Pleasanortmanteaus

Question

A portmanteau word is a combination of two words that takes part of each word and makes them into a single new word. For example, lion + tiger => liger.

Let's write a program to generate portmanteaus from a pair of input words. Computers aren't the best at English, so we'll need to establish some rules to ensure that the output portmanteaus are pleasant to the eye and ear.

^{(Examples here are shown with a separator between the prefix and suffix for clarity: li|ger. However, the program's actual outputs should not have a separator: liger.)}

• Each portmanteau will consist of a nonempty prefix of the first word concatenated to a nonempty suffix of the second word: yes to li|ger, no to |iger.
• If the prefix ends with a vowel, the suffix must start with a consonant, and vice versa: yes to lio|ger or l|er, no to lio|iger or l|ger. You may decide whether to count y as a vowel or a consonant. Your solution must pick one option and stick with it, however.
• The resulting word must not contain either of the original words in full: yes to lio|ger, no to lion|iger or li|tiger.
  • This rule holds even if the portion in question is formed from parts of both words: with input of two + words, output tw|ords is still illegal because it contains the substring words. (The only valid output for this pair would be t|ords.)

Your program or function must take two words and output/return a list of all pleasant portmanteaus that can be formed from those words in that order.

Details

• Standard input and output methods apply. Standard loopholes are forbidden.
• Words will consist only of lowercase letters (or, if you prefer, only of uppercase letters).
• You may take the two input words as a list, tuple, two separate inputs, a single string with a non-letter delimiter, etc.
• Output format is similarly flexible; if you return or output a string, it should be delimited such that it's clear where one portmanteau word ends and the next one begins.
• There should be no delimiters inside a portmanteau word.
• It's okay if your output list includes duplicate results; it's also okay to remove duplicates.

Test cases

> lion, tiger
< liger, ler, liger, lir, lioger, lior

> tiger, lion
< tion, ton, tin, tigion, tigon, tigen

> spoon, fork
< sork, spork, spork, spok, spoork, spook

> smoke, fog
< sog, smog, smog, smokog

> gallop, triumph
< giumph, gumph, gariumph, gamph, gaph, gah, galiumph, galumph, galliumph, gallumph, galloriumph, gallomph, galloh

> breakfast, lunch
< bunch, brunch, brench, brech, breh, breanch, breach, breah, breakunch, breakfunch, breakfanch, breakfach, breakfah, breakfasunch

> two, words
< tords

> harry, ginny (if y is treated as a consonant)
< hinny, hanny, hany, hay, harinny, harrinny

> harry, ginny (if y is treated as a vowel)
> hinny, hy, hanny, hany, harinny, hary, harrinny

Reference solution

Here's a reference solution in Pip (treats y as a consonant).

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This is code-golf: the shortest answer in each language wins!

Answer 1

05AB1E, 28 bytes

y is a vowel (same byte count as consonant though).

нη¨sθ.s¨âʒ`нsθ‚žOsåË_}Jʒs¢Z_

Try it online! or as a slightly modified Test Suite

Answer 2

Retina, 72 bytes

L$w`(?<=[aeiou]()|.())((.+),(.+))\B(?!\4)(?<!\5\3)(?([aeiou])\2|\1)
$`$'

Try it online!

Answer 3

Pyth, 38 bytes

f!s}RTQm+hd_edfxFm}ed"aeiou"T*._hQ.__e

Input is a list of the two words, and y isn't treated as a consonant.

