3.1 KiB

Raw Blame History

5.2. Binary to string

Given a real number between 0 and 1 (0.72), passed in as a double, print the binary representation. If the number can't be represented with less than 32 characters, print "ERROR"

Hints 143, 167, 173, 269, 297

How am I supposed to represent a real number in a binary form?

Hint 1

First do it with integers

str binaryToString(int num) {
    str stringnum = "";
    int biggestexp = 0;
    while (num >= pow(2, biggestexp)) {
        biggestexp += 1;
    }
    while (biggestexp >= 0) {
        if (num >= 0 && num - pow(2, biggestexp) >= 0) {
            num -= pow(2, biggestexp);
            stringnum = "1" + stringnum;
        } else {
            stringnum = "0" + stringnum;
        }
        biggestexp -= 1;
    }
    return stringnum;
}

This should work, takes O(sqrt(num) + sqrt(num)). We need to know the biggest exponent in order to substract it. And then keep substracting it until reaching 0. Maybe some wrong c++ syntax, but should work. Now with decimals??

Hint 2

In a number like .893 (in base 10), what does each digit signify? What then does each digit in .10010 signify in base 2?

So just stringnum = "." + stringnum?

Hint 3

A number such as .893 (in base 10) indicates 8 * 10^-1 + 9 * 10^-2 + 3 * 10^-3. Translate this system into base 2

.10010 = 1 * 2^-1 + 0 * 2^-2 + 0 * 2^-3 + 1 * 2^-4 + 0 * 2^-5

Ok so stringnum in the opposite way, '.' added to it. Can we reverse dividing by pow(2,i) and go from low to high?

str binaryToString(int num) {
    str stringnum = "";
    int i = 0;
    num -= pow(2, i);
    while (num > 0) {
        stringnum = "1" ? num % 2 == 0 : "0" + stringnum;
        i += 1;
        num -= pow(2, i);
    }
    return "." + stringnum;
}

This won't work for 11 -> 1011, it will create 111 and then have 6 left and nowhere to put it. So just do the other way and take longer.

Hint 4

How would you get the first digit in .893? If you multiplied by 10, you'd shift the values over to get 8.93. What happens if you multiply by 2?

Hint 5

Think about what happens for values that can't be represented accurately in binary

I'm officially lost.

Solution

To print the decimal part of the 0.101 number, we multiply by 2 and check if 2n is >= 1. r = 2₁₀ * n = 1.01₂. If r >= 1, we know that n had a 1 right after the decimal point. By doing this continuously, we can check every digit.

public static String printBinary(double num) {
    if (num >= 1 || num <= 0) {
        return "ERROR";
    }

    StringBuilder binary = new StringBuilder();
    binary.append(".");
    while (num > 0) {
        /* Setting a limit on length: 32 characters */
        if (binary.length() > 32) {
            return "ERROR";
        }
        double r = num * 2;
        if (r >= 1) {
            binary.append(1);
            num = r - 1;
        } else {
            binary.append(0);
            num = r;
        }
    }
    return binary.toString();
}

Don't understand this logic at all. Trying with double num = 0.7, I get ".1011001001....".

How to convert decimal number to binary.

3.1 KiB Raw Blame History