Teacup rim text: Difference between revisions
(→Functional Python: Updated word source, formatted output.) |
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# Perhaps add one variant as groupByComparing ? (across JS, AS too ?) |
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# groupBy :: (a -> b) -> [a] -> [[a]] |
# groupBy :: (a -> b) -> [a] -> [[a]] |
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def groupBy(f): |
def groupBy(f): |
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Revision as of 11:32, 5 August 2019
On a set of coasters we have, there's a picture of a teacup. On the rim of the teacup the word "TEA" appears a number of times separated by bullet characters. It occurred to me that if the bullet were removed and the words run together, you could start at any letter and still end up with a meaningful three-letter word. So start at the "T" and read "TEA". Start at the "E" and read "EAT", or start at the "A" and read "ATE".
That got me thinking that maybe there are other words that could be used rather that "TEA". And that's just English. What about Italian or Greek or ... um ... Telugu. For English, use the MIT 10000 word list located at https://www.mit.edu/~ecprice/wordlist.10000
So here's the task: from a web accessible or locally accessible word source, iterate through each word of length 3 or more. With each word, peel off the first letter and put it at the end. Check if the word exists. If it does, keep going with the next letter, repeating the process for as many letters as there are minus one. If all of the words exist store the original word. List each word that survives the filtration process along with all its variants. Having listed a set, for example [ate tea eat], resist displaying permutations of that set, e.g. [eat tea ate] etc.
Factor
<lang factor>USING: fry hash-sets http.client kernel math prettyprint sequences sequences.extras sets sorting splitting ;
"https://www.mit.edu/~ecprice/wordlist.10000" http-get nip "\n" split [ length 2 > ] filter [ [ all-rotations ] map ] [ >hash-set ] bi '[ [ _ in? ] all? ] filter [ natural-sort ] map members .</lang>
- Output:
{
{ "aaa" "aaa" "aaa" }
{ "aim" "ima" "mai" }
{ "arc" "car" "rca" }
{ "asp" "pas" "spa" }
{ "ate" "eat" "tea" }
{ "iii" "iii" "iii" }
{ "ips" "psi" "sip" }
{ "ooo" "ooo" "ooo" }
{ "www" "www" "www" }
{ "xxx" "xxx" "xxx" }
}
Go
<lang go>package main
import (
"bufio" "fmt" "log" "os" "sort" "strings"
)
func check(err error) {
if err != nil {
log.Fatal(err)
}
}
func readWords(fileName string) []string {
file, err := os.Open(fileName)
check(err)
defer file.Close()
var words []string
scanner := bufio.NewScanner(file)
for scanner.Scan() {
word := strings.ToLower(strings.TrimSpace(scanner.Text()))
if len(word) >= 3 {
words = append(words, word)
}
}
check(scanner.Err())
return words
}
func rotate(runes []rune) {
first := runes[0] copy(runes, runes[1:]) runes[len(runes)-1] = first
}
func main() {
words := readWords("mit_10000.txt") // using local copy
n := len(words)
used := make(map[string]bool)
outer:
for _, word := range words {
runes := []rune(word)
variants := []string{word}
for i := 0; i < len(runes)-1; i++ {
rotate(runes)
word2 := string(runes)
if used[word2] {
continue outer
}
ix := sort.SearchStrings(words, word2)
if ix == n || words[ix] != word2 {
continue outer
}
variants = append(variants, word2)
}
for _, variant := range variants {
used[variant] = true
}
fmt.Println(variants)
}
}</lang>
- Output:
