Problem 74: Digit factorial chains

(see projecteuler.net/problem=74)

The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
1! + 4! + 5! = 1 + 24 + 120 = 145

Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:

169 → 363601 → 1454 → 169
871 → 45361 → 871
872 → 45362 → 872

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

69 → 363600 → 1454 → 169 → 363601 (→ 1454)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)

Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.

How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?

Algorithm

A function named fingerprintFactorial counts the number of digits. It's very similar to fingerprint from problem 49, problem 52 and problem 62.
Numbers with the same fingerprint are permutations of each other. The sum of the factorials of their digits will be the same.
This allows for a minor tweak: 0! = 1! = 1. Therefore fingerprintFactorial treats each 1 as a 0.

Whenever the length of a loop has to be determined, the std::vector named loop is filled with the factorial sums until we encounter one for the second time.

cache remembers all loop lengths indexed by their starting number. That cache is important to speed up Hackerrank repeated test cases.

Some numbers "don't fit" into my scheme: those linking back to themselves don't report the correct loop length (they are off-by-one).
That's why I manually adjust the result for those. All such "execeptional numbers" can be found in the problem statement:
145, 169, 871, 872, 1454, 45361, 45362 and 363601

Modifications by HackerRank

I have to print the starting number of such loops (not just count them).

My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.

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 <map>
#include <vector>
#include <algorithm>
 
// count digits, two numbers have the same fingerprint if they are permutations of each other
unsigned long long fingerprintFactorial(unsigned int x)
{
unsigned long long result = 0;
while (x > 0)
{
unsigned int digit = x % 10;
x /= 10;
 
// small optimization: 0! = 1!
if (digit == 1)
digit = 0;
 
// find the right location of the digit's counter
unsigned long long shift = 1;
for (unsigned int i = 0; i < digit; i++)
shift *= 10;
 
result += shift;
}
return result;
}
 
int main()
{
// known loop-sizes
// [first number] => [loop size]
std::map<unsigned long long, unsigned int> cache;
 
// special treatment for those numbers that link back to themselves
// and are encountered before any other number of the same fingerprint
cache[fingerprintFactorial(145)] = 1 + 1;
cache[fingerprintFactorial(169)] = 3 + 1;
 
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int limit = 1000000;
unsigned int loopLength = 60;
std::cin >> limit >> loopLength;
 
// count numbers with a certain loop length
unsigned int result = 0;
 
for (unsigned int i = 0; i <= limit; i++)
{
// determine loop length only for unknown fingerprints
unsigned long long id = fingerprintFactorial(i);
if (cache.count(id) == 0)
{
// all numbers of the current loop
std::vector<unsigned int> loop;
 
// add current number to loop, abort if insert fails (because element already exists)
unsigned int x = i;
while (std::find(loop.begin(), loop.end(), x) == loop.end() && loop.size() <= loopLength)
{
loop.push_back(x);
 
// compute factorial sum of digits
unsigned int facSum = 0;
do
{
// factorials 0! ... 9!
const unsigned int fac[10] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };
 
// add factorial of right-most digit
facSum += fac[x % 10];
// shift all digits to the right by one
x /= 10;
} while (x > 0);
x = facSum;
}
 
// remember that loop length (for all upcoming permutations)
cache[id] = loop.size();
}
 
// now the loop length is available at cache[id]
// correct loop length ?
bool match = (cache[id] == loopLength);
 
// special treatment for those numbers that link back to themselves
// (false negatives and positives) => relevant only for Hackerrank
if (i == 145)
match = (loopLength == 1);
if (i == 169 || i == 1454 || i == 363601)
match = (loopLength == 3);
if (i == 871 || i == 872 || i == 45361 || i == 45362)
match = (loopLength == 2);
 
// count matches
if (match)
result++;
 
//#define ORIGINAL
#ifndef ORIGINAL
if (match)
std::cout << i << " ";
#endif
}
 
#ifdef ORIGINAL
std::cout << result << std::endl;
#else
if (result == 0)
std::cout << "-1";
std::cout << std::endl;
#endif
}
}

This solution contains 18 empty lines, 22 comments and 9 preprocessor commands.

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.

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: Please enter the largest number to be analyzed and the loop length

This is equivalent to
echo "1 200 1" | ./74

Output:

(please click 'Go !')

Note: the original problem's input 1000000 60 cannot be entered
because just copying results is a soft skill reserved for idiots.

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

Benchmark

The correct solution to the original Project Euler problem was found in 0.16 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++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

March 12, 2017 submitted solution
April 27, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler074

My code solves 7 out of 7 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 15% (out of 100%).

Hackerrank describes this problem as easy.

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 never an option.

Links

projecteuler.net/thread=74 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-74-determine-the-number-of-factorial-chains-that-contain-exactly-sixty-non-repeating-terms/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p074.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem074.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler074.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

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

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The 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !