How to Handle Repeated Letter Input in a Tkinter Hangman: Reject and Ignore Duplicate Guesses

Learn to prevent duplicate guesses in a Tkinter Hangman by validating input and checking a seen letters set. Avoid life penalties and keep gameplay predictable.

Duplicate guesses are a common edge case in simple games. In a tkinter-based Hangman, accepting an already tried letter should be a no-op: no state changes, no lives lost, no UI side effects beyond keeping the current state visible. If you don’t short-circuit repeated input, the game keeps processing it and can degrade the experience by acting on the same letter again.

The problem

A Hangman input handler accepts a letter from an Entry, validates it, updates the set of guessed letters, and then either reveals matches or reduces remaining attempts. The missing piece is a check that rejects a letter that was already tried.

Problematic code

The snippet below mirrors the logic that triggers the issue. Names are illustrative, but behavior matches the described flow.

def on_guess(self, evt=None):
    char = self.input_box.get().lower()
    self.input_box.delete(0, tk.END)

    if not char or len(char) != 1 or not char.isalpha():
        return

    self.seen_letters.add(char)
    self.seen_label.config(text="Guessed Letters: " + ", ".join(sorted(self.seen_letters)))

    if char in self.target_word:
        for idx, ch in enumerate(self.target_word):
            if ch == char:
                self.revealed[idx] = char
        self.word_view.config(text=" ".join(self.revealed))
    else:
        self.lives_left -= 1
        self.paint_hangman()

    if "_" not in self.revealed:
        messagebox.showinfo("Hangman", "You won!")
        self.root.quit()
    elif self.lives_left == 0:
        messagebox.showinfo("Hangman", f"You lost! The word was: {''.join(self.target_word)}")
        self.root.quit()

What’s going on and why

The handler never checks whether the current input was already attempted before adding it to the set and proceeding with the core game logic. Because of that, a letter that’s been guessed earlier is processed again as if it were new. The fix is to treat a previously used character as invalid input for the current turn and return early.

The fix

Add a membership check against the set of previously guessed letters as part of the input validation guard. That makes repeated guesses a no-op. Also, the condition that combines “empty string” and “length is not one” can be reduced to just a single length check, since an empty string also fails the length requirement.

def on_guess(self, evt=None):
    char = self.input_box.get().lower()
    self.input_box.delete(0, tk.END)

    if len(char) != 1 or not char.isalpha() or char in self.seen_letters:
        return

    self.seen_letters.add(char)
    self.seen_label.config(text="Guessed Letters: " + ", ".join(sorted(self.seen_letters)))

    if char in self.target_word:
        for idx, ch in enumerate(self.target_word):
            if ch == char:
                self.revealed[idx] = char
        self.word_view.config(text=" ".join(self.revealed))
    else:
        self.lives_left -= 1
        self.paint_hangman()

    if "_" not in self.revealed:
        messagebox.showinfo("Hangman", "You won!")
        self.root.quit()
    elif self.lives_left == 0:
        messagebox.showinfo("Hangman", f"You lost! The word was: {''.join(self.target_word)}")
        self.root.quit()

Why this matters

Maintaining a set of previously guessed letters and rejecting duplicates keeps the game fair and predictable. Players don’t lose lives for repeating input, and the UI remains consistent with the state of the board. This small guard eliminates an entire class of errors that stem from reprocessing the same character over and over.

Conclusion

Make the set of guessed characters a first-class check in your input pipeline. Validate length and alphabetic input, then short-circuit if the character has been used before. The early return prevents state churn, avoids accidental penalties, and keeps the Hangman loop clean and maintainable.