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Minimum weight spanning trees

Learn about Prim's algorithm for finding minimum weight spanning trees, its implementation, and its correctness. Also, explore the Dijkstra's algorithm for shortest path and the Bellman-Ford algorithm for negative edge weights.

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Minimum weight spanning trees

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  1. Minimum weight spanning trees 4 3 2 3 5 7 1 4 6 6 2

  2. Minimum weight spanning trees 4 3 2 3 5 7 1 4 6 6 2

  3. Minimum weight spanning trees 4 3 2 3 5 7 1 4 6 6 2

  4. Minimum weight spanning trees 4 3 2 3 5 7 1 4 6 6 2

  5. Minimum weight spanning trees 4 3 2 3 5 7 1 4 6 6 2

  6. Minimum weight spanning trees 4 3 2 3 5 7 1 4 6 6 2

  7. Minimum weight spanning trees Prim’s algorithm 1) S  {1} 2) add the cheapest edge {u,v} such that uS and vSC S S{v} (until S=V)

  8. Minimum weight spanning trees 1) S  {1} 2) add the cheapest edge {u,v} such that uS and vSC S S{v} (until S=V) P = MST output by Prim T = optimal MST Is P = T ? assume all the edgeweights different

  9. Minimum weight spanning trees P = MST output by Prim T = optimal MST P = T assuming all the edgeweights different v1,v2,...,vn order added to S by Prim smallest i such that the edge e E used by Prim to connect vi+1 is not in T.

  10. Minimum weight spanning trees S S={v1,...,vi} look at T vi+1 f smallest i such that the edge e E used by Prim to connect vi+1 is not in T.

  11. Minimum weight spanning trees S={v1,...,vi} look at T+e S e w(f) > w(e) vi+1 f smallest i such that the edge e E used by Prim to connect vi+1 is not in T.

  12. Minimum weight spanning trees S={v1,...,vi} look at T+e-f S e w(f) > w(e) vi+1 f smallest i such that the edge e E used by Prim to connect vi+1 is not in T.

  13. Prim’s algorithm for i  1 to n do C[i]  C[0]  0 S  {} while S V do j  with minimal C[j] over jSC S S  { j } for u neighbors of j do if w[{j,u}] < C[u] then C[u]w[{j,u}]; P[u] j

  14. Prim’s algorithm for i  1 to n do C[i]  C[0]  0 S  {} while S V do j  with minimal C[j] over jSC S S  { j } for u neighbors of j do if w[{j,u}] < C[u] then C[u]w[{j,u}]; P[u] j Extract-Min O(log n) time O(E+V log V) Decrease-Key O(1) time (amortized)

  15. Shortest path problem 4 3 2 1 2 7 1 A 4 6 8 2 B

  16. Shortest path problem 4 3 2 1 2 7 1 A 4 6 8 2 B

  17. Longest path problem ? 4 3 2 1 2 7 1 A 4 6 8 2 B

  18. Longest path problem ? path from u to v = sequence v1,...,vn such that v1 = u, v2 = v, {vi,vi+1} E no repeated vertices tour from u to v = sequence v1,...,vn such that v1 = u, v2 = v, {vi,vi+1} E no repeated edges walk from u to v = sequence v1,...,vn such that v1 = u, v2 = v, {vi,vi+1} E • walk from u to v  exists path from u to v shortest path = shortest walk longest path  longest walk

  19. Shortest path problem 4 3 2 1 2 7 0 1 A 4 6 8 2 B

  20. Shortest path problem 4 3 2 1 2 7 0 1 1 A 4 6 8 2 B

  21. Shortest path problem 2 4 3 2 1 2 7 0 1 1 A 4 6 8 2 B

  22. Shortest path problem 3 2 4 3 2 1 2 7 0 1 1 A 4 6 8 2 B

  23. Shortest path problem 3 2 4 3 2 1 2 7 0 1 1 A 4 6 4 8 2 B

  24. Shortest path problem 3 2 4 3 2 1 2 7 0 1 1 A 4 6 4 8 2 6 B

  25. Dijkstra’s algorithm for i  1 to n do T[i]  T[0] 0 S {} while S  V do j  with minimal T[j] over jSC S  S  { j } for u neighbors of j do if T[j]+w[{j,u}] < T[u] then T[u]T[j]+w[{j,u}]; P[u] j running time ?

