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Naddika [18.5K]
3 years ago
15

4) Which is a more efficient way to determine the optimal number of multiplications in a matrix-chain multiplication problem: en

umerating all the ways of parenthesizing the product and computing the number of multiplications for each, or running RECURSIVE-MATRIX-CHAIN? Justify your answer

Computers and Technology
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

Running RECURSIVE-MATRIX-CHAIN is asymptotically more efficient than enumerating all the ways of parenthesizing the product and computing the number of multiplications of each.

the running time complexity of enumerating all the ways of parenthesizing the product is n*P(n) while in case of RECURSIVE-MATRIX-CHAIN, all the internal nodes are run on all the internal nodes of the tree and it will also create overhead.

Explanation:

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(a) Store last 7 digits of your student ID in a vector (7 element row or column vector). Write a MATLAB code which creates a 7x7
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Answer:

MATLAB code explained below with appropriate comments for better understanding

Explanation:

clc

clear all

ID = [1 2 3 4 5 6 7]; % Replace this with your student ID

%(a)

A = zeros(7);

for i=1:7

 

A(i,i)= ID(i);

 

end

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%(b)

if(mod(A(6,6),2)==0)

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else if(A(3,3)<0)

fprintf('A(3,3) is negative\n');

else

fprintf('A(3,3)=0\n');

end

end

%(d)

fprintf('\nRequired series : ');

n = 35;

while n>=0

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fprintf(', ');

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n = n - 5;

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n = input('\nInput an integer : ');

fprintf('%i! = ',n);

F = 1;

while n>1

F = F * n;

n = n - 1;

end

fprintf('%i\n',F);

%(f)

clear all

x = [3 7 2 1];

y = [4 3 9 1];

A = 0;

C = x(1);

for i=1:length(x)

A = A + x(i)*y(i);

B(i) = x(i)/y(i);

if(min(x(i),y(i))<C)

C = min(x(i),y(i));

end

end

C = 1/C;

fprintf('\nA = %i\n',A);

fprintf('B = ');

disp(B)

fprintf('C = %0.5g\n',C);

%(g)

clear all

A = randi([5 25],[1,10]);

maxA = A(1);

for i = 2:10

if(maxA<A(i))

maxA = A(i);

end

end

minA = A(1);

i = 2;

while i<11

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i = i+1;

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fprintf('\nA = ');

disp(A);

fprintf('maxA = %i\n',maxA);

fprintf('minA = %i\n',minA);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%output

A =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

B =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

Both A and B are same

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A(3,3) is positive

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Input an integer : 6

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A = 52

B = 0.7500 2.3333 0.2222 1.0000

C = 1

A = 25 14 7 10 13 17 10 17 19 9

maxA = 25

minA = 7

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