1 introduction to algorithms l ecture 17 (chap. 24) shortest paths i 24.2 single-source shortest...

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Introduction to Algorithms

LECTURE 17 (Chap. 24)Shortest paths I• 24.2 Single-source shortest paths

• 24.3 Dijkstra’s algorithm

Edsger Wybe Dijkstra1930-2002

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Edsger Wybe Dijkstra (May 11, 1930 – August 6, 2002)

• Born in Rotterdam, Dijkstra studied theoretical physics at Leiden University, but he quickly realized he was more interested in computer science.

• Originally employed by the Mathematisch Centrum in Amsterdam, he held a professorship at the Eindhoven University of Technology in the Netherlands, worked as a research fellow for Burroughs Corporation in the early 1970s, and later held the Schlumberger Centennial Chair in Computer Sciences at The University of Texas at Austin, in the United States. He retired in 2000

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Paths in graphs

Consider a digraph G = (V, E) with edge-weightfunction w : E → . The weight of path p = v1 →v2 → → vk is defined to be

Example :

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Shortest paths

A shortest path from u to v is a path ofminimum weight from u to v. The shortest-path weight from u to v is defined as

(u, v) = min{w(p) : p is a path from u to v}.

Note: (u, v) = if no path from u to v exists.

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Optimal substructure

Theorem. A subpath of a shortest path is ashortest path.

Proof. Cut and paste:

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Triangle inequality

Theorem. For all u, v, x V, we have(u, v) (u, x) + (x, v).

Proof.

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Well-definedness of shortest paths

If a graph G contains a negative-weight cycle,then some shortest paths may not exist.

Example:

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Single-source shortest pathsProblem. From a given source vertex s V, findthe shortest-path weights (s, v) for all v V.

If all edge weights w(u, v) are nonnegative, allshortest-path weights must exist.

IDEA: Greedy. 1. Maintain a set S of vertices whose shortest-path distances from s are known.2. At each step add to S the vertex v V – S whose distance estimate from s is minimal.3. Update the distance estimates of vertices adjacent to v.

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d[s] ← 0for each v V – {s} do d[v] ← S ← Q ← V ⊳ Q is a priority queue maintaining V – Swhile Q ≠ do u ← Extract-Min(Q) S ← S {u} for each v Adj[u] do

Dijkstra’s algorithm

Implicit DECREASE-KEY

relaxationstepif d[v] > d[u] + w(u, v)

then d[v] ← d[u] + w(u, v)

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Example of Dijkstra’salgorithm

Graph withnonnegativeedge weights:

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Example of Dijkstra’salgorithm

S : {}

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Example of Dijkstra’salgorithm

S : {A}

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Example of Dijkstra’salgorithm

S : {A}

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Example of Dijkstra’salgorithm

S : {A,C}

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Example of Dijkstra’salgorithm

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Example of Dijkstra’salgorithm

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Example of Dijkstra’salgorithm

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Example of Dijkstra’salgorithm

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Example of Dijkstra’salgorithm

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Example of Dijkstra’salgorithm

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Correctness — Part ILemma. Initializing d[s] ← 0 and d[v] ← for allv V – {s} establishes d[v] ≥ (s, v) for all v V,and this invariant is maintained over any sequenceof relaxation steps.Proof. Suppose not. Let v be the first vertex forwhich d[v] < (s, v), and let u be the vertex thatcaused d[v] to change: d[v] = d[u] + w(u, v). Then, d[v] < (s, v) supposition ≤ (s, u) + (u, v) triangle inequality ≤ (s,u) + w(u, v) sh. path ≤ specific path ≤ d[u] + w(u, v) v is first violationContradiction.

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Correctness — Part IITheorem. Dijkstra’s algorithm terminates withd[v] = (s, v) for all v V.Proof. It suffices to show that d[v] = (s, v) for everyv V when v is added to S. Suppose u is the firstvertex added to S for which d[u] ≠ (s, u). Let y be the first vertex in V – S along a shortest path from s to u, and let x be its predecessor:

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Correctness — Part II(continued)

Since u is the first vertex violating the claimed invariant,we have d[x] = (s, x). Since subpaths of shortest pathsare shortest paths, it follows that d[y] was set to (s, x) +w(x, y) = (s, y) when (x, y) was relaxed just after x wasadded to S. Consequently, we have d[y] = (s, y) ≤ (s, u)≤ d[u]. But, d[u] ≤ d[y] by our choice of u, and hence d[y]= (s, y) = (s, u) = d[u]. Contradiction.

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Analysis of Dijkstra

|V|times degree(u

)times

while Q ≠ do u ← EXTRACT-MIN(Q) S ← S {u} for each v Adj[u]

do if d[v] > d[u] + w(u, v) then d[v] ← d[u] + w(u, v)

Handshaking Lemma (E) implicit DECREASE-KEY’s.

Time = (V)·TEXTRACT-MIN + (E)·TDECREASE-KEY

Note: Same formula as in the analysis of Prim’sminimum spanning tree algorithm.

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Analysis of Dijkstra(continued)

Time = (V)·TEXTRACT-MIN + (E)·TDECREASE-KEY

Q TEXTRACT-MIN TDECREASE-KEY Total

array O(V) O(1) O(V2)

binaryheap

O(lg V) O(lg V) O(E lg V)

Fibonacciheap

O(lg V)amortized

O(1)amortized

O(E + V lg V)worst case

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Unweighted graphsSuppose w(u, v) = 1 for all (u, v) E. Can thecode for Dijkstra be improved?

• Use a simple FIFO queue instead of a priority queue.• Breadth-first search

while Q ≠ do u ← DEQUEUE(Q) for each v Adj[u] do if d[v] = then d[v] ← d[u] + 1 ENQUEUE(Q, v)

Analysis: Time = O(V + E).