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You have a function ICKRand7 that generates a random integer from 1 to 7. Use it to write a function ICKRand5 that generates a random integer from 1 to 5.
**

ICKRand7 returns each integer with equal probability. ICKRand5 must also return each integer with equal probability.

Your first thought might be to simply take the result of ICKRand7 and take a modulus:

NSUInteger ICKRand5() {
return ICKRand7() % 5 + 1;
}

**However, this won't give an equal probability for each possible result**. We can write out each possible result from ICKRand7 (each of which is equally probable, per the problem statement) and see that some results for ICKRand5 are more likely because they are caused by more results from ICKRand7:

ICKRand7 | ICKRand5 |
---|---|

1 | 2 |

2 | 3 |

3 | 4 |

4 | 5 |

5 | 1 |

6 | 2 |

7 | 3 |

So we see that there are two ways to get 2 and 3, but only one way to get 1, 4, or 5. This makes 2 and 3 twice as likely as the others.

The answer takes worst-case infinite time. However, we can get away with worst-case space. **Does your answer have a non-constant space cost?** If you're using recursion (and your language doesn't have tail-call optimization), you're potentially incurring a worst-case infinite space cost in the call stack. But replacing your recursion with a loop avoids this.

ICKRand5 must return each integer with equal probability, but we don't need to make any guarantees about its runtime...

In fact, the solution has a small possibility of *never* returning...

We simply "re-roll" whenever we get a number greater than 5.

NSUInteger ICKRand5() {
NSUInteger result;
do {
result = ICKRand7();
} while (result > 5);
return result;
}

So each integer 1,2,3,4, or 5 has a probability \frac{1}{7} of appearing at each roll.

Worst-case time (we might keep re-rolling forever) and space.

Note that if we weren't worried about the potential space cost (nor the potential stack overflow) of recursion, we could use an arguably-more-readable recursive approach with space cost:

NSUInteger ICKRand5() {
NSUInteger roll = ICKRand7();
return (roll <= 5) ? roll : ICKRand5();
}

This kind of math is generally outside the scope of a coding interview, but: if you know a bit of number theory you can *prove* that there exists no solution which is guaranteed to terminate. Hint: it follows from the fundamental theorem of arithmetic.

Making sure each possible result has *the same probability* is a big part of what makes this one tricky.

If you're ever struggling with the math to figure something like that out, don't be afraid to *just count*. As in, write out all the possible results from ICKRand7, and label each one with its final outcome for ICKRand5. Then count up how many ways there are to make each final outcome. If the counts aren't the same, the function isn't uniformly random.

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