some news
Monday, 06-Dec-1999 11:34:10
194.78.234.252 writes:
Hi everybody,
I don't like to start thread (not much luck with that the previous time:(, but as Seeker requested, I do :)
This contest seems to be quite more difficult than a simple playfair one (well, I'm saying this because now we know playfair so
well :).
Anyway, here is a check i did to find possible period length. My thought is as follow :
Imagine the period length is 5. That mean we will have blocks of 10 characters. So the last block will have a 'total' length of the
modolu 520 (length of our cipher) by 10. In this case it's 0, meaning we end on a block bounday. Let's see again with a possible
period of 6. 520%12 = 4. This mean our last black have a length of 4 (2*2 letters).
So I used that to check every possible 'ending' of the plain text, with or without a padding 'X'.
Here are the results (very boring :(
5 :
520%10 = 0
ESSAG SSAGE
EENDS ENDSX
6 :
520%12 = 4
?????M EN ????ME ND
ESSAGE DS SSAGEE SX
7 :
520%14 = 2
?????ME D ????MES S
SSAGEEN S SAGEEND X
8 :
520%16 = 8
SAGE AGEE
ENDS NDSX
9 :
520%18 = 16
?????MES ????MESS
SAGEENDS AGEENDSX
10 :
520%20 = 0
?????????M ????????ME
ESSAGEENDS SSAGEENDSX
11 :
520%22 = 14
???MESS ??MESSA
AGEENDS GEENDSX
12 :
520%24 = 16
?????MES ????MESS
SAGEENDS AGEENDSX
13 :
520%26 = 0
????????????? ?????????????
??MESSAGEENDS ?MESSAGEENDSX
14 :
520%28 = 16
?????MES ????MESS
SAGEENDS AGEENDSX
15 :
520%30 = 10
ESSAG SSAGE
EENDS ENDSX
16 :
520%32 = 8
SAGE AGEE
ENDS NDSX
17 :
520%34 = 10
ESSAG SSAGE
EENDS ENDSX
18 :
520%36 = 16
?????MES ????MESS
SAGEENDS AGEENDSX
19 :
520%38 = 26
????????????? ?????????????
??MESSAGEENDS ?MESSAGEENDSX
Next, we can also notice in the cipher that we have a double repeating digram 8 digram before the end (GG GG). Although we
cannot get much out of this, we can at least be sure the corresponding plaintext digrams must be the same too. So I remove
every possible period which doesn't respect this. For example for a period of 6 we would have a possible ending as :
?????M EN
ESSAGE DS
as 'E' is not equal 'S' we know this can't be good.
So I come to the following results.
6 : only if 'ENDSX'
7 : only if 'ENDS'
9 : not possible
10 : not possible
12 : not possible
13 : not possible
14 : not possible
18 : not possible
19 : not possible
So, we are left with :
5 :
520%10 = 0
ESSAG SSAGE
EENDS ENDSX
6 :
520%12 = 4
????ME ND
SSAGEE SX
7 :
520%14 = 2
?????ME D
SSAGEEN S
8 :
520%16 = 8
SAGE AGEE
ENDS NDSX
11 :
520%22 = 14
???MESS ??MESSA
AGEENDS GEENDSX
15 :
520%30 = 10
ESSAG SSAGE
EENDS ENDSX
16 :
520%32 = 8
SAGE AGEE
ENDS NDSX
17 :
520%34 = 10
ESSAG SSAGE
EENDS ENDSX
Please not that I only checked for a period > 5 and < 19 !!!!!
The question is : what do we do with that ???
Well, personnaly i have no clue :(
What I already tried is to check all possible combinaison of the 4 keyword we have in the 3 possible arrangment we know (row,
col, spiral). That's a total of 12*12 =144 combinaison.
I deciphered the 8 last digram of our cipher with those 144 combinaison and expected (without not much hope) to see some
'ending' as shown above to appear. Helas, nothing showed up. So I suppose maximum one of those 4 keyword was actually
used (maybe none).
Next thing i'll check is to find some information about that 'shotgun hill climbing' algo Jim G. talked about. Anybody got a link
???
All apologize for the 'dirty' writing of this, but I'm in a hurry and wanted to post those result before tomorow, as I'm not sure i
could come online again later.
Just a last note. I did not checked those results very well. So this may be completely wrong. Apologize if it's the case.
