Binary logarithm without java.lang.Math?

Question

How is it possible to perform a binary logarithm without using Math? Now I have this Math.log(x) / Math.log(2) and it works but I can't use java.lang.Math.

What can I do?

int bits_necessaris = (int)(log2(nat + 1)); // This is correct

// The function where I'm having trouble

public static int log2(int x) {
    return (something);
}

IMPORTANT: When I performed Math.log(7) / Math.log(2) I got 2.80XXXXX

So I did this:

(int) Math.ceil(Math.log(7) / Math.log(2));

And I get ceil(2.80XXXXX) = 3.

The return of the function has to be rounded up, for example if the solution is 6.777 the return has to be 7.

So by "Math" you mean the class Math? Can you copy the source of Math and put it in an other class with a different name? — harold, Mar 10 '14 at 22:21
Note that the log-base-2 of a number is approximately the number of significant bits in the number. — Hot Licks, Mar 10 '14 at 22:24
Yes, the class to do SQUAREROOT POW LOG etc.. I can't create another class, I must use the function that I wrote above — user2946696, Mar 10 '14 at 22:25
Make x positive and then perform binary search with x<(1<=(1<<30) then you need 31bit plus the sign bit. — Lutz Lehmann, Mar 10 '14 at 22:25
I don't know how binary search works, nobody taught me that so I asume that I don't need to use this method — user2946696, Mar 10 '14 at 22:36
You can right shift the input number (note: using `>>>`) until it gets to 0, all the while incrementing an integer starting at 0. — fge, Mar 10 '14 at 22:36
You could try [this](http://stackoverflow.com/a/18860385/823393) trick. It uses `BigInteger.TEN.pow` but that's nearly not `Math`. — OldCurmudgeon, Mar 10 '14 at 22:36
Why is everybody answering in the comments? You won't get upvotes that way ;-) — meriton, Mar 10 '14 at 22:48

score 3 · Accepted Answer · edited Aug 25 '20 at 09:15

3

You want to know how many bits are necessary to represent an int? Then there are easier solutions such as:

int bitsNeededFor(int i) {
    int bits = 0;
    while (i > 0) {
        bits++;
        i /= 2;
    }
    return bits;
}

On second thought, the following is faster and easier to understand:

int bitsNeededFor(int i) {
    return 32 - Integer.numberOfLeadingZeros(i);
}

edited Aug 25 '20 at 09:15

Manaus

407
5
9

answered Mar 10 '14 at 22:41

meriton

68,356
14
108
175

Thank you! It worked fine, I was too much focused on using logarithm – user2946696 Mar 10 '14 at 22:52
I think this won't work for powers of two, e.g. 2³=8=1000 will return 4. – Marco13 Mar 10 '14 at 23:00
It correctly computes the number of bits needed (i.e. 1 + floor(log2(i))), which is what OP is ultimately after. – meriton Mar 11 '14 at 02:05

Binary logarithm without java.lang.Math?

1 Answers1