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This is a slight improvement of the ElsieFour cipher as described by Alan Kaminsky [1]. We use 7x7 characters instead of original (barely fitting) 6x6, to be able to encrypt some structured information. We also describe a simple key-expansion algorithm, because remembering passwords is popular. Similar security considerations as with ElsieFour hold.
There's also a 3D-printable SCAD model of the whole thing. Yay!
We suggest printing the model using more than one filament color to make the letters easily recognizable. Thanks go to Martin Ptasek for providing the picture.
If you trust your computer, there are several LS47 implementations around:
- A very simple python implementation in available here in
ls47.py. - A much better version usuable as an actual binary (also supporting several
versions of padding and the original ElsieFour cipher) was supplied by
Bernhard Esslinger from the CrypTool project, available in
lc4.py. - Javascript version (npm-compatible) of the cipher implementations was created by Ulysse McConnell, available at https://github.com/umcconnell/lc4
We have added some real punctuation, basic stuff for writing expressions, punctuation and quotes. The letters of the board now look like this:
_ a b c d e f
g h i j k l m
n o p q r s t
u v w x y z .
0 1 2 3 4 5 6
7 8 9 , - + *
/ : ? ! ' ( )
Zoomed in, it's very practical to have extra position information written on the tiles:
/-----\ /-----\ /-----\ /-----\ /-----\
| | | | | | | | | |
| _ 0| | a 1| | b 2| | c 3| | d 4| ...
| 0 | | 0 | | 0 | | 0 | | 0 |
\-----/ \-----/ \-----/ \-----/ \-----/
/-----\ /-----\
| | | |
| g 0| | h 1| ...
| 1 | | 1 |
\-----/ \-----/
. .
. .
. .
To run (hand-run) the encryption/decryption, you will also need some kind of a marker (e.g. a small shiny stone, bolt nut or similar kind of well-shaped trash).
You may as well see the paper [1], there are also pictures. This is somewhat more concentrated:
- The symmetric key is the initial state of the board. Arrange your tiles to 7x7 square according to the key.
- Put the marker on (0,0).
- Find the next letter you want to encrypt on the board, its position is
P. - Look at the marker; numbers written on the marked tile are coordinates
M. - Compute position of the ciphertext as
C := P + M mod (7,7). Output the letter found on positionCas ciphertext. - Rotate the row that contains the plaintext letter one position to the right, but do not carry the marker if present (it should stay on the same coordinates).
- Rotate the column that now (after the first rotation) contains the ciphertext letter one position down, also not carrying the marker.
- Update the position of the marker:
M := M + C' mod (7,7)whereC'are the numbers written on the ciphertext tile. - Repeat from 3 as many times as needed to encrypt the whole plaintext.
1,2. Symmetric key with 3,4. We want to encrypt 'y'.
marker put on 'e' Look at the marked tile:
[e]f _ a b c d /-----\
l m g h i j k | |
( ) / : ? ! ' | e 5|
s t n o p q r | 0 |
z . u v w x y \-----/
5 6 0 1 2 3 4
+ * 7 8 9 , -
5. Ciphertext is 'w' 6. Rotate the plaintext 1 position
(='y' moved by (5,0)) right, keep marker coordinates.
[e]f _ a b c d [e]f _ a b c d
Output 'w'! l m g h i j k l m g h i j k
( ) / : ? ! ' ( ) / : ? ! '
s t n o p q r s t n o p q r
z . u v w x y >> y z . u v w x
5 6 0 1 2 3 4 5 6 0 1 2 3 4
+ * 7 8 9 , - + * 7 8 9 , -
7. Rotate the ciphertext 1 Now look at the ciphertext tile:
position down.
[e]f _ a b , d /-----\
l m g h i c k | |
( ) / : ? j ' | w 2|
s t n o p ! r | 3 |
y z . u v q x \-----/
5 6 0 1 2 w 4
+ * 7 8 9 3 -
8. Update the marker position 9. GOTO 3.
by ciphertext offset (2,3).
e f _ a b , d
l m g h i c k
( ) / : ? j '
s t[n]o p ! r
y z . u v q x
5 6 0 1 2 w 4
+ * 7 8 9 3 -
Decryption procedure is basically the same, except that in step 5 you know C
and M, and need to produce P by subtraction: P := C - M mod (7,7).
Otherwise (except that you input ciphertext and output plaintext) everything
stays the same.
Grab a bag full of tiles and randomly draw them one by one. Key is the 49-item permutation of them.
Remembering 49-position random permutation that includes weird characters is not very handy. You can instead derive the keys from an arbitrary string of sufficient length.
"Sufficient" means "provides enough entropy". Full keys store around 208 bits of entropy. To reach that, your password should have:
- at least around 61 decimal digits if made only from random decimal digits
- at least around 44 letters if made only from completely random letters
- at least around 40 alphanumeric characters if made randomly only from them
To have the "standard" 128 bits of entropy, the numbers reduce to roughly 39, 28 and 25, respectively.
