Bubble sort in parallel











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I have done bubble sort algorithm on a vector that is filled with randomly generated values. Bubble sort is actually done with odd-even transition method:



#include <iostream>
#include <algorithm>
#include <vector>
#include <random>
#include <thread>
#include <mutex>

void fill_with_random(std::vector<int> &vec)
{
constexpr int lower_bound = 1;
constexpr int upper_bound = 100;

std::random_device rnd_device;
std::mt19937 mersenne_engine(rnd_device());

std::uniform_int_distribution<int> distribution(lower_bound, upper_bound);

auto generator = std::bind(distribution, mersenne_engine);
std::generate(vec.begin(), vec.end(), generator);
}

bool is_odd(int number)
{
return number % 2 != 0;
}

int main(int argc, char *argv)
{
constexpr size_t vector_size = 10;

std::mutex mutex;

std::vector<int> vec(vector_size);
fill_with_random(vec);

std::cout << "Normal vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

for (size_t i = 0; i < vector_size; i++)
{
std::vector<std::thread> threads;
if (is_odd(i))
{
for (size_t j = 1; j < vector_size / 2 + vector_size % 2; j++)
{
size_t second = 2 * j;
size_t first = second - 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}
else
{
for (size_t j = 0; j < vector_size / 2; j++)
{
size_t first = 2 * j;
size_t second = first + 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}

for (auto& thread : threads)
{
thread.join();
}
}

std::cout << "Sorted vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

return 0;
}


Apart from style review, please check functionality of the algorithm itself.
Thanks in advance.










share|improve this question


















  • 1




    Have you tried to test it? Just use std::is_sorted() on your vector. You could write a loop that would check vectors up to certain size.
    – Incomputable
    Jun 5 at 18:47










  • I haven't made a unit test yet but I will.
    – PeMaCN
    Jun 6 at 6:59















up vote
2
down vote

favorite
1












I have done bubble sort algorithm on a vector that is filled with randomly generated values. Bubble sort is actually done with odd-even transition method:



#include <iostream>
#include <algorithm>
#include <vector>
#include <random>
#include <thread>
#include <mutex>

void fill_with_random(std::vector<int> &vec)
{
constexpr int lower_bound = 1;
constexpr int upper_bound = 100;

std::random_device rnd_device;
std::mt19937 mersenne_engine(rnd_device());

std::uniform_int_distribution<int> distribution(lower_bound, upper_bound);

auto generator = std::bind(distribution, mersenne_engine);
std::generate(vec.begin(), vec.end(), generator);
}

bool is_odd(int number)
{
return number % 2 != 0;
}

int main(int argc, char *argv)
{
constexpr size_t vector_size = 10;

std::mutex mutex;

std::vector<int> vec(vector_size);
fill_with_random(vec);

std::cout << "Normal vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

for (size_t i = 0; i < vector_size; i++)
{
std::vector<std::thread> threads;
if (is_odd(i))
{
for (size_t j = 1; j < vector_size / 2 + vector_size % 2; j++)
{
size_t second = 2 * j;
size_t first = second - 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}
else
{
for (size_t j = 0; j < vector_size / 2; j++)
{
size_t first = 2 * j;
size_t second = first + 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}

for (auto& thread : threads)
{
thread.join();
}
}

std::cout << "Sorted vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

return 0;
}


Apart from style review, please check functionality of the algorithm itself.
Thanks in advance.










share|improve this question


















  • 1




    Have you tried to test it? Just use std::is_sorted() on your vector. You could write a loop that would check vectors up to certain size.
    – Incomputable
    Jun 5 at 18:47










  • I haven't made a unit test yet but I will.
    – PeMaCN
    Jun 6 at 6:59













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I have done bubble sort algorithm on a vector that is filled with randomly generated values. Bubble sort is actually done with odd-even transition method:



