Where the dream starts 3

Category: crypto (500 points) Author: william_etotheipi Solves: 58 Challenge: "A modification of a very nice classic." Flag: gaslightCTF{theindecipherablecipher}

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Table of contents

1. Challenge description
2. Series context (parts 1 & 2)
3. The encryption algorithm
4. What the modification does
5. Step 1 — find the key length
6. Step 2 — remove the progressive term
7. Step 3 — recover the keyword (per-column chi-squared)
8. Step 4 — decrypt
9. The recovered plaintext
10. The flag and its meaning
11. Solver
12. Takeaways
13. References

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1. Challenge description

"Upon your successful attacks of the precedent ciphers, Vigenère invited you to his cipher workshop. Friedman and Kasiski are there too."

"A modification of a very nice classic."

The description is a heavy hint package:

• Vigenère — the classic being modified is the Vigenère cipher.
• Friedman — William Friedman, inventor of the index of coincidence (IC), used to estimate key length of a polyalphabetic cipher.
• Kasiski — Friedrich Kasiski, whose Kasiski examination (looking for repeated polygrams in the ciphertext and taking the GCD of their spacings) also recovers key length.

So the intended path is: find the key length, then break a modified Vigenère.

We are given two files:

• encrypt.py — the encryption script (with the algorithm, but not the key).
• output.txt — 547 characters of ciphertext.

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2. Series context (parts 1 & 2)

This is the third in a chain, and the series puns on ciphers that are "variants of classics", with the plaintext describing the cipher itself:

Part	Cipher	Key	Plaintext hint
1	Caesar (shift)	K = 3	"caesar really liked a shift of three"
2	Columnar transposition	key length 3, keyword abc	"The flag is gaslightCTF{tr4nsp0s3-2-th3-k3y-0f-g-fl4t!}"
3	Modified Vigenère	length 5, keyword dream	"...the flag is the indecipherable cipher"

Part 2's key parameters came from part 1's plaintext ("shift of three" → key length 3). For part 3 we must recover everything from the ciphertext alone.

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3. The encryption algorithm

from string import ascii_lowercase

def encrypt(pt: str, keyword: str) -> str:
    m = len(keyword)
    pt = list(map(lambda x: ascii_lowercase.index(x), pt))
    key = list(map(lambda x: ascii_lowercase.index(x), keyword))

    ct = class="hljs-string">""
    for i in range(len(pt)):
        index = (pt[i] + key[i % m] + (i // m)) % 26
        char = ascii_lowercase[index]
        ct += char

    return ct

with open(class="hljs-string">'gaslight-where-the-dream-starts-3/plaintext_and_keyword.txt', class="hljs-string">'r') as file:
    ct = file.readline().strip()
    keyword = file.readline().strip()

print(encrypt(ct, keyword))

Note the ordering: encrypt is actually called with the plaintext read first; the output file is the ciphertext. The plaintext is therefore pure lowercase a-z (any other character would make ascii_lowercase.index(x) raise ValueError), so spaces/punctuation were stripped before encrypting.

The formula

For plaintext index i:

c_i = (p_i + k_{i mod m} + ⌊i/m⌋) mod 26

where:

• p_i = numeric plaintext char (a=0 … z=25),
• k_j = numeric keyword char (so k_{i mod m} is the normal Vigenère key stream),
• ⌊i/m⌋ = the block counter — the number of complete m-letter blocks passed so far.

So compared to classic Vigenère c_i = (p_i + k_{i mod m}) mod 26, every new block of m characters pushes the whole block's shift forward by one extra step. This is a hybrid of Vigenère and a Progressive Caesar (each successive block advances by +1, independent of position within the block).

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4. What the modification does

The term ⌊i/m⌋ is entirely determined by the position and the key length m — it carries no key material. That is the attack: if we can guess m, we can subtract the known progressive offset from the ciphertext and we are left with a plain Vigenère ciphertext.

Define the adjusted stream:

adj_i = (c_i − ⌊i/m⌋) mod 26 = (p_i + k_{i mod m}) mod 26

adj is exactly the output of a standard Vigenère encryption of p under keyword k, which we know how to break (find key length → per-column frequency analysis).

So the whole solve reduces to:

1. Find m.
2. Un-apply ⌊i/m⌋.
3. Break the resulting Vigenère.

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5. Step 1 — find the key length

5a. Index of coincidence (Friedman)

The IC of a text is the probability that two characters picked at random are equal:

IC = Σ f_c(f_c − 1) / (N(N − 1))

Reference values for 26-letter text:

• Random: ≈ 0.038
• English: ≈ 0.066

For a Vigenère-style ciphertext, the IC is close to 0.038. But if we split the text into m columns (taking every m-th character), each column is a Caesar-shifted sample of English, and the average column IC jumps back up toward 0.066 — but only if m is the correct key length.

