Happy New Year!

January 1, 2013

As we begin the new year, we note that 109-8*7+654*3-2/1 = 2013. There are three other combinations of the numbers 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, in order, combined with the five operators NIL, +, -, * and / that also evaluate to 2013.

Your task is to write a program that finds all four expressions that evaluate to 2013. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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11 Responses to “Happy New Year!”

  1. Paul said

    Happy 2013 to you all. Here a version in Python.

    from __future__ import division
    
    from itertools import product
    
    ops =  ["", "+", "-", "*","/"]
    nums = [str(i) for i in range(10, 0, -1)]
        
    def riffle(a, b):
        ita, itb = iter(a), iter(b)
        while 1:
            yield ita.next()
            yield itb.next()
        
    for op in product(ops, repeat=9):
        ev = "".join(riffle(nums, op))
        if eval(ev) == 2013:
            print ev
    
  2. Jussi Piitulainen said

    With Python 3 /, I suppose it’s just luck that some floating point approximation does not hide a proper solution :) Anyway, here’s a denser Python expression.

    from itertools import product, chain, repeat
    print(*(e for e in map(lambda rest : '10' + ''.join(rest).replace('|', ''),
                           product(*chain(*zip(repeat('|+-*/'), '987654321'))))
            if eval(e) == 2013),
           sep = '\n')
    
  3. (defun combine (nums)
      (cond ((null nums) '(""))
            ((null (cdr nums)) nums)
            (t
             (loop
                for rest in (combine (cdr nums))
                nconc (loop
                          for op in '("" "+" "-" "*" "/")
                          collect (concatenate 'string (car nums) op rest)))))) 
    
    (progn
      (assert (equal (combine '("1")) '("1")))
      (assert (equal (combine '("2" "1")) '("21" "2+1" "2-1" "2*1" "2/1")))
      (assert (equal (combine '("3" "2" "1"))
                     '("321" "3+21" "3-21" "3*21" "3/21" "32+1" "3+2+1"
                       "3-2+1" "3*2+1" "3/2+1" "32-1" "3+2-1" "3-2-1" "3*2-1"
                       "3/2-1" "32*1" "3+2*1" "3-2*1" "3*2*1" "3/2*1" "32/1"
                       "3+2/1" "3-2/1" "3*2/1" "3/2/1"))))
    
    (ql:quickload :infix)
    
    (loop
       for iexpr in  (combine '("10" "9" "8" "7" "6" "5" "4" "3" "2" "1"))
       for expr = (infix:string->prefix iexpr)
       for n = (eval expr)
       when (= n 2013)
       do (print iexpr))
    
    
    "10*98/7*6/5*4*3-2-1" 
    "10*9*8*7/6/5*4*3-2-1" 
    "109-8*7+654*3-2*1" 
    "109-8*7+654*3-2/1"
    
    
  4. jpverkamp said

    Here’s my take (in Racket):
    Happy New Year

    As I tend to do, I somewhat over-engineered the solution, allowing for any arbitrary list of numbers (which could actually be expanded to any datatype) and any arbitrary binary infix operators, including precedence. All of that made it a bit slower than it could have been, but it still manages to churn through the roughly 2 million possibilities for this problem in about 20 seconds (I’m using ~ for concatenation):

    > (solve '(10 9 8 7 6 5 4 3 2 1)
             `(((~ ,(lambda (x y)
                      (string->number (string-append 
                                       (number->string x)
                                       (number->string y))))))
               ((* ,*) (/ ,/))
               ((+ ,+) (- ,-)))
             2013)
    (10 ~ 9 - 8 * 7 + 6 ~ 5 ~ 4 * 3 - 2 * 1)
    (10 ~ 9 - 8 * 7 + 6 ~ 5 ~ 4 * 3 - 2 / 1)
    (10 * 9 ~ 8 / 7 * 6 / 5 * 4 * 3 - 2 - 1)
    (10 * 9 * 8 * 7 / 6 / 5 * 4 * 3 - 2 - 1)
    
  5. David said

    In Ruby…

    require 'mathn'
    
    countdown = 10.downto(1).map {|n| n.to_s}
    ['', '+', '-', '*', '/'].repeated_permutation(9) do |perm|
        str = countdown.zip(perm).join
        puts str if eval(str) == 2013
    end
    
    PS C:\Users\dave\Documents\dev\ruby> ruby .\countdown_2013.rb
    109-8*7+654*3-2*1
    109-8*7+654*3-2/1
    10*98/7*6/5*4*3-2-1
    10*9*8*7/6/5*4*3-2-1
    
  6. ardnew said

    It should probably be noted that all five allowable operators may appear 0 or more times in each one of the four expressions. Your first example demonstrates that fact (and that there are 9 required binary operations to occur), but it wasn’t immediately obvious to me

  7. Joe A said

    python with recursion

    def fun_2013(cur=10,expr=''):
    	expr += str(cur)
    
    	if cur == 1:
    		if eval(expr) == 2013:
    			print expr
    		return;
    
    	fun_2013 (cur-1, expr)
    	fun_2013 (cur-1, expr + "+")
    	fun_2013 (cur-1, expr + "-")
    	fun_2013 (cur-1, expr + "/")
    	fun_2013 (cur-1, expr + "*")
    

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