  26. Dijkstra’s algorithm (for s.p.) T[0]  0; S  {}; for i  1 to n do T[i]  while S  V do j  with minimal T[j] over jSC S  S  { j } for u neighbors of j do if T[j]+w[{j,u}] < T[u] then T[u]T[j]+w[{j,u}]; P[u] j Prim’s algorithm (for MST) C[0] 0; S  {}; for i  1 to n do C[i]  while S V do j  with minimal C[j] over jSC S S  { j } for u neighbors of j do if w[{j,u}] < C[u] then C[u]w[{j,u}]; P[u] j

  27. Dijkstra’s algorithm - correctness T[0]  0; S  {}; for i  1 to n do T[i]  while S  V do j  with minimal T[j] over jSC S  S  { j } for u neighbors of j do if T[j]+w[{j,u}] < T[u] then T[u]T[j]+w[{j,u}]; P[u] j Claim: After t steps we have 1. d(0,u) = T[u] for all uS 2. T[u] = dS(0,u) for all u, where dS(0,u) is the length of shortest path from 0 to u, such that all vertices (except possibly u) of the path are in S Proof: induction on t.

  28. Negative edge-weights? 4 3 -2 1 2 7 2 A 4 6 8 2 B

  29. Negative edge-weights? 4 3 -2 1 2 7 2 2 A 4 6 8 2 B

  30. Shortest path problem -4 -3 -2 -1 -2 -7 -1 A -4 -6 -8 -2 B

  31. Shortest path problem -4 4 -3 3 2 -2 1 -1 -2 2 7 -7 -1 1 A A 4 -4 -6 6 8 -8 2 -2 B B Longest path problem

  32. Shortest path problem -4 4 -3 3 2 -2 1 -1 -2 2 7 -7 1 -1 A A 4 -4 -6 6 8 -8 2 -2 B B Longest path problem shortest path  shortest walk (if negative)

  33. Negative edge weights allowed but no negative cycles shortest path = shortest walk (if negative)

  34. The Bellman-Ford algorithm D[v] = estimate on the distance from 0 Initially: D[0] = 0 D[i] =  for all other vertices Relaxation of an edge: if D[u] + w(u,v) < D[v] then D[v]  D[u] + w(u,v)

  35. The Bellman-Ford algorithm D[v] = estimate on the distance from 0 Initially: D[0] = 0 D[i] =  for all other vertices Repeat |V| times relax every edge e running time = O( V.E )

  36. The Bellman-Ford algorithm Let S[u] be the number of edges in the shortest path from 0 to u Claim: After t steps of the algorithm S[u]  t  D[u] = d(0,u)

  37. All-pairs shortest path problem Dijkstra  O( V2 log V + V.E ) (negative edges not allowed)

  38. The Floyd-Warshall algorithm dynamic programming algorithm negative edges allowed, no negative cycles dijk= length of the shortest path from i to j which only uses vertices 1...k dijk= min {dijk-1,dikk-1 + dkjk-1}

  39. The Floyd-Warshall algorithm dij0 wij for all i,j V for k from 1 to n do for i from 1 to n do for j from 1 to n do dijk= min {dijk-1,dikk-1 + dkjk-1} Time = ? Space = ?

  40. The Floyd-Warshall algorithm dij0 wij for all i,j V for k from 1 to n do for i from 1 to n do for j from 1 to n do dijk= min {dijk-1,dikk-1 + dkjk-1} Time = O(n3) Space = O(n3)

  41. The Floyd-Warshall algorithm dij wij for all i,j V for k from 1 to n do for i from 1 to n do for j from 1 to n do dij= min {dij,dik+ dkj} Time = O(n3) Space = O(n2)

  42. Single source shortest paths Dijkstra = O(E+V log V) Bellman-Ford = O(E.V) allows negative edge-weights (but not negative cycles) All-pairs shortest paths Dijkstra = O(EV+V2 log V) Floyd-Warshall = O(V3) allows negative edge-weights (but not negative cycles) Johnson = O(EV+V2 log V) allows negative edge-weights (but not negative cycles)

  43. Johnson’s algorithm Dijkstra = O(E+V log V) Bellman-Ford = O(E.V) allows negative edge-weights (but not negative cycles) w’(u,v) = w(u,v) + h(u) – h(v) doesn’t change the shortest paths use Bellman-Ford to compute h such that w’ is non-negative

  44. M[i-1,j]  d-1 M[i,j]  d  M[i,j-1]  d-1 M[i-1,j-1]  d-1 M[i-1,j]  d-1 and M[i,j-1]  d-1 and M[i-1,j-1]  d-1 and A[i,j]=1  M[i,j]  d

  45. if A[i,j] = 1 then M[i,j]  1+min(M[i-1,j],M[i,j-1],M[i-1,j-1]) else M[i,j]  0

  46. If there exists a subset of size k then the k largest numbers have average at least k

  47. T(n)=T(n/2)+O(n) Find(A[l..r], n, s, B) if r  l+1 then try all possibilities else m  median(A[l..r]); p  partition(A[l..r],m) s’  sum in A[p+1..r]; n’  (r-p) if (s+s’)/(n+n’) < B then Find(A[p+1..r], n, s,B) else Find(A[l..p],n+n’,s+s’,B)

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