Some trivial XOR Cheatsheet and Codecard Generator Scripts, useable as "computerless" one time pad, with pen and paper as a giveaway for friends.
You can download this small PDF Cheatsheet to memorize ASCII table and exclusive-or-calculation for nibbles in hexadecimal notation.
Nibble Hex +++ Rechenbeispiel +++ ----------- XOR 0 1 2 3 4 5 6 7 8 9 A B C D E F XOR 0000 0 Beispiel: Ein Byte in beide Nibble zerlegen #000000 black 0001 1 0 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 also wenn "8a xor d7" #000080 navy 0010 2 1 1 0 3 2 5 4 7 6 9 8 B A D C F E 1 8 xor d = 5 #008000 green 0011 3 2 2 3 0 1 6 7 4 5 A B 8 9 E F C D 2 a xor 7 = d #008080 teal ----------- 3 3 2 1 0 7 6 5 4 B A 9 8 F E D C 3 Gesamtergebnis => 5d 0100 4 4 4 5 6 7 0 1 2 3 C D E F 8 9 A B 4 8a xor d7 = 5d #800000 maroon 0101 5 5 5 4 7 6 1 0 3 2 D C F E 9 8 B A 5 #800080 purple 0110 6 6 6 7 4 5 2 3 0 1 E F C D A B 8 9 6 #808000 olive 0111 7 7 7 6 5 4 3 2 1 0 F E D C B A 9 8 7 #C0C0C0 silver ----------- 8 8 9 A B C D E F 0 1 2 3 4 5 6 7 8 1000 8 9 9 8 B A D C F E 1 0 3 2 5 4 7 6 9 XOR 0 1 AND 0 1 OR 0 1 #808080 gray 1001 9 A A B 8 9 E F C D 2 3 0 1 6 7 4 5 A 0 0 1 0 0 0 0 1 1 #0000FF blue 1010 A B B A 9 8 F E D C 3 2 1 0 7 6 5 4 B 1 1 0 1 0 1 1 1 0 #00FF00 lime 1011 B C C D E F 8 9 A B 4 5 6 7 0 1 2 3 C #00FFFF aqua ----------- D D C F E 9 8 B A 5 4 7 6 1 0 3 2 D 1100 C E E F C D A B 8 9 6 7 4 5 2 3 0 1 E #FF0000 red 1101 D F F E D C B A 9 8 7 6 5 4 3 2 1 0 F #FF00FF fuchsia 1110 E #FFFF00 yellow 1111 F XOR 0 1 2 3 4 5 6 7 8 9 A B C D E F XOR #FFFFFF white --------------------------- --------------------------- --------------------------- ---------------------------------------- Char Binary Code Hex Char Binary Code Hex Char Binary Code Hex Char Binary Code Hexadecimal Code --------------------------- --------------------------- --------------------------- ---------------------------------------- A 0100 0001 41 a 0110 0001 61 0010 0000 20 B 0100 0010 42 b 0110 0010 62 C 0100 0011 43 c 0110 0011 63 ! 0010 0001 21 Deutsche Umlaute in Codepage ISO 8859-1 D 0100 0100 44 d 0110 0100 64 " 0010 0010 22 ---------------------------------------- E 0100 0101 45 e 0110 0101 65 # 0010 0011 23 ö 1111 0110 F6 F 0100 0110 46 f 0110 0110 66 Ö 1101 0110 D6 G 0100 0111 47 g 0110 0111 67 ( 0010 1000 28 ä 1110 0100 E4 H 0100 1000 48 h 0110 1000 68 ) 0010 1001 29 Ä 1100 0100 C4 I 0100 1001 49 i 0110 1001 69 ü 1111 1100 FC J 0100 1010 4A j 0110 1010 6A + 0010 1011 2B Ü 1101 1100 DC K 0100 1011 4B k 0110 1011 6B , 0010 1100 2C ß 1101 1111 DF L 0100 1100 4C l 0110 1100 6C - 0010 1101 2D M 0100 1101 4D m 0110 1101 6D . 0010 1110 2E N 0100 1110 4E n 0110 1110 6E / 0010 1111 2F Deutsche Umlaute in UTF 8 O 0100 1111 4F o 0110 1111 6F ----------------------------------------- P 0101 0000 50 p 0111 0000 70 : 0011 1010 3A ö 1100 0011 1011 0110 c3 b6 Q 0101 0001 51 q 0111 0001 71 ; 0011 1011 3B Ö 1100 0011 1001 0110 c3 96 R 0101 0010 52 r 0111 0010 72 ä 1100 0011 1010 0100 c3 a4 S 0101 0011 53 s 0111 0011 73 ? 