M_ss_ng Lette_s

Question

Given this list of all the words in english and a string with s_me missi_g lett__s, find one way to fill in the letters. Do note that some words on the list contain special characters (! & ' , - . /) and numbers

Input:

This list and a string with missing letters.

The input sentence can contain or be missing any of these characters: ! & ' , - . / 0 1 2 3 4 5 6 7 8 9 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z a b c d e f g h i j k l m n o p q r s t u v w x y z. An underscore is used to mark a missing character. All missing characters have a possible solution.

Output:

One possible solution with all the missing letters fixed. Words split at a space . Case sensitive, includes words with ! & ' , - . /

Test Cases

Input: a string with s_me missi_g lett__s

All possible outputs: a string with [same, seme, sime, some] missing letters

Output Example: a string with same missing letters

Input: A_ 6-po_nt A/_ ada__er

All possible outputs: [A., A1, A4, A5, AA, AB, AF, AG, AH, AI, AY, AJ, AK, AM, AO, AP, AQ, AS, AT, AU, AV, AW, Ax, AZ] 6-point [A/C, A/F, A/O, A/P, A/V]

Output Example: A. 6-point A/C adapter

Input: A_M AT&_ platform_s scyth_'s hee_hee_ he-_e! 2_4,5_t

All possible outputs: [A&M, AAM, ABM, ACM, ADM, AFM, AGM, AIM, ALM, APM, ARM, ASM, ATM, AUM, AVM] AT&T platform's scythe's hee-hee! he-he! 2,4,5.t

Output example: A&M AT&T platform's scythe's hee-hee! he-he! 2,4,5-t

SemicolonSeperatedValue for first possible output:

a string with s_me missi_g lett__s;a string with same missing letters
A_ 6-po_nt A/_ ada__er;A. 6-point A/C adapter
A_M AT&_ platform_s scyth_'s hee_hee_ he-_e! 2_4,5_t;A&M AT&T platform's scythe's hee-hee! he-he! 2,4,5-t

Scoring

As this is code-golf, shortest solution in bytes wins.

Answer 1

Jelly,  18 25 20  19 bytes

...noticed a bug during write-up, cost 7 to fix naively, then saved 5 with another approach to fix it.

LÞṗ⁹ḲL¤K€ḟ”_⁼œ&ʋ@ƇṪ

A dyadic Link accepting a list of lists of characters on the left (words) and a list of characters on the right (string) which returns a list of characters.

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...using a hugely restricted word list due to incredible inefficiency (although it could be far worse!)

How?

LÞṗ⁹ḲL¤K€ḟ”_⁼œ&ʋ@ƇṪ - Link: words, W; string, S
 Þ                  - sort (W) by:
L                   -   length (this is so we get final results which will match
                                in length when we use Ṫ at the end of the Link)
      ¤             - nilad followed by link(s) as a nilad:
   ⁹                -   chain's right argument (S)
    Ḳ               -   split on spaces
     L              -   length
  ṗ                 - Cartesian power (all ways of making that number of words)
       K€           - join €ach with spaces
                 Ƈ  - filter keep those for which:
               ʋ@   -   last four links as a dyad with swapped arguments:
          ”_        -     literal '_' character
         ḟ          -     filter discard (remove all '_'s from S)
             œ&     -     (S) multi-set intersection (current value)
            ⁼       -     equal? (S without '_'s == multi-set intersection?)
                  Ṫ - head (last result)

Answer 2

Python 2, 96 105 99 101 bytes

lambda p,W:' '.join([w for w in W if all(map(lambda c,d:c in['_',d]and d,s,w))][0]for s in p.split())

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W is the list of words which is set up separately; I'm using the unix words file on TIO, which doesn't contain platform's or scythes's.

Added 2 bytes for Shaggy.

Answer 3

Java 10, 127 bytes

L->s->{for(var x:s.split(" "))for(var w:L)if(w.matches(x.replace(".","\\.").replace("_","."))){System.out.print(w+" ");break;}}

Try it online (also uses the /usr/share/dict/words TIO word-library, which doesn't contain platform's or scyth's, but the lambda-function works with any kind of list String-List provided).