[aaa aaa aaa] [aim ima mai] [arc rca car] [asp spa pas] [ate tea eat] [iii iii iii] [ips psi sip] [ooo ooo ooo] [www www www] [xxx xxx xxx]
Haskell
Circular words of more than 2 characters in a local copy of unixdict.txt <lang haskell>import Control.Applicative (liftA2) import qualified Data.Set as S
main :: IO () main = readFile "unixdict.txt" >>= (print . circularWords . lines)
circularWords :: [String] -> [String] circularWords ws =
let lexicon = S.fromList ws in filter (isCircular lexicon) ws
isCircular :: S.Set String -> String -> Bool isCircular lex w =
2 < length w && all (`S.member` lex) (rotations w)
rotations :: [a] -> a rotations = liftA2 fmap rotated (enumFromTo 0 . pred . length)
rotated :: [a] -> Int -> [a] rotated [] _ = [] rotated xs n = zipWith const (drop n (cycle xs)) xs</lang>
- Output:
["aaa","apt","arc","ate","car","eat","iii","pta","rca","tap","tea"]
JavaScript
Reading a local dictionary with a macOS JS for Automation library:
<lang javascript>(() => {
'use strict';
// main :: IO ()
const main = () =>
showLog(
circularWords(
lines(readFile('~/unixdict.txt'))
)
);
// circularWords :: [String] -> [String]
const circularWords = ws =>
ws.filter(isCircular(new Set(ws)), ws);
// isCircular :: Set String -> String -> Bool
const isCircular = lexicon => w => {
const iLast = w.length - 1;
return 1 < iLast && until(
([i, bln, s]) => iLast < i || !bln,
([i, bln, s]) => [1 + i, lexicon.has(s), rotated(s)],
[0, true, w]
)[1];
}
// MAC OS JS FOR AUTOMATION ---------------------------
// readFile :: FilePath -> IO String
const readFile = fp => {
const
e = $(),
uw = ObjC.unwrap,
s = uw(
$.NSString.stringWithContentsOfFileEncodingError(
$(fp)
.stringByStandardizingPath,
$.NSUTF8StringEncoding,
e
)
);
return undefined !== s ? (
s
) : uw(e.localizedDescription);
};
// GENERIC FUNCTIONS ----------------------------------
// lines :: String -> [String] const lines = s => s.split(/[\r\n]/);
// rotated :: String -> String
const rotated = xs =>
xs.slice(1) + xs[0];
// showLog :: a -> IO ()
const showLog = (...args) =>
console.log(
args
.map(JSON.stringify)
.join(' -> ')
);
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};
// MAIN --- return main();
})();</lang>
- Output:
["aaa","apt","arc","ate","car","eat","iii","pta","rca","tap","tea"]
Julia
Using the MIT 10000 word list, and excluding words of less than three letters, to reduce output length. <lang julia>using HTTP
function getwords()
req = HTTP.request("GET", "https://www.mit.edu/~ecprice/wordlist.10000")
Dict{String, Int}((string(x), 1) for x in split(String(req.body), r"\s+"))
end
rotate(s, n) = String(circshift(Vector{UInt8}(s), n))
isliketea(w, d) = (n = length(w); n > 2 && all(i -> haskey(d, rotate(w, i)), 1:n-1))
function getteawords()
wordlistdict = getwords()
for word in collect(keys(wordlistdict))
if isliketea(word, wordlistdict)
println(word, ": ", [rotate(word, i) for i in 1:length(word)-1])
end
end
end
getteawords()
</lang>
- Output:
pas: ["spa", "asp"] xxx: ["xxx", "xxx"] iii: ["iii", "iii"] asp: ["pas", "spa"] tea: ["ate", "eat"] spa: ["asp", "pas"] ate: ["eat", "tea"] aim: ["mai", "ima"] aaa: ["aaa", "aaa"] car: ["rca", "arc"] ooo: ["ooo", "ooo"] sip: ["psi", "ips"] arc: ["car", "rca"] ips: ["sip", "psi"] www: ["www", "www"] mai: ["ima", "aim"] rca: ["arc", "car"] eat: ["tea", "ate"] psi: ["ips", "sip"] ima: ["aim", "mai"]
Lychen
Lychen is V8 JavaScript wrapped in C#, exposing C# into JavaScript.
Using https://www.mit.edu/~ecprice/wordlist.10000 as per the Julia example.
