<< problem 455 - Powers With Trailing Digits | Phigital number base - problem 473 >> |
Problem 461: Almost Pi
(see projecteuler.net/problem=461)
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
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:
This is equivalent toecho 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
September 28, 2017 added comments
Difficulty
Project Euler ranks this problem at 30% (out of 100%).
Links
projecteuler.net/thread=461 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Python github.com/Meng-Gen/ProjectEuler/blob/master/461.py (written by Meng-Gen Tsai)
C++ github.com/roosephu/project-euler/blob/master/461.cpp (written by Yuping Luo)
C++ github.com/steve98654/ProjectEuler/blob/master/461.cpp
Mathematica github.com/steve98654/ProjectEuler/blob/master/461.nb
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 |
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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 !!!
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