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6.006- Introduction to Algorithms

6.006- Introduction to Algorithms. Lecture 16 Alan Deckelbaum. a. j. h. e. f. i. c. d. b. g. Lecture overview. 5. Shor test paths III Bellman-Ford on a DAG (CLRS 24.2) Dijkstra algorithm for the case with non-negative weights (CLRS 24.3). v 1. v 4. 7. 6. 2. -4. 3. 5.

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6.006- Introduction to Algorithms

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  1. 6.006- Introduction to Algorithms Lecture 16 Alan Deckelbaum

  2. a j h e f i c d b g Lecture overview 5 Shortest paths III • Bellman-Ford on a DAG (CLRS 24.2) • Dijkstra algorithm for the case with non-negative weights (CLRS 24.3) v1 v4 7 6 2 -4 3 5 5 5 2 1 2 v3 v2 8 8 1 6 6 2 5 4 4

  3. This graph has a special structure: DAG. How to use it within Bellman-Ford? 5 E={(v1,v2); (v1,v4); (v2,v3); (v4,v2)} v1 v4 2 -4 2 v3 v2

  4. ... first use topological sorting ... 5 5 v1 v1 v4 v2 2 2 -4 -4 2 2 v3 v4 v3 v2 E={(v1,v2); (v1,v4); (v2,v3); (v4,v2)} E={(v1,v2); (v1,v3); (v2,v3); (v3,v4)}

  5. ... Bellman-Ford ... 5 E={(v1,v2); (v1,v3); (v2,v3); (v3,v4)} v1 v2 ∞ 0 2 -4 ⁄ 5 2 ∞ v4 ∞ v3 end of first iteration 1 ⁄ 2 ⁄ ⁄ 3 and we are done ! the shortest paths from v1

  6. d[s] 0; π[s] s • for each vÎV – {s} • dod[v] ¥; π[v]  nil initialization do for each edge (u, v) ÎE do ifd[v] > d[u] + w(u, v) then d[v] d[u] + w(u, v) π[v] u one iteration ofrelaxation steps Bellman-Ford algorithm on DAG topologically sort the vertices V ( f: V → {1, 2, …, |V |} such that (u,v) ∈ E ⇒ f (u) < f (v)) O(n+m) arrange E in lexicographical order of (f(e.a), f(e.b)) O(n) O(m) for each edge (u, v) ÎE do ifd[v] > d[u] + w(u, v) then report a negative cycle final steps not needed

  7. ... why does this work? ... • there are no cycles in a dag => even with negative-weight edges, there are no negative-weight cycles ... • topological ordering implies a linear ordering of the vertices; every path in a dag is a subsequence of topologically sorted vertex order; processing vertices in that order, an edge can’t be relaxed more than once ...

  8. Proof of Correctness • Let tbe an arbitrary vertex. Suffices to show that we compute d[t] properly. • Let s=s0, s1, s2, …, sk=tbe a shortest path to t. Show by induction that we compute each d[si] correctly. • d[si-1] computed correctly by inductive hypothesis. • (si-1,si) relaxed AFTER d[si-1] computed.

  9. Review of Dijkstra(Non-negative Edge Weights) Problem: Given a directed graph G = (V, E) with edge-weight functionw : ER+, and a node s, find the shortest-path weight d(s, v) (and a corresponding shortest path) from sto each v in V. • Greedy iterative approach • maintain a set S of vertices whose shortest-path distances from s are known. • at each step add to S the vertex vÎV – Swhose distance estimate from s is minimal. • update distance estimates of vertices adjacent to v.

  10. Dijkstra’s algorithm • d[s]  0 • for each vÎV – {s} • dod[v] ¥ • S • QV initialization • whileQ ¹ • dou EXTRACT-MIN(Q) • SSÈ {u} • for each vÎAdj[u] • doifd[v] > d[u] + w(u, v) • then d[v] d[u] + w(u, v) (Qmin-priority queue maintainingV – S) relaxation steps (Implicit DECREASE-KEY)

  11. (,-) 7 (,-) 6 initialization (,-) 3 5 5 5 2 (,-) (0,*)+ 1 (,-) a 8 (,-) 8 1 6 6 f e h i g c d b j 2 5 (,-) 4 4 (,-) (,-) Dijkstra: Example Q =V, a = EXTRACT-MIN(Q)

  12. (,-) 7 (3,a)+ 6 1st iteration (,-) 3 5 5 5 2 (,-) (0,*)+ 1 (,-) a 8 (8,a) 8 1 6 6 g j i j h e f i d g c h e d b f c 2 5 (,-) 4 4 (,-) (5,a) (10,b) 7 (3,a)+ 2nd iteration 6 b (,-) 3 5 5 5 2 (,-) (0,*)+ 1 (,-) a 8 (8,a) 8 1 6 6 2 5 (,-) 4 4 (5,a)+ (,-) Dijkstra: Example

