Visualizations
Label Correcting Algorithm
An Example
∞
∞
2
4
2
3
Initialize
d(1) := 0;
d(j) := ∞ for j ≠ 1
0
6
1
3
3
1
∞
∞
7
-2
5
3
2
4
3
-4
3
∞
6
∞
In next slides: the number inside the node will be d(j).
Violating arcs will be in thick lines.
2
2
∞
∞
3
3
Generic Step
6
0
-2
∞
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
∞
-4
∞
3
∞
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
2
∞
3
3
Generic Step
6
0
-2
∞
6
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
∞
-4
∞
3
∞
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
4
2
∞
3
3
Generic Step
-2
6
0
∞
6
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
∞
3
-4
3
∞
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
5
2
∞
5
3
3
Generic Step
-2
6
0
∞
6
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
-4
3
3
∞
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
6
2
5
3
3
Generic Step
-2
6
0
∞
6
4
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
-4
3
3
∞
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
7
2
5
3
3
Generic Step
-2
6
0
∞
6
4
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
-4
3
3
∞
6
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
8
2
5
3
3
Generic Step
-2
6
0
∞
4
6
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
∞
2
3
3
-4
6
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
9
2
5
3
3
Generic Step
-2
6
0
∞
9
4
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
2
4
3
-4
2
6
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
10
2
5
3
3
Generic Step
-2
6
0
4
3
An arc (i,j) is
violating if
d(j) > d(i) + cij.
3
3
1
9
2
4
3
-4
2
6
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
No arc is violating
The distance labels are optimal
We now show the
predecessor arcs.
11
MITOpenCourseWare
http://ocw.mit.edu
15.082J / 6.855J / ESD.78J Network Optimization
Fall 2010
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