<< problem 32 - Pandigital products Digit factorials - problem 34 >>

# Problem 33: Digit cancelling fractions

The fraction frac{49}{98} is a curious fraction, as an inexperienced mathematician in attempting to simplify it
may incorrectly believe that frac{49}{98} = frac{4}{8}, which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, frac{30}{50} = frac{3}{5}, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

# My Algorithm

The original problem can be solved with brute force:

• two nested loops iterate over all numerators n and denominators d such that n < d
• each number is split into its digits
• actually only erasing the lower digit of d and the higher digit of n can produce a valid result
• multiply all ns and ds, then divide by their Greatest Common Divisor
Hackerrank modified the problem such that larger numbers (up to 4 digits) can be found in both numerator and denominator.
On top of that, there is a variable number of digits to be cancelled.
And the worst: Hackerrank's problem description is very vague and doesn't clarify many corner cases.
Nevertheless, it was way more fun than the simple original problem ...

## Modifications by HackerRank

My main insight was that instead of cancelling/removing digits we can do the inverse, too:
iterate over all "small" numbers and insert digits at all possible positions.

Now we have five (instead of two) nested loops:

• the outer loops generate all combinations of n and d such that n < d.
• the "middle" loop generates all potential numbers to be inserted; they may have multiple digits
• the inner loops produce all permutations of the digits to be inserted
I convert my numbers to std::strings (num2str and str2num).
A string can either be a valid number or contain dots which are placeholders and mean "any digit" - inspired by the syntax of regular expressions.
Note: The placeholder must be alphabetically lower than all digits because I use std::next_permutation:

If we cancel two digits, the middle loop emits "..10", "..11" ... "..99" and the inner loops permute them to
(for "..10":) ".10.", ".1.0", ... "10..".

merge combines a mask (like ".1.0") and a number (like "34") to a number 3140.
until the "large" numerator/denominator match the "small" numerator/denominator.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent to
echo "3 1" | ./33

Output:

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <string>
#include <algorithm>
#include <unordered_set>

// convert number to string
std::string num2str(unsigned int x, unsigned int digits)
{
std::string result;
// it's faster to generate the digits in reverse order ...
while (digits-- > 0)
{
auto digit = x % 10;
result += char(digit + '0');
x /= 10;
}
// ... and then reverse them into their original order
std::reverse(result.begin(), result.end());
return result;
}

// ... and back
unsigned int str2num(const std::string& str)
{
unsigned int result = 0;
for (auto s : str)
{
result *= 10;
result += s - '0';
}
return result;
}

// fill all gaps in mask (marked as '.') with the digits found in str and return result as a number
unsigned int merge(const std::string& strFill, const std::string& mask)
{
auto iteFill = strFill.begin();
unsigned int result = 0;
{
result *= 10;
// if placeholder '.' is found, then take next digit from strFill
if (m == '.')
result += *iteFill++ - '0';
else // else take the digit of the mask
result += m - '0';
}
return result;
}

int main()
{
//#define ORIGINAL
#ifdef ORIGINAL
// brute-force solution for the original problem
unsigned int multD = 1;
unsigned int multN = 1;
for (unsigned int d = 10; d <= 99; d++) // denominator
for (unsigned int n = 10; n < d; n++) // numerator
for (unsigned int cancel = 1; cancel <= 9; cancel++)
{
auto lowN  = n % 10;
auto lowD  = d % 10;
auto highN = n / 10;
auto highD = d / 10;
// we could check all combinations:
// 1. cancel low  digit of denominator and low  digit of numerator
// 2. cancel high digit of denominator and low  digit of numerator
// 3. cancel low  digit of denominator and high digit of numerator
// 4. cancel high digit of denominator and low  digit of numerator
// but actually only case 2 is relevant
// (you can prove that but in this problem I focus on the much harder Hackerrank version)
// two fractions a/b and c/d are equal if a*d=b*c
if (highD == cancel && lowN == cancel && lowD * n == highN * d)
{
multN *= n;
multD *= d;
}
}

// shorter code than applying the "least common multiple"
for (unsigned int i = 2; i <= multN; i++)
// remove all common prime factors
while (multN % i == 0 && multD % i == 0)
{
multN /= i;
multD /= i;
}
std::cout << multD << std::endl;

return 0;
#endif

// and now a completely different approach for the Hackerrank version of the problem
unsigned int digits;
unsigned int cancel;
std::cin >> digits >> cancel;
auto keep = digits - cancel;

const unsigned int Tens[] = { 1, 10, 100, 1000, 10000 };
unsigned int sumN  = 0;
unsigned int sumD  = 0;

// don't count fractions twice
std::unordered_set<unsigned int> used;

// I do the inverse:
// "d" and "n" stand for denominator and numerator
// they are small numbers where I insert digits
// let's iterate over all "reduced" number and then iterate over all digits we could insert
// note: initially n and d started at Tens[keep - 1] instead of 1 but I learnt the hard way
//       that Hackerrank thinks 3016/6032 = 01/02 is a valid reduction
for (unsigned int d = 1; d < Tens[keep]; d++)
for (unsigned int n = 1; n < d; n++)
{
// convert to string
auto strN = num2str(n, keep);
auto strD = num2str(d, keep);

// try to insert all combinations
for (auto insert = Tens[cancel - 1]; insert < Tens[cancel]; insert++)
{
// convert to string
auto strInsert = num2str(insert, cancel);

// if number's digits are (partially) descending, then we already saw all its permutations
bool isAscending = true;
for (size_t i = 1; i < strInsert.size(); i++)
if (strInsert[i - 1] > strInsert[i])
{
isAscending = false;
break;
}
if (!isAscending)
continue;

// prepend placeholders (must be alphabetically smaller than '0')
strInsert.insert(0, keep, '.');

// check all permutations
// strInsertN is permutated until we arrive at the original value again
auto strInsertN = strInsert;
do
{
auto newN = merge(strN, strInsertN);

// the leading digit of the not-cancelled fraction must not be zero
// strangely enough, the leading digit of the cancelled fraction can be zero
if (newN < Tens[digits - 1])
continue;

// strInsertD is permutated until we arrive at the original value again
auto strInsertD = strInsert;
do
{
auto newD = merge(strD, strInsertD);

// in case we find the same fraction multiple times
// two fractions a/b and c/d are equal if a*d=b*c
if (newN * d == newD * n)
{
// ensure we haven't seen that fraction yet
auto id = newN * 10000 + newD;
if (used.count(id) == 0)
{
sumN  += newN;
sumD  += newD;

used.insert(id);
}
}
}
while (std::next_permutation(strInsertD.begin(), strInsertD.end()));
}
while (std::next_permutation(strInsertN.begin(), strInsertN.end()));
}
}

std::cout << sumN << " " << sumD << std::endl;
return 0;
}


This solution contains 19 empty lines, 40 comments and 6 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 24, 2017 submitted solution

# Hackerrank

My code solves 6 out of 6 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as hard.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 32 - Pandigital products Digit factorials - problem 34 >>
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