More on Dynamic Programming Bellman-Ford Floyd-Warshall Wednesday, July 30th 1 Outline For Today 1. Bellman-Ford’s Single-Source Shortest Paths 2. Floyd-Warshall’s All-Pairs Shortest Paths 2 Recipe of a DP Algorithm (1) 1: Identify small # of subproblems Ex 1: Find opt independent set in Gi: only n subproblems. Ex 2: Find optimal alignment of Xi and Yj: only mn subproblems 3 Recipe of a DP Algorithm (2) Step 2: quickly + correctly solve “larger” subproblems given solutions to smaller ones, usually via a recursive formula: Ex 1: Linear Ind. Set: A[i] = max {wi + A[i-2], A[i-1]} A[i-1][j-1] + αij Ex 2: Alignment: A[i][j] = min A[i-1][j] + δ A[i][j-1] + δ 4 Recipe of a DP Algorithm (3) 3: After solving all subproblems, can quickly compute final solution Ex 1: Linear Independent Set: return A[n] Ex 2: Alignment: return A[m][n] 5 Outline For Today 1. Bellman-Ford’s Single-Source Shortest Paths 2. Floyd-Warshall’s All-Pairs Shortest Paths 6 Recap: Single-Source Shortest Paths (SSSP) Input: Directed (simple) graph G(V, E), source s, and weights ce on edges. Output: ∀v ∈V, shortest s ⤳ v path. Dijkstra’s Algorithm: O(mlogn); only if ce ≥ 0 7 Pros/Cons of Dijkstra’s Algorithm Pros: O(mlogn) super fast & simple algorithm Cons: 1. Works only if ce ≥ 0 Sometimes need negative weights, e.g. (finance) 2. Not very distributed/parallelizable: Looks very “serial” Need to store large parts of graph in-memory. Therefore not very scalable in very large graphs. Bellman-Ford addresses both of these drawbacks (Last lecture will see parallel version of BF) 8 Preliminary: Negative Weight Cycles Question: How to define shortest paths when G has negative weights cycles? G 4 x s y -6 -5 z 3 t 9 Possible Shortest s⤳v Path Definition 1 Shortest path from s to v with cycles allowed. Problem: Can loop forever in a negative cycle. So there is no “shortest path”! G 4 x s y -6 -5 z 3 t 10 Possible Shortest s⤳v Path Definition 2 Shortest path from s to v, cycles NOT allowed. Problem: Now well-defined. But NP-hard. Don’t expect a “fast” algorithm solving it exactly. G 4 x s y -6 -5 z 3 t 11 Solution: Assume No Negative-Weight Cycles Q: Now, can shortest paths contain cycles? A: No. Assume P is shortest path from s to v, with a cycle. P= s ⤳ x ⤳ x ⤳ v. Then since x ⤳ x is a cycle, and we assumed no negative weight cycles, we could get P`= s ⤳ x ⤳ v, and get a shorter path. 12 Upshot: Bellman-Ford’s Properties Note: Bellman-Ford will be able to detect if there is a negative weight cycle! G(V, E), weights ∃ negativeweight cycle Bellman-Ford Shortest Paths Both outputs computed in reasonable amount of time. 13 Challenge of A DP Approach Need to identify some sub-problems. Linear IS: Line graph was naturally ordered from left to right. Subproblems could be defined as prefix graphs. Sequence Alignment: X, Y strands were naturally ordered strings. Subproblems could be defined as prefix strings. **Shortest Paths’ Input G Has No Natural Ordering** 14 High-level Idea Of Subproblems But the Output Is Paths & Paths Are Sequential! Trick: Impose an Ordering Not On G but on Paths. Larger Paths Will Be Derived By Appending New Edges To The Ends Of Smaller (Shorter) Paths. 15 Subproblems Input: G(V, E) no negative cycles, s, ce arbitrary weights. Output: ∀v, global shortest paths from s to v. Q: Max possible # hops (or # edges) on the shortest paths? A: n-1 (**since there are no negative cycles**) P(v, i) = Shortest path from s to v with at most i edges (& no cycles). 