# M_ss_ng Lette_s

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_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 , shortest solution in bytes wins.

• I tried to made this really clear and well defined unlike the other similar questions, what about it is unclear? – pfg May 14 '18 at 18:51
• @pfg how flexible is the input / output? can we work with a list of the words? – Rod May 14 '18 at 19:11
• For test case purposes, can we have the 'words' list be /usr/share/dict/words instead of something on the web? letters, for example, is not in the standard unix words file (although letter is). That way, on TIO you can use the local file. – Chas Brown May 14 '18 at 19:13
• @pfg i meant the 'Words split at a space' part, the input/output can be separated words? Or we need to split / join the space? – Rod May 14 '18 at 19:16
• @pfg: TIO code is running on a virtual unix instance; so in python (for example) I can use f=open('/usr/share/dict/words');W=f.read() to access that standard unix file. – Chas Brown May 14 '18 at 19:29

# 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.

Try it online!
...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)
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:
”_        -     literal '_' character
ḟ          -     filter discard (remove all '_'s from S)
œ&     -     (S) multi-set intersection (current value)
⁼       -     equal? (S without '_'s == multi-set intersection?)

• Pinging this because I'm curious if theres a way to shorten the naive fix. – Razetime Oct 19 at 11:57
• @Razetime Yes there was a way to do so, I've edited it in. – Jonathan Allan Oct 19 at 12:36

# Python 2, 9610599 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.

• The spec requires you to take the provided dictionary as input. It looks like you're using a different dictionary and assuming it's assigned to a predefined variable. – Shaggy May 14 '18 at 22:42

# 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!

• 211 bytes *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?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));} – ceilingcat Jun 26 '18 at 22:16

# 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


# 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.

• I'm pretty sure it's possible to turn the @ helper into a fold, so if anyone has any ideas about that it'd be wonderful. – Οurous May 14 '18 at 23:02

# 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)