Try it online here, or verify all test cases at once here.

f!s}RTQm+hd_edfxFm}ed"aeiou"T*._hQ.__e   Implicit: Q=eval(input())
                                hQ       First input word
                              ._         All prefixes of the above
                                     e   Second input word (Q inferred)
                                  .__    Reverse, take all prefixes
                             *           Cartesian product of the above
              f                          Filter the above using:
                 m          T              Map d in the current element using:
                   ed                        The last letter of the word part
                  }  "aeiou"                 Is it contained in the vowel list?
               xF                          Take the XOR of the list
                                         (This ensures that the word parts meet at one consonant)
       m                                 Map d in the filtered set using:
        +hd_ed                             Add the first part to the reversed second part
f                                        Filter the above using:
  s}RTQ                                    Does the portmanteau contain either of the input words?
 !                                         Logical NOT (remove from list if the above is true)

Answer 4

Java 8, 228 225 215 bytes

v->w->{String r="",t,p=" aeiou";for(int i=w.length(),j;--i>0;)for(j=1;j<v.length();)r+=(t=v.substring(0,j)+w.substring(i)).matches(v+".*|.*"+w)|p.indexOf(t.charAt(j-1))*p.indexOf(t.charAt(j++))>0?"":t+" ";return r;}

Takes two Strings in currying syntax and returns a String. Treats y as a consonant. Try it online here.

Thanks to DLosc for golfing 2 bytes.

Ungolfed:

v -> w -> { // lambda taking two String parameters in currying syntax
    String r = "", // result
    t, // temporary variable used for storing
       // the portmanteau candidate currently being evaluated
    p = " aeiou"; // vowels for the purposes of this function;
                  // the leading space is so that they all have a positive index
    for(int i = w.length(), j; --i > 0; ) // loop over all proper suffixes
                                          // of the second word
        for(j = 1; j < v.length(); )      // loop over all proper prefixes
                                          // of the first word
            r += // construct the portmanteau candidate
                 (t = v.substring(0, j) + w.substring(i))
                 // if it contains one of the input words ...
                 .matches(v + ".*|.*" + w)
                 // ... or the boundary is consonant-consonant 
                 // or vowel-vowel (here we make use of the facts
                 // that all the vowels have a positive index, and
                 // indexOf() returns -1 in case of no match) ...
                 | p.indexOf(t.charAt(j-1)) * p.indexOf(t.charAt(j++)) > 0
                 ? "" // ... reject it ...
                 : t + " "; // ... else add it to the result
    return r; // return the result
}

Answer 5

Japt, 32 bytes

å+ ïVw å+)f_xè"%v$" uÃmrÈ+YwÃkøN

Japt Interpreter

Saved 10 bytes thanks to Shaggy's clearer understanding of Japt's syntax.

Saved 8 bytes due to a new language feature

Saved 2 byte thanks to some suggestions from ETHproductions

The newest version of Japt introduced the Cartesian Product function, which saved quite a few bytes and allowed me to restore the ordering of inputs (so "lion" "tiger" outputs "liger" and such). "y" is still treated as a consonant.

Explanation:

   ï     )       Cartesian product of...
å+                prefixes of first input
    Vw å+         and suffixes of second input.

f_         Ã     Remove the ones where...
  xè"%v$"         the number of vowels at the joining point
          u       is not 1.

m     Ã          Replace each pair with...
 rÈ+Yw            the prefix and suffix joined together
       køN       then remove the ones that contain either input

Answer 6

Python 3, 156 150 bytes

I've considered y as a consonant.

lambda a,b:{a[:i]+b[j:]for i in range(1,len(a))for j in range(1,len(b))if((a[i-1]in'aeiou')^(b[j]in'aeiou'))*0**(a in a[:i]+b[j:]or b in a[:i]+b[j:])}

-6 bytes thanks to Jonathan Frech

Try it online!

Answer 7

JavaScript (ES6), 124 bytes

Takes the 2 words in currying syntax (a)(b) and prints the results with alert(). Assumes y is a consonant.

a=>b=>[...a].map(c=>[...b].map((C,j)=>!(w=s+b.slice(j)).match(a+'|'+b)&v.test(c)-v.test(C)&&alert(w),s+=c),s='',v=/[aeiou]/)

Try it online!

Answer 8

Python 2, 179 176 166 162 bytes

lambda s,t:[w for w in g(s,t)if(s in w)<1>(t in w)]
g=lambda s,t:s[:-1]and[s[:-1]+t[j:]for j in range(1,len(t))if(s[-2]in'aeiou')^(t[j]in'aeiou')]+g(s[:-1],t)or[]

Try it online!

