Algorithm of Matrix Chain Multiplication
We will use table s to construct an optimal solution. Step 1: Constructing an Optimal Solution: PRINTOPTIMALPARENS (s, i, j) 1. if i=j 2. then print "A" 3. else print "(" 4. PRINTOPTIMALPARENS (s, i, s [i, j]) 5. PRINTOPTIMALPARENS (s, s [i, j] + 1, j) 6. print ")" Analysis: There are three nested loops. Each loop executes a maximum n times.
Body of loop constant complexity Total Complexity is: O (n^{3}) Algorithm with Explained ExampleQuestion: P [7, 1, 5, 4, 2} Solution: Here, P is the array of a dimension of matrices. So here we will have 4 matrices: A_{1}_{7x1} A_{2}_{1x5} A_{3}_{5x4} A_{4}_{4x2} i.e. First Matrix A_{1} have dimension 7 x 1 Second Matrix A_{2} have dimension 1 x 5 Third Matrix A_{3} have dimension 5 x 4 Fourth Matrix A_{4} have dimension 4 x 2 Let say, From P = {7, 1, 5, 4, 2}  (Given) And P is the Position p_{0} = 7, p_{1} =1, p_{2} = 5, p_{3} = 4, p_{4}=2. Length of array P = number of elements in P ∴length (p)= 5 From step 3 Follow the steps in Algorithm in Sequence According to Step 1 of Algorithm MatrixChainOrder Step 1: n ← length [p]1 Where n is the total number of elements And length [p] = 5 ∴ n = 5  1 = 4 n = 4 Now we construct two tables m and s. Table m has dimension [1.....n, 1.......n] Table s has dimension [1.....n1, 2.......n] Now, according to step 2 of Algorithm Case 1: 1. When l  2 for (i ← 1 to n  l + 1) i ← 1 to 4  2 + 1 i ← 1 to 3 When i = 1 do j ← i + l  1 j ← 1 + 2  1 j ← 2 i.e. j = 2 Now, m [i, j] ← ∞ i.e. m [1,2] ← ∞ Put ∞ in m [1, 2] table for k ← i to j1 k ← 1 to 2  1 k ← 1 to 1 k = 1 Now q ← m [i, k] + m [k + 1, j] + p_{i1} p_{k} p_{j} for l = 2 i = 1 j =2 k = 1 q ← m [1,1] + m [2,2] + p_{0}x p_{1}x p_{2} and m [1,1] = 0 for i ← 1 to 4 ∴ q ← 0 + 0 + 7 x 1 x 5 q ← 35 We have m [i, j] = m [1, 2] = ∞ Comparing q with m [1, 2] q < m [i, j] i.e. 35 < m [1, 2] 35 < ∞ True then, m [1, 2 ] ← 35 (∴ m [i,j] ← q) s [1, 2] ← k and the value of k = 1 s [1,2 ] ← 1 Insert "1" at dimension s [1, 2] in table s. And 35 at m [1, 2] 2. l remains 2 L = 2 i ← 1 to n  l + 1 i ← 1 to 4  2 + 1 i ← 1 to 3 for i = 1 done before Now value of i becomes 2 i = 2 j ← i + l  1 j ← 2 + 2  1 j ← 3 j = 3 m [i , j] ← ∞ i.e. m [2,3] ← ∞ Initially insert ∞ at m [2, 3] Now, for k ← i to j  1 k ← 2 to 3  1 k ← 2 to 2 i.e. k =2 Now, q ← m [i, k] + m [k + 1, j] + p_{i1} p_{k} p_{j} For l =2 i = 2 j = 3 k = 2 q ← m [2, 2] + m [3, 3] + p_{1}x p_{2} x p_{3} q ← 0 + 0 + 1 x 5 x 4 q ← 20 Compare q with m [i ,j ] If q < m [i,j] i.e. 20 < m [2, 3] 20 < ∞ True Then m [i,j ] ← q m [2, 3 ] ← 20 and s [2, 3] ← k and k = 2 s [2,3] ← 2 3. Now i become 3 i = 3 l = 2 j ← i + l  1 j ← 3 + 2  1 j ← 4 j = 4 Now, m [i, j ] ← ∞ m [3,4 ] ← ∞ Insert ∞ at m [3, 4] for k ← i to j  1 k ← 3 to 4  1 k ← 3 to 3 i.e. k = 3 Now, q ← m [i, k] + m [k + 1, j] + p_{i1} p_{k} p_{j} i = 3 l = 2 j = 4 k = 3 q ← m [3, 3] + m [4,4] + p_{2} x p_{3} x p_{4} q ← 0 + 0 + 5 x 2 x 4 q 40 Compare q with m [i, j] If q < m [i, j] 40 < m [3, 4] 40 < ∞ True Then, m [i,j] ← q m [3,4] ← 40 and s [3,4] ← k s [3,4] ← 3 Case 2: l becomes 3 L = 3 for i = 1 to n  l + 1 i = 1 to 4  3 + 1 i = 1 to 2 When i = 1 j ← i + l  1 j ← 1 + 3  1 j ← 3 j = 3 Now, m [i,j] ← ∞ m [1, 3] ← ∞ for k ← i to j  1 k ← 1 to 3  1 k ← 1 to 2 Now we compare the value for both k=1 and k = 2. The minimum of two will be placed in m [i,j] or s [i,j] respectively. Now from above Value of q become minimum for k=1 ∴ m [i,j] ← q m [1,3] ← 48 Also m [i,j] > q i.e. 48 < ∞ ∴ m [i , j] ← q m [i, j] ← 48 and s [i,j] ← k i.e. m [1,3] ← 48 s [1, 3] ← 1 Now i become 2 i = 2 l = 3 then j ← i + l 1 j ← 2 + 3  1 j ← 4 j = 4 so m [i,j] ← ∞ m [2,4] ← ∞ Insert initially ∞ at m [2, 4] for k ← i to j1 k ← 2 to 4  1 k ← 2 to 3 Here, also find the minimum value of m [i,j] for two values of k = 2 and k =3 Case 3: l becomes 4 L = 4 For i ← 1 to nl + 1 i ← 1 to 4  4 + 1 i ← 1 i = 1 do j ← i + l  1 j ← 1 + 4  1 j ← 4 j = 4 Now m [i,j] ← ∞ m [1,4] ← ∞ for k ← i to j 1 k ← 1 to 4  1 k ← 1 to 3 When k = 1 q ← m [i, k] + m [k + 1, j] + p_{i1} p_{k} p_{j} q ← m [1,1] + m [2,4] + p_{0}xp_{4}x p_{1} q ← 0 + 28 + 7 x 2 x 1 q ← 42 Compare q and m [i, j] m [i,j] was ∞ i.e. m [1,4] if q < m [1,4] 42< ∞ True Then m [i,j] ← q m [1,4] ← 42 and s [1,4] 1 ? k =1 When k = 2 L = 4, i=1, j = 4 q ← m [i, k] + m [k + 1, j] + p_{i1} p_{k} p_{j} q ← m [1, 2] + m [3,4] + p_{0} xp_{2} xp_{4} q ← 35 + 40 + 7 x 5 x 2 q ← 145 Compare q and m [i,j] Now m [i, j] i.e. m [1,4] contains 42. So if q < m [1, 4] But 145 less than or not equal to m [1, 4] So 145 less than or not equal to 42. So no change occurs. When k = 3 l = 4 i = 1 j = 4 q ← m [i, k] + m [k + 1, j] + p_{i1} p_{k} p_{j} q ← m [1, 3] + m [4,4] + p_{0} xp_{3} x p_{4} q ← 48 + 0 + 7 x 4 x 2 q ← 114 Again q less than or not equal to m [i, j] i.e. 114 less than or not equal to m [1, 4] 114 less than or not equal to 42 So no change occurs. So the value of m [1, 4] remains 42. And value of s [1, 4] = 1 Now we will make use of only s table to get an optimal solution.
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