Is this a word?

Question

Toki pona is a minimalist constructed language, and thus it has minimalist phonotactics (rules describing what sounds make valid words).

Toki Pona has 8 consonant sounds m, n, p, t, k, s, w, l and j and 5 vowel sounds a, e, i, o, u. A single basic syllable in toki pona consists of any 1 consonant, any one vowel and optionally an n. So all of the following are valid:

pi
ko
wa
san
jen

There are four exceptions of sequences that are forbidden they are ji, wu, wo, ti.

With the basic syllables we can build words. A word is just a string of basic syllables with two special rules:

1. When we have an n followed by either n or m we drop the n. e.g. jan + mo is jamo not janmo
2. The initial syllable of a word can drop the initial consonant. e.g. awen and pawen are both legal and distinct words. This does allow for words consisting of just a vowel.

Task

Your task is to take a non-empty string consisting of lowercase alphabetic characters (a through z) and determine if it makes a valid toki pona word.

When the input is a valid word you should output one value and when it is not you should output another distinct value.

This is code-golf so answers will be scored in bytes with the goal being to minimize the size of the source.

Test cases

Accept:

awen
jan
e
monsuta
kepekin
ike
sinpin
pakala
tamako
jamo
pankulato
pawen
an
nene
nenen

Reject:

bimi
ponas
plani
womo
won
tin
wu
miji
sunmo
anna
sam
kain
op
p
n

Answer 1

Retina 0.8.2, 48 47 bytes

A`ji|nm|nn|ti|wu|wo
^((^|[j-npstw])[aeiou]n?)+$

Try it online! Link includes test cases. Edit: Saved 1 obvious byte thanks to @ovs. Explanation:

A`ji|nm|nn|ti|wu|wo

Delete invalid inputs.

^((^|[j-npstw])[aeiou]n?)+$

Match valid inputs that weren't invalidated above.

Answer 2

Jelly,  56  51 bytes

+1 to cater for strict IO (two distinct outputs rather than truthy vs falsey being allowed)

“jtklmnpsw”,ØẹŒpṖṖ¬3,8¦p”n;ƊṗⱮLẎF€⁾mnyw⁾nnƲÐḟḊ€;$e@

A (very inefficiant) monadic Link that yields 0 when the input string is not a Toki Pona word and 1 when it is.

(Don't) Try it online! (it's so inefficient it'll only complete for words of length three or less!)

...but here is a test-suite that has all tests except the four syllable pankulato that (a) limits to three base-syllables, rather than that of the number of characters in the input string and (b) only calls the word-generating code once for all (hence the e@ has been moved out to the footer).

How?

We construct a list containing ALL valid Toki Pona words constructed from at most length(input) syllables and check if the input is in there.

Yep that's soooo nasty, but without easy regex access I imagine it's the golfiest way.

“jtklmnpsw”,ØẹŒpṖṖ¬3,8¦p”n;Ɗṗ - (partial) Link: integer (from below!)
“jtklmnpsw”                   - "jtklmnpsw"
            Øẹ                - "aeiou"
           ,                  - pair
              Œp              - Catesian product
                ṖṖ            - pop off "wu" and "wo"
                   3,8¦       - apply to indices 3 & 8 ("ji" & "ti"):
                  ¬           -   logical NOT (replace these with [0,0] (integers)
                           Ɗ  - last three links as a monad:
                        ”n    -   'n'
                       p      -   Cartesian product (appends 'n' to each)
                          ;   -   concatenate
                            ṗ - Catiasian power (the integer)

...ⱮLẎF€⁾mnyw⁾nnƲÐḟḊ€;$e@ - (continued) Link: string, S
... L                     - length of S
...Ɱ                      - map across [1..length(S)] with:
...                       -   code above -> base-syllable combos of each length
     Ẏ                    - tighten
      F€                  - flatten each
                 Ðḟ       - filter discard those for which:
                Ʋ         -   last four links as a monad:
        ⁾mn               -     "mn"
           y              -     translate (convert ms to ns)
             ⁾nn          -     "nn"
            w             -     index of first occurrence (or zero)
                      $   - last two links as a monad:
                   Ḋ€     -   dequeue each
                     ;    -   concatenate
                        @ - with swapped arguments:
                       e  -   S exists in there?

Answer 3

Perl 5 -p, 64 bytes

$_=!/[jt]i|wu|wo|nm|nn/&&/^([aeiou]n?)?([mnptkswlj][aeiou]n?)*$/

Try it online!

