generating random numbers without consecutive repetition

Question

How can I generate random integers from 0 - 4 in such a way that the same number is not generated twice consecutively? For example, if 3 is the number that is generated the first time then 0,1,2,4 will be the possible numbers for random generation the second time. If 2 is generated the second time then 0,1,3,4 will be the possible numbers for random generation the third time and so on.

You can't make random less random. Use an NSSet for uniquing. — CodaFi, Mar 31 '13 at 20:53
possible duplicate of [Generate random number from given number](http://stackoverflow.com/questions/15680582/generate-random-number-from-given-number) — Martin R, Mar 31 '13 at 20:55
@MartinR - Those aren't dupes of his question. He is not asking for a shuffle. — Hot Licks, Mar 31 '13 at 21:08
@HotLicks: You are right. I overlooked the word *consecutive* when reading the question. Sorry! — Martin R, Mar 31 '13 at 21:13

score 7 · Accepted Answer · answered Mar 31 '13 at 21:06

7

int oldrand = <prior random number>;
int adder = randomNumberGenerator() % 4;
int newrand = (oldrand + adder + 1) % 5;

answered Mar 31 '13 at 21:06

Hot Licks

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I am not sure I understand how this solution works. I would appreciate it if you explain further. – oopology Apr 01 '13 at 03:15
2

@oopology - You have a number 0 .. 4. You want another number in the same range, only not exactly the same number. So add any number between 1 and 4 to the first number, then, if the addition overflows, "wrap it around". So if your first number is 3 and you pick a random value 2, then add 3 + 2 + 1, you get 6, which is larger than your range. But 5 wraps to 0 and 6 wraps to 1, so your result is 1. – Hot Licks Apr 01 '13 at 03:30
thanks for the explanation. I wish I had enough up votes to give your explanation and answer an up vote. – oopology Apr 02 '13 at 07:38

score 1 · Answer 2 · answered Mar 31 '13 at 21:17

uint32_t myRandomNumber(uint32_t upperBound, uint32_t avoid) {
    if (avoid < upperBound) {
        --upperBound;
    }
    uint32_t number = arc4random_uniform(upperBound);
    if (number >= avoid) {
        ++number;
    }
    return number;
}

Call it like this the first time:

uint32_t number = myRandomNumber(5, 5);

Call it like this after the first time:

number = myRandomNumber(5, number);

score 0 · Answer 3 · answered Mar 31 '13 at 22:04

it can be generated with using 2 arrays; 1 is sorted and the other is created by taking random objects form sorted like

-(NSMutableArray *)generateRandomNumbersFrom:(int) a to:(int)b
{

    NSMutableArray * numbersWillBeChosen=[NSMutableArray new];
    NSMutableArray * randomList=[NSMutableArray new];
    //add the numbers that will create your ramdom list in order
    for(int k=0;k<=(b-a);k++)
    {
        [numbersWillBeChosen addObject:@(a+k)];
    }


    //create random list
    for (int p=(b-a)+1; p>0; p--) {

        int rand= arc4random()%p;

        [randomList addObject:[numbersWillBeChosen objectAtIndex:rand]];

        [numbersWillBeChosen removeObjectAtIndex:rand];


    }

    NSLog(@"%@",randomList);

return randomList;
    }

and you can call for your situation

randList=[self generateRandomNumbersFrom:0 to:4];

score 0 · Answer 4 · answered Jan 09 '17 at 05:24

I tried Hot Lick's answer and it ended up generating duplicates for me (or out of range numbers). But it inspired me to come up with a solution that worked for me.

My code is C#, but you can modify it to suit your needs.

//Returns a random number not same as previous random number. ie. no 2 consec random
public static int NonConsecRandom(int max, int? lastNum=null)
{
    if (lastNum == null)
        return Random.Range(0, max);

    int last = (int)lastNum;            
    return Random.Range(last + 1, last + max) % max;
}

Explanation:

first block just returns a random number if there was no previously generated number.

the second part generates a random number from the "other numbers".

For example if my "max" was 5, the possible random numbers are: 0, 1, 2, 3, 4

Lets say my last number was 2, now I want a random number among [3, 4, 0, 1]

So I generate a random between last+1 and last+max. i.e. between 3 (inclusive) and 7 (not inclusive)

[3, 4, 5, 6]

Mod any of those numbers with max (i.e. 5) and you'll get a random number in the desired range: [3 , 4, 0, 1]

generating random numbers without consecutive repetition

4 Answers4

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