Problem 152: Writing 1/2 as a sum of inverse squares

(see projecteuler.net/problem=152)

There are several ways to write the number 1/2 as a sum of inverse squares using distinct integers.

For instance, the numbers { 2,3,4,5,7,12,15,20,28,35 } can be used:

dfrac{1}{2} = dfrac{1}{2^2} + dfrac{1}{3^2} + dfrac{1}{4^2} + dfrac{1}{5^2} + dfrac{1}{7^2} + dfrac{1}{12^2} + dfrac{1}{15^2} + dfrac{1}{20^2} + dfrac{1}{28^2} + dfrac{1}{35^2}

In fact, only using integers between 2 and 45 inclusive, there are exactly three ways to do it, the remaining two being:
{ 2,3,4,6,7,9,10,20,28,35,36,45 } and { 2,3,4,6,7,9,12,15,28,30,35,36,45 }.

How many ways are there to write the number 1/2 as a sum of inverse squares using distinct integers between 2 and 80 inclusive?

My Algorithm

This problem would be extremely easy if we had less numbers. Nevertheless, the search works always the same:

→ that's essentially how the function search() works

Unfortunately this concept can solve the problem for n = 45. Therefore I have to find a clever way how I can skip some numbers.
If I have a (partial) sum dfrac{a}{b} then adding a single term dfrac{1}{x^2}:
(1) dfrac{a}{b} + dfrac{1}{x^2} = dfrac{a x^2 + b}{b x^2}

If b is coprime to x^2 (which implies coprime to x) then I need to add at least one more term to get rid of x^2.
That's a very helpful observation because adding 1 / 41^2 now is impossble - there is just no other term that also has prime factor 41.
So the 41 would be stuck in the denominator - which is not allowed, there must be a single 2 at the end.
The same can be said for all prime numbers p > 80/2 which removes { 41,43,47,53,59,61,67,71,73,79 } from the list of candidates.

Manual calculation showed that even 37 is "forbidden": the only remaining number with the same prime factor is 74.
1 / 37^2 + 1 / 74^2 = 5/5476 but 5476 = 2^2 * 37^2 (I asked Wolfram Alpha). So the 37 is still left in the denominator ...
Then I wrote code to check for each prime number any combination of its multiples, e.g. for p = 11 I checked all 2^7 = 128 combinations of { 11,22,33,44,55,66,77 }.
and looked for a combination where the sum is not a multiple of the prime anymore ("sum mod prime != 0").
This check is too slow for the primes 2 and 3 because there are 2^{floor{80/2}} = 2^40 or 2^{floor{80/3}} = 2^26 combinations.
I just assumed that any number which has only prime factors 2 and/or 3 is "relevant".

The output was astonishing: 11 and all prime numbers above 13 are "forbidden".
The list of candidates can be reduced to all numbers whose prime factors are { 2,3,5,7,13 } and contains these 39 numbers:
{ 2,3,4,5,6, 7,8,9,10,12, 13,14,15,16,18, 20,21,24,27,28, 30,32,35,36,39, 40,42,45,48,52, 54,56,60,63,64, 65,70,72,80 }

Looking at 2^39 approx 55 * 10^10 different combinations is a lot but manageable.
Especially the optimizations mentioned at the beginning reduce this number considerably and I found the correct result in a bit more than 1 minute.

Then I remembered the "trick" from problem 266 which I solved a few weeks ago:
if pre-compute the sum of each subset of the highest numbers (x > 40 turns out to be a good choice) and store them in a container lastNumbers
then I can stop my search after reaching that threshold and lookup in lastNumbers how often the difference dfrac{1}{2} - current can be found.
Now the program finishes after 0.3 seconds !

Alternative Approaches

If you multiply all fractions by lcm(2,3,4,...,80) then you don't need fractions at all.

Modifications by HackerRank

The result doesn't need to be 1/2, it can be any number.
And the number of terms can be changed which causes some problems because then even my 128 bit integers overflow.
In the end about one third of all test cases are correct, one third times out and one third causes overflows.

Note

I kept track of the numbers actually used in all sums and it turns out 8 numbers were "useless": { 16,27,32,48,54,64,65,80 }
With this "optimal" input set the program becomes twice as fast and finishes after 0.13 seconds.

The "optimal" input set reveals an interesting pattern if I write down the prime factorization:
{ 2^4, 3^3, 2^5, 2^4 * 3, 2 * 3^3, 2^6, 5 * 13, 2^4 * 5 }
I don't fully understand why all multiples of 2^4 and 3^3 are "useless" - but apparently there is a good reason.

The Fraction class is part of my toolbox and can add fractions, reduce them, etc.
Some numbers becomes huge (before being reduced to a proper fraction) and I need GCC's 128 bit integer to find the correct result.
That means my code doesn't compile with Visual C++.

I could remove the members containers because it's only useful during debugging (to print exactly which solutions were found).
But this problem was surprisingly hard and I just glad I solved it so I leave the code "as is". Well, I added a few comments afterwards.
With over 300 lines it's one of my longest solutions, see Code Metrics.

Interactive test

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

Input data (separated by spaces or newlines):
Note: Enter the denominator of the sum (e.g. 2) and the highest value of any term (e.g. 45)

This is equivalent to
echo "2 45" | ./152

Output:

(please click 'Go !')

Note: the original problem's input 2 80 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++11 and can be compiled with G++, Clang++. You can download it, too. Or just jump to my GitHub repository. 

       
#include <iostream>

       
#include <vector>

       
#include <algorithm>

       
#include <set>

       
#include <map>

       
#include <cmath>

       
 

       
// uncomment to get more debugging output

       
//#define VERBOSE

       
 

       
// ---------- code copied from my toolbox ----------

       
// (I removed unused features to reduce source code size)

       
 

       
// simple class to representation a fraction

       
// note: no checks for a zero denominator

       
//       signs are not normalized

       
struct Fraction

       
{

       
  // change to long long if you need to support larger values

       
  typedef unsigned __int128 T;

       
 

       
  // numerator

       
  T num;

       
  // denominator

       
  T den;

       
 

       
  Fraction(T numerator = 0, T denominator = 1)

       
  : num(numerator), den(denominator)

       
  {}

       
 

       
  // add

       
  Fraction operator+(const Fraction& other) const

       
  {

       
    if (den == other.den)

       
      return Fraction(num + other.num, den);

       
    // n1/d1 + n2/d2 = (n1*d2 + n2*d1) / d1*d2

       
    return { num * other.den + other.num * den, den * other.den };

       
  }

       
  // subtract

       
  Fraction operator-(const Fraction& other) const

       
  {

       
    if (den == other.den)

       
      return Fraction(num - other.num, den);

       
    // n1/d1 - n2/d2 = (n1*d2 - n2*d1) / d1*d2

       
    return { num * other.den - other.num * den, den * other.den };

       
  }

       
  // sort

       
  bool operator< (const Fraction& other) const

       
  {

       
    // n1/d1 < n2/d2 => n1*d2 < n2*d2

       
    return num * other.den <  other.num * den;

       
  }

       
  // compare

       
  bool operator==(const Fraction& other) const

       
  {

       
    // n1/d1 < n2/d2 => n1*d2 < n2*d2

       
    return num * other.den == other.num * den;

       
  }

       
 

       
  // convert to a proper fraction

       
  Fraction reduced() const

       
  {

       
    // catch simple cases: 0/d and n/1

       
    if (num == 0 || den == 1)

       
      return { num, 1 };

       
 

       
    // divide numerator and denominator by their gcd()

       
    // code taken from my toolbox / gcd()

       
    auto a = num;

       
    auto b = den;

       
    while (a != 0)

       
    {

       
      auto c = a;

       
      a = b % a;

       
      b = c;

       
    }

       
    // now b contains the gcd()

       
    if (b == 1)

       
      return *this;

       
    return { num/b, den/b };

       
  }

       
};

       
 

       
// ---------- problem-specific code ----------

       
 

       
// sum of fractions should be exactly 1/2

       
Fraction sum(1, 2);

       
 

       
// all fractions that could be used

       
std::vector<unsigned int> candidates;

       
// store all combinations of all fractions beyond 1/threshold^2

       
unsigned int lastNumberThreshold = 40;

       
std::multiset<Fraction> lastNumbers; // a simple std::set would suffice for limit = 80 and lastNumberThreshold = 40

       
// sum of all fractions whose denominator exceeds a value (e.g. 40 => all fractions 1/40^2 + 1/42^2 + ... 1/80^2)

       
std::map<unsigned int, Fraction> remaining; // is equal to the largest key in lastNumbers

       
 

       
// track fractions of the current sum (debugging only)

       
std::vector<unsigned char> members;

       
 

       
// count how many combinations add up to 1/2

       
unsigned int search(Fraction current = Fraction(0, 1), size_t next = 0)

       
{

       
  // match !

       
  if (current == sum)

       
  {

       
#ifdef VERBOSE

       
    for (auto x : members)

       
      std::cout << int(x) << " + ";

       
    std::cout << std::endl;