Respects to all and good luck,
Laurent.
**************************
when you're lost ..
Monday, 06-Dec-1999 12:58:58
193.158.164.213 writes:
.. in some dark woods, and you do not which way to go - the best could be, to go back to the starting point and try a new.
What I want to say is : we know - now - a lot about the playfair cipher, but our knowledge about the Double playfair is less - so
maybe we should collect some experience.
(and BTW - with all respects - we do not know, if the cipher code is really working, or if this posting has been just a joke !)
the seeker
**************************
..you are still where you are.
Monday, 06-Dec-1999 13:51:09
212.50.131.2 writes:
I will give hints after Nova's fashion, if necessary. Be assured, the cipher is not a joke. I could easily show you the matrices, but
that would spoil the puzzle. The Seeker is right though. You may find some of your new knowledge working a bit against you,
exempli gratia the fact that double letter pairs are permissible. If you manage to decipher it, you may see one of my avatars.
Good luck.
Oh yes, and I am sorry I didn't format the original message. It looks so untidy.
*******************************
Re: follow up thoughts to add
Monday, 06-Dec-1999 14:24:25
216.224.156.151 writes:
please bare in mind...I ain't (ain't) saying any of the following is correct...but here is where my brain is at this moment:
I sure hope it gets formatted right...sometimes pre works well and others it doesn't i assume according to place ment...Shade I
got your ICQ; we format using the pre tag
like so:
<.pre>...text...<./pre>
okay to recap:
this is what I think to be the correct de-encription from the message by thee rules I read:
.
.
KRQVE XDRAK QNNQE LCZMW XSUAU DYRZC ANMWI XRKMD OWXZW AWWWN EYOKZ OWGXU HXOAW GUUAF CONWM FWNMW
GJWQP EQCGT MDHVY NLULG MPJTP TSCSF GNVMJ OCXGP ZYKXZ GFCBJ HGFXK IWOPK BDHLM NECXT FOGMT HKIVV
.
.
CFPVZ DXDUR YWVKJ MKKCX XJCML DVUEK AISDB RRBPH PUKDW KAHYW JHWZI ZNRZA XZNOM SORKM FXSRN DGPJH
XRKZF ZLLCG NWHZZ NCNFX PQNLX NJIHX TVGMN CSNJK JYLSX XNMHB MEMAS MDRAG NAJPV GICTM ZOPCW NNDKA
.
.
ZWPKA WSWDF WBYRH NRCXU BZMHN VJZSY WOGME DNRZH SMRWO ETGXW ZDPXM ROKRR FHRWC ILVOW RZRET PDFMW
UVIZS VQKPG WWGIM GRFRC OPBCV NBDMC TQGOC QHZFU PBRZN GBLRK NHQLZ RNXNC TBCOZ MCNZY GACOI KVGFG
.
.
? MESSA <<<---this format is subject to ???? certainly
G EENDS
WGCJN BHWZE MLGGM XWNWU
WGFYL PWKSH BKGGV BDWMY
maybe
-MESSA
GEENDS
maybe
messageends
What do we know?
Its not like singel PlayFair
There are no weasels to place in the message
"Message ends" is in some combination at the ending of message.
Since we don't know how its configuration might be I think the only thing we can assume
for sure is the last letters in the botton row no matter what must be either ds or sx
ie: even if they are grouped 10 letters together or 50....ds or sx would always be in same place... (?)
xxxxxxxxxx
xxxxxxxxds
or
xxxxxxxxxx
xxxxxxxxsx
soooooooo.......
xxxxx xxxxx
xxxxx xxxds
MLGGM XWNWU
BKGGV BDWMY
or
xxxxx xxxxx
xxxxx xxxsx <<<<---we might assume also in this case that several of the other letters can be added
MLGGM XWNWU
BKGGV BDWMY
IE:
xxxxx xxxxx
xxxxx xends <<<<---we might assume also in this case that several of the other letters can be added
MLGGM XWNWU
BKGGV BDWMY
xxxxx xxxxx
xxxxx endsx
MLGGM XWNWU
BKGGV BDWMY
If this is correct how does this help us to build a matrix?
I'm not sure BUT I think it can help us IDENTIFY WHICH TWO matrix keywords were used so that we don't have to build any two matrixes!
The keywords OR combinations OF are known instead
So it seems to me that the controller would know which two to use.
So what would I do?
I'd build every combination of the keywords given in a 5x5 and use two of them side by side
to see which two produces any combination of....message ends...
or in example above if correctly thought out...find the two 5x5's that give me:
ds
WU
MY
or
sx
WU
MY
It might be two verticals/ or one spiral and one horizontal/ two spirals/ ect ect
or any combination of any two...when that is found...the rest of the
message is busted... (if my layout is correct)
How i would rebuild if I were the enemy and did not have the keywords I am not sure yet;
But having these several KNOWNS this is the way I might do it...
What do you think?
I would build 5x5's of the KNOWN keywords
Blue Diamonds
Horizontal:
b l u e d
i a m o n
s c f g h
k p q r t
v w x y z
Vertical
b i s k v * note that the horizontal matrix and this vertical matrix are actualy the same;
l a c p w the only real difference being that now the letters are flopped.
u m f q x ie: horizontal: BZ == DV
e o g r y vertical: BZ == VD
d n h t z
Horizontal Vertical
Spiral: Spiral
b l u e d b k h g f * Note here reverse spiral also gives us the same type matrix; again
k p q r i l p y x c the only real difference being that the letters are flopped.
h y z t a u q z w s Left to right row spiral B0 ==DF
g x w v m e r t v n Vertical spiral column B0 ==FD
f c s n o d i a m o
*******************
Hard Like Rock
Horizontal
h a r d l
i k e o c
b f g m n
p q s t u
v w x y z
Vertical
H i b p v
a k f q w
r e g s x
d o m t y
l c n u z
Spirals
h a r d l h p n m g
p q s t i a q y x f
n y z u k r s z w b
m x w v e d t u v c
g f b c o l i k e o
**************************
Of Quartz
Horizontal
o f q u a
r t z b c
d e g h i
k l m n p
s v w x y
Vertical:
o r d k s
f t e l v
q z g m w
u b h n x
a c i p y
Spiral
o f q u a o k i h g
k l m n r f l x w e
i x y p t q m y v d
h w v s z u n p s c
g e d c b a r t z b
**************
Crystal Palace
Horizontal:
c r y s t
a l p e b
d f g h i
k m n o q
u v w x z
Vertical:
c a d k u
r l f m v
y p g n w
s e h o x
t b i q z
Spiral:
c r y s t c k i h g
k m n o a r m x w f
i x z q l y n z v d
h w v u p s o q u b
g f d b e t a l p e
should we not then be able to mix and match two of these to come up with the resulting message ends???
If so
perhaps we can split up the possibilities:
Someone take Blue Diamonds and compare it to all other 5x5's
someone take Hard Like Rock and compare to all other 5x5's including the Blue Diamond possibility...
why?
because I think (but not sure) that example:
Horizontal Horizontal
h a r d l b l u e d
i k e o c i a m o n
b f g m n s c f g h
p q s t u k p q r t
v w x y z v w x y z
MYHOVER FULLO
CRAFTIS FEELS
We encipher each vertical pair. Find the first letter of each pair
in the first square and the second letter in the second square.
.......The plaintext pair, MC, is on the same row. In this case we
take the letters to the left of each plaintext letter, again taking
the first letter from the second key square and the second
letter from the first key square, giving SG.
. . . . . . . . . .
. . . . . . . . . .
. . g m . S c . . .
. . . . . . . . . .
. . . . . . . . . .
The second step is like the first: Encrypt SG by finding S in the
first square and G in the second; The next step is to find the letters at the opposite corners of
the rectangle formed by the plaintext pair, starting with the
ciphertext letter in the second square.
RG are at the opposite
corners. Remember to take the first letter of the ciphertext
from the second key square, and the second letter of
ciphertext from the first key square. This completes the
encryption of the pair MC.
. . . . . . . . . .
. . . . . . . . . .
. . g . . . . . g .
. . s . . . . . r .
. . . . . . . . . .
MYHOVER FULLO
CRAFTIS FEELS
R
G
my question now is: If we switch these will we come up with the same encription?
Move hard Like Rock to right and Blue Diamonds to the left...
Horizontal Horizontal
b l u e d h a r d l
i a m o n i k e o c
s c f g h b f g m n
k p q r t p q s t u
v w x y z v w x y z
MC==OA
OA==KE
they would not encript the same; so 5x5's MUST also be flopped for checking...
so what am I looking for?
