Given a function rand6() that returns random numbers from 1 to 6 with equal probability, implement one-liner function rand12() using rand6() that returns random numbers from 1 to 12 with equal probability. Solution should minimize the number of calls to rand6() method. Use of any other library function and floating point arithmetic are not allowed.
The idea is to use expression rand6() + (rand6() % 2) * 6. It returns random numbers from 1 to 12 with equal probability. The expression is equivalent to –
// if rand6() is even
if (rand6() % 2)
return 6 + rand6();
else // if rand6() is odd
return rand6();
We can also use any one of below expressions that works in a similar way –
- rand6() + !(rand6() % 2) * 6 or
- rand6() + (rand6() & 1) * 6 or
- rand6() + !(rand6() & 1) * 6
Below is C++ implementation of above idea –
// C++ Program to print random numbers from 1 to 12 // with equal probability using a function that returns // random numbers from 1 to 6 with equal probability #include <iostream> using namespace std; // Function that returns random numbers from 1 to 6 // with equal probability int rand6() { // rand() will generate random numbers between 0 and // RAND_MAX with equal probability // rand() % 6 returns number from 0 to 5 equal probability // (rand() % 6) + 1 returns number from 1 to 6 with // equal probability return (rand() % 6) + 1; } // The function uses rand6() to return random numbers // from 1 to 12 with equal probability int rand12() { return rand6() + (rand6() % 2) * 6; } // Driver code to test above functions int main() { // Initialize random number generator srand(time(NULL)); int N = 12; int freq[N + 1] = { 0 }; // call rand12() multiple times and store its results for (int i = 0; i < N * 100000; i++) freq[rand12()]++; // print frequency of numbers 1-12 for (int i = 1; i <= N; i++) cout << freq[i] << " "; return 0; } |
Output:
100237 100202 99678 99867 100253 99929 100287 100449 99827 99298 100019 99954
Another Solution –
int rand12()
{
return (rand6() * 2) - (rand6() & 1);
}
rand6() * 2 will return even numbers 2, 4, 6, 8, 10 and 12 with equal probability and rand6() & 1 will return 0 or 1 based on rand6() is even or odd respectively. So, the expression (rand6() * 2) – (rand6() & 1) will return random numbers from 1 to 12 with equal probability.
Please note above solutions will produce different results every time we run them.
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