Updated April 30, 2024

One-word Solutions to the New York Times’ LetterBoxed Puzzle

Below is a list of one-word solutions to the New York Times’ LetterBoxed puzzle together with possible puzzle configurations for those solutions.

The program that I used to find the solutions is also shown below. I ran the program in Google Colab. The code was written by ChatGPT 4, not by me, and later debugged by Claude 3 Opus. ChatGPT 4 also taught me how to use Google Colab.

The file “wordlistcomplete.txt” referenced in the code is a list of 198,422 English words I compiled several years ago from various word lists on the web. Many of the words on that list and among the one-word solutions below are obscure and would not be allowed as puzzle solutions by the NYT.

People who enjoy the NYT’s Spelling Bee puzzle might also be interested in this list of possible pangram solutions.

Program written by ChatGPT 4

# Import the necessary libraries
import os
from google.colab import drive
from itertools import permutations

# Mount your Google Drive
drive.mount("/content/drive", force_remount=True)

# Get the path to the file
file_path = os.path.join('/content/drive/My Drive/Colab Notebooks/', 'wordlistcomplete.txt')

# Open the file and read its contents
with open(file_path, 'r') as f:
    words = f.read().splitlines()

# Check if each word contains exactly 12 unique letters and has no double letters
def is_valid_word(word):
    if len(set(word)) != 12:
        return False
    for i in range(len(word) - 1):
        if word[i] == word[i + 1]:
            return False
    return True

def no_adjacent_partition(word):
    unique_letters = sorted(set(word))  # Get unique letters and sort them for consistent processing
    for partition in permutations(unique_letters):
        sets = [partition[i*3:(i+1)*3] for i in range(4)]  # Divide the permutation into four sets of three
        
        # Check if any adjacent letters in the word are in the same set
        is_valid = True
        for i in range(len(word) - 1):
            for set_ in sets:
                if word[i] in set_ and word[i+1] in set_:
                    is_valid = False
                    break
            if not is_valid:
                break
        
        if is_valid:
            return sets  # Partition is valid
    
    return None  # No valid partition found

# Process words
for word in words:
    if is_valid_word(word):
        partition = no_adjacent_partition(word)
        if partition:
            print(f"{word} [{'] ['.join(' '.join(set) for set in partition)}]")
        else:
            print(f"{word} (no partitions)")

Output of the above program