Explanation Given information: For m = 4, n = 4 and sets A and B represented as the arrays a [ 1 ] = u , a [ 2 ] = v , a [ 3 ] = w , a [ 4 ] = x , b [ 1 ] = r , b [ 2 ] = u , b [ 3 ] = y , b [ 4 ] = z . Calculation: Input: m = 4 n = 4 a [ 1 ] = u a [ 2 ] = v a [ 3 ] = w a [ 4 ] = x b [ 1 ] = r b [ 2 ] = u b [ 3 ] = y b [ 4 ] = z We also initialize i as 1 and we initialize answer as A ⊆ B . i 1 j found answer A ⊆ B Next, the algorithm will enter the outer while-loop. We first assign the value 1 to j and the value “no” to found. i 1 1 j 1 found no answer A ⊆ B A ⊆ B Since, a [ 1 ] = u ≠ r = b [ 1 ] , the value of found is not changed in the first iteration of the inner while-loop. We increase the value of j by 1. Since a [ 1 ] = u = b [ 2 ] , we change the value of found to “yes”. i 1 1 1 j 1 2 found no yes answer A ⊆ B A ⊆ B A ⊆ B Since, found is set to “yes”, the inner while-loop stops executing and we then increase i by 1. Next, we start the second iteration of the outer while-loop, which re-initializes j as 1 and found as “no”. i 1 1 1 2 j 1 2 1 found no yes no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B Since, a [ 2 ] = v ≠ w = b [ 1 ] , the value of found is not changed in the first iteration of the inner while-loop. We increase the value of j by 1. Since, a [ 2 ] = v ≠ r = b [ 2 ] , the value remains “no” i 1 1 1 2 2 j 1 2 1 2 found no yes no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B We increase the value of j by 1. Since a [ 2 ] = v ≠ y = b [ 3 ] , the value of found remains “no” i 1 1 1 2 2 2 j 1 2 1 2 3 found no yes no no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B We increase the value of j by 1. Since a [ 2 ] = v ≠ z = b [ 4 ] , the value of found remains “no” i 1 1 1 2 2 2 2 j 1 2 1 2 3 4 found no yes no no no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B We increase the value of j by 1. Since j = 5 > 4 = n, the inner while-loop stops executing. Since found is still set to “no”, answer is changed to A ⊆ B . i 1 1 1 2 2 2 2 2 j 1 2 1 2 3 4 5 found no yes no no no no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B Next, we increase i by 1. We then start the third iteration of the outer while-loop, which re-initializes j as 1 and found as “no” i 3 j 1 found no answer A ⊆ B Since a [ 3 ] = w ≠ r = b [ 1 ] , the value of found is not changed in the first iteration of the inner while-loop. We increase the value of j by 1. Since a [ 3 ] = w ≠ u = b [ 2 ] , the value of found remains “no” i 3 3 j 1 2 found no no answer A ⊆ B A ⊆ B We increase the value of j by 1. Since a [ 3 ] = w ≠ y = b [ 3 ] , the value of found remains “no” i 3 3 3 j 1 2 3 found no no no answer A ⊆ B A ⊆ B A ⊆ B We increase the value of j by 1. Since a [ 3 ] = w ≠ z = b [ 4 ] , the value of found remains “no” i 3 3 3 3 j 1 2 3 4 found no no no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B We increase the value of j by 1. Since j = 5 > 4 = n, the inner while-loop stops executing. Since found is still set to “no”, answer is set to A ⊆ B . i 3 3 3 3 3 j 1 2 3 4 5 found no no no no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B Next, we increase i by 1

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ISBN: 9781337694193

Author: EPP, Susanna S.

Publisher: Cengage Learning,

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Chapter 6.1, Problem 37ES

To determine

To trace:

The action of Algorithm 6.1.1 on the variables i, j, found and answer for m = 4,

n = 4.

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Chapter 6.1, Problem 36ES

Chapter 6.1, Problem 38ES

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Discrete Mathematics With Applications

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To trace: The action of Algorithm 6.1.1 on the variables i, j, found and answer for m = 4, n = 4.

Solution Summary: The author traced the action of Algorithm 6.1.1 on the variables i, j, found and answer for m = 4, n = 4.

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To trace: The action of Algorithm 6.1.1 on the variables i, j, found and answer for m = 4, n = 4.

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Chapter 6 Solutions

To trace:

The action of Algorithm 6.1.1 on the variables i, j, found and answer for m = 4,

n = 4.