This puzzle consists of a sequence of encrypted sections of text with spaces preserved (like a cryptogram). The solver may notice that every line within a section of text is the same length, and has its spaces in the same place, which could suggest that each line within a section is actually encoding the same message. The exception is the last line of each section, which has the same pattern of spaces, but is truncated.

There’s also one additional line of text within a box at the bottom of the puzzle that does not follow this pattern (we’ll come back to this later).

There are (at least) two different approaches to deciphering the encrypted sections of text.

Approach 1: The Known Plaintext Attack

The font, formatting and repeated nature of the sections of text may suggest to a solver familiar with The Simpsons that the texts are taken from the set of “chalkboard gags” that Bart is writing near the beginning of the credits of most episodes. These phrases are usually different for each episode, and are well-documented online.

With the set of known plaintexts, the solver can figure out which plaintext corresponds with each ciphertext based on the lengths of the words. The next step is to break the ciphers. With some analysis of the encrypted texts, the solver can determine that they’re not encrypted with a simple single-letter substitution cipher. The general flavor of “repetition” in this puzzle may suggest a Vigenere cipher (where the key is a repeated phrase).

With the plaintext chalkboard gags lined up with their ciphertexts, the Vigenere key for each text can be derived by subtracting the plaintext from the ciphertext (modulo 26).

Approach 2: Ciphertext Cryptanalysis

If the solver guesses (correctly) that each line of an encrypted section of text is encoding the same message, and that a Vigenere cipher is being used, then they can take advantage of the fact that each line has the same plaintext to break the cipher, without knowing the plaintexts.

Here’s a Python program that does this by guessing key lengths, and then setting up the z3 theorem prover to search for keys and plaintexts that would result in the given ciphertexts.

from z3 import *

def CheckKeyLen(ciphertexts, keylen):
    ciphertexts = [ciphertext.replace(" ", "") for ciphertext in ciphertexts]

    solver = Solver()

    k = [Int("k%d" % i) for i in xrange(keylen)]
    for kc in k:
        solver.add(kc >= 0)
        solver.add(kc < 26)

    p = [Int("p%d" % i) for i in xrange(len(ciphertexts[0]))]
    for pc in p:
        solver.add(pc >= 0)
        solver.add(pc < 26)

    key_index = 0
    for ciphertext in ciphertexts:
        for c_index, c_letter in enumerate(ciphertext):
            c = ord(c_letter) - ord('A')
            solver.add((p[c_index] + k[key_index]) % 26 == c)
            key_index = (key_index + 1) % keylen

    result = solver.check()
    if result == unsat:
        return (None, None)
    else:
        m = solver.model()
        key = "".join([chr(m.evaluate(kc).as_long() + ord('A')) for kc in k])
        plaintext = "".join([chr(m.evaluate(pc).as_long() + ord('A')) for pc in p])
        return (key, plaintext)

def PrintRotatedKeyAndPlaintext(key, plaintext):
    def Rotate(s, offset):
        return "".join([chr((ord(c) - ord('A') + offset) % 26 + ord('A')) for c in s])
    for offset in xrange(26):
        print Rotate(key, offset), Rotate(plaintext, (26 - offset) % 26)

def Decipher(ciphertexts):
    for keylen in xrange(1, 30):
        print "Testing key length", keylen
        key, plaintext = CheckKeyLen(ciphertexts, keylen)
        if key:
            PrintRotatedKeyAndPlaintext(key, plaintext)
            return

if __name__ == "__main__":
    ciphertexts = [
        "Z KWQE BTKVF LZV EG SDDM",
        "W BBZA CVJTI EMG OE VAAD",
        "B KXAC BTMMV PWE RB SRFM",
    ]
    Decipher(ciphertexts)

The output of this program (configured to solve the first block of encrypted text):

Testing key length 1
Testing key length 2
Testing key length 3
Testing key length 4
Testing key length 5
Testing key length 6
Testing key length 7
Testing key length 8
Testing key length 9
Testing key length 10
Testing key length 11
Testing key length 12
Testing key length 13
Testing key length 14
Testing key length 15
Testing key length 16
Testing key length 17
ZWWNBWXXZWXZQMBWZ AOADDFWNWJOAFSFWEEQ
AXXOCXYYAXYARNCXA ZNZCCEVMVINZEREVDDP
BYYPDYZZBYZBSODYB YMYBBDULUHMYDQDUCCO
CZZQEZAACZACTPEZC XLXAACTKTGLXCPCTBBN
DAARFABBDABDUQFAD WKWZZBSJSFKWBOBSAAM
EBBSGBCCEBCEVRGBE VJVYYARIREJVANARZZL
FCCTHCDDFCDFWSHCF UIUXXZQHQDIUZMZQYYK
GDDUIDEEGDEGXTIDG THTWWYPGPCHTYLYPXXJ
HEEVJEFFHEFHYUJEH SGSVVXOFOBGSXKXOWWI
IFFWKFGGIFGIZVKFI RFRUUWNENAFRWJWNVVH
JGGXLGHHJGHJAWLGJ QEQTTVMDMZEQVIVMUUG
KHHYMHIIKHIKBXMHK PDPSSULCLYDPUHULTTF
LIIZNIJJLIJLCYNIL OCORRTKBKXCOTGTKSSE
MJJAOJKKMJKMDZOJM NBNQQSJAJWBNSFSJRRD
NKKBPKLLNKLNEAPKN MAMPPRIZIVAMRERIQQC
OLLCQLMMOLMOFBQLO LZLOOQHYHUZLQDQHPPB
PMMDRMNNPMNPGCRMP KYKNNPGXGTYKPCPGOOA
QNNESNOOQNOQHDSNQ JXJMMOFWFSXJOBOFNNZ
ROOFTOPPROPRIETOR IWILLNEVERWINANEMMY
SPPGUPQQSPQSJFUPS HVHKKMDUDQVHMZMDLLX
TQQHVQRRTQRTKGVQT GUGJJLCTCPUGLYLCKKW
URRIWRSSURSULHWRU FTFIIKBSBOTFKXKBJJV
VSSJXSTTVSTVMIXSV ESEHHJARANSEJWJAIIU
WTTKYTUUWTUWNJYTW DRDGGIZQZMRDIVIZHHT
XUULZUVVXUVXOKZUX CQCFFHYPYLQCHUHYGGS
YVVMAVWWYVWYPLAVY BPBEEGXOXKPBGTGXFFR

Solving the rest of the puzzle

The key phrase used to encrypt each chalkboard gag is a clue to a character featured in the episode with which that chalkboard gag first aired. The solver should note the names of each of these characters.

Then, by counting the number of letters used on the last truncated line of each ciphertext, and indexing this many letters into that ciphertext’s character’s name, the solver can extract the word KWYJIBO (famously played by Bart in a Scrabble game in the second episode of the series, “Bart the Genius”).

Decrypting the final boxed line of the puzzle using KWYJIBO as the Vigenere key gives the answer SUSPENDED SENTENCE.