Dijkstra’s Shortest Paths

CS 312Lecture 4

Announcements

• Project 1 comes out Friday– Min spanning trees and scheduling– due 2 weeks from Friday (1 week later than

current calendar)

• Quiz 1 is ready and waiting– Open book– 30 min limit– get a copy on line before you start

Objectives

• Follow up – big-O, big-Theta and big-Omega– Prim’s algorithm: outside of class.

• Understand the shortest paths problem.

• Understand Dijkstra’s algorithm

• Analyze Dijkstra’s algorithm

Big-Letters

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bounds-lower describes -big

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bounds-upper describes O-big

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THE PROBLEMGiven:

•a directed graph G=N,A•each edge has a nonnegative length•one designated source node

Determine the length of the shortest path from the source to each of the other nodes of the graph.

THE SOLUTION

A greedy algorithm called Dijkstra’s Algorithm.

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Dijkstra’s Algorithm: Overview•Maintain

•a set S of vertices whose minimum distance from the source is known, •a set C of candidate vertices (SC=N, SC=).

•Initially, S={s}. (the source)•Keep adding vertices to S, choosing from C the node whose distance from the source is least.•Eventually, S=N.

Special path: a path from ‘s’ with all intermediate nodes in S.

Maintain an array D with:•If w S, then D[w] is the length of the shortest path to w.•If w S, then D[w] is the length of the shortest special path from s to w.

Dijkstra’s algorithmL[i,j]: the weight along path from i to j. D[k]: length of the shortest special path from source to k.j in C : shortest path from source to j is not known. k in S : shortest path from source to k is known.

Dijkstra (L[1…n, 1…n]) : array [2…n] array D [2…n]C {2,3…n} for i 2 to n do D[i] L[1,i] repeat n - 2 times v some element of C minimizing D[v] C C \ {v} for each w in C do D[w] min (D[w] , D[v] + L[v,w])return D

Inner loop

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v-

C{2,3,4,5}

D[50, 30, 100, 10]

Choose node from C minimizing D (node 5)

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C{2,3,4,5}{2,3,4}

D[50, 30, 100, 10][50, 30, 20, 10]

Compute D for another node in C

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C{2,3,4,5}{2,3,4}

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Compute D for another node in C

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C{2,3,4,5}{2,3,4}

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Compute D for another node in C

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C{2,3,4,5}{2,3,4}

D[50, 30, 100, 10][50, 30, 20, 10]

Choose node from C minimizing D (node 4)

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C{2,3,4,5}{2,3,4}{2,3}

D[50, 30, 100, 10][50, 30, 20, 10][50, 30, 20, 10]

Compute D for another node in C

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C{2,3,4,5}{2,3,4}{2,3}

D[50, 30, 100, 10][50, 30, 20, 10][40, 30, 20, 10]

Compute D for another node in C

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C{2,3,4,5}{2,3,4}{2,3}

D[50, 30, 100, 10][50, 30, 20, 10][40, 30, 20, 10]

Choose node from C minimizing D (node 3)

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C{2,3,4,5}{2,3,4}{2,3}{2}

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Compute D for another node in C (last one)

To find not only the length, but where they pass:•add an array P[2..n], where P[v] contains v’s predecessor•follow the pointers P backwards from destination to source.

Modify the code:

•initialize P[i] to 1, for I=2,3,…,n

•replace the contents of the inner for loop by

if D[w]>D[v] + L[v,w] thenD[w]D[v]+L[v,w]

P[w]v

Special Path

A path from the source to node x is special if all nodes in the path from the source to x belong to S.

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We prove by mathematical induction that•(a) If w1 and w S, then D[w] is the length of the shortest path to w from the source.•(b) If w S, then D[w] is the length of the shortest special path from s to w.

Basis: Initially, only s S, so (a) holds (vacuously). (b) is true because the only special path from s is the direct path and D is initialized accordingly.

Induction hypothesis: Conditions (a) and (b) hold before a node v is added to S.

Induction step for (a): For nodes already in S nothing changes, so (a) is true for them. Before adding v to S we must check that D[v] is the length of the shortest path from s to v.By the induction hypothesis, D[v] is the shortest special path, so we must verify that the shortest path from s to v does not go outside of S.

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The shortest path.

The shortest special path.

Suppose the contrary; the shortest path from s is not special, that is it goes outside S, and the first node encountered is x.

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The shortest path.

The shortest special path.

Since the path from 1…x is special, D[x] = length of shortest special pathfrom 1 … x by (b).

D[x], by induction hypothesis (b)

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So total distance from 1…x…v > 1…x (since … what?) and D[x] > D[v] since D[v] was chosen first. Therefore, D[x]+length(x…v) > D[v].

D[x], by induction hypothesis on (b)

Dijkstra (L[1…n, 1…n]) : array [2…n] array D [2…n]C {2,3…n} for i 2 to n do D[i] L[1,i] repeat n - 2 times v some element of C minimizing D[v] C C \ {v} for each w in C do D[w] min (D[w] , D[v] + L[v,w])return D

Analysis

)(nO

Dijkstra (L[1…n, 1…n]) : array [2…n] array D [2…n]C {2,3…n} for i 2 to n do D[i] L[1,i] repeat n - 2 times v some element of C minimizing D[v] C C \ {v} for each w in C do D[w] min (D[w] , D[v] + L[v,w])return D

Analysis

2nO

Therefore: 2nO