0011 1111 3F Ä 1100 0011 1000 0100 c3 84 T 0101 0100 54 t 0111 0100 74 ü 1100 0011 1011 1100 c3 bc U 0101 0101 55 u 0111 0101 75 0 0011 0000 30 Ü 1100 0011 1001 1100 c3 9c V 0101 0110 56 v 0111 0110 76 1 0011 0001 31 ß 1100 0011 1001 1111 c3 9f W 0101 0111 57 w 0111 0111 77 2 0011 0010 32 X 0101 1000 58 x 0111 1000 78 3 0011 0011 33 Warum hat man sich gerade c3 zur Y 0101 1001 59 y 0111 1001 79 4 0011 0100 34 Signalisierung der utf8 folge ausgesucht? Z 0101 1010 60 z 0111 1010 7A 5 0011 0101 35 Weil es binär 1100 0011 schick aussieht. 6 0011 0110 36 Entspricht ja eigentlich ascii der 7 0011 0111 37 A-Tilde Glpyhe. Nur die ersten 128 Chars, 8 0011 1000 38 0-127 also die 7 bit des bytes sind ja 9 0011 1001 39 für die printable chars genutzt.
Important hint for any real world usecase: choose a better random number generator for your keycode. Do NOT USE pseudo random number generator /dev/urandom like in this example. don't use computers at all. use pen, paper and 16 sided dices. xor your random keycode with multiple other random keycodes, to get "better" randomness at all. then use it as key. just for fun.
feel free to use this oneliner to generate some small and convenient one-time-pad codecards for testing purpose, suitable for your wallet.
just generate your pdf file, print it twice, put it in your wallet and give the copy to a friend.
NEVER EVER use a codepage twice!
the following onliner uses the following tools a2ps, xxd, and ps2pdf from ghostscript
and will generate such a codecard_example.pdf (page format A4 landscape). You'll have to use a separate sheet of paper for encrypting/decrypting.
Use text2pdf.c and this oneliner to directly generate a pdf file
this will generate a page like this codesheet_example.pdf. print the page twice. keep one, give the copy to your friend. You can directly encrypt and decrypt on this printed page. nibble by nibble, line by line, page by page. The follwing example is using this method.
In a nutshell: Use the codesheet_example.pdf linked above. Use a red pen to write clear text hex nibbles above the black lined keycode. xor each nibble of each byte vertically and write the result with a green pen below the black lined key hex nibbles. then green code is your encrypted message. send it to your friend. you can use any unsecure channel.
Example: You want to encrypt The following Message "Hello Bob. The Weather is nice"
So look at a ASCII-table. Character by Character. This Message in cleartext ascii hex:
48 65 6C 6C 6F 20 42 6F 62 2E 20 54 68 65 20 57 65 61 74 68 65 72 20 69 73 20 6E 69 63 65.
You write this ASCII HEX with a red pen above the black lined keycode hex.