I'm looking for two 5x5's that will give me
ds
WU
MY
(or)
sx
WU
MY
working backwards...bad example only...
....m .....
..... .....
..... .....
..... .....
..... ....w
....m ....X x's example only; choose any letter at this point for corners
..... .....
..... .....
..... .....
....X ....w
....m ....s x's example only; choose any letter at this point
..... .....
..... .....
..... .....
....y ....w
so
WM==YS
Now find Y in second 5x5 and S in first 5x5
....m ....s x's example only; choose any letter at this point
..... .....
..s.. .....
..... .....
....y ..y.w
what is the result of th corners?
....m ....s x's example only; choose any letter at this point
..... .....
..s.. ..l..
..... .....
..w.y ..y.w
YS== WL
soooooooooo; these two 5x5's do not reveal ds as we needed; move to next two sets of possibilities.
Does any of this make sense or am i on a long bad way?
laurent will be sending u mail in a little while to answer yesterdays questions on decompiler ?
jeff
****************************
Re: Re: follow up thoughts to add
Monday, 06-Dec-1999 14:53:16
195.238.19.64 writes:
Hi Jeff,
I shouldn't be here, but i can't resist :)
First of all, your above post main idea seems very correct to me. Just a few remarks.
1. 'EE' don't have to be 'splitted' with a 'X' in double playfair (I think you did that somewhere in your post).
2. There is one case where DS or SX will not be grouped together. The case is when they grouped the plain text by 7
characters. In this case you end up with 2 character in your last block (DS or SX) and so they appear up/down and not side by
side.
3. you say :
I'm looking for two 5x5's that will give me
ds
WU
MY
(or)
sx
WU
MY
Are you sure your ds (or sx) don't have to appear vertical (above UY) ?
If I'm right, then we don't have to look for the DS or SX pattern, but well for ?X or ?S (? beeing the letter about X or S).
Appart those 3 remarks, I followed the same path as you and tried all the combinaison of the 4 keyword we have
("BLUEDIAMOND", "HARD...) in the 3 arrangment we know (row, col, spiral). This make 12 possible grids for both left and right
square (i agree with you that the order is of importance). This give us 144 combinaison. I tried them all and looked for a
meaningfull pattern ... NADA :( ... (maybe my program is wrong, but i doubt so. I will recheck tomorrow just in case)
If you want i can mail you the result (144 lines, too long to post on here), so you don't have to do it by hand.
Maybe I should try 'reverse spiral' too.
Remember also that with the nova contest, for example, only 1 valid keyword was given.
Got to run now ... Keep thinking, your way seems good to me. We have to search for 2 valid keyword with 2 correspoding
arrangment.
respects,
Laurent.
PS : Half of your 'private' playfair cipher is done :)
Laurent
**************************
Important !
Monday, 06-Dec-1999 17:16:31
193.158.164.195 writes:
Sorry, I was a little bit tooo busy. We should not miss this hint from our unknown cipher-master :
The secret agent's girlfriend likes blue diamonds. Strongly suspect it is the keyword for the second matrix, but he never starts a
matrix in the top left corner.
apologizes
the seeker
*************************
mportant CORRECTION !!
Tuesday, 07-Dec-1999 08:33:12
193.159.101.220 writes:
Sorry for this late posting.
BUT : we have learnt NEVER to let out any part (maybe useful, maybe trash). So here is the complete message :
I am not +HIM though I am flattered! You did some good work on the other thread but I can see that
Department X must pass on the intelligence they have gathered (stingy bastards).
The secret agent's girlfriend likes blue diamonds. Strongly suspect it is the keyword for the second matrix, but he
never starts a matrix in the top left corner.
apologizes once again
the seeker apologizes
************************
My thoughts so far.
Monday, 06-Dec-1999 18:20:29
203.57.68.13 writes:
After reading the double playfair hint it looks like the next step is
a little hit and miss trying to recreate one of the possible squares
from the previous messages and prove it and the message on the
"Message Ends" text.
The variables are frightening.
I notice the four groups of GG one G above the other, is this possible???
. . 1 . . . . 3 . .
. 3 . . . . . . 5 .
. 6 4 . . . . 2 4 .
. . . . . . . . . .
. . . . . . . . . .
