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I want to generate a random number that will be in the range of 1 to 6 (I know how to do that) but won't be equal to 4 other variables (for instance, their names are "num1", "num2", "num3" and "num4") in the algorithm. Can I do that? and how?

RobertS supports Monica Cellio
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shubbbik
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    Let me see if I understand... You want four variables, each with a random value between 1 and 6 (inclusive), but no repeats? – Fiddling Bits Jan 07 '20 at 18:44
  • You can do that. One solution would be to have two loops. Outer loop to set each variable with a random value; inner loop to check that random value is not a repeat. – Fiddling Bits Jan 07 '20 at 18:47
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    Or shuffle the array `{1, 2, 3, 4, 5, 6}` and take the first four values. – M Oehm Jan 07 '20 at 18:48
  • is there any sophisticated way to check that the value is not repeated? that's the reason I asked if there is a way to add conditions... sadly, I cant shuffle an array, I cant submit things we havent learnt in class... – shubbbik Jan 07 '20 at 18:53
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    @Oona why sophisticated? You could just check that your value does not equal to any other random values, and if they are equal, just generate another random number. – Ach113 Jan 07 '20 at 18:57
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    I don't know what you've learned in class, but [shuffle isn't hard to implement](https://stackoverflow.com/q/6127503/10077). – Fred Larson Jan 07 '20 at 18:58
  • Isn't a simpler and less ambiguous way of specifying is "_five_ variables each with a unique value 1 to 6?" It is not clear in your question that the variables `num1` etc. are unique, but your response to @FiddlingBits question would suggest that. You need to fix the question rather then add important information in a comment response. in your question `num1` to `num4` could all have the same value so long as it was not the same as the value of your random value. – Clifford Jan 07 '20 at 19:44
  • Also the bit you say you know how to do; include that in your question so that an answer can build on it rather then conflict with it. – Clifford Jan 07 '20 at 19:44
  • If you want what @FiddlingBits has suggested (and you answered yes to that, even though that is not what your question states), then you need five unique integers in the range 1 to 6, so all you need is to select _one_ random integer from 1 to 6 to _omit_. – Clifford Jan 07 '20 at 19:47
  • With 4 variables each having values [1...6] without repeat leads to 6*5*4*3 or 360 combinations. Try generating a random number [0...359], divide by 6 to find first, then divided remainder by 5 to select one of 5 remaining digits, then to the next ...... – chux - Reinstate Monica Jan 08 '20 at 00:29
  • @chux-ReinstateMonica although he responded "yes" to Fiddling Bits clarification, it seems that 5 unique values are required rather than 4. The question is in dire need of clarification. – Clifford Jan 08 '20 at 08:19

1 Answers1

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It is not (currently) clear in the question, but from the clarification in the comments:

You want four variables, each with a random value between 1 and 6 (inclusive), but no repeats?

Assuming in fact you need 5 variables because the question states:

won't be equal to 4 other variables

then since you need 5 unique variables from 6 possibilities. The simplest solution then is to pick at random just one of the possibilities to omit. Furthermore the sensible means of storing 5 values of the same type is to use a single array variable rather then five separate variables - it makes the code far simpler.

int num[5] ;
int omit = 1 + rand() % 6 ;
for( int i = 0, j = 0; i < sizeof(num) / sizeof(*num); i++, j++ )
{
    if( i != omit ) num[j] = i ;
}

Note that the expression 1 + rand() % 6 will have a bias for some values even if rand() were entirely random. An exercise for the reader if that is an issue.

Clifford
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  • `1 + std::rand() / ((RAND_MAX + 1u) / 6)` can generate [1...7]. e.g. `rand() == 2147483647, RAND_MAX == 2147483647` --> 7. For such learner code, `1 + std::rand()%6` would suffice. – chux - Reinstate Monica Jan 08 '20 at 00:42
  • I cannot see how you get 7 from that (when correctly applying operator precedence). I agree that %6 would suffice, but was trying avoid comments from pedants regarding bias. That is why I asked the OP to include his method, so that the decision was his regarding its importance. – Clifford Jan 08 '20 at 08:13
  • Clever way to rethink it, randomly choosing numbers to omit rather than include. – Fiddling Bits Jan 08 '20 at 13:53
  • "how you get 7" : (2147483647 + 1u)/6 --> 357913941. 2147483647/357913941 --> 6. 1 + 6 --> 7. – chux - Reinstate Monica Jan 08 '20 at 14:28
  • @chux-ReinstateMonica : I knew you had to be right - more coffee needed. Rolled back to original text. If anyone is interested should have been `int omit ; do{ omit = 1 + std::rand() / ((RAND_MAX + 1u) / 6)} while( omit < 7 );` – Clifford Jan 08 '20 at 14:48
  • @Clifford Perhaps `while( omit < 7 )` --> `while (omit == 7)` (and more coffee)? – chux - Reinstate Monica Jan 08 '20 at 16:01
  • @chux-ReinstateMonica yes, that. I'll stop embarrassing myself now! – Clifford Jan 08 '20 at 17:48