CS1022
Computer Programming &
Principles
Lecture 8.2
Digraphs (2)
Plan of lecture
•
•
•
•
•
•
Paths in digraphs
Reachability matrix
Warshall’s algorithm
Shortest path
Weight matrix
Dijkstra’s algorithm
CS1022
2
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!
CS1022
3
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
CS1022
4
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
CS1022
5
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
CS1022
6
Reachability matrix (3)
• Calculate the reachability matrix of digraph
a
b
c
d
CS1022
F
F
M 
F

F
F

F T T
F F F

F T F
T F
7
Reachability matrix (4)
• So we have
F T
F F
2
M 
F F

F F
F  F


T T  F

F F  F
 
T F  F
F
F  F


F T T  F

F F F  F
 
F T F  F
T F
F T T

F T F
F F F

F F F
• The 3 T entries in M2 indeed correspond to paths of
length 2 in G, namely
–abc
–abd
–bdc
CS1022
8
Reachability matrix (5)
• Further calculation gives
F F T F
F F F F
F F F F
F F F F
3
4



M 
M 
F F F F
F F F F




F F F F
F F F F
• Therefore,
F T T T  • For example, T in top-right
F F T T  corner of M* arises from
*
 M2 and corresponds to
M 
F F F F  path a b d
CS1022

F F

T F
9
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
CS1022
10
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*
CS1022
11
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
CS1022
12
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;
CS1022
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))
13
Warshall’s algorithm (4)
To calculate Wk from Wk – 1 proceed as follows
1. Consider column k in Wk – 1
2. For each “F” row in this column, copy that row to Wk
3. 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))
CS1022
14
Warshall’s algorithm (5)
• Calculate reachability matrix of digraph
2
3
4
1
CS1022
5
F
F
W0  T
F
T
T
F
F
F
F
F
T
F
F
T
F
F
T
F
F
F
F
F
F
F
15
Warshall’s algorithm (6)
2
3
4
• We now calculate W1:
– Using step 1 we consider column 1 of W0 1
– Using step 2 we copy rows 1, 2 and 4 directly to W1
5
F T F F F
F F T F F
W1 
F F F F F
CS1022
16
Warshall’s algorithm (7)
3
2
4
• We now use step 3 – row 3 in W1 is
– Logical or of row 3 with row 1 of W0
F
F
W0  T
F
T
CS1022
T
F
F
F
F
F
T
F
F
T
F
F
T
F
F
F
F
F
F
F
F
F
W1  T
F
5
1
T
F
T
F
F
T
F
F
F
F
T
F
F
F
F
F
17
Warshall’s algorithm (8)
3
2
4
• We use step 3 again – row 5 in W1 is
– Logical or of row 5 with row 1 of W0
F
F
W0  T
F
T
CS1022
T
F
F
F
F
F
T
F
F
T
F
F
T
F
F
F
F
F
F
F
F
F
W1  T
F
T
5
1
T
F
T
F
T
F
T
F
F
T
F
F
T
F
F
F
F
F
F
F
18
Warshall’s algorithm (9)
3
2
4
• We now compute W2 from W1
1
– 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
F
W1  T
F
T
CS1022
T
F
T
F
T
F
T
F
F
T
F
F
T
F
F
F
F
F
F
F
F
F
W2  T
F
T
T
F
T
F
T
T
T
T
F
T
5
F
F
T
F
F
F
F
F
F
F
19
Warshall’s algorithm (10)
• Notice entry (3, 3)
– Indicates a cycle from vertex 3 back to itself
– Going via vertices 1 and/or 2
F
F
W2  T
F
T
CS1022
T
F
T
F
T
T
T
T
F
T
F
F
T
F
F
F
F
F
F
F
2
3
4
1
5
20
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
T
T
W3  T
F
T
CS1022
T
T
T
F
T
T
T
T
F
T
T
T
T
F
T
F
F
F
F
F
21
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)
CS1022
22
Shortest paths (2)
• Suppose the following weighted digraph
1
C
B
5
2
4
A
F
3
D
2
1
E
– 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
CS1022
23
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
CS1022
Edsger W. Dijkstra
24
Weight matrix (1)
• A compact way to represent weighted digraphs
• Matrix w, whose entries w(u, v) are given by
0 if u  v

w(u, v)   if uv is not an arc
d if uv is an arc of weight d

CS1022
25
Weight matrix (2)
• Our digraph is represented as
B
1
2
C
5
4
A
F
3
D
2
1
E
A
B
w C
D
E
F
A B
0 2
0




C DE F
 3 
14
0  5
0 2
0 1
  0
0 if u  v

w(u, v)   if uv is not an arc
d if uv is an arc of weight d

CS1022
26
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
CS1022
27
Dijkstra’s algorithm (2)
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
A
B
C
D
E
F
A
0





B
2
0




C

1
0



D
3


0


E

4

2
0

F


5

1
0
Vertex to
Distance to vertex Unmarked
Step be marked A B C D E F
vertices
0
A
0 2  3   B, C, D, E, F
CS1022
28
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
C
1
2
4
A
F
3
1
D
CS1022
5
2
E
29
Dijkstra’s algorithm (4)
Step 1 (Cont’d)
– 2nd row: smallest value of d[v] for
unmarked vertices occurs for C and D
A
B
C
D
E
F
A
0





B
2
0




C

1
0



D
3


0


E

4

2
0

F


5

1
0
Vertex to
Distance to vertex Unmarked
Step be marked A B C D E F
vertices
0
A
0 2  3   B, C, D, E, F
1
B
0 2 3 3 6  C, D, E, F
CS1022
30
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
1
2
C
4
A
F
3
D
CS1022
5
2
1
E
31
Dijkstra’s algorithm (6)
Step 2 (Cont’d)
– Next row generated, in which smallest
value of d[v] for unmarked vertices
occurs for vertex C
Step
0
1
2
CS1022
Vertex to
be marked
A
B
D
A
B
C
D
E
F
Distance to vertex
A B C D E F
0 2  3  
0 2 3 3 6 
0 2 3 3 5 
A
0





B
2
0




C

1
0



D
3


0


E

4

2
0

F


5

1
0
Unmarked
vertices
B, C, D, E, F
C, D, E, F
C, E, F
32
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
Distance to vertex Unmarked
Step be marked A B C D E F
vertices
0
A
0 2  3   B, C, D, E, F
1
B
0 2 3 3 6  C, D, E, F
2
D
0 2 3 3 5  C, E, F
3
C
0 2 3 3 5 8 E, F
CS1022
33
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
Step
0
1
2
3
4
5
CS1022
Vertex to
be marked
A
B
D
C
E
F
Distance to vertex
A B C D E F
0 2  3  
0 2 3 3 6 
0 2 3 3 5 
0 2 3 3 5 8
0 2 3 3 5 6
0 2 3 3 5 6
Unmarked
vertices
B, C, D, E, F
C, D, E, F
C, E, F
E, F
F
34
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
CS1022
35
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
begin
u := unmarked vertex whose distance from A is minimal;
Mark vertex u;
for each unmarked vertex v with uv  E do
begin
d’ := d[u] + w(u, v);
if d’ < d[v] then
begin
d[v] := d’;
PATHTO(v) := PATHTO(u), v;
end
end
CS1022
36
Further reading
• R. Haggarty. “Discrete Mathematics for
Computing”. Pearson Education Ltd.
2002. (Chapter 8)
• Wikipedia’s entry on directed graphs
• Wikibooks entry on graph theory
CS1022
37