1&2 beget 3&4
3&4 beget 5&6
Damm it is possible this thing scares me. It doesn't even prove
anything except that there is an occurance where the same two letters
are on the top row as are on the bottom.
We know we have ten good chars from the phrase wichever format which
then gives us twenty possible formats:
1
MLGGMXWNWU
BKGGVBDWMY
essageends block of 10 or greater
2
MLGGMXWNWU
BKGGVBDWMY
ssageendsx block or 10 or > with offset
3
MLGGMXWNWU
BKGGVBDWMY
e
ssageends block of 9
4
MLGGMXWNWU
BKGGVBDWMY
s
sageendsx block of 9 with offset
5
MLGGMXWNWU
BKGGVBDWMY
es
sageends block of 8
6
MLGGMXWNWU
BKGGVBDWMY
ss
ageendsx block of 8 with offset
7
MLGGMXWNWU
BKGGVBDWMY
ess
ageends block of 7
8
MLGGMXWNWU
BKGGVBDWMY
ssa
geendsx block of 7 with offset
9
MLGGMXWNWU
BKGGVBDWMY
essa
geends block of 6
10
MLGGMXWNWU
BKGGVBDWMY
ssag
eendsx block of 6 with offset
11
MLGGMXWNWU
BKGGVBDWMY
essag
eends block of 5
12
MLGGMXWNWU
BKGGVBDWMY
ssage
endsx block of 5 with offset
13
MLGGMXWNWU
BKGGVBDWMY
sage
esends block of 4
14
MLGGMXWNWU
BKGGVBDWMY
agee
ssndsx block of 4 with offset
15
MLGGMXWNWU
BKGGVBDWMY
egee
ssands block of 3
16
MLGGMXWNWU
BKGGVBDWMY
seen
sagdsx block of 3 with offset
17
MLGGMXWNWU
BKGGVBDWMY
saen
esgeds block of 2
18
MLGGMXWNWU
BKGGVBDWMY
agnd
sseesx block of 2 with offset
19
MLGGMXWNWU
BKGGVBDWMY
esged
saens block of 1
20
MLGGMXWNWU
BKGGVBDWMY
saens
sgedx block of 1 with offset.
Wow the enormity of this dawns on me. These are all the possibles for
our phrase and we have to try to prove one of these using a
hypothetical square of which there could be:
4 possible passphrases * 2 possible squares *
left to right row starting top
right to left row starting top
left to right row starting bottom
right to left row starting bottom
top to bottom starting left
top to bottom startting right
bottom to top starting left
bottom to top starting right
spiral clockwise
spiral counterclockwise =10
(I may have missed some here)
which gives us 80 possible squares.
Multiply this by our possible phrase layout and we have 1600 combinations.
But to add to this what are we trying to prove?
Lets look at option 20
20
MLGGMXWNWU
BKGGVBDWMY
saens
sgedx block of 1 with offset
Are we trying to prove here that (using the last block of 4 letters)
s=U and x=Y? Lets say on our first attempt we by stroke of luck get
the second square completly correct in every way. How many possible
combinations are there for the first square to be in, in order to
produce a U for the s? (I can't figure this but it is a very high
figure) Lets say we take one and form a rectangle.
1 . . . . . . . U .
. s . . . . . 1 . .
. 2 . . . . . x . .
. . . . . . . . . .
Y . . . . . . . 2 .
Success we have a possibility, but to what do we reference it in
order to prove it? We have only four other pairs, no where near
enough! Lets complicate it even more (as if we need it) by using
option 1
1
MLGGMXWNWU
BKGGVBDWMY
essageends block of 10 or greater
Now what are we trying to prove?????????
We don't even know what the top row of the Plaintext is!
I cannot see a feasable way for a human bieng to attack this
problem (I could be wrong) in a reasonable length of time unless
there was a large co-ordinated team. So to solve it and still
maintain some sanity we would need a computer program. What do we
want this program to do??? Defining the rules it should follow are no
small step.
To quote the NOVA cipher solution:
"The mechanics of the Double Playfair are
involved enough that it is expected that solving this cipher would be
most readily achieved by a computer program that recovers the first
square by shifting letters around while testing for common letter
sequences and words in the resulting text."
