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CSE 421 Longest Path in a DAG, LIS, Shortest Path with Negative Weights Shayan Oveis Gharan 1 Longest Path in a DAG Longest Path in a DAG Goal: Given a DAG G, find the longest path. Recall: A directed graph G is a DAG if it has no cycle.

  1. CSE 421 Longest Path in a DAG, LIS, Shortest Path with Negative Weights Shayan Oveis Gharan 1

  2. Longest Path in a DAG

  3. Longest Path in a DAG Goal: Given a DAG G, find the longest path. Recall: A directed graph G is a DAG if it has no cycle. This problem is NP-hard for general 2 3 directed graphs: - It has the Hamiltonian Path as a 6 5 4 special case 7 1 3

  4. DP for Longest Path in a DAG Q: What is the right ordering? Remember, we have to use that G is a DAG, ideally in defining the ordering We saw that every DAG has a topological sorting So, let’s use that as an ordering. 2 3 1 2 3 4 5 6 7 6 5 4 7 1 4

  5. DP for Longest Path in a DAG Suppose we have labelled the vertices such that (𝑗, π‘˜) is a directed edge only if 𝑗 < π‘˜. 1 2 3 4 5 6 7 Let π‘ƒπ‘„π‘ˆ(π‘˜) = length of the longest path ending at π‘˜ Suppose in the longest path ending at π‘˜, last edge is (𝑗, π‘˜) . Then, none of the 𝑗 + 1, … , π‘˜ βˆ’ 1 are in this path since topological ordering. So, π‘ƒπ‘„π‘ˆ π‘˜ = π‘ƒπ‘„π‘ˆ 𝑗 + 1. 5

  6. DP for Longest Path in a DAG Suppose we have labelled the vertices such that (𝑗, π‘˜) is a directed edge only if 𝑗 < π‘˜. Let π‘ƒπ‘„π‘ˆ(π‘˜) = length of the longest path ending at π‘˜ 0 If π‘˜ is a source π‘ƒπ‘„π‘ˆ π‘˜ = 0 1 + 5: 5,7 89 :;<: π‘ƒπ‘„π‘ˆ(𝑗) max o.w. 6

  7. DP for Longest Path in a DAG Let G be a DAG given with a topological sorting: For all edges (𝒋, π’Œ) we have i<j. Compute-OPT(j){ if (in-degree(j)==0) return 0 if (M[j]==empty) M[j]=0; for all edges (i,j) M[j] = max(M[j],1+Compute-OPT(i)) return M[j] } Output max(M[1],…,M[n]) Running Time: 𝑃 π‘œ + 𝑛 Memory: 𝑃 π‘œ Can we output the longest path? 7

  8. Outputting the Longest Path Let G be a DAG given with a topological sorting: For all edges (𝒋, π’Œ) we have i<j. Initialize Parent[j]=-1 for all j. Compute-OPT(j){ if (in-degree(j)==0) return 0 if (M[j]==empty) M[j]=0; Record the entry that for all edges (i,j) we used to compute OPT(j) if (M[j] < 1+Compute-OPT(i)) M[j]=1+Compute-OPT(i) Parent[j]=i return M[j] } Let M[k] be the maximum of M[1],…,M[n] While (Parent[k]!=-1) Print k k=Parent[k] 8

  9. Longest Path in a DAG

  10. Longest Path in a DAG Goal: Given a DAG G, find the longest path. Recall: A directed graph G is a DAG if it has no cycle. This problem is NP-hard for general 2 3 directed graphs: - It has the Hamiltonian Path as a 6 5 4 special case 7 1 10

  11. DP for Longest Path in a DAG Suppose we have labelled the vertices such that (𝑗, π‘˜) is a directed edge only if 𝑗 < π‘˜. 1 2 3 4 5 6 7 Let π‘ƒπ‘„π‘ˆ(π‘˜) = length of the longest path ending at π‘˜ Suppose OPT(j) is 𝑗 A , 𝑗 B , 𝑗 B , 𝑗 C , … , 𝑗 DEA , 𝑗 D , 𝑗 D , π‘˜ , then Obs 1: 𝑗 A ≀ 𝑗 B ≀ β‹― ≀ 𝑗 D ≀ π‘˜. Obs 2: 𝑗 A , 𝑗 B , 𝑗 B , 𝑗 C , … , 𝑗 DEA , 𝑗 D is the longest path ending at 𝑗 D . π‘ƒπ‘„π‘ˆ π‘˜ = 1 + π‘ƒπ‘„π‘ˆ 𝑗 D . 11

  12. DP for Longest Path in a DAG Suppose we have labelled the vertices such that (𝑗, π‘˜) is a directed edge only if 𝑗 < π‘˜. Let π‘ƒπ‘„π‘ˆ(π‘˜) = length of the longest path ending at π‘˜ 0 If π‘˜ is a source π‘ƒπ‘„π‘ˆ π‘˜ = 0 1 + 5: 5,7 89 :;<: π‘ƒπ‘„π‘ˆ(𝑗) max o.w. 12

  13. Outputting the Longest Path Let G be a DAG given with a topological sorting: For all edges (𝒋, π’Œ) we have i<j. Initialize Parent[j]=-1 for all j. Compute-OPT(j){ if (in-degree(j)==0) return 0 if (M[j]==empty) M[j]=0; Record the entry that for all edges (i,j) we used to compute OPT(j) if (M[j] < 1+Compute-OPT(i)) M[j]=1+Compute-OPT(i) Parent[j]=i return M[j] } Let M[k] be the maximum of M[1],…,M[n] While (Parent[k]!=-1) Print k k=Parent[k] 13

