Holiday Gift Exchange

Question

Another problem from our internal golfing...this one around the holidays last year.

PROBLEM

Andy, Barb, Carl, Didi, Earl, and Fran are buying gifts for each other. Draw names for a gift exchange.

1. Each person buys one gift and receives one gift.
2. Nobody buys their own gift.
3. Running the solution multiple times should produce different results (giver-receiver pair should not be predictable or identical from run to run).

INPUT

None.

OUTPUT

Formatted as in this example:

Andy buys for Barb
Barb buys for Carl
Carl buys for Didi
Didi buys for Earl
Earl buys for Fran
Fran buys for Andy

Answer 1

J, 57

(,.' buys for ',"1|.)(?~6){6 4$'AndyBarbCarlDidiEarlFran'

eg

   (,.' buys for ',"1|.)(?~6){6 4$'AndyBarbCarlDidiEarlFran'
Carl buys for Earl
Andy buys for Barb
Fran buys for Didi
Didi buys for Fran
Barb buys for Andy
Earl buys for Carl

Answer 2

Japt -R, 41 bytes

`AÌ)B¼C¤E¤FÎÂDi¹`qe ö¬ê1 ò mq` ¿ys f 

-2 bytes thanks to @Oliver!

Try it!

This is the approach I took at a high level:

• uncompress a string containing participants names
• split string into an array
• shuffle it
• assign each person to the person with the next highest index
• the last person in the array gets assigned to the first

I have a little history with this problem as I created a "secret santa" program for my work years ago. We ended up asking a few job applicants to work through it as well :)

Answer 3

c99 -- 252 characters

#include <stdio.h>
#define G for(i=0;i<6;i++)
char*n="Andy\0Barb\0Carl\0Didi\0Earl\0Fran",*p[7];int i,j;int main()
{FILE*r=fopen("/dev/random","r");G p[i]=n+5*i;G{j=fgetc(r)%6;p[7]=p[j]
;p[j]=p[i];p[i]=p[7];}G printf("%s buys for %s\n",p[i],p[(i+1)%6]);}

Slight improvement by taking advantage of the circular nature of the permutation. This version always builds a loop-like buying strategy, so it is less random than the previous (271 character) version, but I believe it still meets the spec.

Requires a platform that has a working /dev/random. I ought to be able to knock about 8 off by omitting the \0s in the big string, but my libc doesn't seem to be dealing with %4s print specifiers the way the man page says.

The shuffle is bad, but doing it that way prevents me from having to check on "Foo buys for Foo" conditions.

Readable:

#include <stdio.h>

char *n="Andy\0Barb\0Carl\0Didi\0Earl\0Fran",
  *p[7]; /* 7th cell for temp */
int i,j;

int main(){
  FILE*r=fopen("/dev/random","r");
  for(i=0;i<6;i++)
    p[i]=n+5*i;   /* Initialize the pointers */
  for(i=0;i<6;i++){
    j=fgetc(r)%6; /* Poor numeric properties. Cest le Code Golf */
    p[7]=p[j];
    p[j]=p[i];
    p[i]=p[7];
  }
  for(i=0;i<6;i++)
    printf("%s buys for %s\n",p[i],p[(i+1)%6]);
}

Answer 4

Windows PowerShell, 83

$i=random 5
($n=-split'Andy Barb Carl Didi Earl Fran')|%{"$_ buys for "+$n[++$i%6]}

History:

• 2011-02-11 22:01 (136) – First attempt.
• 2011-02-11 22:05 (130) – Inlined a few things. Shuffling the names now, not the indexes.
• 2011-02-13 16:13 (128) – I don't need the modulus as $i will be recreated every time.
• 2011-02-13 16:20   (87) – Borrowed the idea from Anon.'s C# solution. Just generate a random offset and then just let them gift in circles.
• 2011-02-13 16:26   (83) – Changed random number generation and indexing. Pulled $_ into the string to save the +.

Answer 5

Haskell, 241 189 characters

import Data.List
import Random
main=randomRIO(0,719)>>=mapM_ putStrLn.f
f n=map(\(x,y)->x++" buys for "++y).zip(l n).tail$cycle$l n
l=(permutations(words"Andy Barb Carl Didi Earl Fran")!!)

Fully random output (that still satisfies the spec).

This generates all permutations of the list of the names, picks one at random (I think this is the shortest way in Haskell to shuffle a list - if anyone has anything smaller, I'd appreciate it), and then each person then buys a present for the next person in the list.

