CODEWITHSUNDEEP
c++random
user1557910
user1557910

Reputation:

C++ generating random numbers without repetition .output screen is just blank with a blinking cursor

My intention is to generate random numbers from 1 to 9 without repetition

#include<iostream>
#include<ctime>
#include<cstdlib>
using namespace std;
int randrange(int low,int high)   /* generates a random number within given range*/
{
    return rand()%(low+high)+low+1;     
}

int main()
{
    int num[9]={0},i,j;     
    bool check;                         
    for(i=0;i<9;i++)
    {
        check=false;
        do
        {
            num[i]=randrange(1,9);           

            for(j=0;j<i;j++)
            {
                if( num[i]==num[j])    // checks whether number already exists in  the array 
                    check=false;
                else
                    check=true;   
            } 
        } while(check==false);
    }
    
    // the program is working fine without the repetition  check
    // this section prints out the array elements
    for(i=0;i<9;i++)
    {
        cout<<num[i]<<" ";
    }
    return 0;
}

Upvotes: 5

Views: 35354

Answers (7)

Zeeshan Hamid
Zeeshan Hamid

Reputation: 1

This code works fine for me to generate an array of random numbers from 1 to 9 without repetition

#include<iostream>
#include<ctime>
#include<cstdlib>
using namespace std;
int main(){
    srand(time(0));
    int arr[10];
    for(int i=0; i<9; i++){
        arr[i] = i+1;
        cout << arr[i] << " ";
    }
    
    int j=0;
    int temp=0;
    cout << "\n\n\n-------------------------\n\n\n";
    
    for(int i=0; i<9; i++){
        j = (rand()%8)+1;
        temp = arr[i];
        arr[i]=arr[j];
        arr[j]=temp;
    }
    
    
    for(int i=0; i<9; i++){
        cout << arr[i] << "  ";
    }
    
    
}

Upvotes: 0

Dr. S.A. Alavi Fazel
Dr. S.A. Alavi Fazel

Reputation: 1

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


using namespace std;

void rnd(vector<int> &v, const int n){
    for (size_t i=0;i<=n;++i){
        v.push_back(i);
    }
random_shuffle(v.begin(), v.end()); 
}
void PrintVector(const vector<int> &x){
    for (size_t i=0;i<x.size(); ++i){
        cout<<x[i]<<'\t';
    }
}

int main(){

    vector<int> a;
    rnd(a,10);
    PrintVector(a);

}

Upvotes: 0

P0W
P0W

Reputation: 47784

Okay, you already might have figured out the problem by dasbinkenlight's answer

And in addition to Peter's answer, you can use std::map too to achieve unique random numbers:

std::map<int,int> m;
srand (time (NULL));   
 for(i=0;i<9;i++){
  do{ 
   j=randrange(1,9);           
  }while(m.find(j)!=m.end());

 m[j]; //insert into map, no need for value.
 num[i]=j;
}

Upvotes: 0

neonKow
neonKow

Reputation: 29

Your program is probably looping. It's bit hard to read your code because of your weird indentation, but it looks like there's a logic flaw in your for loop here:

check=false;
do
{
    num[i]=randrange(1,9);           

    for(j=0;j<i;j++)
    {
        if( num[i]==num[j])    // checks whether number already exists in  the array 
            check=false;
        else
            check=true;   
    } 
} while(check==false);

You might want to remove the second check=false; line to do what I think you're trying to do.

Upvotes: 0

Mr Lister
Mr Lister

Reputation: 46559

Your program has a number of flaws, one of which is the range of random numbers that the randrange function returns. It's not 1 to 9!

The direct cause of your program (the hanging of the program) is however, that you set check to false, then perform a loop that does nothing (because the first time around, i is 0 and the inner loop with j is never performed), so check will always be false.

Check the other answers for solutions.

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726509

Your duplicate checking loop has a flaw: the check is set to the result of checking the last pair of values, rather than the result of checking all the preceding pairs.

You need to set check = true before the inner loop, then proceed to verifying all items from zero to i-1. If the check becomes false at any point, stop the loop:

check = true;
for (j = 0 ; (check) && (j < i) ; j++) {
    check = (num[i] != num[j]);
}

In addition, you need to fix the randrange, because your current implementation returns values in range 2..11:

int randrange(int low,int high)
{
    return rand()%(high-low+1)+low;     
}

Upvotes: 5

Peter Alexander
Peter Alexander

Reputation: 54270

Just generate the numbers 1 to 9, then shuffle them randomly using std::random_shuffle.

int nums[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
std::random_shuffle(nums, nums + 9);

This will leave nums with the numbers from 1 to 9 in random order, with no repetitions.

Upvotes: 11

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