Two Digits and a Number – Part 1: The Mistake

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As an old wise man used to say: “for every complicated problem, there is always an easy solution, albeit a wrong one”. Indeed.

 
Problem I stumbled upon was the following: take a digit D1. Take another digit D2. You’re allowed to generate numbers using only D1 and D2. Call the numbers that you generate this way M. Take a number N. Find the smallest M that is a multiple of N.

Ok, an example with real numbers will make this easier to visualize.
D1 = 1
D2 = 4
N = 16

What’s the smallest number composed of only “1”s and “4”s that is a multiple of 16? The answer is 144. Also, assume that there is always a solution.

One would be tempted to write a recursive solution like this (sketch, pseudo-code):

FindMinMultiple(int d1, int d2, int n, int generatedNumber){
  If(generatedNumber%n == 0) è found!
  FindMinMultiple(d1, d2, n, 10 * generatedNumber + Min(d1,d2))
  FindMinMultiple(d1, d2, n, 10 * generatedNumber + Max(d1,d2))
}

What’s wrong with this solution, and what would the right solution be? Answers in the next post. By the way, you should try to find a solution for
D1 = 3
D2 = 7
N = 987654321

With Love,
Marcelo

Comments

  1. TarasFebruary 24, 2015 at 9:45 AM

    Thanks Marcelo for an interesting article!

    As for problems with provided solution:
    - it's not going to terminate if there are no multiples. Easy test case would be D1 = 1, D2 = 3, N = 2. Since all numbers created from 1s and 3s will be odd, there will never be a multiple of 2.
    - it will always try to find a multiple that consists of MIN(D1,D2) before checking the other branch.

    Both of these limitations can be avoided by using a BFS instead of using DFS (as in provided implementation). A very simple implementation would look as follows:

    val MAX_DEPTH: Int = ;
    def findMinMultiple(d1: Int, d2: Int, n: Int): Int = {
    lazy val minMultiples = d1.min(d2) #:: d1.max(d2) #:: (minMultiples.flatMap { m => Stream.cons(m * 10 + d1.min(d2), Stream.cons(m * 10 + d1.max(d2), Stream.empty)) } ).filter(m => m % n == 0).take(MAX_DEPTH)
    minMultiples.head
    }

    Note: this solution can be easily improved to use BigInt instead of Int to be able to tackle bigger numbers, but it's suffices to show the solution.

    The solution uses BFS and as such is not going to be stuck on the MIN(D1,D2) branch and it also has a MAX_DEPTH limit to stop if we can't find a multiple after exploring MAX_DEPTH number of solutions from solution space.

    Thanks again Marcelo!

    ReplyDelete
  2. TarasFebruary 24, 2015 at 9:59 AM

    PS.: practically the recursive function provided in your post won't run forever, as it will very quickly overflow the stack and stop :)

    ReplyDelete
  3. Marcelo De BarrosFebruary 24, 2015 at 10:18 AM

    Taras, as always, you're flawless!!! :) I also love your elegant coding style! Now find the solution for:
    D1 = 3
    D2 = 7
    N = 987654321

    Cheers my friend!!!

    ReplyDelete

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