Word wheel

From Rosetta Code
Task
Word wheel
You are encouraged to solve this task according to the task description, using any language you may know.

A "word wheel" is a type of word game commonly found on the "puzzle" page of newspapers. You are presented with nine letters arranged in a circle or 3×3 grid. The objective is to find as many words as you can using only the letters contained in the wheel or grid. Each word must contain the letter in the centre of the wheel or grid. Usually there will be a minimum word length of 3 or 4 characters. Each letter may only be used as many times as it appears in the wheel or grid.


An example
N D E
O K G
E L W

Task

Write a program to solve the above "word wheel" puzzle.

Specifically:

  • Find all words of 3 or more letters using only the letters in the string   ndeokgelw.
  • All words must contain the central letter   K.
  • Each letter may be used only as many times as it appears in the string.
  • For this task we'll use lowercase English letters exclusively.


A "word" is defined to be any string contained in the file located at   http://wiki.puzzlers.org/pub/wordlists/unixdict.txt.
If you prefer to use a different dictionary,   please state which one you have used.

Optional extra

Word wheel puzzles usually state that there is at least one nine-letter word to be found. Using the above dictionary, find the 3x3 grids with at least one nine-letter solution that generate the largest number of words of three or more letters.


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences



11l[edit]

Translation of: Python
V GRID =
‘N   D   E
 O   K   G
 E   L   W’

F getwords()
   V words = File(‘unixdict.txt’).read().lowercase().split("\n")
   R words.filter(w -> w.len C 3..9)

F solve(grid, dictionary)
   DefaultDict[Char, Int] gridcount
   L(g) grid
      gridcount[g]++

   F check_word(word)
      DefaultDict[Char, Int] lcount
      L(l) word
         lcount[l]++
      L(l, c) lcount
         I c > @gridcount[l]
            R 1B
      R 0B

   V mid = grid[4]
   R dictionary.filter(word -> @mid C word & !@check_word(word))

V chars = GRID.lowercase().split_py().join(‘’)
V found = solve(chars, dictionary' getwords())
print(found.join("\n"))
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

8080 Assembly[edit]

This program runs under CP/M, and takes the dictionary file and wheel definition as arguments. The file is processed block by block, so it can be arbitrarily large (the given ~206kb unixdict.txt works fine).

puts:	equ	9		; CP/M syscall to print string
fopen:	equ	15		; CP/M syscall to open a file
fread:	equ	20		; CP/M syscall to read from file
FCB1:	equ	5Ch		; First FCB (input file)
DTA:	equ	80h		; Disk transfer address
	org	100h
	;;;	Make wheel (2nd argument) lowercase and store it
	lxi	d,DTA+1		; Start of command line arguments
scan:	inr	e		; Scan until we find a space
	rz			; Stop if not found in 128 bytes
	ldax	d
	cpi	' '		; Found it?
	jnz	scan		; If not, try again
	inx	d		; If so, wheel starts 1 byte onwards
	lxi	h,wheel		; Space for wheel
	lxi	b,920h		; B=9 (chars), C=20 (case bit)
whlcpy:	ldax	d		; Get wheel character
	ora	c		; Make lowercase
	mov	m,a		; Store
	inx	d		; Increment both pointers
	inx	h
	dcr	b		; Decrement counter
	jnz	whlcpy		; While not zero, copy next character
	;;;	Open file in FCB1
	mvi	e,FCB1		; D is already 0
	mvi	c,fopen
	call	5		; Returns A=FF on error
	inr	a		; If incrementing A gives zero,
	jz	err		; then print error and stop
	lxi	h,word		; Copy into word
	;;;	Read a 128-byte block from the file
block:	push	h		; Keep word pointer
	lxi	d,FCB1		; Read from file
	mvi	c,fread
	call	5
	pop	h		; Restore word pointer
	dcr	a		; A=1 = EOF
	rz			; If so, stop.
	inr	a		; Otherwise, A<>0 = error
	jnz	err
	lxi	d,DTA		; Start reading at DTA
char:	ldax	d		; Get character
	mov	m,a		; Store in word
	cpi	26		; EOF reached?
	rz			; Then stop
	cpi	10		; End of line reached?
	jz	ckword		; Then we have a full word
	inx	h		; Increment word pointer
nxchar:	inr	e		; Increment DTA pointer (low byte)
	jz	block		; If rollover, get next block
	jmp	char		; Otherwise, handle next character in block
	;;;	Check if current word is valid
ckword:	push	d		; Keep block pointer
	lxi	d,wheel		; Copy the wheel
	lxi	h,wcpy
	mvi	c,9		; 9 characters
cpyw:	ldax	d		; Get character
	mov	m,a		; Store in copy
	inx	h		; Increment pointers
	inx	d
	dcr	c		; Decrement counters
	jnz	cpyw		; Done yet?
	lxi	d,word		; Read from current word
wrdch:	ldax	d		; Get character
	cpi	32		; Check if <32
	jc	wdone		; If so, the word is done
	lxi	h,wcpy		; Check against the wheel letters
	mvi	b,9
wlch:	cmp	m		; Did we find it?
	jz	findch
	inx	h		; If not, try next character in wheel
	dcr	b		; As long as there are characters
	jnz	wlch		; If no match, this word is invalid
wnext:	pop	d		; Restore block pointer
	lxi	h,word		; Start reading new word
	jmp	nxchar		; Continue with character following word
findch:	mvi	m,0		; Found a match - set char to 0
	inx	d		; And look at next character in word
	jmp	wrdch
wdone:	lda	wcpy+4		; Word is done - check if middle char used
	ana	a		; If not, the word is invalid
	jnz	wnext
	lxi	h,wcpy		; See how many characters used
	lxi	b,9		; C=9 (counter), B=0 (used)
whtest:	mov	a,m		; Get wheel character
	ana	a		; Is it zero?
	jnz	$+4		; If not, skip next instr
	inr	b		; If so, count it
	inx	h		; Next wheel character
	dcr	c		; Decrement counter
	jnz	whtest
	mvi	a,2		; At least 3 characters must be used
	cmp	b
	jnc	wnext		; If not, the word is invalid
	xchg			; If so, the word _is_ valid, pointer in HL
	mvi	m,13		; add CR
	inx	h
	mvi	m,10		; and LF
	inx	h
	mvi	m,'$'		; and the CP/M string terminator
	lxi	d,word		; Then print the word
	mvi	c,puts
	call	5
	jmp	wnext
err:	lxi	d,errs		; Print file error
	mvi	c,puts
	jz	5
errs:	db	'File error$'	; Error message
wheel:	ds	9		; Room for wheel
wcpy:	ds	9		; Copy of wheel (to mark characters used)
word:	equ	$		; Room for current word
Output:
A>wheel unixdict.txt ndeokgelw
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

APL[edit]

Works with: Dyalog APL
wordwheel{
    words((~∊)⎕TC⊆⊢) 80 ¯1⎕MAP 
    match{
        0=≢⍵:1
        ~()⍺:0
        [(⍳⍴)~⍳⊃]1
    }
    middle(0.5×≢)⊃⊢
    words((middle )¨words)/words
    words(match¨words)/words
    (⍺⍺≤≢¨words)/words
}
Output:
      'ndeokgelw' (3 wordwheel) 'unixdict.txt'
 eke  elk  keel  keen  keg  ken  keno  knee  kneel  knew  know  knowledge  kong  leek  week  wok  woke 

AppleScript[edit]

use AppleScript version "2.4"
use framework "Foundation"
use scripting additions


------------------------ WORD WHEEL ----------------------

-- wordWheelMatches :: NSString -> [String] -> String -> String
on wordWheelMatches(lexicon, wordWheelRows)
    
    set wheelGroups to group(sort(characters of ¬
        concat(wordWheelRows)))
    
    script isWheelWord
        on |λ|(w)
            script available
                on |λ|(a, b)
                    length of a  length of b
                end |λ|
            end script
            
            script used
                on |λ|(grp)
                    w contains item 1 of grp
                end |λ|
            end script
            
            all(my identity, ¬
                zipWith(available, ¬
                    group(sort(characters of w)), ¬
                    filter(used, wheelGroups)))
        end |λ|
    end script
    
    set matches to filter(isWheelWord, ¬
        filteredLines(wordWheelPreFilter(wordWheelRows), lexicon))
    
    (length of matches as text) & " matches:" & ¬
        linefeed & linefeed & unlines(matches)
end wordWheelMatches


-- wordWheelPreFilter :: [String] -> String
on wordWheelPreFilter(wordWheelRows)
    set pivot to item 2 of item 2 of wordWheelRows
    set charSet to nub(concat(wordWheelRows))
    
    "(2 < self.length) and (self contains '" & pivot & "') " & ¬
        "and not (self matches '^.*[^" & charSet & "].*$') "
end wordWheelPreFilter


--------------------------- TEST -------------------------
on run
    set fpWordList to scriptFolder() & "unixdict.txt"
    if doesFileExist(fpWordList) then
        
        wordWheelMatches(readFile(fpWordList), ¬
            {"nde", "okg", "elw"})
        
    else
        display dialog "Word list not found in script folder:" & ¬
            linefeed & tab & fpWordList
    end if
end run



----------- GENERIC :: FILTERED LINES FROM FILE ----------

-- doesFileExist :: FilePath -> IO Bool
on doesFileExist(strPath)
    set ca to current application
    set oPath to (ca's NSString's stringWithString:strPath)'s ¬
        stringByStandardizingPath
    set {bln, int} to (ca's NSFileManager's defaultManager's ¬
        fileExistsAtPath:oPath isDirectory:(reference))
    bln and (int  1)
end doesFileExist


-- filteredLines :: String -> NString -> [a]
on filteredLines(predicateString, s)
    -- A list of lines filtered by an NSPredicate string
    set ca to current application
    
    set predicate to ca's NSPredicate's predicateWithFormat:predicateString
    set array to ca's NSArray's ¬
        arrayWithArray:(s's componentsSeparatedByString:(linefeed))
    
    (array's filteredArrayUsingPredicate:(predicate)) as list
end filteredLines


-- readFile :: FilePath -> IO NSString
on readFile(strPath)
    set ca to current application
    set e to reference
    set {s, e} to (ca's NSString's ¬
        stringWithContentsOfFile:((ca's NSString's ¬
            stringWithString:strPath)'s ¬
            stringByStandardizingPath) ¬
            encoding:(ca's NSUTF8StringEncoding) |error|:(e))
    if missing value is e then
        s
    else
        (localizedDescription of e) as string
    end if
end readFile


-- scriptFolder :: () -> IO FilePath
on scriptFolder()
    -- The path of the folder containing this script
    try
        tell application "Finder" to ¬
            POSIX path of ((container of (path to me)) as alias)
    on error
        display dialog "Script file must be saved"
    end try
end scriptFolder


------------------------- GENERIC ------------------------

-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
    -- Constructor for a pair of values, 
    -- possibly of two different types.
    {type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple


-- all :: (a -> Bool) -> [a] -> Bool
on all(p, xs)
    -- True if p holds for every value in xs
    tell mReturn(p)
        set lng to length of xs
        repeat with i from 1 to lng
            if not |λ|(item i of xs, i, xs) then return false
        end repeat
        true
    end tell
end all


-- concat :: [[a]] -> [a]
-- concat :: [String] -> String
on concat(xs)
    set lng to length of xs
    if 0 < lng and string is class of (item 1 of xs) then
        set acc to ""
    else
        set acc to {}
    end if
    repeat with i from 1 to lng
        set acc to acc & item i of xs
    end repeat
    acc
end concat


-- eq (==) :: Eq a => a -> a -> Bool
on eq(a, b)
    a = b
end eq


-- filter :: (a -> Bool) -> [a] -> [a]
on filter(p, xs)
    tell mReturn(p)
        set lst to {}
        set lng to length of xs
        repeat with i from 1 to lng
            set v to item i of xs
            if |λ|(v, i, xs) then set end of lst to v
        end repeat
        if {text, string} contains class of xs then
            lst as text
        else
            lst
        end if
    end tell
end filter


-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
        end repeat
        return v
    end tell
end foldl


-- group :: Eq a => [a] -> [[a]]
on group(xs)
    script eq
        on |λ|(a, b)
            a = b
        end |λ|
    end script
    
    groupBy(eq, xs)
end group


-- groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
on groupBy(f, xs)
    -- Typical usage: groupBy(on(eq, f), xs)
    set mf to mReturn(f)
    
    script enGroup
        on |λ|(a, x)
            if length of (active of a) > 0 then
                set h to item 1 of active of a
            else
                set h to missing value
            end if
            
            if h is not missing value and mf's |λ|(h, x) then
                {active:(active of a) & {x}, sofar:sofar of a}
            else
                {active:{x}, sofar:(sofar of a) & {active of a}}
            end if
        end |λ|
    end script
    
    if length of xs > 0 then
        set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, rest of xs)
        if length of (active of dct) > 0 then
            sofar of dct & {active of dct}
        else
            sofar of dct
        end if
    else
        {}
    end if
end groupBy


-- identity :: a -> a
on identity(x)
    -- The argument unchanged.
    x
end identity


-- length :: [a] -> Int
on |length|(xs)
    set c to class of xs
    if list is c or string is c then
        length of xs
    else
        (2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
    end if
end |length|


-- min :: Ord a => a -> a -> a
on min(x, y)
    if y < x then
        y
    else
        x
    end if
end min


-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    -- 2nd class handler function 
    -- lifted into 1st class script wrapper. 
    if script is class of f then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn


-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    -- The list obtained by applying f
    -- to each element of xs.
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map


-- nub :: [a] -> [a]
on nub(xs)
    nubBy(eq, xs)
end nub


-- nubBy :: (a -> a -> Bool) -> [a] -> [a]
on nubBy(f, xs)
    set g to mReturn(f)'s |λ|
    
    script notEq
        property fEq : g
        on |λ|(a)
            script
                on |λ|(b)
                    not fEq(a, b)
                end |λ|
            end script
        end |λ|
    end script
    
    script go
        on |λ|(xs)
            if (length of xs) > 1 then
                set x to item 1 of xs
                {x} & go's |λ|(filter(notEq's |λ|(x), items 2 thru -1 of xs))
            else
                xs
            end if
        end |λ|
    end script
    
    go's |λ|(xs)
end nubBy



-- sort :: Ord a => [a] -> [a]
on sort(xs)
    ((current application's NSArray's arrayWithArray:xs)'s ¬
        sortedArrayUsingSelector:"compare:") as list
end sort


-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
    if 0 < n then
        items 1 thru min(n, length of xs) of xs
    else
        {}
    end if
end take


-- unlines :: [String] -> String
on unlines(xs)
    -- A single string formed by the intercalation
    -- of a list of strings with the newline character.
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, linefeed}
    set s to xs as text
    set my text item delimiters to dlm
    s
end unlines


-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
on zipWith(f, xs, ys)
    set lng to min(|length|(xs), |length|(ys))
    if 1 > lng then return {}
    set xs_ to take(lng, xs) -- Allow for non-finite
    set ys_ to take(lng, ys) -- generators like cycle etc
    set lst to {}
    tell mReturn(f)
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs_, item i of ys_)
        end repeat
        return lst
    end tell
end zipWith
Output:
17 matches:

eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

AutoHotkey[edit]

letters := ["N", "D", "E", "O", "K", "G", "E", "L", "W"]

FileRead, wList, % A_Desktop "\unixdict.txt"
result := ""
for word in Word_wheel(wList, letters, 3)
    result .= word "`n"
MsgBox % result
return

Word_wheel(wList, letters, minL){
    oRes := []
    for i, w in StrSplit(wList, "`n", "`r")
    {
        if (StrLen(w) < minL)
            continue
        word := w
        for i, l in letters
            w := StrReplace(w, l,,, 1)
        if InStr(word, letters[5]) && !StrLen(w)
            oRes[word] := true
    }
    return oRes
}
Output:
eke
elk
k
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
ok
week
wok
woke

AWK[edit]

# syntax: GAWK -f WORD_WHEEL.AWK letters unixdict.txt
#         the required letter must be first
#
# example: GAWK -f WORD_WHEEL.AWK Kndeogelw unixdict.txt
#
BEGIN {
    letters = tolower(ARGV[1])
    required = substr(letters,1,1)
    size = 3
    ARGV[1] = ""
}
{   word = tolower($0)
    leng_word = length(word)
    if (word ~ required && leng_word >= size) {
      hits = 0
      for (i=1; i<=leng_word; i++) {
        if (letters ~ substr(word,i,1)) {
          hits++
        }
      }
      if (leng_word == hits && hits >= size) {
        for (i=1; i<=leng_word; i++) {
          c = substr(word,i,1)
          if (gsub(c,"&",word) > gsub(c,"&",letters)) {
            next
          }
        }
        words++
        printf("%s ",word)
      }
    }
}
END {
    printf("\nletters: %s, '%s' required, %d words >= %d characters\n",letters,required,words,size)
    exit(0)
}
Output:
eke elk keel keen keg ken keno knee kneel knew know knowledge kong leek week wok woke
letters: kndeogelw, 'k' required, 17 words >= 3 characters

BASIC[edit]

10 DEFINT A-Z
20 DATA "ndeokgelw","unixdict.txt"
30 READ WH$, F$
40 OPEN "I",1,F$
50 IF EOF(1) THEN CLOSE 1: END
60 C$ = WH$
70 LINE INPUT #1, W$
80 FOR I=1 TO LEN(W$)
90 FOR J=1 TO LEN(C$)
100 IF MID$(W$,I,1)=MID$(C$,J,1) THEN MID$(C$,J,1)="@": GOTO 120
110 NEXT J: GOTO 50
120 NEXT I  
130 IF MID$(C$,(LEN(C$)+1)/2,1)<>"@" GOTO 50
140 C=0: FOR I=1 TO LEN(C$): C=C-(MID$(C$,I,1)="@"): NEXT
150 IF C>=3 THEN PRINT W$,
160 GOTO 50
Output:
eke           elk           keel          keen          keg
ken           keno          knee          kneel         knew
know          knowledge     kong          leek          week
wok           woke

BCPL[edit]

get "libhdr"

// Read word from selected input
let readword(v) = valof 
$(  let ch = ?
    v%0 := 0
    $(  ch := rdch()
        if ch = endstreamch then resultis false
        if ch = '*N' then resultis true
        v%0 := v%0 + 1
        v%(v%0) := ch
    $) repeat
$)

