1. Notes This set of slides (Handout #7) is missing the material on Reductions and Greed that was presented in class. Those slides are still under construction, and will be posted as Handout #6. The quiz on Thursday of 8 th week will cover Greed and Dynamic Programming, and through HW #4

2. Topological Sorting Prelude to shortest paths Generic scheduling problem Input: –Set of tasks {T 1, T 2, T 3, …, T n } Example: getting dressed in the morning: put on shoes, socks, shirt, pants, belt, … –Set of dependencies {T 1  T 2, T 3  T 4, T 5  T 1, …} Example: must put on socks before shoes, pants before belt, … Want: –ordering of tasks which is consistent with dependencies Problem representation: Directed Acyclic Graph –Vertices = tasks; Directed Edges = dependencies –Acyclic: if  cycle of dependencies, no solution possible

3. Topological Sorting TOP_SORT PROBLEM: Given a DAG G=(V,E) with |V|=n, assign labels 1,...,n to v i  V s.t. if v has label k, all vertices reachable from v have labels > k “Induction idea”: –Know how to label DAG’s with < n vertices Claim: A DAG G always has some vertex with indegree = 0 –Take an arbitrary vertex v. If v doesn’t have indegree = 0, traverse any incoming edge to reach a predecessor of v. If this vertex doesn’t have indegree = 0, traverse any incoming edge to reach a predecessor, etc. –Eventually, this process will either identify a vertex with indegree = 0, or else reach a vertex that has been reached previously (a contradiction, given that G is acyclic). “Inductive approach”: Find v with indegree(v) = 0, give it lowest available label, then delete v (and incident edges), update degrees of remaining vertices, and repeat

4. Dynamic Programming Key Phrases As we discuss shortest paths, and then the dynamic programming approach, keep in mind the following “key phrases” –“Principle of Optimality”: Any subsolution of an optimal solution is itself an optimal solution –“Overlapping Subproblems”: Can exploit a polynomially bounded number of possible subproblems –“Bottom-Up D/Q” / “Table (cache) of Subproblem Solutions”: avoid recomputation by tabulating subproblem solutions –“Relaxation / Successive Approximation”: Often, the DP approach makes multiple passes, each time solving a less restricted version of the original problem instance Eventually, solve completely unrestricted = original problem instance

5. Single-Source Shortest Paths Given G=(V,E) a directed graph, L : E   + a length function, and a distinguished source s  V. Find the shortest path in G from s to each v i  V, v i  s. Let L*(i,j) = length of SP from v i to v j Lemma: Suppose S   V contains s = v i and L*(1,w) is known  w  S. For v i  S: –Let D(v i ) = min w  S L*(1,w) + L(w,i) (*) L*(1,w) = path length, with path restricted to vertices of S L(w,i) = edge length –Let v v minimize D(v i ) over all nodes v i  S (**) Then L*(1,v) = D(v v ). Notation: D(v) = length of the 1-to-v SP that uses only vertices of S (except for v). (D(v) is not necessarily the same as L*(1,v), because the path is restricted!)

6. Single-Source Shortest Paths Proof of Lemma: To prove equality, prove  and  L*(1,v)  D(v) is obvious because D(v) is the minimum length of a restricted path, while L*(1,v) is unrestricted Let L*(1,v) be the shortest path s=v 1, v 2, …, v r = v from the source v 1 to v. Let v j be the first vertex in this SP that is not in S Then: L*(1,v) = L*(1, v j-1 ) + L(v j-1, v j ) + L*(v j,v) // else, not a shortest path  D(v j ) + L*(v j,v) // D(v j ) minimizes over all w in S, including v j-1  D(v) + L*(v j,v) // because both v j and v v  S, but v v chosen first  D(v) // since L*(v j,v)  0. Lemma  Dijkstra’s Algorithm

7. A Fact About Shortest Paths Triangle Inequality:  (u,v)   (u,x) +  (x,v) (shortest path distances induce a metric) u x v

8. Shortest Path Formulations Given a graph G=(V,E) and w: E   –(1 to 2) “s-t”: Find a shortest path from s to t –(1 to all) “single-source”: Find a shortest path from a source s to every other vertex v  V –(All to all) “all-pairs”: Find a shortest path from every vertex to every other vertex Weight of path =  w(v[i],v[i+1]) Sometimes: no negative edges –Examples of “negative edges”: travel inducements, exothermic reactions in chemistry, unprofitable transactions in arbitrage, … Always: no negative cycles –Makes the shortest-path problem well-defined <0

