Created
November 23, 2022 05:53
-
-
Save StephanTLavavej/726b4f8139f8b45fb2f05b5d29e9eedd to your computer and use it in GitHub Desktop.
Boost.uBLAS solution for parquet tiles coding exercise
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
// Boost.uBLAS solution by Stephan T. Lavavej | |
// for https://www.reddit.com/r/cpp/comments/z1xich/c_interview_coding_exercise_with_solution/ | |
#include <algorithm> | |
#include <cassert> | |
#include <chrono> | |
#include <cstddef> | |
#include <cstdint> | |
#include <format> | |
#include <iostream> | |
#include <numeric> | |
#include <vector> | |
#include <boost/numeric/ublas/matrix_sparse.hpp> | |
#include <boost/numeric/ublas/vector.hpp> | |
using namespace std; | |
constexpr int total_length{30}; | |
constexpr int num_strips{11}; | |
[[nodiscard]] constexpr uint32_t encode_strip(const vector<int>& strip) { | |
uint32_t bits{0u}; | |
int sum{0}; | |
assert(!strip.empty()); | |
for (auto it = strip.begin(), stop = strip.end() - 1; it != stop; ++it) { // stop before the final (outer) edge | |
sum += *it; | |
assert(sum < 32); | |
bits |= 1u << sum; | |
} | |
return bits; | |
} | |
constexpr void add_permutations(vector<uint32_t>& encoded_strips, const int twos, const int threes) { | |
vector<int> strip(twos + threes); | |
fill(strip.begin(), strip.begin() + twos, 2); | |
fill(strip.begin() + twos, strip.end(), 3); | |
assert(is_sorted(strip.begin(), strip.end())); | |
do { | |
encoded_strips.push_back(encode_strip(strip)); | |
} while (next_permutation(strip.begin(), strip.end())); | |
} | |
[[nodiscard]] constexpr vector<uint32_t> get_all_encoded_strips() { | |
vector<uint32_t> encoded_strips; | |
for (int threes = 0; threes * 3 <= total_length; threes += 2) { // micro-optimization, due to parity | |
const int remaining_length = total_length - threes * 3; | |
assert(remaining_length % 2 == 0); | |
add_permutations(encoded_strips, remaining_length / 2, threes); | |
} | |
return encoded_strips; | |
} | |
int main() { | |
const auto start = chrono::steady_clock::now(); | |
const vector<uint32_t> encoded_strips = get_all_encoded_strips(); | |
const size_t possible_strips = encoded_strips.size(); | |
namespace ublas = boost::numeric::ublas; | |
ublas::compressed_matrix<uint64_t> compatible_strips(possible_strips, possible_strips); | |
// Inefficient O(N^2) solution, where N is the number of possible strips. | |
// I'm sure there are cleverer ways to do this. | |
for (size_t i = 0; i < possible_strips; ++i) { | |
for (size_t k = i + 1; k < possible_strips; ++k) { | |
if ((encoded_strips[i] & encoded_strips[k]) == 0u) { // they're compatible! | |
compatible_strips(i, k) = 1; | |
compatible_strips(k, i) = 1; // this is a symmetric matrix | |
} | |
} | |
} | |
// For the first row of the floor, we can choose any possible strip. | |
// After completing it, we record how many choice-paths led to it. | |
// By definition, there's 1 path for each possible choice for the first row. | |
ublas::vector<uint64_t> valid_paths{ublas::scalar_vector<uint64_t>{possible_strips, 1}}; | |
for (int row_idx = 1; row_idx < num_strips; ++row_idx) { // we've completed row 0, start with row 1 | |
valid_paths = ublas::prod(compatible_strips, valid_paths); | |
} | |
const uint64_t grand_total = accumulate(valid_paths.begin(), valid_paths.end(), uint64_t{0}); | |
cout << format("Boost.uBLAS Grand Total: {}\n", grand_total); | |
const auto finish = chrono::steady_clock::now(); | |
cout << chrono::duration_cast<chrono::milliseconds>(finish - start) << "\n"; | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment