Can anyone can solve this code

easqlmkwzptylzfsxejwdlxvcwnghejxpbxxzqjsfimxdayeqnbizmrlcakensqstmyajwsj

Hello Cipher Challengers...

I was trying to push my Cherokee G-BMFP out
of a rut in the grass at Blackbushe airfield
last Saturday 22nd April, when my foot went
down a rabbit hole. Sadly I did not fall into
Wonderland, so I am not going to wake up and
find it is all a bad dream. I've snapped my
Achilles tendon, and I'm on crutches now
until about midsummer. So...

I have a ticket for the lecture at
Bletchley Park by John Herivel at 6pm next
Wednesday 10th but I can no longer drive there.
If anyone else from around Fleet Hampshire
is going and could give me a lift I would
be grateful; failing that, if any "friend of BP"
would like to go on my ticket, let me know.

Please email me (not the group) at
TJFROGGATT at COMPUSERVE dot COM
ideally giving me your UK phone number.
Regards, Terry Froggatt.

I'm sorry if I have spoiled it for anyone.
I hadn't catch the beginning of the conversation. so I thought it was open
to comments.
I didn't realize an invitation to have it a go had been made.
Sorry...

Aurélio Moreira


>From: "Alexandra Fiona Dixon" <alexandra@...>
>Reply-To: CipherChallenge@yahoogroups.com
>To: <cipherchallenge@yahoogroups.com>
>Subject: [CipherChallenge] DaVinci Code
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>
>Apparently the judge in the Dan Brown / DaVinci Code plagiarism trial in
>London inserted a code into his 71-page ruling.  He italicized some
>letters, the first 10 of which read plainly "SMITHY CODE," (The judge's
>name is Smith), and the rest of which were goobledy-gook.
>
>There's also a hint on (page 52?) saying that to solve the code you need to
>consult both books on which the plagiarism trial was contested.
>
>It's been cracked, by a London law firm.
>
>But apparently the code went unnoticed for three weeks until a London
>barrister noticed the anomalous fonts.
>
>I'm sure the internet is rife with stories about it, so I won't cite any
>(my info comes from an American news show called "Nightline" which just ran
>a piece on it).
>
>And I have no idea what encryption method he used - I'm guessing it would
>be child's play for most on this list, though!
>
>Alexandra
>
>[Non-text portions of this message have been removed]
>

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I believe the correct sequence of letters was found using fibonacci numbers.

Aurélio Moreira

>From: "Alexandra Fiona Dixon" <alexandra@...>
>Reply-To: CipherChallenge@yahoogroups.com
>To: <cipherchallenge@yahoogroups.com>
>Subject: [CipherChallenge] DaVinci Code
>Date: Fri, 28 Apr 2006 00:03:51 -0700
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>
>Apparently the judge in the Dan Brown / DaVinci Code plagiarism trial in
>London inserted a code into his 71-page ruling.  He italicized some
>letters, the first 10 of which read plainly "SMITHY CODE," (The judge's
>name is Smith), and the rest of which were goobledy-gook.
>
>There's also a hint on (page 52?) saying that to solve the code you need to
>consult both books on which the plagiarism trial was contested.
>
>It's been cracked, by a London law firm.
>
>But apparently the code went unnoticed for three weeks until a London
>barrister noticed the anomalous fonts.
>
>I'm sure the internet is rife with stories about it, so I won't cite any
>(my info comes from an American news show called "Nightline" which just ran
>a piece on it).
>
>And I have no idea what encryption method he used - I'm guessing it would
>be child's play for most on this list, though!
>
>Alexandra
>
>[Non-text portions of this message have been removed]
>

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Just in case anyone feels like having a go.

SMITHYCODE JAEIEXTOSTGPSACGREAMQWFKADPMQZVZ

Having looked at the solution, it may help having read
the book, though its not too difficult to work out.
The solution does contain two spelling mistakes.

The judge gave hints towards a solution, that can be
found on the (u.k.) times website.

Simon
PS. Mr Singh made an appearance regarding this on
Channel 4 news last night.