Answer 6

Golfscript: 72 64 57 chars

"AndyBarbCarlDidiEarlFran"4/{;9rand}${.n+\' buys for '}%(

Tests

$ golfscript codegolf-838.gs 
Fran buys for Carl
Carl buys for Andy
Andy buys for Barb
Barb buys for Didi
Didi buys for Earl
Earl buys for Fran

$ golfscript codegolf-838.gs 
Didi buys for Earl
Earl buys for Andy
Andy buys for Barb
Barb buys for Carl
Carl buys for Fran
Fran buys for Didi

• Thanks gnibbler for "AndyBarbCarlDidiEarlFran"4/, updated and got 7 chars less
• 57 chars solution is basically by Nabb :D, and also noticed that ;9rand is more random than my 6rand*

Answer 7

Python - 118 chars

from random import*;L="Andy Barb Carl Didi Earl Fran".split()
for i in sample(range(6),6):print L[i-1],"buys for",L[i]

Python - 120 chars

import random as R;L="Andy Barb Carl Didi Earl Fran".split();R.shuffle(L)
for i in range(6):print L[i-1],"buys for",L[i]

Answer 8

R - 85 characters

paste(n<-sample(c('Andy','Barb','Carl','Didi','Earl','Fran')),'buys for',n[c(6,1:5)])

Answer 9

Python - 154 chars

import random as R;L="Andy Barb Carl Didi Earl Fran".split();M=L[:]
while any(map(str.__eq__,L,M)):R.shuffle(M) 
for i in zip(L,M):print"%s buys for %s"%i

Answer 10

D: 233 Characters

import std.random,std.stdio;void main(){auto p=["Andy","Barb","Carl","Didi","Earl","Fran"];auto q=p.dup;o:while(1){for(int i;i<6;++i)if(p[i]==q[i]){randomShuffle(q);continue o;}break;}foreach(i,a;p)writefln("%s buys for %s",a,q[i]);}

More Legibly:

import std.random, std.stdio;

void main()
{
    auto p = ["Andy", "Barb", "Carl", "Didi", "Earl", "Fran"];
    auto q = p.dup;

    o:while(1)
    {
        for(int i; i < 6; ++i)
            if(p[i] == q[i])
            {
                randomShuffle(q);
                continue o;
            }

        break;
    }

    foreach(i, a; p)
        writefln("%s buys for %s", a, q[i]);
}

Answer 11

Python (175)

import random as r
n=['Andy','Barb','Carl','Didi','Earl','Fran']
m=n[:]
r.shuffle(m)
b=' buys for '
for i in n:
 h=m.pop()
 while h==i:
  m.append(h)
  h=m.pop()
 print(i+b+h)

Answer 12

Scheme, 173

Gives one of two solutions.

(define(m lst)
    (printf"~v buys for ~v~n"(car lst)(cadr lst))
    (if(eq?(cadr lst)'Andy)0(m(cdr lst)))
)
(m((if(odd?(random 2))reverse values)'(Andy Barb Carl Didi Earl Fran Andy)))

Answer 13

C#, 210 183 characters

using System;class a{static void Main(){var n="Andy Barb Carl Didi Earl Fran".Split();var c=0,i=new Random().Next(1,6);for(;c<6;c++)Console.WriteLine(n[c]+" buys for "+n[(c+i)%6]);}}

Heaps of boilerplate :(

This solution isn't totally random - there are always one or more "loops" of people e.g. A->C->E->A, and the offsets are always the same in the loops. However, it is not possible to predict the output of a particular run unless you have part of that output.

Answer 14

Ruby - 89 chars

(a=%w(Andy Barb Carl Didi Earl Fran).shuffle).zip(a.reverse).each{|e|puts e*' buys for '}

Output:

Andy buys for Didi
Barb buys for Earl
Fran buys for Carl
Carl buys for Fran
Earl buys for Barb
Didi buys for Andy

Answer 15

MathGolf, 41 bytes

"δ%è╘+µ√♂JÇ"2/$╦╕ää▐δáw_╪" buys for "+m+n

Try it online!

Explanation

This is not guaranteed to produce each case with equal probability, but it does produce different results each run. One byte could be removed if I had a shuffle operator, but that's for another day.

"δ%è╘+µ√♂JÇ"                                push the string "δ%è╘+µ√♂JÇ"
            2/                              split into segments of two characters
              $                             transform to ordinals using base 256
               ╦                            fetch dictionary words (['Andy', 'barb', 'Carl', 'Earl', 'Fran'])
                ╕ää                         Push "didi"
                   ▐                        append to end of list
                    δ                       capitalize all strings in list
                     áw                     sort by random character in each string (shuffle)
                       _                    duplicate TOS
                        ╪                   right-rotate bits in int, list, str
                         " buys for "       push the string " buys for "
                                     +      Add to all strings in list
                                      m+    zip add the two arrays
                                        n   join array with newline