Looping and printing matrix

if A=[2;4;6;8;6;7]. B=1

C=A-B,

C will become, [1;3;5;7;5;6], since one of the elements of C is 1 or less than one, Z1=B(current)

again i will increase B=2

now C = [0;2;4;6;4;5] since it is the same element which is zero, i want to ignore B(current).

again i will increase B=3

now C = [-1;1;3;5;3;4]. since one more element is equal or less than 1, Z2=B(current)

like this i want to do continously.

How to achieve this.

Any help will be appreciated.

Thanks.

Answer 1

[Image Analyst]

When do you stop? Eventually all elements will be 1 or less so you'll just keep increasing B forever and all elements of C will end up moving towards minus infinity. Do you stop when all elements are 1 or less? If so, why not just do

C = A - max(A(:)) + 1;

and not use a loop at all?

[Jagadesh Rao]

Hi image analyst, Thanks for your attention.

i will stop if all of it are 1. my real question is how to save the values of B as Z1 and Z2 variables with or without loop.

Thanks Jagadesh Rao Thalur

[Image Analyst]

You will never have all the elements be 1 if you are subtracting the same number from each element. Just think about it. You will never ever arrive at C=[1,1,1,1,1] by subtracting the same number from [2;4;6;8;6;7].