16 Example Let P(v, i) be the shortest s-v path with ≤ i edges. x 1 s -1 w 2 1 z 3 -6 t 2 a 1 b 0 c Q: P(t,1)? A: Does not exist (assume such paths have ∞ weights.) 17 Example Let P(v, i) be the shortest s-v path with ≤ i edges. x 1 s -1 w 2 1 z 3 -6 t 2 a 1 b 0 c Q: P(t,2)? A: s->x->t with weight 2. 18 Example Let P(v, i) be the shortest s-v path with ≤ i edges. x 1 s -1 w 2 1 z 3 -6 t 2 a 1 b 0 c Q: P(t,3)? A: s->w->z->t with weight -1. 19 Example Let P(v, i) be the shortest s-v path with ≤ i edges. x 1 s -1 w 2 1 z 3 -6 t 2 a 1 b 0 c Q: P(t,4)? A: s->w->z->t with weight -1. 20 Solving P(v,i) In Terms of “Smaller” Subproblems Let P=P(v, i) be the shortest s-v path with ≤ i edges Note: For some v, an s-v path with ≤ i edges may not exist. Assume v has such a path. A Claim that does not require a proof: |P| ≤ i-1 OR |P| = i 21 Case 1: |P=P(v, i)| ≤ i-1 s v Q: What can we assert about P(v, i-1)? A: P(v, i-1) = P(v, i) (by contradiction) (P(v, i) is also the shortest s⤳v path with at most i-1 edges) 22 Case 2: |P=P(v, i)| = i s u v P`=>|s⤳u| = i-1 Q: What can we assert about P`? Claim: P` = P(u, i-1) (P` is shortest s⤳u path in with ≤ i-1 edges) 23 Proof that P`=P(u, i-1) Assume ∃a better s⤳u path Q with ≤ i-1 edges s u v Q Q had ≤ i-1 edges, then Q ∪ (u,v) has ≤ i edges. cost(Q) < cost(P`), cost(Q ∪ (u,v)) < cost(P). Q.E.D 24 Summary of the 2 Cases Case 1: |P(v, i)| ≤ i-1 => P(v, i-1) = P(v, s v Case 2: |P(v, i)| = i => P` = P(u, s i) u i-1) v P`=>|s⤳u| = i-1 25 Our Subproblems Agains ∀ v, and for i={1, …, n} P(v, i): shortest s ⤳ v path with ≤ i edges (or null) L(v, i): w(P(v, i)) (and +∞ for null paths) L(v, i-1) L(v, i) = min minu: ∃(u,v)∈E : L(u, i-1) + c(u,v) 26 Bellman-Ford Algorithm L(v, i): w(P(v, i)) Let A be an nxn 2D array. A[i][v] = shortest path to vertex v with ≤ i edges. procedure Bellman-Ford(G(V,E), weights C): Base Cases: A[0][s] = 27 Bellman-Ford Algorithm L(v, i): w(P(v, i)) Let A be an nxn 2D array. A[i][v] = shortest path to vertex v with ≤ i edges. procedure Bellman-Ford(G(V,E), weights C): Base Cases: A[0][s] = 0 A[0][j] = 28 Bellman-Ford Algorithm L(v, i): w(P(v, i)) Let A be an nxn 2D array. A[i][v] = shortest path to vertex v with ≤ i edges. procedure Bellman-Ford(G(V,E), weights C): Base Cases: A[0][s] = 0 A[0][j] = +∞ where j ≠ s for i = 1, …, n-1: for v ∈ V: A[i][v] = min {A[i-1][v] min(u,v)∈E A[i-1][u]+c(u,v) 29 Correctness of BF By induction on i and correctness of the recurrence for L(v, i) (exercise) 30 Runtime of BF # entries in A is n2. Q: How much time for computing each A[i][v]? A: in-deg(v) For each i, total work for all A[i][v] entries is: å in - deg(v) vÎV Total Runtime: O(nm) … for i = 1, …, n-1: for v ∈ V: A[i][v] = min {A[i-1][v] min(u,v)∈E A[i-1][u]+c(u,v) 31 Runtime Optimization: Stopping Early Suppose for some i ≤ n: A[i][v] = A[i-1][v] ∀ v. Q: What does this mean? A: Values will not change in any later iteration => We can stop! … Values only depend on the previous iteration! for i = 1, …, n-1: for v ∈ V: A[i][v] = min {A[i-1][v] min(u,v)∈E A[i-1][u]+c(u,v) 32 Negative Cycle Checking Consider any graph G(V, E) with arbitrary edge weights. =>There may be negative cycles. Claim: If BF stabilizes at some iteration i > 0, then G has no negative cycles. (i.e., negative cycles implies BF never stabilizes!) 