Find transitive-closure using Warshall's algorithm a presentation for MSc program Advanced Graph Theory course

1. Name: Md Mamun Hasan
Roll: 1907555 (KUET)
Course: CSE 6101
Find Transitive Closure using
Warshall’s Algorithm

2. Transitive Closure
• Given a digraph G, the transitive
closure of G is the digraph G* such
that
• G* has the same vertices as G
• if G has a directed path from u
to v (u  v), G* has a directed
edge from u to v
• The transitive closure provides
reachability information about a
digraph
2
1
4
3
5
2
1
4 5
G
3

3. Warshall’s Algorithm
Main Idea
Suppose two nodes 1 & 5
A path exists between two vertices 1, 5, iff
• there is an edge from 1 to 5; or
• there is a path from 1 to 5 going through vertex 2; or
• there is a path from 1 to 5 going through vertex 2 and/or
3; or
• there is a path from 1 to 5 going through vertex 2, 3,
and/or 4
…
So, (1, 5) is a transitive closure
2
1
4
3
5
2
1
4
3
5
2
1
4
3
5

4. Warshall’s Algorithm
On the kth iteration, the algorithm determine if a path exists between two
vertices i, j using just vertices among 1,…,k allowed as intermediate
𝑅 𝑘
[i, j]
= {𝑅 𝑘−1
[i, j] (path using just 1 ,…,k-1)
Or
𝑅 𝑘−1
[i, k] and 𝑅 𝑘−1
[k, j] (path from i to k and from k to i using just 1
,…,k-1)
i
j
k
𝑘 𝑡ℎ iteration

5. Warshall’s Algorithm
Algorithm Warshall
Input: Adjacency matrix A of relation R on a set of n elements
Output: Adjacency matrix T of the transitive closure of R.
Algorithm Body:
T := A [initialize T to A]
for j := 1 to n
for i := 1 to n
if 𝑻𝒊,𝒋 = 1 then
𝑇𝑖 := 𝑇𝑖∨ 𝑇𝑗[form the Boolean OR of row iand row j, store it in 𝑇𝑖]
next i
next j
end Algorithm Warshall

6. Warshall’s Algorithm
2
1
4
3
5
0 0 1 0 0
0 0 1 1 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
Here A = T =
1 2 3 4
5 j1
2
3
4
5
i
j = 1
i = 1 𝑇𝑖,𝑗 = 0 no action
i = 2 𝑇𝑖,𝑗 = 0 no action
i = 3 𝑇𝑖,𝑗 = 0 no action
i = 4 𝑇𝑖,𝑗 = 0 no action
i = 5 𝑇𝑖,𝑗 = 0 no action
𝐴𝑖,𝑗 =
1 𝑖𝑓 𝑡ℎ𝑒 𝑑𝑖𝑔𝑟𝑎𝑝ℎ ℎ𝑎𝑠 𝑎𝑛 𝑒𝑑𝑔𝑒 𝑓𝑟𝑜𝑚 𝑣𝑖 𝑡𝑜 𝑣𝑗
0 𝑖𝑓 𝑡ℎ𝑒 𝑑𝑖𝑔𝑟𝑎𝑝ℎ ℎ𝑎𝑠 𝑛𝑜 𝑒𝑑𝑔𝑒 𝑓𝑟𝑜𝑚 𝑣𝑖 𝑡𝑜 𝑣𝑗

7. Warshall’s Algorithm
0 0 1 0 0
0 0 1 1 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
1 2 3 4
51
2
3
4
5
𝑇1 =
j= 2
i = 1 𝑇𝑖,𝑗 = 0 no action
i = 2 𝑇𝑖,𝑗 = 0 no action
i = 3 𝑇𝑖,𝑗 = 0 no action
i = 4 𝑇𝑖,𝑗 = 0 no action
i = 5 𝑇𝑖,𝑗 = 0 no action
2
1
4
3
5

8. Warshall’s Algorithm
0 0 1 0 0
0 0 1 1 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
1 2 3 4
51
2
3
4
5
j=3
i = 1 𝑇𝑖,𝑗 = 1 𝑇1 = 𝑇1 ∨ 𝑇3 = (0 0 1 0 0) ∨ (0 0 0 1 0) = (0 0 1 1 0)
i = 2 𝑇𝑖,𝑗 = 1 𝑇2 = 𝑇2 ∨ 𝑇3 = (0 0 1 1 0) ∨ (0 0 0 1 0) = (0 0 1 1 0)
i = 3 𝑇𝑖,𝑗 = 0 no action
i = 4 𝑇𝑖,𝑗 = 0 no action
i = 5 𝑇𝑖,𝑗 = 0 no action
𝑇2 =
2
1
4
3
5

9. Warshall’s Algorithm
0 0 1 1 0
0 0 1 1 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
1 2 3 4
51
2
3
4
5
j=4
i = 1 𝑇𝑖,𝑗 = 1 𝑇1 = 𝑇1 ∨ 𝑇4 = (0 0 1 1 0) ∨ (0 0 0 0 1) = (0 0
1 1 1)
i = 2 𝑇𝑖,𝑗 = 1 𝑇2 = 𝑇2 ∨ 𝑇4 = (0 0 1 1 0) ∨ (0 0 0 0 1) = (0 0 1 1
1)
i = 3 𝑇𝑖,𝑗 = 1 𝑇3 = 𝑇3 ∨ 𝑇4 = (0 0 0 1 0) ∨ (0 0 0 0 1) = (0 0 0 1
1)
i = 4 𝑇𝑖,𝑗 = 0 no action
i = 5 𝑇𝑖,𝑗 = 0 no action
𝑇3 =
2
1
4
3
5

10. Warshall’s Algorithm
0 0 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 1
0 0 0 0 0
1 2 3 4
51
2
3
4
5
j=5
i = 1 𝑇𝑖,𝑗 = 1 𝑇1 = 𝑇1 ∨ 𝑇5 = (0 0 1 1 1) ∨ (0 0 0 0 0) = (0 0
1 1 1)
i = 2 𝑇𝑖,𝑗 = 1 𝑇2 = 𝑇2 ∨ 𝑇5 = (0 0 1 1 1) ∨ (0 0 0 0 0) = (0 0 1 1
1)
i = 3 𝑇𝑖,𝑗 = 1 𝑇3 = 𝑇3 ∨ 𝑇5 = (0 0 0 1 1) ∨ (0 0 0 0 0) = (0 0 0 1
1)
i = 4 𝑇𝑖,𝑗 = 1 𝑇4 = 𝑇4 ∨ 𝑇5 = (0 0 0 0 1) ∨ (0 0 0 0 0) = (0 0 0 0
1)
i = 5 𝑇𝑖,𝑗 = 0 no action
𝑇4 =
2
1
4
3
5

11. Warshall’s Algorithm
0 0 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 1
0 0 0 0 0
1 2 3 4
51
2
3
4
5
T = 𝑇5 =
2
1
4
3
5
G*

12. Warshall’s Algorithm
Time Complexity
This algorithm has two nested loops containing
a Θ(n) core, so it takes Θ (𝑛3
) time.
Time complexity = n * Θ (𝑛2) = Θ (𝑛3)
Space Complexity
At any point in the algorithm, we only need the
last two matrices computed, so we can re-use
the storage from the other matrices.
space complexity = Θ (𝑛2
)

13. Thank You
Any Question?