Lecture 10 The Label Correcting Algorithm.
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Lecture 10 The Label Correcting Algorithm
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Transcript of Lecture 10 The Label Correcting Algorithm.
Lecture 10
The Label Correcting Algorithm
Label Correcting Algorithm
An Example
3
Initialize
d(1) := 0; d(j) := for j 1
1
2 4
5
3 6
7
2
331
6 -2
32
-443
0
In next slides: the number inside the node will be d(j).
Violating arcs will be in thick lines.
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
5
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
5
4
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
5
4
6
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
5
4
62
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
5
4
62
9
An Example
3
Generic Step
An arc (i,j) is violating if d(j) > d(i) + cij.
0
2
331
6 -2
32
-443
Pick a violating arc (i,j) and replace d(j) by d(i) + cij.
3
6
3
5
4
62
9
No arc is violating
The distance labels are optimal
We now show the predecessor arcs.
Modified Label Correcting Algorithm
The Modified Label Correcting Algorithm
3
Initialize
d(1) := 0; d(j) := for j 1
1
2 5
4
3 6
7
2
331
6 -2
32
-443
0
In next slides: the number inside the node will be d(j).
LIST := {1}
LIST := { 1 }
An Example
3
Generic Step
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3LIST := {1}LIST := { }
6
LIST := { 2 }LIST := { 2, 3 }
3
LIST := { 2, 3, 4 }
0
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3LIST := {1}LIST := { }
6
LIST := { 2 }LIST := { 2, 3 }
3
LIST := { 2, 3, 4 } 3LIST := { 3, 4 }
4
5LIST := { 3, 4, 5 }
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3
6
3
4
5LIST := { 3, 4, 5 }
3
LIST := { 4, 5 }
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3
6
3
4
5LIST := { 3, 4, 5 }LIST := { 4, 5 }
4
LIST := { 5 }
6
LIST := { 5, 6 }
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3
6
3
4
5LIST := { 3, 4, 5 }LIST := { 4, 5 }LIST := { 5 }
6
LIST := { 5, 6 } 5LIST := { 6 }
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3
6
3
4
5LIST := { 3, 4, 5 }LIST := { 4, 5 }LIST := { 5 }
6
LIST := { 5, 6 }LIST := { 6 }
6
LIST := { }
2
LIST := { 3 }
9
LIST := { 3, 7 }
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3
6
3
4
5LIST := { 3, 4, 5 }LIST := { 4, 5 }LIST := { 5 }
6
LIST := { 5, 6 }LIST := { 6 }LIST := { }
2
LIST := { 3 }
9
LIST := { 3, 7 }
2
LIST := { 7 }
An Example
3
Take a node i from LIST
0
2
331
6 -2
32
-443
Update(i): for each arc (i,j) with d(j) > d(i) + cij
replace d(j) by d(i) + cij.
3
6
3
4
5LIST := { 3, 4, 5 }LIST := { 4, 5 }LIST := { 5 }
6
LIST := { 5, 6 }LIST := { 6 }LIST := { }
2
LIST := { 3 }
9
LIST := { 3, 7 }LIST := { 7 }
9
LIST := { }
An Example
3
LIST is empty.
The distance labels are optimal
0
2
331
6 -2
32
-443
3
6
3
4
5LIST := { 3, 4, 5 }LIST := { 4, 5 }LIST := { 5 }
6
LIST := { 5, 6 }LIST := { 6 }LIST := { }
2
LIST := { 3 }
9
LIST := { 3, 7 }LIST := { 7 }LIST := { }
Here are the predecessors
