Regular Expression for Binary Numbers Divisible by 3

Question

I am self-studying regular expressions and found an interesting practice problem online that involves writing a regular expression to recognize all binary numbers divisible by 3

Answer 1

Binary numbers divisible by 3 fall into 3 categories:

1. Numbers with two consecutive 1's or two 1's separated by an even number of 0's. Effectively every pair "cancels" itself out.

(ex. 11, 110, 1100,1001,10010, 1111)

(decimal: 3, 6, 12, 9, 18, 15)

2. Numbers with three 1's each separated by an odd number of 0's. These triplets also "cancel" themselves out.

(ex. 10101, 101010, 1010001, 1000101)

(decimal: 21, 42, 81, 69)

3. Some combination of the first two rules (including inside one another)

(ex. 1010111, 1110101, 1011100110001)

(decimal: 87, 117, 5937)

So a regular expression that takes into account these three rules is simply:

0*(1(00)*10*|10(00)*1(00)*(11)*0(00)*10*)*0*

How to read it:

() encapsulate

* means the previous number/group is optional

| indicates a choice of options on either side within the parentheses