6. DYNAMIC PROGRAMMING II
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1 6. DYNAMIC PROGRAMMING II sequence alignmen Hirschberg s algorihm Bellman Ford algorihm disance vecor proocols negaive cycles in a digraph 6. DYNAMIC PROGRAMMING II sequence alignmen Hirschberg s algorihm Bellman Ford algorihm disance vecor proocols negaive cycles in a digraph Lecure slides by Kevin Wayne Copyrigh 00 Pearson-Addison Wesley hp:// Las updaed on 8//7 :6 PM Sring similariy Edi disance Q. How similar are wo srings? Ex. ocurrance and occurrence. o c u r r a n c e o c c u r r e n c e 6 mismaches, gap o c u r r a n c e o c c u r r e n c e 0 mismaches, gaps o c u r r a n c e o c c u r r e n c e mismach, gap Edi disance. [Levenshein 966, Needleman Wunsch 970] Gap penaly δ; mismach penaly α pq. Cos = sum of gap and mismach penalies. C T G A C C T A C G C T G G A C G A A C G cos = δ + αcg + αta Applicaions. Unix diff, speech recogniion, compuaional biology,...
2 Sequence alignmen Sequence alignmen: problem srucure Goal. Given wo srings x x... x m and y y... y n find a min-cos alignmen. Def. An alignmen M is a se of ordered pairs x i y j such ha each iem occurs in a mos one pair and no crossings. Def. The cos of an alignmen M is: cos(m ) = α xi y j + δ + δ (x i, y j ) M i : x!# "## i unmached j : y j unmached!### #" #### mismach gap x x x x x x6 C T A C C G T A C A T G y y y y y y6 an alignmen of CTACCG and TACATG: M = { x y, x y, x y, x y, x 6 y 6 } xi yj and xiʹ yj cross if i < i, bu j > j ʹ Def. OPT(i, j) = min cos of aligning prefix srings x x... x i and y y... y j. Goal. OPT(m, n). Case. OPT(i, j) maches x i y j. Pay mismach for x i y j + min cos of aligning x x... x i and y y... y j. Case a. OPT(i, j) leaves x i unmached. Pay gap for x i + min cos of aligning x x... x i and y y... y j. Case b. OPT(i, j) leaves y j unmached. Pay gap for y j + min cos of aligning x x... x i and y y... y j. " OPT(i, j) = # % jδ if i = 0 " α xi y j +OPT(i, j ) min # δ +OPT(i, j) oherwise % δ +OPT(i, j ) iδ if j = 0 opimal subsrucure propery (proof via exchange argumen) 6 Sequence alignmen: boom-up algorihm Sequence alignmen: analysis SEQUENCE-ALIGNMENT (m, n, x,, x m, y,, y n, δ, α) FOR i = 0 TO m M [i, 0] i δ. FOR j = 0 TO n M [0, j] j δ. FOR i = TO m FOR j = TO n M [i, j] min { α[xi, yj] + M [i, j ], δ + M [i, j], δ + M [i, j ]). already compued Theorem. The dynamic programming algorihm compues he edi disance (and opimal alignmen) of wo srings of lenghs m and n in Θ(mn) ime and Θ(mn) space. Algorihm compues edi disance. Can race back o exrac opimal alignmen iself. Q. Can we avoid using quadraic space? A. Easy o compue opimal value in O(mn) ime and O(m + n) space. Compue OPT(i, ) from OPT(i, ). Bu, no longer easy o recover opimal alignmen iself. RETURN M [m, n]. 7 8
3 Sequence alignmen in linear space 6. DYNAMIC PROGRAMMING II sequence alignmen Hirschberg s algorihm Bellman Ford algorihm disance vecor proocols negaive cycles in a digraph Theorem. There exiss an algorihm o find an opimal alignmen in O(mn) ime and O(m + n) space. Clever combinaion of divide-and-conquer and dynamic programming. Inspired by idea of Savich from complexiy heory. Programming Techniques G. Manacher Edior A Linear Space Algorihm for Compuing Maximal Common Subsequences D.S. Hirschberg Princeon Universiy The problem of