CareerCup

switch(rand(3))
case 1:
rand(3)
case 2:
3 + rand(3)
case 3:
6 + rand(3)

return (3*rand(3) + rand(3) -3)

Define a 3x3 matrix called A. Assign a value in range [1...9] to each entry of the matrix. Call rand(3) twice, and call the results i and j. Return the value corresponding to A[i][j]. Each outcome has probability 1/9.

public static int Rand_3(){
return (int)(Math.random()*3)+1;
}
public static int Rand_9(){
return (Rand_3()*3)-(Rand_3()-1);
}

no_of_times = rand(3);
if(no_of_times == 1){
    return rand(3);
}
else if(no_of_times == 2){
    return rand(3) + rand(3);
}
else{
    return rand(3) + rand(3) + rand(3);
}

(rand(3) -1)*3 + (rand(3) - 1)*1 + 1
Do it as you would generate a number in base 3 number.

Example:
Just for the example lets say rand(3) generates numbers 0,1,2 and we are looking form numbers from 0-8 (rand(9)). In base 3 system the max number with length 2 will be 22 = 2*3^1 + 2*3^0 = 2*3 + 2*1 = 8 (eq1). So we just do generate the number in base 3 and find what it is:
in eq1 we change the numbers: rand(9) = rand(3)*3 + rand(3)*1

so now changing rand(3) return 1-3, and rand(9) must return 1-9
and final equation is: rand(9) = (rand(3) -1)*3 + (rand(3) - 1)*1 + 1

the solution will need to call the function random(3) 2 times......
first divide numbers from 1 to 9 into three equal size buffers......1-3...4-6....7-9.
Each buffer is containing exactly 3 numbers.
Now write the function random(9) as:
int random(9)
{
int bufferNo=random(3);
// if bufferNo=1,choose first else if 2 choose buffer 2 else choose buffer 3.
// again call random(3) to choose number in the chosen buffer and return that number
}

Basically we are looking for :
{1,2,3} + 3*0 = {1,2,3}
{1,2,3} + 3*1 = {4,5,6}
{1,2,3} + 3*2 = {7,8,9}

Soln : RAND(3) + 3 * (RAND(3) - 1)

why not rand(3)+rand(3)+rand(3)