Word ladder: Difference between revisions
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It also assumes that the dictionary file is in mixed case as well as the words entered on the CL.
To treat the dictionary and input words as caseless, all words are translated to lowercase.
Programming note: this REXX program uses the '''lower''' BIF which Regina has).
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L= length(base) /*length of the BASE (in characters). */
if L<2 then call err 'base word is too small or missing' /*oops, too small*/
if length(targ)\==L then call
call letters /*assign letters, faster than SUBSTR. */
#= 0 /*# of words whose length matches BASE.*/
@.= /*default value of any dictionary word.*/
!.= 0
say copies('─', 30) recs "words in the dictionary file: " iFID
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rung= targ
$= base
do f=1 for
end /*f*/
say
if f>
do f-2;
end /*forever*/
end /*f-2*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
show:
letters:
/*──────────────────────────────────────────────────────────────────────────────────────*/
look: procedure expose @. !. a. $ abc
$$=;
do i=1 for words($);
do k=1 for L
do n=1 for 26; z= overlay(a.n, y, k) /*change a letter*/
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if !.z then iterate /* " " a repeat? " " " */
if z==rungs then rung= y rung /*prepend a word to the rung list. */
if z==rungs &
if z==targ then return z
$$= $$ z /*append a possible ladder word to $$*/
Line 357 ⟶ 356:
<pre>
────────────────────────────── 25104 words in the dictionary file: unixdict.txt
────────────────────────────── 2187 words in the dictionary file of length:
────────────────────────────── base word is: girl
────────────────────────────── target word is: lady
Line 373 ⟶ 372:
<pre>
────────────────────────────── 25104 words in the dictionary file: unixdict.txt
────────────────────────────── 2187 words in the dictionary file of length:
────────────────────────────── base word is: john
────────────────────────────── target word is: jane
Line 387 ⟶ 386:
<pre>
────────────────────────────── 25104 words in the dictionary file: unixdict.txt
────────────────────────────── 3161 words in the dictionary file of length:
────────────────────────────── base word is: child
────────────────────────────── target word is: adult
|
Revision as of 16:58, 6 June 2021
Yet another shortest path problem. Given two words of equal length the task is to transpose the first into the second.
Only one letter may be changed at a time and the change must result in a word in unixdict, the minimum number of intermediate words should be used.
Demonstrate the following:
A boy can be made into a man: boy -> bay -> ban -> man
With a little more difficulty a girl can be made into a lady: girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady
For the wokes, john can be made into jane: john -> cohn -> conn -> cone -> cane -> jane
A child can not be turned into an adult.
Optional transpositions of your choice.
C++
This borrows heavily from Wren and a bit from Raku. <lang cpp>#include <algorithm>
- include <fstream>
- include <iostream>
- include <map>
- include <string>
- include <vector>
using word_map = std::map<size_t, std::vector<std::string>>;
// Returns true if strings s1 and s2 differ by one character. bool one_away(const std::string& s1, const std::string& s2) {
if (s1.size() != s2.size()) return false; int sum = 0; for (size_t i = 0, n = s1.size(); i != n; ++i) { if (s1[i] != s2[i]) { if (++sum > 1) return false; } } return sum == 1;
}
// Join a sequence of strings into a single string using the given separator. template <typename iterator_type, typename separator_type> std::string join(iterator_type begin, iterator_type end,
separator_type separator) { std::string result; if (begin != end) { result += *begin++; for (; begin != end; ++begin) { result += separator; result += *begin; } } return result;
}
// Return true if v contains e. template <typename vector_type, typename element_type> bool contains(const vector_type& v, const element_type& e) {
return std::find(v.begin(), v.end(), e) != v.end();
}
// If possible, print the shortest chain of single-character modifications that // leads from "from" to "to", with each intermediate step being a valid word. // This is an application of breadth-first search. bool word_ladder(const word_map& words, const std::string& from,
