CS1022 Computer Programming & Principles Lecture 8.2 Digraphs (2)

CS1022 Computer Programming &

Principles

Lecture 8.2Digraphs (2)

Plan of lecture• Paths in digraphs• Reachability matrix• Warshall’s algorithm• Shortest path• Weight matrix• Dijkstra’s algorithm

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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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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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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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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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Reachability matrix (3)• Calculate the reachability matrix of digraph

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a

c

b

d

FF

F

F

FTFFFFFFTTFFFFTF

M

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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FF

F

F

FFFFFFFFFTFFTTFF

FTFFFFFFTTFFFFTF

FTFFFFFFTTFFFFTF

2M

Reachability matrix (5)• Further calculation gives

• Therefore,

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FF

F

F

FFFFFFFFFFFFFTFF

3M

FFFFFFFFFFFFFFFF

4M

FTFFFFFFTTFFTTTF

*M

• For example, T in top-right corner of M* arises from M2 and corresponds to path a b d

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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FF

F

F

Warshall’s algorithm (1)• Let G be a digraph with vertices v1, v2, , vn

• Warshall’s algorithm generates sequenceW0, 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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FF

F

F

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

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FF

F

Fbegin

W := M; for k 1 to n do

for i 1 to n dofor j 1 to n do

W(i, j) := W(i, j) or (W(i, k) and W(k, j))end

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

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FF

F

F

W := M; for k 1 to n do

for i 1 to n dofor j 1 to n do

W(i, j) := W(i, j) or (W(i, k) and W(k, j))

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

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W := M; for k 1 to n dofor i 1 to n do

for j 1 to n doW(i, j) := W(i, j) or (W(i, k) and W(k, j))

Warshall’s algorithm (5)• Calculate reachability matrix of digraph

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2

1

3

5

4 W0

F T F F FF F T F FT F F T FF F F F FT F T F F

Warshall’s algorithm (6)• 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

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2

1

3

5

4

W1

F T F F FF F T F F

F F F F F

Warshall’s algorithm (7)• We now use step 3 – row 3 in W1 is – Logical or of row 3 with row 1 of W0

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W1

F T F F FF F T F F

F F F F FW0


W1

F T F F FF F T F FTF F F F F

W1

F T F F FF F T F FT TF F F F F

W1

F T F F FF F T F FT T FF F F F F

W1

F T F F FF F T F FT T F T FF F F F F

W1


2

1

3

5

4

Warshall’s algorithm (8)• We use step 3 again – row 5 in W1 is – Logical or of row 5 with row 1 of W0

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W0


W1


2

1

3

5

4

F T F F FF F T F FT T F T FF F F F FT

F T F F FF F T F FT T F T FF F F F FT T

F T F F FF F T F FT T F T FF F F F FT T T

F T F F FF F T F FT T F T FF F F F FT T T F F


Warshall’s algorithm (9)• 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



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W1


W2 F F T F F

F F F F F

2

1

3

5

4


F T T F FF F T F F

F F F F F


F T T F FF F T F FT T T T FF F F F F


F T T F FF F T F FT T T T FF F F F FT T T F F

Warshall’s algorithm (10)• Notice entry (3, 3)– Indicates a cycle from vertex 3 back to itself– Going via vertices 1 and/or 2

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W2

F T T F FF F T F FT T T T FF F F F FT T T F F

2

1

3

5

4

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

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W3

T T T T FT T T T FT T T T FF F F F FT T T T F

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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Shortest paths (2)• Suppose the following weighted digraph

– 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

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B

D

C

E

FA

1

2

3

4

21

5

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

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Weight matrix (1)• A compact way to represent weighted digraphs• Matrix w, whose entries w(u, v) are given by

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duv

uv

vu

d

vuw

weightof arc an is if arc an not is if

if

0

),(

Weight matrix (2)• Our digraph is represented as

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duv

uv

vu

d

vuw

weightof arc an is if arc an not is if

if

0

),(

B

D

C

E

FA

1

2

3

4

21

5

w

A B C D E FABCDEF

A B C D E FA 0B 0C 0D 0E 0F 0

A B C D E FA 0 B 0 C 0 D 0 E 0F 0

A B C D E FA 0 2 3 B 0 C 0 D 0 E 0F 0

A B C D E FA 0 2 3 B 0 1 4 C 0 D 0 E 0F 0

A B C D E FA 0 2 3 B 0 1 4 C 0 5D 0 2 E 0 1F 0

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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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

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A B C D E FA 0 2 3 B 0 1 4 C 0 5D 0 2 E 0 1F 0

Vertex tobe marked

Distance to vertexUnmarked

verticesStep A B C D E F

0 A 0 2 3 B, C, D, E, F

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

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B

D

C

E

FA

1

2

3

4

21

5

Dijkstra’s algorithm (4)Step 1 (Cont’d)– 2nd row: smallest value of d[v] for

unmarked vertices occurs for C and D

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A B C D E FA 0 2 3 B 0 1 4 C 0 5D 0 2 E 0 1F 0

Vertex tobe marked



0 A 0 2 3 B, C, D, E, F1 B 0 2 3 3 6 C, D, E, F

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

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B

D

C

E

FA

1

2

3

4

21

5

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

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A B C D E FA 0 2 3 B 0 1 4 C 0 5D 0 2 E 0 1F 0

Vertex tobe marked



0 A 0 2 3 B, C, D, E, F1 B 0 2 3 3 6 C, D, E, F2 D 0 2 3 3 5 C, E, F

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

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Vertex tobe marked



0 A 0 2 3 B, C, D, E, F1 B 0 2 3 3 6 C, D, E, F2 D 0 2 3 3 5 C, E, F3 C 0 2 3 3 5 8 E, F

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

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Vertex tobe marked



0 A 0 2 3 B, C, D, E, F1 B 0 2 3 3 6 C, D, E, F2 D 0 2 3 3 5 C, E, F3 C 0 2 3 3 5 8 E, F4 E 0 2 3 3 5 6 F5 F 0 2 3 3 5 6

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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Dijkstra’s algorithm (10)

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for each v V dobegin

d[v] := w(A, v);PATHTO(v) := A;

endMark vertex A;while unmarked vertices remain dobegin

u := unmarked vertex whose distance from A is minimal;Mark vertex u;for each unmarked vertex v with uv E dobegin

d’ := d[u] + w(u, v);if d’ < d[v] thenbegin

d[v] := d’;PATHTO(v) := PATHTO(u), v;

endend

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

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http://en.wikipedia.org/wiki/Directed_graph

http://en.wikibooks.org/wiki/Graph_Theory

CS1022 Computer Programming & Principles Lecture 8.2 Digraphs (2)

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