Check if a number is divisible by 3

Question

Write code to determine if a number is divisible by 3. The input to the function is a single bit, 0 or 1, and the output should be 1 if the number received so far is the binary representation of a number divisible by 3, otherwise zero.

Examples:

input  "0":       (0)  output 1
inputs "1,0,0":   (4)  output 0
inputs "1,1,0,0": (6)  output 1

This is based on an interview question. I ask for a drawing of logic gates but since this is stackoverflow I'll accept any coding language. Bonus points for a hardware implementation (verilog etc).

Part a (easy): First input is the MSB.

Part b (a little harder): First input is the LSB.

Part c (difficult): Which one is faster and smaller, (a) or (b)? (Not theoretically in the Big-O sense, but practically faster/smaller.) Now take the slower/bigger one and make it as fast/small as the faster/smaller one.

Answer 1

Actually the LSB method would actually make this easier. In C:

MSB method:

/* 
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_msb(char *input) {
  unsigned value = 0;
  char *p = input;
  if (*p == '1') {
    value &= 1;
  }
  p++;
  while (*p) {
    if (*p != ',') {
      value <<= 1;
      if (*p == '1') {
        ret &= 1;
      }
    }
    p++;
  }
  return (value % 3 == 0) ? 1 : 0;
}

LSB method:

/* 
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_lsb(char *input) {
  unsigned value = 0;
  unsigned mask = 1;
  char *p = input;
  while (*p) {
    if (*p != ',') {
      if (*p == '1') {
        value &= mask;
      }
      mask <<= 1;
    }
    p++;
  }
  return (value % 3 == 0) ? 1 : 0;
}

Personally I have a hard time believing one of these will be significantly different to the other.

Answer 2

You need to do all calculations using arithmetic modulo 3. This is the way

MSB:

number=0
while(!eof)
    n=input()
    number=(number *2 + n) mod 3

if(number == 0)
    print divisible

LSB:

number = 0;
multiplier = 1;
while(!eof)
    n=input()
    number = (number + multiplier * n) mod 3
    multiplier = (multiplier * 2) mod 3

if(number == 0)
   print divisible

This is general idea...

Now, your part is to understand why this is correct.

And yes, do homework yourself ;)

Answer 3

The idea is that the number can grow arbitrarily long, which means you can't use mod 3 here, since your number will grow beyond the capacity of your integer class.

The idea is to notice what happens to the number. If you're adding bits to the right, what you're actually doing is shifting left one bit and adding the new bit.

Shift-left is the same as multiplying by 2, and adding the new bit is either adding 0 or 1. Assuming we started from 0, we can do this recursively based on the modulo-3 of the last number.

last | input || next | example
------------------------------------
0    | 0     || 0    | 0 * 2 + 0 = 0
0    | 1     || 1    | 0 * 2 + 1 = 1
1    | 0     || 2    | 1 * 2 + 0 = 2
1    | 1     || 0    | 1 * 2 + 1 = 0 (= 3 mod 3)
2    | 0     || 1    | 2 * 2 + 0 = 1 (= 4 mod 3)
2    | 1     || 2    | 2 * 2 + 1 = 2 (= 5 mod 3)

---

Now let's see what happens when you add a bit to the left. First, notice that:

2²ⁿ mod 3 = 1

and

2²ⁿ⁺¹ mod 3 = 2

So now we have to either add 1 or 2 to the mod based on if the current iteration is odd or even.

last | n is even? | input || next | example
-------------------------------------------
d/c  | don't care | 0     || last | last + 0*2^n = last
0    | yes        | 1     || 0    | 0 + 1*2^n = 1 (= 2^n mod 3)
0    | no         | 1     || 0    | 0 + 1*2^n = 2 (= 2^n mod 3)
1    | yes        | 1     || 0    | 1 + 1*2^n = 2
1    | no         | 1     || 0    | 1 + 1*2^n = 0 (= 3 mod 3)
1    | yes        | 1     || 0    | 2 + 1*2^n = 0
1    | no         | 1     || 0    | 2 + 1*2^n = 1

Answer 4

Read the following two bit-twiddling hacks:

• Compute modulus division by (1 << s) - 1 without a division operator
• Compute modulus division by (1 << s) - 1 in parallel without a division operator

They are not the answers you seek, but is probably of interest to you.

Answer 5

I think Nathan Fellman is on the right track for part a and b (except b needs an extra piece of state: you need to keep track of if your digit position is odd or even).

I think the trick for part C is negating the last value at each step. I.e. 0 goes to 0, 1 goes to 2 and 2 goes to 1.

