# Puzzle #3. WBAN

By shuttlecock | Shuttlecock Puzzles | 11 Oct 2021

UPDATE: All puzzles have been solved. Please refer to the solutions at the end of the post.

This puzzle contains information about either the seed or mnemonic phrase for three Banano wallets. Information about the mnemonic phrase is encoded within both the puzzle title and the image.

Title: WBAN / WBANWBANWBANWBANWBANWB

NOTE: While there are three wallets, please do not claim more than one prize. This is to ensure that everyone has a fair chance of winning something. Participants who have claimed one prize should refrain from participating in teams to win the other prizes.

The first person to gain access to Wallet #3 also wins the following NFT (sponsored by Potassions#3443):

To claim the NFT, please DM @shuttlecock#3376 on Discord. You will then receive further instructions on how to proceed.

NOTE: One hint will be provided here, for every 24 hours that the puzzle remains unsolved. Come back here to find out what they are!

[NOTE: Looking through the solutions of my previous puzzles may help you figure out a solution to this puzzle: https://www.publish0x.com/shuttlecock-puzzles.]

All the best, and most importantly, have fun!

DAY 1 UPDATE:

- Wallet 2 has been solved by Arkhath#3845.

Updated image, with irrelevant information removed:

Hint for Wallet 1:

The battleship moves as a clock does.

Solution is a 64-character seed.

Hint for Wallet 3:

The Polygon logo encodes some important information; we need a key to unlock it.

Solution is a 64-character seed.

DAY 2 UPDATE:

- Wallet 1 has been solved by an unknown person.

Hint for Wallet 3:

The Polygon may be connected to the name of a cipher. Where else in the title or image may we find the key?

Solution is a 64-character seed.

DAY 3 UPDATE:

Wallet 3’s prize has been increased to 1000 BAN!

Updated image, with irrelevant information removed:

Final hint for Wallet 3:

The solution is a 64-character hash of whatever you encode or decode from the string of text.

## SOLUTIONS

This puzzle comprises 3 distinct components. We consider them in turn.

QUESTION: How do I know which information is for which puzzle?

Notice that both the “inner grid” and the “outer ring” contain 64 boxes. A wallet seed comprises 64 characters. This is confirmed by the fact that the characters in each of the boxes range from 0 to 9, and A to F. This strongly suggests that these elements are part of the same puzzle.

Notice that the “middle ring” contains 24 grids. A mnemonic has 24 words. This strongly suggests that the middle “ring” contains information about the mnemonic phrase for a wallet, and is part of a puzzle.

The coloured boxes, which make up the logo for Polygon, are clearly distinguished from the rest. There are 22 boxes — which makes it unclear whether it contains information about a 64-character seed or a 24-word mnemonic.

Finally, we have the “side panels”; it is unclear which puzzle the side panels belong to.

Side panels

The side panels contain 12 “boxes” each, on the North, South, East and West positions. Each of these boxes contain a letter, in HEXADECIMAL. When decoded, we get the following:

* North panel = 43 4F 52 4E 45 52 53 46 49 52 53 54 = CORNERSFIRST

* West panel = 41 4E 54 49 43 4C 4F 43 4B 57 49 53 = ANTICLOCKWIS

* South panel = 53 54 41 52 54 54 4F 50 4C 45 46 54 = STARTTOPLEFT

* East panel = 48 45 58 54 4F 44 45 43 49 4D 41 4C = HEXTODECIMAL

We need to figure out what puzzle these side panels belong to.

### Puzzle 1 (Wallet containing 200 ban)

The inner grid comprises 8 rows and 8 columns, and contains 64 characters. It contains all the characters of a wallet seed.

The outer ring contains information about the position of the characters.

The rows of the inner grid are labelled 1 to 8, and the columns are labelled A to H.

Start from the top left of the outer ring, then move clockwise.

We get the following:

First character is in position G6 = 0;

Second character is in position D5 = 5;

Third character is in position E2 = 8;

And so on.

When solved, you will receive the following 64-character seed: 058F420A9C202D0657EF6FD3D506F6F977B62053C76DCC6B2780DCEB0B78AEE3.

How do we know that the side panels don’t belong to this puzzle? Recall that the side panels contain the following information:

* North panel = 43 4F 52 4E 45 52 53 46 49 52 53 54 = CORNERSFIRST

* West panel = 41 4E 54 49 43 4C 4F 43 4B 57 49 53 = ANTICLOCKWIS

* South panel = 53 54 41 52 54 54 4F 50 4C 45 46 54 = STARTTOPLEFT

* East panel = 48 45 58 54 4F 44 45 43 49 4D 41 4C = HEXTODECIMAL

What is crucial for us is the East panel, which says HEXTODECIMAL. The information in the inner grid and outer ring are in hexademical. But if we convert the information on the inner grid or outer ring from hexadecimal to decimal, we will get no useful information for a wallet seed. This indicates that the side panels do not belong to this puzzle.

### Puzzle 2 (Wallet containing 300 ban)

There are 24 “grids” in this picture, each containing 4 “boxes”. This contains information about the 24-word mnemonic phrase. We need to find information about how to put them in order.

This is where the side panels enter. Recall that the side panels contain the following information:

* North panel = 43 4F 52 4E 45 52 53 46 49 52 53 54 = CORNERSFIRST

* West panel = 41 4E 54 49 43 4C 4F 43 4B 57 49 53 = ANTICLOCKWIS

* South panel = 53 54 41 52 54 54 4F 50 4C 45 46 54 = STARTTOPLEFT

* East panel = 48 45 58 54 4F 44 45 43 49 4D 41 4C = HEXTODECIMAL

This provides information about how to arrange the 4-box grids above: We start from the corners of the image (CORNERSFIRST), at the top left (STARTTOPLEFT), and move in an anti-clockwise position (ANTICLOCKWIS). The information in the boxes is encoded in decimal, from hexadecimal (HEX2DEC).

1st grid (outer, top left) = 77 65 78 85

2nd grid: (outer, bottom left) = 67 73 86 73

3rd grid = 67 82 79 83

4th grid = 69 76 73 84

5th grid = 69 89 69 66

6th grid = 65 71 69 32

24th grid = 69 78 65 67

These are decimal values. When converted to hexadecimal, they become:

1st grid (outer, top left) = 77 65 78 85 = 4D 41 4E 55

2nd grid: (outer, bottom left) = 67 73 86 73 = 43 49 56 49

3rd grid = 67 82 79 83 = 43 52 4F 53

4th grid =: 69 76 73 84 = 45 4C 49 54

5th grid = 69 89 69 66 = 45 59 45 42

6th grid = 65 71 69 32 = 41 47 45 20

24th grid = 69 78 65 67 = 45 4E 41 43

1st grid (outer, top left) = 77 65 78 85 = 4D 41 4E 55 = MANU

2nd grid: (outer, bottom left) = 67 73 86 73 = 43 49 56 49 = CIVI

3rd grid = 67 82 79 83 = 43 52 4F 53 = CROS

4th grid =: 69 76 73 84 = 45 4C 49 54 = ELIT

5th grid = 69 89 69 66 = 45 59 45 42 = EYEB

6th grid = 65 71 69 32 = 41 47 45 20 = AGE

24th grid = 69 78 65 67 = 45 4E 41 43 = ENAC

Search for these words on the BIP-39 list. You’ll end up with the following mnemonic:

MANUAL CIVIL CROSS ELITE EYEBROW AGE LEAVE MUSEUM AISLE WIDTH FLIP PATTERN AISLE NAIVE AIM PHOTO CLUSTER CARBON BIND TOAST PULSE AMOUNT OYSTER ENACT

### Puzzle 3 (Wallet containing 1000 ban)

This puzzle contains 22 coloured boxes, each containing either decimal or hexadecimal. When combined, we get either of these two strings (depending on where we start):

Start from left, move anti-clockwise: 7 8 0 D 86 69 69 B F C 5 D D 2 3 65 83 76 5 7 4 7 (call this “String A”)

Start from right, move clockwise: 7 4 7 5 76 83 65 3 2 D D 5 C F B 69 69 86 D 0 8 7 (call this “String B”)

From these strings, we need to get either a 64-character seed, or a 24-word mnemonic phrase.

There is an obvious way of getting to a 64-character seed — to convert the strings to SHA-256 (https://emn178.github.io/online-tools/sha256.html), which always returns a 64-character string.

Check if String A and String B return the correct 64-character string. They do not. Something else must be done.

Two pieces of information in this puzzle remain unused: the Polygon logo, and the title: WBAN / WBANWBANWBANWBANWBANWB.

Notice that the second part of the title is simply a repeat of “WBAN”. This is something that the poly-alphabetic cipher (http://pi.math.cornell.edu/~mec/2003-2004/cryptography/polyalpha/polyalpha.html) uses. The polyalphabetic cipher uses a “key” which is repeated as many times as necessary, up to the length of the string to be encoded. The most famous poly-alphabetic cipher is the Vigenère cipher.

Apply the Vigenère cipher to String A and B. Since this is a poly-alphabetic cipher, the numbers are irrelevant and left untouched. Only the letters change. We get the following for String A.

This ciphertext, when converted to SHA-256, gives us the following 64-character seed: C8CF7F7F726F92816A93264A21758DF518BD740FC552972247FF59EB15463E03.

This is the correct seed for the wallet containing the 1000 ban prize.

shuttlecock