Generate integer from 1 to 7 with equal probability

21 Oct, 2021
Given a function foo() that returns integers from 1 to 5 with equal probability, write a function that returns integers from 1 to 7 with equal probability using foo() only. Minimize the number of calls to foo() method. Also, use of any other library function is not allowed and no floating point arithmetic allowed.Solution : We know foo() returns integers from 1 to 5. How we can ensure that integers from 1 to 7 occur with equal probability? If we somehow generate integers from 1 to a-multiple-of-7 (like 7, 14, 21, ...) with equal probability, we can use modulo division by 7 followed by adding 1 to get the numbers from 1 to 7 with equal probability.We can generate from 1 to 21 with equal probability using the following expression.
 5*foo() + foo() -5 
Let us see how above expression can be used. 1. For each value of first foo(), there can be 5 possible combinations for values of second foo(). So, there are total 25 combinations possible. 2. The range of values returned by the above equation is 1 to 25, each integer occurring exactly once. 3. If the value of the equation comes out to be less than 22, return modulo division by 7 followed by adding 1. Else, again call the method recursively. The probability of returning each integer thus becomes 1/7.The below program shows that the expression returns each integer from 1 to 25 exactly once.  
C++
Java
Python3
C#
PHP
Javascript
#include <stdio.h> int main(){    int first, second;    for ( first=1; first<=5; ++first )        for ( second=1; second<=5; ++second )            printf ("%d \n", 5*first + second - 5);    return 0;}
// Java code to demonstrate// expression returns each integer// from 1 to 25 exactly once class GFG {    public static void main(String[] args)    {        int first, second;        for ( first=1; first<=5; ++first )            for ( second=1; second<=5; ++second )            System.out.printf ("%d \n", 5*first + second - 5);    }} // This code is contributed by// Smitha Dinesh Semwal
# Python3 code to demonstrate# expression returns each integer# from 1 to 25 exactly once if name == '__main__':         for first in range(1, 6):        for second in range(1, 6):                print(5 * first + second - 5) # This code is contributed by Smitha Dinesh Semwal.
// C# code to demonstrate expression returns// each integer from 1 to 25 exactly onceusing System; class GFG {         public static void Main()    {        int first, second;         for ( first = 1; first <= 5; ++first )            for ( second = 1; second <= 5; ++second )                Console.WriteLine ( 5*first + second - 5);    }} // This code is contributed by Sam007.
<?php// PHP program to Generate integer// from 1 to 7 with equal probability     $first;    $second;    for ( $first = 1; $first <= 5; ++$first )        for ( $second = 1; $second <= 5; ++$second )            echo 5 * $first + $second - 5, "\n"; // This code is contributed by ajit.?>
<script> // Javascript Program to demonstrate// expression returns each integer// from 1 to 25 exactly once  // Driver Code     let first, second;    for ( first=1; first<=5; ++first )        for ( second=1; second<=5; ++second ) {             document.write( 5*first + second - 5);             document.write("<br />");        } </script>
Output :  
1
2
.
.
24
25
Time Complexity: O(1)
Auxiliary Space: O(1)
The below program depicts how we can use foo() to return 1 to 7 with equal probability.  
C++
Java
Python3
C#
PHP
Javascript
// C++ program to Generate integer from// 1 to 5 with equal probability#include <stdio.h> // given method that returns 1 to 5 with equal probabilityint foo(){    // some code here} int my_rand() // returns 1 to 7 with equal probability{    int i;    i = 5*foo() + foo() - 5;    if (i < 22)        return i%7 + 1;    return my_rand();}// Driver codeint main(){    printf ("%d ", my_rand());    return 0;}
// Java program to Generate integer from// 1 to 5 with equal probabilityclass GfG{ // given method that returns 1 to 5 with equal probabilitystatic int foo(){    // some code here    return 0;} // returns 1 to 7 with equal probabilitypublic static int my_rand() {    int i;    i = 5*foo() + foo() - 5;    if (i < 22)        return i%7 + 1;    return my_rand();} // Driver codepublic static void main (String[] args) {     System.out.println(my_rand());}}
# Python3 program to Generate integer# from 1 to 5 with equal probability# given method that returns 1 to 5# with equal probabilitydef foo():     # some code here    return 0; # returns 1 to 7 with equal probabilitydef my_rand():    i = 0;    i = (5 * foo()) + (foo() - 5);    if (i < 22):        if(i < 0):            return (i % 7 - 7) + 1;        else:            return (i % 7) + 1;                 return my_rand(); # Driver codeif __name__ == '__main__':    print(my_rand()); # This code contributed by PrinciRaj1992
// C# program to Generate integer from// 1 to 5 with equal probabilityusing System;class GfG{  // given method that returns 1 to 5 with equal probabilitystatic int foo(){    // some code here    return 0;}  // returns 1 to 7 with equal probabilitypublic static int my_rand() {    int i;    i = 5*foo() + foo() - 5;    if (i < 22)        return i%7 + 1;    return my_rand();}  // Driver codepublic static void Main () {      Console.Write(my_rand()+"\n");}}
<?php// PHP program to Generate integer from// 1 to 5 with equal probability// given method that returns 1 to 5// with equal probabilityfunction foo(){    // some code here} // returns 1 to 7 with equal probabilityfunction my_rand(){    $i;    $i = 5 * foo() + foo() - 5;    if ($i < 22)        return $i % 7 + 1;    return my_rand();} // Driver codeecho my_rand(); // This code is contributed by Tushil.?>
<script> // Javascript program to Generate integer from// 1 to 5 with equal probability     // given method that returns 1 to 5 with equal probability    function foo()    {        // some code here        return 0;    }         // returns 1 to 7 with equal probability    function my_rand()    {        let i;        i = 5*foo() + foo() - 5;        if (i < 22)            return i%7 + 1;        return my_rand();    }         // Driver code    document.write(my_rand());         // This code is contributed by rag2127     </script>
Output: 
-4
Time Complexity: O(1)
Auxiliary Space: O(1)
This article is compiled by Aashish Barnwal and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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