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#1 2009-02-23 09:23:43

Z
Member
Registered: 2009-02-23
Posts: 1

Encryption 3

Dear fellow thisislegal members smile

I have some tiny problems with that challs. Got the clue, but:

1. because there are only numbers in the ciphertext, the encryption is done only in half, am I right?
2. because of the very nature of the cipher, the numbers should be in pairs, but there is a ? in the middle of the ciphertext. Does it mean that the number before the ? is not sure?
3. because the challenge page said that no key is needed, is the assumption right that the fill-in of the square is default?

Thanks for helping
Z

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#2 2009-02-23 09:35:06

t0mmy9
Administrator
Registered: 2005-01-07
Posts: 21

Re: Encryption 3

1. the encryption uses numbers only
2. since this encryption only has the alphabet and not a "?" sybmol, treat that as a normal question mark.
3. this uses the default values for this cipher

if you got the clue, you shouldnt be too far from doing this, good luck


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#3 2009-06-07 20:42:21

WingeDD
Member
Registered: 2009-06-07
Posts: 7

Re: Encryption 3

Are you 100% sure this challenge used the default Polybius square? I don't think so.


Ciphertext: ymv rsyADMIN EDIT, some removed


Plaintext using default PS: xunkyd...

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#4 2009-06-08 02:27:31

t0mmy9
Administrator
Registered: 2005-01-07
Posts: 21

Re: Encryption 3

You seem to be close but making a mistake on the Y somehow. The encryption should be:

543251...etc

but yours comes out as:

553352...etc


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#5 2009-06-12 14:14:27

s_ha_dum
Member
Registered: 2009-06-12
Posts: 4

Re: Encryption 3

Got it, but I can't make sense of the hint that its a bifid or of the claim that it uses 'default values'. I solved it only after tossing both of those assumptions.

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#6 2009-07-01 23:18:40

busyr
Member
Registered: 2009-07-01
Posts: 5

Re: Encryption 3

I guess I'm making the same mistake as WingeDD, since I'm coming up with the same plaintext...  weird...

hackin on..

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#7 2009-07-19 07:19:23

WingeDD
Member
Registered: 2009-06-07
Posts: 7

Re: Encryption 3

But that's the point - 55 is Z, not Y. Maybe you were using an unknown J-included non-square?

P.S. since we have 11, I cannot use your hint. I can't shift THAT up.

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#8 2009-07-19 09:34:09

t0mmy9
Administrator
Registered: 2005-01-07
Posts: 21

Re: Encryption 3

I understand your question. There are 2 defaults for this type of cypher. This one is the "J included" version. The other default may share a square, but this one doesnt. So in this case 55 is not Z. I hope it isnt too hard to guess the other default from this information.


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#9 2009-07-20 13:13:48

busyr
Member
Registered: 2009-07-01
Posts: 5

Re: Encryption 3

hmm.. I've found another 'default' that doesn't share a square, but Z is still 55 in that version...

Is there a Z at all in your default?

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#10 2009-07-20 15:50:48

t0mmy9
Administrator
Registered: 2005-01-07
Posts: 21

Re: Encryption 3

that would be giving away a lot tongue


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#11 2009-07-22 15:24:56

Fuckdarock
Member
Registered: 2008-06-12
Posts: 13

Re: Encryption 3

wel hav got this 1 as a plaintext:

"XUNKYD RISPYU VCOUQO YEYUVY ADBOSY BU "

but im making some mistake; what would i do now????????

