Regex Counting By 3s

Question

I'm teaching myself regular expressions, and found a quizzing site that has been helping me find more applications for them and has been helping me expand my knowledge of how they work.

I found a question asking me to form a regex to match 10 digit numbers that are multiples of 3s. The only way I can think of doing this is by having the regex recognise numbers' values and be able to manipulate them mathematically. How is this possible?

In other words, what regex would match

but not match

Is it regex.alf.nu? There's a regex for this puzzle that doesn't entirely rely on multiplicity of numbers which I believe would be the shortest regex. Otherwise, you could check out [this site](http://quaxio.com/triple/). — Jerry, Dec 27 '13 at 06:08
This is in fact for regex.alf.nu. I tried using hobbs' explanation to work on a solution, and ended up with something slightly more robust than wcp's, but still wrong. Reading your (Jerry's) link helped me notice I was ignoring complex mod 0 cases. Took me a while, but I found it. Thanks guys! — Humus, Dec 27 '13 at 07:37
Neat! If you're using the generic regex for multiples of three, it should work. Maybe you forgot to use the anchors `^$` to make sure the whole string is checked? — Jerry, Dec 27 '13 at 07:40

hobbs · Accepted Answer · 2013-12-27T06:59:23.483

First you need to start with the rule that a number is divisble by three if and only if the sum of its digits is divisible by three (proving this takes a little number theory, but it helps to see that 9, 99, 999 etc. are all multiples of three and therefore 1, 10, 100, 1000, etc. all contribute the same amount to the remainder of a number when divided by three).

Then, notice that there are three kinds of digits:

The multiples of three: 0, 3, 6, and 9. These are equivalent to 0 (mod 3).
One more than multiples of three: 1, 4, and 7. These are equivalent to 1 (mod 3).
Two more than multiples of three: 2, 5, and 8. These are equivalent to -1 (mod 3). Conventionally this class is named 2, but -1 is more useful to us. Because 1 + 2 = 0 (mod 3), -1 is a legitimate name for 2.

Then, the numbers 0, 3, 6, and 9 are multiples of three. If we add any number of digits from class 0 anywhere, the number remains a multiple of three (so 33, 999, and 963 are all multiples of three). If we add a digit from class 1 anywhere, we need to either add another digit of class -1, or add two more digits of class 1 to bring the remainder back to 0. Likewise if we add a digit from class -1 anywhere, we either need to add another digit of class 1, or add two more digits of class -1 to bring the remainder back to 0.

Here's wcp's answer, formatted as a perl /x regex for readability:

/
(   [0369] # 0
  | [147] [0369]* [258] # 1 + 0 + -1 = 0
  | (   [258] # -1
      | [147] [0369]* [147] # 1 + 0 + 1 = -1
    ) # -1
    (   [0369] # 0
      | [258] [0369]* [147] # -1 + 0 + 1 = 0
    )* # 0
    (   [147] # 1
      | [258] [0369]* [258] # -1 + 0 + -1 = 1
    ) # 1 ... -1 + 0 + 1 = 0
)+
/x

The regex matches groups of digits having remainder of 0. The first branch matches only digits of class 0; the second branch matches groups where the remainder goes up to 1 and then back to 0; and the third branch matches groups where the remainder goes down to -1 and then back to 0. There's some cleverness in how it's constructed (a less-golfed regex would have five major branches instead of three, I think), but the comments should be enough to let you follow it.

Thank you so much. I hadn't considered forming the strings by mod remainder. I had found a couple articles talking about basic math that could be used in regexes and got focused in that direction. — Humus, Dec 27 '13 at 07:18

score 2 · Answer 2 · answered Dec 27 '13 at 06:26

2

Just like @Jerry say in comment, you can use this:

([0369]|[147][0369]*[258]|([258]|[147][0369]*[147])([0369]|[258][0369]*[147])*([147]|[258][0369]*[258]))+

more short

answered Dec 27 '13 at 06:26

wcp

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Regex Counting By 3s

2 Answers2

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