Minimum perimeter of an area

Question

Just a simple code golf function for fun, intentionally left open with few rules to see what creativity comes up.

Input: An integer representing the area of a rectangle.

Output: Two integers representing the side lengths of the rectangle that have the least perimeter for the area. (In any order.)

Test cases:

25 => 5, 5
53 => 53, 1
4294967295 => 65537, 65535

Answer 1

GolfScript (21 chars)

:^,{)^\%!},.,2/=)^1$/

This takes the input as a number on the stack and leaves the result as two numbers on the stack.

For fair comparison with Howard's solution, taking input on stdin and giving output on stdout separated by newline is 23 chars:

~:^,{)^\%!},.,2/=)n^2$/

It works because this problem is trivial: it's just looking for the pair of factors closest to sqrt(area).

Online demo for a square; online demo for a non-square.

Answer 2

Mathematica 34 26

Besides the explicit search there is a nice convergent series:

n = 27

{i=√n//.i_:>n/⌈n/⌊i⌋⌉,n/i}

{3, 9}

Three previous approaches with 34 characters:

{#,n/#}&@FixedPoint[n/⌈n/⌊#⌋⌉&,√n]

For[i=√n,i>(i=n/⌈n/⌊i⌋⌉),];{i,n/i}

f@i_:=f[f@i=n/⌈n/⌊i⌋⌉]
{i=f@√n,n/i}

ClearAll[f]

Visualization:

p = FixedPointList[n/⌈n/⌊#⌋⌉ &, Sqrt[n]];

Plot[n/x, {x, 0, 11}, GridLines -> {Range@n, Range@n}, 
 AspectRatio -> Automatic, PlotRange -> {{0, 10.2}, {0, 7.2}}, 
 Epilog -> {Red, Thickness[0.005], 
   Arrow[Transpose[{{n/p, ⌈p⌉}, {n/p, ⌊p⌋}, {⌈n/⌊p⌋⌉, ⌊p⌋}}, {2, 3, 1}]], 
  PointSize[0.02], Black, Point[{n/p[[-1]], p[[-1]]}]}]

Answer 3

Python 2 (63 62)

n=input()
print[(i,n/i)for i in range(1,n)if i*i>=n/i*i==n][0]

This produces all pairs of integers (i, n/i) that could be the sides of the rectangle, starting from the first one greater or equal to the square root of n. It prints the first one.

Answer 4

Golfscript (46 43 40)

~1{..*2$>!}{1$1$%!{.@@}*)}while;1$/p p];

No way to beat the math oriented languages at this challenge I suspect :)

Somewhat "long winded", it would be shorter to work with arrays, sadly they get a bit large with the last test case.

Basically what it does is similar to the Python solution, it loops from 1..sqrt(n), testing for an even multiplier, then just displaying the last value found.

Answer 5

GolfScript, 30 characters

~:t,{[.t\/].~*t=1$~>!&\@if}*n*

Does a test on all numbers as many other solutions.

Answer 6

C# (178)

int[] R(int n){var a=Enumerable.Range(1,n);var b=a.SelectMany(x=>a.SelectMany(y=>a.Select(_=>new{x,y}))).Where(f=>f.x*f.y== n).OrderBy(f=>f.x+f.y).First();return new[]{b.x,b.y};}

Pretty

int[] R(int n)
{
    var a = Enumerable.Range(1, n);
    var b = a.SelectMany(x => a.SelectMany(y => a.Select(_ => new { x, y }))).Where(f => f.x * f.y == n).OrderBy(f => f.x + f.y).First();
    return new[] { b.x, b.y };
}

Answer 7

Java 8, 73 72 bytes

Saved one byte due to @ThomasKwa!

n->{for(int i=(int)Math.sqrt(n);;i--)if(n%i<1)return new int[]{n/i,i};};

Lambda function, test with:

public class Rectangle {
    interface Test {
        int[] run(int v);
    }
    public static void main (String[] args){
        Test test = n->{for(int i=(int)Math.sqrt(n);;i--)if(n%i<1)return new int[]{n/i,i};};

        int[] testCases = {1, 4, 8, 15, 47, 5040, 40320, 25, 53};
        for (int i : testCases) {
            int[] result = test.run(i);
            System.out.println(i + ": " + result[0] + ", " + result[1]);
        }
    }
}

Finds the greatest divisor less than or equal to the square root of the area. Returns an int[].

Answer 8

C, 54 bytes

f(x,y){for(y=sqrt(x);x%y;y--);printf("%d, %d",y,x/y);}

Just some general silliness.

Test if you like:

int main() {
    printf("Enter a number\n");
    int a;
    scanf("%d", &a);
    f(a);
    printf("\n");
    return 0;
}

Answer 9

Jelly, 8 bytes

÷Ɱ½ĖḞƑƇṪ

Try it online!

Use a bunch of new quicks. (ƇⱮƑ)

---

Jelly, 9 bytes

ọⱮ½TṪð,÷@

Try it online!

A port of my other answer.

Answer 10

C# 4.0: 79 characters

long[] P(long a){long r=0,s=0;while(++s<a/s)if(a%s==0)r=s;return new[]{r,a/r};}

If it weren't for the last test case, I could save 3 characters naming long to int.

In pretty format:

long[] Perimeter(long area)
{
    long result = 0;
    long side = 0;
    while (++side < area / side)
    {
        if (area % side == 0)
            result = side;
    }
    return new long[] { result, area / result };
}

Answer 11

C, 71 bytes

Nothing fancy, find the pair of integers (a, b) closest to sqrt(n) such that a*b==n:

f;x;L(n,a,b)int*a,*b;{for(f=0,x=sqrt(n);x;--x)n%x||f++?:(*a=x,*b=n/x);}

If the function is allowed to work only on first invocation, then we can shrink it to 67 bytes:

f;x;L(n,a,b)int*a,*b;{for(x=sqrt(n);x;--x)n%x||f++?:(*a=x,*b=n/x);}

Test main:

#include <stdio.h>

int main() {
  int testdata[] = {1, 4, 8, 15, 47, 5040, 40320};
  int x,y;
  for (int i = 0; i < 7; ++i) {
     L(testdata[i], &x, &y);
     printf("%5d >  %dx%d\n", testdata[i], x, y);
  }
}

The code was originally written for another golf, which is a duplicate of this.