<lang javascript> const wc = new CS.System.Net.WebClient(); const lines = wc.DownloadString("https://www.mit.edu/~ecprice/wordlist.10000"); const words = lines.split(/\n/g); const collection = {}; words.filter(word => word.length > 2).forEach(word => {
let allok = true;
let newword = word;
for (let i = 0; i < word.length - 1; i++) {
newword = newword.substr(1) + newword.substr(0, 1);
if (!words.includes(newword)) {
allok = false;
break;
}
}
if (allok) {
const key = word.split("").sort().join("");
if (!collection[key]) {
collection[key] = [word];
} else {
if (!collection[key].includes(word)) {
collection[key].push(word);
}
}
}
}); Object.keys(collection) .filter(key => collection[key].length > 1) .forEach(key => console.log("%s", collection[key].join(", "))); </lang>
aim, ima, mai arc, car, rca asp, pas, spa ate, eat, tea ips, psi, sip
Perl 6
Much of the previous exposition here is superfluous as the reference word list has changed.
There doesn't seem to be any restriction that the word needs to consist only of lowercase letters, so words of any case are included. Since the example code specifically shows the example words (TEA, EAT, ATE) in uppercase, I elected to uppercase the found words.
<lang perl6>my %words; './wordlist.10000'.IO.slurp.words.map: { .chars < 3 ?? (next) !! %words{.uc.comb.sort.join}.push: .uc };
my @teacups; my %seen;
for %words.values -> @these {
MAYBE: for @these {
my $maybe = $_;
next if %seen{$_};
my @print;
for ^$maybe.chars {
if $maybe ∈ @these {
@print.push: $maybe;
$maybe = $maybe.comb.list.rotate.join;
} else {
@print = ();
last MAYBE
}
}
if @print.elems {
@teacups.push: @print;
%seen{$_}++ for @print;
}
}
}
say .join(", ") for sort @teacups;</lang>
- Output:
AAA, AAA, AAA AIM, IMA, MAI ARC, RCA, CAR ASP, SPA, PAS ATE, TEA, EAT III, III, III IPS, PSI, SIP OOO, OOO, OOO WWW, WWW, WWW XXX, XXX, XXX
Python
Functional
Composing generic functions, and taking only circular words of more than two characters. <lang python>Teacup rim text
from itertools import groupby from os.path import expanduser
- main :: IO ()
def main():
Circular words of more than two characters
in a local copy of unixdict.txt
print(
showGroups(
circularWords(
# Local copy of
# https://www.mit.edu/~ecprice/wordlist.10000
lines(readFile('~/mitWords.txt'))
)
)
)
- circularWords :: [String] -> [String]
def circularWords(ws):
The subset of those words in the given list
which are circular.
lexicon = set(ws)
return list(filter(isCircular(lexicon), ws))
- isCircular :: Set String -> String -> Bool
def isCircular(lexicon):
True if the given word contains more than
two characters, and all of its rotations
are found in the lexicon.
def go(w):
def f(tpl):
(i, _, x) = tpl
return (1 + i, x in lexicon, rotated(x))
iLast = len(w) - 1
return 1 < iLast and until(
lambda tpl: iLast < tpl[0] or (not tpl[1])
)(f)(
(0, True, w)
)[1]
return lambda s: go(s)
- DISPLAY -------------------------------------------------
- showGroups :: [String] -> String
def showGroups(xs):
List of groups of circular words.
return unlines([
gp[0][0] + (
' -> ' + ', '.join(
[snd(x) for x in gp[1:]]
) if 1 < len(gp) else
)
for gp in groupBy(fst)(
sorted(
[(.join(sorted(x)), x) for x in xs],
key=fst
)
)
])
- GENERIC -------------------------------------------------
- fst :: (a, b) -> a
def fst(tpl):
First member of a pair. return tpl[0]
- groupBy :: (a -> b) -> [a] -> a
def groupBy(f):
The elements of xs grouped,
preserving order, by equality
in terms of the key function f.
return lambda xs: [
list(x[1]) for x in groupby(xs, key=f)
]
- lines :: String -> [String]
def lines(s):
A list of strings,
(containing no newline characters)
derived from a single new-line delimited string.
return s.splitlines()
- 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()
- rotated :: String -> String
def rotated(s):
A list rotated 1 character to the right. return s[1:] + s[0]
- snd :: (a, b) -> b
def snd(tpl):
Second member of a pair. return tpl[1]
- unlines :: [String] -> String
def unlines(xs):
A single string formed by the intercalation
of a list of strings with the newline character.
return '\n'.join(xs)
- until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
The result of repeatedly applying f until p holds.
