<< problem 100 - Arranged probability Triangle containment - problem 102 >>

# Problem 101: Optimum polynomial

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

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.

Input data (separated by spaces or newlines):
Note: Enter the number of coefficients and then the coefficients of the generating polynomial

This is equivalent to
echo "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. Or just jump to my GitHub repository.

The code contains #ifdefs 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 #include // compute sum(coeff[i] * x^i) template T sequence(T x, const std::vector& 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 T lagrange(const std::vector& 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 T newton(std::vector 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 coefficients = { +1, -1, +1, -1, +1, -1, +1, -1, +1, -1, +1 }; #else size_t numCoefficients; std::cin >> numCoefficients; std::vector coefficients(numCoefficients + 1); for (auto& i : coefficients) std::cin >> i; #endif long long sum = 0; std::vector 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

# Difficulty

Project Euler ranks this problem at 35% (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

I stopped working on Project Euler problems around the time they released 617.
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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 100 - Arranged probability Triangle containment - problem 102 >>
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