Explanation:

L->s->{                   // Method with String-List & String parameters and no return-type
  for(var x:s.split(" ")) //  Loop over the parts of the input-String, split by spaces
    for(var w:L)          //   Inner loop over the words
      if(w.matches(x.replace(".","\\.")
                          //    Escape all dots
                    .replace("_","."))){ 
                          //    Replace all underscores with regex-dots
                          //    And if the current word matches it:
        System.out.print(w+" ");
                          //     Print the word, plus a space
        break;}}          //     And stop the inner loop

Answer 4

C (gcc), 232 226 bytes

• -23 bytes from @ceilingcat
• +17 bytes for required header file

Like the other submissions, I used /usr/share/dict/words; change the first argument to select another wordlist and the second argument to specify the size (the function itself just needs an array of strings as the dictionary.) On a non-matched word with a _, the original word is returned.

#import<string.h>
*g(s,t,u,v,w)char*s,**t,*u,*v,*w;{for(w=index(s,95)?0:s;(v=*t)&&!w;w=*u+*v?0:*t,t++)for(u=s;*u==95|*v==*u&&*v++&&*u++;);s=w?:s;}h(s,t,c,u)int*s,*t,*c,*u;{for(u=s;c=strtok(u," ");u=0)printf(" %s"+!!u,g(c,t));}

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Answer 5

Clean, 174 171 bytes

import StdEnv,Text
$l=join" "o map(hd o filter(\w=any((==)w)l)o@o split"_")o split" "
@[h:t]|t>[]=[h<+c<+v\\c<-['0'..'9']++['a'..'z']++['A'..'Z']++['!&\',-./'],v<- @t]=[h]

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TIO link uses the dictionary in /usr/share/dict/words but it'll work with any list of words.

Answer 6

05AB1E (legacy), 27 bytes

#εD'_åiUʒX‚ζεD'_åsË~}P}Ω]ðý

Can definitely be golfed a bit more, but this will do for now.

First input is the string, second a list of words. If multiple words are possible, it picks a random one.

Assumes the sentence always contains at least a single space. If this is not allowed, the # has to be replaced with ð¡.

Uses the legacy version because the zip-builtins work on strings, whereas these have to be character-lists in the new version of 05AB1E. In the new version, the ʒX‚ζεD'_åsË~}P} would have been 1 byte longer because of that: ʒSXS‚ζε'_åyË~}P}.

Try it online (with a small sample dictionary list).

Explanation:

#                   # Split the (implicit) first input-String on spaces
 ε                  # Map over each word:
  D                 #  Duplicate the current word
   '_åi            '#  If the word contains an underscore:
       U            #   Pop the copy and store it in variable `X`
        ʒ           #   Filter over the (implicit) second input-list of words:
         X‚         #    Pair `X` with the current word
           ζ        #    Zip/transpose, swapping rows/columns with spaces as default
                    #    filler if the strings are of unequal length
            ε       #    Map over each pair of characters
             D      #     Duplicate it
              '_å  '#     Check if the pair contains an underscore
             s      #     Swap to get the map-value again
              Ë     #     Check that both characters are the same
                 ~  #     Check that either of these were truthy
            }P      #    After the map: check if all pairs were truthy
        }Ω          #   After the filter: pop and push a random valid word
                    #  (implicit else: use the unmodified word from the duplicate)
 ]                  # Close both the if-statement and map
  ðý                # And join the list by spaces
                    # (after which the result is output implicitly)

Answer 7

Husk, 22 bytes

wm`₁⁰w²
ḟoΛ='_żμ?'_³=⁰

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For each word in query text, goes through all words in the dictionary substituting matching characters for '_'. If, after doing this, the word is only '_'s, then it matched at all non-'_' characters, so it's a valid replacement.

 ḟoΛ='_żμ?'_³=⁰     # helper function: 
                    # finds word in arg2 that matches arg1 
                    # except at '_' characters.
ḟ                   # get first element that satisfies
 oΛ='_              # all elements are equal to '_'
                    # of
       żμ           # zip togethter arg2 and arg1 with
         ?   =⁰     # if they're equal
            ³       # then arg 2
          '_        # otherwise '_'
          
wm`₁⁰w²             # main program:
w                   # join with spaces
 m   w²             # map arg2 split on spaces 
  `₁⁰               # using helper function with arg1
```

Answer 8

Ruby, 53 bytes

->d,s{s.gsub(/\S+/){d.grep(/#{$&.tr ?_,?.}/)[0]||_1}}

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