Answer 6

Here is an simple way to do it by hand. Since 1 = 2² mod 3, we get 1 = 2²ⁿ mod 3 for every positive integer. Furthermore 2 = 2²ⁿ⁺¹ mod 3. Hence one can determine if an integer is divisible by 3 by counting the 1 bits at odd bit positions, multiply this number by 2, add the number of 1-bits at even bit posistions add them to the result and check if the result is divisible by 3.

Example: 57₁₀=111001₂. There are 2 bits at odd positions, and 2 bits at even positions. 2*2 + 2 = 6 is divisible by 3. Hence 57 is divisible by 3.

Here is also a thought towards solving question c). If one inverts the bit order of a binary integer then all the bits remain at even/odd positions or all bits change. Hence inverting the order of the bits of an integer n results is an integer that is divisible by 3 if and only if n is divisible by 3. Hence any solution for question a) works without changes for question b) and vice versa. Hmm, maybe this could help to figure out which approach is faster...

Answer 7

Heh

State table for LSB:

S I S' O
0 0 0  1
0 1 1  0
1 0 2  0
1 1 0  1
2 0 1  0
2 1 2  0

Explanation: 0 is divisible by three. 0 << 1 + 0 = 0. Repeat using S = (S << 1 + I) % 3 and O = 1 if S == 0.

State table for MSB:

S I S' O
0 0 0  1
0 1 2  0
1 0 1  0
1 1 0  1
2 0 2  0
2 1 1  0

Explanation: 0 is divisible by three. 0 >> 1 + 0 = 0. Repeat using S = (S >> 1 + I) % 3 and O = 1 if S == 0.

S' is different from above, but O works the same, since S' is 0 for the same cases (00 and 11). Since O is the same in both cases, O_LSB = O_MSB, so to make MSB as short as LSB, or vice-versa, just use the shortest of both.

Answer 8

There's a fairly well-known trick for determining whether a number is a multiple of 11, by alternately adding and subtracting its decimal digits. If the number you get at the end is a multiple of 11, then the number you started out with is also a multiple of 11:

47278    4 - 7 + 2 - 7 + 8 = 0, multiple of 11     (47278 = 11 * 4298)
52214    5 - 2 + 2 - 1 + 4 = 8, not multiple of 11 (52214 = 11 * 4746 + 8)

We can apply the same trick to binary numbers. A binary number is a multiple of 3 if and only if the alternating sum of its bits is also a multiple of 3:

4   = 100       1 - 0 + 0 = 1, not multiple of 3
6   = 110       1 - 1 + 0 = 0, multiple of 3
78  = 1001110   1 - 0 + 0 - 1 + 1 - 1 + 0 = 0, multiple of 3
109 = 1101101   1 - 1 + 0 - 1 + 1 - 0 + 1 = 1, not multiple of 3

It makes no difference whether you start with the MSB or the LSB, so the following Python function works equally well in both cases. It takes an iterator that returns the bits one at a time. multiplier alternates between 1 and 2 instead of 1 and -1 to avoid taking the modulo of a negative number.

def divisibleBy3(iterator):

    multiplier = 1
    accumulator = 0

    for bit in iterator:
        accumulator = (accumulator + bit * multiplier) % 3
        multiplier = 3 - multiplier

    return accumulator == 0

Answer 9

A number is divisible by 3 if the sum of it's digits is divisible by 3.
So you can add the digits and get the sum:
- if the sum is greater or equal to 10 use the same method
- if it's 3,6,9 then it is divisible
- if the sum is different than 3,6,9 then it's not divisible

Answer 10

input "0": (0) output 1 inputs "1,0,0": (4) output 0 inputs "1,1,0,0": (6) output 1

shouldn't this last input be 12, or am i misunderstanding the question?

Answer 11

Here... something new... how to check if a binary number of any length (even thousands of digits) is divisible by 3.

-->((0))<---1--->()<---0--->(1)        ASCII representation of graph

From the picture.

1) You start in the double circle. 2) When you get a one or a zero, if the digit is inside the circle, then you stay in that circle. However if the digit is on a line, then you travel across the line. 3) Repeat step two until all digits are comsumed. 4) If you finally end up in the double circle then the binary number is divisible by 3.

You can also use this for generating numbers divisible by 3. And i wouldn't image it would be hard to convert this into a circuit.

1 example using the graph...

11000000000001011111111111101 is divisible by 3 (ends up in the double circle again)

Try it for yourself.

You can also do similar tricks for performing MOD 10, for when converting binary numbers into base 10 numbers. (10 circles, each doubled circled and represent the values 0 to 9 resulting from the modulo)

EDIT: This is for digits running left to right, its not hard to modify the finite state machine to accept the reverse language though.

NOTE: In the ASCII representation of the graph () denotes a single circle and (()) denotes a double circle. In finite state machines these are called states, and the double circle is the accept state (the state that means its evenually divisible by 3)