Selective Caesar Cipher

Question

Input

A string of text.

Output

The string of text, however, every consonant is the next consonant in the alphabet, and every vowel is the next vowel in the alphabet (z becomes b, u becomes a). The case of the new letter should be the same as the letter it is replacing (a becomes e, A becomes E). If there are two consonants in a row, they will be separated by the characters 'ei', the case of 'ei' can be whatever you please. Characters outside of the 26 character Latin alphabet should remain unchanged. This includes whitespace and punctuation.

(For clarity, the vowels are aeiou. Y is not considered a vowel)

Examples: The phrase "Hello World" becomes "Jimeimu Xuseimeif" The phrase "codegolf" becomes "dufihumeig"

• 97.100.97.109: I'd recommend you add more test cases -- for example, does bcd become beiceid?
• chunes: @97.100.97.109 I believe bcd would become ceideif.
• Ismael Miguel: Is "Y" a vowel or a consonant, for this challenge?
• Ismael Miguel: "If there are two consonants in a row, they will be separated by the characters 'ei'." <-- Does this apply to alternating case? For example, would "Bc" be "Cd" or "Ceid"?
• qwr: why was ei added?

Answer 1

Vyxal, 27 bytes

k0kv":ɾJJƒ*‛ß+k0⇧1Ṁ$‡‛eijøṙ

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This is a bit of a mess.

k0kv" # Consonants and vowels, paired
 :ɾ # Duplicate and make an uppercase copy of each
 J # Concatenate to the original
 J # Concatenate to the input
 ƒ* # Reduce by ring translation, ring translating by each string
 ‛ß+ # "[bcdfghjklmnpqrstvwxyz]+"
 1Ṁ # Insert at position 1...
 k0⇧ # "BCDFGHJKLMNPQRSTVWXYZ"
 $ # Put that under the value
 øṙ # Regex replace that with...
 ‡ # A function that...
 ‛eij # Joins by "ei"

Answer 2

JavaScript (Node.js), 131 bytes

s=>s.replace(/[a-z]/gi,c=>(s+g(i=B(c)[0]-1)?"":"ei")+(h=j=>g(i)^g(j=-~j%26)?h(j):B([j+1|i&96]))(i&31),B=Buffer,g=i=>s=1065233>>i&1)

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Commented

s => s.replace( // replace in the input string s ...
 /[a-z]/gi, c => // ... each letter c (case insensitive)
 ( //
 s + // if the previous letter was a vowel (or this
 // is the 1st iteration and s is still a string)
 g( // or the current letter
 i = // whose ASCII code - 1 is loaded in i
 B(c)[0] - 1 // 
 ) ? // is a vowel:
 "" // append nothing
 : // else:
 "ei" // append "ei"
 ) + ( //
 h = j => // h is a recursive function looking for the
 // replacement letter
 g(i) ^ // if the type of the current letter
 g( // does not match the type of
 j = -~j % 26 // the next letter in the alphabet
 // obtained by incrementing j modulo 26
 ) ? // then:
 h(j) // keep advancing by doing a recursive call
 : // else:
 B([ // output the letter
 j + 1 | // whose ASCII code is j + 1
 i & 96 // with bits #5 and #6 taken from i
 ]) //
 )(i & 31), // initial call to h with j = i mod 32
 B = Buffer, // define B for Buffer
 g = i => // g is a helper function taking an integer
 s = 1065233 // representing an ASCII code minus 1,
 >> i & 1 // returning 0 for consonant / 1 for vowel,
 // and also saving the result in s
) // end of replace()

Answer 3

sed -E, 131 bytes

s/[A-Z]/@\l&/g
y/bcdfghjklmnpqrstvwxyzaeiou/cdfghjklmnpqrstvwxyzbeioua/
:a
s/@(.)/\u1円/
s/[b-df-hj-np-tv-z]{2}/!&/i
s/!(.)/1円ei/
ta

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This code expect that there won't be characters @ and ! input. Any two characters could be used instead of those, I think that even unprintable ones should work (at least using escape codes), but I chose to keep it as is to make it readable. Also this program could be made without this restriction, but it would be longer and less interesting.