33 How To Check For Cycles If Claim is True Run BF just one extra iteration! Check if A[n][v] = A[n-1][v] for all v. If so, no negative cycles, o.w. there is a negative cycle. Running n iterations is the general form of BF: G(V, E), weights ∃ negativeweight cycle Bellman-Ford Shortest Paths 34 Proof of Claim: BF Stabilizes => G has no negative cycles Assume BF has stabilized in iteration i. Notation: d(v) = A[i][v] = A[i-1][v] (by above assumption) A[i][v] = min {A[i-1][v] min(u,v)∈E A[i-1][u] + c(u,v) d(v) = min {d(v) min(u,v)∈E d(u) + c(u,v) d(v) ≤ d(u) + c(u,v) Let’s argue that every cycle C has non-negative weight... 35 Proof of Claim (continued) d(v) ≤ d(u) + c(u,v) Fix a cycle C: y z x t 36 Proof of Claim (continued) d(v) ≤ d(u) + c(u,v) => d(v) – d(u) ≤ c(u,v) Fix a cycle C: y z x t 37 Proof of Claim (continued) d(v) ≤ d(u) + c(u,v) => d(v) – d(u) ≤ c(u,v) Fix a cycle C: y z x t d(x) – d(t) ≤ c(t,x) 38 Proof of Claim (continued) d(v) ≤ d(u) + c(u,v) => d(v) – d(u) ≤ c(u,v) Fix a cycle C: y z x t d(x) – d(t) ≤ c(t,x) d(y) – d(x) ≤ c(x,y) 39 Proof of Claim (continued) d(v) ≤ d(u) + c(u,v) => d(v) – d(u) ≤ c(u,v) Fix a cycle C: y z x t d(x) – d(t) ≤ c(t,x) d(y) – d(x) ≤ c(x,y) d(z) – d(y) ≤ c(y,z) 40 Proof of Claim (continued) d(v) ≤ d(u) + c(u,v) => d(v) – d(u) ≤ c(u,v) Fix a cycle C: y z x t d(x) – d(t) ≤ c(t,x) d(y) – d(x) ≤ c(x,y) d(z) – d(y) ≤ c(y,z) d(t) – d(z) ≤ c(z,t) 0 ≤ w(C) Same algebra and result for any cycle (exercise). Q.E.D. 41 Space Optimization (1) Only need A[i-1][v]’s to compute A[i][v]s. Only need O(n) space; i.e., O(1) per vertex. Q: By throwing things out, what do we lose in general? A: Reconstruction of the actual paths. But with only O(n) more space, we can actually … reconstruct the paths! for i = 1, …, n-1: for v ∈ V: A[i][v] = min {A[i-1][v] min(u,v)∈E A[i-1][u]+c(u,v) 42 Space Optimization (1) Fix: Each v stores a predecessor pointer (initially null) Whenever A[i][v] is updated to A[i-1][u]+c(u,v), we set the Pred[v] to u. Claim: At termination, tracing pointers back from v yields the shortest s-v path. … (Details in the book, by induction on i) for i = 1, …, n-1: for v ∈ V: A[i][v] = min {A[i-1][v] min(u,v)∈E A[i-1][u]+c(u,v) 43 Summary of BF Runtime: O(nm), not as fast as Dijkstra’s O(mlogn). But works with negative weight edges. And is distributable/parallelizable. Will see its distributed version last lecture of class. 44 Outline For Today 1. Bellman-Ford’s Single-Source Shortest Paths 2. Floyd-Warshall’s All-Pairs Shortest Paths 45 All-Pairs Shortest Paths (APSP) Input: Directed G(V, E), arbitrary edge weights, no neg. cycles. Output: ∀ u, v, d(u, v): shortest (u, v) path in G. (no fixed source s) Q: What’s a poly-time algorithm to solve APSP? A: for each s, run BF from s => O(n2m) Better Algorithm: Floyd-Warshall 46 Floyd-Warshall Idea Linear IS: input graph naturally ordered sequentially Seq. Alignment: strings naturally ordered sequentially Bellman-Ford: output paths naturally ordered sequentially FW imposes sequentiality on the vertices order vertices from 1 to n only use the first i vertices in each subproblem (Same idea works for SSSP, but less efficient than BF.) 47 Floyd-Warshall Subproblems V= {1, …, n}, ordered completely arbitrarily Vk = {1, …, k} Original Problem: ∀(u, v) shortest u, v path. We need to define the subproblems. Subproblem P(i, j, k) = shortest i, j path that uses only Vk as intermediate nodes (excluding i and j). 