Locate a number in an array with a special ordering

Question

Below are two examples of arrays (with a special ordering with odd numbers on one side, and even number on the other side) I have, with some given len:

9  7  5  3  1  2  4  6  8

11  9  7  5  3  1  2  4  6  8  10

For some value z, I want to find the column the value is in. MATLAB code I wrote:

if z == 1
   column = ceil(len/2);   
elseif logical(mod(z,2))
   column = ceil(len/2) - (z-1)/2;
else
   column = ceil(len/2) + z/2;
end

Does this above locate the correct column?

Answer 1

I suspect that the better solution is to compute the column index directly, without the if-else conditions. You can do this by using integer division, and modulo. Consider the following computation breakdown:

% declare an int32 value 2
itwo = int32(2)
% compute the position of the column with the value 1
onecol = idivide(len + 1, itwo)
% compute how far away the desired value is from the 1 column
distance = idivide(z, itwo)
% compute the direction of the desired value from the 1 column (-1 is left (odd numbers), +1 is right (even numbers))
direction = 1 - 2 * mod(z, itwo) 

You can string this together as a 1-liner:

column = idivide(len + 1, itwo) + idivide(z, itwo) * (1 - 2 * mod(z, itwo))

You can also put it in a function should you choose. I have put together this example here:

https://goo.gl/Oenp85 (and updated for int32 conversion here: https://goo.gl/NgJJhN

Note that it has been a long time since I played with MATLAB.

indexOfVal = @(mat, val) idivide(numel(mat) + 1, int32(2)) + idivide(val, int32(2)) * (1 - 2 * mod(val, int32(2)));

x = [9 7 5 3 1 2 4 6 8];

result1 = indexOfVal(x, 9)
result2 = indexOfVal(x, 7)
result3 = indexOfVal(x, 5)
result4 = indexOfVal(x, 3)
result5 = indexOfVal(x, 1)
result6 = indexOfVal(x, 2)
result7 = indexOfVal(x, 4)
result8 = indexOfVal(x, 6)
result9 = indexOfVal(x, 8)

Produces:

result1 =  1                                                                                                                                                                                                                
result2 =  2                                                                                                                                                                                                                
result3 =  3                                                                                                                                                                                                                
result4 =  4                                                                                                                                                                                                                
result5 =  5                                                                                                                                                                                                                
result6 =  6                                                                                                                                                                                                                
result7 =  7                                                                                                                                                                                                                
result8 =  8                                                                                                                                                                                                                
result9 =  9                        

Answer 2

Obviously, the if z == 1 special case can be eliminated. When z is 1, then the logical(mod(z,2)) case applies, and the - (z-1)/2 term is 0.