In the game
Wordle , there is a grid of 6 rows of 5 columns. You have 6 attempts to guess the secret 5-letter word. As you type a word, it appears in the next unfilled row. Once entered, the background color of each letter becomes either black (if that letter is not in the secret word), yellow (if the letter is in the secret word, but not in that position), or green (if the letter is in the word and in that same position).
If your guess word contains multiples of the same letter, the color rules are slightly different. For example if you guess SHEET and one E becomes green and the other black, then there is only one E in the secret word and you found it. But if one E becomes green and the other yellow, then there are (at least) 2 copies of the letter E, but not in the yellow location. If one is yellow and one is black, you know there is exactly one E but not in either of your guessed locations.
In Hard Mode, once you have found a green letter, subsequent guesses must use that letter in that position. Similarly yellow letters must be used in each subsequent guess though not necessarily in the same location.
At the conclusion of the game, the background color of each square in the 6 by 5 grid is either black, yellow, green, or clear for any unused guesses.
For Wordle Hard Mode, considering only the background colors, how many color patterns are possible once the game is over?
Break down your answer into groups of games won in 1 guess, in 2 guesses, ... , in 6 guesses, and finally games lost (in which the 6th guess was still not the secret word).
(In reply to
re: proposed computer solution by Jer)
The 238 was indeed an overcount by 1; it had included the possibility that the first line had been all green. Only on subsequent lines did I zero out this state before proceeding to the next row.
The 5 in row 1 column 20 apparently has no effect on the totals. I just replaced all of column 20 by zeros and get the same totals:
1
237
12483
352009
7141380
119136806
1753981922
The only difference is in my zeroing out the all-green possible beginning for row 2 wins.
I see:
states=startNo;
for turn=2:6
states=states*trans;
states(20)=0; % impossible state: 4 green, 1 yellow
count(turn)=states(21);
states(21)=0; % once finished, it's out of the counting
end
so even if state 20 gets populated by the multiplication by the matrix, the "impossible state" line zeros it out.
Edited on February 26, 2025, 3:55 pm
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Posted by Charlie
on 2025-02-26 15:53:03 |
re(2): proposed computer solution