  14. Longest Increasing Subsequence

  15. Longest Increasing Subsequence Given a sequence of numbers Find the longest increasing subsequence 41, 22, 9, 15, 23, 39, 21, 56, 24, 34, 59, 23, 60, 39, 87, 23, 90 41, 22, 9 , 15 , 23 , 39, 21, 56, 24 , 34 , 59 , 23, 60 , 39, 87 , 23, 90 15

  16. DP for LIS Let OPT(j) be the longest increasing subsequence ending at j. Observation: Suppose the OPT(j) is the sequence 𝑦 5 I , 𝑦 5 J , … , 𝑦 5 K , 𝑦 7 Then, 𝑦 5 I , 𝑦 5 J , … , 𝑦 5 K is the longest increasing subsequence ending at 𝑦 5 K , i.e., π‘ƒπ‘„π‘ˆ π‘˜ = 1 + π‘ƒπ‘„π‘ˆ(𝑗 D ) If 𝑦 7 > 𝑦 5 for all 𝑗 < π‘˜ 1 o.w. π‘ƒπ‘„π‘ˆ π‘˜ = L 1 + max 5:M N OM P π‘ƒπ‘„π‘ˆ(𝑗) Remark: This is a special case of Longest path in a DAG: Construct a 16 graph 1,…n where (𝑗, π‘˜) is an edge if 𝑗 < π‘˜ and 𝑦 5 < 𝑦 7 .

  17. Shortest Paths with Negative Edge Weights

  18. Shortest Paths with Neg Edge Weights Given a weighted directed graph 𝐻 = π‘Š, 𝐹 and a source vertex 𝑑, where the weight of edge (u,v) is 𝑑 W,X Goal: Find the shortest path from s to all vertices of G. Recall that Dikjstra’s Algorithm fails when weights are negative 1 1 -2 -2 3 3 3 source 3 2 2 s s -1 2 -1 2 4 4 18

  19. Impossibility on Graphs with Neg Cycles Observation: No solution exists if G has a negative cycle. This is because we can minimize the length by going over the cycle again and again. So, suppose G does not have a negative cycle. 1 3 -2 2 3 s 2 -1 4 19

  20. DP for Shortest Path Def: Let π‘ƒπ‘„π‘ˆ(𝑀, 𝑗) be the length of the shortest 𝑑 - 𝑀 path with at most 𝑗 edges. Let us characterize π‘ƒπ‘„π‘ˆ(𝑀, 𝑗). Case 1: π‘ƒπ‘„π‘ˆ(𝑀, 𝑗) path has less than 𝑗 edges. Then, π‘ƒπ‘„π‘ˆ 𝑀, 𝑗 = π‘ƒπ‘„π‘ˆ 𝑀, 𝑗 βˆ’ 1 . β€’ Case 2: π‘ƒπ‘„π‘ˆ(𝑀, 𝑗) path has exactly 𝑗 edges. Let 𝑑, 𝑀 A , 𝑀 B , … , 𝑀 5EA , 𝑀 be the π‘ƒπ‘„π‘ˆ(𝑀, 𝑗) path with 𝑗 edges. β€’ Then, 𝑑, 𝑀 A , … , 𝑀 5EA must be the shortest 𝑑 - 𝑀 5EA path with at β€’ most 𝑗 βˆ’ 1 edges. So, π‘ƒπ‘„π‘ˆ 𝑀, 𝑗 = π‘ƒπ‘„π‘ˆ 𝑀 5EA , 𝑗 βˆ’ 1 + 𝑑 X NZI ,X 20

  21. DP for Shortest Path Def: Let π‘ƒπ‘„π‘ˆ(𝑀, 𝑗) be the length of the shortest 𝑑 - 𝑀 path with at most 𝑗 edges. 0 if 𝑀 = 𝑑 ∞ if 𝑀 β‰  𝑑, 𝑗 = 0 π‘ƒπ‘„π‘ˆ 𝑀, 𝑗 = [ min(π‘ƒπ‘„π‘ˆ 𝑀, 𝑗 βˆ’ 1 , W: W,X 89 :;<: π‘ƒπ‘„π‘ˆ 𝑣, 𝑗 βˆ’ 1 + 𝑑 W,X ) min So, for every v, π‘ƒπ‘„π‘ˆ 𝑀, ? is the shortest path from s to v. But how long do we have to run? Since G has no negative cycle, it has at most π‘œ βˆ’ 1 edges. So, π‘ƒπ‘„π‘ˆ(𝑀, π‘œ βˆ’ 1) is the answer. 21

  22. Bellman Ford Algorithm for v=1 to n if π’˜ β‰  𝒕 then M[v,0]= ∞ M[s,0]=0. for i=1 to n-1 for v=1 to n M[v,i]=M[v,i-1] for every edge (u,v) M[v,i]=min(M[v,i], M[u,i-1]+c u,v ) Running Time: 𝑃 π‘œπ‘› Can we test if G has negative cycles? Yes, run for i=1…3n and see if the M[v,n-1] is different from M[v,3n] 22

  23. DP Techniques Summary Recipe: β€’ Follow the natural induction proof. β€’ Find out additional assumptions/variables/subproblems that you need to do the induction β€’ Strengthen the hypothesis and define w.r.t. new subproblems Dynamic programming techniques. β€’ Whenever a problem is a special case of an NP-hard problem an ordering is important: β€’ Adding a new variable: knapsack. β€’ Dynamic programming over intervals: RNA secondary structure. Top-down vs. bottom-up: β€’ Different people have different intuitions β€’ Bottom-up is useful to optimize the memory 23

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