const std::string& to) { auto w = words.find(from.size()); if (w != words.end()) { auto poss = w->second; std::vector<std::vector<std::string>> queueTemplate:From; while (!queue.empty()) { auto curr = queue.front(); queue.erase(queue.begin()); std::vector<std::string> next; for (const std::string& str : poss) { if (one_away(str, curr.back())) next.push_back(str); } if (contains(next, to)) { curr.push_back(to); std::cout << join(curr.begin(), curr.end(), " -> ") << '\n'; return true; } poss.erase(std::remove_if(poss.begin(), poss.end(), [&next](const std::string& str) { return contains(next, str); }), poss.end()); for (const auto& str : next) { std::vector<std::string> temp(curr); temp.push_back(str); queue.push_back(std::move(temp)); } } } std::cout << from << " into " << to << " cannot be done.\n"; return false;
}
int main() {
word_map words; std::ifstream in("unixdict.txt"); if (!in) { std::cerr << "Cannot open file unixdict.txt.\n"; return EXIT_FAILURE; } std::string word; while (getline(in, word)) words[word.size()].push_back(word); word_ladder(words, "boy", "man"); word_ladder(words, "girl", "lady"); word_ladder(words, "john", "jane"); word_ladder(words, "child", "adult"); word_ladder(words, "cat", "dog"); word_ladder(words, "lead", "gold"); word_ladder(words, "white", "black"); word_ladder(words, "bubble", "tickle"); return EXIT_SUCCESS;
}</lang>
- Output:
boy -> bay -> ban -> man girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady john -> cohn -> conn -> cone -> cane -> jane child into adult cannot be done. cat -> cot -> cog -> dog lead -> load -> goad -> gold white -> whine -> chine -> chink -> clink -> blink -> blank -> black bubble -> babble -> gabble -> garble -> gargle -> gaggle -> giggle -> jiggle -> jingle -> tingle -> tinkle -> tickle
F#
<lang fsharp> // Word ladder: Nigel Galloway. June 5th., 2021 open System.Text.RegularExpressions let fN g=List.init(String.length g)(fun n->let g=g.ToCharArray() in g.[n]<-'.'; g|>System.String)|>String.concat "|" let fG n g=let g=fN g in n|>List.partition(fun n->Regex.IsMatch(n,g)) let wL n g=let dict=seq{use n=System.IO.File.OpenText("unixdict.txt") in while not n.EndOfStream do yield n.ReadLine()}|>Seq.filter(Seq.length>>(=)(Seq.length n))|>List.ofSeq|>List.except [n]
let (|Done|_|) n=n|>List.tryFind((=)g) let rec wL n g l=match n with h::t->let i,e=fG l (List.head h) in match i with Done i->Some((i::h)|>List.rev) |_->wL t ((i|>List.map(fun i->i::h))@g) e |_->match g with []->None |_->wL g [] l let i,e=fG dict n in match i with Done i->Some([n;g]) |_->wL(i|>List.map(fun g->[g;n])) [] e
[("boy","man");("girl","lady");("john","jane");("child","adult")]|>List.iter(fun(n,g)->printfn "%s" (match wL n g with Some n->n|>String.concat " -> " |_->n+" into "+g+" can't be done")) </lang>
- Output:
boy -> bay -> ban -> man girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady john -> cohn -> conn -> cone -> cane -> jane child into adult can't be done
Julia
<lang julia>const dict = Set(split(read("unixdict.txt", String), r"\s+"))
function targeted_mutations(str::AbstractString, target::AbstractString)
working, tried = str, Set{String}() while all(a -> a[end] != target, working) newworking = Vector{Vector{String}}() for arr in working s = arr[end] push!(tried, s) for j in 1:length(s), c in 'a':'z' w = s[1:j-1] * c * s[j+1:end] if w in dict && !(w in tried) push!(newworking, [arr; w]) end end end isempty(newworking) && return "This cannot be done." working = newworking end return filter(a -> a[end] == target, working)
end
println("boy to man: ", targeted_mutations("boy", "man")) println("girl to lady: ", targeted_mutations("girl", "lady")) println("john to jane: ", targeted_mutations("john", "jane")) println("child to adult: ", targeted_mutations("child", "adult"))
</lang>
- Output:
boy to man: [["boy", "bay", "may", "man"], ["boy", "bay", "ban", "man"], ["boy", "bon", "ban", "man"]] girl to lady: [["girl", "gill", "gall", "gale", "gaze", "laze", "lazy", "lady"]] john to jane: [["john", "cohn", "conn", "cone", "cane", "jane"]] child to adult: [["This cannot be done."]]
Phix
sequence words = get_text("demo/unixdict.txt",GT_LF_STRIPPED) function right_length(string word, integer l) return length(word)=l end function function one_away(string a, b) return sum(sq_ne(a,b))=1 end function procedure word_ladder(string a, b) sequence poss = filter(words,right_length,length(a)), todo = {{a}}, curr -- aka todo[1], word chain starting from a while length(todo) do {curr,todo} = {todo[1],todo[2..$]} sequence next = filter(poss,one_away,curr[$]) if find(b,next) then printf(1,"%s\n",{join(append(curr,b),"->")}) return end if poss = filter(poss,"out",next) todo &= apply(true,append,{{curr},next}) end while printf(1,"%s into %s cannot be done\n",{a,b}) end procedure word_ladder("boy","man") word_ladder("girl","lady") word_ladder("john","jane") word_ladder("child","adult")
Aside: an initial poss = filter(poss,"out",{a}) might be prudent, but would only prevent a single next:={} step, at about the same cost as the initial filter anyway.