The initial seed value is x.
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
aaa acr -> car, rca aet -> eat, tea aim -> ima, mai aps -> pas, spa iii ips -> psi, sip ooo www xxx
REXX
All words that contained non─letter (Latin) characters (periods, decimal digits, minus signs, underbars, or embedded blanks) weren't considered as candidates for circular words.
The dictionary wasn't assumed to be sorted in any way. <lang rexx>/*REXX pgm finds circular words (length>2), using a dictionary, suppress permutations.*/ parse arg iFID L . /*obtain optional arguments from the CL*/ if iFID==|iFID=="," then iFID= 'wordlist.10k' /*Not specified? Then use the default.*/ if L==| L=="," then L= 3 /* " " " " " " */
- = 0 /*number of words in dictionary, Len>L.*/
@.= /*stemmed array of non─duplicated words*/
do r=0 while lines(iFID)\==0 /*read all lines (words) in dictionary.*/
z= space( linein(iFID) ) /*obtain get a word from the dictionary*/
if length(z)<L | @.z\== then iterate /*length must be L or more, no dups.*/
if \datatype(z, 'M') then iterate /*Word contains non-letters? Then skip*/
@.z = z /*assign a word from the dictionary. */
#= # + 1; $.#= z /*bump word count; append word to list.*/
end /*r*/ /* [↑] dictionary need not be sorted. */
say "There're " r ' entries in the dictionary (of all types): ' iFID say "There're " # ' words in the dictionary of at least length ' L say cw= 0 /*the number of circular words (so far)*/
do j=1 for #; x= $.j; y= x /*obtain the Jth word in the list. */
if x== then iterate /*if a null, don't show variants. */
yy= y /*the start of a list of the variants. */
do k=1 for length(x)-1 /*"circulate" the litters in the word. */
y= substr(y, 2)left(y, 1) /*add the left letter to the right end.*/
if @.y== then iterate j /*if not a word, then skip this word. */
yy= yy',' y /*append to the list of the variants. */
if y\==x then @.y= /*nullify word to suppress permutations*/
end /*k*/ /* [↓] ··· except for monolithic words.*/
cw= cw + 1 /*bump counter of circular words found.*/
say 'circular word: ' yy /*display a circular word to the term. */
end /*j*/
say say cw ' circular words were found.' /*stick a fork in it, we're all done. */</lang>
- output when using the default inputs:
There're 10000 entries in the dictionary (of all types): wordlist.10k There're 9578 words in the dictionary of at least length 3 circular word: aaa, aaa, aaa circular word: aim, ima, mai circular word: arc, rca, car circular word: asp, spa, pas circular word: ate, tea, eat circular word: iii, iii, iii circular word: ips, psi, sip circular word: ooo, ooo, ooo circular word: www, www, www circular word: xxx, xxx, xxx 10 circular words were found.
zkl
<lang zkl>// this is limited to the max items a Dictionary can hold words:=File("mit_wordlist_10000.txt").pump(Dictionary().add.fp1(True),"strip"); seen :=Dictionary(); foreach word in (words.keys){
rots,w,sz := List(), word, word.len();
if(sz>2 and not seen.holds(word)){
do(sz-1){
w=String(w[-1],w[0,-1]); // rotate one character if(not words.holds(w)) continue(2); // not a word, do next word rots.append(w); // I'd like to see all the rotations
}
println(rots.append(word).sort().concat(" "));
rots.pump(seen.add.fp1(True)); // we've seen these rotations
}
}</lang>
- Output:
www www www asp pas spa ips psi sip iii iii iii ate eat tea aaa aaa aaa aim ima mai arc car rca xxx xxx xxx ooo ooo ooo