Problem 71: Ordered fractions

(see projecteuler.net/problem=71)

Consider the fraction, dfrac{n}{d}, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d <= 8 in ascending order of size, we get:

dfrac{1}{8}, dfrac{1}{7}, dfrac{1}{6}, dfrac{1}{5}, dfrac{1}{4}, dfrac{2}{7}, dfrac{1}{3}, dfrac{3}{8}, dfrac{2}{5}, dfrac{3}{7}, dfrac{1}{2}, dfrac{4}{7}, dfrac{3}{5}, dfrac{5}{8}, dfrac{2}{3}, dfrac{5}{7}, dfrac{3}{4}, dfrac{4}{5}, dfrac{5}{6}, dfrac{6}{7}, dfrac{7}{8}

It can be seen that dfrac{2}{5} is the fraction immediately to the left of dfrac{3}{7}.

By listing the set of reduced proper fractions for d <= 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of dfrac{3}{7}.

Algorithm

While researching several approaches I read something about Farey sequences (see en.wikipedia.org/wiki/Farey_sequence):
If you know two neighboring fractions dfrac{a}{b} and dfrac{c}{d}, then there exists a so-called mediant m=dfrac{a+c}{b+d}.

dfrac{0}{1} and dfrac{1}{1} are neighboring fractions because the fulfil the condition bc - ad = 1.

My program starts with these two fractions. In each iteration it checks whether the mediant is less or greater than the user input (3/7 by default).
- m < 3/7 → continue searching in interval a/b and m
- m < 3/7 → continue searching in interval m and c/d
- if denominator of m exceeds 1000000 then stop

The first steps will be:
interval 0/1 and 1/1 => (0+1)/(1+1) = 1/2 which is larger than 3/7
interval 0/1 and 1/2 => (0+1)/(1+2) = 1/3 which is smaller than 3/7
interval 1/3 and 1/2 => (1+1)/(3+2) = 2/5 which is smaller than 3/7
interval 2/5 and 1/2 => (2+1)/(5+2) = 3/7 which is equal to 3/7 (treat the same as if larger)
interval 2/5 and 3/7 => (2+3)/(5+7) = 7/12 ... and so on ...

The algorithm finds the correct result but might be too slow.
Therefore, I stop whenever I my right interval border becomes a/b (which is 3/7 in the example) because after that only the left interval border can change.
Then a much faster explicit computation is possible:
- each iteration would adjust the left border: leftN += rightN and leftD += rightD until leftN + leftD > limit.
- there would be 1 + (limit - (leftD + rightD)) / rightD iterations

I wrote a function isLess that compares two fractions and returns true if dfrac{a}{b} < dfrac{c}{d}:
if a/b < c/d (and all numbers are positive) then ad < cb
This can be done with precise integer arithmetic. However, Hackerrank's denominators are up to 10^15 and therefore the product exceeds 64 bits.
The template multiply is a portable way to multiply two 64 bits and return the 128 bit result as two 64 bit variables.
Note: GCC supports __int128 and there is no need to have such a template on Linux/Apple/MinGW.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
 
// multiply two unsigned number where the result may exceed the largest native data type
template <typename T>
void multiply(T a, T b, T& result_high, T& result_low)
{
const T Shift = 4 * sizeof(a);
const T Mask = T(~0) >> Shift;
 
auto a_high = a >> Shift;
auto a_low = a & Mask;
auto b_high = b >> Shift;
auto b_low = b & Mask;
 
auto c_0 = a_low * b_low;
auto c_1a = a_high * b_low;
auto c_1b = a_low * b_high;
auto c_2 = a_high * b_high;
 
auto carry = ((c_0 >> Shift) + (c_1a & Mask) + (c_1b & Mask)) >> Shift;
 
result_high = c_2 + (c_1a >> Shift) + (c_1b >> Shift) + carry;
result_low = c_0 + (c_1a << Shift) + (c_1b << Shift);
}
 
// if a/b < c/d (and all numbers are positive)
// then a*d < c*b
bool isLess(unsigned long long a, unsigned long long b, unsigned long long c, unsigned long long d)
{
// GCC has 128-bit support:
//return (unsigned __int128)x1 * y2 < (unsigned __int128)y1 * x2;
 
unsigned long long r1_high, r1_low;
unsigned long long r2_high, r2_low;
multiply(a, d, r1_high, r1_low);
multiply(c, b, r2_high, r2_low);
// compare high bits
if (r1_high < r2_high)
return true;
if (r1_high > r2_high)
return false;
// compare low bits
return (r1_low < r2_low);
}
 
int main()
{
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int a = 3;
unsigned int b = 7;
unsigned long long limit = 1000000;
std::cin >> a >> b >> limit;
 
// generate all numbers in the Farey sequence using binary subdivision
// for each step decide whether the number was left or right of the desired fraction
// start with 0/1 and 1/1
unsigned long long leftN = 0;
unsigned long long leftD = 1;
unsigned long long rightN = 1;
unsigned long long rightD = 1;
while (leftD + rightD <= limit)
{
auto mediantN = leftN + rightN;
auto mediantD = leftD + rightD;
 
// if x1/y1 < x2/y2 (and all numbers are positive)
// then x1*y2 < x2*y1
if (isLess(mediantN, mediantD, a, b))
{
// adjust left border
leftN = mediantN;
leftD = mediantD;
}
else
{
// adjust right border
rightN = mediantN;
rightD = mediantD;
// did we "hit the spot" ?
if (rightN == a && rightD == b)
break;
}
}
 
// now the right border is the fraction we read from input
// and we only have to adjust the left border from here on
 
//while (leftD + rightD <= limit) // gets the job done, but still slow ...
//{
// leftN += rightN;
// leftD += rightD;
//}
 
// "instant" result
if (limit >= leftD + rightD)
{
auto difference = limit - (leftD + rightD);
auto repeat = 1 + difference / rightD;
leftN += repeat * rightN;
leftD += repeat * rightD;
}
 
std::cout << leftN << " " << leftD << std::endl;
}
return 0;
}

This solution contains 14 empty lines, 23 comments and 1 preprocessor command.

Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 3 7 8" | ./71

Output:

(please click 'Go !')

Note: the original problem's input 1 3 7 1000000 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 less than 0.01 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 13, 2017 submitted solution
May 2, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler071

My code solved 21 out of 21 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 10% (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=71 - 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-71-proper-fractions-ascending-order/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p071.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem071.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler071.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 already 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.

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