3 bytes from Jonathan Frech. And 10 bytes thx to The Matt.

In my world, y is not a vowel. (It's a yowl!)

Answer 9

Jelly, 27 bytes

¹Ƥp¹ÐƤ}Ø.ị"e€Øc⁻/ƲƇẎ€wÐḟƒ@,

Try it online!

Yy is a consonant. Both cases supported. Returns duplicates.

Output has been prettified over TIO. Remove +/€ from the footer to see the actual output.

Answer 10

C++ 11, 217 202 bytes

[](auto v,auto w){auto r=v,t=v,p=v;r="",p="aeiou";for(int i=w.size(),j;--i;)for(j=v.size();j;)(t=v.substr(0,j)+w.substr(i)).find(v)+1|t.find(w)+1|p.find(t[j-1])<5==p.find(t[j--])<5?v:r+=t+" ";return r;}

Makes heavy use of std::string#find. Treats y as a consonant. Try it online here.

Ungolfed:

// lambda; relies on auto to keep declarations short
[] (auto v, auto w) {
    // let's declare some strings. To keep it terse, we're using auto and the type of the arguments.
    auto r = v, // result string
    t = v,      // temporary string for storing the portmanteau candidate
    p = v;      // vowels string
    // now assign them their values
    r = "",    // result starts empty
    p = "aeiou"; // vowels don't include 'y'
    for(int i = w.size(), j; --i; ) // suffixes of the second word
        for(j = v.size(); j; ) // prefixes of the first word
            // create the portmanteau candidate
            (t = v.substr(0, j) + w.substr(i))
            // if it includes one of the input words ...
            .find(v) + 1 | t.find(w) + 1
            // ... or the boundary is consonant-consonant or vowel-vowel ...
            | p.find(t[j - 1]) < 5 == p.find(t[j--]) < 5
            ? v // ... discard it ...
            : r += t + " "; // ... else add it to the result.
    return r; // return the result
}

Answer 11

Ruby, 113 112 109 104 bytes

y is a consonant

This outputs the same duplicates as the examples in the question, I must be using the same loop

->a,b,i=j=1{r=a[0,i]+b[j..-1];g=:aeiou;!g[a[i-1]]^g[b[j]]|r[a]|r[b]||z=[*z,r];b[j+=1]||a[i+=j=1]?redo:z}

Try it online!

Answer 12

Emacs Lisp, 306 + 13 = 319 bytes

+13 for (require'seq)

(require'seq)(lambda(a b)(dotimes(i(1-(length b)))(dotimes(j(1-(length a)))(progn(setq w(substring a 0(1+ j))x(substring b(1+ i))c(concat w x))(defun V(c)(seq-contains"aeiou"(elt c 0)'char-equal))(if(not(or(string-prefix-p a c)(string-suffix-p b c)))(if(V(substring w -1))(if(not(V x))(print c))(if(V x)(print c))))))))

Try It Online!

Defines an anonymous lambda function. Outputs a sequence of newline-separated portmanteaus with each one surrounded by quotes. Golfing tips are welcome. The letter y is considered a consonant.

Ungolfed

(require 'seq)                                                                                                                                                           
(defun Portmanteus(word1 word2)
  "Find all valid portmanteus of the two given words"
  (dotimes (i (1- (length word2)))
    (dotimes (j (1- (length word1)))
      (progn
        (setq w (substring word1 0 (1+ j)) w2 (substring word2 (1+ i)) comb (concat w w2))
        (defun isVowel (c) (seq-contains "aeiou" (elt c 0) 'char-equal))
        (if (not (or (string-prefix-p word1 comb) (string-suffix-p word2 comb)))
          (if (isVowel (substring w -1))
            (if (not (isVowel w2))
              (princ (format "%s\n" comb))
            )
            (if (isVowel w2)
              (princ (format "%s\n" comb))
            )
          )
        )
      )
    )
  )
)