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

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())

Try it online!

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 2

C (gcc), 232 bytes

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.

char*g(s,t,u,v,w)char*s,**t,*u,*v,*w;{for(w=strchr(s,95)?0:s;(v=*t++)&&!w;w=(*u+*v)?0:*(t-1))for(u=s;*u&&*v&&*v==*(*u==95?v:u);u++,v++);return w?w:s;}h(s,t,c,u)char*s,**t,*c,*u;{for(u=s;c=strtok(u," ");u=0)printf(" %s"+!!u,g(c,t));}

Try it online!

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

Jelly,  18  25 bytes

...noticed a bug during write-up, cost 7 to fix naively. There must be better!

ṗ⁹ḲL¤K€œ&⁼⁹ḟ”_¤ɗƇ⁹,L€EɗƇḢ

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 lists of characters.

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

How?

This is the 18-byte code...

ṗ⁹ḲL¤K€œ&⁼⁹ḟ”_¤ɗƇḢ - Link: words, W; string, S
    ¤              - 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:
               ɗ   -   last three links as a dyad:
       œ&          -   multi-set intersection
              ¤    -   nilad followed by link(s) as a nilad:
          ⁹        -     chain's right argument (S)
            ”_     -     literal '_' character
           ḟ       -     filter discard
         ⁼         -   equal? (multi-set intersection == string without '_'s?)
                 Ḣ - head (first result) ... UGH BUG.

...the current fix is to filter the list to only keep those with a length equal to the input before heading:

...⁹,L€EɗƇḢ - (continued)
   ⁹        - use the chain's right argument as the right
            -   argument of the first 17 bytes of the code above.
         Ƈ  - filter keep those for which:
        ɗ   -   last three links as a dyad:
    ,       -   pair left and right
     L€     -   length of €ach
       E    -   equal?
          Ḣ - head

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]

Try it online!

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