9. Shortest Paths First case: All edges have positive length –Length of (v i,v j ) edge = d ij Condition 1: d ij > 0 Condition 2: d i + d jk  d ik for some i,j,k –else shortest-path problem would be trivial Observation 1: Length of a path > length of any of its subpaths Observation 2: Any subpath of a shortest path is itself a shortest path Principle of Optimality Observation 3: Any shortest path contains  n-1 edges pigeonhole principle; assumes no negative cycles; n nodes total

10. Shortest Paths Scenario: All shortest paths from v 0 = source to other nodes are ordered by increasing length: –|P 1 |  |P 2 |  …  |P n-1 | –Index nodes accordingly Algorithm: Find P 1, then find P 2, etc. Q: How many edges are there in P 1 ? –Exactly 1 edge, else can find a subpath that is shorter Q: How many edges are there in P k ? –At most k edges, else can find k (shorter) subpaths, which would contradict the definition of P k Observation 4: P k contains  k edges To find P 1 : only look at one-edge paths (min = P 1 ) To find P 2 : only look at one- and two-edge paths –But, need only consider two-edge paths of form d 01 + d 1i –Else would have  1 paths shorter than P 2, a contradiction

11. Another Presentation of Dijkstra’s Algorithm Terminology –Permanent label: true SP distance from v 0 to v i –Temporary label: restricted SP distance from v 0 to v i (going through only existing permanently-labeled nodes) Permanently labeled nodes = set S in previous development Dijkstra’s Algorithm 0. All vertices v i, i = 1,…, n-1, receive temporary labels l i with value d 0i LOOP: 1. Among all temporary labels, pick l k = min I l i and change l k to l k * (i.e., make l k ‘s label permanent) // stop if no temporary labels left 2. Replace all temporary labels of v k ‘s neighbors, using l i  min (l i, l k * + d ki )

12. Prim’s Algorithm vs. Dijkstra’s Algorithm Prim: Iteratively add edge e ij to T, such that v i  T, v j  T, and d ij is minimum Dijkstra: Iteratively add edge e ij to T, such that v i  T, v j  T, and l i + d ij is minimum Both are building trees, in very similar ways! –Prim: Minimum Spanning Tree –Dijkstra: Shortest Path Tree What kind of tree does the following algorithm build? Prim-Dijkstra: Iteratively add edge e ij to T, such that v i  T, v j  T, and c  l i + d ij is minimum // 0  c  1

13. Bellman-Ford Algorithm Idea: Successive Approximation / Relaxation –Find SP using  1 edges –Find SP using  2 edges –… –Find SP using  n-1 edges  have true shortest paths Let l j (k) denote shortest v 0 – v j pathlength using  k edges Then, l i (1) = d 0j  j = 1, …, n-1 // d ij =  if no i-j edge In general, l j (k+1) = min { l j (k), min i (l i (k) + d ij ) } –l j (k) : don’t need k+1 arcs –min i (l i (k) + d ij ) : view as length-k SP plus a single edge

14. Bellman-Ford vs. Dijkstra B S C D A 2 2 13 8 4 Pass 1 2 3 4 Label A 8 min(8, 3+4,  +1, 2+  ) = 7 min(7, 3+4, 4+1, 2+  ) = 5 min(5, 3+4, 4+1, 2+  )= 5 B 3 min(3, 8+4,  + , 2+  ) = 3 min(3, 7+4, 4+ , 2+  ) = 3 min(3, 5+4, 4+ , 2+  ) = 3 C  min( , 8+1, 3+ , 2+2) = 4 min(4, 7+1, 3+ , 2+2) = 4 min(4, 5+1, 3+ , 2+2) = 4 D 2 min(2, 8+ , 3+ ,  +2) = 2 min(2, 7+ , 3+ , 4+2) = 2 min(2, 5+ , 3+ , 4+2) = 2

15. Bellman-Ford vs. Dijkstra B S C D A 2 2 13 8 4 Pass 1 2 3 4 Label A 8 min([8], 2+  ) = 8 min([8], 3+4) = 7 min([7], 4+1) = 5* B 3 min([3], 2+  ) = 3* C  min([  ], 2+2) = 4 min(4, 3+  ) = 4* D 2*