// Test word against wheel
let match(wheel, word) = valof
$(  let wcopy = vec 2+9/BYTESPERWORD
    for i = 0 to wheel%0 do wcopy%i := wheel%i
    for i = 1 to word%0 do
    $(  let idx = ?
        test valof
        $(  for j = 1 to wcopy%0 do
                if word%i = wcopy%j then
                $(  idx := j
                    resultis true
                $)
            resultis false
        $)  then wcopy%idx := 0 // we've used this letter
            else resultis false // word cannot be made
    $)
    resultis 
        wcopy%((wcopy%0+1)/2)=0 & // middle letter must be used
        3 <= valof                // at least 3 letters must be used
        $(  let count = 0
            for i = 1 to wcopy%0 do
                if wcopy%i=0 then count := count + 1
            resultis count
        $)
$)

// Test unixdict.txt against ndeokgelw
let start() be
$(  let word = vec 2+64/BYTESPERWORD
    let file = findinput("unixdict.txt")
    let wheel = "ndeokgelw"
    
    selectinput(file)
    while readword(word) do
        if match(wheel, word) do
            writef("%S*N", word)
    endread()
$)
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

C[edit]

#include <stdbool.h>
#include <stdio.h>

#define MAX_WORD 80
#define LETTERS 26

bool is_letter(char c) { return c >= 'a' && c <= 'z'; }

int index(char c) { return c - 'a'; }

void word_wheel(const char* letters, char central, int min_length, FILE* dict) {
    int max_count[LETTERS] = { 0 };
    for (const char* p = letters; *p; ++p) {
        char c = *p;
        if (is_letter(c))
            ++max_count[index(c)];
    }
    char word[MAX_WORD + 1] = { 0 };
    while (fgets(word, MAX_WORD, dict)) {
        int count[LETTERS] = { 0 };
        for (const char* p = word; *p; ++p) {
            char c = *p;
            if (c == '\n') {
                if (p >= word + min_length && count[index(central)] > 0)
                    printf("%s", word);
            } else if (is_letter(c)) {
                int i = index(c);
                if (++count[i] > max_count[i]) {
                    break;
                }
            } else {
                break;
            }
        }
    }
}

int main(int argc, char** argv) {
    const char* dict = argc == 2 ? argv[1] : "unixdict.txt";
    FILE* in = fopen(dict, "r");
    if (in == NULL) {
        perror(dict);
        return 1;
    }
    word_wheel("ndeokgelw", 'k', 3, in);
    fclose(in);
    return 0;
}
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

C++[edit]

Library: Boost

The puzzle parameters can be set with command line options. The default values are as per the task description.

#include <array>
#include <iostream>
#include <fstream>
#include <map>
#include <string>
#include <vector>
#include <boost/program_options.hpp>

// A multiset specialized for strings consisting of lowercase
// letters ('a' to 'z').
class letterset {
public:
    letterset() {
        count_.fill(0);
    }
    explicit letterset(const std::string& str) {
        count_.fill(0);
        for (char c : str)
            add(c);
    }
    bool contains(const letterset& set) const {
        for (size_t i = 0; i < count_.size(); ++i) {
            if (set.count_[i] > count_[i])
                return false;
        }
        return true;
    }
    unsigned int count(char c) const {
        return count_[index(c)];
    }
    bool is_valid() const {
        return count_[0] == 0;
    }
    void add(char c) {
        ++count_[index(c)];
    }
private:
    static bool is_letter(char c) { return c >= 'a' && c <= 'z'; }
    static int index(char c) { return is_letter(c) ? c - 'a' + 1 : 0; }
    // elements 1..26 contain the number of times each lowercase
    // letter occurs in the word
    // element 0 is the number of other characters in the word
    std::array<unsigned int, 27> count_;
};

template <typename iterator, typename separator>
std::string join(iterator begin, iterator end, separator sep) {
    std::string result;
    if (begin != end) {
        result += *begin++;
        for (; begin != end; ++begin) {
            result += sep;
            result += *begin;
        }
    }
    return result;
}

using dictionary = std::vector<std::pair<std::string, letterset>>;

dictionary load_dictionary(const std::string& filename, int min_length,
                           int max_length) {
    std::ifstream in(filename);
    if (!in)
        throw std::runtime_error("Cannot open file " + filename);
    std::string word;
    dictionary result;
    while (getline(in, word)) {
        if (word.size() < min_length)
            continue;
        if (word.size() > max_length)
            continue;
        letterset set(word);
        if (set.is_valid())
            result.emplace_back(word, set);
    }
    return result;
}

void word_wheel(const dictionary& dict, const std::string& letters,
                char central_letter)  {
    letterset set(letters);
    if (central_letter == 0 && !letters.empty())
        central_letter = letters.at(letters.size()/2);
    std::map<size_t, std::vector<std::string>> words;
    for (const auto& pair : dict) {
        const auto& word = pair.first;
        const auto& subset = pair.second;
        if (subset.count(central_letter) > 0 && set.contains(subset))
            words[word.size()].push_back(word);
    }
    size_t total = 0;
    for (const auto& p : words) {
        const auto& v = p.second;
        auto n = v.size();
        total += n;
        std::cout << "Found " << n << " " << (n == 1 ? "word" : "words")
            << " of length " << p.first << ": "
            << join(v.begin(), v.end(), ", ") << '\n';
    }
    std::cout << "Number of words found: " << total << '\n';
}

void find_max_word_count(const dictionary& dict, int word_length) {
    size_t max_count = 0;
    std::vector<std::pair<std::string, char>> max_words;
    for (const auto& pair : dict) {
        const auto& word = pair.first;
        if (word.size() != word_length)
            continue;
        const auto& set = pair.second;
        dictionary subsets;
        for (const auto& p : dict) {
            if (set.contains(p.second))
                subsets.push_back(p);
        }
        letterset done;
        for (size_t index = 0; index < word_length; ++index) {
            char central_letter = word[index];
            if (done.count(central_letter) > 0)
                continue;
            done.add(central_letter);
            size_t count = 0;
            for (const auto& p : subsets) {
                const auto& subset = p.second;
                if (subset.count(central_letter) > 0)
                    ++count;
            }
            if (count > max_count) {
                max_words.clear();
                max_count = count;
            }
            if (count == max_count)
                max_words.emplace_back(word, central_letter);
        }
    }
    std::cout << "Maximum word count: " << max_count << '\n';
    std::cout << "Words of " << word_length << " letters producing this count:\n";
    for (const auto& pair : max_words)
        std::cout << pair.first << " with central letter " << pair.second << '\n';
}

constexpr const char* option_filename = "filename";
constexpr const char* option_wheel = "wheel";
constexpr const char* option_central = "central";
constexpr const char* option_min_length = "min-length";
constexpr const char* option_part2 = "part2";

int main(int argc, char** argv) {
    const int word_length = 9;
    int min_length = 3;
    std::string letters = "ndeokgelw";
    std::string filename = "unixdict.txt";
    char central_letter = 0;
    bool do_part2 = false;
    
    namespace po = boost::program_options;
    po::options_description desc("Allowed options");
    desc.add_options()
        (option_filename, po::value<std::string>(), "name of dictionary file")
        (option_wheel, po::value<std::string>(), "word wheel letters")
        (option_central, po::value<char>(), "central letter (defaults to middle letter of word)")
        (option_min_length, po::value<int>(), "minimum word length")
        (option_part2, "include part 2");

    try {
        po::variables_map vm;
        po::store(po::parse_command_line(argc, argv, desc), vm);
        po::notify(vm);

        if (vm.count(option_filename))
            filename = vm[option_filename].as<std::string>();
        if (vm.count(option_wheel))
            letters = vm[option_wheel].as<std::string>();
        if (vm.count(option_central))
            central_letter = vm[option_central].as<char>();
        if (vm.count(option_min_length))
            min_length = vm[option_min_length].as<int>();
        if (vm.count(option_part2))
            do_part2 = true;

        auto dict = load_dictionary(filename, min_length, word_length);
        // part 1
        word_wheel(dict, letters, central_letter);
        // part 2
        if (do_part2) {
            std::cout << '\n';
            find_max_word_count(dict, word_length);
        }
    } catch (const std::exception& ex) {
        std::cerr << ex.what() << '\n';
        return EXIT_FAILURE;
    }
    return EXIT_SUCCESS;
}
Output:

Output including optional part 2:

Found 5 words of length 3: eke, elk, keg, ken, wok
Found 10 words of length 4: keel, keen, keno, knee, knew, know, kong, leek, week, woke
Found 1 word of length 5: kneel
Found 1 word of length 9: knowledge
Number of words found: 17

Maximum word count: 215
Words of 9 letters producing this count:
claremont with central letter a
spearmint with central letter a

Delphi[edit]

Translation of: Wren
program Word_wheel;

{$APPTYPE CONSOLE}

{$R *.res}

uses
  System.SysUtils,
  System.Classes;

function IsInvalid(s: string): Boolean;
var
  c: char;
  leters: set of char;
  firstE: Boolean;
begin
  Result := (s.Length < 3) or (s.IndexOf('k') = -1) or (s.Length > 9);
  if not Result then
  begin
    leters := ['d', 'e', 'g', 'k', 'l', 'n', 'o', 'w'];
    firstE := true;
    for c in s do
    begin
      if c in leters then
        if (c = 'e') and (firstE) then
          firstE := false
        else
          Exclude(leters, AnsiChar(c))
      else
        exit(true);
    end;
  end;
end;

var
  dict: TStringList;
  i: Integer;
begin
  dict := TStringList.Create;
  dict.LoadFromFile('unixdict.txt');

  for i := dict.count - 1 downto 0 do
    if IsInvalid(dict[i]) then
      dict.Delete(i);

  Writeln('The following ', dict.Count, ' words are the solutions to the puzzle:');
  Writeln(dict.Text);

  dict.Free;
  readln;
end.

F#[edit]

// Word Wheel: Nigel Galloway. May 25th., 2021
let fG k n g=g|>Seq.exists(fun(n,_)->n=k) && g|>Seq.forall(fun(k,g)->Map.containsKey k n && g<=n.[k])
let wW n g=let fG=fG(Seq.item 4 g)(g|>Seq.countBy id|>Map.ofSeq) in seq{use n=System.IO.File.OpenText(n) in while not n.EndOfStream do yield n.ReadLine()}|>Seq.filter(fun n->2<(Seq.length n)&&(Seq.countBy id>>fG)n)
wW "unixdict.txt" "ndeokgelw"|>Seq.iter(printfn "%s")
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Factor[edit]

Works with: Factor version 0.99 2020-07-03
USING: assocs io.encodings.ascii io.files kernel math
math.statistics prettyprint sequences sorting ;

! Only consider words longer than two letters and words that
! contain elt.
: pare ( elt seq -- new-seq )
    [ [ member? ] keep length 2 > and ] with filter ;

: words ( input-str path -- seq )
    [ [ midpoint@ ] keep nth ] [ ascii file-lines pare ] bi* ;

: ?<= ( m n/f -- ? ) dup f = [ nip ] [ <= ] if ;

! Can we make sequence 1 with the elements in sequence 2?
: can-make? ( seq1 seq2 -- ? )
    [ histogram ] bi@ [ swapd at ?<= ] curry assoc-all? ;

: solve ( input-str path -- seq )
    [ words ] keepd [ can-make? ] curry filter ;

"ndeokgelw" "unixdict.txt" solve [ length ] sort-with .
Output:
{
    "eke"
    "elk"
    "keg"
    "ken"
    "wok"
    "keel"
    "keen"
    "keno"
    "knee"
    "knew"
    "know"
    "kong"
    "leek"
    "week"
    "woke"
    "kneel"
    "knowledge"
}

FreeBASIC[edit]

#include "file.bi"

Function String_Split(s_in As String,chars As String,result() As String) As Long
    Dim As Long ctr,ctr2,k,n,LC=Len(chars)
    Dim As boolean tally(Len(s_in))
    #macro check_instring()
    n=0
    While n<Lc
        If chars[n]=s_in[k] Then 
            tally(k)=true
            If (ctr2-1) Then ctr+=1
            ctr2=0
            Exit While
        End If
        n+=1
    Wend
    #endmacro
    
    #macro split()
    If tally(k) Then
        If (ctr2-1) Then ctr+=1:result(ctr)=Mid(s_in,k+2-ctr2,ctr2-1)
        ctr2=0
    End If
    #endmacro
    '==================  LOOP TWICE =======================
    For k  =0 To Len(s_in)-1
        ctr2+=1:check_instring()
    Next k
     if ctr=0 then
         if len(s_in) andalso instr(chars,chr(s_in[0])) then ctr=1':beep
         end if
    If ctr Then Redim result(1 To ctr): ctr=0:ctr2=0 Else  Return 0
    For k  =0 To Len(s_in)-1
        ctr2+=1:split()
    Next k
    '===================== Last one ========================
    If ctr2>0 Then
        Redim Preserve result(1 To ctr+1)
        result(ctr+1)=Mid(s_in,k+1-ctr2,ctr2)
    End If
   
    Return Ubound(result)
End Function

Function loadfile(file As String) As String
	If Fileexists(file)=0 Then Print file;" not found":Sleep:End
    Dim As Long  f=Freefile
    Open file For Binary Access Read As #f
    Dim As String text
    If Lof(f) > 0 Then
        text = String(Lof(f), 0)
        Get #f, , text
    End If
    Close #f
    Return text
End Function

Function tally(SomeString As String,PartString As String) As Long
    Dim As Long LenP=Len(PartString),count
    Dim As Long position=Instr(SomeString,PartString)
    If position=0 Then Return 0
    While position>0
        count+=1
        position=Instr(position+LenP,SomeString,PartString)
    Wend
    Return count
End Function

Sub show(g As String,file As String,byref matches as long,minsize as long,mustdo as string)
    Redim As String s()
    Var L=lcase(loadfile(file))
    g=lcase(g)
    string_split(L,Chr(10),s())
    For m As Long=minsize To len(g)
        For n As Long=Lbound(s) To Ubound(s)
            If Len(s(n))=m Then
                For k As Long=0 To m-1
                    If Instr(g,Chr(s(n)[k]))=0 Then Goto lbl 
                Next k
                If Instr(s(n),mustdo) Then 
                    For j As Long=0 To Len(s(n))-1
                        If tally(s(n),Chr(s(n)[j]))>tally(g,Chr(s(n)[j])) Then Goto lbl
                    Next j
                    Print s(n)
                    matches+=1
                End If
            End If
            lbl:
        Next n
    Next m
End Sub

dim as long matches
dim as double t=timer
show("ndeokgelw","unixdict.txt",matches,3,"k")
print
print "Overall time taken ";timer-t;" seconds"
print matches;" matches"
Sleep
Output:
eke
elk
keg
ken
wok
keel
keen
keno
knee
knew
know
kong
leek
week
woke
kneel
knowledge

Overall time taken  0.02187220007181168 seconds
 17 matches

Go[edit]

Translation of: Wren
package main

import (
    "bytes"
    "fmt"
    "io/ioutil"
    "log"
    "sort"
    "strings"
)

func main() {
    b, err := ioutil.ReadFile("unixdict.txt")
    if err != nil {
        log.Fatal("Error reading file")
    }
    letters := "deegklnow"
    wordsAll := bytes.Split(b, []byte{'\n'})
    // get rid of words under 3 letters or over 9 letters
    var words [][]byte
    for _, word := range wordsAll {
        word = bytes.TrimSpace(word)
        le := len(word)
        if le > 2 && le < 10 {
            words = append(words, word)
        }
    }
    var found []string
    for _, word := range words {
        le := len(word)
        if bytes.IndexByte(word, 'k') >= 0 {
            lets := letters
            ok := true
            for i := 0; i < le; i++ {
                c := word[i]
                ix := sort.Search(len(lets), func(i int) bool { return lets[i] >= c })
                if ix < len(lets) && lets[ix] == c {
                    lets = lets[0:ix] + lets[ix+1:]
                } else {
                    ok = false
                    break
                }
            }
            if ok {
                found = append(found, string(word))
            }
        }
    }
    fmt.Println("The following", len(found), "words are the solutions to the puzzle:")
    fmt.Println(strings.Join(found, "\n"))

    // optional extra
    mostFound := 0
    var mostWords9 []string
    var mostLetters []byte
    // extract 9 letter words
    var words9 [][]byte
    for _, word := range words {
        if len(word) == 9 {
            words9 = append(words9, word)
        }
    }
    // iterate through them
    for _, word9 := range words9 {
        letterBytes := make([]byte, len(word9))
        copy(letterBytes, word9)
        sort.Slice(letterBytes, func(i, j int) bool { return letterBytes[i] < letterBytes[j] })
        // get distinct bytes
        distinctBytes := []byte{letterBytes[0]}
        for _, b := range letterBytes[1:] {
            if b != distinctBytes[len(distinctBytes)-1] {
                distinctBytes = append(distinctBytes, b)
            }
        }
        distinctLetters := string(distinctBytes)
        for _, letter := range distinctLetters {
            found := 0
            letterByte := byte(letter)
            for _, word := range words {
                le := len(word)
                if bytes.IndexByte(word, letterByte) >= 0 {
                    lets := string(letterBytes)
                    ok := true
                    for i := 0; i < le; i++ {
                        c := word[i]
                        ix := sort.Search(len(lets), func(i int) bool { return lets[i] >= c })
                        if ix < len(lets) && lets[ix] == c {
                            lets = lets[0:ix] + lets[ix+1:]
                        } else {
                            ok = false
                            break
                        }
                    }
                    if ok {
                        found = found + 1
                    }
                }
            }
            if found > mostFound {
                mostFound = found
                mostWords9 = []string{string(word9)}
                mostLetters = []byte{letterByte}
            } else if found == mostFound {
                mostWords9 = append(mostWords9, string(word9))
                mostLetters = append(mostLetters, letterByte)
            }
        }
    }
    fmt.Println("\nMost words found =", mostFound)
    fmt.Println("Nine letter words producing this total:")
    for i := 0; i < len(mostWords9); i++ {
        fmt.Println(mostWords9[i], "with central letter", string(mostLetters[i]))
    }
}
Output:
The following 17 words are the solutions to the puzzle:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Most words found = 215
Nine letter words producing this total:
claremont with central letter a
spearmint with central letter a

Haskell[edit]

import Data.Char (toLower)
import Data.List (sort)
import System.IO (readFile)

------------------------ WORD WHEEL ----------------------

gridWords :: [String] -> [String] -> [String]
gridWords grid =
  filter
    ( ((&&) . (2 <) . length)
        <*> (((&&) . elem mid) <*> wheelFit wheel)
    )
  where
    cs = toLower <$> concat grid
    wheel = sort cs
    mid = cs !! 4

wheelFit :: String -> String -> Bool
wheelFit wheel = go wheel . sort
  where
    go _ [] = True
    go [] _ = False
    go (w : ws) ccs@(c : cs)
      | w == c = go ws cs
      | otherwise = go ws ccs

--------------------------- TEST -------------------------
main :: IO ()
main =
  readFile "unixdict.txt"
    >>= ( mapM_ putStrLn
            . gridWords ["NDE", "OKG", "ELW"]
            . lines
        )
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

J[edit]

require'stats'
wwhe=: {{
   ref=. /:~each words=. cutLF tolower fread 'unixdict.txt'
   y=.,y assert. 9=#y
   ch0=. 4{y
   chn=. (<<<4){y
   r=. ''
   for_i.2}.i.9 do.
     target=. <"1 ~./:~"1 ch0,.(i comb 8){chn
     ;:inv r=. r,words #~ ref e. target
   end.
}}
Task example:
   wwhe'ndeokgelw'
eke elk keg ken wok keel keen keno knee knew know kong leek week woke kneel knowledge

JavaScript[edit]

A version using local access to the dictionary, through the macOS JavaScript for Automation API.

Works with: JXA
(() => {
    "use strict";

    // ------------------- WORD WHEEL --------------------

    // gridWords :: [String] -> [String] -> [String]
    const gridWords = grid =>
        lexemes => {
            const
                wheel = sort(toLower(grid.join(""))),
                wSet = new Set(wheel),
                mid = wheel[4];

            return lexemes.filter(w => {
                const cs = [...w];

                return 2 < cs.length && cs.every(
                    c => wSet.has(c)
                ) && cs.some(x => mid === x) && (
                    wheelFit(wheel, cs)
                );
            });
        };


    // wheelFit :: [Char] -> [Char] -> Bool
    const wheelFit = (wheel, word) => {
        const go = (ws, cs) =>
            0 === cs.length ? (
                true
            ) : 0 === ws.length ? (
                false
            ) : ws[0] === cs[0] ? (
                go(ws.slice(1), cs.slice(1))
            ) : go(ws.slice(1), cs);

        return go(wheel, sort(word));
    };


    // ---------------------- TEST -----------------------
    // main :: IO ()
    const main = () =>
        gridWords(["NDE", "OKG", "ELW"])(
            lines(readFile("unixdict.txt"))
        )
        .join("\n");


    // ---------------- GENERIC FUNCTIONS ----------------

    // lines :: String -> [String]
    const lines = s =>
        // A list of strings derived from a single string
        // which is delimited by \n or by \r\n or \r.
        Boolean(s.length) ? (
            s.split(/\r\n|\n|\r/u)
        ) : [];


    // readFile :: FilePath -> IO String
    const readFile = fp => {
        // The contents of a text file at the
        // given file path.
        const
            e = $(),
            ns = $.NSString
            .stringWithContentsOfFileEncodingError(
                $(fp).stringByStandardizingPath,
                $.NSUTF8StringEncoding,
                e
            );

        return ObjC.unwrap(
            ns.isNil() ? (
                e.localizedDescription
            ) : ns
        );
    };


    // sort :: Ord a => [a] -> [a]
    const sort = xs =>
        Array.from(xs).sort();


    // toLower :: String -> String
    const toLower = s =>
        // Lower-case version of string.
        s.toLocaleLowerCase();


    // MAIN ---
    return main();
})();
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Julia[edit]

using Combinatorics

const tfile = download("http://wiki.puzzlers.org/pub/wordlists/unixdict.txt")
const wordlist = Dict(w => 1 for w in split(read(tfile, String), r"\s+"))

function wordwheel(wheel, central)
    returnlist = String[]
    for combo in combinations([string(i) for i in wheel])
        if central in combo && length(combo) > 2
            for perm in permutations(combo)
                word = join(perm)
                if haskey(wordlist, word) && !(word in returnlist)
                    push!(returnlist, word)
                end
            end
        end
    end
    return returnlist
end

println(wordwheel("ndeokgelw", "k"))
Output:
["ken", "keg", "eke", "elk", "wok", "keno", "knee", "keen", "knew", "kong", "know", "woke", "keel", "leek", "week", "kneel", "knowledge"]

Faster but less general version[edit]

const tfile = download("http://wiki.puzzlers.org/pub/wordlists/unixdict.txt")
const wordarraylist = [[string(c) for c in w] for w in split(read(tfile, String), r"\s+")]

function wordwheel2(wheel, central)
    warr, maxlen = [string(c) for c in wheel], length(wheel)
    returnarraylist = filter(a -> 2 < length(a) <= maxlen && central in a &&
            all(c -> sum(x -> x == c, a) <= sum(x -> x == c, warr), a), wordarraylist)
    return join.(returnarraylist)
end

println(wordwheel2("ndeokgelw", "k"))
Output:
["eke", "elk", "keel", "keen", "keg", "ken", "keno", "knee", "kneel", "knew", "know", "knowledge", "kong", "leek", "week", "wok", "woke"]

Lua[edit]

LetterCounter = {
  new = function(self, word)
    local t = { word=word, letters={} }
    for ch in word:gmatch(".") do t.letters[ch] = (t.letters[ch] or 0) + 1 end
    return setmetatable(t, self)
  end,
  contains = function(self, other)
    for k,v in pairs(other.letters) do
      if (self.letters[k] or 0) < v then return false end
    end
    return true
  end
}
LetterCounter.__index = LetterCounter

grid = "ndeokgelw"
midl = grid:sub(5,5)
ltrs = LetterCounter:new(grid)
file = io.open("unixdict.txt", "r")
for word in file:lines() do
  if #word >= 3 and word:find(midl) and ltrs:contains(LetterCounter:new(word)) then
    print(word)
  end
end
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Mathematica / Wolfram Language[edit]

ClearAll[possible]
possible[letters_List][word_String] := Module[{c1, c2, m},
  c1 = Counts[Characters@word];
  c2 = Counts[letters];
  m = Merge[{c1, c2}, Identity];
  Length[Select[Select[m, Length /* GreaterThan[1]], Apply[Greater]]] == 0
  ]
chars = Characters@"ndeokgelw";
words = Import["http://wiki.puzzlers.org/pub/wordlists/unixdict.txt", "String"];
words = StringSplit[ToLowerCase[words], "\n"];
words //= Select[StringLength /* GreaterEqualThan[3]];
words //= Select[StringContainsQ["k"]];
words //= Select[StringMatchQ[Repeated[Alternatives @@ chars]]];
words //= Select[possible[chars]];
words
Output:
{eke,elk,keel,keen,keg,ken,keno,knee,kneel,knew,know,knowledge,kong,leek,week,wok,woke}

Nim[edit]

import strutils, sugar, tables

const Grid = """N D E
                O K G
                E L W"""

let letters = Grid.toLowerAscii.splitWhitespace.join()

let words = collect(newSeq):
              for word in "unixdict.txt".lines:
                if word.len in 3..9:
                  word

let midLetter = letters[4]

let gridCount = letters.toCountTable
for word in words:
  block checkWord:
    if midLetter in word:
      for ch, count in word.toCountTable.pairs:
        if count > gridCount[ch]:
          break checkWord
      echo word
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Pascal[edit]

Works with: Free Pascal
program WordWheel;

{$mode objfpc}{$H+}

uses
  SysUtils;

const
  WheelSize = 9;
  MinLength = 3;
  WordListFN = 'unixdict.txt';

procedure search(Wheel : string);
var
  Allowed, Required, Available, w : string;
  Len, i, p : integer;
  WordFile : TextFile;
  Match : boolean;
begin
  AssignFile(WordFile, WordListFN);
  try
    Reset(WordFile);
  except
    writeln('Could not open dictionary file: ' + WordListFN);
    exit;
  end;
  Allowed := LowerCase(Wheel);
  Required := copy(Allowed, 5, 1);  { central letter is required }
  while not eof(WordFile) do
    begin
      readln(WordFile, w);
      Len := length(w);
      if (Len < MinLength) or (Len > WheelSize) then continue;
      if pos(Required, w) = 0 then continue;
      Available := Allowed;
      Match := True;
      for i := 1 to Len do
        begin
          p := pos(w[i], Available);
          if p > 0 then
            { prevent re-use of letter }
            delete(Available, p, 1)
          else
            begin
              Match := False;
              break;
            end;
        end;
      if Match then
        writeln(w);
    end;
  CloseFile(WordFile);
end;

{ exercise the procedure }
begin
  search('NDE' + 'OKG' + 'ELW');  
end.
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Perl[edit]

UPDATED: this version builds a single regex that will select all valid words straight from the file string.

