How to resolve issue while dealing with larger matrices?

A=randi(16,[4,4]);

B=A(:)';

P=randi(16,[1 4]);

V=randi(16,[1 4]);

B(P)=V; % replace entries at position P with values in V

C=B(1,P)

C=V satisfied in this case

But I'm not getting the right answer in case of larger size matrices.

A=randi(255,[512,512]);

>> B=A(:)';

>> P=randi(255,[1 256]);

>> V=randi(256,[1 256]);

>> B(P)=V; %

>> C=B(1,P);

C should be equal to B , but I'm not getting it right,

Request for help regarding this.

[Arif Hoq]

No, its not same even for a smaller matrix. try to run several times for case 1. C and V are not same every time as you are working with random number.

A=randi(16,[4,4]);
B=A(:)';
P=randi(16,[1 4]);
V=randi(16,[1 4]);
B(P)=V; % replace entries at position P with values in V
C=B(1,P) ;
isequal(C,V)

ans = logical
   1

%%
A=randi(16,[4,4]);
B=A(:)';
P=randi(16,[1 4]);
V=randi(16,[1 4]);
B(P)=V; % replace entries at position P with values in V
C=B(1,P) ;
isequal(C,V)

ans = logical
   0

[Ammy]

@Arif Hoq thank you, but is there any way to get back V from B?

[Arif Hoq]

I am not sure about your expectation. but as you are working with random number you need to run several times to get the V from B. still need to develop more.

A=randi(16,[4,4]);
B=A(:)';
P=randi(16,[1 4]);
V=randi(16,[1 4]);
[C out]=ismember(V,B);
if numel(out(out~=0)) ==4
    disp(out)
else
    disp('V is not the member of B')
end

[Ammy]

Sorry, I'm not getting your point , while using random numbers the output varies with in each iteration.

But when I run only one time, and name them with a varable then it is fixed, until I run it next time. So in this case there should be a posibiliy to get V back with B.

A=randi(16,[4,4]);

B=A(:)';

P=randi(16,[1 4]);

V=randi(16,[1 4]);

B(P)=V;

[Arif Hoq]

But when I run only one time, and name them with a varable then it is fixed, until I run it next time

can you please show the result ?

[Ammy]

A=randi(16,[4,4])

B=A(:)'

P=randi(16,[1 4])

V=randi(16,[1 4])

B(P)=V;

A =

1 3 11 13

4 7 12 7

7 3 13 6

4 13 2 8

B =

1 4 7 4 3 7 3 13 11 12 13 2 13 7 6 8

P =

16 16 13 2

V =

2 13 8 7

Now, I wanted to say that these are B,P,V, are fixed now,how to further operate these for required purpose

[Arif Hoq]

You want to get V from B, right ? only the the value of V(here 2 13 8 7) ? and how about other values in B( here 1 4 7 4 3 7 3 13 11 12 13 2 13 7 6 8 ) ? others would be 0 ?

[Ammy]

@Arif Hoq Got the point why I'm facing the problem, thank you very much. How can I vote your answers?

[Arif Hoq]

i am posting my previous response as an answer

Answer 1

[Voss]

As @Arif Hoq points out, the problem is not with the size of the matrices. The problem is that the elements of P are not guaranteed to be unique. Look at what happens when a number occurs more than once in P, in this case 8 appears twice in P:

A=randi(16,[4,4]);
B=A(:)'

B = 1×16

8 9 11 8 9 4 5 11 1 14 13 5 1 15 13 5

P=randi(16,[1 4])

P = 1×4

7 1 8 8

V=randi(16,[1 4])

V = 1×4

2 11 13 4

B(P)=V % replace entries at position P with values in V

B = 1×16

11 9 11 8 9 4 2 4 1 14 13 5 1 15 13 5

C=B(1,P) 

C = 1×4

2 11 4 4

C==V 

ans = 1×4 logical array
   1   1   0   1

In this case, B(P)=V is B([7 1 8 8]) = [2 11 13 4]; which means that B(8) is assigned the value 13 and then immediately assigned the value 4, so that B(8) is 4 after that step.

Then C = B(1,P) is C = B(1,[7 1 8 8]) = [2 11 4 4] not [2 11 13 4] = V

Any time P has repeated values, then V will not be recoverable from B and P.

[Ammy]

Got the point, thank you very much.

Answer 2

[Arif Hoq]

A=randi(16,[4,4]);
B=A(:)';
P=randi(16,[1 4]);
V=randi(16,[1 4]);
[C out]=ismember(V,B);
if numel(out(out~=0)) ==4
    disp(out)
else
    disp('V is not the member of B')
end