how to find the next minimum value using if..else only?

let the given values are, b1(k)=195, b1(k+1)=190,and p1(k)=204,p1(k+1)=187,

p11(k)=220,p111(k)=188,p11(k+1)=203,p111(k+1)=171.

I need to find the optimal value of p1(k) and p1(k+1).

suppose,

     ax= abs(b1(k)-p1(k)) + abs(b1(k+1)-p11(k+1))= 22

     bx= abs(b1(k)-p1(k)) + abs(b1(k+1)-p111(k+1))=28

     cx= abs(b1(k)-p11(k)) + abs(b1(k+1)-p1(k+1))=28

       dx= abs(b1(k)-p11(k)) + abs(b1(k+1)-p11(k+1))=55

      ex= abs(b1(k)-p11(k)) + abs(b1(k+1)-p111(k+1))=44

          fx=abs(b1(k)-p111(k)) + abs(b1(k+1)-p1(k+1))=10

           gx=abs(b1(k)-p111(k)) + abs(b1(k+1)-p11(k+1))=20

          hx=abs(b1(k)-p111(k)) + abs(b1(k+1)-p111(k+1))=26

I need to find the optimal value of p1(k) and p1(k+1) out of available 8 combinations with 3 conditions. The 8 combinations are

1. p1(k),p11(k+1)

2. p1(k),p111(K+1)

3. p11(k),p1(k+1)

4. p11(k),p11(k+1)

6. p11(k),p111(k+1)

7. p111(k),p1(k+1)

8. p111(k),p11(k+1)

9. p111(k),p111(k+1).

and the 3 conditions are

1. The absolute difference between p1(k),p1(k+1) must be between 0 to 15.

2. Any one value out of p1(k) or p1(k+1) must be greater than 192.

3. I should consider first the minimum value among 'ax' to 'hx'

The actual probelem i am facing is, when i choose the smallest value 'fx'=10 then its corresponding values are p1(k)=p111(k)=188 and

p1(k+1)=p1(k+1)=187 ,,,here both the values are less than 192.i dont know how to check for the next lowest value i.e 'gx'=20.

[Image Analyst]

Why the restriction that you can only use if/else in your code? Don't you just want to solve the problem no matter how? Why limit yourself?

[aditya sahu]

Dear sir, it is also ok by using any techhnique.