1. A Shortest Path Algorithm

2. Motivation Given a connected, positive weighted graph
Find the length of a shortest path from vertex a to vertex z.

3. Dijkstra’s Shortest Path Algorithm
Input: A connected, positive weighted graph,vertices a and z
Output: L(z), the length of a shortest path from a to z
Dijkstra(w,a,z,L){
L(a)=0
for all vertices x ≠a
L(x)=∞
T=set of all vertices
while(z є T){
choose v є T with minimum L(v)
T=T-{v}
for each x є T adjacent to v
L(x)=min{L(x),L(v)+w(v,x)} }

4. Example 8.4.2 b 2 c 1 2 4 2 3 a d e z 4 3 7 1 6 f g 5
Find L(z)

5. 1.L(a)=0 2.L(x)=∞,x≠a 3.T={a,b,c,d,e,f,g,z}
Initialization ∞ ∞ b 2 c 1 2 4 2 3 ∞ ∞ a d e z ∞ 4 3 7 1 6 f g 5 ∞ ∞
1.L(a)=0
2.L(x)=∞,x≠a
3.T={a,b,c,d,e,f,g,z}

6. 1.L(a)=min{L(x),xєT} 2.T={b,c,d,e,f,g,z} 3.L(b)=2,L(f)=1
Iteration 1 ∞ 2 ∞ b 2 c 1 2 4 2 3 ∞ ∞ a d e z ∞ 4 3 7 1 6 f g 5 ∞ ∞ 1
1.L(a)=min{L(x),xєT}
2.T={b,c,d,e,f,g,z}
3.L(b)=2,L(f)=1

7. 1.L(f)=min{L(x),xєT} 2.T={b,c,d,e,g,z} 3.L(d)=4,L(g)=6
Iteration 2 2 ∞ b 2 c 1 2 4 2 3 4 ∞ ∞ a d e z ∞ 4 3 7 1 6 f g 5 ∞ 1 6
1.L(f)=min{L(x),xєT}
2.T={b,c,d,e,g,z}
3.L(d)=4,L(g)=6

8. 1.L(b)=min{L(x),xєT} 2.T={c,d,e,g,z} 3.L(c)=4,L(e)=6,L(d)=4
Iteration 3 2 4 ∞ b 2 c 1 2 4 2 3 4 6 ∞ a d e z ∞ 4 3 7 1 6 f g 5 6 1
1.L(b)=min{L(x),xєT}
2.T={c,d,e,g,z}
3.L(c)=4,L(e)=6,L(d)=4

9. Iteration 4 1.L(c)=L(d)=min{L(x),xєT},begin with c first 2.T={d,e,g,z}6 4 6 a d e z 5 ∞ 4 3 7 1 6 f g 5 6 1
1.L(c)=L(d)=min{L(x),xєT},begin with c first
2.T={d,e,g,z}
3.L(e)=6,L(z)=5 ……
We can continue, but L(z) will not change in this example.

10. Proof of Dijkstra’s Algorithm
Basic Step(i=1):
we set L(a)=0, and L(a) is sure the length of a shortest path from a to a.
Inductive step: For an arbitrary step i
Suppose for step k<i, L(v) is the length of a shortest path from a to v.
Next, suppose that at the ith step we choose v in T with minimum L(v). We will seek a contradiction that if there is a w whose length is less than L(v) then w is not in T.
By way of contradiction, suppose there is a w with L(w)<L(v), wєT. Then, let P be the shortest path from a to w, and let x be the vertex nearest to a on P that is in T and let u be x’s predecessor. The node u must not be in T (because x was the nearest node to a that was in T). By assumption, L(u) was the length of the shortest path from a to u.
Then, L(x) ≤ L(u)+w(u,x) ≤ length of P < L(v). This is a contradiction. So w is not in T.
According to our assumption, every path from a to v has length at least L(v). a u x w … P

11. Example 2 b 3 c 2 2 a 1 z 1 2 d e 1
Find L(z)

12. 1.L(a)=0 2.L(x)=∞,x≠a 3.T={a,b,c,d,e,z} Initialization ∞ ∞ b 3 c 2 2 1∞ 1 2 d e 1 ∞ ∞
1.L(a)=0
2.L(x)=∞,x≠a
3.T={a,b,c,d,e,z}

13. 1.L(a)=min{L(x),xєT} 2.T={b,c,d,e,z} 3.L(b)=2,L(d)=1
Iteration 1 ∞
a,2 ∞ b 3 c 2 2 1 a z ∞ 1 2 d e 1 ∞ ∞
a,1
1.L(a)=min{L(x),xєT}
2.T={b,c,d,e,z}
3.L(b)=2,L(d)=1

14. 1.L(d)=min{L(x),xєT} 2.T={b,c,e,z} 3.L(e)=2
Iteration 2
a,2 ∞ b 3 c 2 2 1 a z ∞ 1 2 d e 1 ∞
a,1
d,2
1.L(d)=min{L(x),xєT}
2.T={b,c,e,z}
3.L(e)=2

15. 1.L(b)=min{L(x),xєT},(a,2)<(d,2)
Iteration 3
a,2
b,5 ∞ b 3 c 2 2 1 a z ∞ 1 2 d e 1
d,2
a,1
1.L(b)=min{L(x),xєT},(a,2)<(d,2)
2.T={c,e,z}
3.L(c)=5,L(e)=2

16. 1.L(e)=min{L(x),xєT} 2.T={c,z} 3.L(z)=4
Iteration 4
a,2
b,5 b 3 c 2 2 1 a z ∞
e,4 1 2 d e 1
d,2
a,1
1.L(e)=min{L(x),xєT}
2.T={c,z}
3.L(z)=4

17. 1.L(z)=min{L(x),xєT} 2.T={c} 3.L(c)=5.Stop.
Iteration 5
a,2
b,5 b 3 c 2 2 1 a z
e,4 1 2 d e 1
d,2
a,1
1.L(z)=min{L(x),xєT}
2.T={c}
3.L(c)=5.Stop.

18. Theorem 8.4.5
For input consisting of an n-vertex, simple, connected, weighted graph, Dijkstra’s algorithm has worst-case run time Ѳ(n2).
Proof: The while loop will take Ѳ(n2) worst-case running time.