--- Alexandra Fiona Dixon <alexandra@...>
wrote:

> Apparently the judge in the Dan Brown / DaVinci Code
> plagiarism trial in London inserted a code into his
> 71-page ruling.  He italicized some letters, the
> first 10 of which read plainly "SMITHY CODE," (The
> judge's name is Smith), and the rest of which were
> goobledy-gook.
>
> There's also a hint on (page 52?) saying that to
> solve the code you need to consult both books on
> which the plagiarism trial was contested.
>
> It's been cracked, by a London law firm.
>
> But apparently the code went unnoticed for three
> weeks until a London barrister noticed the anomalous
> fonts.
>
> I'm sure the internet is rife with stories about it,
> so I won't cite any (my info comes from an American
> news show called "Nightline" which just ran a piece
> on it).
>
> And I have no idea what encryption method he used -
> I'm guessing it would be child's play for most on
> this list, though!
>
> Alexandra
>
> [Non-text portions of this message have been
> removed]
>
>


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Apparently the judge in the Dan Brown / DaVinci Code plagiarism trial in London
inserted a code into his 71-page ruling.  He italicized some letters, the first
10 of which read plainly "SMITHY CODE," (The judge's name is Smith), and the
rest of which were goobledy-gook.

There's also a hint on (page 52?) saying that to solve the code you need to
consult both books on which the plagiarism trial was contested.

It's been cracked, by a London law firm.

But apparently the code went unnoticed for three weeks until a London barrister
noticed the anomalous fonts.

I'm sure the internet is rife with stories about it, so I won't cite any (my
info comes from an American news show called "Nightline" which just ran a piece
on it).

And I have no idea what encryption method he used - I'm guessing it would be
child's play for most on this list, though!

Alexandra

[Non-text portions of this message have been removed]

That is correct.  These dang puzzle caches,  I dont know where to start...


John Lobert <johnlobert@...> wrote:
   I'd say we're being asked to help on a geocaching quest...

John Lobert

--- Andrew Bruno wrote:


---------------------------------
Hello Spencer

I've read your couple of posts.

Well without too much thinking I can guess they are map latitude/longitudes
for a place of interest.

But that is just from initial observations.


On 12-Apr-06, you wrote:

> Here is antoher one as well: any hints are appreciated:
>
>
> N 40.2 _ . _ _ _
> W 111.4 _ . _ _ _
>
> OAOIEO OOEEON CTTETC HRRTEI UNSIHO CUSSCC

Regards
--
Look into his eyes, and see if someone else is driving.

Andrew Bruno
abruno@...



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[Non-text portions of this message have been removed]

Yes they are,  the question is  what is the 6x6 cipher.  It should decode to
fill in the missing coords. The decoding is what i need help with.

Andrew Bruno <abruno@...> wrote:  Hello Spencer

I've read your couple of posts.

Well without too much thinking I can guess they are map latitude/longitudes
for a place of interest.

But that is just from initial observations.


On 12-Apr-06, you wrote:

> Here is antoher one as well: any hints are appreciated:
>
>
> N 40.2 _ . _ _ _
> W 111.4 _ . _ _ _
>
> OAOIEO OOEEON CTTETC HRRTEI UNSIHO CUSSCC

Regards
--
Look into his eyes, and see if someone else is driving.

Andrew Bruno
abruno@...




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- Unknown

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[Non-text portions of this message have been removed]

I'd say we're being asked to help on a geocaching quest...

John Lobert

--- Andrew Bruno <abruno@...> wrote:


---------------------------------
Hello Spencer

I've read your couple of posts.

Well without too much thinking I can guess they are map latitude/longitudes
for a place of interest.

But that is just from initial observations.


On 12-Apr-06, you wrote:

> Here is antoher one as well:  any hints are appreciated:
>
>
> N 40.2 _ . _ _ _
> W 111.4 _ . _ _ _
>
> OAOIEO OOEEON CTTETC HRRTEI UNSIHO CUSSCC

Regards
--
Look into his eyes, and see if someone else is driving.

Andrew Bruno
abruno@...



---------------------------------
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     Visit your group "CipherChallenge" on the web.

     To unsubscribe from this group, send an email to:
  CipherChallenge-unsubscribe@yahoogroups.com

     Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.


---------------------------------

Hello Spencer

I've read your couple of posts.

