<< problem 73 - Counting fractions in a range | Singular integer right triangles - problem 75 >> |
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?
My 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).
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.
This is equivalent toecho "1 200 1" | ./74
Output:
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)
My code
… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.
The code contains #ifdef
s 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.
Benchmark
The correct solution to the original Project Euler problem was found in 0.16 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
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 rarely 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:
C# 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)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 73 - Counting fractions in a range | Singular integer right triangles - problem 75 >> |