Can You Decrypt This Encrypted Text Using a Numerical Technique?

[al-mahed]

I didin't find any material regarding the frequency of syllables, digraphs or trigraphs, so it's a little bit difficult go any further.

I was considering if the text is chiphered with another kind of chiper, like the DES kind, RAS, or something like that... I am not an expert, far away of that, but my guess is: because of the patherns is very unlike a strong chiper like those generates patherns like these, so I still think that the chiper is some kind of more simple substitution chiper concerning syllables, digraphs or trigraphs

 

[ramsey2879]

Since this has been kicked around without it being broken, maybe (Mr) Barovdeh could give us more infro on the level of education of the student and any relevant course of study.

 

[ramsey2879]

I googled "73 75 78 76 63 64 68 66 53 54 58 56 52 74 89 77 72 63 79 57 00000000" and the only hit was this thread so I guess it is of little relevance. Just something to kick around when one is not busy. Or could this be something to worry over, like a cipher to start some event to happen around the world. It is not fair that they chose a student to write this! Egads, look at all those who are reading this cipher even now.

 

[ktoz]

This doesn't solve it, but if you break the cypher on the zero blocks and delete the digit on every even index, you get the following

35 86 35 86 23 87
35 86 34 86 34 86 24 97 23 97
23 76 23 76 23 76 23 76
35 86 35 97 56 86 35 86 35 97 56 86 34 98 45 87
35 86 35 87 35 97 34 86 23 97
23 76 23 76 23 76 23 76
35 86 35 87 35 86 34 97 23 98 32 87
35 86 34 87 35 86
35 86 35 86 23 87 23 97 22 98 22 87
35 87 24
23 76 23 76 23 76 23 76
35 86 35 86 34 87 24 97 22 98 22 98
35 87 35 86 35 87 23 97
34 87 34 86 34 87 45 97 45 98 45
35 87 56 86 35 87 56 65 35 87 35 98 56
35 86 35 86 35 87 35 76 35

Which seems to reveal some tantalizing patterns.

I ran that through http://www.paulschou.com/tools/xlate/" translator which yielded the following ascii text

#V#VW#V"V"VaaLLLL#V#a8V#V#a8V"b-W#V#W#a"VaLLLL#V#W#V"ab W#V"W#V#V#VWabW#WLLLL#V#V"Wabb#W#V#Wa"W"V"W-a-b-#W8V#W8A#W#b8#V#V#W#L#

not much help but I think the cypher might interlace irrelevant data in the even positions to throw would be decoders off the track.

 

[Dragonfall]

You know, one-time-pads and RSA are "not highly sophisticated" either.

 

[ktoz]

As hints go, "eight(8)" is pretty worthless. Looks like a function call to me, but a function call can mask enormous complexity. As revealed in my last post, there is some sort of prefixing going on where values on many even indices form groups that differ by only the prefix

43 45 48 46
63 65 68 66
...
73 75 78 76

etc. Remove every digit on even indices and you get the table in my last post. As that reveals, the patterns just seem too repetitive to be meaningful text. Perhaps there is some kind of sorting going on.

 

[ThsTorturedSo]

My guess as to the pattern I have outlined here (I've bracketed all the common prefix numbers per line:


[43 45 48 46] [63 65 68 66] [72] [63] [58 57]

[73 75 78 76] [63 64 68 66] [53 54 58 56 52] [74] [89] [77 72] [63] [79] [57]

[12 13] [87 86] [22 23] [77 76] [32 33] [67 66] [42 43] [57 56]

[43 45 48 46] [53 55] [39] [47] [55 56] [68 66] [43 45 48 46] [53 55] [39] [47] [55 56] [68 66 63 64] [59] [68 64 65 68 67]

[43 45 48 46] [53 55 58 57] [63 65] [39] [47] [73 74] [68 66] [72] [63] [49] [67]

[12 13] [87 86] [22 23] [77 76] [32 33] [67 66] [42 43] [57 56]

[33 35 38 36] [43 45 48 47] [53 55 58 56] [63 64] [59 57] [72] [63] [39] [58] [63] [72] [58] [37]

[43 45 48 46] [53 54 58 57] [63 65 68 66]

[33 35 38 36] [53 55 58 56] [62] [53] [48 47 42] [33] [59 57 52] [42] [39] [58] [42] [52 58] [37]

[53 55 58 57 52 54]

[12 13] [87 86] [22 23] [77 76] [32 33] [67 66] [42 43] [57 56]

[63 65 68 66] [53 55 58 56] [73 74 78 77 72] [54 59 57] [62] [72] [69 68] [72] [62] [49] [58]

[63 65 68 67] [53 55] [48 46 43 45] [58 57] [72] [63] [39] [47]

[73 74 78 77] [63 64 68 66] [53 54 58 57] [74 75] [49] [67 64 65] [79] [68] [54] [5]

[33 35 38 37 35 36] [48 46] [53 55 58 57 55 56] [46 45] [63 65] [78 77 73 75] [49] [38] [75 76]

[33 35] [78 76] [53 55 58 56] [63 65] [38 37] [43 45] [37 36] [73 75]

My guesses as to how the words are made:

1) The possibility of a base-8 cypher is pretty good.
2) A combination of a base-8 cypher and a letter-number cypher is possible as well. What I mean by this is that the coder may have taken the code and used a base-8 cypher to code the text, then coded it again by attaching numbers to each letter (a-1, b-2, c-3, etc.) and then used a shift to code the text again.

Then, the [5] at the end of line 14 is still slightly peculiar, but could possibly be an "E" if the student was lazy and didn't use a shift, but just used the a-1, b-2 scale.

3) Finally, my last guess, which is highly unlikely as it would be quite easy to figure out, is the possibility that the student used a letter-number cypher, attaching 1-a, 2-b, etc. and then multiplied by a random amount to get a separate number, then divided that number by another random number, although not so random as to get a decimal, so that the numbers came out to be 2-digit numbers which need to be multiplied then counted to find the letter.

Those are my current guesses.

 

[al-mahed]

seems that no one ever have decrypted this text

 

[kenewbie]

seems that no one ever have decrypted this text

The interesting part, however, is how hard it is to crack once the algorithm is known.

What immediately springs to mind here is that if you separate every octet and then divide this into 4 two digit numbers, you get a pretty sequential set, also, the octets themselves form something which appears to be sequential.

At the digit level:

4345484663656866
=
[43454846] [63656866]
=
43 45 48 46 , 63 65 68 66
=
+2, +3, -4 , +2 +3, -4

At the octet level:
[4] [6] [7] [6] [5]

+2 +1 -1 -1

This suggests a positional increment as a way of turning a mono-alphabetic cipher into a poly-alphabetical one.

My guess (after looking at this for 5 minutes and not knowing the background of the student) would be that this algorithm works something similar to this:

Let a be 1
Take the first block of plain text, P_a (Assume a block is 4 chars long)
Take each char in the block, C_n and pass it through the cipher along with the increments:
Cipher(C_n, n, a);
Increment a until you are out of text.

In other words, "a" and "n" are double running increments, which account for the two intervals of symmetry (the one at the double digit level and the one at the octet level).

So the complete mask of the cipher would be the function f(a,n). If you strip that away you are left with a mono-alphabetic cipher.

This might be wrong of course, but as I said earlier, the real test of a cipher is how hard it is to break once the algorithm is known.

k

 

[kenewbie]

You know, one-time-pads and RSA are "not highly sophisticated" either.

http://en.wikipedia.org/wiki/Egg_of_Columbus

k