#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Word_wheel
use warnings;

$_ = <<END;
                    N  D  E
                    O  K  G
                    E  L  W
END

my $file = do { local(@ARGV, $/) = 'unixdict.txt'; <> };
my $length = my @letters = lc =~ /\w/g;
my $center = $letters[@letters / 2];
my $toomany = (join '', sort @letters) =~ s/(.)\1*/
  my $count = length "$1$&"; "(?!(?:.*$1){$count})" /ger;
my $valid = qr/^(?=.*$center)$toomany([@letters]{3,$length}$)$/m;

my @words = $file =~ /$valid/g;

print @words . " words for\n$_\n@words\n" =~ s/.{60}\K /\n/gr;
Output:
17 words for
                    N  D  E
                    O  K  G
                    E  L  W

eke elk keel keen keg ken keno knee kneel knew know knowledge
kong leek week wok woke

Phix[edit]

with javascript_semantics
requires("1.0.1") -- (fixed another glitch in unique())
constant wheel = "ndeokgelw",
         musthave = wheel[5]
sequence words = unix_dict(),
         word9 = {} -- (for the optional extra part)
integer found = 0
for i=1 to length(words) do
    string word = lower(words[i])
    integer lw = length(word)
    if lw>=3 then
        if lw<=9 then
            word9 = append(word9,word)
        end if
        if find(musthave,word) then
            string remaining = wheel
            while lw do
                integer k = find(word[lw],remaining)
                if k=0 then exit end if
                remaining[k] = '\0' -- (prevent re-use)
                lw -= 1
            end while
            if lw=0 then
                found += 1
                words[found] = word
            end if
        end if
    end if
end for
string jbw = join_by(words[1..found],1,9," ","\n ")
printf(1, "The following %d words were found:\n %s\n",{found,jbw})
 
-- optional extra
if platform()!=JS then  -- (works but no progress/blank screen for 2min 20s)
                        -- (the "working" won't show, even w/o the JS check)
    integer mostFound = 0
    sequence mostWheels = {},
             mustHaves = {}
    for i=1 to length(word9) do
        string try_wheel = word9[i]
        if length(try_wheel)=9 then
            string musthaves = unique(try_wheel)
            for j=1 to length(musthaves) do
                found = 0
                for k=1 to length(word9) do
                    string word = word9[k]
                    if find(musthaves[j],word) then
                        string rest = try_wheel
                        bool ok = true
                        for c=1 to length(word) do
                            integer ix = find(word[c],rest)
                            if ix=0 then
                                ok = false
                                exit
                            end if
                            rest[ix] = '\0'
                        end for
                        found += ok
                    end if
                end for
                if platform()!=JS then -- (wouldn't show up anyway)
                    printf(1,"working (%s)\r",{try_wheel})
                end if
                if found>mostFound then
                    mostFound = found
                    mostWheels = {try_wheel}
                    mustHaves = {musthaves[j]}
                elsif found==mostFound then
                    mostWheels = append(mostWheels,try_wheel)
                    mustHaves = append(mustHaves,musthaves[j])
                end if
            end for
        end if
    end for
    printf(1,"Most words found = %d\n",mostFound)
    printf(1,"Nine letter words producing this total:\n")
    for i=1 to length(mostWheels) do
        printf(1,"%s with central letter '%c'\n",{mostWheels[i],mustHaves[i]})
    end for
end if
Output:

(Only the first three lines are shown under pwa/p2js)

The following 17 words were found:
 eke elk keel keen keg ken keno knee kneel
 knew know knowledge kong leek week wok woke

Most words found = 215
Nine letter words producing this total:
claremont with central letter 'a'
spearmint with central letter 'a'

Picat[edit]

main =>
  MinLen = 3,
  MaxLen = 9,
  Chars = "ndeokgelw",
  MustContain = 'k',

  WordList = "unixdict.txt",
  Words = read_file_lines(WordList),
  Res = word_wheel(Chars,Words,MustContain,MinLen, MaxLen),
  println(Res),
  println(len=Res.len),
  nl.

word_wheel(Chars,Words,MustContain,MinLen,MaxLen) = Res.reverse => 
  Chars := to_lowercase(Chars),
  D = make_hash(Chars), 
  Res = [],
  foreach(W in Words, W.len >= MinLen, W.len <= MaxLen, membchk(MustContain,W))
    WD = make_hash(W),
    Check = true,    
    foreach(C in keys(WD), break(Check == false))
      if not D.has_key(C) ; WD.get(C,0) > D.get(C,0) then
        Check := false
      end
    end,
    if Check == true then
      Res := [W|Res]
    end
  end.

% Returns a map of the elements and their occurrences
% in the list L.
make_hash(L) = D => 
  D = new_map(),
  foreach(E in L)
    D.put(E,D.get(E,0)+1)
  end.
Output:
[eke,elk,keel,keen,keg,ken,keno,knee,kneel,knew,know,knowledge,kong,leek,week,wok,woke]
len = 17

Optimal word(s):

main =>
  WordList = "unixdict.txt",
  MinLen = 3,
  MaxLen = 9,
  Words = [Word : Word in read_file_lines(WordList), Word.len >= MinLen, Word.len <= MaxLen],
  TargetWords = [Word : Word in Words, Word.len == MaxLen],
  MaxResWord = [],
  MaxResLen = 0,
  foreach(Word in TargetWords)
    foreach(MustContain in Word.remove_dups)
       Res = word_wheel(Word,Words,MustContain,MinLen, MaxLen),
       Len = Res.len,
       if Len >= MaxResLen then
         if Len == MaxResLen then
            MaxResWord := MaxResWord ++ [[word=Word,char=MustContain]]
         else
            MaxResWord := [[word=Word,char=MustContain]],
            MaxResLen := Len
         end
       end
    end
  end,
  println(maxLResen=MaxResLen),  
  println(maxWord=MaxResWord).
Output:
maxReLen = 215
maxWord = [[word = claremont,char = a],[word = spearmint,char = a]]

PureBasic[edit]

Procedure.b check_word(word$)
  Shared letters$  
  If Len(word$)<3 Or FindString(word$,"k")<1
    ProcedureReturn #False
  EndIf  
  For i=1 To Len(word$)    
    If CountString(letters$,Mid(word$,i,1))<CountString(word$,Mid(word$,i,1))
      ProcedureReturn #False
    EndIf
  Next
  ProcedureReturn #True  
EndProcedure

If ReadFile(0,"./Data/unixdict.txt")  
  txt$=LCase(ReadString(0,#PB_Ascii|#PB_File_IgnoreEOL))
  CloseFile(0)
EndIf

If OpenConsole()    
  letters$="ndeokgelw"
  wordcount=1
  Repeat
    buf$=StringField(txt$,wordcount,~"\n")
    wordcount+1
    If check_word(buf$)=#False
      Continue
    EndIf
    PrintN(buf$) : r+1
  Until buf$=""
  PrintN("- Finished: "+Str(r)+" words found -")
  Input()
EndIf
End
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke
- Finished: 17 words found -

Python[edit]

import urllib.request
from collections import Counter


GRID = """
N 	D 	E
O 	K 	G
E 	L 	W
"""


def getwords(url='http://wiki.puzzlers.org/pub/wordlists/unixdict.txt'):
    "Return lowercased words of 3 to 9 characters"
    words = urllib.request.urlopen(url).read().decode().strip().lower().split()
    return (w for w in words if 2 < len(w) < 10)

def solve(grid, dictionary):
    gridcount = Counter(grid)
    mid = grid[4]
    return [word for word in dictionary
            if mid in word and not (Counter(word) - gridcount)]


if __name__ == '__main__':
    chars = ''.join(GRID.strip().lower().split())
    found = solve(chars, dictionary=getwords())
    print('\n'.join(found))
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke


Or, using a local copy of the dictionary, and a recursive test of wheel fit:

'''Word wheel'''

from os.path import expanduser


# gridWords :: [String] -> [String] -> [String]
def gridWords(grid):
    '''The subset of words in ws which contain the
       central letter of the grid, and can be completed
       by single uses of some or all of the remaining
       letters in the grid.
    '''
    def go(ws):
        cs = ''.join(grid).lower()
        wheel = sorted(cs)
        wset = set(wheel)
        mid = cs[4]
        return [
            w for w in ws
            if 2 < len(w) and (mid in w) and (
                all(c in wset for c in w)
            ) and wheelFit(wheel, w)
        ]
    return go


# wheelFit :: String -> String -> Bool
def wheelFit(wheel, word):
    '''True if a given word can be constructed
       from (single uses of) some subset of
       the letters in the wheel.
    '''
    def go(ws, cs):
        return True if not cs else (
            False if not ws else (
                go(ws[1:], cs[1:]) if ws[0] == cs[0] else (
                    go(ws[1:], cs)
                )
            )
        )
    return go(wheel, sorted(word))


# -------------------------- TEST --------------------------
# main :: IO ()
def main():
    '''Word wheel matches for a given grid in a copy of
       http://wiki.puzzlers.org/pub/wordlists/unixdict.txt
    '''
    print('\n'.join(
        gridWords(['NDE', 'OKG', 'ELW'])(
            readFile('~/unixdict.txt').splitlines()
        )
    ))


# ------------------------ GENERIC -------------------------

# readFile :: FilePath -> IO String
def readFile(fp):
    '''The contents of any file at the path
       derived by expanding any ~ in fp.
    '''
    with open(expanduser(fp), 'r', encoding='utf-8') as f:
        return f.read()


