Algorithm to deal with set intervals - c++

What is the best way to go about implementing a class which needs to keep track of set intervals in C++? I was hoping to leverage existing STL or Boost libraries but other than using standard containers, I had to resort to implementing the algorithm by hand which was surprisingly tricky to get right and I had to simplify it at the cost of some performance because of a work deadline. There must be a better way.
For example, I need the following type of behaviour:
class SubRange
{
public:
SubRange(const T& lower_bound, const T& upper_bound);
...
};
Subrange r1(1,3);
Subrange r2(5,6);
Subrange r3(6,9);
Subrange tot = r1+r2+r3;
cout << tot; //displays (1,3) (5,9)
Subrange gaps = SubRange(1,9) - tot;
cout << gaps; // displays (4,4)
Notes A Subrange consists of a lower bound and an upper bound and denotes a contiguous set of elements from the lower bound to the upper bound, where both bounds are included in the set. The set is constructed starting from the lower bound and incrementing (operator++) until the upper bound is reached.
I have looked at Boost Interval which deals with interval arithmetic and doesn't seem to solve my problem but then the documentation is not easy for me to follow. If someone thinks that it will help, then could they please show me how to implement the above example.
For the curious. My use case was for a caching algorithm in a proxy server which needed to keep track of time-intervals of data requested by clients and optimally only ask for those portions of time-intervals from the server that had not already been cached. If client A asks for data from 1-Jan-2016 to 3-Jan-2016, client B asks for data from 5-Jan-2016 to 6-Jan-2016, client C asks for data from 6-Jan-2016 to 9-Jan-2016, then if client D asks for data from 1-Jan-2016 to 9-Jan-2016, then the proxy should only ask the server for 4-Jan-2016 because the rest of the dates are already cached.

You could use the Boost Interval Container Library (ICL) for this task:
#include <iostream>
#include <boost/icl/interval_set.hpp>
int main()
{
using IntervalSet = boost::icl::interval_set<int>;
using Interval = boost::icl::interval<int>;
IntervalSet set;
set.insert(Interval::closed(1,3));
set.insert(Interval::closed(5,6));
set.insert(Interval::closed(6,9));
std::cout << set << std::endl;
IntervalSet total;
total.insert(Interval::closed(1,9));
std::cout << total << std::endl;
IntervalSet diff_set = total - set;
std::cout << diff_set << std::endl;
for (auto it : diff_set)
{
// convert (possibly) open interval into closed
std::cout << "[" << boost::icl::first(it) << "," << boost::icl::last(it) << "]"<< std::endl;
}
}
output:
{[1,3][5,9]}
{[1,9]}
{(3,5)}
[4,4]
live example

try this It worked for me
Note: you can use min and max to get the max sub interval.
#include <iostream>
#include <vector>
using namespace std;
struct val{
int start = 0;
int end = 0;
};
void FindSubSortbyStart(val *x,unsigned long n){
for(int i = 1;i < n;i++)
for(int j = i;j > 0;j--)
if(x[j].start < x[j - 1].start )
swap(x[j],x[j-1]);
else
break;
}
void FindSub(val *x,unsigned long n){
vector<val> subval;
int min = x[0].start;
int max = x[0].end;
for(int i = 1;i <n;i++){
if(max + 1>= x[i].start){
if(x[i].end >= max)
max = x[i].end;
}else{
val temp;
temp.start = max + 1;
temp.end = x[i].start - 1;
max = x[i].end;
min = x[i].start;
subval.push_back(temp);
cout << subval[subval.size() - 1].start << "\t"<< subval[subval.size() - 1].end << endl;
}
}
}
int main(int argc, const char * argv[]) {
unsigned long n = 3;
val *x = new val[n];
x[0].start = 1;
x[0].end = 3;
x[1].start = 5;
x[1].end = 5;
x[2].start = 7;
x[2].end = 9;
FindSubSortbyStart(x,n);
FindSub(x,n);
return 0;
}

Related

When I try to decompose a double number into an array as elements, why do I get an offset preventing me from getting the last digit?

I have a problem. I want from a given number to get each digit as an element in the same array.
But when I compile, if I extend the range from one iteration above the size of the given number, I get a corrupted data exception from Visual Studio in Debug mode as an exception.
I thought first that was because the int type is only 4 digit max length as a 4 bytes entity because I used to get only one digit for greater number above 9999. But I noticed that my number starts at an iteration value one too late...which makes it impossible to show the last digit.
If I add a zero to my given number, I can manually offset in the opposite direction, but that doesn't work with my original number.
But, I can't find out how to fix that...Here is my code.
Before asking for help, here is a screenshot explaining the principle which is used to convert the number into an array: math theory formula
I wish to solve it with the number type only because the char type involves another way managing the memory with buffers...which I don't really know how to handle right know.
Can someone help me to complete the debugging please ?
#include <iostream>
#include <math.h>
//method to convert user number entry to array of digits
long long numToArray(double num,double arrDigits[], const long long n) {
//instanciate variables
//array of with m elements
arrDigits[n];
double* loopValue = new double(0);
//extract the digits and store them into arrDigits array
for (long long i = 0; i < n; i++) {
long temp = 0;
for (long k = 0; k < i + 1; k++) {
//mathematical general formula
temp += arrDigits[i - k] * pow(10, k);
loopValue = new double(0);
*loopValue = floor(num / pow(10, n - i)) - temp;
arrDigits[i] = *loopValue;
}
std::cout << "digits array value at " << i << " is " << arrDigits[i] << " \n";
}
return 0;
}
//main program interacting with the user
int main()
{
std::cout << "please type an integer: ";
double num;
const long long n = sizeof(num);
double array[n]{};
std::cin >> num;
//call the method to test if all values are in the array
numToArray(num, array, n);
return 0;
}
Explaining the troubleshoot
Note : Visual Studio shows error if I extend from n to n+1. If I let the type int or long, sizeof(num) is all the time 4...
Then, I had to set it as double and to extract it from the main scope, which makes it ...double...
People asking to remove pointer, it is impossible to run the program if I do so.
I want from a given number to get each digit as an element in the same array.
If you want to simply get each number into an array, it takes only a few lines of code to convert the decimal to a string, remove the decimal point (if it exists), and then copy the string to a buffer:
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <algorithm>
#include <iterator>
#include <iomanip>
int main()
{
double d = 1.45624234;
std::ostringstream strm;
strm << std::setprecision(12);
// copy this to a string using the output stream
strm << d;
std::string s = strm.str();
// remove the decimal point
s.erase(std::remove(s.begin(), s.end(), '.'), s.end());
// Now copy each digit to a buffer (in this case, vector)
std::vector<int> v;
std::transform(s.begin(), s.end(), std::back_inserter(v), [&](char ch) { return ch - '0';});
// output the results
for (auto c : v )
std::cout << c;
}
Output:
145624234
All of the work you were doing is already done for you by the standard library. In this case the overloaded operator << for double when streamed to a buffer creates the string. How it does it? That is basically what your code is attempting to do, but obviously safely and correctly.
Then it's just a matter of transforming each digit character into an actual integer that represents that digit, and that is what std::transform does. Each digit character is copied to the vector by subtracting the character 0 from each char digit.
#include <iostream>
#include <math.h>
#include <list>
int main()
{
//Entry request of any natural integer within the range of double type
std::cout << "Please type a natural integer from 1 to 99999999\n";
double num;
std::cin >> num;
//counting the number of digits
int count = 0;
long long CountingNum = static_cast<long long>(num);
while (CountingNum != 0) {
CountingNum = CountingNum/10;
++count;
}
std::cout << "number of digits compositing your natural integer: " << count<<std::endl;
//process the value for conversion to list of digits, so you can
//access each digits by power and enhance your calculus operations
double converternum = num * 10;//removing the right offset to keep the last digit
const int containerSize = sizeof(double); //defining array constant size
int sizeRescale = containerSize - count;//set general offset to handle according to the user entry
double arrDigits[containerSize] = {};//initialize array with a sufficient size.
double* loopValue = new double(0); //define pointer variable to make to operation possible
//extract the digits and store them into arrDigits array
for (long long i = 0; i < containerSize; i++) {
long temp = 0;
for (long k = 0; k < i + 1; k++) {
//mathematical general formula adapted to the computation
temp += arrDigits[i - k] * pow(10, k);
loopValue = new double(0); //reinitialize the pointer
*loopValue = floor(converternum / pow(10, containerSize - i)) - temp; //assign the math formula to the pointer
arrDigits[i] = *loopValue;//assigne the formula for any i to the array relatively to k
}
std::cout << "digits array value at " << i << " is " << arrDigits[i] << " \n";
}
//convert array to a list
std::list<double> listDigits(std::begin(arrDigits), std::end(arrDigits));
//print the converted list
std::cout << "array converted to list: ";
for (double j : listDigits) {
std::cout << j << " ";
}
std::cout << std::endl;
//remove the zeros offset and resize the new converted list
for (int j = 0; j < sizeRescale; j++) {
listDigits.pop_front();
}
std::cout << "removed zero element to the list\n";
for (double i : listDigits) {
std::cout << i << " ";
}
std::cout << "natural integer successfully converted into list digits data\n";
return 0;
}
an example on debug mode in Visual Studio 2019
I finally encapsulated the whole code into two functions. But I have an extra value at first and last iteration...
The answer is almost complete, just need to solve the offset from inside the main moved to it's owned function. I finally added a new array variable with the exact size I want from the two new functions, so we get the array which will be possible to manipulate so far away.
#include <iostream>
#include <math.h>
#include <list>
int CountNumberDigits(int num) {
int count = 0;
long long CountingNum = static_cast<long long>(num);
while (CountingNum != 0) {
CountingNum = CountingNum / 10;
++count;
}
return count;
}
double* NumToArray(double num) {
double converternum = num * 10;//removing the right offset to keep the last digit
const int containerSize = sizeof(double); //defining array constant size
int sizeRescale = containerSize - CountNumberDigits(num);//set general offset to handle according to the user entry
double arrDigits[containerSize] = {};//initialize array with a sufficient size.
double* loopValue = new double(0); //define pointer variable to make to operation possible
//extract the digits and store them into arrDigits array
for (long long i = 0; i < containerSize; i++) {
long temp = 0;
for (long k = 0; k < i + 1; k++) {
//mathematical general formula adapted to the computation
temp += arrDigits[i - k] * pow(10, k);
loopValue = new double(0); //reinitialize the pointer
*loopValue = floor(converternum / pow(10, containerSize - i)) - temp; //assign the math formula to the pointer
arrDigits[i] = *loopValue;//assigne the formula for any i to the array relatively to k
}
}
//convert array to a list
std::list<double> listDigits(std::begin(arrDigits), std::end(arrDigits));
for (double j : listDigits) {
std::cout << j << " ";
}
//remove the zeros offset and resize the new converted list
for (int j = 0; j < sizeRescale; j++) {
listDigits.pop_front();
}
//convert list to array
double* arrOutput = new double[listDigits.size()]{};
std::copy(listDigits.begin(), listDigits.end(), arrOutput);
double* ptrResult = arrOutput;
return ptrResult;
}
int main()
{
//Entry request of any natural integer within the range of double type
std::cout << "Please type a natural integer from 1 to 99999999\n";
double num;
std::cin >> num;
int count = CountNumberDigits(num);
std::cout << "number of digits compositing your natural integer: " << count << std::endl;
double* ptrOutput = NumToArray(num);
//reduce the array to the num size
double* shrinkArray = new double[CountNumberDigits(num)];
for (int i = 0; i < CountNumberDigits(num); i++) {
*(shrinkArray+i) = ptrOutput[i];
std::cout << *(shrinkArray+i) << " ";
}

It is normal that my code is faster without multithreading o am i doing something wrong?

i'm trying to optimize my code using multithreading and is not just that the program is not the double speed as is suposed to be in this dual-core computer, it is SO MUCH SLOW. And i just wanna know if i'm doing something wrong or is pretty normal that in this case use multithreading does not help. I make this recreation of how i used the multithreading, and in my computer the parallel versions take's 4 times the time in the comparation of the normal version:
#include <iostream>
#include <random>
#include <thread>
#include <chrono>
using namespace std;
default_random_engine ran;
inline bool get(){
return ran() % 3;
}
void normal_serie(unsigned repetitions, unsigned &result){
for (unsigned i = 0; i < repetitions; ++i)
result += get();
}
unsigned parallel_series(unsigned repetitions){
const unsigned hardware_threads = std::thread::hardware_concurrency();
cout << "Threads in this computer: " << hardware_threads << endl;
const unsigned threads_number = (hardware_threads != 0) ? hardware_threads : 2;
const unsigned its_per_thread = repetitions / threads_number;
unsigned *results = new unsigned[threads_number]();
std::thread *threads = new std::thread[threads_number - 1];
for (unsigned i = 0; i < threads_number - 1; ++i)
threads[i] = std::thread(normal_serie, its_per_thread, std::ref(results[i]));
normal_serie(its_per_thread, results[threads_number - 1]);
for (unsigned i = 0; i < threads_number - 1; ++i)
threads[i].join();
auto result = std::accumulate(results, results + threads_number, 0);
delete[] results;
delete[] threads;
return result;
}
int main()
{
constexpr unsigned repetitions = 100000000;
auto to = std::chrono::high_resolution_clock::now();
cout << parallel_series(repetitions) << endl;
auto tf = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(tf - to).count();
cout << "Parallel duration: " << duration << "ms" << endl;
to = std::chrono::high_resolution_clock::now();
unsigned r = 0;
normal_serie(repetitions, r);
cout << r << endl;
tf = std::chrono::high_resolution_clock::now();
duration = std::chrono::duration_cast<std::chrono::milliseconds>(tf - to).count();
cout << "Normal duration: " << duration << "ms" << endl;
return 0;
}
Things that i already know, but i didn't to make this code shorter:
I should set a max_iterations_per_thread because you don't wanna make 10 iterations per thread, but in this case we are doing one billion iterations so that is not gonna happend.
The number of iterations must be divisible by the number or threads, otherwise the code will not do an effective work.
This is the output that i get in my computer:
Threads in this computer: 2
66665160
Parallel duration: 4545ms
66664432
Normal duration: 1019ms
(Solved partially doing this changes: )
inline bool get(default_random_engine &ran){
return ran() % 3;
}
void normal_serie(unsigned repetitions, unsigned &result){
default_random_engine eng;
unsigned saver_result = 0;
for (unsigned i = 0; i < repetitions; ++i)
saver_result += get(eng);
result += saver_result;
}
All your threads are tripping over each other fighting for access to ran which can only perform one operation at a time because it only has one state and each operation advances its state. There is no point in running operations in parallel if the vast majority of each operation involves a choke point that cannot support any concurrency.
All elements of results are likely to share a cache line, which means there is lots of inter-core communication going on.
Try modifying normal_serie to accumulate into a local variable and only write it to results in the end.

Finding the median value of a vector using C++

I'm a programming student, and for a project I'm working on, on of the things I have to do is compute the median value of a vector of int values and must be done by passing it through functions. Also the vector is initially generated randomly using the C++ random generator mt19937 which i have already written down in my code.I'm to do this using the sort function and vector member functions such as .begin(), .end(), and .size().
I'm supposed to make sure I find the median value of the vector and then output it
And I'm Stuck, below I have included my attempt. So where am I going wrong? I would appreciate if you would be willing to give me some pointers or resources to get going in the right direction.
Code:
#include<iostream>
#include<vector>
#include<cstdlib>
#include<ctime>
#include<random>
#include<vector>
#include<cstdlib>
#include<ctime>
#include<random>
using namespace std;
double find_median(vector<double>);
double find_median(vector<double> len)
{
{
int i;
double temp;
int n=len.size();
int mid;
double median;
bool swap;
do
{
swap = false;
for (i = 0; i< len.size()-1; i++)
{
if (len[i] > len[i + 1])
{
temp = len[i];
len[i] = len[i + 1];
len[i + 1] = temp;
swap = true;
}
}
}
while (swap);
for (i=0; i<len.size(); i++)
{
if (len[i]>len[i+1])
{
temp=len[i];
len[i]=len[i+1];
len[i+1]=temp;
}
mid=len.size()/2;
if (mid%2==0)
{
median= len[i]+len[i+1];
}
else
{
median= (len[i]+0.5);
}
}
return median;
}
}
int main()
{
int n,i;
cout<<"Input the vector size: "<<endl;
cin>>n;
vector <double> foo(n);
mt19937 rand_generator;
rand_generator.seed(time(0));
uniform_real_distribution<double> rand_distribution(0,0.8);
cout<<"original vector: "<<" ";
for (i=0; i<n; i++)
{
double rand_num=rand_distribution(rand_generator);
foo[i]=rand_num;
cout<<foo[i]<<" ";
}
double median;
median=find_median(foo);
cout<<endl;
cout<<"The median of the vector is: "<<" ";
cout<<median<<endl;
}
The median is given by
const auto median_it = len.begin() + len.size() / 2;
std::nth_element(len.begin(), median_it , len.end());
auto median = *median_it;
For even numbers (size of vector) you need to be a bit more precise. E.g., you can use
assert(!len.empty());
if (len.size() % 2 == 0) {
const auto median_it1 = len.begin() + len.size() / 2 - 1;
const auto median_it2 = len.begin() + len.size() / 2;
std::nth_element(len.begin(), median_it1 , len.end());
const auto e1 = *median_it1;
std::nth_element(len.begin(), median_it2 , len.end());
const auto e2 = *median_it2;
return (e1 + e2) / 2;
} else {
const auto median_it = len.begin() + len.size() / 2;
std::nth_element(len.begin(), median_it , len.end());
return *median_it;
}
There are of course many different ways how we can get element e1. We could also use max or whatever we want. But this line is important because nth_element only places the nth element correctly, the remaining elements are ordered before or after this element, depending on whether they are larger or smaller. This range is unsorted.
This code is guaranteed to have linear complexity on average, i.e., O(N), therefore it is asymptotically better than sort, which is O(N log N).
Regarding your code:
for (i=0; i<len.size(); i++){
if (len[i]>len[i+1])
This will not work, as you access len[len.size()] in the last iteration which does not exist.
std::sort(len.begin(), len.end());
double median = len[len.size() / 2];
will do it. You might need to take the average of the middle two elements if size() is even, depending on your requirements:
0.5 * (len[len.size() / 2 - 1] + len[len.size() / 2]);
Instead of trying to do everything at once, you should start with simple test cases and work upwards:
#include<vector>
double find_median(std::vector<double> len);
// Return the number of failures - shell interprets 0 as 'success',
// which suits us perfectly.
int main()
{
return find_median({0, 1, 1, 2}) != 1;
}
This already fails with your code (even after fixing i to be an unsigned type), so you could start debugging (even 'dry' debugging, where you trace the code through on paper; that's probably enough here).
I do note that with a smaller test case, such as {0, 1, 2}, I get a crash rather than merely failing the test, so there's something that really needs to be fixed.
Let's replace the implementation with one based on overseas's answer:
#include <algorithm>
#include <limits>
#include <vector>
double find_median(std::vector<double> len)
{
if (len.size() < 1)
return std::numeric_limits<double>::signaling_NaN();
const auto alpha = len.begin();
const auto omega = len.end();
// Find the two middle positions (they will be the same if size is odd)
const auto i1 = alpha + (len.size()-1) / 2;
const auto i2 = alpha + len.size() / 2;
// Partial sort to place the correct elements at those indexes (it's okay to modify the vector,
// as we've been given a copy; otherwise, we could use std::partial_sort_copy to populate a
// temporary vector).
std::nth_element(alpha, i1, omega);
std::nth_element(i1, i2, omega);
return 0.5 * (*i1 + *i2);
}
Now, our test passes. We can write a helper method to allow us to create more tests:
#include <iostream>
bool test_median(const std::vector<double>& v, double expected)
{
auto actual = find_median(v);
if (abs(expected - actual) > 0.01) {
std::cerr << actual << " - expected " << expected << std::endl;
return true;
} else {
std::cout << actual << std::endl;
return false;
}
}
int main()
{
return test_median({0, 1, 1, 2}, 1)
+ test_median({5}, 5)
+ test_median({5, 5, 5, 0, 0, 0, 1, 2}, 1.5);
}
Once you have the simple test cases working, you can manage more complex ones. Only then is it time to create a large array of random values to see how well it scales:
#include <ctime>
#include <functional>
#include <random>
int main(int argc, char **argv)
{
std::vector<double> foo;
const int n = argc > 1 ? std::stoi(argv[1]) : 10;
foo.reserve(n);
std::mt19937 rand_generator(std::time(0));
std::uniform_real_distribution<double> rand_distribution(0,0.8);
std::generate_n(std::back_inserter(foo), n, std::bind(rand_distribution, rand_generator));
std::cout << "Vector:";
for (auto v: foo)
std::cout << ' ' << v;
std::cout << "\nMedian = " << find_median(foo) << std::endl;
}
(I've taken the number of elements as a command-line argument; that's more convenient in my build than reading it from cin). Notice that instead of allocating n doubles in the vector, we simply reserve capacity for them, but don't create any until needed.
For fun and kicks, we can now make find_median() generic. I'll leave that as an exercise; I suggest you start with:
typename<class Iterator>
auto find_median(Iterator alpha, Iterator omega)
{
using value_type = typename Iterator::value_type;
if (alpha == omega)
return std::numeric_limits<value_type>::signaling_NaN();
}

C++ Advice on manipulating output Matrix data

I have the following code.
Essentially it is creating N random normal variables, and running through an equation M times for a simulation.
The output should be an NxM matrix of data, however the only way I could do the calculation has the output as MxN. ie each M run should be a column, not a row.
I have attempted in vain to follow some of the other suggestions that have been posted on previous similar topics.
Code:
#include <iostream>
#include <time.h>
#include <random>
int main()
{
double T = 1; // End time period for simulation
int N = 4; // Number of time steps
int M = 2; // Number of simulations
double x0 = 1.00; // Starting x value
double mu = 0.00; // mu(x,t) value
double sig = 1.00; // sigma(x,t) value
double dt = T/N;
double sqrt_dt = sqrt(dt);
double** SDE_X = new double*[M]; // SDE Matrix setup
// Random Number generation setup
double RAND_N;
srand ((unsigned int) time(NULL)); // Generator loop reset
std::default_random_engine generator (rand());
std::normal_distribution<double> distribution (0.0,1.0); // Mean = 0.0, Variance = 1.0 ie Normal
for (int i = 0; i < M; i++)
{
SDE_X[i] = new double[N];
for (int j=0; j < N; j++)
{
RAND_N = distribution(generator);
SDE_X[i][0] = x0;
SDE_X[i][j+1] = SDE_X[i][j] + mu * dt + sig * RAND_N * sqrt_dt; // The SDE we wish to plot the path for
std::cout << SDE_X[i][j] << " ";
}
std::cout << std::endl;
}
std::cout << std::endl;
std::cout << " The simulation is complete!!" << std::endl;
std::cout << std::endl;
system("pause");
return 0;
}
Well why can't you just create the transpose of your SDE_X matrix then? Isn't that what you want to get?
Keep in mind, that presentation has nothing to do with implementation. Whether to access columns or rows is your decision. So you want an implementation of it transposed. Then quick and dirty create your matrix first, and then create your number series. Change i and j, and N and M.
