CareerCup

This problem could be regarded as fair coin and biased coin problem, please refer to: wiki- fair coin problem

while (true)
{
	int c0 = 0;
	int c1 = 0;

	for (int i=0; i < 3; ++i)
	{
		if (f() == 1)
			c1++;
	}

	for (int i=0;i<2;i++)
	{
		if (f() == 0)
			c0++;
	}

	return (c1 > c0) ? 1 : 0;
}

Can you explain your logic?

How about this?

int f1()
{
    static int k = 0;
    return k++ % 2 ? f() : 1 - f();
}

This problem could be regarded as fair coin and biased coin problem, please refer to: wiki- fair coin problem

fdsdfssdf

f1():
if [f() + {1 - f()}] == 2:
	return 1
else
	return 0

Lets draw the truth table for it

f	1	f	sum		f1
0	1	0	1		0
0	1	1	2		1
1	1	0	2		1
1	1	1	3		0

f1(x) = f(x) + (-1)^x * 0.1

void f1(){
int first = -1, second = -1;
//toss twice
//00-4/25,11-4/25, 01-6/25,10-9/25
//check for 00 and 11
do{
	first = f();
	second = f();
}while(first!=second) //only return 01 or 10
return first;
}

Data:
P(0) = 0.4
P(1) = 0.6

Observation:
P(0) * P(1) = P(1) * P(0) = 0.4 * 0.6 = 0.6 * 0.4
The probability of generating pairs (0, 1) and (1, 0) is equal.

Solution:
Generate the pair of numbers:
a) num1 = f();
b) num2 = f();
Keep generating num2 till num1 == num2. Once you generated two different numbers return num1. Hence P(1) = P(0) = 0.5

Code:

public int f1() {
        int v = f();
        int v2 = v;
        while(v == v2) {
            v2 = f();
        }
        return v;
    }