# Modify the code from the previous example

2 vues (au cours des 30 derniers jours)
Abigail McGahey le 19 Fév 2019
Commenté : Walter Roberson le 22 Fév 2019
clear, close all;
% Activity problem 6
% First example. With a loop, no ligical indexing.
N=6;
%NxN magic square
A=magic(N);
%Initialise B to be the same size as A with ones
B=ones(N);
%Set B to one where A is divisible by 3
%Loop over all elements
for row=1:N
for col=1:N
%Get current elemnet's value.
value=A(row,col);
%Is it a multiple of 3:
if (mod(value,3)==0)
B(row,col)=0;
end
end
end
disp(B)
% Second example, use logical indexing.
N=6;
A=magic(N)
B=ones(N);
indices=mod(A,3)==0;
B(indices)=0;
disp(B)
% part c
B(mod(A,3)==0)=0;
Now, based on the practice above, modify the code so that it starts with a magic square of size 5x5 and produces a second array that has a value of one where the magic square element is odd and a value of 2 where the corresponding magic square element is even.
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Abigail McGahey le 19 Fév 2019
okay, so to be more clear, I am struggling to figure out how to replace the odd elements with one and the even elements with the value of two.
Walter Roberson le 22 Fév 2019
"I am struggling" is not a question. A question might be something like how to write loops, or how to detect that a value is odd or even.

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### Réponses (1)

Pranjal Priyadarshi le 22 Fév 2019
The code below takes a magic square of size 6 and replaces the even elements with 2 and replaces the odd elements with 1.
N=6;
A=magic(N);
evenindex=(mod(A,2) == 0); %finds the index for even elements
finalans=A;
finalans(evenindex)=2; %replaces even elements with 2
finalans(~evenindex)=1; %replaces odd elements with value 1
disp(A);
disp(finalans);
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