Lecture 23: Composition of Relations, Transitive
Closure and Warshall’s Algorithm
Rasha Eqbal
07CS1039
Teacher: Prof. Niloy Ganguly
Department of Computer Science and Engineering
IIT Kharagpur
7thSeptember, 2008
1
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Lecture 23: Composition of Relations, Transitive
Closure and Warshall’s Algorithm
Rasha Eqbal
07CS
Teacher: Prof. Niloy Ganguly
Department of Computer Science and Engineering
IIT Kharagpur
7 thSeptember, 2008
CONTENTS CONTENTS
Contents
1 Composition of Relations 3
2 Transitive Closure 7
3 Warshall’s Algorithm 12
1 COMPOSITION OF RELATIONS
Proof If an element z ∈ C is in (S ◦ R)(A 1 ), then x(S ◦ R)z for some x ∈ A 1. By the definition of composition, this means that xRy and ySz for some y ∈ B. So now we have z ∈ S(y) and y ∈ R(x) ⇒ z ∈ S(R(x)). Since {x} ⊆ A 1 , we can also say that S(R(x)) ⊆ S(R(A 1 )). Now, since z ∈ S(R(x)), therefore z ∈ S(R(A 1 )) also. This means that (S ◦ R)(A 1 ) ⊆ S(R(A 1 )). Conversely, suppose z ∈ S(R(A 1 )). Then z ∈ S(y) for some y ∈ R(A 1 ). Similarly, y ∈ R(x) for some x ∈ A 1. This means that xRy and ySz. So from the definition of composition we can say x(S ◦R)z. Thus z ∈ (S ◦R)(x). Since {x} ⊆ A 1 , we can say that (S ◦R)(x) ⊆ (S ◦R)(A 1 ). Hence z also belongs to (S ◦ R)(A 1 ). So S(R(A 1 )) ⊆ (S ◦ R)(A 1 ). Since (S ◦ R)(A 1 ) ⊆ S(R(A 1 )) and S(R(A 1 )) ⊆ (S ◦ R)(A 1 ), we can say that (S ◦ R)(A 1 ) = S(R(A 1 )). This proves the theorem.
† Example 2: Let A = {a, b, c} and let R and S be relations on A whose matrices are:
MR =
MS =
Find S ◦ R. Solution: We see from the matrices that:
- (a, a) ∈ R and (a, a) ∈ S ⇒ (a, a) ∈ S ◦ R
- (a, c) ∈ R and (c, a) ∈ S ⇒ (a, a) ∈ S ◦ R
- (a, c) ∈ R and (c, c) ∈ S ⇒ (a, c) ∈ S ◦ R
It is easily seen that (a, b) ∈/ S ◦ R since, if we had (a, x) ∈ R and (x, b) ∈ S, then from Matrix MR we know that x would have to be either a or c; but from matrix MS we know that neither (a, b) nor (c, b) is an element of S. Hence we see that the first row of MS◦R is 1 0 1. Proceeding in a similar manner we get:
- (b, a) ∈ R and (a, a) ∈ S ⇒ (b, a) ∈ S ◦ R
1 COMPOSITION OF RELATIONS
- (b, b) ∈ R and (b, b) ∈ S ⇒ (b, b) ∈ S ◦ R
- (b, b) ∈ R and (b, c) ∈ S ⇒ (b, c) ∈ S ◦ R
- (b, c) ∈ R and (c, a) ∈ S ⇒ (b, a) ∈ S ◦ R
- (b, c) ∈ R and (c, c) ∈ S ⇒ (b, c) ∈ S ◦ R
Hence the second row of MS◦R is 1 1 1.
- (c, b) ∈ R and (b, b) ∈ S ⇒ (c, b) ∈ S ◦ R
- (c, b) ∈ R and (b, c) ∈ S ⇒ (c, c) ∈ S ◦ R
Hence the third row of MS◦R is 0 1 1. Therefore the composition matrix is
MS◦R =
Now we shall deduce an important and useful result. Let us consider three sets, A = {a 1 ,... , an}, B = {b 1 ,... , bp} and C = {c 1 ,... , cm}; and relation R defined from A to B, and S defined from B to C. The Boolean matrices MR and MS are of sizes n × p and p × m respectively. Let us represent MR as [rij ], MS as [sij ] and MS◦R as [tij ]. Now tij = 1 ⇔ (ai, cj ) ∈ S ◦ R, which means that for some k between 1 and p, (ai, bk) ∈ R and (bk, cj ) ∈ S, that is, rik = 1 and skj = 1. In other words, if rik = 1 and skj = 1 then tij ← 1. This is identical to the condition needed for MR Ø MS to have a 1 in position (i,j). Hence we can say that: MS◦R = MR Ø MS. In the special case when we have S = R, S ◦ R = R^2 and MS◦R = MR^2 = MR Ø MR.