Answer 4

TypeScript type system, 313 bytes

type v="a"|"e"|"i"|"o"|"u";type i<T>=T extends""?1:T extends`${Exclude<`${"m"|"n"|"p"|"t"|"k"|"s"|"w"|"l"|"j"}${v}`,"ji"|"wu"|"wo"|"ti">}${infer r}`?i<r>extends 1?1:r extends`n${infer e}`?e extends`${"n"|"m"}${any}`?0:i<e>:0:0;type o<T>=T extends`${v}${infer p}`?i<p>extends 1?1:p extends`n${infer r}`?i<r>:0:i<T>

This is written entirely with TypeScript types - the o type outputs 1 if the input parameter is a valid word and 0 if it is not. There's probably some room for further golfing.

Answer 5

Charcoal, 59 58 55 bytes

∧θ¬⊙⪪”&↧q1o⁺VＰα”²№θι≔aeiouηＦ⮌θ¿№ηι≔⁻”&↧ï⁸t∕p№t⟦”ηη¿⁻ιn⎚

Try it online! Link is to verbose version of code. Explanation:

∧θ¬⊙⪪”&↧q1o⁺VＰα”²№θι

Check that the word doesn't contain any of the illegal letter pairs contained in the compressed string.

≔aeiouη

Start by expecting the last character to be a vowel.

Ｆ⮌θ

Loop over the word in reverse.

¿№ηι

If we see an expected letter, ...

≔⁻”&↧ï⁸t∕p№t⟦”ηη

... then flip the set of expected letters by subtracting it from the string all the legal Toki Pona letters grouped into vowels and consonants.

¿⁻ιn

Otherwise, if the current letter is not an n, ...

⎚

... then erase any previous validity there might have been.

Answer 6

05AB1E, 49 bytes

„nn„nm‚åà≠×ε.•2Ñ|qγù•žM⨨D27SèKD'n««N>ãJ}˜D€¦«Iå

Port of @JonathanAllan's Jelly answer, but even slower.. :/
Outputs 1/0 for accept/reject respectively.

Try it online.
As is it's too slow for a test suite, but by adding 2äн between the × and ε (map over halve the input-length instead), we can verify all but the longest few truthy test cases and falsey test cases respectively, in separated test suites.

Explanation:

„nn„nm‚               # Push pair ["nn","nm"]
       åà≠            # Check that NEITHER is present in the (implicit) input
          ×           # 'Multiply' it by the (implicit) input-string
                      # (the input if truthy; "" if falsey)
ε                     # Map over the characters:
 .•2Ñ|qγù•            #  Push compressed string "jtklmnpsw"
          žM          #  Push builtin vowels "aeiou"
            â         #  Pop both, and create a list of all possible char-pairs
             ¨¨       #  Remove the last two ("wu" and "wo")
               D      #  Duplicate the list
                27S   #  Push pair [2,7]
                   è  #  Index those into the copy: ["ji","ti"]
                    K #  Remove those as well
 D                    #  Duplicate the list again
  'n«                '#  Append an "n" to each string
     «                #  Merge the two lists together
 N                    #  Push the 0-based map-index
  >                   #  Increase it by 1 to make it 1-based
   ã                  #  Cartesian product this index on the list of syllables
    J                 #  Join each inner list together to a string
}˜                    # After the map: flatten the list of lists
  D                   # Duplicate the list
   €¦                 # Remove the first consonant from each
     «                # Merge the two lists together
Iå                    # Check if the input-string is in this list
                      # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•2Ñ|qγù• is "jtklmnpsw".

Answer 7

Pip, 56 53 47 bytes

-3 bytes by porting Neil's Retina answer

X<>"jiwuwotinnnm"NIa&a~=+:`^|[j-nptsw]`+XV.`n?`

Returns 1 for a valid word, 0 for an invalid word. Attempt This Online!

Verify all test cases

Explanation

At its core, this solution works similarly to Neil's Retina answer:

• The input does not contain any of the illegal sequences ji, wu, wo, ti, nn, or nm; AND
• The input fully matches the regex ((^|[j-nptsw])[aeiou]n?)+

First half:

X<>"jiwuwotinnnm"NIa
   "jiwuwotinnnm"     That string
 <>                   Grouped into pairs of characters
X                     Converted to a regex that matches any of those pairs
                 NI   Does not match in
                   a  The command-line argument

Second half:

a~=+:`^|[j-nptsw]`+XV.`n?`
     `^|[j-nptsw]`          That regex
                  +         Wrapped in a non-capturing group and followed by
                   XV       Built-in regex `[aeiou]`
                     .      Followed by
                      `n?`  That regex
   +:                       Apply the + quantifier to the above wrapped in n.c. group
a~=                         Command-line argument fully matches that regex