Here, the first line:
H e l l o B o b . T h e W 48 65 6C 6C 6F 20 42 6F 62 2E 20 54 68 65 20 57 cleartext
93 31 4F 1D D1 0F F1 7D 47 A6 F7 AF 3F 7B AD 37 keycode
now, do the xor boogie. calculate nibble by nibble of each byte vertically:
First byte of the cleartext (48) xored with the first byte of the keycode (93) means
individually xoring the nibbles:
Have a Look at the XOR Table. X-Axis and Y-Axis for better understanding:
XOR 0 1 2 3 4 5 6 7 8 9 A B C D E F XOR 0 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 1 1 0 3 2 5 4 7 6 9 8 B A D C F E 1 2 2 3 0 1 6 7 4 5 A B 8 9 E F C D 2 3 3 2 1 0 7 6 5 4 B A 9 8 F E D C 3 4 4 5 6 7 0 1 2 3 C D E F 8 9 A B 4 5 5 4 7 6 1 0 3 2 D C F E 9 8 B A 5 6 6 7 4 5 2 3 0 1 E F C D A B 8 9 6 7 7 6 5 4 3 2 1 0 F E D C B A 9 8 7 8 8 9 A B C D E F 0 1 2 3 4 5 6 7 8 9 9 8 B A D C F E 1 0 3 2 5 4 7 6 9 A A B 8 9 E F C D 2 3 0 1 6 7 4 5 A B B A 9 8 F E D C 3 2 1 0 7 6 5 4 B C C D E F 8 9 A B 4 5 6 7 0 1 2 3 C D D C F E 9 8 B A 5 4 7 6 1 0 3 2 D E E F C D A B 8 9 6 7 4 5 2 3 0 1 E F F E D C B A 9 8 7 6 5 4 3 2 1 0 F XOR 0 1 2 3 4 5 6 7 8 9 A B C D E F XOR
Now we xor the second byte. First Nibble 6 XOR 3 = 5 and second nibble
5 XOR 1 = 4, resulting in the second encrypted byte 54. and so on...
H e l l o B o b . T h e W 48 65 6C 6C 6F 20 42 6F 62 2E 20 54 68 65 20 57 cleartext
93 31 4F 1D D1 0F F1 7D 47 A6 F7 AF 3F 7B AD 37 keycode
↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓
DB 54 23 71 BE 2F B3 12 25 88 D7 FB 57 1E 8D 60 encrypted
For the whole message this is:
H e l l o B o B . T h e W 48 65 6C 6C 6F 20 42 6F 62 2E 20 54 68 65 20 57 cleartext line 1 ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ 93 31 4F 1D D1 0F F1 7D 47 A6 F7 AF 3F 7B AD 37 keycode ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ DB 54 23 71 BE 2F B3 12 25 88 D7 FB 57 1E 8D 60 encrypted e a t h e r i s n i c e 65 61 74 68 65 72 20 69 73 20 6E 69 63 65 cleartext line 2 ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ 82 D5 82 FB 7A C8 43 00 68 E8 38 5E FE 6B 4B 23 keycode ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ E7 B4 F6 93 1F BA 63 69 1B C8 56 37 9D 0E encrypted
So your encrypted message is: DB 54 23 71 BE 2F B3 12 25 88 D7 FB 57 1E 8D 60 E7 B4 F6 93 1F BA 63 69 1B C8 56 37 9D 0E
Just like Encrypting, but in reverse. Use a green pen to write the encrypted hex above over the black line of keycode. do the xor nibble boogie and take a red pen to write the clear text hex nibbles below the black lined hex codes.
DB 54 23 71 BE 2F B3 12 25 88 D7 FB 57 1E 8D 60 encrypted ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ 93 31 4F 1D D1 0F F1 7D 47 A6 F7 AF 3F 7B AD 37 keycode ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ 48 65 6C 6C 6F 20 42 6F 62 2E 20 54 68 65 20 57 cleartext line 1 H e l l o B o B . T h e W E7 B4 F6 93 1F BA 63 69 1B C8 56 37 9D 0E encrypted ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ ⊻⊻ 82 D5 82 FB 7A C8 43 00 68 E8 38 5E FE 6B 4B 23 keycode ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ ↓↓ 65 61 74 68 65 72 20 69 73 20 6E 69 63 65 cleartext line 2 e a t h e r i s n i c e
It is safe, if and because:
Math Formular is:
x xor y = y xor x
x xor (y xor z) = (x xor y) xor z
x xor 0 = x
x xor y xor y = x
x xor x = 0
x xor 0xFFFF = not x
Additionally, XOR can be composed using the 3 basic operations (AND, OR, NOT)
a xor b = (a or b) and ((not a) or (not b))
a xor b = (a and (not b)) or ((not a) and b)
You can quickly lookup xor of two hex nibbles using the following 3-tuples list. Memorize the following combinations or use it as a short lookup table:
-1- -2- -3- -4- -5- -6- -7- 123 | 246 | 347 | 48C | 58D | 68E | 78F 145 | 257 | 356 | 49D | 59C | 69F | 79E 167 | 28A | 38B | 4AE | 5AF | 6AC | 7AD 189 | 29B | 39A | 4BF | 5BE | 6BD | 7BC 1AB | 2EC | 3CF 1CD | 2DF | 3DE 1EF | if x==y then x xor y = 0 x xor 0 = x ( and 0 xor x = x)
RULE 1: "Same Hex?> Result is Zero (0)" - Example: So F xor F is 0. 1 xor 1 is 0, 2 xor 2 is 0... and so on... so just remember: same hex? result is 0.
