Brilliant Cryptarithm

Computer Science Level 4

$\large \overline{BRI}\times\overline{LL}=\overline{IANT}$

Above shown is a cryptarithm where each letter represents a distinct single digit non-negative integer where $B,L,I$ are non-zero. How many solutions are there to the cryptarithm above?

The answer is 6.

2 solutions

Chew-Seong Cheong
Jul 27, 2015

I use Python coding as follows:

 from itertools import permutations
 n = 0
 for p in permutations("0123456789",7):
     B = p[0]
     R = p[1]
     I = p[2]
     L = p[3]
     A = p[4]
     N = p[5]
     T = p[6]
     if B != '0':
         if int(B+R+I)*int(L+L)==int(I+A+N+T):
             n += 1
             print n, B+R+I+L+A+N+T

Moderator note:

Simple standard approach.

Is there a non-tedious case checking way to solve this problem? (I don't think so)

Matthew Christopher
Aug 13, 2020

Generalised solver in C#.

static void Main(string[] args)
{
    Console.WriteLine("Multiplication cryptarithm");

    Console.Write("Enter first number: ");
    string num1 = Console.ReadLine();

    Console.Write("Enter second number: ");
    string num2 = Console.ReadLine();

    Console.Write("Enter result: ");
    string result = Console.ReadLine();

    Console.WriteLine($"Found {(CountCryptarithmSolutions(num1, num2, result))} solution(s).");
}

private static int CountCryptarithmSolutions(string num1, string num2, string result)
{
    int numberOfSolutions= 0;
    char[] uniqueChars = UniqueChars(num1, num2, result);

    List<int[]> digitPermutations = new List<int[]>();
    List<int[]> validDigitPermutations = new List<int[]>();

    foreach (IEnumerable<int> i in GetPermutations(Enumerable.Range(0, 10), uniqueChars.Length))
    {
        List<int> j = new List<int>();

        foreach(int k in i)
        {
            j.Add(k);
        }

        digitPermutations.Add(j.ToArray());
    }

    int num1LeadIndex = Array.IndexOf(uniqueChars, num1[0]);
    int num2LeadIndex = Array.IndexOf(uniqueChars, num2[0]);
    int resultLeadIndex = Array.IndexOf(uniqueChars, result[0]);
    foreach (int[] perm in digitPermutations)
    {
        if (!(perm[num1LeadIndex] == 0 || perm[num2LeadIndex] == 0 || perm[resultLeadIndex] == 0))
        {
            validDigitPermutations.Add(perm);
        }
    }

    foreach (int[] validPerm in validDigitPermutations)
    {
        if (IntFromString(num1, validPerm, uniqueChars) * IntFromString(num2, validPerm, uniqueChars) == IntFromString(result, validPerm, uniqueChars))
        {
            ++numberOfSolutions;

            Console.WriteLine($"{IntFromString(num1, validPerm, uniqueChars)} * {IntFromString(num2, validPerm, uniqueChars)} = {IntFromString(result, validPerm, uniqueChars)}");
        }
    }

    return numberOfSolutions;
}

private static int IntFromString(string input, int[] digits, char[] letters)
{
    string output = string.Empty;

    foreach (char c in input)
    {
        output += digits[Array.IndexOf(letters, c)].ToString();
    }

    return int.Parse(output);
}

private static char[] UniqueChars(string num1, string num2, string result)
{
    string output = num1 + num2 + result;

    return String.Concat(output.OrderBy(c => c)).Distinct().ToArray();
}

private static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length)
{
    if (length == 1) return list.Select(t => new T[] { t });

    return GetPermutations(list, length - 1)
        .SelectMany(t => list.Where(e => !t.Contains(e)),
            (t1, t2) => t1.Concat(new T[] { t2 }));
}

Brilliant Cryptarithm

The answer is 6.

2 solutions

Moderator note:

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