<< problem 100 - Arranged probability | Triangle containment - problem 102 >> |
Problem 101: Optimum polynomial
(see projecteuler.net/problem=101)
If we are presented with the first k terms of a sequence it is impossible to say with certainty the value of the next term,
as there are infinitely many polynomial functions that can model the sequence.
As an example, let us consider the sequence of cube numbers. This is defined by the generating function,
u_n = n^3: 1, 8, 27, 64, 125, 216, ...
Suppose we were only given the first two terms of this sequence. Working on the principle that "simple is best" we should assume a linear relationship
and predict the next term to be 15 (common difference 7). Even if we were presented with the first three terms, by the same principle of simplicity, a quadratic relationship should be assumed.
We shall define OP(k, n) to be the nth term of the optimum polynomial generating function for the first k terms of a sequence.
It should be clear that OP(k, n) will accurately generate the terms of the sequence for n <= k, and potentially the first incorrect term (FIT) will be OP(k, k+1);
in which case we shall call it a bad OP (BOP).
As a basis, if we were only given the first term of sequence, it would be most sensible to assume constancy; that is, for n >= 2, OP(1, n) = u_1.
Hence we obtain the following OPs for the cubic sequence:
OP(1, n) = 1
1, 1, 1, 1, ...
OP(2, n) = 7n-6
1, 8, 15, ...
OP(3, n) = 6n^2-11n+6
1, 8, 27, 58, ...
OP(4, n) = n^3
1, 8, 27, 64, 125, ...
Clearly no BOPs exist for k >= 4.
By considering the sum of FITs generated by the BOPs (indicated in bold above), we obtain 1 + 15 + 58 = 74.
Consider the following tenth degree polynomial generating function:
u_n = 1 - n + n^2 - n^3 + n^4 - n^5 + n^6 - n^7 + n^8 - n^9 + n^10
Find the sum of FITs for the BOPs.
My Algorithm
When I read this problem, I didn't know anything about polynomial interpolation - and started reading the Wikipedia page: en.wikipedia.org/wiki/Polynomial interpolation
The Lagrange polynomial caught my attention but my implementation just couldn't find the correct solution.
Then I tried the Newton polynomial and succeeded. Even more, I was able to fix the off-by-one error in my Lagrange code ...
Nevertheless, I kept both routines in my solution. They return the same result.
Modifications by HackerRank
[TODO] I modified my code to accept arbitrary coefficients for the polynomial generating function (originally { +1, -1, +1, -1, +1, -1, +1, -1, +1, -1, +1 }
)
but still fail to solve anything else than the default input.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This live test is based on the Hackerrank problem.
This is equivalent toecho "0 0 0 1" | ./101
Output:
Note: the original problem's input 11 1 -1 1 -1 1 -1 1 -1 1 -1 1
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.
The code contains #ifdef
s to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL
to produce the result for the original problem (default setting for most problems).
//#define ORIGINAL
#include <iostream>
#include <vector>
// compute sum(coeff[i] * x^i)
template <typename T>
T sequence(T x, const std::vector<T>& coefficients)
{
T xx = 1;
T result = 0;
for (auto i : coefficients)
{
result += i * xx;
xx *= x;
}
return result;
}
// given f(1),f(2),f(3),...f(n) then find f(n+1)
// (Lagrange polynomials)
template <typename T>
T lagrange(const std::vector<T>& known)
{
T result = 0;
size_t next = known.size() + 1;
for (size_t i = 1; i < next; i++)
{
// build Lagrange polynomials
// n = numerator, d = denominator
T n = 1;
T d = 1;
for (size_t j = 1; j < next; j++)
{
if (i == j)
continue;
n *= next - j;
d *= i - j;
}
// evaluate
result += known[i - 1] * (n / d);
}
return result;
}
// given f(1),f(2),f(3),...f(n) then find f(n+1)
// (Newton divided differences)
template <typename T>
T newton(std::vector<T> known)
{
T result = known[0];
size_t j = 1;
size_t k = known.size();
for (size_t last = known.size() - 1; last > 0; last--)
{
for (size_t i = 0; i < last; i++)
known[i] = (known[i + 1] - known[i]) / j;
T multDiff = 1;
for (size_t i = 0; i < j; i++)
multDiff *= k - i;
result += known[0] * multDiff;
j++;
}
return result;
}
int main()
{
// read coefficients
#ifdef ORIGINAL
std::vector<long long> coefficients = { +1, -1, +1, -1, +1, -1, +1, -1, +1, -1, +1 };
#else
size_t numCoefficients;
std::cin >> numCoefficients;
std::vector<long long> coefficients(numCoefficients + 1);
for (auto& i : coefficients)
std::cin >> i;
#endif
long long sum = 0;
std::vector<long long> data;
// iterate over 10 points
for (long long x = 1; x < (long long)coefficients.size(); x++)
{
// add the next point
data.push_back(sequence(x, coefficients));
// estimate next point
long long next = lagrange(data);
//long long next = newton(data);
sum += next;
#ifndef ORIGINAL
std::cout << (next % 1000000007) << " ";
#endif
}
#ifdef ORIGINAL
std::cout << sum << std::endl;
#endif
return 0;
}
This solution contains 16 empty lines, 13 comments and 9 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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
May 23, 2017 submitted solution
May 23, 2017 added comments
Difficulty
Project Euler ranks this problem at 35% (out of 100%).
Links
projecteuler.net/thread=101 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-101-optimum-polynomial-function/ (written by Kristian Edlund)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p101.java (written by Nayuki)
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 a new problem is published | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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 !!!
<< problem 100 - Arranged probability | Triangle containment - problem 102 >> |