RULE 2: "Any Zero? Result is the other part" - Example: 0 xor A is A, 4 xor 0 is 4, 7 xor 0 is 7.... so just just remember: any of them is 0, just omit 0 and take the non-zero part as result.
RULE 3: Any other combination? -> Three-Tuple-Lookup-Method! Example 8 xor 1 is 9.
8 xor 1? All you have to do: choose the least number, which is 1, and look in column 1 above. pick the tuple "189", since there you'l find 8. so 8 xor 1 is 9, since it is the third and last element in this 3-tuple. This also means: 8 xor 9 is 1. and 9 xor 1 is 8, and 1 xor 8 is 9, and 1 xor 9 is 8, ...
Example: D xor E is 3. since there is no D or E in lookuptable, begin in column 7, see there is no DE in any tuple, go back to column 6, se there is also no DE, go back to column 5, so there is also no DE, go back to column 4, see there is also no DE, go to column 3 and see there is 3DE, so D xor E is 3. Same for E xor D.
Just convert the Nibble from HEX to Binary Notation, xor them and convert it back to HEX Notation!
XOR means EXLUSIVE OR, so you can use the nmemonic: "the result is only 1 if there is no part exclusively!". Meaning 0 xor 0 is 0, and 1 xor 1 is 0, since yin and yang are never together. every part IS present exclusively". Another nmemonic:
"Only if there exists a yin (0) AND a yang (1), the result is fine (1)". So only 1 xor 0 is 1. 0 xor 1 is 1.
You can also nmemonic "FAIL, if same input!" or another nmemonic: "It just WINS, if there a no TWINS.".
Just have a look at this so called Truth table for binary logic operation:
0 xor 0 = 0 input is the same. only yin twins, fail. 0 xor 1 = 1 input is yin and yang! bingo! no twins. fine. 1 xor 0 = 1 input is yang any yin! bingo! no twins. fine. 1 xor 1 = 0 input is the same. only yang twins, fail.
Just Write any HEX Nibble in Binary notation. (Have a look a the cheat sheet table, upper left corner) Example: 2 (0010) xor D (1101) is F (1111)...
now just do binary xor (from top to bottom) with the following rule: So the result for 2 xor D is.. 0 xor 1 = 1, 0 xor 1 = 1, 1 xor 0 = 1, 0 xor 1 = 1, so 1111, so F.
2 = 0010
D = 1101
--------
↓↓↓↓
1111 = F
to memorize the xor table. symmetry is clearly recognizable.
P2 16 16 16 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 0 3 2 5 4 7 6 9 8 11 10 13 12 15 14 2 3 0 1 6 7 4 5 10 11 8 9 14 15 12 13 3 2 1 0 7 6 5 4 11 10 9 8 15 14 13 12 4 5 6 7 0 1 2 3 12 13 14 15 8 9 10 11 5 4 7 6 1 0 3 2 13 12 15 14 9 8 11 10 6 7 4 5 2 3 0 1 14 15 12 13 10 11 8 9 7 6 5 4 3 2 1 0 15 14 13 12 11 10 9 8 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 9 8 11 10 13 12 15 14 1 0 3 2 5 4 7 6 10 11 8 9 14 15 12 13 2 3 0 1 6 7 4 5 11 10 9 8 15 14 13 12 3 2 1 0 7 6 5 4 12 13 14 15 8 9 10 11 4 5 6 7 0 1 2 3 13 12 15 14 9 8 11 10 5 4 7 6 1 0 3 2 14 15 12 13 10 11 8 9 6 7 4 5 2 3 0 1 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
You can find sourcecode xor.c right here. this tool will xor stdin or infile against keyfile and will output the xored result to stdout.