EMIS 8374 Dijkstra

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EMIS 8374 Dijkstras Algorithm Updated 18
February 2008
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Dijkstras Algorithm

• Network may have cycles, but all arc lengths must
  be nonnegative.
• Maintains a partition of N into two subsets
• Set P permanently labeled nodes
• Set T temporarily labeled nodes
• Moves nodes from T into P one at a time in
  non-decreasing order by shortest-path distance
  from the source node

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Dijkstras Algorithm (pg 109)
begin P T N d(i) ? for each
node i in N d(s) 0 and pred(s) 0
while P lt n do begin pick i in T with
minimum d(i) value move i from T to P
for each (i, j) in A(i) do if d(j) gt
d(i) cij then d(j) d(i) cij
pred(j) i end end
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Dijkstras Algorithm Example
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P , T 1, 2, 3, 4, 5, 6
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Dijkstras Algorithm Example
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P 1, T 2, 3, 4, 5, 6
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Dijkstras Algorithm Example
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P 1,4, T 2, 3, 5, 6
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Dijkstras Algorithm Example
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P 1,4,3, T 2, 5, 6
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Dijkstras Algorithm Example
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P 1,4,3,5, T 2, 6
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Dijkstras Algorithm Example
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P 1,4,3,5,2, T 6
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Dijkstras Algorithm Example
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P 1,4,3,5,2,6, T
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Shortest Path Tree
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Shortest Path Tree
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Correctness of Dijkstras Algorithm

• At the end of any iteration the following
  statements hold
• The distance label d(i) is optimal for any node i
  in the set P. That is, d(i) d(i) for all i in
  P.
• The distance label d(i) for any node i in T is
  the length of a shortest path from the source to
  i such that all internal nodes are in P.

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Statement 2 Example d(6) after scanning node 3
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Shortest path in Statement 2 uses internal nodes
1, 4, and 3
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Statement 2 Example d(6) after scanning node 5
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Shortest path in Statement 2 uses internal nodes
1, 4, and 5
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Proof

• Statements 1 and 2 are clearly true after the
  first iteration.
• Assume they are true after iteration k and prove
  that they hold after iteration k1

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Base Cases for Proof by Induction

• After the first iteration the only node in P is
  the source with d(s) 0.
• After the first iteration there are two cases for
  a node j in T
• If j is in A(s), then d(j) csj
• If j is not in A(s), then d(j) ?.

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Illustration of Base Cases
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P 1, T 2, 3, 4, 5, 6
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Proof that Statement 1 Holds after Iteration k1

• Let i be the node moved from T to P in iteration
  k1
• Need to show d(i) d(i)
• Idea
• Let B be a path from s to i
• Let L(B) be the length of B
• Show that d(i) L(B)

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Proof that Statement 1 Holds after Iteration k1

• There are two cases to consider for B
• All internal nodes of B are in P
• At least one internal node of B is in T
• Case 1 Statement 2 implies that d(i) L(B)

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Proof that Statement 1 Holds after Iteration k1

• Case 2
• Let j be first internal node of B that is in T
• By Statement 2, the length of the subpath of B
  from s to j is at least d(j)
• Since all arc lengths are non-negative, the
  length of the subpath of B from j to i is at
  least zero
• Thus, L(B) d(j)
• Since node i was picked to move to P, d(i) d(j)
• Therefore, d(i) L(B)

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Diagram for Case 2
P
i
s
j
Since Statement 2 holds for the first k
iterations, d(j) length of the subpath of B
from s to j.
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Diagram for Case 2
P
i
s
j
Since the arc lengths are non-negative, the
length of the subpath of B from j to i is at
least zero.
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Diagram for Case 2
P
i
s
j
The length of path B from s to i is at least d(j).
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Diagram for Case 2
P
i
s
j
Since d(i) d(j), the length of path B from s to
i is at least d(i).
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Diagram for Case 2
P
i
s
j
Since d(i) L(B) for any path B from s to i ,
d(i) d(i).
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Proof of Statement 2

• When node i moves from T to P, the algorithm
  changes pred(j) to i and sets d(j) d(i) cij
  for all j in T such that d(j) gt d(i) cij .
• Thus, the path defined by the predecessor indices
  from j to s satisfies Property 4.2.
• Therefore, d(j) is the length of a shortest path
  from s to j subject to the restriction that each
  internal node in the path is in P.

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Diagram for Statement 2
T
P
j
pred(j)
s
i
Move node i from T to P
If d(j) ? d(i) cij then Statement 2 still holds
for node j.
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Diagram for Statement 2
T
P
old pred(j)
j
s
i
new pred(j)
Move node i from T to P
If d(j) gt d(i) cij then d(j) gets reset to d(i)
cij
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Complexity of Dijkstras Algorithm

• Node selections
• Must compare the distance labels of all nodes in
  T O(n (n-1) (n-2) 1) O(n2)
• Total work for scanning nodes is O(m)
• Overall complexity is O(n2 m) O(n2)
• Can be improved to O(m n log n) with Fibonacci
  heaps