# Challenge description

We've had a few challenges involving the Look-and-say sequence. Quick reminder:

• The sequence starts with 1,
• Subsequent terms of this sequence are generated by enumerating each group of repeating digits in the previous term,

So the first few terms are:

1        "one"
11       "one one" (we look at the previous term)
21       "two ones"
1211     "one two, one one"
111221   "one one, one two, two ones"
312211   "three ones, two twos, one one"


Now let's do the same thing, but use Roman Numerals instead. We start with I and follow the same rules (we apply the digit-counting rule to characters instead, so we read IVX as one one, one five, one ten instead of one four, one ten or some other way):

I           "one"
II          "one one"
III         "two ones" = "II" + "I"
IIII        "three ones" = "III" + "I"
IVI         "four ones" = "IV" + "I"
IIIVII      "one one, one five, one one"
IIIIIVIII   "three ones, one five, two ones" = ("III" + "I") + ("I" + "V") + ("II" + "I")


Given a positive integer N, either:

• Output first N numerals of this sequence (any reasonable separator is fine, as well as ["I", "II", "III", ...]
• Output Nth term of this sequence (it may be 0-indexed).

Remember to make your code as short as possible, since this is a challenge!

EDIT: I believe that there is always one standard/preferred way of expressing integers as roman numerals, (like 95 -> XCV instead of VC). Couple of Roman numeral converters I found online corroborate my opinion. If in doubt, use an online converter, as listing all the possible edge-cases and specific rules of writing Roman numerals is not the point of this challenge.

EDIT2: @PeterTaylor and @GregMartin pointed out that only numbers less or equal to 5 appear in the sequence, so you don't have to worry about the ambiguity of Roman numerals (numbers 1 - 8 are I, II, III, IV, V, VI, VII, and VIII)

• There isn't a unique Roman numeral expression for each integer. Which numbers might it be necessary to express, and which expressions of those numbers are valid? Aug 30, 2016 at 15:13
• What do you mean by "there isn't a unique Roman numeral expression for each integer"? Like 4 / IV / IIII ? Or 95 / XCV / VC ? There might not always be a unique way to express an integer, but I'm pretty sure there's always a preferred (standard) one - correct me if I'm wrong. Aug 30, 2016 at 15:17
• how far do we have to go with our roman numberals? Aug 30, 2016 at 15:19
• Yes, both of those cases. In the second case, I think it's very much a matter of opinion which is preferable. Aug 30, 2016 at 15:21
• @shooqie if these details weren't clarified, how would you compare answers? If there are certain edge cases left up to interpretation the actual scores become meaningless because they might make a bigger difference than any golfing tricks you could come up with. Aug 30, 2016 at 16:41

# Perl, 49 bytes

Includes +1 for -p

Run with the 0-based index on STDIN, e.g.

ecce.pl <<< 14


ecce.pl:

#!/usr/bin/perl -p
s,(.)\1*,$&/$1%182 .$1,eg for($_=/$/)x$;y;19;IV


Magic formulas are so magic.

Normally I would use ($_=//)x$' to make the loop control one byte shorter, but scoring on this site gives that a handicap of 2 so it ends up 1 byte longer. On older perls you can drop the space before for. Some versions of perl force you to add a final ; to close the transliteration. But what is given above is the code that works on my system.

## Explanation

Working backwards from solution to code:

The string transformations we need:

I     -> II
II    -> III
III   -> IIII
IIII  -> IVI
IIIII -> VI

V     -> IV
VV    -> IIV


Each replacement ends with the repeated character. I will get a sequence of the same characters using regex /(.)\1*/, so this can be done by appending $1. The part before the -> is in $&. With that I still need:

I     -> I
II    -> II
III   -> III
IIII  -> IV
IIIII -> V

V     -> I
VV    -> II


Write I as 1 and V as 9:

1     -> 1
11    -> 11
111   -> 111
1111  -> 19
11111 -> 9

9     -> 1
99    -> 11


By dividing the part before -> by the repeated digit this becomes:

1     -> 1
11    -> 11
111   -> 111
1111  -> 19
11111 -> 9

1     -> 1
11    -> 11


So now the original repeated V is not an exception anymore. So I want an expression that makes this happen:

1     -> 1
11    -> 11
111   -> 111
1111  -> 19
11111 -> 9


And this can be done by a simple modulo 182:

1     % 182 = 1
11    % 182 = 11
111   % 182 = 111
1111  % 182 = 19
11111 % 182 = 9


(this even gets IIIIII to VI right though it isn't needed here)

All that is left is initializing the working variable to 1 for index 0, repeat this transformation in a loop and at the end replace 1 by I and 9 by V

1, 9 and 182 is the only parameter combination for which this simple formula works.

• This is genius! :)
– Lynn
Sep 2, 2016 at 11:15

# Mathematica, 11390 83 bytes

Thanks to Martin Ender for suggestions that reduced the length by over 25%!

Showing off the high-level commands in Mathematica.

Nest[Flatten[Characters@{RomanNumeral@#,#2}&@@@Reverse@@@Tally/@Split@#]&,{"I"},#]&


A pure function, taking an argument N and outputting the Nth element of this (0-indexed) sequence, as a list of characters. Spread out a bit:

Nest[
Flatten[
Characters @ {RomanNumeral@#,#2}& @@@
Reverse @@@ Tally /@ Split@ #
]& ,
{"I"}, #]&


The outer Nest iterates the middle four-line function, starting on {"I"}, N times. Line 4 splits the character list of the input Roman numeral into runs of like characters, counts each run with Tally, and puts the counts before the characters they're counting. Line 3 renders the counts as Roman numerals, then splits those Roman numerals up into lists of characters. The Flatten command reduces the whole list-of-lists to a one-dimensional list.

Here's the initial version:

Nest[
"" <> Flatten[{RomanNumeral@#[[1]], #[[2]]} & /@
(Reverse@#[[1]] & /@
Tally /@
Split@Characters@#)] &,
"I", #] &

• Grrr Mathematica ;) Aug 30, 2016 at 18:50
• If you use @@@ instead of /@ you can use # and #2 instead of #[[1]] and #[[2]]. Also, lists of characters are acceptable string types, so you can work with those and avoid using Characters@. Aug 30, 2016 at 21:11
• @MartinEnder Aha, I knew there must have been a @@@-like shortcut! As for lists of characters being acceptable string types (which I agree would shorten the code): is there a post on this site you can point me to that describes the community standard(s)? Aug 30, 2016 at 22:22
• meta.codegolf.stackexchange.com/a/2216/8478 Aug 30, 2016 at 22:24
• A few more savings: Characters threads automatically so you can use @, Reverse@#& is of course the same as plain Reverse, in which case you also don't need those parentheses. And prefix notation (in the case of Flatten) doesn't save anything if you need to add parentheses to make it work. Combining all of those: Nest[Flatten[Characters@{RomanNumeral@#,#2}&@@@Reverse@@@Tally/@Split@#]&,{"I"},#]& Aug 31, 2016 at 7:58