48 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 1 6 0 7 Q: P(6,4,1)? A: null (weight of +∞) 49 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 1 6 0 7 Q: P(2,4,1)? A: 2->1->4 (weight of 2) 50 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 1 6 0 7 Q: P(2,6,0)? A: 2-6 (weight of 2) (no intermediate nodes needed) 51 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 1 6 0 7 Q: P(2,6,1)? A: 2-6 (still weight of 2) (without intermediate nodes) 52 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 1 6 0 7 Q: P(2,6,2)? A: 2-6 (still weight of 2) (without intermediate nodes) 53 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 1 6 0 7 Q: P(2,6,3)? A: 2->3->6 (weight of 0) (now with intermediate node 3) 54 Floyd-Warshall Subproblems 1 1 2 1 4 5 -1 2 3 2 6 7 1 0 Final shortest i⤳j path is P(i, j, n) when we’re allowed to use any vertices as intermediate nodes. 55 Claim That Doesn’t Require A Proof Fix source i, and destination j k ∉ P(i, j, k) OR k∈P(i, j, k) i j (either one of the intermediate vertices is k or it’s not) 56 Case 1: k ∉ P(i, j, k) Then all internal nodes are from 1,…,k-1. Q: What can we assert about P(i, j, k)? i j all intermediate vertices are from Vk-1 A: P(i, j, k) = P(i, j, k-1) (proof by contradiction) 57 Case 2: k ∈ P(i, j, k) Q: What can we assert about P1 and P2? i P1 k j P2 one (and only one) of the int. nodes is k. (why only one?) A1: P1 & P2 only contain int. nodes 1,…,k-1 A2: P1 = P(i,k,k-1) & P2 = P(k,j,k-1) (proof by contradiction) 58 Summary of the 2 Cases Case 1: k ∉ P(i, j, k) => P(i, j, k) = P(i, j, k-1) i j Case 2: k ∈ P(i, j, k) => P1 = P(i,k,k-1) & P2 = P(k,j,k-1) i P1 k j P2 59 Recurrence for Larger Subproblems ∀ i, j, k and where i,j,k={1, …, n} P(i, j, k): shortest i ⤳ j path with all intermediate nodes from Vk={1,…,k} (or null) L(i, j, k): w(P(i, j, k)) (and +∞ for null paths) L(i, j, k) = min L(i, j, k-1) With appropriate base cases. L(i, k, k-1) + L(k, j, k-1) 60 Floyd-Warshall Algorithm Let A be an nxnxn 3D array. A[i][j][k] = shortest i⤳j path with Vk as intermediate nodes procedure Floyd-Warshall(G(V,E), weights C): Base Cases: A[i][i][0] 61 Floyd-Warshall Algorithm Let A be an nxnxn 3D array. A[i][j][k] = shortest i⤳j path with Vk as intermediate nodes procedure Floyd-Warshall(G(V,E), weights C): Base Cases: A[i][i][0] = 0 A[i][j][0] = 62 Floyd-Warshall Algorithm Let A be an nxnxn 3D array. A[i][j][k] = shortest i⤳j path with Vk as intermediate nodes procedure Floyd-Warshall(G(V,E), weights C): Base Cases: A[i][i][0] = 0 A[i][j][0] = Ci,j if (i,j) ∈ E +∞ if (i,j) ∉ E for k = 1, …, n: for i = 1, …, n: for j = 1, …, n: A[i][j][k] = min {A[i][j][k-1], A[i][k][k-1] + A[k][j][k-1]} 63 Correctness & Runtime Correctness: induction on i,j,k & correctness of recurrence Runtime: O(n3) (b/c n3 subproblems, O(1) for each one) procedure Floyd-Warshall(G(V,E), weights C): Base Cases: A[i][i][0] = 0 A[i][j][0] = Ci,j if (i,j) ∈ E +∞ if (i,j) ∉ E for k = 1, …, n: for i = 1, …, n: for j = 1, …, n: A[i][j][k] = min {A[i][j][k-1], 64 A[i][k][k-1] + A[k][j][k-1]} Detecting Negative Cycles Just check the A[i][i][n] for each i! Let C be a negative cycle with |C| = l =>for any vertex j on C, A[j][j][l] ≤ 0 => therefore A[j][j][n] will be negative 65 Path Reconstruction Similar to BF but keep successors instead of predecessors. (exercise) 66 Dijkstra, BF, FW Dijkstra BF FW Single-Source /All Pairs SingleSource SingleSource All Pairs Run-time O(mlog(n)) O(mn) O(n3) Negative Edges No Yes Yes Negative Cycles No No, but No, but can detect can detect 67 Will see a parallel version of BF in last lecture. Next Week: Intractability, P vs NP & What to Do for NP-hard Problems? 68

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