  13. (10,b) 7 (3,a)+ 6 3rd iteration b (,-) 3 5 5 5 2 (,-) (0,*)+ 1 (,-) a 8 (7,d)+ 8 1 6 6 g f j i h e j g c i e h f 2 5 d (,-) 4 4 (5,a)+ (9,d) (10,b) 7 (3,a)+ 4th iteration 6 b (,-) 3 5 5 5 2 (15,c) (0,*)+ 1 (,-) a c 8 (7,d)+ 8 1 6 6 2 5 d (,-) 4 4 (9,d)+ (5,a)+ Dijkstra: Example

  14. (10,b)+ 7 (3,a)+ 6 5th iteration b (,-) 3 5 5 5 2 (15,c) (0,*)+ 1 (,-) a c 8 (7,d)+ 8 1 6 6 j e f e h h i j i 2 5 d g (13,g) 4 4 (5,a)+ (9,d)+ (10,b)+ 7 (3,a)+ 6th iteration 6 b f (16,f) 3 5 5 5 2 (15,c) (0,*)+ 1 (,-) a c 8 (7,d)+ 8 1 6 6 2 5 d g (13,g)+ 4 4 (9,d)+ (5,a)+ Dijkstra: Example

  15. (10,b)+ 7 (3,a)+ 6 7th iteration b (16,f) 3 5 5 5 2 (14,i)+ (0,*)+ 1 (19,i) a c 8 (7,d)+ 8 1 6 6 h j i h f j e 2 5 i d g (13,g)+ 4 4 (5,a)+ (9,d)+ (10,b)+ 7 (3,a)+ 8th iteration 6 b f (15,e)+ 3 5 5 5 2 (14,i)+ (0,*)+ 1 (19,i) a c e 8 (7,d)+ 8 1 6 6 2 5 d g (13,g)+ 4 4 (9,d)+ (5,a)+ Dijkstra: Example

  16. (10,b)+ 7 (3,a)+ 6 9th iteration b (15,e)+ 3 5 5 h 5 2 (14,i)+ (0,*)+ 1 a c e (17,h)+ 8 (7,d)+ 8 1 6 6 g a f e d f j h b i j c 2 5 i d g (13,g)+ 4 4 (5,a)+ (9,d)+ (10,b)+ 7 (3,a)+ Shortest-path tree (15,e)+ 3 2 (14,i)+ (0,*)+ 1 (17,h)+ (7,d)+ 1 2 5 (13,g)+ 4 4 (5,a)+ (9,d)+ Dijkstra: Example

  17. d[u] u w(u, v) s v d[v] Correctness — Part I Lemma. Initializing d[s]  0 and d[v] ¥for all vÎV – {s} establishes d[v] ³d(s, v) for all vÎV, and this invariant is maintained over any sequence of relaxation steps. Proof. Recall relaxation step: ifd[v] > d[u] + w(u, v)set d[v] d[u] + w(u, v)

  18. Correctness — Part II Theorem.Dijkstra’s algorithm terminates with d[v] = d(s, v) for all vÎV. • Proof. • It suffices to show that d[v] = d(s, v) for every vÎVwhen v is added to S • Suppose u is the first vertex added to S for which d[u]>d(s, u) . Let y be the first vertex in V – Salong a shortest path from s to u, and let x be its predecessor: u s S, just before adding u. x y

  19. S s x u Correctness — Part II (continued) Case 1: y =u • Since u is the first vertex violating the claimed invariant, we have d[x] = d(s, x) at the time xwas added to S. • Just after xwas added to S, we therefore set d[u] = d(s,u) • This is a contradiction, since d[u] is never increased by edge relaxation.

  20. S u s x y Correctness — Part II (continued)Case 2: y != u • Since u is the first vertex violating the claimed invariant, we have d[x] = d(s, x) • Since subpaths of shortest paths are shortest paths, it follows that d[y] was set to d(s, x) + w(x, y) = d(s, y) just after x was added to S • Consequently, we have d[y] = d(s, y) £d(s, u)<d[u] • But, d[y] ≥ d[u] since the algorithm chose ufirst => a contradiction

  21. |V|times degree(u) times DECREASE-KEY Analysis of Dijkstra • whileQ ¹ • dou EXTRACT-MIN(Q) • SSÈ {u} • for each vÎAdj[u] • doifd[v] > d[u] + w(u, v) • then d[v] d[u] + w(u, v) Time=Q(n)·TEXTRACT-MIN + Q(m)·TDECREASE-KEY

  22. array O(n) O(1) O(n2) binary heap O(lgn) O(lgn) O(mlgn) Fibonacci heap O(lgn) amortized O(1) amortized O(m+nlgn) worst case Analysis of Dijkstra (continued) Time=Q(n)·TEXTRACT-MIN + Q(m)·TDECREASE-KEY Q TEXTRACT-MIN TDECREASE-KEY Total

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