For our ciphertext (N = 547):

IC(raw ciphertext) = 0.0385     ← polyalphabetic / flat, as expected

After removing the progressive term ⌊i/m⌋ and computing the average IC of the m columns:

m	avg column IC	per-column ICs
2	0.0395	0.041, 0.038
3	0.0383	0.039, 0.039, 0.038
4	0.0375	0.040, 0.036, 0.040, 0.035
5	0.0608	0.061, 0.064, 0.060, 0.057, 0.062
6	0.0394	0.044, 0.035, 0.040, 0.034, 0.040, 0.044
7	0.0394	0.037, 0.050, 0.036, 0.040, 0.039, 0.039, 0.036
8	0.0388	0.034, 0.040, 0.033, 0.041, 0.047, 0.039, 0.037, 0.040

m = 5 jumps to 0.061 (≈ English 0.066) while every other candidate sits at ~0.039 (≈ random). This is a slam-dunk: key length m = 5.

5b. Kasiski examination (bonus cross-check)

Repeated ciphertext trigrams and their spacings:

"bxx" spaced 30 apart
"qbv" spaced 63 apart
...

For a proper Vigenère, repeated trigrams arise from repeated plaintext trigrams aligned with the same key letters, so their spacings are multiples of m. (With only 547 chars the sample is sparse; IC gave us the answer more cleanly.) GCD-like reasoning also points to 5 — e.g. spacings of 30 and 65 are both multiples of 5.

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6. Step 2 — remove the progressive term

With m = 5, build the adjusted stream:

adj = [(ascii_lowercase.index(ct[i]) - (i // 5)) % 26 for i in range(len(ct))]

Example: the ciphertext starts l n e s e m l y ... (indexes 11, 13, 4, 18, 4, 12, 11, 24, …). For i = 0..4, ⌊i/5⌋ = 0, so the first block is unchanged Vigenère; from i = 5..9, ⌊i/5⌋ = 1, so each of those chars is shifted back by one, and so on.

After this the stream is ordinary Vigenère under a length-5 keyword.

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7. Step 3 — recover the keyword (per-column chi-squared)

Now adj is Vigenère with period 5. Every character at index i satisfies:

p_i = (adj_i − k_{i mod 5}) mod 26

For each residue column r ∈ {0,1,2,3,4} (positions i ≡ r (mod 5)), all characters are shifted by the same Caesar shift k_r. Each column is a Caesar-shifted sample of English. We find the shift by chi-squared minimisation against English letter frequencies:

χ²(k) = Σ_{c=0}^{25} (obs_c − 26·N·freq_c)² / (26·N·freq_c)

for each candidate shift k, where obs_c is the count of letter c in the decrypted column.

Recovered key, column by column:

r:  0  1  2  3  4
k:  3  17  4  0  12
    d  r  e  a  m

Keyword: dream — which is, of course, the point of a challenge titled "where the dream starts".

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8. Step 4 — decrypt

pt = class="hljs-string">''.join(
    chr((adj[i] - k[i % 5]) % 26 + 97)
    for i in range(len(adj))
)

where k = [3, 17, 4, 0, 12] (i.e. d r e a m).

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9. The recovered plaintext

iwassittingwritingonmytextbookbuttheworkdidnotprogressmythoughtswereelsewhereiturned
mychairtothefireanddozedagaintheatomsweregambolingbeforemyeyesthistimethesmallergroups
keptmodestlyinthebackgroundmymentaleyerenderedmoreacutebytherepeatedvisionsofthekind
couldnowdistinguishlargerstructuresofmanifoldconformationlongrowssometimesmoreclosely
fittedtogetheralltwiningandtwistinginsnakelikemotionbutlookwhatwasthatoneofthesnakes
hadseizedholdofitsowntailandtheformwhirledmockinglybeforemyeyesasifbyaflashoflightning
iawoketheflagistheindecipherablecipher

With spaces restored:

I was sitting writing on my textbook but the work did not progress. My thoughts were elsewhere. I turned my chair to the fire and dozed again. The atoms were gamboling before my eyes; this time the smaller groups kept modestly in the background. My mental eye, rendered more acute by the repeated visions of the kind, could now distinguish larger structures of manifold conformation; long rows, sometimes more closely fitted together, all twining and twisting in snakelike motion. But look! What was that? One of the snakes had seized hold of its own tail, and the form whirled mockingly before my eyes. As if by a flash of lightning I awoke. The flag is the indecipherable cipher.

This is the famous anecdote of August Kekulé, who claimed to have discovered the ring structure of benzene after a daydream of a snake seizing its own tail (the Ouroboros) — a textbook example of insight arriving in a dream. The story literally begins "I was sitting writing on my textbook", continuing the series' dream theme.

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10. The flag and its meaning

The final line:

the flag is the indecipherable cipher

"Le chiffre indéchiffrable" ("the indecipherable cipher") is the historical nickname for the Vigenère cipher — perfectly on-theme since this challenge is a modified Vigenère.

gaslightCTF{theindecipherablecipher}

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11. Solver

python3 solver.py:

from string import ascii_lowercase

ct = open(class="hljs-string">'output.txt').read().strip()
n = len(ct)

ENGF = [0.08167,0.01492,0.02782,0.04253,0.12702,0.02228,0.02015,0.06094,0.06966,
        0.00153,0.00772,0.04025,0.02406,0.06749,0.07507,0.01929,0.00095,0.05987,
        0.06327,0.09056,0.02758,0.00978,0.02360,0.00150,0.01974,0.00074]

def chi2(col):
    counts = [0]*26
    for x in col: counts[x] += 1
    L = len(col)
    return sum(((counts[i] - ENGF[i]*L)**2)/(ENGF[i]*L if ENGF[i] else 1) for i in range(26))

def solve_for_m(m):
    # c_i = p_i + k_{i mod m} + floor(i/m)  ->  remove floor(i/m)
    adj = [(ascii_lowercase.index(ch) - (i//m)) % 26 for i, ch in enumerate(ct)]
    key = []
    for r in range(m):
        col = adj[r::m]
        best = min(range(26), key=lambda s: chi2([(x-s) % 26 for x in col]))
        key.append(best)
    pt = class="hljs-string">''.join(chr((adj[i] - key[i % m]) % 26 + 97) for i in range(n))
    return class="hljs-string">''.join(ascii_lowercase[k] for k in key), pt

def score_pt(pt):
    words = sum(pt.count(w) for w in
                [class="hljs-string">"the",class="hljs-string">"and",class="hljs-string">"ing",class="hljs-string">"that",class="hljs-string">"with",class="hljs-string">"this",class="hljs-string">"cipher",class="hljs-string">"flag",class="hljs-string">"key",
                 class="hljs-string">"shift",class="hljs-string">"classic",class="hljs-string">"progressive",class="hljs-string">"keyword",class="hljs-string">"dream",class="hljs-string">"lightning"])
    counts = [0]*26
    for ch in pt: counts[ascii_lowercase.index(ch)] += 1
    L = len(pt)
    c = sum(((counts[i]-ENGF[i]*L)**2)/(ENGF[i]*L if ENGF[i] else 1) for i in range(26))
    return c, words

results = []
for m in range(1, 41):
    key, pt = solve_for_m(m)
    c, w = score_pt(pt)
    results.append((m, c, w, key, pt))

m, c, w, key, pt = max(results, key=lambda x: x[2])
print(fclass="hljs-string">"key length: {m}, keyword: {key!r}")
print(class="hljs-string">"plaintext:")
print(pt)
idx = pt.find(class="hljs-string">"theflagis")
print(class="hljs-string">"FLAG:", class="hljs-string">"gaslightCTF{" + pt[idx+len(class="hljs-string">"theflagis"):] + class="hljs-string">"}")

Output:

key length: 5, keyword: 'dream'
plaintext:
iwassittingwritingonmytextbook...iawoketheflagistheindecipherablecipher
FLAG: gaslightCTF{theindecipherablecipher}

The brute force over m ∈ [1,40] is fast (547 chars, tiny key space); the correct m=5/dream decryption is unmistakable by common-word count and chi², while every other m decrypts to gibberish.

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12. Takeaways

• The modification was separable from the key. Because ⌊i/m⌋ depends only on the index and key length, guessing m lets you strip it off entirely, collapsing the "new" cipher back into the classic you already know how to break.
• Friedman/Kasiski still work because the progressive term is constant within each block — it shifts whole blocks, so the periodic structure (and thus the column IC peak) is preserved.
• Chi-squared column analysis is a general Vigenère key-recovery technique that avoids needing a full dictionary.
• Series trend: the plaintext narrates its own cipher ("the indecipherable cipher"), rewarding the solver who connects the cryptanalysis with the story.

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13. References

• William F. Friedman — The Index of Coincidence and Its Applications in Cryptography (1920).
• Friedrich Kasiski — Die Geheimschriften und die Dechiffrir-Kunst (1863).
• Kekulé's benzene dream / Ouroboros anecdote.
• English letter frequencies (approx.) used for chi-squared key recovery.