<< problem 70 - Totient permutation | Counting fractions - problem 72 >> |

# 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:

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

Output:

*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 solves **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 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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My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

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