1. CS1022 Computer Programming & Principles
Lecture 8.2
Digraphs (2)

2. Plan of lecture Paths in digraphs Reachability matrix
Warshall’s algorithm
Shortest path
Weight matrix
Dijkstra’s algorithm
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3. Paths in digraphs (1) Directed graphs used to represent
Routes of airlines
Connections between networked computers
What would happen if a link (vertex or arc) is lost?
If city is unreachable (due to poor weather) and a plane needs refuelling, it may be impossible to re-route plane
If a path in a computer network is lost, users may no longer be able to access certain file server
Problem: is there a path between two vertices of a digraph?
Solution: try every combination of edges...
We can do better than this!
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4. Paths in digraphs (2)
Let G  (V, E) be a digraph with n vertices (|V|  n)
Let M be its adjacency matrix
A T entry in the matrix represents an arc in G
An arc is a path of length 1
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5. Reachability matrix (1)
The logical Boolean product of M with itself is M2
A T entry indicates a path of length 2
M3  M.M.M records all paths of length 3
Mk records all paths of length k
Finally, the reachability matrix:
M*  M or M2 or ... or Mn
Records the existence of paths of some length between vertices
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6. Reachability matrix (2)
Logical or of two matrices is the result of forming the logical or of corresponding entries
This requires that both matrices have the same number of rows and same number of columns
Reachability matrix of G  (V, E) is in fact adjacency matrix of the transitive closure E* on E
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7. Reachability matrix (3)
Calculate the reachability matrix of digraph a c b d
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8. Reachability matrix (4)
So we have
The 3 T entries in M2 indeed correspond to paths of length 2 in G, namely
a b c
a b d
b d c
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9. Reachability matrix (5)
Further calculation gives
Therefore,
For example, T in top-right corner of M* arises from M2 and corresponds to path a b d
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10. Reachability matrix (6)
For large digraphs, calculating M* via higher and higher powers of M is laborious and inefficient
A more efficient way is Warshall’s algorithm
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11. Warshall’s algorithm (1)
Let G be a digraph with vertices v1, v2, , vn
Warshall’s algorithm generates sequence
W0, W1, W2, , Wn (where W0 = M)
For k  1, entries in matrix Wk are Wk(i, j) = T if, and only if, there is a path (of any length) from vi to vj
Intermediary vertices in path are in v1, v2, , vk
Matrix W0 is the original adjacency matrix M
Matrix Wn is the reachability matrix M*
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12. Warshall’s algorithm (2)
Clever use of nested for-loops – very elegant
Each pass of outer loop generates matrix Wk
This is done by updating entries in matrix Wk– 1
begin
W := M;
for k  1 to n do
for i  1 to n do
for j  1 to n do
W(i, j) := W(i, j) or (W(i, k) and W(k, j))
end
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13. Warshall’s algorithm (3)
To find row i of Wk we evaluate
W(i, j) := W(i, j) or (W(i, k) and W(k, j))
If W(i, k) = F then (W(i, k) and W(k, j)) = F
So expression depends on W(i, j)
Row i remains unchanged
Otherwise, W(i, k) = T
Expression reduces to (W(i, j) or W(k, j))
Row i becomes the logical or of the current row i with current row k
W := M;
for k  1 to n do
for i  1 to n do
for j  1 to n do
W(i, j) := W(i, j) or (W(i, k) and W(k, j))
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14. Warshall’s algorithm (4)
To calculate Wk from Wk – 1 proceed as follows
Consider column k in Wk – 1
For each “F” row in this column, copy that row to Wk
For each “T” row in this column, form the logical or of that row with row k, and write the resulting row in Wk
W := M;
for k  1 to n do
for i  1 to n do
for j  1 to n do
W(i, j) := W(i, j) or (W(i, k) and W(k, j))
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15. Warshall’s algorithm (5)
Calculate reachability matrix of digraph
W0  F T 2 1 3 5 4
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16. Warshall’s algorithm (6)2 1 3 5 4
We now calculate W1:
Using step 1 we consider column 1 of W0
Using step 2 we copy rows 1, 2 and 4 directly to W1
W1  F T
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17. Warshall’s algorithm (7)2 1 3 5 4
We now use step 3 – row 3 in W1 is
Logical or of row 3 with row 1 of W0
W0  F T
W1  F T
W1  F T
W1  F T
W1  F T
W1  F T
W1  F T
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18. Warshall’s algorithm (8)2 1 3 5 4
We use step 3 again – row 5 in W1 is
Logical or of row 5 with row 1 of W0
W0  F T F T F T F T F T F T
W1  F T
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19. Warshall’s algorithm (9)2 1 3 5 4
We now compute W2 from W1
Copy rows 2 and 4 to W2
Row 1 in W2 is logical or of rows 1 and 2 of W1
Row 3 in W2 is logical or of rows 3 and 2 of W1
Row 5 in W2 is logical or of rows 5 and 2 of W1 F T F T F T
W1  F T F T F T
W2  F T F T
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20. Warshall’s algorithm (10)
Notice entry (3, 3)
Indicates a cycle from vertex 3 back to itself
Going via vertices 1 and/or 2
W2  F T 2 1 3 5 4
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21. Warshall’s algorithm (11)
W3 is computed similarly:
Since no arcs lead out of vertex 4, W4 is the same as W3
For a similar reason, W5 is the same as W4
So W3 is reachability matrix
W3  T F
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22. Shortest paths (1) Given a digraph with weighted arcs
Find a path between two vertices which has the lowest sum of weights of the arcs travelled along
Weights can be costs, petrol, etc.
