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What I am trying to do is calculate the row and column of which an id is in. For example I have the grid

0  1  2  3
4  5  6  7
8  9  10 11
12 13 14 15

As you can see this is a 4x4 grid. I work out the row number by getting the id then dividing by the number of columns, and rounding this number down. For example rownumber = 5(id value)/4(number of columns) = 1.25 then round down = 1. But what I can't figure out is how to get the column number of say the number 6.

Any help?

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closed as off-topic by Len, Ric, leoll2, Mike Earnest, Engineer Toast Apr 28 '15 at 17:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Len, Ric, leoll2, Mike Earnest, Engineer Toast
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ The operation you want is "mod", or possibly "rem" for "remainder", depending on the specific program you're using. If you want the column number of 5, you evaluate 5 mod 4 = 1 (sometimes written as mod(5,4)). If you don't have access to "mod" or "rem", you can do it by getting the row number, multiplying by the number of columns, and subtracting it from the id. So 5 - 4*1 = 1. $\endgroup$ – Glen O Apr 28 '15 at 13:33
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If $n$ is the number in the table

the row is so calculated:

$n/4$ rounded to the lower integer if decimal.

while the column is:

$n \bmod(4)$ where $\bmod(4)$ means the remainder of the division by 4.

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  • $\begingroup$ Rounding is often not an inbuilt function, so $n-n \bmod(4)$ will directly give the row number. $\endgroup$ – ghosts_in_the_code Apr 28 '15 at 15:05
  • $\begingroup$ @ghosts_in_the_code Good point. If this has to be implemented, he could simply do (int)(n/4) in C++ or anything equivalent in other programming languages. Btw, this is a bit off-topic, as the question is... $\endgroup$ – leoll2 Apr 28 '15 at 15:09

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