finding a longes common subsequence of wo srings has been solved in quadraic ime and space. An algorihm is presened which will solve his problem in quadraic ime and in linear space. Key Words and Phrases: subsequence, longes common subsequence, sring correcion, ediing CR Caegories:.6,.7,.79,.,. Inroducion 0 Hirschberg s algorihm Edi disance graph. Le f (i, j) be shores pah from (0,0) o (i, j). Lemma: f (i, j) = OPT(i, j) for all i and j. Hirschberg s algorihm Edi disance graph. Le f (i, j) be shores pah from (0,0) o (i, j). Lemma: f (i, j) = OPT(i, j) for all i and j. y y y y y y 6 Pf of Lemma. [ by srong inducion on i + j ] Base case: f (0, 0) = OPT (0, 0) = 0. Inducive hypohesis: assume rue for all (iʹ, jʹ) wih iʹ + jʹ < i + j. Las edge on shores pah o (i, j) is from (i, j ), (i, j), or (i, j ). Thus, x f(i, j) = min{ xiy j + f(i,j ), + f(i,j), + f(i, j )} { } α xi y j δ = min{ xiy j + OPT(i,j ), + OPT(i,j), + OPT(i, j )} { } x δ = OPT(i, j) α xi y j δ x δ
4 Hirschberg s algorihm Edi disance graph. Le f (i, j) be shores pah from (0,0) o (i, j). Lemma: f (i, j) = OPT(i, j) for all i and j. Can compue f (, j) for any j in O(mn) ime and O(m + n) space. Hirschberg s algorihm Edi disance graph. Le g (i, j) be shores pah from (i, j) o (m, n). Can compue by reversing he edge orienaions and invering he roles of (0, 0) and (m, n). j y y y y y y 6 y y y y y y 6 x x δ δ x i+y j+ x x x x Hirschberg s algorihm Hirschberg s algorihm Edi disance graph. Le g (i, j) be shores pah from (i, j) o (m, n). Can compue g(, j) for any j in O(mn) ime and O(m + n) space. Observaion. The cos of a shores pah ha uses (i, j) is f (i, j) + g(i, j). j y y y y y y 6 y y y y y y 6 x x x x x x 6
5 Hirschberg s algorihm Hirschberg s algorihm Observaion. le q be an index ha minimizes f (q, n/) + g (q, n/). Then, here exiss a shores pah from (0, 0) o (m, n) ha uses (q, n/). Divide. Find index q ha minimizes f (q, n/) + g(q, n/); align x q and y n /. Conquer. Recursively compue opimal alignmen in each piece. n / n / y y y y y y 6 y y y y y y 6 x q x q x x x x 7 8 Hirschberg s algorihm: running ime analysis warmup Hirschberg s algorihm: running ime analysis Theorem. Le T(m, n) = max running ime of Hirschberg s algorihm on srings of lenghs a mos m and n. Then, T(m, n) = O(m n log n). Theorem. Le T(m, n) = max running ime of Hirschberg s algorihm on srings of lenghs a mos m and n. Then, T(m, n) = O(m n). T(m, n) T(m, n / ) + O(m n) T(m, n) = O(m n log n). Remark. Analysis is no igh because wo subproblems are of size (q, n/) and (m q, n / ). In nex slide, we save log n facor. [ by inducion on n ] O(m n) ime o compue f (, n / ) and g (, n / ) and find index q. T(q, n / ) + T(m q, n / ) ime for wo recursive calls. Choose consan c so ha: T(m, ) c m T(, n) c n T(m, n) c m n + T(q, n / ) + T(m q, n / ) Claim. T(m, n) c m n. Base cases: m = or n =. Inducive hypohesis: T(m, n) c m n for all (mʹ, nʹ) wih mʹ + nʹ < m + n. T(m, n) T(q, n / ) + T(m q, n / ) + c m n c q n / + c (m q) n / + c m n = c q n + c m n c q n + c m n = c m n 9 0