# MAIN ---
if __name__ == '__main__':
    main()
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

q[edit]

ce:count each
lc:ce group@                                              / letter count
dict:"\n"vs .Q.hg "http://wiki.puzzlers.org/pub/wordlists/unixdict.txt"
// dictionary of 3-9 letter words
d39:{x where(ce x)within 3 9}{x where all each x in .Q.a}dict

solve:{[grid;dict]
  i:where(grid 4)in'dict;
  dict i where all each 0<=(lc grid)-/:lc each dict i }[;d39]
q)`$solve "ndeokglew"
`eke`elk`keel`keen`keg`ken`keno`knee`kneel`knew`know`knowledge`kong`leek`week`wok`woke

A naive solution to the second question is simple

bust:{[dict]
  grids:distinct raze(til 9)rotate\:/:dict where(ce dict)=9;
  wc:(count solve@)each grids;
  grids where wc=max wc }

but inefficient. Better:

best:{[dict]
  dlc:lc each dict;                                      / letter counts of dictionary words
  ig:where(ce dict)=9;                                   / find grids (9-letter words)
  igw:where each(all'')0<=(dlc ig)-/:\:dlc;              / find words composable from each grid (length ig)
  grids:raze(til 9)rotate\:/:dict ig;                    / 9 permutations of each grid
  iaz:(.Q.a)!where each .Q.a in'\:dict;                  / find words containing a, b, c etc
  ml:4 rotate'dict ig;                                   / mid letters for each grid
  wc:ce raze igw inter/:'iaz ml;                         / word counts for grids
  distinct grids where wc=max wc }                       / grids with most words
q)show w:best d39
"ntclaremo"
"tspearmin"

q)ce solve each w
215 215

Full discussion at code.kx.com

Raku[edit]

Works with: Rakudo version 2020.05

Everything is adjustable through command line parameters.

Defaults to task specified wheel, unixdict.txt, minimum 3 letters.

Using Terminal::Boxer from the Raku ecosystem.

use Terminal::Boxer;

my %*SUB-MAIN-OPTS = :named-anywhere;

unit sub MAIN ($wheel = 'ndeokgelw', :$dict = './unixdict.txt', :$min = 3);

my $must-have = $wheel.comb[4].lc;

my $has = $wheel.comb».lc.Bag;

my %words;
$dict.IO.slurp.words».lc.map: {
    next if not .contains($must-have) or .chars < $min;
    %words{.chars}.push: $_ if .comb.Bag$has;
};

say "Using $dict, minimum $min letters.";

print rs-box :3col, :3cw, :indent("\t"), $wheel.comb».uc;

say "{sum %words.values».elems} words found";

printf "%d letters:  %s\n", .key, .value.sort.join(', ') for %words.sort;
Output:
Using defaults
raku word-wheel.raku
Using ./unixdict.txt, minimum 3 letters.
	╭───┬───┬───╮
	│ N │ D │ E │
	├───┼───┼───┤
	│ O │ K │ G │
	├───┼───┼───┤
	│ E │ L │ W │
	╰───┴───┴───╯
17 words found
3 letters:  eke, elk, keg, ken, wok
4 letters:  keel, keen, keno, knee, knew, know, kong, leek, week, woke
5 letters:  kneel
9 letters:  knowledge
Larger dictionary

Using the much larger dictionary words.txt file from https://github.com/dwyl/english-words

raku word-wheel.raku --dict=./words.txt
Using ./words.txt, minimum 3 letters.
	╭───┬───┬───╮
	│ N │ D │ E │
	├───┼───┼───┤
	│ O │ K │ G │
	├───┼───┼───┤
	│ E │ L │ W │
	╰───┴───┴───╯
86 words found
3 letters:  dkg, dkl, eek, egk, eke, ekg, elk, gok, ked, kee, keg, kel, ken, keo, kew, kln, koe, kol, kon, lek, lgk, nek, ngk, oke, owk, wok
4 letters:  deek, deke, doek, doke, donk, eked, elke, elko, geek, genk, gonk, gowk, keel, keen, keld, kele, kend, keno, keon, klee, knee, knew, know, koel, koln, kone, kong, kwon, leek, leke, loke, lonk, okee, oken, week, welk, woke, wolk, wonk
5 letters:  dekle, dekow, gleek, kedge, kendo, kleon, klong, kneed, kneel, knowe, konde, oklee, olnek, woken
6 letters:  gowked, keldon, kelwen, knowle, koleen
8 letters:  weeklong
9 letters:  knowledge
Top 5 maximum word wheels with at least one 9 letter word

Using unixdict.txt:

Wheel		words
eimnaprst:	215
celmanort:	215
ceimanrst:	210
elmnaoprt:	208
ahlneorst:	201

Using words.txt:

Wheel		words
meilanrst:	1329
deilanrst:	1313
ceilanrst:	1301
peilanrst:	1285
geilanrst:	1284

REXX[edit]

Quite a bit of boilerplate was included in this REXX example.

No assumption was made as the "case" of the words (upper/lower/mixed case).   Duplicate words were detected and
eliminated   (god and God),   as well as words that didn't contain all Roman (Latin) letters.

The number of minimum letters can be specified,   as well as the dictionary fileID and the letters in the word wheel (grid).

Additional information is also provided concerning how many words have been skipped due to the various filters.

/*REXX pgm finds (dictionary) words which can be found in a specified word wheel (grid).*/
parse arg grid minL iFID .                       /*obtain optional arguments from the CL*/
if grid==''|grid==","  then grid= 'ndeokgelw'    /*Not specified?  Then use the default.*/
if minL==''|minL==","  then minL= 3              /* "      "         "   "   "     "    */
if iFID==''|iFID==","  then iFID= 'UNIXDICT.TXT' /* "      "         "   "   "     "    */
oMinL= minL;                minL= abs(minL)      /*if negative, then don't show a list. */
gridU= grid;  upper gridU                        /*get an uppercase version of the grid.*/
Lg= length(grid);           Hg= Lg % 2  +  1     /*get length of grid & the middle char.*/
ctr= substr(grid, Hg, 1);   upper ctr            /*get uppercase center letter in grid. */
wrds= 0                                          /*# words that are in the dictionary.  */
wees= 0                                          /*"   "     "   "  too short.          */
bigs= 0                                          /*"   "     "   "  too long.           */
dups= 0                                          /*"   "     "   "  duplicates.         */
ills= 0                                          /*"   "     "   contain  "not" letters.*/
good= 0                                          /*"   "     "   contain center letter. */
nine= 0                                          /*" wheel─words that contain 9 letters.*/
say '                                Reading the file: ' iFID         /*align the text. */
@.= .                                            /*uppercase non─duplicated dict. words.*/
$=                                               /*the list of dictionary words in grid.*/
     do recs=0  while lines(iFID)\==0            /*process all words in the dictionary. */
     u= space( linein(iFID), 0);   upper u       /*elide blanks;  uppercase the word.   */
     L= length(u)                                /*obtain the length of the word.       */
     if @.u\==.           then do; dups= dups+1; iterate; end  /*is this a duplicate?   */
     if L<minL            then do; wees= wees+1; iterate; end  /*is the word too short? */
     if L>Lg              then do; bigs= bigs+1; iterate; end  /*is the word too long?  */
     if \datatype(u,'M')  then do; ills= ills+1; iterate; end  /*has word non─letters?  */
     @.u=                                        /*signify that  U  is a dictionary word*/
     wrds= wrds + 1                              /*bump the number of "good" dist. words*/
     if pos(ctr, u)==0        then iterate       /*word doesn't have center grid letter.*/
     good= good + 1                              /*bump # center─letter words in dict.  */
     if verify(u, gridU)\==0  then iterate       /*word contains a letter not in grid.  */
     if pruned(u, gridU)      then iterate       /*have all the letters not been found? */
     if L==9  then nine= nine + 1                /*bump # words that have nine letters. */
     $= $ u                                      /*add this word to the "found" list.   */
     end   /*recs*/
say
say '    number of  records (words) in the dictionary: '   right( commas(recs), 9)
say '    number of ill─formed words in the dictionary: '   right( commas(ills), 9)
say '    number of  duplicate words in the dictionary: '   right( commas(dups), 9)
say '    number of  too─small words in the dictionary: '   right( commas(wees), 9)
say '    number of  too─long  words in the dictionary: '   right( commas(bigs), 9)
say '    number of acceptable words in the dictionary: '   right( commas(wrds), 9)
say '    number center─letter words in the dictionary: '   right( commas(good), 9)
say '    the minimum length of words that can be used: '   right( commas(minL), 9)
say '                the word wheel (grid) being used: '   grid
say '      center of the word wheel (grid) being used: '   right('↑', Hg)
say;  #= words($);   $= strip($)
say '    number of word wheel words in the dictionary: '   right( commas(#   ), 9)
say '    number of   nine-letter   wheel words  found: '   right( commas(nine), 9)
if #==0  |  oMinL<0  then exit #
say
say '    The list of word wheel words found:';   say copies('─', length($));  say lower($)
exit #                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
lower:  arg aa; @='abcdefghijklmnopqrstuvwxyz'; @u=@; upper @u;  return translate(aa,@,@U)
commas: parse arg _;  do jc=length(_)-3  to 1  by -3; _=insert(',', _, jc); end;  return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
pruned: procedure; parse arg aa,gg               /*obtain word to be tested, & the grid.*/
           do n=1  for length(aa);    p= pos( substr(aa,n,1), gg);  if p==0  then return 1
           gg= overlay(., gg, p)                 /*"rub out" the found character in grid*/
           end   /*n*/;               return 0   /*signify that the  AA  passed the test*/
output   when using the default inputs:
                                Reading the file:  UNIXDICT.TXT

    number of  records (lines) in the dictionary:     25,105
    number of ill─formed words in the dictionary:        123
    number of  duplicate words in the dictionary:          0
    number of  too─small words in the dictionary:        159
    number of  too─long  words in the dictionary:      4,158
    number of acceptable words in the dictionary:     20,664
    number center─letter words in the dictionary:      1,630
    the minimum length of words that can be used:          3
                the word wheel (grid) being used:  ndeokgelw
      center of the word wheel (grid) being used:      ↑

    number of word wheel words in the dictionary:         17
    number of   nine-letter   wheel words  found:          1

    The list of word wheel words found:
─────────────────────────────────────────────────────────────────────────────────────
eke elk keel keen keg ken keno knee kneel knew know knowledge kong leek week wok woke

Note:   my "personal" dictionary that I built   (over   915,000   947,359   words),   there are   178   words that are in the (above) word wheel.


output   when using the inputs:     satRELinp   -3

(I am trying for a maximum word wheel count for the   UNIXDICT   dictionary;
the negative minimum word length indicates to   not   list the words found.)
Thanks to userid   Paddy3118,   a better grid was found.

                                Reading the file:  UNIXDICT.TXT

    number of  records (lines) in the dictionary:     25,105
    number of ill─formed words in the dictionary:        123
    number of  duplicate words in the dictionary:          0
    number of  too─small words in the dictionary:        159
    number of  too─long  words in the dictionary:      4,158
    number of acceptable words in the dictionary:     20,664
    number center─letter words in the dictionary:     11,623
    the minimum length of words that can be used:          3
                the word wheel (grid) being used:  satRELinp
      center of the word wheel (grid) being used:      ↑

    number of word wheel words in the dictionary:        234
    number of   nine-letter   wheel words  found:          0
output   when using the inputs:     setRALinp   -3

Thanks to userid   Simonjsaunders,   a better grid was found.

                                Reading the file:  UNIXDICT.TXT

    number of  records (words) in the dictionary:     25,104
    number of ill─formed words in the dictionary:        123
    number of  duplicate words in the dictionary:          0
    number of  too─small words in the dictionary:        159
    number of  too─long  words in the dictionary:      4,158
    number of acceptable words in the dictionary:     20,664
    number center─letter words in the dictionary:     10,369
    the minimum length of words that can be used:          3
                the word wheel (grid) being used:  setRALinp
      center of the word wheel (grid) being used:      ↑

    number of word wheel words in the dictionary:        248
    number of   nine-letter   wheel words  found:          0

Ruby[edit]

wheel  = "ndeokgelw"
middle, wheel_size = wheel[4], wheel.size

res = File.open("unixdict.txt").each_line.select do |word|
  w = word.chomp
  next unless w.size.between?(3, wheel_size)  
  next unless w.match?(middle)
  wheel.each_char{|c| w.sub!(c, "") } #sub! substitutes only the first occurrence (gsub would substitute all)
  w.empty?
end

puts res
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Transd[edit]

#lang transd

MainModule: {
  maxwLen: 9,
  minwLen: 3,
  dict: Vector<String>(),
  subWords: Vector<String>(),

  procGrid: (λ grid String() cent String() subs Bool()
    (with cnt 0 (sort grid)
    (for w in dict where (and (neq (find w cent) -1)
                              (match w "^[[:alpha:]]+$")) do
      (if (is-subset grid (sort (cp w))) (+= cnt 1) 
        (if subs (append subWords w))
      )
    )
    (ret cnt)
  )),

  _start: (λ locals: res 0 maxRes 0
    (with fs FileStream() 
      (open-r fs "/mnt/proj/res/unixdict.txt")
      (for w in (read-lines fs)
             where (within (size w) minwLen maxwLen) do
        (append dict w))
    )

    (lout "Main part of task:\n")
    (procGrid "ndeokgelw" "k" true)
    (lout "Number of words: " (size subWords) ";\nword list: " subWords)

    (lout "\n\nOptional part of task:\n")
    (for w in dict where (eq (size w) maxwLen) do
      (for centl in (split (unique (sort (cp w))) "") do
        (if (>= (= res (procGrid (cp w) centl false)) maxRes) (= maxRes res)
          (lout "New max. number: " maxRes ", word: " w ", central letter: " centl)
  ) ) ) )
}
Output:
Main part of task:

Number of words: 17;
word list: ["eke", "elk", "keel", "keen", "keg", "ken", "keno", "knee", "kneel", "knew", "know", "knowledge", "kong", "leek", "week", "wok", "woke"]


Optional part of task:

New max. number: 100, word: abdominal, central letter: a
New max. number: 117, word: abernathy, central letter: a
New max. number: 119, word: abhorrent, central letter: r
New max. number: 121, word: absorbent, central letter: e
New max. number: 123, word: adsorbate, central letter: a
New max. number: 125, word: adventure, central letter: e
New max. number: 155, word: advertise, central letter: e
New max. number: 161, word: alongside, central letter: a
New max. number: 170, word: alongside, central letter: l
New max. number: 182, word: ancestral, central letter: a
New max. number: 182, word: arclength, central letter: a
New max. number: 185, word: beplaster, central letter: e
New max. number: 215, word: claremont, central letter: a
New max. number: 215, word: spearmint, central letter: a

Wren[edit]

Library: Wren-sort
Library: Wren-seq
import "io" for File
import "/sort" for Sort, Find
import "/seq" for Lst

var letters = ["d", "e", "e", "g", "k", "l", "n", "o","w"]

var words = File.read("unixdict.txt").split("\n")
// get rid of words under 3 letters or over 9 letters
words = words.where { |w| w.count > 2 && w.count < 10 }.toList
var found = []
for (word in words) {
    if (word.indexOf("k") >= 0) {
        var lets = letters.toList
        var ok = true
        for (c in word) {
            var ix = Find.first(lets, c)
            if (ix == - 1) {
                ok = false
                break
            }
            lets.removeAt(ix)
        }
        if (ok) found.add(word)
    }
}

System.print("The following %(found.count) words are the solutions to the puzzle:")
System.print(found.join("\n"))

// optional extra
var mostFound = 0
var mostWords9 = []
var mostLetters = []
// iterate through all 9 letter words in the dictionary
for (word9 in words.where { |w| w.count == 9 }) {
    letters = word9.toList
    Sort.insertion(letters)
    // get distinct letters
    var distinctLetters = Lst.distinct(letters)
    // place each distinct letter in the middle and see what we can do with the rest
    for (letter in distinctLetters) {
        found = 0
        for (word in words) {
            if (word.indexOf(letter) >= 0) {
                var lets = letters.toList
                var ok = true
                for (c in word) {
                    var ix = Find.first(lets, c)
                    if (ix == - 1) {
                        ok = false
                        break
                    }
                    lets.removeAt(ix)
                }
                if (ok) found = found + 1
            }
        }
        if (found > mostFound) {
            mostFound = found
            mostWords9 = [word9]
            mostLetters = [letter]
        } else if (found == mostFound) {
            mostWords9.add(word9)
            mostLetters.add(letter)
        }
    }
}
System.print("\nMost words found = %(mostFound)")
System.print("Nine letter words producing this total:")
for (i in 0...mostWords9.count) {
    System.print("%(mostWords9[i]) with central letter '%(mostLetters[i])'")
}
Output:
The following 17 words are the solutions to the puzzle:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke

Most words found = 215
Nine letter words producing this total:
claremont with central letter 'a'
spearmint with central letter 'a'

XPL0[edit]

string  0;      \use zero-terminated strings
int     I, Set, HasK, HasOther, HasDup, ECnt, Ch;
char    Word(25);
def     LF=$0A, CR=$0D, EOF=$1A;
[FSet(FOpen("unixdict.txt", 0), ^I);
OpenI(3);
repeat  I:= 0;  HasK:= false;  HasOther:= false;
        ECnt:= 0;  Set:= 0;  HasDup:= false;
        loop    [repeat Ch:= ChIn(3) until Ch # CR;     \remove possible CR
                if Ch=LF or Ch=EOF then quit;
                Word(I):= Ch;
                I:= I+1;
                if Ch = ^k then HasK:= true;
                case Ch of ^k,^n,^d,^e,^o,^g,^l,^w: []  \assume all lowercase
                other HasOther:= true;
                if Ch = ^e then ECnt:= ECnt+1
                else    [if Set & 1<<(Ch-^a) then HasDup:= true;
                        Set:= Set ! 1<<(Ch-^a);
                        ];
                ];
        Word(I):= 0;                                    \terminate string
        if I>=3 & HasK & ~HasOther & ~HasDup & ECnt<=2 then
                [Text(0, Word);  CrLf(0);
                ];
until   Ch = EOF;
]
Output:
eke
elk
keel
keen
keg
ken
keno
knee
kneel
knew
know
knowledge
kong
leek
week
wok
woke