## CJam (33 30 bytes)

"I"{e{(4md1$^'I*\'V*@}%e_}ri*  Online demo Key to the correctness of the implementation is the following theorem: ### If the first generation is I, no run length is ever greater than five Lemma: if the first generation is I, no string ever contains VVV. Proof is by contradiction. Suppose that there is a first index n for which the nth generation contains VVV. If that VVV breaks down as (a)V VV then the conversion from the previous generation is bad: it should have been (a+5)V. So it must be VV V(d), and the previous generation contained VVVVV, contradicting the choice of n. Now, suppose there is a first index m for which the mth generation contains ...IIIIII.... Note that there can be no digits other than I and V in the string, because no previous generation has had a run of nine Is or nine Vs. At most four of the Is come from a run of Is in the previous string, so the corresponding section of the previous string must be ...IIIVV... giving ... IIII IIV .... Since the VV in generation m-1 doesn't come from VVVVV (see lemma), the second V must be a run-length of digit I, so in generation m-1 we have ...IIIVVI.... And since we want the initial Is to give IIII and not IVI or VI, it is preceded either by the start of the string or by a V. If we have (...V)?IIIVVI... in generation m-1, what do we have in generation m-2? We've already observed that the VV of gen. m-1 must be parsed as (a)V V(I). Suppose we take a=2: (...V)?I IIV VI... Actually it must be ...VI IIV VI..., although that leading V might be part of IV; so in the previous generation we have either (...V)? IIII VV IIIII... or (...V)? IIIII VV IIIII. Either way we run into trouble with VVIIIII: the second V must be a run-length, but then ...VI IIII... requires a following (run-length, digit) pair with the same digit. So it must be a=1: (...V)?II IV VI.... Since generation m is the first with a run of six Is, that must be (...V)? II IV VI..., so that generation m-2 is (...V)? I V IIIII.... ...VIVIIIII... is impossible: however we choose to interpret the second V we end up with two consecutive (run-length, digit) pairs with the same digit. Therefore generation m-2 must be ^IVIIIII..., parsed as ^IV IIII I(V)... or ^IV III II(V).... These give respectively generation m-3 as ^V III V ... or ^V II VV.... But if we look at the start of the strings beginning with the first one that starts with V, we get a cycle:  VI IV I... IV III IV ... II IV IVI ... IIII IV II IV ...  and so no generation ever starts with either VIIIV or VIIVV. We must conclude that there is no such m. ### Dissection "I" e# Initial generation { e# Loop... e e# Run-length encode { e# Foreach [run-length char] pair... ( e# Extract the run-length r 4md1$^   e#     Get the number of Vs and the number of Is
e#     The number of Vs is r/4 ; the number of Is is (r%4)^(r/4)
'I*\'V*@ e#     Repeat strings the appropriate number of times and reorder
}%
e_         e#  Flatten to a simple string
}ri*         e# ... n times, where n is taken from stdin


# Python 3, 195 bytes

There's a lot of bytes wasted on the roman numerals, so there's likely some golfing to be done there.

Thanks to @El'endiaStarman, @Sherlock9 and @Shooqie

import re
def f(x,r=""):
for v,i in(5,"V"),(4,"IV"),(1,"I"):a,x=divmod(x,v);r+=i*a
return r
s="I"
for i in[0]*int(input()):print(s);s=re.sub(r'(.)\1*',lambda m:f(len(m.group()))+m.group()[0],s)


Ideone it!

• You can omit square brackets: for v,i in(5,"V"),(4,"IV"),(1,"I") Aug 30, 2016 at 17:40
• @shooqie I had no idea that you could do that :D Aug 30, 2016 at 17:53
• for v,i in(5,"V"),(4,"IV"),(1,"I"):a,x=divmod(x,v);r+=i*a saves a byte. Aug 30, 2016 at 18:12
• @βετѧΛєҫαγ: Also, you're not seem to be using i (as in for i in range(...)). I tried dabbling with exec but this escaped 1 in the 'sub' method seems to be messing up the code, I haven't been able to find a workaround. Aug 30, 2016 at 18:22
• @shooqie I shortened it a bit by getting rid of range Aug 30, 2016 at 18:26

## R, 110 107 Bytes

as.roman combined with rle makes this easy. Scoping abuse and built in cat behavior of <<- saves a few bytes.

x="I"
replicate(scan(),{r=rle(strsplit(x,"")[[1]])
x<<-paste(rbind(paste(as.roman(r$l)),r$v),collapse="")})


Takes in N from console. Outputs first 2 to N terms of sequence (which I believe is within spec...)

