<< problem 458 - Permutations of Project Phigital number base - problem 473 >>

# Problem 461: Almost Pi

Let f_n(k) = e^{k/n} - 1, for all non-negative integers k.

Remarkably, f_200(6) + f_200(75) + f_200(89) + f_200(226) = 3.141592644529 ... approx pi.

In fact, it is the best approximation of pi of the form f_n(a) + f_n(b) + f_n(c) + f_n(d) for n = 200.

Let g(n) = a^2 + b^2 + c^2 + d^2 for a, b, c, d that minimize the error: |f_n(a) + f_n(b) + f_n(c) + f_n(d) - pi|
(where |x| denotes the absolute value of x).

You are given g(200) = 6^2 + 75^2 + 89^2 + 226^2 = 64658.

Find g(10000).

# Very inefficient solution

My code needs more than 256 MByte RAM to find the correct result. (scroll down to the benchmark section)
Apparantly a much smarter algorithm exists - or my implementation is just inefficient.

# My Algorithm

I tried to find the result with a randomized genetic search but failed because I was always stuck in local minimums - and finally gave up.

Surprisingly, pretty much everybody else solved the problem with something similar to my current approach:

• create all pairs f_10000(a) + f_10000(b)
• for each such pair find a second pair f_10000(c) + f_10000(d) such that |f_10000(a) + f_10000(b) + f_10000(c) + f_10000(d) - pi| is minimized
The upper limit for a, b, c, d is 14211 because e^{14211/10000} - 1 > pi.
I can assume pair-wise that a < b and c < d without loss of generality.
Therefore I have to analyze up to T(14211) = 14211 * 14210 / 2 = 100969155 pairs.
However, the sum of a few pairs exceeds pi so that only 72299877 pairs need to be considered (see container pairs).

To find the optimal partner for each pairs[i] I perform a binary search for PI - pairs[i].
That's the same concept I used in solving problem 266 (and in problem 152).

The STL's std::upper_bound returns the first value that is too large. This number and its predecessor need to be analyzed.
If their error is less than before then I remember their indices left and right.

The final step is to map left and right back to a, b, c, d.
Maybe there is a smarter way than just two nested loops - but they are pretty fast, so that's okay for me.

## Alternative Approaches

I'm quite worried that I exceed the memory limit of 256 MByte: my program needs about 640 MByte.
After submitting my result to the Project Euler forum I saw that a clever use of priority queues could reduce memory consumption from O(n^2) to O(n).
Those guys are brilliant !

I thought a while about a segmented algorithm, where pairs represents only a part of all pairs.
Thus I could stay within the 256 MByte limit at the expense of a severely increased execution time.
However, the main reason against such an approach is that the code would increase a lot, too, and readability suffer massively.

# 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):

This is equivalent to
echo 200 | ./461

Output:

Note: the original problem's input 10000 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++, Visual C++. You can download it, too.

       #include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

const double PI = 3.1415926535897932;

int main()
{
// example was n = 200, problem asks for n = 10000
unsigned int order = 10000;
std::cin >> order;

// find maximum value such that e^(maximum/n) - 1 <= PI
// ... and while at it: store all values in f
std::vector<double> f;
unsigned int maximum = 0;
while (true)
{
auto current = exp(maximum / double(order)) - 1;
// too large ?
if (current > PI)
break;

f.push_back(current);
maximum++;
}

// generate all pairs f[i] + f[j] <= PI (with i <= j)
std::vector<double> pairs;
pairs.reserve(maximum * maximum / 10); // some heuristic to avoid too frequent allocations
for (size_t i = 0; i < maximum; i++)
for (size_t j = i; j < maximum; j++)
{
// abort, sum exceeds 3.1415
if (f[i] + f[j] > PI)
break;

pairs.push_back(f[i] + f[j]);
}

// sort all values
//std::sort(pairs.begin(), pairs.end());
// a bit faster: compare double as 64 bit integers => works on CPU, not FPU
// only possible because all values are positive and there are no NaN or other special values
std::sort(pairs.begin(), pairs.end(), [] (double a, double b)
{
// cast from 64 bit double to 64 bit integer
auto aa = (const long long*) &a;
auto bb = (const long long*) &b;
return *aa < *bb;
});

// optimal value for f_a + f_b
size_t left  = 0;
// optimal value for f_c + f_d
size_t right = 0;
// minimize error |f_a + f_b - PI|
double minError  = PI;
for (size_t i = 0; i < pairs.size(); i++)
{
// binary search for the best match
auto current = pairs[i];
auto need    = PI - current;
if (need < current) // same as need < PI/2
break;

// this value and its predecessor are candidates
auto match = std::upper_bound(pairs.begin(), pairs.end(), need); // actually "sums.begin() + i" works, too

// "match" is a value slightly too large
auto error = fabs(need - *match);
if (error < minError)
{
minError = error;
left  = i;
right = std::distance(pairs.begin(), match);
}

// now "match" is a value slightly too small (or an exact match => spoiler alert: there is no perfect match)
match--;
error = fabs(need - *match);
if (error < minError)
{
minError = error;
left  = i;
right = std::distance(pairs.begin(), match);
}
}

// resolve the left pair f_a and f_b
unsigned int result = 0;
bool resolvedLeft = false;
for (size_t a = 0; a < maximum && !resolvedLeft; a++)
for (size_t b = a; b < maximum; b++)
if (pairs[left]  == f[a] + f[b])
{
result += a*a + b*b;
resolvedLeft  = true;
break;
}

// resolve the left pair f_c and f_d
bool resolvedRight = false;
for (size_t c = 0; c < maximum && !resolvedRight; c++)
for (size_t d = c; d < maximum; d++)
if (pairs[right] == f[c] + f[d])
{
result += c*c + d*d;
resolvedRight = true;
break;
}

std::cout << result << std::endl;
return 0;
}


This solution contains 14 empty lines, 20 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 6.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about  634 MByte , thus exceeding the limit of 256 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 28, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 30% (out of 100%).

# 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
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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.

 << problem 458 - Permutations of Project Phigital number base - problem 473 >>
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