1. Lecture 10 The Label Correcting Algorithm

8. Label Correcting Algorithm

9. An Example 3 Initialize   d(1) := 0; d(j) :=  for j  1 1 2 4 5 36 7 2 3 3 1 6 -2 3 2 -4 43   0  In next slides: the number inside the node will be d(j). Violating arcs will be in thick lines.

10. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3

11. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6

12. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3

13. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3 5

14. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3 5 4

15. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3 5 4 6

16. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3 5 4 62

17. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3 5 4 62 9

18. An Example 3 Generic Step An arc (i,j) is violating if d(j) > d(i) + c ij. 0      2 3 3 1 6 -2 3 2 -4 43 Pick a violating arc (i,j) and replace d(j) by d(i) + c ij. 3 6 3 5 4 62 9 No arc is violating The distance labels are optimal We now show the predecessor arcs.

30. Modified Label Correcting Algorithm

31. The Modified Label Correcting Algorithm 3 Initialize   d(1) := 0; d(j) :=  for j  1 1 2 5 4 36 7 2 3 3 1 6 -2 3 2 -4 43   0  In next slides: the number inside the node will be d(j). LIST := {1}

32. An Example 3 Generic Step Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 LIST := {1}LIST := { } 6 LIST := { 2 }LIST := { 2, 3 } 3 LIST := { 2, 3, 4 } 0

33. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 LIST := {1}LIST := { } 6 LIST := { 2 }LIST := { 2, 3 } 3 LIST := { 2, 3, 4 } 3 LIST := { 3, 4 } 4 5 LIST := { 3, 4, 5 }

34. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 6 3 4 5 LIST := { 3, 4, 5 } 3 LIST := { 4, 5 }

35. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 6 3 4 5 LIST := { 3, 4, 5 } LIST := { 4, 5 } 4 LIST := { 5 } 6 LIST := { 5, 6 }

36. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 6 3 4 5 LIST := { 3, 4, 5 } LIST := { 4, 5 }LIST := { 5 } 6 LIST := { 5, 6 } 5 LIST := { 6 }

37. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 6 3 4 5 LIST := { 3, 4, 5 } LIST := { 4, 5 }LIST := { 5 } 6 LIST := { 5, 6 }LIST := { 6 } 6 LIST := { } 2 LIST := { 3 } 9 LIST := { 3, 7 }

38. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 6 3 4 5 LIST := { 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 }

39. An Example 3 Take a node i from LIST 0      2 3 3 1 6 -2 3 2 -4 43 Update(i): for each arc (i,j) with d(j) > d(i) + c ij replace d(j) by d(i) + c ij. 3 6 3 4 5 LIST := { 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 := { }

40. An Example 3 LIST is empty. The distance labels are optimal 0      2 3 3 1 6 -2 3 2 -4 43 3 6 3 4 5 LIST := { 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