Encryption Test :D

ok guys.

i know i have done this in the past but i like to keep upto date on all of ya. i usually give out a string for you peoples to decrypt, with a few clues and see how long it takes you guys to decrypt it. now, because i change the algorithm constantly, i dont write a program for me to do this. but i write it out manually, so human error might be a possibility but unlikely. so, here are the clues and the string.

After a week or so i will post how i did it :) have fun

If you really want more clues or different ones, you gonna have to ask.... but dont ask stupid ones like, what does the a stand for. :P

[Clues]
1) 3 Matrices were used
2) All letters are simple substitution with another letter
3) spaces are one of 3 nominated (3 letter) combinations. the combinations will not be found as a part of any word. example: ter (combination) word: letter. they will be like zbm (combination) so that it doesnt fit into any word.....sorta hard to explain.
4) There are no numbers used. hence no algorithm for them.
5) There are a total of 16 whole words

[String]

jvukjklypalfnjuoptfcgckqhdqczyqhdgujvjkljvlyytfcqcjluayk.tfcfoijklqikjqhdbyqhdwoozjklcjtfcjvukqhdjojklalcasqhduj!

I'm confused lol . I will be interested to find out how this is done

Okay okay,

obviously im not detailed enough, so here is the "source code" that i used to make the code with.

download it and open it up and read through as i explain it.

At the very top you will see 3 matrices, all with the alphabet running along the top. if i want to letter A, then i start at the top, and work my way to the next matrix. so, it goes like this.

a=z

now move the the second matrix with the letter z.

z=i

now move to the third matrix with the letter i

i=c

so, that for the letter a. every a used, becomes c through that process. work it all out if ya dont believe me.

Space, now i nominated 3 strings to represent coded spaces. these strings dont contain letters that would show up normally, so hopefully and decryption program can notice that it doesnt form a word and that it is a space.

anyways, once you see the source i spose you will understand better. i usually post these when i get around to making them.

*choke*... I was on the wrong course here. I thought that it was just one substitution for the whole thing... Anyways, I thought that the "j" was definately "e", because it showed up so many times. I had just started trying to deduce what the vu was, and I came up with one of "cubfgmpwy" for v, and "hrsdl" for u... ...Just letter frequency checking.

Well anyways, neat encryption test. It had me fooled. :D


-Tim_axe

yea, when i said that it was letter substitution, i wasnt lying, but i also said it went through 3 matrices and im not sure how many people knew what that meant. anyways, theres always next time i do it :P

Do you always use all three matrices as they stand or do you shift them about at all?
If you don't do any shifting or reordering or using only some of them or the like then there really wouldn't be any difference between using one matrix and any number of matrices. for instance:

in the first matrix:
a always becomes z

in the second
z always becomes i

in the third
i always becomes c

so therefore
a always becomes c and the intermediate steps are irrelevant and it's just a simple newspaper cryptogram

now granted, the random choice of a three letter combination for the spaces would complicate this somewhat, but once that was figured out, it becomes fairly simple. With a long message this would become even easier since the space replacements would start to repeat and a pattern would emerge and also using letter frequency is more likely to break it.

Now a good alteration to this algorithm might be to use the first matrix on the first letter, then the first+second on the second letter, and the first+second+third on the third letter, or some combination like this.

I'm working on another Encryption Scheeme which should by nature be impossible to crack...well.....very damn difficult.....hehehe...I love stuff like this, but your encryption challenge was rather difficult to get to grip's with....try maing it a bit easier next time, and then build it up from that... :) hehe, looks like you read my Tut on Marix Encryption though

- Noia

Noia,

actually i didnt know you had a tut on matrix encryption...i already knew.... but now that you mention you have one me might have a look ;)

Trust_Not_123; The following is informative, not insultive (I enjoyed the puzzle, but have some issues with your method).

First, you can't encode unless you know how to spell.

"This encrxption was made..."?

Second, you needed to state that the triple strings (for spaces) were inserted into the code after the encoding process. Insertion prior to the encoding could result in a triple string that did not meet your set criteria. You should have also stated that each string was repetative, as after encoding many single strings occur that meet your criteria. As the puzzles stands in the original post, no key was provided, and the final encoded string contains too many variables to be broken within a reasonable time frame with the limits you set.

Third, you didn't have to give the full key in your follow up post. Perhaps a word problem as a clue would have been more helpful. For example, you first matrice could have been expressed with "The boy looks into a mirror." The second matrice could have been stated, "He will be ten years old next month." (The boy is currently 9, but progressing in a positive numeric direction, thus giving the alphanumeric key A=J, or each letter equals a letter removed 9 progressions in a positive direction.

Fourth, and most importantly, you used a sum pattern in all three matrices that can easily be condensed into a single matrice based off the first matrice. The second and third matrice was just extra work, each adding a sum to the base matrice. ABC=CBA solves your puzzle. Perhaps a matrice based upon a repeating pattern in the third decryption sequence would have been a little more difficult, after giving out a key.

Keep these posts coming, just refine them a little please!!!!

(Im really not being picky here..being picky would entail me pointing out that the first, and easiest matrice to attempt, being a common matrice, was posted incorrectly.)

Let's work through it a little:
jvukjklypalfnjuoptfcgckqhdqczyqhdgujvjkljvlyytfcqcjluayk. tfcfoijklqikjqhdbyqhdwoozjklcjtfcjvukqhdjojklalcasqhduj!

We know the spaces between words are represented by the letter groups jkl, qhd or tfc, so let's remove those:
jvuk ypalfnjuop gck qczy gujv jvlyy qcjluayk. foi qikj by wooz cj jvuk jo alcas uj!

Trust_Not_123 first matrice (key) to decoding is the following, as listed above:
a=z b=y c=x d=w e=v f=u g=t h=s i=r j=q k=p l=o m=n n=m o=l p=k q=j r=i s=h t=g u=f v=e w=d x=c y=b z=a
So letters change according to the key:
qefp bkzoumqflk txp jxab (first four letter groups only, you can do the rest yourself)

Trust_not_123 second matrice (key) to decoding our new letters is the following, as listed above:
a=j b=k c=l d=m e=n f=o g=p h=q i=r j=s k=t l=u m=v n=w o=x p=y q=z r=a s=b t=c u=d v=e w=f x=g y=h z=i
So the letter groups change again, having new values assigned to them by a new key:
znoy ktixdvzout cgy sgjk (I told you Im only doing the first four letter groups)

Then Trust_Not_123 gives us the final key, to change our new letters once more, hopefully into something understandable:
a=u b=v c=w d=x e=y f=z g=a h=b i=c j=d k=e l=f m=g n=h 0=i p=j q=k r=l s=m t=n u=o v=p w=q x=r y=s z=t
revealing:
this encrxption was made (The word encryption was obviously mispelled before being encoded by Trust_Not_123).

Now after combining the three keys, I came up with a single key that follows a simple pattern:

a=c b=b c=a d=z e=y f=x g=w h=v i=u j=t k=s l=r m=q n=p o=o p=n q=m r=l s=k t=j u=i v=h w=g x=f y=e z=d
This key will unlock the original code, after the letter groups jkl, qhd or tfc are removed.

Code with spelling corrected:
jvuk ypalenjuop gck qczy gujv jvlyy qcjluayk. eoi qikj by wooz cj jvuk jo alcas uj!


Answer: This encryption was made with the matrices. You must be good at this to crack it.

have a good day

um, no. i spelt it correctly, you have just checked that letter back to front. you will see im right :D

im pretty sure that everything is spelt correctly

Usually i'd be the first to jump at a chance to do some basic decryption but somehow i seemed to have missed this when it was first posted. Anyway, phazeout420 is correct, you misspelled or miscrypted(that is not even a word but it should be) the word "encryption". Also, using multiple caesar ciphers yields no greater degree of difficulty than using one because the result is a single caesar cipher with a shift equal to the net of each that you used. For instance a caesar cipher shift 7 followed by a caesar cipher shift 5 is the equivilant of a caesar cipher shift 12. That in mind, i obtained your original mesage as follows:

jvukjklypalfnjuoptfcgckqhdqczyqhdgujvjkljvlyytfcqcjluayk.tfcfoijklqikjqhdbyqhdwoozjklcjtfcjvukqhdjojklalcasqhduj!

look for recuring groups of 3 letters and replace them with a space character.
qhd, jkl, tfc removed

jvuk ypalfnjuop gck qczy gujv jvlyy qcjluayk. foi qikj by wooz cj jvuk jo alcas uj!

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
d e f g h i j k l m n o p q r s t u v w x y z a b c
(caesar cipher shift 3)

myxn bsdoiqmxrs jfn tfcb txmy myobb tfmoxdbn. irl tlnm eb grrc fm myxn mr dofdv xm!

d e f g h i j k l m n o p q r s t u v w x y z a b c
c b a z y x w v u t s r q p o n m l k j i h g f e d
(inversion)

this encrxption was made with three matrices. you must be good at this to crack it!

trust_not_123
i hope you will continue posting
i enjoyed decrypting it
take care for now