For any integer n and where k is power of 2,
n % k = n & (k-1)
Example
I wasn't able to proof this statement, how could it be proofed without taking number as example?
Thanks in Advance!
# | User | Rating |
---|---|---|
1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 173 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 160 |
5 | nor | 157 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | pajenegod | 145 |
For any integer n and where k is power of 2,
n % k = n & (k-1)
Example : Let n = 26 and k = 2 ^ 3 = 8 , so 26 % 8 = 2
bin(26) = 11010
bin(8-1) = 00111
26 & 7 = 00010 = 2, which is CORRECT
...
I wasn't able to proof this statement, how could it be proofed without taking number as example?
Thanks in Advance!
Name |
---|
Please see the definition of binary numbers.
I think I know the definition of binary number (number in base 2), but I wasn't able to relate this the modulo operation.
I am beginner here, can u help?
You know (a+b)%m = a%m + b%m. Since you also know the definition of binary numbers, you can write it as a sum of some powers of 2 and you'll know why.
Say you want to find n%(2^x), note that in binary representation it denotes the last 'x' bits of 'n'(from LSB) because higher bits are powers of 2 with exponent >= x, next note that 2^x — 1 is just an 'x' bit number with all '1', so if you operate & between n and (2^x — 1), you'll get the last 'x' bits of 'n'(since higher bits become '0' and 1&y = y) which was the expected answer. Did you understand it?