16 replies to this topic

[phil67rpg]

I have done a lot of research on this problem. I am trying to store random values in a vector. I want to generate a random value from 1 to 9. Then I want to fill a vector with 5 unique random values. Then I want to output the values.
here is the code I am using.

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

using namespace std;

int main() {
    vector<int> v;


    for(int i=1;i<10;i++)
        v.push_back(i);

    random_shuffle(v.begin(), v.end());

    for(vector<int>::iterator itr=v.begin(); itr != v.end(); ++itr)
        cout << *itr << endl;


}

basically this code takes the numbers from 1 to 9 and shuffles them and then outputs them.my question is how do I shuffle a sequence of random numbers that do not repeat. Let me know if you need more explanation.

[frob]

It looks like your code is doing what you expect; you feed in ten digits, shuffle them, and output the result.


Can you give some examples of input and output? What do you mean by "a sequence of random numbers that do not repeat" ? Individual digits frequently repeat themselves in random number sequences.

[fastcall22]

How do I shuffle a sequence of random numbers that do not repeat

Hi Phil,

By default, the seed used by the random number generator is set to a default value each time the program starts. This causes the same sequence of random numbers. You will need to reseed the random number generator before using it.

Use the time function under the ctime header to get the current time (number of seconds since the epoch), then use that to seed the random number generator using srand:

srand( (unsigned)time(0) );

Edited by fastcall22, 15 November 2012 - 05:47 PM.

[Servant of the Lord]

What you have *shouldn't* repeat, except for one thing: You never seed the random number generator.

std::random_shuffle, if I recall correctly, uses rand() under the hood (unless you pass a different generator) - though I'm not sure if that changes for C++11, nor if it's standardized that rand() is default.

Since you never call srand(), rand() is seeded with 0 - thus the random shuffle repeats each time.
Instead, seed srand() with time(NULL) at the start of your program, and hopefully the result will be more suiting to your tastes.

[Edit:] Three posts within 60 seconds of each other. =)

Edited by Servant of the Lord, 15 November 2012 - 05:48 PM.

[phil67rpg]

It looks like your code is doing what you expect; you feed in ten digits, shuffle them, and output the result.


Can you give some examples of input and output? What do you mean by "a sequence of random numbers that do not repeat" ? Individual digits frequently repeat themselves in random number sequences.

well I made some changes to my code.
input should be a random number from 1 to 9
output should be a list of random numbers that do not repeat.
such as: 3 7 8 9 1 2 5 4 6

#include <iostream>
#include <vector>
#include <algorithm>
#include <time.h>

using namespace std;

int main() 
{
	srand(time(NULL));

	vector<int> v;

	for(int i=1;i<10;i++)
	{
	int j=rand()%9+1;  
	v.push_back(j);
	}

    random_shuffle(v.begin(), v.end());

    for(vector<int>::iterator itr=v.begin(); itr != v.end(); ++itr)
        cout << *itr << endl;


}

[Álvaro]

input should be a random number from 1 to 9
output should be a list of random numbers that do not repeat.
such as: 3 7 8 9 1 2 5 4 6

What's the connection between the input and the output? You want to only produce 9 different sequences?

[phil67rpg]

input should be a list of numbers from 1 to 9 generated randomly that can repeat
output should be a list of numbers from 1 to 9 that do not repeat
input: 1,3,4,2,1,5,5,9,8
output: 3,1,4,2,5,9,8

[Álvaro]

Are you trying to skip numbers that have appeared already? You example isn't quite right...

[rip-off]

The output list contains a randomised ordering of the unique set of elements in the input list?

[Servant of the Lord]

input should be a list of numbers from 1 to 9 generated randomly that can repeat
output should be a list of numbers from 1 to 9 that do not repeat
input: 1,3,4,2,1,5,5,9,8
output: 3,1,4,2,5,9,8

Your code should already do that. What is the current result that you are getting that you don't want?
Why does your input have non-unique variables? Where is this input coming from - the player or is it generated by your code?

Since we are all failing to understand, could you give a bigger-picture description of why you want this, and how you intend to use it?

[fastcall22]

input should be a list of numbers from 1 to 9 generated randomly that can repeat
output should be a list of numbers from 1 to 9 that do not repeat
input: 1,3,4,2,1,5,5,9,8
output: 3,1,4,2,5,9,8

If this is the case Phil, then you need to be generating a random set of integers from 1 to 9, not a sequence of integers from 1 to 9:

    // v is always:  1 2 3 4 5 6 7 8 9      (no duplicates, ever)
    for(int i=1;i<10;i++)
        v.push_back(i);
    random_shuffle(v.begin(),v.end());   // (elements might be in a random order, but there still aren't any duplicates)

    // v could be:  9 1 8 6 7 4 6 3 2 
    for(int i=1;i<10;i++)
        v.push_back( rand()%9 + 1 );     // (may have duplicates)

(See the reference on rand. rand() returns an integer between 0 and RAND_MAX -- usually 32767 -- and the modulus wraps that around the range between 0 and 8, and adding one brings it between 1 and 9.)

Then, you'll need to include algorithm, such that you can use sort to move duplicates together, then use unique to move the duplicate values to the back. Once the unique values are in front, you can remove the duplicates from v by using the iterator returned by unique:

    sort( v.begin(), v.end() );
    v.erase( unique(v.begin(),v.end()), v.end() );

Edit:
Hurr durr, I read can documentation good.

Edited by fastcall22, 16 November 2012 - 12:03 PM.

[Servant of the Lord]

@fastcall: Ah, good comprehension. I was misunderstanding what he meant.

[BitMaster]

Unless I'm mistaken, std::unique only removes consecutive duplicates. That means a sequence like "1 2 1" is still completely possible. Or am I misunderstanding some constraint as well?

[Khatharr]

Unless I'm mistaken, std::unique only removes consecutive duplicates. That means a sequence like "1 2 1" is still completely possible. Or am I misunderstanding some constraint as well?

I was thinking that too, but deferred since I'm too lazy to look it up.

...

Nope. Still too lazy. Sry

[KnolanCross]

The easiest way to do it is to insert the numbers in a std::set (so there will be no duplicates), then copy to a std::vector and finally shuffle then.

[rip-off]

Yes, std::unique only handles consecutive duplicates. The input list could be sorted beforehand, or alternatively it could be changed to a std::set, preventing duplicates occurring in the first place.

[Servant of the Lord]

Unless I'm mistaken, std::unique only removes consecutive duplicates. That means a sequence like "1 2 1" is still completely possible.

Which is what I think the OP was wanting. His current method (generate array from 0 to 9, shuffle array) provides unique elements already.

If the OP *wants* repeats, but just not two in a row of the same number, std::unique would help with that.
Otherwise, I have no clue what the OP wants, since his original code already gives him only uniques.