 [1] "II"
[2] "III"
[3] "IIII"
[4] "IVI"
[5] "IIIVII"
[6] "IIIIIVIII"
[7] "VIIVIIII"
[8] "IVIIIIVIVI"
[9] "IIIVIVIIVIIIVII"
[10] "IIIIIVIIIVIIIIVIIIIIVIII"
[11] "VIIVIIIIIVIVIIVVIIVIIII"
[12] "IVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[13] "IIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
[14] "IIIIIVIIIVIIIIIVIVIIVVIIIVIIIIVIIIVIIIIVIIIIIVIII"
[15] "VIIVIIIIIVVIIVIIIVIIIIIVIIIIIVIVIIVIIIIIVIVIIVVIIVIIII"
[16] "IVIIIIVVIIIVIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[17] "IIIVIVIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
[18] "IIIIIVIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVVIIVIVIIVVIIVIIIVIIIIIVIVIIVVIIIVIIIIVIIIVIIIIVIIIIIVIII"
[19] "VIIVIIIIIVVIIIVIIIIVIIIIIVVIIVVIIIVIIIIVIIIVIIIIIVIIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVIVIIVIIIIIVIVIIVVIIVIIII"
[20] "IVIIIIVVIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIIVIVIIVIIIIIVVIIVIVIIVVIIIVIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[21] "IIIVIVIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIIVIIIIVIIIVIIIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"
[22] "IIIIIVIIIVIIIIIVVIIIVIIIIVIIIIIVVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIIVIVIIVIIIIIVVIIVVIIVIIIVIIIIIVIIIIIVVIIVIVIIVVIIVIIIVIIIIIVIVIIVVIIIVIIIIVIIIVIIIIVIIIIIVIII"
[23] "VIIVIIIIIVVIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIIVVIIVIVIIVVIIVIIIVIIIIIVVIIVIIIVIIIIVVIIIVIIIIIVIIIIVIIIIIVVIIVVIIIVIIIIVIIIVIIIIIVIIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVIVIIVIIIIIVIVIIVVIIVIIII"
[24] "IVIIIIVVIIIVIIIIIVVIIVIIIVIIIIIVIIIIIVVIIVVIIIVIIIIVIIIVIIIIIVIIIIVIIIIIVVIIIVIIIIVIIIIIVIVIIIVIIIIIVVIIVIVIIVVIIIVIIIIIVIIIIIVIVIIVIIIIIVVIIVIVIIVVIIIVIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIVIIIVIIIIIVIIIIVIVI"
[25] "IIIVIVIIIVIIIIIVVIIIVIIIIVIIIIIVVIIVVIIIVIIIIIVIIIIIVIVIIVIIIIIVVIIVIVIIVVIIIVIIIIIVIVIIVVIIVIIIVIIIIIVVIIIVIIIIVIIIVIIIIIVIIIIIVVIIVVIIVIIIVIIIIVVIIIVIIIIVIIIVIIIIIVIIIIIVIVIIVVIIIVIIIIIVIIIIVIIIIIVIVIIIVIIIIVIIIIIVVIIVIVIIVIIIVII"


# JavaScript (ES6), 107

Recursive function returning the Nth term 0 based

f=(n,r='I')=>n?f(n-1,r.match(/I+|V+/g).map(x=>((n=x.length)-4?'VIII'.slice(n<5,1+n%5):'IV')+x[0]).join):r


Test

f=(n,r='I')=>n?f(n-1,r.match(/I+|V+/g).map(x=>((n=x.length)-4?'VIII'.slice(n<5,1+n%5):'IV')+x[0]).join):r

function update() {
O.textContent=f(I.value)
}

update()
<input id=I value=25 type=number oninput='update()'><pre id=O></pre>

# Perl 6, 62 bytes

{("I",{S:g/(.)$0*/{<I II III IV V>[$/.chars-1]~$0}/}...*)[$_]}


Anonymous function that accepts a zero-based index.

Makes use of the fact that roman numbers higher than 5 aren't needed, because the only groups of repeating digits that can occur, are:

I     -> II
II    -> III
III   -> IIII
IIII  -> IVI
IIIII -> VI

V     -> IV
VV    -> IIV
`