So the program would have to go through each of the forty possible
second square variations. For each variation it would have to try
shifting the letters in the first square and decrypting the text
based on each of the combinations.(What rules should it use for
shifting the letters in the first square?) It would then have to
search this text for common words. Bearing in mind that the shorter
the blocks that were used for splitting the original text the less
likely we are to find common words. Take as an example of this option
20. There are no recognisable words here.
So we also need the program to re-construct the original text in all
of the possible twenty options as per the known Phrase and then check
for common words. There will also be depending on our list of words a
high number of false words. Take as an example of this option 16
where we see the word seen. This of course is not a valid decryption,
so we would have to sort throught all the programs successful outputs
and visually check the word arrangement.
We could reduce this by requiring a certain number of successful word
hits in the text.
This is some incredible brute force program we are talking about here .
These were my thoughts of last night and I have since read the
seekers pointer that the second square is highly likely to be
"blue diamonds" and that the orientation is not started in
the top left corner. This drastically reduces the size of the problem
but it is still a huge mountain.
The rules for the swapping of the first square have to be built and
also what keywords would we search for?
shade
**************************
Re: Shade has pointed out a major flaw in my thinking
Monday, 06-Dec-1999 18:31:13
216.224.156.151 writes:
returning to NOVA I must agree with Shade
they did NOT NESSARILY USE TWO of the presented Keywords...they only used ONE of the documented Keywords....the
other matrix was a totaly new and different word...
this would mean that a silent 5x5 would still need to be re-constructed (if this holds true)....this gets more harder to put a fix on
as we discuss it....but more worthwhile when we do it!
jeff
***********************
Re: Re: Last four Doublets Corrected format
Monday, 06-Dec-1999 21:37:22
216.58.8.109 writes:
Greetings
Looking at Laurent's posting that means there are only 8 variations (or 16 when we considder the last character to be a null
character)of the last four doublets.
If the cipher is broken into lengths of 5 or 15 or 17 the last 4 doublets are the same.
Last four Doublets
5, 15, 17 SSAG SAGE
ENDS NDSX
6 ?M EN ME ND
GE DS EE SX
7 ?ME D MES S
EEN S END X
8, 16 SAGE AGEE
ENDS NDSX
9, 12, 14, 18 ?MES MESS
ENDS NDSX
10 ???M ??ME
ENDS NDSX
11 MESS ESSA
ENDS NDSX
13, 19 ???? ????
ENDS NDSX
I do not know yet but I hope this will make our work easier.
Princess
***************************
A few costly ideas
Tuesday, 07-Dec-1999 07:55:54
207.230.57.214 writes:
Costly as in a lot of sweat for few results.
IF I get the idea, the last block is whatever is left after you have broken the original message into whatever size chunks. So,
since the message is written over/under, the number to be divided is 260. I ran through the possible remainders and got (1-25 for
the chunks) 1,2,4,5,7,8,10,13,18,&20. So the last piece has to be one of those, again, if I've got this evil beastie straight in my
head.
How'm I doin'? Comments and critiques needed.
Dan
****************************+
A suspicion
Tuesday, 07-Dec-1999 08:43:26
207.230.57.216 writes:
The last block has to be shorter than 8, because at 8 you run into the quad Gs. That leaves the chunk size at 2-9, 11, 15-17, &
23(1-25 checked).
Unfortunately, my ideas have run out of gas there. Time for trial and error?
Dan
*****************************
Re: A suspicion
Tuesday, 07-Dec-1999 10:41:36
195.238.20.154 writes:
Hi Dan,
I agree with you that the last block have to be less than 16 char (so, max 14 (2*7) characters).
I also agree with the different possible block size you gave (see the first post above for my results).
From the list of '4 ending doublets' Princess posted, I think we can remove some of them
and I come up with the following list :
Last four Doublets for given block size :
5, 15, 17 SSAG SAGE
ENDS NDSX
6 ?MEN MEND
GEDS EESX
7 ?MED MESS
EENS ENDX
8, 16 SAGE AGEE
ENDS NDSX
11 MESS ESSA
ENDS NDSX
Now, I'll try to explain what I'm doing right now.
We know that the right square is done with 'BLUEDIAMOND' (not starting in top left corner).
We know the possible ending for the 4 last doublets (except for a length 7, what really bother me).
So, I start with a left square made without keyword so, my square is 'clear' :
A B C D E
F G H I K
L M N O P
Q R S T U
V W X Y Z
I run this square along 8 possible orientation of the right square. I check for the last 2 doublets to be in our ending list.
