Dijkstra

1
Dijkstras Shortest Paths
CS 312 Lecture 4
2
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

3
Objectives

• Follow up
• big-O, big-Theta and big-Omega
• Prims algorithm outside of class.
• Understand the shortest paths problem.
• Understand Dijkstras algorithm
• Analyze Dijkstras algorithm

4
Big-Letters
5
THE PROBLEM

• Given
• 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 Dijkstras Algorithm.
6

• Dijkstras Algorithm Overview
• Maintain
• a set S of vertices whose minimum distance from
  the source is known,
• a set C of candidate vertices (S?CN, S?C?).
• Initially, Ss. (the source)
• Keep adding vertices to S, choosing from C the
  node whose distance from the source is least.
• Eventually, SN.
• Special path a path from s with all
  intermediate nodes in S.
• Maintain an array D with
• If w ? S, then Dw is the length of the shortest
  path to w.
• If w ? S, then Dw is the length of the shortest
  special path from s to w.

1
10
50
30
2
5
100
5
10
20
3
4
50
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Dijkstras algorithm
Li,j the weight along path from i to j.
Dk 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 (L1n, 1n)
array 2n array D 2n C ? 2,3n for i ? 2
to n do Di ? L1,i repeat n - 2 times v
? some element of C minimizing Dv C ? C \
v for each w in C do Dw ? min
(Dw , Dv Lv,w) return D
Inner loop
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Choose node from C minimizing D (node 5)
1
Step Init.
v -
C 2,3,4,5
D 50, 30, 100, 10
10
50
30
2
5
100
5
10
20
3
4
50
9
Compute D for another node in C
1
Step Init. 1
v - 5
C 2,3,4,5 2,3,4
D 50, 30, 100, 10 50, 30, 20, 10
10
50
30
2
5
100
5
10
20
3
4
50
10
Compute D for another node in C
1
Step Init. 1
v - 5
C 2,3,4,5 2,3,4
D 50, 30, 100, 10 50, 30, 20, 10
10
50
30
2
5
100
5
10
20
3
4
50
11
Compute D for another node in C
1
Step Init. 1
v - 5
C 2,3,4,5 2,3,4
D 50, 30, 100, 10 50, 30, 20, 10
10
50
30
2
5
100
5
10
20
3
4
50
12
Choose node from C minimizing D (node 4)
1
Step Init. 1
v - 5
C 2,3,4,5 2,3,4
D 50, 30, 100, 10 50, 30, 20, 10
10
50
30
2
5
100
5
10
20
3
4
50
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Compute D for another node in C
1
Step Init. 1 2
v - 5 4
C 2,3,4,5 2,3,4 2,3
D 50, 30, 100, 10 50, 30, 20, 10 50, 30, 20,
10
10
50
30
2
5
100
5
10
20
3
4
50
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Compute D for another node in C
1
Step Init. 1 2
v - 5 4
C 2,3,4,5 2,3,4 2,3
D 50, 30, 100, 10 50, 30, 20, 10 40, 30, 20,
10
10
50
30
2
5
100
5
10
20
3
4
50
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Choose node from C minimizing D (node 3)
1
Step Init. 1 2
v - 5 4
C 2,3,4,5 2,3,4 2,3
D 50, 30, 100, 10 50, 30, 20, 10 40, 30, 20,
10
10
50
30
2
5
100
5
10
20
3
4
50
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Compute D for another node in C (last one)
1
Step Init. 1 2 3
v - 5 4 3
C 2,3,4,5 2,3,4 2,3 2
D 50, 30, 100, 10 50, 30, 20, 10 40, 30, 20,
10 35, 30, 20, 10
10
50
30
2
5
100
5
10
20
3
4
50
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• To find not only the length, but where they pass
• add an array P2..n, where Pv contains vs
  predecessor
• follow the pointers P backwards from destination
  to source.

• Modify the code
• initialize Pi to 1, for I2,3,,n
• replace the contents of the inner for loop by
• if DwgtDv Lv,w thenDw?DvLv,w
• 
  Pw?v

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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.
1
10
50
S 4,5 then path 1,5,4 is special and path
1,4,3 is special but path 1,3,2 is not.
30
2
5
100
5
10
20
3
4
50
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• We prove by mathematical induction that
• (a) If w?1 and w ? S, then Dw is the length of
  the shortest path to w from the source.
• (b) If w ? S, then Dw 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 Dv is the
  length of the shortest path from s to v.
• By the induction hypothesis, Dv 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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Suppose the contrary the shortest path from s is
not special, that is it goes outside S, and the
first node encountered is x.
x
s
v
The shortest path.
The shortest special path.
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Since the path from 1x is special, Dx length
of shortest special path from 1 x by (b).
Dx, by induction hypothesis (b)
x
s
v
The shortest path.
The shortest special path.
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So total distance from 1xv gt 1x (since
what?) and Dx gt Dv since Dv was chosen
first. Therefore, Dxlength(xv) gt Dv.
Dx, by induction hypothesis on (b)
x
s
v
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Dx, by induction hypothesis on (b)
x
s
v
Total distance to v via x gt Dx, (lengths are
positive) Dx gt Dv, (the algorithm chose v
before x)
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Suppose the contrary the shortest path from s
goes outside S, and the first node encountered is
x.
Dx, by induction hypothesis on (b)
x
s
v
Total distance to v via x gt Dx, (lengths are
positive) Dx gt Dv, (the algorithm chose v
before x) ? Total distance to v via x gt Dv,
(transitivity) Contradiction! (The
shortest path is longer than the shortest special
path ???)
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Analysis
Dijkstra (L1n, 1n) array 2n array D
2n C ? 2,3n for i ? 2 to n do Di ?
L1,i repeat n - 2 times v ? some element
of C minimizing Dv C ? C \ v for
each w in C do Dw ? min (Dw , Dv
Lv,w) return D
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Analysis
Dijkstra (L1n, 1n) array 2n array D
2n C ? 2,3n for i ? 2 to n do Di ?
L1,i repeat n - 2 times v ? some element
of C minimizing Dv C ? C \ v for
each w in C do Dw ? min (Dw , Dv
Lv,w) return D
Therefore