Code Golf: Countdown Number Game

Python

Number of characters: 548 482 425 421 416 413 408

from operator import *
n=len
def C(N,T):
 R=range(1<<n(N));M=[{}for i in R];p=1
 for i in range(n(N)):M[1<<i][1.*N[i]]="%d"%N[i]
 while p:
  p=0
  for i in R:
   for j in R:
    m=M[i|j];l=n(m)
    if not i&j:m.update((f(x,y),"("+s+o+t+")")for(y,t)in M[j].items()if y for(x,s)in M[i].items() for(o,f)in zip('+-*/',(add,sub,mul,div)))
    p|=l<n(m)
 return min((abs(x-T),e)for t in M for(x,e)in t.items())[1]

you can call it like this:

>>> print C([50, 100, 4, 2, 2, 4], 203)
((((4+2)*(2+100))/4)+50)

Takes about half a minute on the given examples on an oldish PC.

Here’s the commented version:

def countdown(N,T):
  # M is a map: (bitmask of used input numbers -> (expression value -> expression text))                  
  M=[{} for i in range(1<<len(N))]

  # initialize M with single-number expressions                                                           
  for i in range(len(N)):
    M[1<<i][1.0*N[i]] = "%d" % N[i]

  # allowed operators                                                                                     
  ops = (("+",lambda x,y:x+y),("-",lambda x,y:x-y),("*",lambda x,y:x*y),("https://stackoverflow.com/",lambda x,y:x/y))

  # enumerate all expressions                                                                             
  n=0
  while 1:

    # test to see if we're done (last iteration didn't change anything)                                   
    c=0
    for x in M: c +=len(x)
    if c==n: break
    n=c

    # loop over all values we have so far, indexed by bitmask of used input numbers                       
    for i in range(len(M)):
      for j in range(len(M)):
        if i & j: continue    # skip if both expressions used the same input number                       
        for (x,s) in M[i].items():
          for (y,t) in M[j].items():
            if y: # avoid /0 (and +0,-0,*0 while we're at it)                                             
              for (o,f) in ops:
                M[i|j][f(x,y)]="(%s%s%s)"%(s,o,t)

  # pick best expression                                                                                  
  L=[]
  for t in M:
    for(x,e) in t.items():
      L+=[(abs(x-T),e)]
  L.sort();return L[0][1]

It works through exhaustive enumeration of all possibilities. It is a bit smart in that if there are two expressions with the same value that use the same input numbers, it discards one of them. It is also smart in how it considers new combinations, using the index into M to prune quickly all the potential combinations that share input numbers.