To show the number of possible seams grows exponential.

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Answer 1

Question

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .

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Answer 2

Question

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .

Answer 6

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

Answer 7

Chapter 15, Problem 8P

(a)

Program Plan Intro

To show the number of possible seams grows exponential.

Answer 8

(a)

Expert Solution

Explanation of Solution

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .

Answer 13

(b)

Program Plan Intro

To give an algorithm that finds a seam with lowest disruption measure.

Answer 14

(b)

Expert Solution

Explanation of Solution

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .