An acquaintance of mine pointed me to a comic stip that had a task to perform cryptoanalyzis on the string !35452555844414548495D4. Stupid as I am I went ahead with the challenge.

The paper I used look like:

Cipher

! 3 5 4 5 2 5 5 5 8 4 4 4 1 4 5 4 8 4 9 5 D 4



! is possibly a padding



Frequency analyzis

! |

1 |

2 |

3 |

4 |||||||

5 ||||||

6

7

8 ||

9 |

A

B

C

D |



Guess, starting point: 5 is E

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 A B C D E F



!3

54

52

55

58

44

41

45

48

49

5D

4



In blocks of two, every pair start off with ! 5 or 3



  1 2 3 4 5 8 9 D

! a b c d e f g h

5 i j k l m n o p

4 q r s t u v w x



cljmntquvwp



  1 2 3 4 5 8 9 D

! a b c d e f g h

4 i j k l m n o p

5 q r s t u v w x



ctruvlimnwx



35

45

25

55

84

44

14

54

84

95

D4



  1 2 3 4 5 8 9 d

4 a b c d e f g h

5 i j k l m n o p



kljmfdaefop



  1 2 3 4 5 8 9 d

4 t e q u i l a k

5 b c d f g h j l



dfcglutil



frequency of pairs

14

25

35

44

45

55

84 - twise

54

95

D4



84 = e ?



35 y

45 o

25 u 

55 n

84 e

44 e

14 d

54 h

84 e

95 l

D4 p



  1 2 3 4 5 8 9 d

4           e

5 



----e---e--



my guess, 54 minutes after seeing the cipher is:

youneedhelp



in which case the matrix is



  1 2 3 4 5 8 9 d

4 d     e h e l p

5   u y o n