I was making a puzzle game and i want to generate a list of random numbers between some limit. I have already used rand and srand function but it gives me duplicate value also. I want to generate a random list without duplication how would i do that?
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Are you saying it's generating the same numbers on every call? Or just that occasionally it repeats numbers? If the latter, I know of no better way than keeping track of the numbers you've already generated and maintaining that information as you generate unique values. – prelic Jan 19 '12 at 17:23
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3http://stackoverflow.com/a/196065/857994 actually is better :) – John Humphreys Jan 19 '12 at 17:24
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@w00te: This is not exactly the same question as here an small upper limit is not specified for the random numbers. – salva Jan 19 '12 at 18:53
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Yeah, I realized that but I thought it was helpful. Notice I didn't cast a close vote :) – John Humphreys Jan 19 '12 at 18:58
5 Answers
The usual approach for this is something like:
populate an array source_array of size <n> with numbers from 0 to n-1
while n > 0
use rand to generate a random number x in the range 0..n-1
add source_array[x] to the result list
source_array[x] = source_array[n-1]; // replace number just used with last value
--n; // next time, one less number

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and see http://stackoverflow.com/questions/196017/unique-random-numbers-in-o1/196065#196065 for a more detailed explanation, with pictures – David Gelhar Jan 19 '12 at 17:29
I'll presume you are using an array to store the numbers, you could use something like this inArray()
function and use it with a do ... while
loop which would generate a random number until it generates one that is not in the array
Edit: w00te's comment links to this answer which I believe is better then my method and definitely worth a read. This is also what David Gelhar is briefly suggesting.

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This method results appropriate when the limit is high and you only want to generate a few random numbers.
#!/usr/bin/perl
($top, $n) = @ARGV; # generate $n integer numbers in [0, $top)
$last = -1;
for $i (0 .. $n-1) {
$range = $top - $n + $i - $last;
$r = 1 - rand(1.0)**(1 / ($n - $i));
$last += int($r * $range + 1);
print "$last ($r)\n";
}
Note that the numbers are generated in ascending order, but you can shuffle then afterwards.

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To avoid duplication, you could store all generated numbers in a list and on each iteration, check if the number was already generated; If yes, regenerate, if not add it to the list and show it.

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You can do this:
#include <stdio.h>
#include <stdlib.h>
#define MAX_NUMBER_VALUE 60
#define HOW_MANY_NUMBERS 10
int main ( void )
{
int numbers[HOW_MANY_NUMBERS];
int counter = 0;
while ( counter < HOW_MANY_NUMBERS)
{
int tempRandom = rand ( MAX_NUMBER_VALUE + 1 );
int check = 0;
while ( check <= counter )
{
if ( number[counter] == number[check] )
tempRandom = rand ( MAX_NUMBER_VALUE + 1 );
}
printf("Random number #%d - Value: %d", counter, tempRandom);
}
}
You need perform this, but it works.

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