Well without too much thinking I can guess they are map latitude/longitudes
for a place of interest.

But that is just from initial observations.


On 12-Apr-06, you wrote:

> Here is antoher one as well:  any hints are appreciated:
>
>
> N 40.2 _ . _ _ _
> W 111.4 _ . _ _ _
>
> OAOIEO OOEEON CTTETC HRRTEI UNSIHO CUSSCC

Regards
--
Look into his eyes, and see if someone else is driving.

Andrew Bruno
abruno@...

I was hoping someone here could help me ge some direction as to how
to solve this below.   Any help is appreciated.


I'm not providing additional clues to solve the puzzle at this time,
but I may in the future depending on how difficult it is to crack.

This type of code was developed years ago to provide better
encryption and make it more difficult for the cryptologist to crack.
Not fool proof by any means, but goes beyond the basic substitution
of characters to figure out the plaintext. Good luck.

N 40. _ _ . _ _ _
W 111. _ _ . _ _ _

Here is the code:

NKKSFFQLWS
TTSWTUZZLG
FFTIOKEUFF
LRHIOUTXQY
GEAYYABEDR
INYABEDRDK
QUDKQUSDWM
CNEUXL

Here is antoher one as well:  any hints are appreciated:


N 40.2 _ . _ _ _
W 111.4 _ . _ _ _

OAOIEO OOEEON CTTETC HRRTEI UNSIHO CUSSCC

I was hoping someone here could help me ge some direction as to how
to solve this below.   Any help is appreciated.


I'm not providing additional clues to solve the puzzle at this time,
but I may in the future depending on how difficult it is to crack.

This type of code was developed years ago to provide better
encryption and make it more difficult for the cryptologist to crack.
Not fool proof by any means, but goes beyond the basic substitution
of characters to figure out the plaintext. Good luck.

N 40. _ _ . _ _ _
W 111. _ _ . _ _ _

Here is the code:

NKKSFFQLWS
TTSWTUZZLG
FFTIOKEUFF
LRHIOUTXQY
GEAYYABEDR
INYABEDRDK
QUDKQUSDWM
CNEUXL

> Would it be logical to computer an RSA signature on a long message
> by
> first finding what the message equals, mod n, and signing that?

No it wouldn't. It would be logical finding that message's collision
resistant HASH value (i recommend Tiger-192 or SHA-512 for now),
concatenating it with current time value and signing all that. The
whole thing would always be less than n, so no mod n is required.



__________________________________
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http://www.yahoo.com/r/hs

Would it be logical to computer an RSA signature on a long message by
first finding what the message equals, mod n, and signing that?

Did you try a Google search to answer these questions? The answers are all over
the
internet. Try that first, and if you can't find the answers, then let us know.

Having said that, if you had to ask 1 through 5, the you'll probably never get 6
on your
own, so the answer is....

SPOILER






SPOILER









SPOILER

Answer: PT One Owe Nine lost in action in blacke

Are you sure you have all the letters, because PT 109 was sunk in the Blackett
Strait.

--- In CipherChallenge@yahoogroups.com, "to_sultan911" <to_sultan911@y...>
wrote:
> please help me in these quastions:
> 1. What is the difference between an unconditionally secure
> cipher and a computationally secure cipher?
>
> 2. Briefly define the Caesar cipher.
>
> 3. Briefly define the monoalphapetic.
>
> 4. What is the difference between a monoalphapetic cipher and a
> polyalphapetic cipher?
>
> 5. Briefly define the Playfair cipher.
>
> 6. Decrypt the following message using the Playfair cipher. The
> key used was royal new Zealand navy.
> The encrypted message:
> KXJEY UREBE ZWEHE WRYTU HEYFS KREHE GOYFIW
>
> please i want the solution, of curse i have some of them but the
> last quastion is the imolrtant one.
>
> to_sultan911@y...

please help me in these quastions:
1. What is the difference between an unconditionally secure
cipher and a computationally secure cipher?

2. Briefly define the Caesar cipher.

3. Briefly define the monoalphapetic.

4. What is the difference between a monoalphapetic cipher and a
polyalphapetic cipher?

5. Briefly define the Playfair cipher.

6. Decrypt the following message using the Playfair cipher. The
key used was royal new Zealand navy.
The encrypted message:
KXJEY UREBE ZWEHE WRYTU HEYFS KREHE GOYFIW

please i want the solution, of curse i have some of them but the
last quastion is the imolrtant one.

to_sultan911@...

If anyone in this group is interested in trying to solve the Dorabella
cipher, composed by the English composer Edward Elgar in 1897, and
still unsolved, please read the messages in the group Elgar-
Cipher@yahoogroups.com, or contact me at t.roberts@....

Tim

Some body took my book and didn't yet give it back...

For the first point :
-------------------
Let's take an equivalent :
Alice and Bob share the number N which can be public
Alice chooses X and doen't communicate it
Bob chooses Y and doesn't communicate it

Alice sends to Bob N1 = N power X (Bob doesn't know X and can't calculate it
!)
Bob sends Alice N2= N power Y (Alice doesn't know Y and can't calculate it
!)

Bob calculates N3 = N1 power Y
Alice calculates N4 = N2 power X

N3 = N1 power Y = (N power X) power Y = N power (XY)
N4 = N2 power X = (N power Y) power X = N power (YX)
So, N3 = N4

Alice and Bob have choosen together a secret number (N3 or N4)
With this process, all information exchanged between Alice and Bob can be
public, it's not possible for somebody else to calculate  X, Y or N3

The power function can be replaced by any commutative function provided tha
that function is not inversible (thas is : when Y = f(X), it's not possible
to calculate X whenyou know Y, impossible meaning impossible for the
duration where the message must remain secret...)

For the second point :
--------------------
In RSA, as in most codes used by computers, the message can always be seen
by the computer as a serie of numbers. For encoding and decoding, the
message is always cut into successive blocks of a given value. (usually, the
length of a block is the length of the key) The last block is completed to
have the same length.
So, when you have a very very short message, it's a padded value that you
use. By this way, the padded value is different of 0 or 1, and the result of
the power function used by RSA is something that cant be used to know the
plain text value if you don't have the key. No matter how short is the
initial message, you need the key to decode. And nothing can allow you to
guess that the initial message is very simple : one block gives one block,
no less, no more.
The problem of RSA is that it requieres a lot of calculation. So, it's
usually not rapid enough.
The normal uses of RSA is for coding the keys. The messages are coded with
triple des, using the key that has been exchanged using RSA.
And for the keys, the standard use is to share a 'master key' used only once
for exchanging a key which will be used for exchanging keys...

I hope this is clear, despite of my poor english practice...

Didier

Thanks to everybody who explained that.

As for my first question ("staged" example giving common shared key),
I'll have to go back and work some examples but when I tried one set of
values other than those supplied in the exhibit, I got different answers
for Alice and Bob.  Granted, I did this by hand so human error may have
had something to do with it.  I'll try some more exammples when I get
the time.

Alexandra


----- Original Message -----
From: "David A. Darbyshire" <david.darbyshire1@...>
To: <CipherChallenge@yahoogroups.com>
Sent: Thursday, August 18, 2005 5:58 AM
Subject: [CipherChallenge] Re: Question re: RSA


> Hello there,
>
> I can not help withe your first point but I can with your second
> point.
>
> I read Singh's and and had exactley the same thought, then I realised
> something very simple:- 'X' is the entire message.  If the message
> was 'The quick brown fox jumpes over the lazy dog.' then this is the
> entire message.  you have to convert each letter to its numerical
> value to produce one very long number and that is what you work with.
>
> If you have a long message then the number could be thousands of
> characers long but that is not a problem for a computer. I hope this
> makes sense.
> David.
>>
>> The second question I have concerns the structure of the ciphertext
> in
>> the RSA method.  Singh goes through an example in the book where
> person
>> A wants to send person B a message - say, just the letter X.  He
> shows
>> how the X is encrypted, based on public and private keys, and
> becomes an
>> 11.  Then the recipient decrypts it, and it becomes X again.
>>
>> So my question is - how exactly is a long message enciphered?  I
> mean,
>> if all X's become 11, and all Y's become ##, then all you have is a
> very
>> weak monoalphabetic substitution cipher!
>>
>> I'm definitely missing something here.
>>
>> Anybody care to bring me up to speed?
>>
>> Thanks,
>>
>> Alexandra
>
>
>
>
>
>
> ------------------------ Yahoo! Groups
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>

Hello there,

I can not help withe your first point but I can with your second
point.

I read Singh's and and had exactley the same thought, then I realised
something very simple:- 'X' is the entire message.  If the message
was 'The quick brown fox jumpes over the lazy dog.' then this is the
entire message.  you have to convert each letter to its numerical
value to produce one very long number and that is what you work with.

If you have a long message then the number could be thousands of
characers long but that is not a problem for a computer. I hope this
makes sense.
David.
>
> The second question I have concerns the structure of the ciphertext
in
> the RSA method.  Singh goes through an example in the book where
person
> A wants to send person B a message - say, just the letter X.  He
shows
> how the X is encrypted, based on public and private keys, and
becomes an
> 11.  Then the recipient decrypts it, and it becomes X again.
>
> So my question is - how exactly is a long message enciphered?  I
mean,
> if all X's become 11, and all Y's become ##, then all you have is a
very
> weak monoalphabetic substitution cipher!
>
> I'm definitely missing something here.
>
> Anybody care to bring me up to speed?
>
> Thanks,
>
> Alexandra

> After finding my copy of the code book again and having some fun
> looking this stuff up I agree with Tom's answer on both questions.
>
> On question 2, think back to these exciting days of the cipher
> challenge. (Were you part of one of the stage 9 solving groups?)
> As can be read from the plain text of Stage 9 (see also the
> Stockholm document on www.codebook.org) Tom's scenario is exactly
what
> Simon used in stage 10.
>
> In stage 10 you are challenged to factor a large N, so that you can
> decipher one given RSA-encoded 155 digit number (Simon calls this
the
> 'shorter message'). The plain text result of that exercise forms the
> DES key that you are then supposed to use on the encrypted message
> (the 'longer message').

Hi, Rolf

I debated whether being this specific would constitute "hinting" but,
quite frankly, without some heavy computing iron, and even heavier
maths (mowing the Galois Field with a Sieve?), stages 9 and 10 are
pretty hint-proof.  IIRC there were a few wrinkles in the actual path
to solving stage 10 but the general concept you describe is correct.

And even if one didn't crack stage 9, it was fairly obvious that
stage 10 would be an RSA-enciphered challenge.

I came along too late to paticipate in the stage 9 cracking fun,
although I DID code up the DES block-enciphering engine and wrapped a
key-search around it.  I had a better than even chance of finding the
key before the Sun's core ran out of hydrogen and began to expand off
the main sequence...

The maths involved in RSA and stage 10, have continued to interest
me, although I will readily admit that I can't even spell "GNFS".  I
was in a bookstore in Sydney, Australia, in 2001 and I stumbled
across a book entitled Cryptological Mathematics.  It was written by
Robert Edward Lewand- who turned out to be a teacher at Goucher
College in my hometown of Baltimore, Maryland!  Small world, eh?  The
book, somewhere on a shelf at home, did a pretty good job of
explaining modulo arithmetic, Galois fields, and the number
theoretical underpinnings behind RSA and the challenge of factoring
large integers.  And Mr. Lewand was a pretty decent chap about
responding pleasantly to my questions via email.  I might get
motivated enough to pick up the book again, and have another go at
trying to understand what those 40-lb brains up in Stockholm managed
to accomplish.

Cheers!

Tom Moloney-Harmon

After finding my copy of the code book again and having some fun
looking this stuff up I agree with Tom's answer on both questions.

On question 2, think back to these exciting days of the cipher
challenge. (Were you part of one of the stage 9 solving groups?)
As can be read from the plain text of Stage 9 (see also the
Stockholm document on www.codebook.org) Tom's scenario is exactly what
Simon used in stage 10.

In stage 10 you are challenged to factor a large N, so that you can
decipher one given RSA-encoded 155 digit number (Simon calls this the
'shorter message'). The plain text result of that exercise forms the
DES key that you are then supposed to use on the encrypted message
(the 'longer message').

