# Problem 154: Exploring Pascal's pyramid

A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level. Then, we calculate the number of paths leading from the apex to each position:

A path starts at the apex and progresses downwards to any of the three spheres directly below the current position.

Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it
(depending on the position, there are up to three numbers above it).

The result is Pascal's pyramid and the numbers at each level n are the coefficients of the trinomial expansion (x + y + z)^n.

How many coefficients in the expansion of (x + y + z)^200000 are multiples of 10^12?

# My Algorithm

Wikipedia told me that a value of Pascal's pyramid can be derived from Pascal's triangle (en.wikipedia.org/wiki/Pascal's_pyramid)
P(level, i, j) = T(level, i) * T(i, j)

And a value of Pascal's triangle is computed with the binomial coefficient:
T(n, k) = dfrac{n!}{(n-k)! * k!}

Those values get huge - but I don't need to know their true value.
All I need to know is whether P(level, i, j) can be factorized into at least 12x 2 and 12x 5 because 2^12 * 5^12 = 10^12.

My program stores for each number between 1 and 200000 (= layer) how often it contains 2 and 5 (see mulPrime1 and mulPrime2).
Repeatedly dividing the current number by 2 and 5 produces the desired result.

Then it stores for each factorial between 1 and 200000 how often it contains 2 and 5 (see sum1 and sum2).
Since s^x * s^y = s^{x+y} it boils down to summing the exponents up to the current point.

The same idea make choose very simple:
choose(T(n, k)) = sums[n] - (sums[n-k] + sums[k])
When I call choose with sums = sum1 then I know how many 2s T(n,k) contains.
And the same happens with sums = sum2 for the number of 5s.

Whenever I have at least 12x 2s and 12x 5s then I found a multiple of 10^12.

There a two simple optimizations:

• if T(level, i) is a multiple of 10^12 then any T(level, i) * T(i, j) is as well
• T(n, k) = T(n, n - k) → the pyramid is symmetric and I only need to compute its left half

## Modifications by HackerRank

Hackerrank wants you to count all numbers that are multiples of p_1^{a_1} * p_2^{a_2} where p_1 and p_2 are distinct prime numbers.
10^12 = 2^12 * 5^12, therefore my program finds the correct solution for the original problem when entering 200000 2 12 5 12.

## Note

There are more symmetries in the pyramid that I didn't exploit: I can see six "areas" with identical values.
Proper use would make my program probably 3 times faster.

# 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):
Note: Enter the level of the pyramid, then the first prime and its exponent, then the second prime and its exponent.

This is equivalent to
echo "3 2 1 3 1" | ./154

Output:

Note: the original problem's input 200000 2 12 5 12 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.

       #include <iostream>
#include <vector>

typedef std::vector<unsigned int> Exponents;

// return exponent of a prime factor of C(n,k)
// looks a bit like the logarithm:
// C(n,k) = n! / ((n-k)! * k!)
unsigned int choose(const Exponents& sums, unsigned int n, unsigned int k)
{
return sums[n] - (sums[n - k] + sums[k]);
}

int main()
{
unsigned int layer     = 200000;
// 10^12 = 2^12 * 5^12
unsigned int prime1    =  2;
unsigned int exponent1 = 12;
unsigned int prime2    =  5;
unsigned int exponent2 = 12;
std::cin >> layer >> prime1 >> exponent1 >> prime2 >> exponent2;

// analyze for each number between 0 and layer how often they contain prime1 and prime2
Exponents mulPrime1 = { 0 };
Exponents mulPrime2 = { 0 };
for (unsigned int x = 1; x <= layer; x++)
{
auto current = x;
unsigned int count = 0;
// extract first prime (=2) as often as possible
while (current % prime1 == 0)
{
current /= prime1;
count++;
}
mulPrime1.push_back(count);

count = 0;
// extract second prime (=5) as often as possible
while (current % prime2 == 0)
{
current /= prime2;
count++;
}
mulPrime2.push_back(count);
}

// sum1[x] = sum of mulPrime1[0 ... x]
Exponents sum1;
unsigned int count = 0;
for (auto x : mulPrime1)
{
count += x;
sum1.push_back(count);
}

// the same stuff for the other prime
Exponents sum2;
count = 0;
for (auto x : mulPrime2)
{
count += x;
sum2.push_back(count);
}

unsigned long long result = 0;
for (unsigned int i = 0; i <= layer; i++)
{
// how often is each prime used by C(layer, i) ?
auto found1 = choose(sum1, layer, i);
auto found2 = choose(sum2, layer, i);

if (found1 >= exponent1 && found2 >= exponent2)
{
// no need to enter the inner-most loop, each iteration would succeed
result += i + 1;
continue;
}

// note: abort early because of mirrored values
for (unsigned int j = 0; j <= (i+1) / 2; j++)
{
if (found1 + choose(sum1, i, j) >= exponent1 &&
found2 + choose(sum2, i, j) >= exponent2)
{
// found a match
result++;
// left and right side are identical
if (j < i / 2)
result++;
}
}
}

// and we're done !
std::cout << result << std::endl;
return 0;
}


This solution contains 11 empty lines, 16 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 16.5 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 6 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

July 13, 2017 submitted solution

# Hackerrank

My code solves 11 out of 21 test cases (score: 45%)

I failed 0 test cases due to wrong answers and 10 because of timeouts

# Difficulty

Project Euler ranks this problem at 60% (out of 100%).

Hackerrank describes this problem as hard.

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 rarely an option.

# 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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !