2021 April Fool's Puzzle Competition Summary & Solutions

The clue is Base64 encoded. Running it through a decoder such as base64 -d gives:

whatisakey

The answer is whatisakey.

Running the clue through ROT13 gives:

I2yfoTyuoFOTVRMlnJIxoJShPt==

Running that through base64 -d gives:

William F Friedman

William F Friedman was a US Army cryptographer, hinting at the theme, cryptography.

The answer is williamffriedman.

5-letter groupings are common in crypto--we can ignore the spaces as irrelevant.

This is an ADFGX cipher, a Playfair variant.

Using the base grid (no rearrangement) and the key "CRYPTO", we decrypt the above to:

gabrieldeluetzbaronetseigneurdaramonetdevallabregues

or

Or Gabriel de Luetz, Baron et Seigneur d'Aramon et de Vallabregues. Also known as Gabriel d'Aramon.

The answer is gabrieldaramon.

The page is from the Voynich Manuscript, but a reverse image search shows the brighter text on the bottom-left has been added.

The "First Study Group", led by the previously mentioned William F. Friedman, devised an ASCII equivalent for the text. The added text would read:

nothing is / more beautiful / than to kno(u/w) all

A quote by AthanasiusKircher.

The message is run-length encoded. The clue, 1H1E2L1O means "1H, 1E, 2L's, 1O". It decompresses as:

HELLO

The full message decompresses as:

.##..###..###.#...#.####..#.#..#..#..##.##.#..#....###...#..#.#.#.##.#..#.#.#...#..#...#.#...##..#..#.###.#...#.###

The length is 115 or 5x23 characters. Reshaping it into a rectangle (for example with the unix tool fold(1)) gives:

.##..###..###.#...#.###
#..#.#..#..#..##.##.#..
#....###...#..#.#.#.##.
#..#.#.#...#..#...#.#..
.##..#..#.###.#...#.###

CRIME is a security vulnerability against encrypted protocols like HTTPS that use compression.

BREACH is a variant attack of CRIME. The answer is BREACH.

The audio clip is 4.643 seconds of mono audio. It's in Vorbis audio format inside an Ogg file. Sounds funky.

$ ogginfo disco_55c8jPxXdoJ2VLSj.ogg
...
User comments section follows...
        ARTIST=ircpuzzles
        TITLE=Disco
        DATE=2021
        GENRE=Crypto
        COMMENTS=There's no hint in here, but have an easter egg instead.
Vorbis stream 1:
        Total data length: 159698 bytes
        Playback length: 0m:04.643s
...

OK, great. Let's convert that to a spectrogram.

ffmpeg -i disco_55c8jPxXdoJ2VLSj.ogg -f s16le -acodec pcm_s16le - | specgram --live

ffmpeg -i ~/Downloads/disco_55c8jPxXdoJ2VLSj.ogg -f s16le -acodec pcm_s16le - | specgram -f 2048 -g256 spectrogram.jpg

This code is the same as the one from the Perseverance rover parachute.

In Perseverance the codes read counter-clockwise in binary. We'll treat bright green as 1 and dark blue as 0.

Reading clockwise instead, the binary patterns are:

• Inner: 1111111111 1111111111 0000001101 0000000001 0000000111 0000001001 0000000011 0001111111
• Middle: 0000000010 0000001001 0000001110 0001111111 1111111111 1111111111 0000010010 0000001111
• Outer: 0001111111 1111111111 1111111111 1111111111 0000001000 0000001111 0000001111 0000000100
• Rim: 0000011111 0000010111 0000110011 0000111011 0000110010 0000001110 0000000000 0000101100

Note that each segment starts with 000 (or is 1111111111, which we ignore), as in the real parachute.

(We cheated so this would work out--if it doesn't, shift where you started the circle a couple digits.)

Next, convert binary to decimal:

• Inner: -, -, 13, 1, 7, 9, 3, -
• Middle: 2, 9, 14, 127, -, -, 18, 15
• Outer: 127, -, -, -, 8, 15, 15, 4
• Rim: 31, 23, 51, 59, 50, 14, 0, 44

Let's interpret the first three rings as letters (A=1) as in Perseverance. For the rim, perseverance used degree notation for GPS, and we'll do the same. 14=N, 23=W, and the rest are literal numbers in degree notation.

• Inner: magic
• Middle: robin?
• Outer: hood?
• Rim: 51°59'50" N, 0°44'31" W

The answer is a magician. And someone who, like Robin Hood, was born in Nottingham. The coordinates point to his workplace, Hut 8 of Bletchley Park.

The answer is the real name of A.J. Alan, LeslieHarrisonLambert.

The intended solution was to use a single column transposition cipher. The key is "DIGITAL". Duplicate letters are forbidden in keys. The standard is to instead use DIGJTAL, confirmed by the hint / (increasing).

Sorting the key letters:

DIGJTAL
1234567

ADGIJLT
6132475

The deciphering writes columns from left to right, top to bottom.

ADGIJLT
6132475
_CaluSs
rcnho*r

We needed one padding *. Why does it get added there? Well, encrypt any phrase with padding at the end, and that's where it shows up. Or, it's the highest-numbered (7th) column.

Returning the key to original order:

DIGJTAL
1234567
Claus_S
chnorr*

Claus Schnorr was a cryptographer.

The answer is ClausSchnorr.

The cryptocurrency icons are, in order:

1. cosmos
2. siacoin
3. filecoin
4. decred
5. zcash
6. uniswap
7. waves
8. tron
9. monero
10. ontology
11. chainlink
12. nano

Filling in the crossword:

   d      m  t     
   e  f   o  r C   
 z cHainlinK O o   
 c r  l   E uniswap
waves e   r    m   
 s D  c   onTology 
 H    o    A   s   
     siAcoin       
      n    o       

The numbered cells read THEDAOHACK.

In 2016, The DAO was hacked, controversially leading to a hard-fork of the Ethereum blockchain.

The answer is TheDAOHack.

tri hints at the Trifid Cipher

As usual, the key is CRYPTO. The trigram grid for CRYPTO is:

CRY EFG NQS
PTO HIJ UVW
ABD KLM XZ+

Under each letter of the ciphertext, we write the layer, then row, then column:

bjepnjgtifovufpaqblofzkhiupfk
12213221221332113121232223122
32121212212221231332133222213
23111332223212112223221121121

Using the second hint, we split the ciphertext into groups of 6.

bjepnj gtifov ufpaqb lofzkh iupfk
122132 212213 321131 212322 23122
321212 122122 212313 321332 22213
231113 322232 121122 232211 21121

For each group of 6, we read the numbers in row-order.

bjepnj: 122 132 321 212 231 113
gtifov: 212 213 122 122 322 232
ufpaqb: 321 131 212 313 121 122
lofzkh: 212 322 321 332 232 211
iupfk : 231 222 221 321 121

We use the chart to decode the trigrams:

bjepnj: onefis
gtifov: htwofi
ufpaqb: shredf
lofzkh: ishblu
iupfk : efish

The decoded text is the rhyming phrase:

One Fish, Two Fish, Red Fish, Blue Fish

One Fish, Two Fish, Red Fish, Blue Fish is a well-known book by Theodor Seuss Geisel, pen name Dr. Seuss.

The answer is DrSeuss.

Let's call the numbers:

• A=159082528975647841273288552666760531261
• B=243322489515519334824631119429144576991
• C=3

A and B are roughly 128-bit integers.

A isn't prime:

$ factor 159082528975647841273288552666760531261
159082528975647841273288552666760531261: 7 43 1279 121061 2447948497 1394372834168827027

B isn't prime. Testing whether something is prime is fast; finding the factors is slow.

$ openssl prime 243322489515519334824631119429144576991 # Takes 0.02 second
B70E34F56FFEEFFFEB395DB286A283DF (243322489515519334824631119429144576991) is not prime
$ factor 243322489515519334824631119429144576991 # Takes 20 minutes
243322489515519334824631119429144576991: 14117854731608700389 17235089476501026419

The these numbers are for the public-key cryptography scheme, RSA.

• C is e=C=3, is the public exponent
• p=14117854731608700389, is a secret prime
• q=17235089476501026419, is a secret prime
• B is n=pq=243322489515519334824631119429144576991 is a public semiprime
• A is c=m^e = m^3 = 159082528975647841273288552666760531261 (mod n), the encrypted message.
• λ(n)= lcm(p-1,q-1)=243322489515519334824631119429144576991 is secret. λ(n) and e are coprime.
• If ed≡1 (mod λ(n)), then d=40553748252586555798879695870172475031 (calculated using the fairly simple "extended gcd algorithm")

Here's a python session illustrating the above:

import math

e=3
p=14117854731608700389
q=17235089476501026419
c=159082528975647841273288552666760531261
n=243322489515519334824631119429144576991
n==p*q
> True
th = math.lcm(p-1,q-1) # Can also use (p-1)*(q-1)

def gcdExtended(a, b):
    if a==0: return b,0,1
    gcd,x,y = gcdExtended(b%a, a)
    x,y=y-(b//a)*x,x
    return gcd,x,y
d=gcdExtended(3, th)[1]
d*e % th
> 1

pow(pow(100, d, n), e, n)
> 100
pow(pow(100, e, n), d, n)
> 100

Yep, everything looks good.

To decrypt it, we calculate: m≡c^d (mod λ(n))

m=pow(c,d,n); m
> 215232259421775365151868797862979120299

import codecs
codecs.decode(hex(m)[2:],"hex")
> b'Pyvssbeq Pbpxf'

Passing Pyvssbeq Pbpxf through ROT13, we get Clifford Cocks. Clifford Cocks was the (top-secret) inventor of the RSA algorithm before Rivest, Shamir, and Adleman.

The answer is Clifford_Cocks.