Thanks for triggering me to read up on this again - I had a good time.
Rolf

> In, I think it was Exhibit 76, Simon
> Singh is talking about the initial
> concept of the modular function and
> the public key, <snip>
>
> It seems to me, this was a forced
> example, because you can take other
> examples where they do not arrive at
> the same shared key.

Really?  Try it with different values of A and/or B.  The resulting
key isn't always 9, but Alice and Bob always arrive at the same key.

That said, I should point out that the scheme Simon was describing --
the Diffie-Hellman-Merkle key exchange system -- is AFAIK not central
to public key cryptography and the RSA method.


> The second question I have concerns
> the structure of the ciphertext in
> the RSA method.  <snip>
>
> So my question is - how exactly is a
> long message enciphered?  I mean, if
> all X's become 11, and all Y's become
> ##, then all you have is a very weak
> monoalphabetic substitution cipher!

I think the general approach is to use RSA to encipher a key and then
use the key to encipher the long message.  (See pp 298-299 of the
Code Book, which describes how this is done in PGP).  The key should
be for a strong symmetric cipher, otherwise the ciphertext can be
directly attacked without bothering with deciphering the key.  Using
RSA to directly encode text symbols, or blocks of symbols, would
expose the ciphertext to the same direct attack methods that any
substitution code suffers from, as you correctly point out.

Tom Moloney-Harmon
Baltimore, Maryland

So, maybe I'm missing something...I never really had a great grasp on
the RSA algorithm - I mean, I knew it involved one way functions and two
huge primes, yada yada...but I really wanted to understand it in detail
so I've been reading _The Code Book_ chapter about it again after a few
years!

And, I have some questions.

In , I think it was Exhibit 76, Simon Singh is talking about the initial
concept of the modular function and the public key, and he gives an
example in which Alice has a private key  of 4, and Bob has a private
key  of 2, and then they both share the function c^key(mod (p*q)) where
p and q are the two primes...and then Bob sends Alice his result (6) and
she sends him hers (3) and voila - she takes her private key to the
exponent of his result, and he does the same, and she gets 4^3 and he
gets 2^6 and they both mod (whatever) and they arrive at a common key of
9.

It seems to me, this was a forced example, because you can take other
examples where they do not arrive at the same shared key.

So, what am I missing?

The second question I have concerns the structure of the ciphertext in
the RSA method.  Singh goes through an example in the book where person
A wants to send person B a message - say, just the letter X.  He shows
how the X is encrypted, based on public and private keys, and becomes an
11.  Then the recipient decrypts it, and it becomes X again.

So my question is - how exactly is a long message enciphered?  I mean,
if all X's become 11, and all Y's become ##, then all you have is a very
weak monoalphabetic substitution cipher!

I'm definitely missing something here.

Anybody care to bring me up to speed?

Thanks,

Alexandra

Hello Tom

On 12-Aug-05, you wrote:

> I might add/amplify:
> (5) Treat this more as a puzzle than a cipher (although, yes, it is
> one of those, too). The book is called "Sit and Solve Ciphers" and
> its design and presentation implies the ability to solve them with
> all the tools available, ah, throne-side ;)  Do letter-frequency
> counts and pattern analyses if you must, but then set them all aside
> and...as Tim suggests...just stare at the text.  A purely mechanistic
> approach will not get you there.

Yeah, but I impulsivly bought the book.

I hope that one day I will have time to invest in trying some of them.

Doing letter frequency tables while on the "Throne" is kind of something I
am not wanting to think about.

I prefere concentrating on the business at hand.   (sorry.)    ;-)

> I wonder if the other ciphers in the book show the same level of
> gentle wit as this one?  Might be worth a peek...

I think they do.

I have nearly got one out by (What's his name....?)   Woddy Allen,

Something about making a U-turn and driving.

Alas I can't quite finish it, as one of the words I don't know - a big one
too - which kind of stumps my ability to recognise which letter is what.




Regards
--
Philosophers DO IT with their minds.

Andrew Bruno
abruno@...

I finally solved it.  Thanks for all the extra clues.

