Recently I solved a challenge (problem 1 of the 3rd easy problem set of phase 2 of the '19 APL Competition) on abbreviating and expanding IP v6 addresses. For that matter, I had to write two functions, AbbreviateIPv6 and ExpandIPv6. The first function ought to take a character vector of a v6 IP address and abbreviate it like this online tool. The second function ought to undo exactly what the first one did.

I need a community review because these two were among a series of problems rated as easy and while most of them were fairly easy to solve and could be decently solved in a couple of lines, I found really hard to write the AbbreviateIPv6 function so I think I overlooked something. What is more, I failed to find a solution that didn't use regular expressions, which bothered me even more.

First, a helper function:

:Namespace ReplaceFirst
    _ReplaceFirst_ ← {
        ⍝ Operator that replaces the first occurrence of the left operand with the right operand.
        ⍝ e.g. ('ui' ReplaceFirst 'fafa') 'this ui is ui' gives 'this fafa is ui'

        startAt ← ⊃⍸⍺⍺⍷⍵
        stopAt ← startAt + (¯1+≢⍺⍺)×startAt>0
        pruned ← ⍵/⍨ ~(startAt∘<∧≤∘stopAt)⍳≢⍵
        replaceAt ← (>∘0⍴⊢) startAt
        ∊ ((⊂,⍵⍵)@replaceAt) pruned

that I used in the AbbreviateIPv6 function:

AbbreviateIPv6 ← {
  ⍝ Monadic function taking character vector as input and returning character vector.
  ⍝ Abbreviates an IP v6 address.
  ⍝ e.g. '2001:0DB8:0000:0042:0000:8A2E:0370:7334' becomes '2001:DB8:0:42:0:8A2E:370:7334'
  ReplFirst←⎕fix 'file://path/to/ReplaceFirst.dyalog'

  reduced ← ('0000' ':0{1,3}' ⎕R (,¨'0' ':'))⍵
  runs ← ⌽ ⍴∘'0:'¨ 1+2×⍳7
  shortened ← {(⍵ ReplFirst._ReplaceFirst_ '') reduced}¨ runs
  ((⊃∘⍋≢¨)⊃⊢) shortened

I know I could've defined the _ReplaceFirst_ as a function inside AbbreviateIPv6 specialized for the case I wanted, but I figured the function was getting so long I might as well write a general operator that I might reuse later.

Finally, just for comparison, I was able to write the ExpandIPv6 in a much more compact way:

ExpandIPv6 ← {
    ⍝ Monadic function expecting and returning character vector.
    ⍝ Expands an abbreviated IP v6 address.
    ⍝ e.g. '2041:0:140F::875B:131B' gives '2041:0000:140F:0000:0000:0000:875B:131B'.

    colons ← +/':'=⍵
    intermediate ← ('::' ⎕R(':0'⍴⍨ 3+2×7-colons))⍵
    splits ← (':'∘≠⊆⊢) intermediate
    LeftZeroPadder ← (⍴∘'0')∘(4∘-)∘≢,⊢
    ⊃ {∊⍺':'⍵}/ LeftZeroPadder¨ splits
  • \$\begingroup\$ Maybe you could link directly to the specific problem instead of the general page? \$\endgroup\$
    – Adám
    Apr 17, 2020 at 2:03
  • \$\begingroup\$ @Adám I can link to the year but I can only find the problem in a pdf, not in a specific URL. \$\endgroup\$
    – RGS
    Apr 17, 2020 at 6:37
  • \$\begingroup\$ You can link to the right page though. \$\endgroup\$
    – Adám
    Apr 17, 2020 at 7:44
  • \$\begingroup\$ @Adám done! I linked to the challenge page where you can test phase 1 submissions and included a descriptive path to the problem at hand \$\endgroup\$
    – RGS
    Apr 17, 2020 at 8:19
  • 1
    \$\begingroup\$ I meant link straight to the right page in the pdf. \$\endgroup\$
    – Adám
    Apr 17, 2020 at 9:50

1 Answer 1


I think using regular expressions is a perfectly sensible approach. What you're missing is probably that Dyalog APL allows setting a Match Limit with ⍠'ML' n where a positive n limits to the first n matches and a negative n limits to the (absolute value of) nth match. With this in mind, I'd use regular expressions extensively:

AbbreviateIPv6 ← {
    collapsed0s ← '\b0{1,3}' ⎕R '' ⊢ ⍵      ⍝ remove up to 3 leading 0s
    runsOf0s ← '\b(0:)*0\b'                 ⍝ 0:0:0:…:0
    nth ← - ⊃⍒ runsOf0s ⎕S 1 ⊢ collapsed0s  ⍝ 1: lengths, ⊃⍒: index of first max
    abbreviated ← runsOf0s ⎕R ':' ⍠'ML' nth ⊢ collapsed0s
    ':::' '^:$' ⎕R '::' ⊢ abbreviated       ⍝ exactly two

Try it online!


  • The linked online IPv6 compressor gives wrong results, as do many others. This one seems to work, even though it misses the occasional shortening opportunity. I still have not found an online tool that compresses both fully and correctly.
  • If runsOf0s ⎕S 1 doesn't find any runs of 0s, it returns so nth becomes 0, which is fine even though 'ML' 0 means "no limit" because there are no matches.
  • \$\begingroup\$ Lovely, thanks! The ⍠'ML' and the negative argument to it were exactly what I was missing... Also, \b in the regex also matches the beginning of the string, right? Finally, I don't understand if the "edge cases" you mention correspond to maximal compression when the whole IP address is just 0s. \$\endgroup\$
    – RGS
    Apr 17, 2020 at 16:11
  • 1
    \$\begingroup\$ @RGS Yes, \b matches any word boundary, even at the edges of a string. The comment was imprecise: Since we substitute 0:0 with : we end up with ::: when counting the surrounding colons (except at beginning or end where we get two, and for all-zero where we get one). \$\endgroup\$
    – Adám
    Apr 17, 2020 at 17:37

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