comparing intervals using integers

Question

First off, I know there are many different easier/reliable ways to perform what I am about to ask, but that is not the point of my question.

I am attempting to compare two sets of integers as if they were fractions. What I mean by this is suppose I have a 2d array:

int array[2][2];

array[0][0] = 2;
array[0][1] = 3;
array[1][0] = 1;
array[1][1] = 50;

How I want to treat these numbers is that the number in:

array[0][0] = 2 <--- is the numerator

array[0][1] = 3 <--- is the denominator

Or just 2/3 in this case. What I want to do is then compare the two fractions;

if(2/3 < 1/50){
  //blah blah blah code here
}

The caveat here is that I can not convert the numbers to floating point numbers to retain their accuracy or create a temporary floating point placeholder. Is there any way to compare these using only integer values?

I also don't know exactly what I should tag for this question, if you think of something let me know and I'll tag it.

You should change the title to say "rational fractions" rather than "intervals" -- there are no intervals here. — Jim Balter, Jun 28 '12 at 05:21

score 5 · Accepted Answer · answered Jun 28 '12 at 04:16

5

Cross multiply the two numerators by one another's denominators

IE

2/3 vs 1/50th: multiply 50 and 1 by 3 and multiply 2 and 3 by 50.

Then you can compare the numerator without having to convert to a float.

answered Jun 28 '12 at 04:16

argentage

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The example is correct but the wording in the first sentence is not. Should be "cross multiply the two fractions by one another's denominators". – Jon Burgess Jun 28 '12 at 04:18
very nice, creating a common denominator didn't even cross my mind, thanks! – Jason M. Jun 28 '12 at 04:26

score 1 · Answer 2 · edited Jun 28 '12 at 20:37

1

if(array[0][0]*array[1][1])<array[0][1]*array[1][0])
{
    // your code here

}

edited Jun 28 '12 at 20:37

Praveen Kumar Purushothaman

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answered Jun 28 '12 at 04:19

Y2theZ

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score 0 · Answer 3 · answered Jun 28 '12 at 04:20

0

The most straightforward way is to find the least common multiple, and then convert the numerator. After that you could compare the numerator as integers.

i.e. 2*50 = 100; 1 * 3 = 3; ==> 100 > 3

answered Jun 28 '12 at 04:20

Hao Liu

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Least common multiple? No, you just multiply them. – flight Jun 28 '12 at 04:21
For the example asked, I simply multiply them. But the array could be more than 2 rows, it has to given a more generic solution. [2,3][1,50][1,25] ==> [100,150][3,150][6,150], 150 is the least common multiple for 3, 50, 25. – Hao Liu Jun 28 '12 at 04:34

comparing intervals using integers

3 Answers3