1. Discrete Math 2 Shortest Path Using Matrix
2001
Discrete Math 2 Shortest Path Using Matrix
CIS112
February 10, 2007
Daniel L. Silver

2. Overview Previously: Now: In weighted graph . .
Shortest path from one vertex to another
Search tree method
Now:
Same problem
Matrix method
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3. Strategy Represent weighted graph as matrix Create search matrix . .
with entries matching
nodes expansion
node production
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4. Matrix for Weighted Graph
vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 17 16 19 15 24 20 14 21 25 22
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5. Step #0 Create a search matrix Layout same as weighted graph matrix
Entries will hold path information
Vertices along path
Total cost of path
Path info built up step by step
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6. Search Matrix
vtx 1 2 3 4 5 6 7 8 9 10 11 12
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7. Step #1 Enter information for first path segment
Initial entry goes in row #7 . .
Since #7 is starting vertex
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8. Search Matrix – Step #1 vtx 1 2 3 4 5 6 7 8 9 10 11 12 15 25 2007
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9. Comment Notice similarity to search tree Choose lowest cost entry
Which is in column #3
Corresponds to expanding lowest cost node
Next entry goes in row #3
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10. Search Matrix – Step #2 vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 34 15 25
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11. Comment Entries represent path cost Path elements stored implicitly
15+8=23
15+19=34
Path elements stored implicitly
Look in row #7 (starting point)
Find lowest cost entry (column #3)
In corresponding row
Find lowest cost entry (column #1)
Path is: 7  1
Again, note similarity to search tree
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12. Step #3 Among all leaves . . Find lowest cost entry [7,10]
“Expand” “node” #10
I.e., compute path cost + edge cost
Enter into matrix
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13. Search Matrix – Step #3 vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 34 15 25 2421
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14. Comment No entry is made for starting point – #7
There are two entries in column #8
Correspondence to search tree
Two nodes for #8
So we mark higher one as deleted
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15. Search Matrix – Step #3b vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 34 15 25 2421
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16. Continuing . . Keep choosing “nodes” Keep “expanding” them
Until there are no more to “expand”
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17. Search Matrix – Step #4 vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 34 15 25 2421 44 43 26
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18. Search Matrix – Step #4b vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 34 15 25 2421 44 43 26
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19. Comment Reached #12, but not yet finished
Other less costly path possible
But not through #4 & #9
Why not?
So mark #4 & #9 as dead ends
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20. Search Matrix – Step #4c vtx 1 2 3 4 5 6 7 8 9 10 11 12 23 34x 15 2524 21 44
43x 26
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21. Step #5 Now expand #1 . . Marking dead ends, if they occur 2007
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22. Search Matrix – Step #5 vtx 1 2 3 4 5 6 7 8 9 10 11 12 46x 28x 23 34x15 25 24 21 44
43x 26
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23. Comment All entries in row #1 are dead ends
So the column is also a dead end
Then the same happens . .
with row #3
and column #3
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24. Search Matrix – Step #5b vtx 1 2 3 4 5 6 7 8 9 10 11 12 46x 28x 23x25 15 24 21 44
43x 26
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25. Step #6 And expand #8 . . Marking dead ends, if they occur 2007
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26. Search Matrix – Step #6 vtx 1 2 3 4 5 6 7 8 9 10 11 12 46x 28x 23x 34x25 15 35
48x 47
24x 21 44
43x 26
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27. Comment We are finished . . since there are no more nodes to expand
We already have the path cost = 26
How do we get the path?
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28. To Get the Path Start at row 7 Column 10 has an entry . . Go to row 10
giving us 7  10
Go to row 10
Column has 11 an entry . .
giving us 7  10  11
Go to row 11
Column has 12 an entry . .
giving us 7  10  11  12
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29. Shortest Path from 7 to 12 7  10  11  12 :: 26 2007
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30. Final Comments
We see how the 1-1 search tree can be implemented as a matrix
We look next at how the 1-many can also
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