- Output:
boy->bay->ban->man girl->gill->gall->gale->gaze->laze->lazy->lady john->cohn->conn->cone->cane->jane child into adult cannot be done
Raku
<lang perl6>constant %dict = 'unixdict.txt'.IO.lines
.classify(*.chars) .map({ .key => .value.Set });
sub word_ladder ( Str $from, Str $to ) {
die if $from.chars != $to.chars;
my $sized_dict = %dict{$from.chars}; my @workqueue = (($from,),); my $used = ($from => True).SetHash; while @workqueue { my @new_q; for @workqueue -> @words { my $last_word = @words.tail; my @new_tails = gather for 'a' .. 'z' -> $replacement_letter { for ^$last_word.chars -> $i { my $new_word = $last_word; $new_word.substr-rw($i, 1) = $replacement_letter;
next unless $new_word ∈ $sized_dict and not $new_word ∈ $used; take $new_word; $used{$new_word} = True; return |@words, $new_word if $new_word eq $to; } } push @new_q, ( |@words, $_ ) for @new_tails; } @workqueue = @new_q; }
} for <boy man>, <girl lady>, <john jane>, <child adult> -> ($from, $to) {
say word_ladder($from, $to) // "$from into $to cannot be done";
}</lang>
- Output:
(boy bay may man) (girl gill gall gale gaze laze lazy lady) (john cohn conn cone cane jane) child into adult cannot be done
REXX
This REXX entry does a little more error checking.
It also assumes that the dictionary file is in mixed case as well as the words entered on the CL.
To treat the dictionary and input words as caseless, all words are translated to lowercase.
Programming note: this REXX program uses the lower BIF which Regina has).
If your REXX doesn't support that BIF, here is an equivalent function:
<lang rexx>lower: procedure; parse arg a; @= 'abcdefghijklmnopqrstuvwxyz'; @u= @; upper @u
return translate(a, @, @u)</lang>
<lang rexx>/*REXX program finds words (within an identified dict.) to solve a word ladder puzzle.*/ parse arg base targ iFID . /*obtain optional arguments from the CL*/ if base== | base=="," then base= 'boy' /*Not specified? Then use the default.*/ if targ== | targ=="," then targ= 'man' /* " " " " " " */ if iFID== | iFID=="," then iFID='unixdict.txt' /* " " " " " " */ abc= 'abcdefghijklmnopqrstuvwxyz' /*the lowercase (Latin) alphabet. */ abcU= abc; upper abcU /* " uppercase " " */ base= lower(base); targ= lower(targ) /*lowercase the BASE and also the TARG.*/
L= length(base) /*length of the BASE (in characters). */
if L<2 then call err 'base word is too small or missing' /*oops, too small*/ if length(targ)\==L then call msg , "target word isn't the same length as the base word" call letters /*assign letters, faster than SUBSTR. */
- = 0 /*# of words whose length matches BASE.*/
@.= /*default value of any dictionary word.*/
do recs=0 while lines(iFID)\==0 /*read each word in the file (word=X).*/ x= lower(strip( linein( iFID) ) ) /*pick off a word from the input line. */ if length(x)\==L then iterate /*Word not correct length? Then skip. */ #= # + 1; @.x= 1 /*bump # words with length L; semaphore*/ end /*recs*/ /* [↑] semaphore name is uppercased. */
!.= 0 say copies('─', 30) recs "words in the dictionary file: " iFID say copies('─', 30) # "words in the dictionary file of length: " L say copies('─', 30) ' base word is: ' base say copies('─', 30) 'target word is: ' targ rung= targ $= base
do f=1 for m; call look; if result\== then leave /*Found? Quit.*/ end /*f*/
say if f>m then call msg 'no word ladder solution possible for ' base " ──► " targ
do f-2; $= base; !.= 0 /*process all the rungs that were found*/ do forever; call look; if result\== then leave /*Found? Quit.*/ end /*forever*/ end /*f-2*/
call show words(rung) exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ msg: say; if arg()==2 then say '***error*** ' arg(2); else say arg(1); say; exit 13 show: say 'a solution: ' base; do j=1 to arg(1); say left(,12) word(rung,j); end; return letters: do m=1 for length(abc); a.m= substr(abc, m, 1); end; return /*──────────────────────────────────────────────────────────────────────────────────────*/ look: procedure expose @. !. a. $ abc base L rung targ search; rungs= word(rung, 1)