       
#endif

       
    return 1;

       
  }

       
 

       
  // sum already too high ?

       
  if (sum < current)

       
    return 0;

       
  // no more terms ?

       
  if (next == candidates.size())

       
    return 0;

       
 

       
  // get next denominator (actually the whole fraction will be 1/number^2)

       
  auto number = candidates[next];

       
 

       
  // even if I choose all remaining number will it be enough to "reach" 1/2 ?

       
  auto maximum = current + remaining[number];

       
  if (maximum < sum)

       
    return 0;

       
 

       
  // lookup difference in all pre-computed sums of the last values

       
  if (number >= lastNumberThreshold)

       
  {

       
    // is there any sum matching the difference ?

       
    auto difference = sum - current;

       
    auto solutions  = lastNumbers.count(difference);

       
 

       
#ifdef VERBOSE

       
    if (solutions > 0)

       
    {

       
      for (auto x : members)

       
        std::cout << int(x) << " + ";

       
      std::cout << " (" << solutions << " solutions)" << std::endl;

       
    }

       
#endif

       
 

       
    return solutions;

       
  }

       
 

       
  // keep searching ...

       
  unsigned long long result = 0;

       
  // try to build the sum without number

       
  result += search(current, next + 1);

       
 

       
  // try to build the sum with    number

       
  Fraction add(1, number*number);

       
  members.push_back(number);

       
  current = current + add;

       
  current = current.reduced();

       
  result += search(current, next + 1);

       
  members.pop_back();

       
 

       
  return result;

       
}

       
 

       
int main()

       
{

       
  unsigned int denominator = 2;

       
  unsigned int limit = 80;

       
  std::cin >> denominator >> limit;

       
  sum.den = denominator;

       
 

       
  // find primes 2..80

       
  std::vector<unsigned int> primes;

       
  for (unsigned int i = 2; i <= limit; i++)

       
  {

       
    // basic trial division

       
    bool isPrime = true;

       
    for (unsigned int j = 2; j*j <= i; j++)

       
      if (i % j == 0)

       
      {

       
        isPrime = false;

       
        break;

       
      }

       
    if (isPrime)

       
      primes.push_back(i);

       
  }

       
 

       
  // determine relevant numbers

       
  std::set<unsigned int> relevantPrime;

       
  std::vector<bool> found(limit + 1, false);

       
  for (auto p : primes)

       
  {

       
    // enumarate 1/p^2, 1/(2p)^2, 1/(3p)^2, ...

       
    std::vector<Fraction> multiples;

       
    for (unsigned int m = p; m <= limit; m += p)

       
      multiples.push_back({ 1, m*m });

       
 

       
    // now check whether any combination is able to get rid of the prime in the denominator

       
 

       
    // go through all 2^n subsets

       
    auto combinations = 1 << multiples.size();

       
    for (auto mask = 1; mask < combinations; mask++)

       
    {

       
      // add all fractions

       
      Fraction current(0, 1);

       
      for (unsigned int pos = 0; pos < multiples.size(); pos++)

       
      {

       
        auto bit = 1 << pos;

       
        if ((bit & mask) == 0)

       
          continue;

       
        current = current + multiples[pos];

       
        current = current.reduced();

       
      }

       
      // and reduce them to a proper fraction

       
      current = current.reduced();

       
 

       
      // prime still there ? try another combination ...

       
      if (current.den % p == 0)

       
        continue;

       
 

       
      // same loop again, now mark all involved numbers as "found"

       
      for (unsigned int pos = 0; pos < multiples.size(); pos++)

       
      {

       
        auto bit = 1 << pos;

       
        if ((bit & mask) == 0)

       
          continue;

       
 

       
        // p is at position 0, 2p at 1, 3p at 2, ...

       
        auto isGood = (pos + 1) * p;

       
        found[isGood] = true;

       
      }

       
 

       
      // yes, it's a relevant prime

       
      relevantPrime.insert(p);

       
      found[p] = true;

       
 

       
      // analyze all combinations only for p >= 5 (special treatment for p=2 and p=3 follows below)

       
      if (p < 5)

       
        break;

       
    }

       
  }

       
 

       
  // assume that all numbers 2^n * 3^m need to be considered

       
  // (I only have to do this because the previous loop would be waaaay to slow for full evaluation of all cases)

       
  for (unsigned int two = 1; two <= limit; two *= 2)

       
    for (unsigned int three = 1; two*three <= limit; three *= 3)

       
      found[two * three] = true;

       
 

       
  // each suitable number between 2..80 must only consist of relevant primes

       
  for (unsigned int i = 2; i <= limit; i++)