6 Shores pahs 6. DYNAMIC PROGRAMMING II Shores-pah problem. Given a digraph G = (V, E), wih arbirary edge weighs or coss c vw, find cheapes pah from node s o node. sequence alignmen Hirschberg s algorihm Bellman Ford disance vecor proocols negaive cycles in a digraph source s cos of pah = = 8 desinaion Shores pahs: failed aemps Negaive cycles Dijksra. May no produce shores pahs when edge weighs are negaive. s u v 8 w Reweighing. Adding a consan o every edge weigh does no necessarily make Dijksra s algorihm produce shores pahs. Def. A negaive cycle is a direced cycle such ha he sum of is edge weighs is negaive. s u a negaive cycle W : c(w ) = e W c e < 0 v w
7 Shores pahs and negaive cycles Shores pahs and negaive cycles Lemma. If some pah from v o conains a negaive cycle, hen here does no exis a cheapes pah from v o. If here exiss such a cycle W, hen can build a v pah of arbirarily negaive weigh by deouring around cycle as many imes as desired. Lemma. If G has no negaive cycles, hen here exiss a cheapes pah from v o ha is simple (and has n edges). Consider a cheapes v pah P ha uses he fewes edges. If P conains a cycle W, can remove porion of P corresponding o W wihou increasing he cos. v v W W c(w) < 0 c(w) 0 6 Shores-pahs and negaive-cycle problems Shores pahs: dynamic programming Single-desinaion shores-pahs problem. Given a digraph G = (V, E) wih edge weighs c vw and no negaive cycles and a disinguished node, find cheapes v pah for each node v. Negaive-cycle problem. Given a digraph G = (V, E) wih edge weighs c vw, find a negaive cycle (if one exiss). shores-pahs ree negaive cycle Def. OPT(i, v) = cos of shores v pah ha uses i edges. Case : Cheapes v pah uses i edges. - OPT(i, v) = OPT(i, v) Case : Cheapes v pah uses exacly i edges. - if (v, w) is firs edge, hen OPT uses (v, w), and hen selecs bes w pah using i edges * OPT(i, v) = % &* 0 if i = 0 ' min% OPT(i, v), min { OPT(i, w)+ c vw } ( & (v, w) E ) oherwise Observaion. If no negaive cycles, OPT(n, v) = cos of cheapes v pah. By Lemma, cheapes v pah is simple. opimal subsrucure propery (proof via exchange argumen) 7 8
8 Shores pahs: implemenaion Shores pahs: implemenaion SHORTEST-PATHS (V, E, c, ) FOREACH node v V M [0, v]. M [0, ] 0. FOR i = TO n FOREACH node v V M [i, v] M [i, v]. FOREACH edge (v, w) E M [i, v] min { M [i, v], M [i, w] + c vw }. Theorem. Given a digraph G = (V, E) wih no negaive cycles, he dynamic programming algorihm compues he cos of a cheapes v pah for each node v in Θ(mn) ime and Θ(n ) space. Table requires Θ(n ) space. Each ieraion i akes Θ(m) ime since we examine each edge once. Finding he shores pahs. Approach : Mainain a successor(i, v) ha poins o nex node on cheapes v pah using a mos i edges. Approach : Compue opimal coss M[i, v] and consider only edges wih M[i, v] = M[i, w] + c vw. 9 0 Shores pahs: pracical improvemens Bellman Ford: efficien implemenaion Space opimizaion. Mainain wo d arrays (insead of d array). d(v) = cos of a cheapes v pah ha we have found so far. successor(v) = nex node on a v pah. Performance opimizaion. If d(w) was no updaed in ieraion i, hen no reason o consider edges enering w in ieraion i. BELLMAN FORD (V, E, c, ) FOREACH node v V d(v). successor(v) null. d() 0. FOR i = TO n FOREACH node w V IF (d(w) was updaed in previous ieraion) FOREACH edge (v, w) E IF (d(v) > d(w) + cvw) d(v) d(w) + cvw. successor(v) w. IF no d(w) value changed in ieraion i, STOP. pass
