Thread: List of numbers in random order

If it is in fact a list of numbers including every number between X and Y, then populate an array with them and shuffle it.

Quzah.

How do you shuffle cards? ........ man, it's not brain surgery. Just jump around in the array and switch the places of variables until you get tired of doing it.

Quzah.

The most straight forward way is to use the random_shuffle() function defined in the algorithms header.

http://www.sgi.com/tech/stl/random_shuffle.html

EDIT: Damn, shouldn't code at night. Realized I was using the wrong header name and a non-standard extension function.
The following should work fine.

Code:

#include <cstdlib>
#include <vector>
#include <algorithm>
#include <ctime>
#include <iostream>

using namespace std;

int main()
{
  const size_t CONTAINER_SIZE = 1001;  
    
  srand( time(NULL) ); // seed rand
  vector<int> nums(CONTAINER_SIZE);

  for(size_t ix = 0; ix < nums.size(); ix++)    // fill from 1 to 1000
  {
     nums[ix] = ix + 1;
  }
  
  random_shuffle(nums.begin(), nums.end() );   // shuffle elements

  for(size_t ix = 0; ix < nums.size(); ix++)
  {
     cout << nums[ix] << '\n';
  }

  return 0;
}

to avoid repeats, you could also try:

Code:

#include<iostream>
#include<cstdlib>
#include<ctime>

int main()
{
    int cards&#091;1000&#093;;                //this array is to hold the random cards
    unsigned short int taken&#091;1000&#093;; //this array is to keep track of used cards
    srand(static_cast<unsigned int>(time(0))); //don't forget the seed

    for(register int i=0;i<1000;i++)
        taken&#091;i&#093;=0;                     //initialize the 'used' array

    for(register int i=0;i<1000;i++)
    {
        int nextCard=1+rand()%1000;     //pull a random # in the range (1..1000)
        if(taken&#091;nextCard-1&#093;==1)        //if the card's been used already
        {
            --i;                        //decrement the loop control variable
            continue;                   //and start the loop from the top
        }
        cards&#091;i&#093;=nextCard;              //if not, use that card
        taken&#091;nextCard-1&#093;=1;            //and mark it as used
    }
    
    for(register int i=0;i<1000;i++)    //output loop
        std::cout<<cards&#091;i&#093;<<", ";
        
    std::cin.get();
    return 0;
}

heh... got a little into it, so there's some working code...

That code might take a loooooooong time to execute. Think about it: once you're down to the last available card, each iteration has a 1/1000th chance of hitting it. In other words, the average runtime of this method is O(n²). Worst case - if rand() were truly random - would be forever, in the weird case that this last number just won't appear.

> Worst case - if rand() were truly random - would be forever,
Not so - true random numbers are allowed to repeat (just look at a coin toss).
Specifically excluding all previous results is not a random stream.

But yes, the solution is horrible - use a shuffle like Quzah suggested.

Originally Posted by Salem

> Worst case - if rand() were truly random - would be forever,
Not so - true random numbers are allowed to repeat (just look at a coin toss).
Specifically excluding all previous results is not a random stream.

Exactly what I'm saying. What if the random stream happens to repeat just one number for a year? It will always try to select the same card, thus looping and looping without ever getting anywhere.

rand() on the other hand is guaranteed to visit every number of its range once in a while.