Least amount of voters, given two halves

Question

One of my former students sent me a message about this interview question he got while applying for a job as a Junior Developer.

There are two candidates running for president in a mock classroom election. Given the two percentages of voters, find out the least amount of possible voters in the classroom.

Examples:

Input: 50.00,50.00
Output: 2

Input: 25.00,75.00
Output: 4

Input: 53.23, 46.77
Output: 124 // The first value, 1138 was wrong. Thanks to Loïc for the correct value

Note: The sum of the input percentages are always 100.00%, two decimal places

The last example got me scratching my head. It was the first time I heard about this problem, and I'm kindof stumped on how to solve this.

EDIT: I called my student about the problem, and told me that he was not sure about the last value. He said, and I quote, "It was an absurdly large number output" :( sorry! I should've researched more before posting it online~ I'm guessing 9,797 is the output on the last example though..

Answer 1

You can compute these values by using the best rational approximations of the voter percentages. Wikipedia describes how to obtain these values from the continued fraction (which can be computed these using the euclidean algorithm). The desired result is the first approximation which is within 0.005% of the expected value.

Here's an example with 53.23%:

10000 = 1 * 5323 + 4677
5323  = 1 * 4677 + 646
4677  = 7 * 646  + 155
646   = 4 * 155  + 26
155   = 5 * 26   + 25
26    = 1 * 25   + 1
25    = 25* 1    + 0

Approximations:
1:   1 / 1
 -> 1 = 100%
2:   1 / (1 + 1/1) 
 -> 1/2 = 50%
2.5: 1 / (1 + 1 / (1 + 1/6))
 -> 7/1 = 53.75%
3:   1 / (1 + 1 / (1 + 1/7))
 -> 8/15 = 53.33%
3.5: 1 / (1 + 1 / (1 + 1 / (7 + 1/3)))
 -> 25/47 = 53.19%
4:   1 / (1 + 1 / (1 + 1 / (7 + 1/4)))
 -> 33/62 = 53.23%

The reason we have extra values before the 3rd and 4th convergents is that their last terms (7 and 4 respectively) are greater than 1, so we must test the approximation with the last term decremented.

The desired result is the denominator of the first value which rounds to the desired value, which in this vase is 62.

Sample Ruby implementation available here (using the formulae from the Wikipedia page here, so it looks slightly different to the above example).

Answer 2

First you can notice that a trivial solution is to have 10.000 voters. Now let's try to find something lower than that.

For each value of N starting à 1
  For Each value of i starting à 1
     If i/N = 46.77
       return N

Always choose the minimum of the two percentages to be faster.

Or faster :

For each value of N starting à 1
  i = floor(N*46.77/100)
  For j = i or i+1 
     If round(j/N) = 46.77 and round((N-j)/N) = 53.23
       return N

---

For the third example :

• 605/1138 = .5316344464
• (1138-605)/1138 = .4683655536

but

• 606/1138 = .5325131810
• (1138-606)/1138 = .4674868190

It can't be 1138...

But 62 is working :

• 33/62 = .5322580645
• (62-33)/62 = .4677419355

Rounded it's giving you the good values.