Explanation:

s/[A-Z]/@\l&/g # lowercase all characters and make note which characters were lowercased
 # this have to be done because transliteration (following command) is case sensitive
 # otherwise there would have to also be all uppercase characters
y/bcdfghjklmnpqrstvwxyzaeiou/cdfghjklmnpqrstvwxyzbeioua/ # change character to next one
:a # beginning of loop
s/@(.)/\u1円/ # uppercase back all characters where we made note
 # it is inside of loop, to save 1B by not using g flag
s/[b-df-hj-np-tv-z]{2}/!&/i # make note where to place string ei
 # this is done so that the long bracket expression doesn't need to be repeated twice
 # also this command is the reason for loop as sed doesn't supported overlapping replacements
s/!(.)/1円ei/ # put ei after noted character
ta # repeat till there is nothing to replace

Answer 4

Goruby -p, 99 bytes

c="aeiou"+b="bcdfghj-np-tv-z"
d="eioua"+b[1..]+?b
$_.t!c+c.up,d+d.up
gsub(/[#{b}]{2}/i){_1.t.j"ei"}

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Ruby, 110 bytes

c="aeiou"+b="bcdfghj-np-tv-z"
d="eioua"+b[1..]+?b
$_.tr!c+c.upcase,d+d.upcase
gsub /#{"([#{b}])"*2}/i,'1円ei2円'

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Answer 5

Retina 0.8.2, 85 bytes

T`_o`A\EIOUA-DF-HJ-NP-TV-ZBaei\oua-df-hj-np-tv-zb
i`[b-z-[eiou]](?=[b-z-[eiou]])
$&ei

Try it online! Link includes test cases. Edit: Fixed the character classes so that digits were not counted as consonants. (Character class subtractions allowed me to avoid increasing the byte count.) Explanation:

T`_o`A\EIOUA-DF-HJ-NP-TV-ZBaei\oua-df-hj-np-tv-zb

Translate each letter in the string to the next letter in the string. E and o are special so have to be quoted in the replacement string (o in the source string refers to a copy of the replacement string) while - indicates a character range (also avoids having to quote H and h). _ in the source string is just a filler to align the strings.

i`[b-z-[eiou]](?=[b-z-[eiou]])
$&ei

Case-insensitively find pairs of consecutive consonants and insert ei after the first.

Answer 6

C (clang), (削除) 152 (削除ここまで) 149 bytes

-3 bytes thanks to ceilingcat!!

*i,c,u,p;z(s){i=wcschr(s,c-u);c=i?i[1]:c;}f(*s){for(p=0;c=*s++;p=i)u=c/91*32,z(L"AEIOUA"),z(L"BCDFGHJKLMNPQRSTVWXYZB"),printf("ei%c"+(!p|!i)*2,c+u);}

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• Rhaixer: Haha when I remove the argument -w it generates 17 warnings along with output
• jdt: @py3programmer, only losers cares about warnings ;-)
• Rhaixer: true, atleast the output comes

Answer 7

Python, 166 bytes

f=lambda s,v='UOIEA',c='BZYXWVTSRQPNMLKJHGFDC':s and[y:=(q:=[x:=(t:=s.upper())[0],[v,c][x in c]][x in v+c])[~-q.find(x)],y.lower()][y<s]+'ei'*({*t[:2]}<{*c})+f(s[1:])

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Answer 8

Charcoal, 57 bytes

≔⪪aei×ばつei×ばつ⊟υΣ⊞OυNoη↧ι⊟EΦ⟦ιθη↥θ↥η⟧Noλι§λ⊕⌕λι

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

≔⪪aeiou1θ

Get a list of the vowels in lowercase.

≔−⪪β1θη

Subtract them from the lowercase alphabet to leave the consonants.s

⊞υ0

Start off with no previous consonant.

⭆S+

Map over each input character, joining together...

×ばつei×ばつ⊟υΣ⊞OυNoη↧ι

... "ei" repeated the number of times the previous character was a consonant times the number of times the current character is a consonant, saving the result for the next iteration, ...

⊟EΦ⟦ιθη↥θ↥η⟧Noλι§λ⊕⌕λι

... taking any of the uppercase consonants, uppercase vowels, lowercase consonants or lowercase vowels that contain the character, or as a last resort the original character, cyclically rotating the character in that list.

Answer 9

05AB1E, (削除) 28 (削除ここまで) 26 bytes

žNÃü2D€S„aeý:žNžM‚Du«vyDÀ‡

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ü2D€S could alternatively be SãDJs: try it online.

Explanation:

žN # Push the consonants constant
 Ã # Only keep those characters from the (implicit) input-string
 ü2 # Pop and push its overlapping pairs as strings
 D # Duplicate this list of string-pairs
 €S # Convert each 2-letter string to a character-pair
 „aeý # Join each pair with "ae" delimiter
 : # Replace all pairs with the 4-letter strings in the (implicit) input
žN # Push the consonants constant
 žM # Push the vowels constant
 ‚ # Pair them together
 Du« # Merge an uppercase copy of this pair
 vy # Loop over all four strings:
 D # Duplicate the current string
 À # Rotate its copy once towards the left
 ‡ # Transliterate the characters in the input-string
 # (after which the result is output implicitly)
 S # Convert it to a list of characters
 ã # Create all possible character-pairs with the cartesian power of 2
 D # Duplicate this list of character-pairs
 J # Join each inner character-pair to a 2-letter string
 s # Swap so the character-pair list is at the top of the stack again