Imaginary Numbers

The puzzle consists of four minipuzzles. The answers, in order, are DAVE, BARRY, RIDLEY, and PEARSON. Dave Barry and Ridley Pearson are the authors of the novel Peter and the Starcatchers. The text at the bottom of the puzzle reads “That which is caught”, so the answer is STAR.

Each of the minipuzzles has flavortext referencing a book which features a cipher. We must understand each cipher and use it (or a related idea) to solve the minipuzzle. Below are solutions to the individual minipuzzles.

Minipuzzle #1

In Cryptonomicon by Neal Stephenson, whose main character is named Randy Waterhouse, a playing card cipher called the Pontifex is used. This was later developed into a real-world cipher called Solitaire. The cipher involves shifting each letter based on an algorithm, which is determined by the order of a deck of cards, including the two jokers.

The flavortext “Si Stebbins” clues the order the deck of cards should come in. The Si Stebbins stack is a cyclical card ordering for a standard 52-card deck. Using this as a starting point, we only need to determine one card’s fixed position (e.g. what the first card is), as well as where the two jokers are inserted, in order to determine the whole deck. These can be deduced by the given clues. Note that all the numbers in the clues are given in the Blobbish language, so we will have to translate those first (highlighted below).

Thus, the first card is the 6 of diamonds.

Thus, Joker A is between the 5 of spades and 8 of diamonds.

Thus, Joker B comes between the 7 of clubs and 10 of hearts.

With this, we have the order of the entire deck, including the jokers. The keystream is as follows: 19, 9, 38, 41, 18, 8, 37, 40, 17, 7, B, 36, 52, 16, 6, 35, 51, 15, 5, 34, 50, 14, 4, 33, 49, 26, 3, 32, 48, 25, 2, 31, 47, 24, 1, 30, 46, 23, 13, 29, 45, 22, 12, 28, 44, A, 21, 11, 27, 43, 20, 10, 39, 42.

Then applying the Solitaire cipher on the given text, WBCD, gives DAVE, the answer to this minipuzzle.

Minipuzzle #2

In Arthur Conan Doyle’s The Valley of Fear, Sherlock and Watson (but mostly Sherlock) decode a note encoded using a book cipher. In this cipher, the words of a message are encoded by specifying their indices in the second column of page 534 of the Whitaker’s Almanac. However, Watson’s initial guess is that the book is the Bible, and that is what we should use for this puzzle. The flavortext, “... use the most popular version,” indicates that we should use the King James Version of the Bible.

Passages in the Bible are typically specified by a book, a chapter, and a verse, written with a colon between the chapter and verse. Examples include Isaiah 41:10, Jeremiah 29:11, etc.

Each row of the puzzle has four entries. Three of them are numbers, and the fourth uses some strange symbols. The unknown symbols are numerals in the Blobbish language.

This suggests that each row is a sequence of four numbers, specifying a book (by its index in the list of books), chapter, verse, and word. If we have already solved Minipuzzle #1, we will have learned some of the numbers, and can use this to understand how to express numbers in the Blobbish language. Here’s the decoded table of numbers, with the numbers given in Blobbish highlighted, and the extracted word on the left.

We notice that there are five words, each repeated four times. Grouping them and reordering within each group, we get the following table.

We notice that there is one Blobbish number in each position within each group. Thus, by reading only the Blobbish numbers, each group gives us a book index, chapter, verse, and word number. If we repeat the procedure with these sets, we find that the words we extract are actually numbers!

Indexing these numbers into the corresponding group words, and sorting by book index, gives us BARRY, the answer to this minipuzzle.

Minipuzzle #3

In Shadow of the Hegemon, by Orson Scott Card, Petra Arkanian sends a message to Bean from captivity, by encoding it in the pixels of an image of a dragon. In this puzzle, we must do the same, following the given steps. Here are the steps in plain English, with Blobbish numbers highlighted:

1. Checksum: The number of 0s in the ciphertext is 270 and the number of 1s in the ciphertext is 242.
2. Flip each bit whose index (zero-indexed) is congruent to zero modulo 7, 11,13, or 19.
3. Checksum: The number of nibbles that don’t represent one of Dragon Army’s toons (in hexadecimal) is 117.
4. Figure out a circular bit shift to apply to the entire string.
5. Figure out a circular bit shift that you can apply to every byte (the same shift for each byte).
6. Checksum: YOU SHOULD GET A 64 CHARACTER (INCLUDING SPACES) CLUEPHRASE, STARTING WITH <Start of Text> (0x02) AND ENDING WITH <End of Text> (0x03).

The image is 512 x 512 pixels. Looking at the image, it’s clear that a row of pixels around the middle has been modified somehow. The 256th row of the image,

11010101|01011110|01010110|00001001|00011101|01101000|01100000|10000010|01100000|00011100|01001101|11000001|01111100|10000011|11100110|00000111|01101100|00111101|01011000|11101101|00000100|11110000|11000010|00111111|01010100|01010000|11010010|00000011|11001011|01100111|01110111|01011000|01011000|10000101|00010000|10101011|01000111|00010001|11001000|10110100|01100011|00100101|11101011|01001111|11010111|00011000|00111001|00100001|00001110|11000101|01010110|01010110|11111011|01010100|00010111|10110101|11000011|11001100|01001101|00010101|00011101|01100011|01110111|11111000

is the 512-bit ciphertext, and it also satisfies the checksum.