Just as Tom and Tim suspected, I was approaching this one in a very
logical and analytical fashion, which was exactly my undoing.  Here
is a case where digraph frequency analysis only threw me off track.
And I had developed a big list of four letter words with the ABBC
pattern (e.g., cook) and cycled through them all.  Also, I managed to
misinterpret every hint that was posted (I've documented my
misinterpretations, in fact).  Of course, it all came to me while I
was driving to work, without the "benefit" of all my home grown
software and freqency tables in front of me.

It is interesting how the mind works.

I agree that it would be interesting to see a few more problems from
the same book.

Kind regards.....Bill Swope




--- In CipherChallenge@yahoogroups.com, "Tom Moloney-Harmon"
<tom.harmon@n...> wrote:
>
> > Hi Bill,
> >
> > I suspect you're being *far* too logical!
> > (1) Guess the first two words VWOX OX...
> > (2) Guess the two words after the comma ZLV Q...
> > (3) Fill in the corresponding letters throughout the ciphertext.
> > (4) Do NOT rule things out if they seem impossible at first (this
> is probably what you've done so far)....just stare at the text.
> >
> > Tim
>
> I might add/amplify:
> (5) Treat this more as a puzzle than a cipher (although, yes, it is
> one of those, too). The book is called "Sit and Solve Ciphers" and
> its design and presentation implies the ability to solve them with
> all the tools available, ah, throne-side ;)  Do letter-frequency
> counts and pattern analyses if you must, but then set them all
aside
> and...as Tim suggests...just stare at the text.  A purely
mechanistic
> approach will not get you there.
>
> I wonder if the other ciphers in the book show the same level of
> gentle wit as this one?  Might be worth a peek...
>
> Tom Moloney-Harmon
> infrequent poster from
> Baltimore, Maryland

> Hi Bill,
>
> I suspect you're being *far* too logical!
> (1) Guess the first two words VWOX OX...
> (2) Guess the two words after the comma ZLV Q...
> (3) Fill in the corresponding letters throughout the ciphertext.
> (4) Do NOT rule things out if they seem impossible at first (this
is probably what you've done so far)....just stare at the text.
>
> Tim

I might add/amplify:
(5) Treat this more as a puzzle than a cipher (although, yes, it is
one of those, too). The book is called "Sit and Solve Ciphers" and
its design and presentation implies the ability to solve them with
all the tools available, ah, throne-side ;)  Do letter-frequency
counts and pattern analyses if you must, but then set them all aside
and...as Tim suggests...just stare at the text.  A purely mechanistic
approach will not get you there.

I wonder if the other ciphers in the book show the same level of
gentle wit as this one?  Might be worth a peek...

Tom Moloney-Harmon
infrequent poster from
Baltimore, Maryland

Hi again folks.

Ok, I have already mentioned my inablility to complete the cipher.

I mentioned I had made a spreadsheet to help with mono-alphabetical ciphers.

As I now know how to do it, here is what they look like:

(friend's site)
http://www.zeta.org.au/~doghouse/takingthefifth.xls

(I hope that works)

You enter the CT in the lines and on the left (from memory) you see the
alphabet.

There you match the CT and PT letters and see the result on the main
part of the
sheet.


Enjoy..

Hi Bill,

I suspect you're being *far* too logical!
(1) Guess the first two words VWOX OX...
(2) Guess the two words after the comma ZLV Q...
(3) Fill in the corresponding letters throughout the ciphertext.
(4) Do NOT rule things out if they seem impossible at first (this is probably
what you've done so far)....just stare at the text.

Tim

	 -----Original Message-----
	 From: CipherChallenge@yahoogroups.com on behalf of Bill Swope
	 Sent: Wed 10/08/2005 2:17 PM
	 To: CipherChallenge@yahoogroups.com
	 Cc:
	 Subject: [CipherChallenge] Re: Some puzzles I with which I am playing - VWOX OX
QJ ....




	 OK, I have been working on this one for a very long time and have not been
	 able to solve it, even with the couple of clues that have been posted.
	 Would someone please post another clue for me?  When (or if) I solve this
	 one, I would like to understand how various people approached it.  (Perhaps
	 "off line", i.e., not on the forum, since it might spoil it for others who
	 are still working on it.)

	 Regards....Bill







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