CareerCup

My solution is similar to dalvik_king's solution:
It generates 25 numbers (0 - 24) with same probability and since 21 = 3*7 and 25-21 = 4 we discard these 4 numbers (by trying again) and return x%7 where x is between 0 and 20 and for each possible result modulus 7 (0-6) we have 3 possibilities, so the probability is the same.

static public int rand7() {
	result = 0;
	while (result >= 21) 
        	result = rand5()*5 + rand5();
	return result%7;
}

static public int rand7() {

    // This will gives us a result of 
    // 0 to 7 (binary 000 to 111)
    // which should be evenly distributed.
    // reject 7 so we have 0 to 6
    int result = 7;
    while (result == 7) {
        result = randBit() * 4 + randBit() * 2 + randBit();
    }
    
    return result;
}

// gets a random 0 or 1
static public int randBit() {

    // Eliminate zero so we have an equal number 
    // of even/odd values (1, 2, 3, or 4)
    int result = 0;
    while (result == 0) {
        result = rand5();
    }

    return result % 2;
}

int rand7()
{
int vals[5][5] = {
{ 1, 2, 3, 4, 5 },
{ 6, 7, 1, 2, 3 },
{ 4, 5, 6, 7, 1 },
{ 2, 3, 4, 5, 6 },
{ 7, 0, 0, 0, 0 }
};

int result = 0;
while (result == 0)
{
int i = rand5();
int j = rand5();
result = vals[i-1][j-1];
}
return result;
}

I feel it should be something like

int rand7()
{
   return ( rand5()*7 + rand5() )%7;
}

This should preserve the uniformity. In fact, rand5()*7 + rand5() provides a number between 0 and 34 and there's only one way to generate each of these 35 numbers. Also, since 0%7=0 and 34%7=6, you have that all numbers between 0 and 34 are transformed to a number between 0 and 6 with the same probability.

public class Random7 {

    public static int val=0;
    public int random7(){
        return (new Random().nextInt(5)*5 +new Random().nextInt(5) )%7;
    }
}

There are many ways to solve this problem, below is my attempt

int rand7()
{
  int k = rand5() + rand5() + rand5() + rand5() + rand5() + rand5() + rand5();
    return k % 7;
}

does rand(5) generate real numbers between 0 and 4, or only integers? the below works for real numbers:

return (((rand(5) + 1) * 7) - 1) / 5;

(note that the division by 5 is integer division)

This is a general solution that works for any +ve integer with a equal dristribution.
for this question n is 7 and m is 5

so rand_5() is given, which has been represented as rand_m below. The point is we need to find random of N where random of m is given.
Calculating the factor value is what trivial here...
for this example
7x-1 has to be divisible by 4 for a smallest integer value of x. and then (7x-1)/4 = 5 (where x = 3), so 5 is the factor.
so (rand_5 + rand_5 + rand_5 + rand_5 + rand_5)%7 is the result.
above sum can generate values in range of "0 to 20" with equal distribution, that means equal number of change (3 chances of each value) of getting 0 to 6.


static int getRandom(int n, int m) {
int i = 1;
while((n*i-1)%(m-1) != 0) i++;

int factor = (n*i-1)/(m-1);

int randomSum = 0;


while(factor > 0){
randomSum += rand_m();
factor--;
}
return randomSum%n;
}

Use the previous ideas only add rejection sampling. S that the number taken into acciunt be divisible by 7

rand(5) + rand(5)%2

num = rand(5) + rand(5);
return(num%7);

I think we can use rand5 to implementation rand2 and rand10 first.
int rand2()
{
int k = 0;
do
{
k= rand5();
}
while(k > 2);
return k;
}

int rand10()
{
return rand2() * 5 + rand5();
}

for the rand7(). we can use rand10 to implementation:
int rand7()
{ k = 0;
do
{
k = rand10();
}while(k > 7);
return k;

}

Doesn't this work?

int rand7() {
    return rand(3) + rand(4);
}

@All, this question has been discussed quite well on Stack Overflow, just google for "rand 5 to rand 7"

Thanks hulk for posting this problem. It refreshes some discrete math skills, which is great!

int randomNumber = random.nextInt(max - min) + min;

int rand7()
{
	int x,y,z,n=0;
	while((x=rand5())==2);
	x/=2;
	while((y=rand5())==2);
	y/=2;
	while((z=rand5())==2);
	z/=2;
	n=4*x+2*y+1*z;
	return n;
}

package com.test.careercup;

import java.util.Random;

public class Test {

	public static void main(String[] arg)
	{
		System.out.println("int : "+rand(7));
	}
	public static int rand(int maxNum)
	{
		Random r=new Random();
		return r.nextInt(maxNum);
		
	}
}

rand(5) + (int)rand(5)/2

I believe it should be simple. Just get sum of two rand5 values and find its mod according to 7;

int rand7()
{
return (rand5() + rand5())%7;
}

lets test it with extreme and normal values;
both rand5 functions generate 0;
(0 + 0) % 7 = 0 <--- in the range
both rand5 functions generate 4;
(4 + 4) % 7 = 1 <--- in the range
both rand5 functions generate 3;
(3 + 3) % 7 = 6 <--- in the range


hmm looks ok but lets test if we can generate every number in the range;

2 and 3 generated and we got 5
2 and 2 generated and we got 4
1 and 2 generated and we got 3

looks like we can generate every number in the range. Only problem here is that the possibilities of the outcomes are not same. Since the question is not telling about the possibilities, I believe this should be the simplest answer.

int rand7()
{
return ( ((float)rand5() / 5.0f ) * 7.0f);
}

int get_rand7_from_rand5()
{
	int rnd; 
	do { 
    		int x = rand5(); //generates one of 0,1,2,3,4 
    		int y = rand5(); //generates one of 0,1,2,3,4 
    		rnd = 5*x + y; // generates a number between 0 to 24 
	} while (rnd > 20)
	return rnd % 7; 
    //for numbers between 0 to 20 : if we take modulo 7 we can see that probability of picking 0 is 3/21 (for 0, 7 and 14), probability of picking 1 is 3/21 (for 1, 8, 15). Similarly for 2, 3, 4, 5 and 6. Hence any number between 0 to 6 has equal probability.
}
//probability of the while loop running n times before it returns is (4/25)^n

How about

int rand7(){
	int num=rand5()+1;
	int result=0;

	for(int i=0;i<num;i++){
		result += rand5();
		}

return result%7;
}

How about

int rand7(){
	int num=rand5()+1;
	int result=0;

	for(int i=0;i<num;i++){
		result += rand5();
		}

return result%7;
}

rand5() generates the values from 0 to 4 i.e., 0,1,2,3,4
expected is 0 to 6, i.e., 0,1,2,3,4,5,6 . using rand5().

My solution is:
rand7() = rand5() + rand5()/2


Example: Assume first rand5 is generating 4, and second rand5 is generating 0 to 4, then
rand5() + rand5()/2 will have following values for rand7() i.e, 4,5,6


4, ==> 4,5,6

4 + 0/2 = 4
4 + 1/2 = 4
4 + 2/2 = 5
4 + 3/2 = 5
4 + 4/2 = 6


repeat for 1, ==> 1,2,3

1 + 0/2 = 1
1 + 1/2 = 1
1 + 2/2 = 2
1 + 3/2 = 2
1 + 4/2 = 3


repeat for 1, ==> 1,2,3

0 + 0/2 = 0


Example: Assume rand5 is generating 4, then
rand