9 Bellman Ford: analysis Bellman Ford: analysis Claim. Afer he i h pass of Bellman Ford, d(v) equals he cos of a cheapes v pah using a mos i edges. Counerexample. Claim is false! d(v) = d(w) = d() = 0 Lemma. Throughou Bellman Ford algorihm, d(v) is he cos of some v pah; afer he i h pass, d(v) is no larger han he cos of a cheapes v pah using i edges. [by inducion on i] Assume rue afer i h pass. Le P be any v pah wih i + edges. Le (v, w) be firs edge on pah and le Pʹ be subpah from w o. By inducive hypohesis, d(w) c(pʹ) since Pʹ is a w pah wih i edges. Afer considering v in pass i + : d(v) c vw + d(w) c vw + c(pʹ) = c(p) v w if node w considered before node v, hen d(v) = afer pass Theorem. Given a digraph wih no negaive cycles, Bellman Ford compues he coss of he cheapes v pahs in O(mn) ime and Θ(n) exra space. Lemmas +. can be subsanially faser in pracice Bellman Ford: analysis Bellman Ford: analysis Claim. Throughou he Bellman Ford algorihm, following successor(v) poiners gives a direced pah from v o of cos d(v). Claim. Throughou he Bellman Ford algorihm, following successor(v) poiners gives a direced pah from v o of cos d(v). Counerexample. Claim is false! Cos of successor v pah may have sricly lower cos han d(v). Counerexample. Claim is false! Cos of successor v pah may have sricly lower cos han d(v). consider nodes in order:,,, consider nodes in order:,,, s() = d() = 0 s() = d() = 0 d() = 0 s() = s() = d() = 0 d() = d() = s() = d() = s() = d() = 6
10 Bellman Ford: analysis Bellman Ford: analysis Claim. Throughou he Bellman Ford algorihm, following successor(v) poiners gives a direced pah from v o of cos d(v). Counerexample. Claim is false! Cos of successor v pah may have sricly lower cos han d(v). Successor graph may have cycles. Claim. Throughou he Bellman Ford algorihm, following successor(v) poiners gives a direced pah from v o of cos d(v). Counerexample. Claim is false! Cos of successor v pah may have sricly lower cos han d(v). Successor graph may have cycles. consider nodes in order:,,,, consider nodes in order:,,,, d() = 0 d() = 8 d() = 0 d() = 8 9 d() = 0 9 d() = d() = d() = 7 d() = d() = 8 Bellman Ford: finding he shores pahs Bellman Ford: finding he shores pahs Lemma. If he successor graph conains a direced cycle W, hen W is a negaive cycle. If successor(v) = w, we mus have d(v) d(w) + c vw. (LHS and RHS are equal when successor(v) is se; d(w) can only decrease; d(v) decreases only when successor(v) is rese) Le v v vk be he nodes along he cycle W. Assume ha (v k, v ) is he las edge added o he successor graph. Jus prior o ha: d(v ) d(v ) + c(v, v ) d(v) d(v) + c(v, v) d(vk ) d(vk) + c(vk, vk) d(vk) > d(v) + c(vk, v) Adding inequaliies yields c(v, v) + c(v, v) + + c(vk, vk) + c(vk, v) < 0. W is a negaive cycle holds wih sric inequaliy since we are updaing d(vk) 9 Theorem. Given a digraph wih no negaive cycles and a node, Bellman Ford finds cheapes v pahs for each node v in O(mn) ime and Θ(n) exra space. The successor graph canno have a negaive cycle. [Lemma ] Thus, following he successor poiners from v yields a direced pah o. Le v = v v v k = be he nodes along his pah P. Upon erminaion, if successor(v) = w, we mus have d(v) = d(w) + c vw. (LHS and RHS are equal when successor(v) is se; d( ) did no change) Thus, d(v) = d(v) + c(v, v) d(v) = d(v) + c(v, v) d(vk ) = d(vk) + c(vk, vk) Adding equaions yields d(s) = d() + c(v, v) + c(v, v) + + c(vk, vk). min cos of any v pah (Theorem ) 0 cos of pah P since algorihm erminaed 0
