Corrcey my Soellign

Question

Introduciton

Some of you might have realised that I am a sloppy typer when using my phone. That's why I want you to write a program which corrects my typos.

Chalkrnge

Given a misspelled word, output all of the possible words which I meant to write.

Typso

The main cause of my typos is down to the fact that I hit the wrong keys and often hit the key next door. The following is my keyboard layout:

q w e r t y u i o p
 a s d f g h j k l
   z x c v b n m
   , [ space ] .

Note that the bottom row, , [ space ] . will never be used in this challenge

For some reason, I only make mistakes horizontally: I would never hit the n instead of the j, but I might hit an f instead of a d.

For example, I could end up spelling the word sloppy as:

akioot

Where I've gone left of each key.

However, don't forget that I may not necessarily make a mistake on every letter in the word.

Ezsmple

Let's say the input is:

vid

The possibilities that the word could have been are:

vid cid bid
vis cis bis
vif cif bif
vod cod bod
vos cos bos
vof cof bof
vud cud bud
vus cus bus
vuf cuf buf

Out of those, the following are in the dictionary:

cod
cud
bid
bud
bus

So that should be your output.

Rulws

You should only use the text file found here as your dictionary: http://mieliestronk.com/corncob_lowercase.txt. You do not have to count this file as part of your byte count.

All input will be a single word. You may display your output in any way you wish (as long as there is some kind of separator).

Assume that with all input, you will find a variant which is in the dictionary.

Wibninf

The shortest code in bytes wins.

Answer 1

Python 2.7, 161 159 bytes

from itertools import*
lambda a,k=' qwertyuiop asdfghjkl zxcvbnm ':set(map(''.join,product(*[k[k.index(l)-1:k.index(l)+2].strip()for l in a])))&set("<dictionary>".split())

readable version

from itertools import *
dictionary=set("<dictionary>".split())
keyboard=' qwertyuiop asdfghjkl zxcvbnm '
x=[]
for letter in input():
 index=keyboard.index(letter)
 x.append(keyboard[index-1:index+2].strip())

words=set(map(''.join,product(*x)))
print words&dictionary

• Saved 1 byte thanks to @TuukkaX

Answer 2

Japt, 50 47 bytes

;D=R+Dv;W=3pUl¹óW ®s3 s1 £DgDbUgY)-X+1}Ãf@~V·bX

Input is the word to fix, and the dictionary as a string. Test it online! (Note: you'll have to manually paste the dictionary into the string.)

How it works

;D=R+Dv;W=3pUl¹óW ®s3 s1 £DgDbUgY)-X+1}Ãf@~V·bX  // Implicit: U = input, V = dictionary, R = newline
;                                                // Re-assign preset variables. D = "QWERTYUIOP\nASDFGHJKL\nZXCVBNM";
 D=R+Dv;                                         // Convert D to lowercase and prepend a newline.
        W=3pUl¹                                  // Set W to 3 to the power of U.length.
               óW                                // Create the range [W, W+W).
                  ®                       Ã      // Map each item Z in the range by this function:
                   s3                            //  Take Z.toString(3).
                      s1                         //  Remove the first character.
                                                 //  If the input is two chars long, e.g. "id", the array is now
                                                 //  ["00", "01", "02", "10", "11", "12", "20", "21", "22"].
                         £            }          //  Map each char X and index Y in the string by this function:
                              UgY                //   Get the char at position Y in U.
                            Db   )               //   Take the index of the char in D.
                                  -X+1           //   Subtract X and add 1.
                          Dg                     //   Get the char at that position in D.
                                                 // This maps our array for "id" to:
                                                 // ["of", "od", "os", "if", "id", "is", "uf", "ud", "us"].
                                        f@       // Filter to only the items X where
                                             bX  //  the index of X in
                                           V·    //  the dictionary, split at newlines,
                                          ~      //  is not -1.
                                                 // This filters our array for "id" to:
                                                 // ["of", "if", "id", "is", "us"].
                                                 // Implicit: output last expression