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Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193

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Section
BuyFindarrow_forward

Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193
Chapter 6.1, Problem 37ES
Textbook Problem
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Trace the action of Algorithm 6,1,1 on the variables I, j, found, and answer for m = 4, n = 4, and sets A and B represented as the arrays
a [ 1 ] = u , a [ 2 ] = v , a [ 3 ] = w ; b [ 1 ] = w ; b [ 2 ] = u .

To determine

To trace:

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

n = 4.

Explanation of Solution

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 AB.

i 1
j
found
answer AB

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 AB AB

Since, a[1]=ur=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 AB AB AB

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 AB AB AB AB

Since, a[2]=vw=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]=vr=b[2], the value remains “no”

i 1 1 1 2 2
j 1 2 1 2
found no yes no no
answer AB AB AB AB AB

We increase the value of j by 1.

Since a[2]=vy=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 AB AB AB AB AB AB

We increase the value of j by 1.

Since a[2]=vz=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 AB AB AB AB AB AB AB

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 AB.

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 AB AB AB AB AB AB AB AB

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 AB

Since a[3]=wr=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]=wu=b[2], the value of found remains “no”

i 3 3
j 1 2
found no no
answer AB AB

We increase the value of j by 1.

Since a[3]=wy=b[3], the value of found remains “no”

i 3 3 3
j 1 2 3
found no no no
answer AB AB AB

We increase the value of j by 1.

Since a[3]=wz=b[4], the value of found remains “no”

i 3 3 3 3
j 1 2 3 4
found no no no no
answer AB AB AB AB

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 AB.

i 3 3 3 3 3
j 1 2 3 4 5
found no no no no no
answer AB AB AB AB AB

Next, we increase i by 1

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Chapter 6 Solutions

Discrete Mathematics With Applications
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