If I found a 'working' combinaison i got to the next step. If I don't foun one, I SWAP 2 random letters from my left square and try
again.
Next step : I remember the orientation of the right square AND the set of letters which have to remain in their position in the left
square to have the 2 ending doublets (this can be max 8 letters). I then check to have the 3 ending doublets from our list.
If it doesn't work, I SWAP 2 random letters from my left square. BUT I don't SWAP the letters that have to remain in place. In
this way, I can expect to have in the best case only 17 letters to move (25-8).
When I got a working left square i go to the next step, which is the same except that I now check for the last 4 doublets while
not moving up to 12 letters (3*4) in the best case.
I think this method is close to what they call shotgun hill climbing algo (but Im not sure cause I still didn't found anything on that
algo).
Anyway, with max 10000 random swap for both step 2 and 3, I still come up with a lot of left square that return 4 last correct
doublets when using with a right made of 'BLUEDIAMOND'.
Helas, I have no idea on how to sort all those result, neither how to try to go a step further wihout leaving out some possible
ending (the unknow one).
I think now i'll try to make another check as follow. Once I got a square which return 4 valid ending doublets I'll decrypt the
whole cipher and count the occurence of the letter 'E'. As my left square can have up to 16 right positionned letter (in the best
case again). The letter 'E' should appear many times with the correct left square.
I could maybe also check for presence of duet like 'ST' 'OP', 'TO'.
If this explanation is clear, can you give me your opinion on this method ?
Regards,
Laurent.
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My opinion(worth what you payfor)
Tuesday, 07-Dec-1999 13:23:09
207.230.57.225 writes:
You're losing me on the 'last four doublets' part. As far as I can tell, the last block doesn't have to be four wide. It's just how
many you have left after breaking the rest of the message, put in two rows.
For the remainder 5 I got:
ESSAG
EENDS
or
SSAGE
ENDSX
I think I've figured out a way to eliminate some more of the possibilities. Looking at that quad G, and trying out the different
endings, I got:
2= ------ EN or ------ ND
ESSAGE DS SSAGEE SX
The first block puts ES where the two Gs should be, so, it's out. the second, however, puts two Ss there so it's still a
possibility. 4 worked out to be
4= ---- SAGE or ---- AGEE
-MES ENDS MESS NDSX
The ME means it has to be the first choice. Unfortunately, I couldn't ruled any on 5 or 7.
I checked through 50, and the whole list of possible groupings (1-50) is 1-9,11,15-17,23,32,37,&43.
Okay, I ruled out three, but I also added 21. So, time to brute force?
Dan
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e: My opinion(worth what you payfor)
Tuesday, 07-Dec-1999 13:59:09
194.78.233.242 writes:
Hi Dan,
You are right, the last block doesn't have to be 4 wide. But :
Let's say that the grouping is done by 11.
That is :
xxxxxxxxxxx xxxxxxxxxxx
xxxxxxxxxxx xxxxxxxxxxx
In this case we will end up with 14 (520%22) characters for the last block. Thus we will have 2 lines of 7 char after the last full
block of 2 lines of 11 char.
In this last block we have to fit the last 14 character of the plain text. We know the 11 last ones. So let's pad with '???'. Thus
our last block will be :
???MESS
AGEENDS
or if we have a padding 'X' :
??MESSA
GEENDSX
In both case we know at least the 4 last doublets (ME EN SD SS or EN SD SS AX).
The only 'hard' case is if the group are made of 7 letters :
xxxxxxx xxxxxxx
xxxxxxx xxxxxxx
In this case we end up with 520%14 = 2 characters. Thus 2 lines of 1 characters.
Let's fill our x's :
?????ME D
SSAGEEN S
So we do have a last full block of 7 followed by a last block of 1. But in this case (and only this case) we don't know that 4th
doublet :(
Now the same case with a 'padding X'.
????MES S
SAGEEND X
This case has to be rejected. Why ? cause on the 7th and 8th last doublet we have our 'quad G'. That mean we must have twice
the same doublets, which cannot be the case here because S != A
Hope this is clearer and error-free :)
Btw, I don't take into account grouping value below 5 or above 19 (improbable IMHO, but could be worth to check).
Respects,
Laurent.
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