TheAlgorithms/Java · error · IllegalArgumentException

Invalid Baconian code: {}

Error message

Invalid Baconian code: {}

What it means

Thrown by BaconianCipher.decrypt(String) when a 5-character group sliced from the ciphertext is not present in REVERSE_BACONIAN_MAP. Bacon's cipher encodes each letter as a 5-symbol group (traditionally 'a'/'b' or '0'/'1'); the decryptor walks the input in 5-char chunks and looks each up, throwing if a chunk is unknown.

Source

Thrown at src/main/java/com/thealgorithms/ciphers/BaconianCipher.java:65

        return ciphertext.toString();
    }

    /**
     * Decrypts the given ciphertext encoded in binary (A/B) format using the Baconian cipher.
     *
     * @param ciphertext The ciphertext to decrypt.
     * @return The decrypted plaintext message.
     */
    public String decrypt(String ciphertext) {
        StringBuilder plaintext = new StringBuilder();

        for (int i = 0; i < ciphertext.length(); i += 5) {
            String code = ciphertext.substring(i, i + 5);
            if (REVERSE_BACONIAN_MAP.containsKey(code)) {
                plaintext.append(REVERSE_BACONIAN_MAP.get(code));
            } else {
                throw new IllegalArgumentException("Invalid Baconian code: " + code);
            }
        }

        return plaintext.toString();
    }
}

View on GitHub (pinned to fdfb9a395b)

Solutions

1. Verify ciphertext length is an exact multiple of 5 before decrypting.
2. Confirm the symbol alphabet matches the cipher's REVERSE_BACONIAN_MAP (check whether it uses a/b or 0/1).
3. Reject or pad truncated ciphertext rather than passing it through.

Example fix

// before
String plain = cipher.decrypt(text);

// after
if (text.length() % 5 != 0) {
    throw new IllegalArgumentException("Ciphertext length must be a multiple of 5");
}
String plain = cipher.decrypt(text);

if (ciphertext == null || ciphertext.length() % 5 != 0
        || !ciphertext.matches("[ab01]+")) {
    throw new IllegalArgumentException("Invalid Baconian ciphertext");
}
String plain = cipher.decrypt(ciphertext);

static boolean isPlausibleBaconian(String s) {
    return s != null && s.length() % 5 == 0;
}

try {
    String plain = cipher.decrypt(ciphertext);
} catch (IllegalArgumentException e) {
    // a 5-char group was not in the map; check alphabet/length
}

When it happens

Trigger: Ciphertext whose length is not a multiple of 5 (the final chunk is short/garbled), ciphertext using a different symbol alphabet than the map expects (e.g. 0/1 when the map uses a/b), or corrupted characters.

Common situations: Encrypt/decrypt alphabet mismatch; transmission corruption that drops a character (length no longer divisible by 5); assuming a 0/1 encoding when the map uses a/b or vice versa.

Related errors

• DES key must be supplied as a 64 character binary string
• Encrypted message should be a multiple of 64 characters in l
• Base, secret, and prime must be non-null and positive values
• Other public value must be non-null and positive.
• Bit length must be at least {} for security.

AI-assisted analysis of TheAlgorithms/Java@fdfb9a395b (2026-08-13). Data as JSON: /api/errors/ddee74ee0a64d9b5. Report an issue: GitHub.