§ Theorem 2 Let A, B, C and D be sets, R a relation from A to B, S a relation from B to C and T a relation from C to D. Then T ◦ (S ◦ R) = (T ◦ S) ◦ R Proof Let the Boolean matrices for the relations R, S and T be MR, MS and MT respec- tively. As was shown in Example 2, the Boolean matrix product represents the matrix of composition, i.e. MS◦R = MR Ø MS. Thus we have:
2 TRANSITIVE CLOSURE
2 Transitive Closure
A relation R is said to be transitive if for every (a, b) ∈ R and (b, c) ∈ R there is a (a, c) ∈ R. A transitive closure of a relation R is the smallest transitive relation containing R. Suppose that R is a relation defined on a set A and that R is not transitive. Then the transitive closure of R is the connectivity relation R∞. We will now try to prove this claim.
§ Theorem 4 Let R be a relation on a set A. Then R∞^ is the transitive closure of R. Proof We need to prove that R∞^ is transitive and also that it is the smallest transitive relation containing R. If a and b ∈ A, then aR∞b if and only if there exists a path in R from a to b. If aR∞b and bR∞c, then we can say that aR∞c. This is because aR∞b means that there exists a path from a to b in R, similarly bR∞c means that there exists a path from b to c in R. Hence there will also exist a path form a to c in R. (We can simply start along the path from a to b and then continue along the path from b to c. This will give us the path from a to c.) This proves that R∞^ is transitive. Now let us consider a transitive relation S (containing R) i.e. R ⊆ S. Since S is transitive we can say that Sn^ ⊆ S ∀ n. (This means that if there is a path of length n from a to b, then aSb, which is true as S is a transitive relation.) Now, S∞^ =
n=1 S
n.
Hence S∞^ ⊆ S. Since R ⊆ S, therefore R∞^ ⊆ S∞, and as S∞^ ⊆ S, we can say that R∞^ ⊆ S. This means that R∞^ is the smallest of all transitive relations on A that contain R. As R∞^ satisfies both the properties, we can say that R∞^ is the transitive closure of R on set A. This completes our proof.
zNote: If we include the identity relation ∆ then R∞^ ∪ ∆ is the reachability relation R∗.
† Example 3: Let A = { 1 , 2 , 3 , 4 }, and let R = {(1, 2), (2, 3), (3, 4), (2, 1)}. Find the transitive closure of R.
Solution: Method 1: Using Digraph
2 TRANSITIVE CLOSURE
Figure 1: Digraph of R
We can determine R∞^ by geometrically computing all paths from the digraph. From vertex 1 we have paths to vertices 1, 2, 3 and 4. So the ordered pairs (1, 1), (1, 2), (1,
- and (1,4) ∈ R∞. From vertex 2 we have paths to vertices to vertices 1, 2, 3 and 4. This gives us the ordered pairs (2, 1), (2, 2), (2, 3) and (2, 4). From vertex 3 we have only one path to vertex 4. This gives us the ordered pair (3, 4). From vertex 4 we do not have any paths. So we have R∞^ = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 4)} Method 2: Using Matrices Writing down the Boolean matrix we get
MR =
2 TRANSITIVE CLOSURE
§ Theorem 5 Let A be a set with |A| = n, and let R be a relation on A. Then R∞^ = R ∪ R^2 ∪... ∪ Rn i.e. powers of R greater than n need not be considered to compute R∞. Proof Let a and b ∈ A, and let a, x 1 , x 2 ,... , xm, b be a path from a to b in R; i.e. (a, x 1 ), (x 1 , x 2 ),... , (xm, b) ∈ R. Now if xi and xj correspond to the same vertex for some i < j, then the path from a to b can be distinctly divided into three regions. First, a path from a to xi; second, a path from xi to xj ; and lastly, a path from xj to b. Here we see that the second path forms a closed loop as xi = xj. So we can eliminate it altogether and put the first and third paths together to give us a shorter path. In a similar manner we can keep eliminating all the loops we get later on in the path too, to give us a path a, x 1 , x 2 ,... , xk, b where all of x 1 , x 2 ,... , xk are distinct. This path is the shortest one possible from a to b.
2 TRANSITIVE CLOSURE
Figure 2: Shows how a closed loop can be eliminated to give the shortest path. Here, i = 2 and j = 5.
