6 ms·

Does anybody know how to do math with regex? (triples) And conditionals don't seem to work?

• danielweber 13y ago
  My guess is that 147 must appear the same number of times as 258, but I'm not sure if that's even expressible in regexp.
  • recursive 13y ago
    No. 111 is a multiple of 3.
  • johnlbevan2 13y ago
    Sadly that wouldn't work for: 140091876 147 = 4 times 258 = 1 time
  • johnlbevan2 13y ago
    Refined: [0369] can appear any number of times [147] and [258] must appear an equal number of times, or for any remaining: [147] must appear a multiple of 3 times [258] must appear a multiple of 3 times
• dudus 13y ago
  From: http://quaxio.com/triple/ http://quaxio.com/triple/ ^([0369]|[258][0369]*[147]|[147]([0369]|[147][0369]*[258])*[258]|[258][0369]*[258]([0369]|[147][0369]*[258])*[258]|[147]([0369]|[147][0369]*[258])*[147][0369]*[147]|[258][0369]*[258]([0369]|[147][0369]*[258])*[147][0369]*[147])*$
  • moron4hire 13y ago
    569pts: ([^31]0|31|[017]2|[03]03|[^1]4|(900|01|7)5|6|[48]7|[57]8|09)$
    • chch 13y ago
      Originally had 568, then saw this and improved. :) 580pts: 00(0$|3|6|9|12|15)|[^0]14|.53|^3[^38]|55|43|23|9.7
      • 3JPLW 13y ago
        Nice. A smidge better at 582: 5[54]|2[437]|00($|[369]|1[25])|^8[17]|^3[29]|9.7
        • jaytaylor 13y ago
          One more smidgen, behold 584: 5[54]|2[437]|00($|[369]|1[25])|^[83][1729]|9.7
          • grobie 13y ago
            586: ^[378][12479]|00($|[369]|1[25])|5[45]|2[347]
            • nwellnhof 13y ago
              589: ^[378][12479]|00($|[369]|1[25])|55|2[347]
  • raverbashing 13y ago
    The idea is as thus: Remember division rules? If the sum of digits is divisible by 3, then the number is divisible by 3 So, if you have 0, 3, 6, 9, it's as you can remove them, the value won't change Remaining are the 2, 3, etc digit numbers divisible by 3
• tcfunk 13y ago
  I got 187 with this [369](?![7238])|(?:0)12 It's sloppy and certainly not correct, but a decent score.
• galen_tyrol 13y ago
  ^[01349][064] works for 197