Reputation: 874
For context Wordle is game where you have to decipher a 5 letter word in 6 or less guesses based on certain hints. The hints you get are as follows:
I am making a wordle solver program that takes in an array of word attempts and eliminates them from a list of the possible words.
I feel that the best algorithm to solve this problem is a black list where a word that breaks one of the rules is eliminated from the array. But if there is a better alternative I am open to suggestion.
const text =
[
[
["N","black"],["i","black"],["g","black"],
["h","black"],["t","green"]
],
[
["b","black"],["e","black"],["l","orange"],
["o","orange"],["w","black"]
]
]
const words = "dozen,brave,apple,climb,outer,pitch,ruler,holds,fixed,costs,calls, ...etc"
const solver = (text: any) => {
this.resultingWords = words.split(",").filter(word => {
word = word.toUpperCase()
for (var i = 0; i < text.length; i++) {
for (var j = 0; j < 5; j++) {
let currentText = text[i][j]
currentText[0] = currentText[0].toUpperCase()
if (currentText[0] == '') { continue }
if (currentText[1] == "green" && (word[j] != currentText[0])) {
return false
}
if (currentText[1] == "black" && word.includes(currentText[0])) {
return false;
}
if (currentText[1] == "orange" &&
(word[j] == currentText[0] || !word.includes(currentText[0]))) {
return false
}
}
}
return true
})
}
The issue I am having is if a word has multiple of the same letter and one of them is a green or orange match but the other is black. I get no results because of the way I've written my algorithm.
What would be the way to correctly solve this problem?
Is a black list style of filtering the best solution? (as opposed to white list).
Upvotes: 3
Views: 830
Reputation: 8773
You are building a list of candidates, what is a good start I think. It doesn't really matter whether you white- or blacklist the result is the list of candidates. The only concern is that you could get the solution faster or more reliably by guessing words that are not on the candidates list. Why? Because that way you can introduce more new letters at once to check if the word contains them. Maybe a mix between the two strategies is best, hard to tell without testing it first.
There are a lot of ideas for improvements. First I would create the filter before going through the words. Using similar logic as above you would get a collection of four different rule types: A letter has to be or cannot be at a specific position or a letter has to appear exactly (possibly 0) or at least a specific number of times. Then you go through the words and filter using those rules. Otherwise some work might be done multiple times. It is easiest to create such a filter by collecting the same letters in the guess first. If there is an exact number of appearence rule then you can obviously drop a minimal number of appearance rule for the same letter.
To guess the word fast I would create an evaluation function to find the most promising next guess among the candidates. Possible values to score:
Upvotes: 1
Reputation: 350841
You need to match pairs of letters (one from the guess, one from the secret word) and strike them out, so they don't serve for any other match.
You cannot conclude from a "black" that the corresponding letter is not occurring at all in the solution. When the same letter was used twice in the guess, one may get a match (orange or green), while the other may get black.
I would suggest to manage the number of occurrences in the targeted word. A black indication means you know the maximum number of occurrences of a certain letter (which may be 0). When you get a green/orange indication, you know about a minimum number of occurrences (possibly updating the maximum so it is not inconsistent).
I made a little implementation with the following characteristics:
A Game
class manages the choice of the secret word; giving feedback on a guess; and giving an indication on whether the game is over (after 6 guesses or a correct guess)
A play
function which chooses a word based on previous feedback from the Game
instance, and submits it as a guess, and repeats the process until the game is over.
The play
logic will always make a guess (randomly) that is in line with all the feedback received during the game. NB: This is not the optimal strategy. For instance, 4 letters out of 5 may be green, but then many attempts may be needed to find that remaining character.
The play
logic builds a regular expression from all the feedback it got, and only filters words with that regex.
Here is the code:
class Game {
#solution; // Private
constructor(words) {
this.#solution = words[Math.floor(Math.random() * words.length)];
this.turnCount = 0;
this.state = "playing";
}
guess(guessed) {
if (guessed.length !== 5) throw "Invalid length";
if (this.state !== "playing") throw "Game is already over";
this.turnCount++;
let attempt = [...guessed];
let correct = [...this.#solution];
attempt.forEach((ch, i) => {
if (correct[i] === ch) correct[i] = attempt[i] = null;
});
this.state = !attempt.some(Boolean) ? "won" : this.turnCount >= 6 ? "lost" : "playing";
return attempt.map((ch, i) => {
if (ch == null) return [guessed[i], "green"];
let j = correct.indexOf(ch);
if (j >= 0) {
correct[j] = null;
return [ch, "orange"];
}
return [ch, "black"];
});
}
}
function play(words) {
let game = new Game(words);
let allowed = Array(5).fill("abcdefghijklmnopqrstuvwxyz");
let letters = {}; // Letters with their minimum and maximum frequency
while (game.state === "playing") {
let regex = RegExp(Object.entries(letters).map(([ch, {min,max}]) =>
max ? `(?=^[^${ch}]*(?:${ch}[^${ch}]*){${min},${max}}$)`
: `(?!.*${ch})`
).join("") + allowed.map(s => "[" + s + "]").join(""), "i");
words = words.filter(word => regex.test(word));
// Pick a random word from the list of potential candidates
let word = words[Math.floor(Math.random() * words.length)];
let result = game.guess(word);
console.log(words.length + " options. Guessing: " + word + " - feedback: " + result.map(([ch, color]) => color[0]).join(""));
letters = {}; // This always starts from scratch
result.forEach(([ch, color], i) => {
if (letters[ch]) {
letters[ch].max = color === "black" ? letters[ch].min : Math.max(++letters[ch].min, letters[ch].max);
} else {
letters[ch] = color === "black" ? { min: 0, max: 0 } : { min: 1, max: 5 };
}
allowed[i] = color === "green" ? ch : allowed[i].replace(ch, "");
});
}
console.log("Game over. I " + game.state + " after " + game.turnCount + " attempts.");
}
const words = "dozen,brave,apple,climb,outer,pitch,ruler,holds,fixed,costs,calls"
.match(/\w+/g);
play(words);
Upvotes: 1