeled trees ( Σ = )

70 Open Problems TED to APSP reduction? Largest common subforest: unlabeled trees ( Σ = ) Levenshtein distance: every elementary edit operation costs

71 Open Problems TED to APSP reduction? Largest common subforest: unlabeled trees ( Σ = ) Levenshtein distance: every elementary edit operation costs log shaves?

72 Than You!