       
  {

       
    // I don't need this number ?

       
    if (!found[i])

       
      continue;

       
 

       
    // factorize

       
    auto reduce = i;

       
    for (auto p : relevantPrime)

       
      while (reduce % p == 0)

       
        reduce /= p;

       
 

       
    if (reduce != 1)

       
    {

       
      found[i] = false;

       
      continue;

       
    }

       
 

       
    // yes, all prime factors found in relevantPrimes

       
    candidates.push_back(i);

       
  }

       
 

       
  // my code finds 2,3,4,5,6,7,8,9,10,12,13,14,15,16,18,20,21,24,27,28,30,32,35,36,39,40,42,45,48,52,54,56,60,63,64,65,70,72,80

       
  // but I analyzed all valid sums and found that only these numbers are actually needed:

       
  //candidates = { 2,3,4,5,6,7,8,9,10,12,13,14,15,   18,20,21,24,   28,30,   35,36,39,40,42,45,   52,   56,60,63,      70,72 };

       
 

       
#ifdef VERBOSE

       
  for (auto x : candidates)

       
    std::cout << x << ",";

       
  std::cout << std::endl;

       
#endif

       
 

       
  // for each number n compute the sum of 1/n^2 from n to 80

       
  // e.g. then I know for n = 9 whether I can reach 1/2 (or more) if I would use all remaining fractions

       
  // => it becomes obvious that I always need 2

       
  // => and I need 3 or 4

       
  // => helps me to skip analyzed many cases where the sum is just too small

       
  Fraction sum(0, 1);

       
  for (auto i = candidates.rbegin(); i != candidates.rend(); i++)

       
  {

       
    auto current = *i;

       
    sum = sum + Fraction(1, current * current);

       
    sum = sum.reduced();

       
    remaining[current] = sum;

       
  }

       
 

       
  // precompute sums of all combinations of the last numbers

       
  lastNumberThreshold = limit / 2; // arbitrary heuristic: all relevant numbers >= 40

       
                                   // it just turned out to be the fastest/best threshold ...

       
  // get range of the last numbers

       
  auto iteLastNumbers = candidates.begin();

       
  while (iteLastNumbers != candidates.end() && *iteLastNumbers < lastNumberThreshold)

       
    iteLastNumbers++;

       
  auto numLastNumbers = std::distance(iteLastNumbers, candidates.end());

       
 

       
  // go through all 2^numLastNumbers combinations

       
  unsigned int combinations = 1 << numLastNumbers;

       
  for (unsigned int i = 0; i < combinations; i++)

       
  {

       
    // add all fractions for each subset

       
    Fraction current(0, 1);

       
    for (unsigned int pos = 0; pos < numLastNumbers; pos++)

       
    {

       
      auto bit = 1 << pos;

       
      if ((i & bit) == 0)

       
        continue;

       
 

       
      auto add = *(iteLastNumbers + pos);

       
      current  = current + Fraction(1, add * add);

       
      current  = current.reduced();

       
    }

       
    // store for later use in search()

       
    lastNumbers.insert(current);

       
  }

       
 

       
  // let's go !

       
  std::cout << search() << std::endl;

       
  return 0;

       
}

This solution contains 44 empty lines, 74 comments and 12 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.28 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 4 MByte.

(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

September 17, 2017 submitted solution
September 17, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler152

My code solves 10 out of 31 test cases (score: 30%)

I failed 9 test cases due to wrong answers and 12 because of timeouts

Difficulty

65% Project Euler ranks this problem at 65% (out of 100%).

Hackerrank describes this problem as advanced.

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=152 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python github.com/hughdbrown/Project-Euler/blob/master/euler-152.py (written by Hugh Brown)
Python github.com/Meng-Gen/ProjectEuler/blob/master/152.py (written by Meng-Gen Tsai)
Python github.com/roosephu/project-euler/blob/master/152.py (written by Yuping Luo)
Python github.com/smacke/project-euler/blob/master/python/152.py (written by Stephen Macke)
C++ github.com/roosephu/project-euler/blob/master/152.cpp (written by Yuping Luo)
Java github.com/HaochenLiu/My-Project-Euler/blob/master/152.java (written by Haochen Liu)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem152.java (written by Magnus Solheim Thrap)
Go github.com/frrad/project-euler/blob/master/golang/Problem152.go (written by Frederick Robinson)
Perl github.com/shlomif/project-euler/blob/master/project-euler/152/euler-152.pl (written by Shlomi Fish)

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
[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.

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 !!!

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 !