16. Special Case: DAGs (24.2) Longest-Path Problem: well-defined only when there are no cycles DAG: topologically sort the vertices  labels v 1, …, v n s.t. all edges directed from v i to v j, i < j Let l i denote longest v 0 – v j pathlength –l 0 = 0 –l 1 = d 01 // d ij = -  if no i-j edge –l 2 = max(d 01 + d 12, d 02 ) –In general, l k = max j<k (l j + d jk ) Shortest pathlength in DAG: replace max by min, use d ij = +  if no i-j edge

17. DAG Shortest Paths Complexity (24.2) Bellman-Ford = O(VE) Topological sort O(V+E) (DFS) Will never relax edges out of vertex v until have done all edges in to v –Runtime O(V+E) Application: PERT ( program evaluation and review technique ) – critical path is the longest path through the DAG

18. All-Pairs Shortest Paths (25.1-25.2) Directed graph G = (V,E), weight E   Goal: Create n  n matrix of SP distances  (u,v) Running Bellman-Ford once from each vertex O( ) = O( ) on dense graphs Adjacency-matrix representation of graph: – n  n matrix W = (w ij ) of edge weights –assume w ii = 0  i, SP to self has no edges, as long as there are no negative cycles

19. Simple APSP Dynamic Programming (25.1) d ij (m) = weight of s-p from i to j with  m edges d ij (0) = 0 if i = j and d ij (0) =  if i  j d ij (m) = min k {d ik (m-1) + w kj } Runtime = O( n 4 ) n-1 passes, each computing n 2 d’s in O(n) time j i  m-1

20. Matrix Multiplication (25.1) Similar: C = A  B, two n  n matrices c ij =  k a ik  b kj O(n 3 ) operations replacing: ‘‘ + ’’  ‘‘ min ’’ ‘‘  ’’  ‘‘ + ’’ –gives c ij = min k {a ik + b kj } –D (m) = D (m-1) ‘‘  ’’ W –identity matrix is D (0) –Cannot use Strassen’s because no subtraction Time is still O(n  n 3 ) = O(n 4 ) Repeated squaring: W 2n = W n  W n (addition chains) Compute W, W 2, W 4,..., W 2 k, k = log n  O(n 3 log n)

21. Floyd-Warshall Algorithm (26.2/25.2) Also DP, but even faster (by another log n actor  O(n 3 )) c ij (m) = weight of SP from i to j with intermediate vertices in the set {1, 2,..., m}   (i, j)= c ij (n) DP: compute c ij (n) in terms of smaller c ij (n-1) – c ij (0) = w ij –c ij (m) = min {c ij (m), c im (m-1) + c mj (m-1) } intermediate nodes in {1, 2,..., m} j i c im (m-1) c mj (m-1) m c ij (m-1)

22. Floyd-Warshall Algorithm (26.2/25.2) Difference from previous: we do not check all possible intermediate vertices. for m=1..n do for i=1..n do for j = 1..n do c ij (m) = min {c ij (m-1), c im (m-1) + c mj (m-1) } Runtime O(n 3 ) Transitive Closure G* of graph G: –(i,j)  G* iff  path from i to j in G –Adjacency matrix, elements on {0,1} –Floyd-Warshall with ‘‘ min ’’  ‘‘OR’’, ‘‘+’’  ‘‘ AND ’’ –Runtime O(n 3 ) –Useful in many problems

23. BEGIN DIGRESSION (alternative presentation, can be skipped)

24. CLRS Notation: Bellman-Ford (24.1) SSSP in General Graphs Essentially a BFS based algorithm Shortest paths (tree) easy to reconstruct for each v  V do d[v]   ; d[s]  0 // initialization for i =1,...,|V|-1 do for each edge (u,v)  E do d[v]  min{d[v], d[u]+w(u,v)} // relaxation for each v  V do if d[v]> d[u] + w(u,v) then no solution // negative cycle checking for each v  V, d[v]=  (s,v) // have true shortest paths

25. Bellman-Ford Analysis (CLRS 24.1) Runtime = O(VE) Correctness –Lemma: d[v]   (s,v) Initially true Let d[v] = d[u] +w(u,v) by triangle inequality, for first violation d[v] <  (s,v)   (s,u)+w(u,v)  d(u)+w(u,v) –After |V|-1 passes all d values are  ’s if there are no negative cycles s  v[1]  v[2] ...  v (some shortest path) After i-th iteration d[s,v[i]] is correct and final