Now let us consider the case when a 6 = b. Since the total number of elements in the set A is n, the maximum path length we can get is n − 1. If a = b, then we get the maximum path length as n (as |A| = n and all the vertices except a and b are distinct). ’There is a path from a to b in R’ is equivalent to aR∞b. And if aR∞b (i.e. there is a path from a to b in R), from the preceding discussion we know that aRkb for some k, 1 ≤ k ≤ n (as the maximum path length possible is n). Thus R∞^ = R ∪ R^2 ∪... ∪ Rn. Hence the theorem is proved.
3 WARSHALL’S ALGORITHM
- Put 1’s at all positions (yi, zj ) of Wk (if not already present).
† Example 4: Find the transitive closure of R defined in Example 3. Solution:
W 0 = MR =
Here we have n = 4. To find W 1 , k = 1. We can see that W 0 has 1’s in column 1 at location 2, and in row 1 at location 2. Thus W 1 has a new 1 at position (2, 2).
W 1 =
For W 2 , k = 2. W 1 has 1’s in column 2 at locations 1 and 2, and in row 2 at locations 1, 2 and 3. So the new 1’s would go to positions (1, 1), (1, 2), (1, 3), (2, 1), (2, 2) and (2, 3) (if not already there).
W 2 =
For W 3 , k = 3. W 2 has 1’s in column 3 at locations 1 and 2, and in row 3 at location
- So the new 1’s would come at positions (1, 4) and (2, 4) (if not already there).
W 3 =
For W 4 , k = 4. W 3 has 1’s in column 4 at locations 1, 2 and 3 but no 1’s in row 4. So no new 1’s are added. Hence W 4 = W 3. This gives us the matrix representation of R∞^ which is the same as that obtained in Example 3.
3 WARSHALL’S ALGORITHM
The Algorithm To find the matrix CLOSURE of the transitive closure of a relation R whose n × n matrix representation is MAT.
WARSHALL
- CLOSURE ← MAT
- for k = 1 to n
- for i = 1 to n
- for j = 1 to n
- CLOSURE(i, j) ← CLOSURE(i, j)∨(CLOSURE(i, k)∧CLOSURE(k, j))
This algorithm involves three for loops, with two of them nested. Each loop iterates from 1 to n. This gives us a time complexity of O(n^3 ). If we were to find the transitive closure using the matrix multiplication method we would get a time complexity of O(n^4 ). Each time a matrix multiplication is performed the time complexity is O(n^3 ) as there are three loops running (two nested) from 1 to n. The matrix multiplications are carried out a total of n − 1 times to find matrices (MR)^2 Ø, (MR)^3 Ø,... , (MR)n Ø, since MR∞ = MR ∨ (MR)^2 Ø ∨... ∨ (MR)n Ø. So the number of steps involved are n^3 (n − 1), giving us a time complexity of O(n^4 ). Thus we see that Warshall’s algorithm is surely simpler and more efficient than the matrix multiplication method.
An Application of Warshall’s Algorithm § Theorem 6 If R and S are equivalence relations on a set A, then the smallest equivalence relation containing both R and S is (R ∪ S)∞.
Proof We know that a relation is reflexive if and only if it contains the identity or equality relation ∆. Since both R and S are reflexive, ∆ ⊆ R and ∆ ⊆ S. This implies that ∆ ⊆ R ∪ S ⊆ (R ∪ S)∞. So (R ∪ S)∞^ is also reflexive. R and S are symmetric. Let us consider (a, b) ∈ R ⇒ (b, a) ∈ R (as R is symmetric). Since R ⊆ (R ∪ S) ⊆ (R ∪ S)∞, therefore (a, b) and (b, a) ∈ (R ∪ S)∞. This implies that (R ∪ S)∞^ is also symmetric. This can be proved in a similar manner by taking (a, b) ∈ S instead of R. The property of transitive closure tells us that any relation T ∞^ is the smallest transi- tive relation containing the relation T. Applying this property to R ∪ S, we can conclude that (R ∪ S)∞^ is the smallest transitive relation containing (R ∪ S). Since it is transitive
3 WARSHALL’S ALGORITHM
Now we compute W 4 (k = 4). W 3 has 1’s at locations 3, 4 and 5 of column 4, and at locations 3, 4 and 5 of row 4. So new 1’s need to be added at positions (3, 5) and (5, 3) of W 3 to get W 4. Thus
W 4 =
To compute W 5 (k = 5), we see that, since W 4 has 1’s at locations 3, 4 and 5 of column 5, and at locations 3, 4 and 5 of row 5, no new 1’s need to be added. So W 5 = W 4.
∴ (R ∪ S)∞^ = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5), (5, 3), (5, 4), (5, 5)}.
The corresponding partition is {{ 1 , 2 }, { 3 , 4 , 5 }}.