Answer 8

Python 3, 97 88 86 bytes

lambda x:re.sub("((?!ji|wu|wo|ti|.*n[nm])(^|[j-npstw])[aeiou]n?)*$","",x)>""
import re

Try it online!

return False for valid word, True for invalid

Thanks to @14m2 for -2 bytes

How it works:

• at each syllable, we chek for ji|wu|wo|ti and prevent any capture if it is present. We also chek for the presence of either nn or nm further in the word.
• if it was absent, we capture the syllable (consonant + voyel (+ n))
• All the syllables captured are replaced by the empty string
• We then check if the result is greater than the empty string (falsey) or equal to the empty string (thruthy)

Answer 9

Lexurgy, 195 bytes

Lexurgy is a tool made for conlangers for applying sound changes, so this is perfect for this challenge! (and here I am bashing it into code golf)

Outputs the original word if it's valid Toki Pona, and an empty string otherwise.

Extremely slow version:

Class c {m,n,p,t,k,s,w,l,j}
Class v {a,e,i,o,u}
a:
{({j,t} i),(w {o,u}),({m,n} {m,n}),!@c&!@v}=>`
{(!n&@c @c),(@v @v)}=>` *
!@v&!n=>`/_ $
n=>`/$ _ $
c propagate:
[]=>`/{` _,_ `}
d:
`=>*

Much faster version, 199 bytes:

Class c {m,n,p,t,k,s,w,l,j}
Class v {a,e,i,o,u}
a:
{j,t} i=>`
w {o,u}=>`
{m,n} {m,n}=>`
!n&@c @c=>` *
@v @v=>` *
!@v&!n=>`/_ $
n=>`/$ _ $
!@c&!@v=>`
c propagate:
[]=>`/{` _,_ `}
d:
`=>*

Ungolfed:

Class cons {m,n,p,t,k,s,w,l,j}
Class vow {a,e,i,o,u}

remove-forbidden:
 {j,t} i => ` # ji, ti
 w {o,u} => ` # wo, wu
 {m,n} {m,n} => ` # mn, mm, etc
 !n&@cons @cons => ` * # no consecutive consonants
 @vow @vow => ` * # no consecutive vowels
 !@vow&!n => ` / _ $ # ending with a vowel or n
 n => ` / $ _ $ # nothing of length 1
Then:
 !@cons&!@vow => ` # convert any invalid character
Then propagate:
 [] => ` / {` _, _ `} # spread the invalid
Then:
 ` => * # delete the invalid

Answer 10

C (gcc), 438 bytes

#define R return
int c(l){char a[]={'n','m','p','t','k','s','w','l','j'};for(int i=0;i<9;i++)if(l==a[i])R 1;R 0;}
int v(l){R l==97||l==101||l==105||l==111||l==117?1:0;}
int f(char* s){int i,a,b;for(i=0;*s!=0;s++,i++){a =*s;b=*(s+1);if(!(c(a)||v(a))||((a=='j'||a=='t')&&b=='i'||a=='w'&&(b=='u'||b=='o')||a=='n'&&(b=='n'||b=='m'))||(c(a)&&c(b)&&a!='n')||(v(a)&&v(b))) R 0;}if(i==1&&c(*(s-1))) R 0;if(*s==0&&v(*(s-2))&&*(s-1)!='n') R 0;R 1;}

Try it online!

Explanations :

#define R return
// function to detect a consonant
int c(l){char a[]={'n','m','p','t','k','s','w','l','j'};for(int i=0;i<9;i++)if(l==a[i])R 1;R 0;}
// function to detect a vowel
int v(l){R l==97||l==101||l==105||l==111||l==117?1:0;}

int f(char* s){int i,a,b;for(i=0;*s!=0;s++,i++)
{
    a =*s;b=*(s+1);
    if(!(c(a)||v(a))||      // detect if characters are allowed
    ((a=='j'||a=='t')&&b=='i'||a=='w'&&(b=='u'||b=='o')||a=='n'&&(b=='n'||b=='m'))|| // detect if sequences ji, wu, wo & ti are not used
    (c(a)&&c(b)&&a!='n')||  // detect if there are not 2 consecutives consonants
    (v(a)&&v(b)))           // detect if there are not 2 consecutives vowels
    R 0;
    if(i==1&&c(*(s-1))) R 0;    // detect if it a single letter word & a vowel
    if(*s==0&&v(*(s-2))&&*(s-1)!='n') R 0;  // test if the last character is not a consonant except 'n'
    R 1;
}
```