I said quick and dirty, because the program at all is bad:
why don't you just keep it simple and use a better data structure for your matrix? If you know the size: compile-time array or dynamic vectors at runtime? Maybe there are some nicer implementation for 2d array.
There is a bug I think: you create N doubles and access index 0 to N inclusive.
In every iteration you set index 0 to x0 what is also needless.
I would change your code a bit make more clear:
create your matrix at first
initialize the first value of the matrix
provide an algorithm function calculating a target cell taking the matrix and the parameters.
Go through each cell and invoke your function for that cell
Thank you all for your input. I was able to implement my code and have it displayed as needed.
I added a second for loop to rearrange the matrix rows and columns.
Please feel free to let me know if you think there is anyway I can improve it.
#include <iostream>
#include <time.h>
#include <random>
#include <vector>
int main()
{
double T = 1; // End time period for simulation
int N = 3; // Number of time steps
int M = 2; // Number of simulations
int X = 100; // Max number of matrix columns
int Y = 100; // Max number of matrix rows
double x0 = 1.00; // Starting x value
double mu = 0.00; // mu(x,t) value
double sig = 1.00; // sigma(x,t) value
double dt = T/N;
double sqrt_dt = sqrt(dt);
std::vector<std::vector<double>> SDE_X((M*N), std::vector<double>((M*N))); // SDE Matrix setup
// Random Number generation setup
double RAND_N;
srand ((unsigned int) time(NULL)); // Generator loop reset
std::default_random_engine generator (rand());
std::normal_distribution<double> distribution (0.0,1.0); // Mean = 0.0, Variance = 1.0 ie Normal
for (int i = 0; i <= M; i++)
{
SDE_X[i][0] = x0;
for (int j=0; j <= N; j++)
{
RAND_N = distribution(generator);
SDE_X[i][j+1] = SDE_X[i][j] + mu * dt + sig * RAND_N * sqrt_dt; // The SDE we wish to plot the path for
}
}
for (int j = 0; j <= N; j++)
{
for (int i = 0; i <=M; i++)
{
std::cout << SDE_X[i][j] << ", ";
}
std::cout << std::endl;
}
std::cout << std::endl;
std::cout << " The simulation is complete!!" << std::endl;
std::cout << std::endl;
system("pause");
return 0;
}

Using pow() for large number

I am trying to solve a problem, a part of which requires me to calculate (2^n)%1000000007 , where n<=10^9. But my following code gives me output "0" even for input like n=99.
Is there anyway other than having a loop which multilplies the output by 2 every time and finding the modulo every time (this is not I am looking for as this will be very slow for large numbers).
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
int main()
{
unsigned long long gaps,total;
while(1)
{
cin>>gaps;
total=(unsigned long long)powf(2,gaps)%1000000007;
cout<<total<<endl;
}
}
You need a "big num" library, it is not clear what platform you are on, but start here:
http://gmplib.org/
this is not I am looking for as this will be very slow for large numbers
Using a bigint library will be considerably slower pretty much any other solution.
Don't take the modulo every pass through the loop: rather, only take it when the output grows bigger than the modulus, as follows:
#include <iostream>
int main() {
int modulus = 1000000007;
int n = 88888888;
long res = 1;
for(long i=0; i < n; ++i) {
res *= 2;
if(res > modulus)
res %= modulus;
}
std::cout << res << std::endl;
}
This is actually pretty quick:
$ time ./t
./t 1.19s user 0.00s system 99% cpu 1.197 total
I should mention that the reason this works is that if a and b are equivalent mod m (that is, a % m = b % m), then this equality holds multiple k of a and b (that is, the foregoing equality implies (a*k)%m = (b*k)%m).
Chris proposed GMP, but if you need just that and want to do things The C++ Way, not The C Way, and without unnecessary complexity, you may just want to check this out - it generates few warnings when compiling, but is quite simple and Just Works™.