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#12 2009-09-14 23:59:35

xyberz09
Member
Registered: 2009-06-10
Posts: 46

Re: Encryption 3

Hey I'm pretty sure I solved this encryption but my answer just wont work!!<br />
frown<br />
I found a famous quote to be the ciphertext<br />
(which if you ask me didn't look very much like ciphertext tongue)<br />
<br />
Now i assume the answer is the name of th person, but it didn't work. I tried all variations in caps since the challenge said it was case sensitive. It still didn't work frown<br />
<br />
Please Help!!
<br />
<br />
EDIT: Never mind, i solved it biggrin
But I need to give all you guys a big hint. The statement "remember, this is also case sensitive." is misdirecting cos when you're solving this on your own, you probably don't have CAPS LOCK on. Anyway, the answer is in lowercase
<br />
<br />
BIG HINT: All those kiddies out there trying to solve this by heading to http://www.braingle.com/brainteasers/codes/polybius.php are in for a big surprise. To complete this challenge, you need to make your own Polybius Square encryption scheme that's a little different from the onef on the site. Look closely at the scheme on the site. It's probably missing something. Include this 'something' in your home-brewed Polybius cipher and Lo and Behold!! You've cracked it!! biggrin
<br />
<br />
TIP: The ? is just there to confuse you a bit. Just ignore it and solve the challenge smile
<br />
<br />
<br />
I hope my post has been helpful and not spoiler-ish.
tommy9 is requested to modify my post as much as necessary since I have no idea how much I've given away smile

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#13 2009-09-23 09:31:13

WingeDD
Member
Registered: 2009-06-07
Posts: 7

Re: Encryption 3

Ok t0mmy9, now my square has M as 33 and Y as 55, however... I'll try to decrypt it manually, but I really doubt the current ciphertext (which is valid according to your hint about what is the diff between my old and the correct sheme) will decrypt into anything meaningful.
The plaintext is WTMJX[...] and considering the clue is fulfilled, I'm quite out of ideas.

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#14 2009-09-23 13:30:55

t0mmy9
Administrator
Registered: 2005-01-07
Posts: 21

Re: Encryption 3

There are 2 default values to this encryption. Most people are using the first, this is the second. Maybe try change I/J


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#15 2009-09-23 21:09:52

WingeDD
Member
Registered: 2009-06-07
Posts: 7

Re: Encryption 3

check my PM

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#16 2009-09-25 02:08:15

xyberz09
Member
Registered: 2009-06-10
Posts: 46

Re: Encryption 3

@WingeDD: no matter in whichever fashion you arrange your Polybius Square you'll ALWAYS(I think) have M at 33 and Y at 55
(I'm talking about the 'J included' version)

I think your square has either J or I
(which according to Wikipedia is the way it should be)
but in this case, you have to construct such a square where 'J is included and nothing is left out'
(except Z maybe) smile

I hope I didn't give away much. BTW, this is NOT a conventional Polybius Square
(if you define 'conventional' by the way Wikipedia thinks about it)




PS: A [*] M [*] Y are in their original positions no matter whichever way I try to create my square

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#17 2009-09-26 06:16:47

WingeDD
Member
Registered: 2009-06-07
Posts: 7

Re: Encryption 3

Oh so you mean I just have to rotate the square... Thanks, i'll see about that.

Also I've sent the letters and their codes in my psquare and it has both J and I, your assumption is wrong. M is 32 in "conventional" squares. By the way, this square (I'm posting it because it isn't related to the solution) is which I used:
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

(letter Z is translated to letter I)


By the way, it was noted that the assumption this is Bifid cipher mustn't be used to solve this challenge. After reading how bifid is performed (rows <-> cols) I really believe that, unless a really heavily modified square is used (although tommy9's hint denies that).

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#18 2009-09-29 02:36:42

xyberz09
Member
Registered: 2009-06-10
Posts: 46

Re: Encryption 3

@WingeDD:

I checked your profile and it seems you haven't solved the challenge yet, so here's another hint:

> There are two ways to denote 'M' (or any other letter) in the 'conventional' Polybius Square. Using the first method, you'd get 32(as you said earlier) but using the second, you'd get something else. And I think it's the second way we're concerned about in here smile




PS: By the way, your square works perfectly fine. I checked it and there seems to be no problem. So  suppose you've got the co-ords wrong. It's a 5x5 square. Check out Wikipedia's article on Polybius Square. Hope this helps.

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#19 2009-10-17 06:56:43

tolito
Member
Registered: 2009-10-17
Posts: 1

Re: Encryption 3

I have the text decipher, and i try with the autor to the quote...but don't work.

Any hint?

