Mod of power 2 on bitwise operators?

Question

1. How does mod of power of 2 work on only lower order bits of a binary number (1011000111011010)?
2. What is this number mod 2 to power 0, 2 to power 4?
3. What does power of 2 have to do with the modulo operator? Does it hold a special property?
4. Can someone give me an example?

The instructor says "When you take something mod to power of 2 you just take its lower order bits". I was too afraid to ask what he meant =)

Solution 1

He meant that taking number mod 2^n is equivalent to stripping off all but the n lowest-order (right-most) bits of number.

For example, if n == 2,

number      number mod 4
00000001      00000001
00000010      00000010
00000011      00000011
00000100      00000000
00000101      00000001
00000110      00000010
00000111      00000011
00001000      00000000
00001001      00000001
etc.

So in other words, number mod 4 is the same as number & 00000011 (where & means bitwise-and)

---

Note that this works exactly the same in base-10: number mod 10 gives you the last digit of the number in base-10, number mod 100 gives you the last two digits, etc.

Solution 2

What he means is that :

x modulo y = (x & (y − 1))

When y is a power of 2.

Example:

0110010110 (406) modulo
0001000000 (64)  =
0000010110 (22)
^^^^<- ignore these bits

Using your example now :

1011000111011010 (45530) modulo
0000000000000001 (2 power 0) =
0000000000000000 (0)
^^^^^^^^^^^^^^^^<- ignore these bits

1011000111011010 (45530) modulo
0000000000010000 (2 power 4) =
0000000000001010 (10)
^^^^^^^^^^^^<- ignore these bits

Solution 3

Consider when you take a number modulo 10. If you do that, you just get the last digit of the number.

  334 % 10 = 4
  12345 % 10 = 5

Likewise if you take a number modulo 100, you just get the last two digits.

  334 % 100 = 34
  12345 % 100 = 45

So you can get the modulo of a power of two by looking at its last digits in binary. That's the same as doing a bitwise and.

Solution 4

Modulo in general returns the remainder of a value after division. So x mod 4, for example, returns 0, 1, 2 or 3 depending on x. These possible values can be represented using two bits in binary (00, 01, 10, 11) - another way to do x mod 4 is to simply set all the bits to zero in x except the last two ones.

Example:

      x = 10101010110101110
x mod 4 = 00000000000000010

Solution 5

Answering your specific questions:

1. mod is a remainder operator. If applied to a series of numbers x in 0, 1, ..., then x mod n will be 0, 1, ..., n-1, 0, 1, ..., n-1, ad infinitum. When your modulus n is a power of 2, then x mod n will count up in binary from 0 to n-1, back to 0, to n-1, etc; for modulus n that looks like binary 01xxxxx, x mod n will cycle through every of those low-order bits xxxxx.
2. binary 1011000111011010 mod 1 is 0 (mod 2^0 yields the last zero bits; everything mod 1 is zero). binary 1011000111011010 mod binary 10000 is 1010 (mod 2^4 yields the last four bits).
3. Division and remainder of binary number by powers of two is particularly efficient because it's just shifting and masking; mathematically it's nothing special.
4. Example: See answer to question 2.

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Comments

[starblue]

Why don't you try a few example calculations by hand, then you'll see what happens.

[Zo Has]

Hey thanks for the reply. I did not got your example completely ?

[Zo Has]

Would that hold for powers of 2 as well ? 54 % 32 which is 2^5 gives 22.

[Brian]

Popo: Yes. The number of bits at the end is determined by which power of two you're using. As described by Cicada, you would calculate it as 54 & (32-1).

[calandoa]

This is only the case when all operands are positive! Depending on the language the behavior will differ. For instance in C, -5 % 4 == -1 despite the fact that in algebra we usually expect that -5 mod 4 is 3 (and in C: -5 & (4 - 1) == 3. This means for instance that a compiler will not optimize a literal % 4 with a & if the left operand is not unsigned.

[BlueRaja - Danny Pflughoeft]

@calandoa: We are discussing binary as a number system here, not bit-encoding of numbers. -5, for instance, is written -101.

[calandoa]

@BlueRaja: Of course we are talking about bit encoding : & is not defined for mathematical -. I guess your remark is more "we are not talking about negative numbers". Maybe we are not, but this is not clear in the question nor in your answer, so I made it clearer.

[Sandburg]

Do you thing at something like number MOD 256 <=> (number & (0xFF00)) >> 8 (for 2^8 = 256)