We make it simple and think of distances
Hence the “shortest path”, but it could be “cheapest”
You might also want to maximise sum (e.g., taxi drivers)
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23. Shortest paths (2) Suppose the following weighted digraph B D C E F A
Not so many vertices and arcs
We could list all possible paths between any two vertices
We then pick the one with lowest sum of weights of arcs
In real-life scenarios there are too many possibilities
We need more efficient ways to find shortest path B D C E F A 1 2 3 4 5
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24. Shortest paths (3) Dijkstra’s algorithm
Let’s see its “effect” with previous digraph
Problem:
Find shortest path between A and other vertices
Shortest = “minimal total weight” between two vertices
Total weight is sum of individual weights of arcs in path
Edsger W. Dijkstra
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25. Weight matrix (1) A compact way to represent weighted digraphs
Matrix w, whose entries w(u, v) are given by
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26. Weight matrix (2) A B C D E F 2  3 A B C D E F 2  3 1 4 A B C D E F2  3 A B C D E F 2  3 1 4 A B C D E F 2  3 1 4 5 A B C D E F 
w  A B C D E F A B C D E F
Our digraph is represented as B D C E F A 1 2 3 4 5
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27. Dijkstra’s algorithm (1)
For each vertex v in digraph we assign a label d[v]
d[v] represents distance from A to v
Initially d[v] is the weight of arc (A, v) if it exists
Otherwise d[v]  
We traverse vertices and improve d[v] as we go
At each step of the algorithm a vertex u is marked
This is done when we are sure we found a best route to it
For remaining unmarked vertices v,
Label d[v] is replaced by minimum of its current value and distance to v via last marked vertex u
Algorithm terminates when
All vertices have received their final label and
All possible vertices have been marked
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28. Dijkstra’s algorithm (2)B C D E F 2  3 1 4 5
Step 0
We start at A so we mark it and use first row of w for initial values of d[v]
Smallest value is d[B] = 2
Vertex to
be marked
Distance to vertex
Unmarked
vertices
Step A B C D E F 2  3
B, C, D, E, F
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29. Dijkstra’s algorithm (3)
Step 1
Mark B since it is the closest unmarked vertex to A
Calculate distances to unmarked vertices via B
If a shorter distance is found use this
In our case,
A  B  C has weight 3
A  B  E has weight 6
Notice that previously d[C] and d[E] were  B D C E F A 1 2 3 4 5
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30. Dijkstra’s algorithm (4)B C D E F 2  3 1 4 5
Step 1 (Cont’d)
2nd row: smallest value of d[v] for unmarked vertices occurs for C and D
Vertex to
be marked
Distance to vertex
Unmarked
vertices
Step A B C D E F 2  3
B, C, D, E, F 1 6
C, D, E, F
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31. Dijkstra’s algorithm (5)
Step 2
Of remaining unmarked vertices D and C are closest to A
Choose one of them (say, D)
Path A  D  E has weight 5, so current value of d[E] can be updated to 5 B D C E F A 1 2 3 4 5
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32. Dijkstra’s algorithm (6)B C D E F 2  3 1 4 5
Step 2 (Cont’d)
Next row generated, in which smallest value of d[v] for unmarked vertices occurs for vertex C
Vertex to
be marked
Distance to vertex
Unmarked
vertices
Step A B C D E F 2  3
B, C, D, E, F 1 6
C, D, E, F 5
C, E, F
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33. Dijkstra’s algorithm (7)
Step 3
We mark vertex C and recompute distances
Vertex F can be accessed for the first time via path
A  B  C  E
So d[F] = 8, and two unmarked vertices remain
Vertex to
be marked
Distance to vertex
Unmarked
vertices
Step A B C D E F 2  3
B, C, D, E, F 1 6
C, D, E, F 5
C, E, F 8
E, F
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34. Dijkstra’s algorithm (8)
Step 4 and 5
We mark vertex E next, which reduces the distance to F from 8 to 6
Finally, mark F
Vertex to
be marked
Distance to vertex
Unmarked
vertices
Step A B C D E F 2  3
B, C, D, E, F 1 6
C, D, E, F 5
C, E, F 8
E, F 4
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35. Dijkstra’s algorithm (9)
Input: G = (V, E) and A  V
Finds shortest path from A to all v  V
For any u and v, w(u, v) is the weight of the arc uv
PATHTO(v) stores the current shortest path from A to v
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36. Dijkstra’s algorithm (10)
for each v  V do
begin
d[v] := w(A, v);
PATHTO(v) := A;
end
Mark vertex A;
while unmarked vertices remain do
u := unmarked vertex whose distance from A is minimal;
Mark vertex u;
for each unmarked vertex v with uv  E do
d’ := d[u] + w(u, v);
if d’ < d[v] then
d[v] := d’;
PATHTO(v) := PATHTO(u), v;
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37. Further reading
R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd (Chapter 8)
Wikipedia’s entry on directed graphs
Wikibooks entry on graph theory
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