You can split your 2^n into chunks of 2^m. You need to find: `
2^m * 2^m * ... 2^(less than m)
Number m should be 31 is for 32-bit CPU. Then your answer is:
chunk1 % k * chunk2 * k ... where k=1000000007
You are still O(N). But then you can utilize the fact that all chunk % k are equal except last one and you can make it O(1)
I wrote this function. It is very inefficient but it works with very large numbers. It uses my self-made algorithm to store big numbers in arrays using a decimal like system.
mpfr2.cpp
#include "mpfr2.h"
void mpfr2::mpfr::setNumber(std::string a) {
for (int i = a.length() - 1, j = 0; i >= 0; ++j, --i) {
_a[j] = a[i] - '0';
}
res_size = a.length();
}
int mpfr2::mpfr::multiply(mpfr& a, mpfr b)
{
mpfr ans = mpfr();
// One by one multiply n with individual digits of res[]
int i = 0;
for (i = 0; i < b.res_size; ++i)
{
for (int j = 0; j < a.res_size; ++j) {
ans._a[i + j] += b._a[i] * a._a[j];
}
}
for (i = 0; i < a.res_size + b.res_size; i++)
{
int tmp = ans._a[i] / 10;
ans._a[i] = ans._a[i] % 10;
ans._a[i + 1] = ans._a[i + 1] + tmp;
}
for (i = a.res_size + b.res_size; i >= 0; i--)
{
if (ans._a[i] > 0) break;
}
ans.res_size = i+1;
a = ans;
return a.res_size;
}
mpfr2::mpfr mpfr2::mpfr::pow(mpfr a, mpfr b) {
mpfr t = a;
std::string bStr = "";
for (int i = b.res_size - 1; i >= 0; --i) {
bStr += std::to_string(b._a[i]);
}
int i = 1;
while (!0) {
if (bStr == std::to_string(i)) break;
a.res_size = multiply(a, t);
// Debugging
std::cout << "\npow() iteration " << i << std::endl;
++i;
}
return a;
}
mpfr2.h
#pragma once
//#infdef MPFR2_H
//#define MPFR2_H
// C standard includes
#include <iostream>
#include <string>
#define MAX 0x7fffffff/32/4 // 2147483647
namespace mpfr2 {
class mpfr
{
public:
int _a[MAX];
int res_size;
void setNumber(std::string);
static int multiply(mpfr&, mpfr);
static mpfr pow(mpfr, mpfr);
};
}
//#endif
main.cpp
#include <iostream>
#include <fstream>
// Local headers
#include "mpfr2.h" // Defines local mpfr algorithm library
// Namespaces
namespace m = mpfr2; // Reduce the typing a bit later...
m::mpfr tetration(m::mpfr, int);
int main() {
// Hardcoded tests
int x = 7;
std::ofstream f("out.txt");
m::mpfr t;
for(int b=1; b<x;b++) {
std::cout << "2^^" << b << std::endl; // Hardcoded message
t.setNumber("2");
m::mpfr res = tetration(t, b);
for (int i = res.res_size - 1; i >= 0; i--) {
std::cout << res._a[i];
f << res._a[i];
}
f << std::endl << std::endl;
std::cout << std::endl << std::endl;
}
char c; std::cin.ignore(); std::cin >> c;
return 0;
}
m::mpfr tetration(m::mpfr a, int b)
{
m::mpfr tmp = a;
if (b <= 0) return m::mpfr();
for (; b > 1; b--) tmp = m::mpfr::pow(a, tmp);
return tmp;
}
I created this for tetration and eventually hyperoperations. When the numbers get really big it can take ages to calculate and a lot of memory. The #define MAX 0x7fffffff/32/4 is the number of decimals one number can have. I might make another algorithm later to combine multiple of these arrays into one number. On my system the max array length is 0x7fffffff aka 2147486347 aka 2^31-1 aka int32_max (which is usually the standard int size) so I had to divide int32_max by 32 to make the creation of this array possible. I also divided it by 4 to reduce memory usage in the multiply() function.
- Jubiman