26. Dijkstra Yet Again (CLRS 24.3) Faster than Bellman-Ford because can exploit having only non-negative weights Like BFS, but uses priority queue for each v  V do d[v]   ; d[s]  0 S   ; Q  V While Q   do u  Extract-Min(Q) S  S + u for v adjacent to u do d[v]  min{d[v], d[u]+w(u,v)} (relaxation = Decrease-Key)

27. Dijkstra Runtime (24.3) Extract-Min executed |V| times Decrease-Key executed |E| times Time = |V|  T(Extract-Min) // find+delete = O(log V)) + |E|  T(Decrease-Key) // delete+add =O(log V)) Binary Heap = E  log V (30 years ago) Fibonacci Heap = E + V  log V (10 years ago) Optimal time algorithm found 1 year ago; runs in time O(E) (Mikel Thorup)

28. Same Lemma as for Bellman-Ford: d[v]   (s,v) Thm: Whenever u is added to S, d[u] =  (s,u) Proof: –Assume that u is the first vertex s.t. d[u] >  (s,u) –Let y be first vertex  V-S that is on actual shortest s-u path  d[y] =  (s,y) –For y’s predecessor x, d[x] =  (s,x) –At the moment that we put x in S, d[y] gets value  (s,y) –d[u] >  (s,u) =  (s,y) +  (y,u) = d[y] +  (y,u)  d[y] Dijkstra Correctness (24.3) s x y u S Q

29. END DIGRESSION (alternative presentation, can be skipped)

30. Dynamic Programming (CLRS 15) Third Major Paradigm so far –DQ, Greed are previous paradigms Dynamic Programming = metatechnique (not a particular algorithm) “Programming” refers to use of a “tableau” in the method (cf. “mathematical programming”), not to writing of code

31. Longest Common Subsequence (CLRS 15.4) Problem: Given x[1..m] and y[1..n], find LCS x: A B C B D A B  B C B A y: B D C A B A Brute-force algorithm: –for every subsequence of x, check if it is in y –O(n2 m ) time 2 m subsequences of x (each element is either in or out of the subsequence) O(n) for scanning y with x-subsequence (m  n)

32. Recurrent Formula for LCS (15.4) Let c[i,j] = length of LCS of X[i]=x[1..i],Y[j]= y[1..j] Then c[m,n] = length of LCS of x and y Theorem: if x[i] = y[j] otherwise Proof: x[i] = y[j]  LCS([X[i],Y[j]) = LCS(X[i-1],Y[j-1]) + x[i]

33. DP Properties(15.3) Any part of the optimal answer is also optimal –A subsequence of LCS(X,Y) is the LCS for some subsequences of X and Y. Subproblems overlap –LCS(X[m],Y[n-1]) and LCS(X[m-1],Y[n]) have common subproblem LCS(X[m-1],Y[n-1]) –There are polynomially few subproblems in total = mn for LCS –Unlike divide and conquer

34. DP (CLRS 15.3) After computing solution of a subproblem, store in table Time = O(mn) When computing c[i,j] we need O(1) time if we have: –x[i], y[j] –c[i,j-1] –c[i-1,j] –c[i-1,j-1]

35. DP Table for LCS y B D C A B A xABCBDABxABCBDAB

36. DP Table for LCS y B D C A B A xABCBDABxABCBDAB 0 0 0 0 1 1 0 1

37. DP Table for LCS y B D C A B A xABCBDABxABCBDAB 0 0 0 0 1 1 0 1 2 0 1

38. DP Table for LCS y B D C A B A xABCBDABxABCBDAB 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 2 2 0 1 1 2 2 2 2 0 1 1 2 2 3 3 0 1 2 2 2 3 3 0 1 2 2 3 3 4 0 1 2 2 3 4 4

39. Optimal Polygon Triangulation Polygon has sides and vertices Polygon is simple = not self-intersecting. Polygon P is convex if any line segment with ends in P lies entirely in P Triangulation of P is partition of P with chords into triangles. Problem: Given a convex polygon and weight function defined on triangles (e.g. the perimeter). Find triangulation of minimum weight (of minimum total length). # of triangles: Always have n-2 triangles with n-3 chords

40. Optimal Polygon Triangulation Optimal sub-triangulation of optimal triangulation Recurrent formula: t[i,j] = the weight of the optimal triangulation of the polygon ; If i=j, then t[i,j]=0; else Runtime O(n 3 ) and space is O(n 2 ) v[i-1] v[i] v[j] v[i-1] v[i] v[j] v[k]