Once we flip the bits with indices congruent to 0 modulo 7, 11, 13, and 19, we obtain the string

01010100|01001000|01000000|00100001|01001110|01000000|00101001|01000011|00000010|00010010|01000101|01010000|01001100|01000001|01000011|01000101|01000000|00110100|01001000|01000101|01000000|00100001|01000000|00101001|01001110|01000000|00110010|01001111|01000010|01000101|01010010|01010100|01000000|00100101|01000000|00101100|01000101|01000101|01000000|00100101|01010111|01000101|01001100|01001100|01010011|01000000|00101011|01001001|01001100|01001100|01000101|01010010|01010011|01000000|00110011|01010101|01010010|01001110|01000001|01001101|01000101|01000000|00110111|01001001

The checksum here refers to the toons in Dragon Army in Ender's Game, which were labeled A, B, C, D, and E. This means that we should count the set of nibbles (half-bytes) which are not 1010, 1011, 1100, 1101, or 1110.

We have reached the main part of this puzzle - identifying the correct shifts. Because the start-of-text character (0x02) in ASCII is 00000010 and the end-of-text character (0x03) is 00000011, we can search for adjacent 8-bit substrings with exactly two and one 1s, respectively, to find a valid global shift. The 16-bit substring giving rise to the start and end characters is bolded in the above string.

There are exactly two global circular shifts which satisfy this property, and only one of them - once locally shifted so that the first bit is 0x02 - gives a string of uppercase letters (other than start and end):

00000010|01010010|01000101|01010000|01001100|01000001|01000011|01000101|00100000|01010100|01001000|01000101|00100000|01000001|00100000|01001001|01001110|00100000|01010010|01001111|01000010|01000101|01010010|01010100|00100000|01000101|00100000|01001100|01000101|01000101|00100000|01000101|01010111|01000101|01001100|01001100|01010011|00100000|01001011|01001001|01001100|01001100|01000101|01010010|01010011|00100000|01010011|01010101|01010010|01001110|01000001|01001101|01000101|00100000|01010111|01001001|01010100|01001000|00100000|01000001|01001110|00100000|01001001|00000011

Converting each byte to ASCII, we obtain (removing the start and end of text characters) the phrase “REPLACE THE A IN ROBERT E LEE EWELLS KILLERS SURNAME WITH AN I”. The killer of Robert E. Lee (Bob) Ewell in To Kill a Mockingbird is Boo Radley, giving the answer RIDLEY.

It's a fun exercise to show that if there exists a global shift followed by a local shift to the post-bitflips string which extracts to a string of uppercase letters and spaces, then there's another global shift and local shift which extracts to the same string but where spaces are replaced with nulls and the first letter after each space is lowercase,“REPLACE tHE a iN rOBERT e LEE eWELLS kILLERS sURNAME wITH aN i". This is because of two simultaneous coincidences. The first is that while 00100000 encodes a space in ASCII, 00000000 (the null character) is also displayed as a space by most code which prints ASCII. The second is that every uppercase ASCII letter has a 0 in its third bit, and its corresponding lowercase letter has a 1 in its third bit.

Here's a sketch of the idea. First, if Y is the correct (uppercase with spaces) string and Z is the other (inverted title case with nulls) string, then we want to show that there's a sequence of local, global, and local shifts taking Y to Z. It turns out that a local shift of 8-i, a global shift of 1, and then a local shift of i-1, works. These operations replace the ith bit of each 8-bit block with the ith bit of the preceding block. In this way, each block receives a 0 from the preceding block unless the preceding block is a space, in which case it receives a 1 (and becomes lowercase).

Minipuzzle #4

In Gregor and the Code of Claw, by Suzanne Collins, the rats use a cipher called the Code of Claw for wartime communications. The humans eventually realize that the Code of Claw is a modified Caesar shift, where the letters in GORE are invariant but all other letters shift forwards by one. For example, the word “FORT” would become “HORU”. In this puzzle, we start with some ciphertext in the Code of Claw, and need to figure out what the invariant keyword is.

The given numbers are all between 1 and 26, suggesting that these should be converted to letters via A1Z26.

After playing with this for a little while, we should realize that the plaintext is

“May the odds be ever in your favor”
-Effie Trinket.

This is a famous phrase from a better-known series by Suzanne Collins! Once this is understood, the set of fixed letters is seen to be {A, E, N, O, P, R, S}. We know that the letter 'P' is fixed because the 'M' in "MAY" became a 'Q', skipping over 'N', 'O', and 'P'.

The underscores and greater-than/less-than signs give us a partial ordering of the letters (interpreted as A1Z26 numbers), and the answer is

P > E > A < R < S > O > N,

or PEARSON.