#include <iostream>
#include <algorithm>
#include <vector>
#include <random>
#include <thread>
#include <mutex>

void fill_with_random(std::vector<int> &vec)
{
constexpr int lower_bound = 1;
constexpr int upper_bound = 100;

std::random_device rnd_device;
std::mt19937 mersenne_engine(rnd_device());

std::uniform_int_distribution<int> distribution(lower_bound, upper_bound);

auto generator = std::bind(distribution, mersenne_engine);
std::generate(vec.begin(), vec.end(), generator);
}

bool is_odd(int number)
{
return number % 2 != 0;
}

int main(int argc, char *argv)
{
constexpr size_t vector_size = 10;

std::mutex mutex;

std::vector<int> vec(vector_size);
fill_with_random(vec);

std::cout << "Normal vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

for (size_t i = 0; i < vector_size; i++)
{
std::vector<std::thread> threads;
if (is_odd(i))
{
for (size_t j = 1; j < vector_size / 2 + vector_size % 2; j++)
{
size_t second = 2 * j;
size_t first = second - 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}
else
{
for (size_t j = 0; j < vector_size / 2; j++)
{
size_t first = 2 * j;
size_t second = first + 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}

for (auto& thread : threads)
{
thread.join();
}
}

std::cout << "Sorted vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

return 0;
}


Apart from style review, please check functionality of the algorithm itself.
Thanks in advance.










share|improve this question













I have done bubble sort algorithm on a vector that is filled with randomly generated values. Bubble sort is actually done with odd-even transition method:



#include <iostream>
#include <algorithm>
#include <vector>
#include <random>
#include <thread>
#include <mutex>

void fill_with_random(std::vector<int> &vec)
{
constexpr int lower_bound = 1;
constexpr int upper_bound = 100;

std::random_device rnd_device;
std::mt19937 mersenne_engine(rnd_device());

std::uniform_int_distribution<int> distribution(lower_bound, upper_bound);

auto generator = std::bind(distribution, mersenne_engine);
std::generate(vec.begin(), vec.end(), generator);
}

bool is_odd(int number)
{
return number % 2 != 0;
}

int main(int argc, char *argv)
{
constexpr size_t vector_size = 10;

std::mutex mutex;

std::vector<int> vec(vector_size);
fill_with_random(vec);

std::cout << "Normal vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

for (size_t i = 0; i < vector_size; i++)
{
std::vector<std::thread> threads;
if (is_odd(i))
{
for (size_t j = 1; j < vector_size / 2 + vector_size % 2; j++)
{
size_t second = 2 * j;
size_t first = second - 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}
else
{
for (size_t j = 0; j < vector_size / 2; j++)
{
size_t first = 2 * j;
size_t second = first + 1;

threads.emplace_back(
[&vec, first, second, &mutex]()
{
if (vec[first] > vec[second])
{
std::lock_guard<std::mutex> lock(mutex);
std::iter_swap(vec.begin() + first, vec.begin() + second);
}
}
);
}
}

for (auto& thread : threads)
{
thread.join();
}
}

std::cout << "Sorted vector: ";
for (auto item : vec)
{
std::cout << item << " ";
}
std::cout << std::endl;

return 0;
}


Apart from style review, please check functionality of the algorithm itself.
Thanks in advance.







c++ multithreading






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share|improve this question











share|improve this question




share|improve this question










asked Jun 5 at 18:14









PeMaCN

1404




1404








  • 1




    Have you tried to test it? Just use std::is_sorted() on your vector. You could write a loop that would check vectors up to certain size.
    – Incomputable
    Jun 5 at 18:47










  • I haven't made a unit test yet but I will.
    – PeMaCN
    Jun 6 at 6:59














  • 1




    Have you tried to test it? Just use std::is_sorted() on your vector. You could write a loop that would check vectors up to certain size.
    – Incomputable
    Jun 5 at 18:47










  • I haven't made a unit test yet but I will.
    – PeMaCN
    Jun 6 at 6:59








1




1




Have you tried to test it? Just use std::is_sorted() on your vector. You could write a loop that would check vectors up to certain size.
– Incomputable
Jun 5 at 18:47




Have you tried to test it? Just use std::is_sorted() on your vector. You could write a loop that would check vectors up to certain size.
– Incomputable
Jun 5 at 18:47












I haven't made a unit test yet but I will.
– PeMaCN
Jun 6 at 6:59




I haven't made a unit test yet but I will.
– PeMaCN
Jun 6 at 6:59










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    How much energy can I measure for a 500-dimensional array?





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      How much energy can I measure for a 500-dimensional array?





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        answered 4 mins ago









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