Three Wise Men

December 28, 2012

It is easier to convert all the prices to pennies and work with a multiplicative total of 65.52 * 1003. Here’s a function using a list comprehension that solves the problem for any given total:

(define (magi n)
  (list-of (list a b c)
    (a range 1 (quotient n 2))
    (b range (+ a 1) (quotient (- n a) 3))
    (c is (- n a b))
    (= (* a b c) (* n 10000))))

> (magi 6552)
((52 200 6300))

The three prices are $0.52, $2.00 and $63.00. You can run the program at http://programmingpraxis.codepad.org/VpH8ZI25.

Pages: 1 2

Posted by programmingpraxis
Filed in Exercises
11 Comments »

11 Responses to “Three Wise Men”

  1. Paul said
    December 28, 2012 at 10:22 AM

    A Python version.

    def three_wise_men(sumabc):
    prodabc = sumabc * 10000
    for a in xrange(1,sumabc // 3):
        if prodabc % a == 0:
            prodbc = prodabc // a
            sumbc =sumabc - a
            for b in xrange(a, sumbc // 2):
                if prodbc % b == 0:
                    c = prodbc // b
                    if b + c == sumbc:
                        return a, b, c
            
print three_wise_men(6552)
  2. Programming Praxis – Three Wise Men « Bonsai Code said
    December 28, 2012 at 11:39 AM

    […] today’s Programming Praxis exercise, our goal is to solve a puzzle in which both the addition and […]

  3. Remco Niemeijer said
    December 28, 2012 at 11:40 AM

    My Haskell solution (see http://bonsaicode.wordpress.com/2012/12/28/programming-praxis-three-wise-men/ for a version with comments):

    main :: IO ()
main = print $ head [(a,b,c) | a <- [1..6550], b <- [1..a], let c = 6552 - a - b, a * b * c == 65520000]
  5. Mark Henderson said
    December 28, 2012 at 4:12 PM

    Takes a few minutes:

    def three_wise():
    r = [x for x in range(1, 6553)]
    return next((a, b, c) for a in r for b in r for c in reversed(r)
            if a + b + c == 6552
            if a * b * c == 65520000)

print three_wise()
  6. Overpants said
    December 28, 2012 at 10:02 PM

    This is an easier version of an earlier problem (https://programmingpraxis.com/2009/11/27/7-11/). Unlike the other problem, it seems that you can just brute force this one and still have it run fairly quickly.

  7. cosmin said
    December 29, 2012 at 6:19 AM

    Another Python solution:

    from math import sqrt, ceil

def divisors(n):
	f = [i for d in xrange(1, 1 + int(sqrt(n))) if n%d == 0 for i in [d, n/d]]
	if len(f) >= 2 and f[-1] == f[-2]: f.pop()
	return f

def three_wise_men(x):
	s, p = int(ceil(x*100)), int(ceil(x*1000000))
	return [(a, b, c) for a in divisors(p) if a <= s/3 for b in divisors(p/a) for c in [p/a/b] if a<=b<=c and a+b+c == s]

print three_wise_men(65.52)
  10. Heather said
    December 31, 2012 at 3:33 AM

    My fastest Ruby solution:

    $ time ruby wisemen.rb
    [0.52, 2.0, 63.0]

    real 0m1.919s
    user 0m1.795s
    sys 0m0.120s

    dollars = 65.52
cents = (dollars * 100).to_i

trips = []

i = 1
j = 1
halfcents = cents / 2
while true
  k = cents - i - j
  if k < j
    i += 1
    j = i
    if (i >= halfcents) || (k < 2)
      break
    else
      next
    end
  end
  trips << [i, j, k]
  j += 1
end

trips.each do |trip|
  trip = trip.map { |t| t / 100.0 }
  if trip.inject(:*) == dollars
    puts trip.inspect
    break
  end
end

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