Note that you can save the expanded tile board for later if you don't want to expand the passwords before each encryption/decryption.
The actual expansion can be as simple as this:
- initialize
I:=0, put the tiles on the board sorted by their numbers (i.e. as on the picture above) - Take the first letter of the password and see the numbers on its tile; mark them
Px, Py. - Rotate
I-th rowPxpositions right - Rotate
Ith columnPypositions down I := I + 1 mod 7, repeat from 2 with next letter of the password.- Resulting tile positions are the expanded key
To get a different ciphertext even if the same plaintext is encrypted repeatedly; prepend it with a nonce. A nonce is a completely random sequence of letters of a pre-negotiated length (e.g. N tiles drawn randomly from a bag, adviseable value of N is at least 10).
You may also want to add a random number of spaces to the end of the ciphertext
-- it prevents the enemy from seeing the difference between ciphertexts of 'yes
please' and 'no', which would otherwise encrypt to gibberish that is easily
distinguishable by length, like qwc3w_cs'( and +v.
Because ciphertext may be altered in the transfer or during the error-prone
human processing, it is advised to append a simple "signature" to the end of
the message; which may look as simple as __YourHonorableNameHere. If the
signature doesn't match expectations (which happens with overwhelming
probability if there was any error in the process), either try again to see if
you didn't make a mistake, or discard the message and ask the sender to
re-transmit.
This works because the cipher output is message-dependent: Having a wrong bit somewhere in the middle causes avalanche effect and erases any meaning from the text after several characters.
If you find the above tiles complicated to obtain or create, you can very easily use playing cards to do the same, similarly as with the Solitaire cipher by Schneier. (Moreover, use of playing cards could be more innocuous and easily explainable to a suddenly appearing adversary. Playing cards is sure less suspicious than playing some peculiarly numbered tiles! :] )
To simplify things a bit, we will use the following layout:
a b c d e f g
h i j k l m n
o p q r s t u
v w x y z _ .
, - + * / : ?
! ' ( ) 1 2 3
4 5 6 7 8 9 0
This maps nicely to the playing card suits:
| Card value | ♦ | ♣ | ♥ | ♠ |
|---|---|---|---|---|
| A | a | n | _ | 1 |
| 2 | b | o | . | 2 |
| 3 | c | p | , | 3 |
| 4 | d | q | - | 4 |
| 5 | e | r | + | 5 |
| 6 | f | s | * | 6 |
| 7 | g | t | / | 7 |
| 8 | h | u | : | 8 |
| 9 | i | v | ? | 9 |
| 10 | j | w | ! | 0 |
| J | k | x | ' | |
| Q | l | y | ( | |
| K | m | z | ) |
The last 3 cards are not used. (Actually, you are free to use the suits in whatever order you like and discard the last 3 of the suit you like the least.)
Getting the X and Y offsets from the cards is a slightly more challenging than
with the tiles that have the offsets marked, but still doable. Calculate
13*suit + card value - 1, then compute the standard quotient and remainder
for division by 7. I settled with having the second card set laid out next to
the "active" board, serving as a reference.
The result may look like this:
In this example, the marker is on J♣ (at the top left), and we want to encrypt
the plaintext character e. The image includes the character, index, and
offsets for the relevant cards.
- From the card mapping,
emaps to 5♦ (lettersathroughmmap to Tiles), which is on the second row. - The marker card, J♣, has:
- an index of 24 (Clubs are suit 1, and the Jack is the 11th card in the
suit):
i = 1 * 13 + 11 = 24 - an x-offset of 3:
x = i % 7 = 24 % 7 = 3 - a y-offset of 3:
y = i / 7 = 24 / 7 = 3
- an index of 24 (Clubs are suit 1, and the Jack is the 11th card in the
suit):
- Using the marker offsets, the ciphertext card is three rows down and three columns to the right of 5♦, which is 6♠.
- From the card mapping, 6♠ maps to
6(Spades map directly to digits), and has:- an index of 45 (Spades are suit 3, and 6 is the 6th card in the suit):
i = 3 * 13 + 6 = 45 - an x-offset of 3:
x = i % 7 = 45 % 7 = 3 - a y-offset of 6:
y = i / 7 = 45 / 7 = 6
- an index of 45 (Spades are suit 3, and 6 is the 6th card in the suit):
- We then complete the cipher as normal:
- Output
6as the ciphertext - Rotate the row containing the plaintext card, 5♦
- Rotate the column containing the ciphertext card, 6♠
- Move the marker according to the offsets of the ciphertext card, to the right 3 and down 6
- Output
For the original LC4, you need just 3 suits, modifying the board to "align" to the suits e.g. as follows:
a b c d e f
g h i j k l
m n o p q r
s t u v w x
y z _ 2 3 4
5 6 7 8 9 #
[1] Kaminsky, Alan. "ElsieFour: A Low-Tech Authenticated Encryption Algorithm For Human-to-Human Communication." IACR Cryptology ePrint Archive 2017 (2017): 339.