Your program is given a string consisting entirely of lowercase letters at STDIN (or closest alternative). The program must then output a truthy or falsey value, depending on whether the input is valid romaji.


  • It must be possible to divide the entire string into a sequence of kana without any leftover characters.
  • Each kana can be a single vowel (aeiou)
  • Each kana can also be a consonant p, g, z, b, d, k, s, t, n, h, m, or r followed by a vowel. For example, ka and te are valid kana, but qa is not.
  • The exceptions to the above rule are that zi, di, du, si, ti, and tu are not valid kana.
  • The following are also valid kana: n, wa, wo, ya, yu, yo, ji, vu, fu, chi, shi, tsu.
  • If a particular consonant is valid before an i (i.e ki, pi), the i can be replaced by a ya, yu, or yo and still be valid (i.e kya, kyu, kyo)
  • Exceptions to the above rule are chi and shi, for which the y has to be dropped too (i.e cha, chu, cho, sha, shu, sho)
  • It is also valid to double consonants if they are the first character of a kana (kka is valid but chhi is not)
  • Shortest answer wins. All regular loopholes are disallowed.

List of all valid kana:

Can have double consonant:

ba, bu, be, bo, bi
ga, gu, ge, go, gi
ha, hu, he, ho, hi
ka, ku, ke, ko, ki
ma, mu, me, mo, mi
na, nu, ne, no, ni
pa, pu, pe, po, pi
ra, ru, re, ro, ri
sa, su, se, so,
za, zu, ze, zo,
da,     de, do,
ta,     te, to,
wa,         wo,
ya, yu,     yo,

Can not have double consonant:

a, i, u, e, o, 
chi, cha, cho, chu,
shi, sha, sho, shu,

Test cases




  • \$\begingroup\$ How do we win? Is this a code golf? \$\endgroup\$ May 24, 2017 at 0:09
  • 2
    \$\begingroup\$ Need test cases. Also could do with a list of all valid kana instead of the rules \$\endgroup\$ May 24, 2017 at 0:11
  • 1
    \$\begingroup\$ @RobertFraser both is preferred - test cases are not rules \$\endgroup\$
    – Stephen
    May 24, 2017 at 0:12
  • 2
    \$\begingroup\$ n cannot be doubled. I know enough about the Japanese alphabets to say that. If n was doubled, it would need to have a vowel after, but then it wouldn't be n. So if kanna was a word (just making it up), it'd actually be ka n na. \$\endgroup\$
    – mbomb007
    May 24, 2017 at 14:16
  • 1
    \$\begingroup\$ You know, I wanted to make a solution using unicodedata, but it'll definitely be longer than a regex solution. Partial program \$\endgroup\$
    – mbomb007
    May 24, 2017 at 14:30

2 Answers 2


Ruby, 96 149 bytes

Regex solution to match all the valid kana. Interestingly, "ecchi" is not valid according to the current rules, but perhaps it's for the best.


Try it online! feat. Cruel Angel's Thesis

  • \$\begingroup\$ It failes on simple tests zi and zye \$\endgroup\$ May 24, 2017 at 13:32
  • \$\begingroup\$ @DeadPossum fixed. \$\endgroup\$
    – Value Ink
    May 24, 2017 at 23:01

Python 2, 166 bytes

Long regex solution
Try it online

I think that f-strings from 3.[something] python can help to shorten it by replacing repeated [auo and {1,2}.
Unfortunatetly I can't check it by myself now :c

import re
lambda x:re.sub('[bghkmnpr]~([auoei]|y[auo])|[sz]~[auoe]|[dt]~[aeo]|w~[ao]|([fv]~|ts)u|(j~|[cs]h)(i|y[auo])|y~[auo]|[auoien]'.replace('~','{1,2}'),'',x)==''
  • \$\begingroup\$ re.sub('~','{1,2}',(your regex) is shorter than (your regex).replace('~','{1,2}') by 1 byte. \$\endgroup\$
    – Value Ink
    May 24, 2017 at 23:03
  • \$\begingroup\$ Your regex is also failing on a simple test case: bku. Doubled consonants have to be the same consonant. \$\endgroup\$
    – Value Ink
    May 24, 2017 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.