$$=; rung#= words(rungs) do i=1 for words($); y= word($, i); !.y= 1 do k=1 for L do n=1 for 26; z= overlay(a.n, y, k) /*change a letter*/ if @.z== then iterate /*Is this not a word? Then skip it. */ if !.z then iterate /* " " a repeat? " " " */ if z==rungs then rung= y rung /*prepend a word to the rung list. */ if z==rungs & rung#>1 then return z /*short─circuit. */ if z==targ then return z $$= $$ z /*append a possible ladder word to $$*/ end /*n*/ end /*k*/ end /*i*/ $= $$; return </lang>
- output when using the default inputs:
────────────────────────────── 25104 words in the dictionary file: unixdict.txt ────────────────────────────── 796 words in the dictionary file of length: 3 ────────────────────────────── base word is: boy ────────────────────────────── target word is: man a solution: boy bay may man
- output when using the inputs of: girl lady
────────────────────────────── 25104 words in the dictionary file: unixdict.txt ────────────────────────────── 2187 words in the dictionary file of length: 4 ────────────────────────────── base word is: girl ────────────────────────────── target word is: lady a solution: girl gill gall gale gaze laze lazy lady
- output when using the inputs of: john jane
────────────────────────────── 25104 words in the dictionary file: unixdict.txt ────────────────────────────── 2187 words in the dictionary file of length: 4 ────────────────────────────── base word is: john ────────────────────────────── target word is: jane a solution: john cohn conn cone cane jane
- output when using the inputs of: child adult
────────────────────────────── 25104 words in the dictionary file: unixdict.txt ────────────────────────────── 3161 words in the dictionary file of length: 5 ────────────────────────────── base word is: child ────────────────────────────── target word is: adult no word ladder solution possible for child ──► adult
Swift
<lang swift>import Foundation
func oneAway(string1: String, string2: String) -> Bool {
if string1.count != string2.count { return false } var sum = 0 for (c1, c2) in zip(string1, string2) { if c1 != c2 { sum += 1 if sum > 1 { return false } } } return sum == 1
}
func wordLadder(words: [String], from: String, to: String) {
var poss = words.filter{$0.count == from.count} var queue: String = from while !queue.isEmpty { var curr = queue[0] let last = curr[curr.count - 1] queue.removeFirst() let next = poss.filter{oneAway(string1: $0, string2: last)} if next.contains(to) { curr.append(to) print(curr.joined(separator: " -> ")) return } poss.removeAll(where: {next.contains($0)}) for str in next { var temp = curr temp.append(str) queue.append(temp) } } print("\(from) into \(to) cannot be done.")
}
do {
let words = try String(contentsOfFile: "unixdict.txt", encoding: String.Encoding.ascii) .components(separatedBy: "\n") .filter{!$0.isEmpty} wordLadder(words: words, from: "man", to: "boy") wordLadder(words: words, from: "girl", to: "lady") wordLadder(words: words, from: "john", to: "jane") wordLadder(words: words, from: "child", to: "adult")
} catch {
print(error.localizedDescription)
}</lang>
- Output:
man -> ban -> bay -> boy girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady john -> cohn -> conn -> cone -> cane -> jane child into adult cannot be done.
Wren
<lang ecmascript>import "io" for File import "/sort" for Find import "/seq" for Lst
var words = File.read("unixdict.txt").trim().split("\n")
var oneAway = Fn.new { |a, b|
var sum = 0 for (i in 0...a.count) if (a[i] != b[i]) sum = sum + 1 return sum == 1
}
var wordLadder = Fn.new { |a, b|
var l = a.count var poss = words.where { |w| w.count == l }.toList var todo = a while (todo.count > 0) { var curr = todo[0] todo = todo[1..-1] var next = poss.where { |w| oneAway.call(w, curr[-1]) }.toList if (Find.first(next, b) != -1) { curr.add(b) System.print(Lst.flatten(curr).join(" -> ")) return } poss = poss.where { |p| !next.contains(p) }.toList for (i in 0...next.count) { var temp = [curr] temp.add(next[i]) todo.add(temp) } } System.print("%(a) into %(b) cannot be done.")
}
var pairs = [
["boy", "man"], ["girl", "lady"], ["john", "jane"], ["child", "adult"]
] for (pair in pairs) wordLadder.call(pair[0], pair[1])</lang>
- Output:
boy -> bay -> ban -> man girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady john -> cohn -> conn -> cone -> cane -> jane child into adult cannot be done.