11 Disance vecor proocols 6. DYNAMIC PROGRAMMING II sequence alignmen Hirschberg s algorihm Bellman Ford disance vecor proocols negaive cycles in a digraph Communicaion nework. Node rouer. Edge direc communicaion link. Cos of edge delay on link. Dijksra s algorihm. Requires global informaion of nework. Bellman Ford. Uses only local knowledge of neighboring nodes. Synchronizaion. We don expec rouers o run in locksep. The order in which each foreach loop execues is no imporan. Moreover, algorihm sill converges even if updaes are asynchronous. naurally nonnegaive, bu Bellman Ford used anyway! Disance vecor proocols Disance vecor proocols. [ rouing by rumor ] Each rouer mainains a vecor of shores-pah lenghs o every oher node (disances) and he firs hop on each pah (direcions). Algorihm: each rouer performs n separae compuaions, one for each poenial desinaion node. Ex. RIP, Xerox XNS RIP, Novell s IPX RIP, Cisco s IGRP, DEC s DNA Phase IV, AppleTalk s RTMP. Cavea. Edge coss may change during algorihm (or fail compleely). Pah vecor proocols Link sae rouing. Each rouer also sores he enire pah. Based on Dijksra s algorihm. Avoids couning-o-infiniy problem and relaed difficulies. Requires significanly more sorage. no jus he disance and firs hop Ex. Border Gaeway Proocol (BGP), Open Shores Pah Firs (OSPF). deleed s v d(s) = d(v) = d() = 0 couning o infiniy
12 Deecing negaive cycles 6. DYNAMIC PROGRAMMING II Negaive cycle deecion problem. Given a digraph G = (V, E), wih edge weighs c vw, find a negaive cycle (if one exiss). sequence alignmen Hirschberg s algorihm Bellman Ford disance vecor proocol negaive cycles in a digraph 6 6 Deecing negaive cycles: applicaion Deecing negaive cycles Currency conversion. Given n currencies and exchange raes beween pairs of currencies, is here an arbirage opporuniy? Remark. Fases algorihm very valuable! 0.7 *.66 *.99 = EUR USD GBP Lemma. If OPT(n, v) = OPT(n, v) for all v, hen no negaive cycle can reach. Bellman Ford algorihm. Lemma 6. If OPT(n, v) < OPT(n, v) for some node v, hen (any) cheapes pah from v o conains a cycle W. Moreover W is a negaive cycle. [by conradicion] Since OPT(n, v) < OPT(n, v), we know ha shores v pah P has exacly n edges. By pigeonhole principle, P mus conain a direced cycle W. Deleing W yields a v pah wih < n edges W has negaive cos. v W CAD 0.9 CHF 7 c(w) < 0 8
13 Deecing negaive cycles Deecing negaive cycles Theorem. Can find a negaive cycle in Θ(mn) ime and Θ(n ) space. Add new node and connec all nodes o wih 0-cos edge. G has a negaive cycle iff Gʹ has a negaive cycle han can reach. If OPT(n, v) = OPT(n, v) for all nodes v, hen no negaive cycles. If no, hen exrac direced cycle from pah from v o. (cycle canno conain since no edges leave ) 6 G Theorem. Can find a negaive cycle in O(mn) ime and O(n) exra space. Run Bellman Ford for n passes (insead of n ) on modified digraph. If no d(v) values updaed in pass n, hen no negaive cycles. Oherwise, suppose d(s) updaed in pass n. Define pass(v) = las pass in which d(v) was updaed. Observe pass(s) = n and pass(successor(v)) pass(v) for each v. Following successor poiners, we mus evenually repea a node. Lemma his cycle is a negaive cycle. Remark. See p. 0 for improved version and early erminaion rule. (Tarjan s subree disassembly rick) 9 0
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