   H   S   I   b   A   G   
GYQyvluEYIXuvcrLjyelApkhEig
   c   x   Y   E   M   w   
iOxfFkmMeCfTuXeFMlhfcuvmfLk
   L   d   b   k   G   I   
xziLdjhNlfifaBelYDMuCFbuDOe
   h   C   l   c   S   B   
oiHsgBmihQyOXapKIhcwhmLXvKM
   Z   m   D   M   F   w   
sqwAbtJNUZABJZsGzCsMfPUZqze
   Z   F   m   O   o   N   

The grid shape is a hint at grilles from cryptography.

This should be combined with level 7's completed crossword:

The crossword is 19 across and 9 tall. There are 171 letters in the grille--exactly 19x9. We rearrange the grille:

HSIbAGGYQyvluEYIXuv
crLjyelApkhEigcxYEM
wiOxfFkmMeCfTuXeFMl
hfcuvmfLkLdbkGIxziL
djhNlfifaBelYDMuCFb
uDOehClcSBoiHsgBmih
QyOXapKIhcwhmLXvKMZ
mDMFwsqwAbtJNUZABJZ
sGzCsMfPUZqzeZFmOoN

Next, we'll take only the lowercase letters:

   b     yvlu    uv
cr jyel pkh igcx   
wi xf km e f u e  l
hfcuvmf k dbk  xzi 
djh lfifa el   u  b
u  eh lc  oi sg mih
 y  ap  hcwhm  v   
m   wsqw bt        
s z s f   qze  m o 

We'll place this as a grille over the crossword

   X     XXXX    XX
XX XXXX XXX XXXX
XX XX XX X X X X  X
XXXXXXX X XXX  XXX
XXX XXXXX XX   X  X
X  XX XX  XX XX XXX
 X  XX  XXXXX  X
X   XXXX XX
X X X X   XXX  X X
-------- + --------
   d      m  t     
   e  f   o  r c   
 z chainlink o o   
 c r  l   e uniswap
waves e   r    m   
 s d  c   ontology 
 h    o    a   s   
     siacoin       
      n    o       
-------- = --------
   X     XXXXt   XX
XX XXXX XXX XXXX
XX XXaXXlXnX X X  X
XXXXXXX X XXXniXXXp
XXXeXXXXX XX   X  X
Xs XX XX  XXtXXoXXX
 X  XXo XXXXX  X
X   XXXXcXXn
X X X X   XXX  X X
-------------------
             t     

     a  l n        
             ni   p
   e               
 s          t  o   
      o         
        c  n

But talnnipestoocn is nothing. What gives?

Let's look at the corresponding letters in the grille:

             E     

     F  M C        
             GI   L
   N               
 D          H  B   
      K            
        A  J       

Taking the pairs, we get:

ABCDEFGHIJKLMN
constantinople

The answer is constantinople

Hint(s):

• 🦄: Symbol from crossword
• ⚛: Symbol from crossword
• If you can't escape from Verðandi's house, it's necessary to consult Urðr to meet Skuld: If you can't escape from the present, consult the past to meet the future.
• image #1: Combine [harvester]
• grille: Cryptographic grille
• [a-z] = 🧱, [A-Z] = 🕳️︎: A-Z are the 'holes' in the grille
• tr -d ' ': Try removing whitespace in the grille
• 171: There are 171 letters in the grille
• {B="i",A="h",D="t",C="n"}: Re-ordering at the end

Hint(s):

• image #1: Clocktower in Bletchley Park
• (2 + 10) * 5: Combine the first 2 groups of five letters, then the remaining 10.

The first 10 letters CIKIIIWIIQ are settings for an Enigma M3 machine.

• Reflector is UKW C
• First rotor is I in position K
• Second rotor is III in position W
• Third rotor is II in position Q
• There are no plugs in the plugboard

The remainder are the ciphertext: FHYFWWXNRYUAAIILYGWVWZQWWEWQRQQBYJYPDGJCBDRIWCZKJO.

Using an online solver gives the output:

xxxxn otxxx xthex xxxfl amexx xxthr owing xxxxt rumpe txxxx

'not the flame throwing trumpet'

Claude Shannon invented a flame-throwing trumpet and a roman-numeral computer called THROBAC.

The answer is THROBAC.

The intended reading of the pictures is:

• Not "The Raven"
• Bug
• Gold (medal)
• Skull

The four images indicate Edgar Allen Poe's The Gold-Bug. (He also wrote The Raven)

In the Gold Bug, there is a secret hidden in the left eye of the skull. It also mentions invisible ink.

If we brighten the left eye socket of the skull, we see a secret code:

In The Gold-Bug, the decoding alphabet is:

52-†81346,709*‡.$();?¶]¢:[
ABCDEFGHIJKLMNOPQRSTUVWXYZ

Decoding )63#!# with this substitution, we get SIGABA. SIGABA was a WW2-era cipher machine:

The answer is SIGABA.

Hint(s):

• What do https://youtu.be/zA52uNzx7Y4 (Eiffel 65 - Blue (Da Ba Dee)) and https://youtu.be/62EjFgPECgo (The Fresh Prince of Bel Air) have in common?: Starting lyrics "a story"
• https://youtu.be/qHepKd38pr0: "Enhance" scene. A hint to zoom in and increase contrast.
• image #2: This is the character Poe, an Uglydoll toy.