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 .

Question

Still sussing out bartleby?

Answer 1

BuyFind

Discrete Mathematics With Applicat...

5th Edition

EPP + 1 other

Publisher: Cengage Learning,

ISBN: 9781337694193

Solutions

Chapter

Section

Answer 2

BuyFind

Discrete Mathematics With Applicat...

5th Edition

EPP + 1 other

Publisher: Cengage Learning,

ISBN: 9781337694193

Solutions

Chapter

Section

Answer 3

Solutions

Chapter

Section

Answer 4

Chapter

Section

Answer 5

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 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

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 .

Discrete Mathematics With Applicat...

Solutions

Discrete Mathematics With Applicat...

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 .

Explanation of Solution

Still sussing out bartleby?

The Solution to Your Study Problems

Chapter 6 Solutions

Additional Math Textbook Solutions