Write an algorithm to generate 0 and 1 with 75% and 25% probability respectively using a specified function which produces either 0 or 1 each with 50% probability.

Below function generates 0 or 1 with 50% probability each, which we will be using to generate 0 and 1 with 75% and 25% probability.

1 2 3 4 5 6 7 8 9 |
int random() { // rand() produces a random number int random = rand(); // if the random number is even, return 0 // if the random number is odd, return 1 return (random % 2); } |

### 1. Using Bitwise AND Operator (or Logical AND Operator)

We can use bitwise or logical AND operator to solve this problem. The idea is two make two calls to the `random()` function and return AND of results returned by the individual calls.

1 2 3 4 5 6 7 8 9 |
// Return 0 and 1 with 75% and 25% probability respectively using the // specified function and bitwise AND operator int generate() { int x = random(); int y = random(); return (x & y); } |

###### Code Explanation:

x can be either {0, 1}

y can be either {0, 1}

(x & y) can be either {0, 0, 0, 1}

### 2. Using Bitwise OR Operator (or Logical OR Operator)

We can also use bitwise or logical OR operator. The idea remains similar. First make two calls to the `random()` function and then return negation of OR of results returned by the individual calls as shown below:

1 2 3 4 5 6 7 8 9 |
// Return 0 and 1 with 75% and 25% probability respectively using the // specified function and bitwise OR operator int generate() { int x = random(); int y = random(); return !(x | y); } |

###### Code Explanation:

x can be either {0, 1}

y can be either {0, 1}

(x | y) can be either {1, 1, 1, 0}

!(x | y) can be either {0, 0, 0, 1}

### 3. Using Left Shift Operator and Bitwise XOR Operator

The idea is to use this expression: `(random() << 1) ^ random()`

###### How this expression works?

random() returns either 0000 or 0001 (in binary)

(random() << 1) can be either 0000 or 0010

(random() << 1) ^ random() can be either {0001, 0011, 0000, 0010}

1 2 3 4 5 6 |
// Return 0 and 1 with 75% and 25% probability respectively using the // specified function, left shift operator and bitwise XOR operator int generate() { return ((random() << 1) ^ random()) == 0; } |

**Author:** Aditya Goel

**Thanks for reading.**

Please use our online compiler to post code in comments. To contribute, get in touch with us.

Like us? Please spread the word and help us grow. Happy coding 🙂

## Leave a Reply