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#20 2009-10-19 13:22:14

Glasklar
Member
Registered: 2008-12-29
Posts: 27

Re: Encryption 3

hrmm... i dont mean to be a douche, but Google Bifid and try for a bit, if it dont work try something simpler, the challenge is ranked easy

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#21 2010-02-03 17:52:38

dot_Cipher
Member
Registered: 2009-11-10
Posts: 13

Re: Encryption 3

Having a little trouble with this one too..  Trying doing multiple things to achieve the end plaintext.  I got the quote and the authors name correct so i am assuming that my polybus sqaure was correct as well in the decryption, i just dont know what do to from there since it wont let me enter in the authors name...<br />
<br />
EDIT:  I hope this doesnt give too much away but I also used bifid in both cases..  the plaintext quote and the plaintext author and reversed them again but only got gibberish from the name of the author like:<br />
<br />
QGBYDLHPP<br />
<br />

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#22 2012-08-30 22:19:52

Admiral Obvious
Member
Registered: 2012-06-02
Posts: 11

Re: Encryption 3

I am a bit frustated with this challenge tongue

[spoiler]
This is my Polybius square.
   1     2      3       4      5
1 A      B      C       D      E
2 F      G      H       I      J
3 K      L      M       N      O
4 P      Q      R       S      T
5 U      V      W       X      Y

First I tried splitting it in half:

5 4 3 2 5 1 4 2 4 3 5 4 5 1 3 4 4 3 5 1 5 4 3 5 5 1 1 1 3 4 5 1 4
3 5 3 5 4 4 2 4 3 5 4 5 1 3 4 5 1 4 4 5 4 5 1 4 1 4 2 4 3 4 2 5 4

WTMJXD (gibberish)
Then I tried taking each word individually:

543         42435451
251         34435154
VTK         RISMYPYI  (bullshit)

Then I decided to split the text in half(the question and the other part)
5 4 3 2 5 1 4 2 4 3 5
4 5 1 3 4 4 3 5 1 5 4
XTKHXDRJPOX
Again I got bullshit.

Before trying all this, I tried to do this:

543251 4243545134435154? 3551 113451 435354 42435451345144545141 4243 4254 (this is the original)
Then I formed the pairs:
54 32 51 42 43 54 51 34 43 51 54 35 51 11 34 51 43 53 54 42 43 54 51 34 51 44 54 51 41 42 43 42 54
But there is a problem. These are 33 and I cannot split them in half..
[/spoiler]

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#23 2012-09-15 07:12:18

Admiral Obvious
Member
Registered: 2012-06-02
Posts: 11

Re: Encryption 3

Funny it turned out that the way I was looking at the coordinates was wrong(a online tut told me the exact opposite). I have to note something about this challenge. This is not bifid, this is a polybius square. Bifid requires you to split them and decrypt usually 5 characters at a time(depending on the period of the current crypt).

Anyway I am still working on it because I cannot find the right answer but I think I am close.

[spoiler]
Fuck math for liberal studies tutorial.
[/spoiler]

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#24 2013-06-06 05:20:53

Backbite
Member
Registered: 2013-04-26
Posts: 66

Re: Encryption 3

You seem to be close but making a mistake on the Y somehow. The encryption should be:

543251...etc

but yours comes out as:

553352...etc

by t0mmy9

He has hints about the square table.



@WingeDD: no matter in whichever fashion you arrange your Polybius Square you'll ALWAYS(I think) have M at 33 and Y at 55
(I'm talking about the 'J included' version)

I think your square has either J or I
(which according to Wikipedia is the way it should be)
but in this case, you have to construct such a square where 'J is included and nothing is left out'
(except Z maybe)

I hope I didn't give away much. BTW, this is NOT a conventional Polybius Square
(if you define 'conventional' by the way Wikipedia thinks about it)




PS: A [*] M [*] Y are in their original positions no matter whichever way I try to create my square

by xyberz09

He tell the word in decrypted.


Good luck.

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#25 2014-01-11 14:36:09

agentm
Member
Registered: 2012-03-23
Posts: 11

Re: Encryption 3

ok I got the quote but I have no idea what to do with it.

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