Cipher Puzzle
Mike Mc
message :-
227 111 193 149 189 194 46 157 23 227 5 254 86 204 27 188 227 194 77 229 193
245 135 109 220 67 176 68 170 25 130 79 244 166 144 12 176 258 86 194
Mark Spahn
"Mike Mc" <apa...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote in
message news:40ed2...@mk-nntp-2.news.uk.tiscali.com...
(227 occurs 3 times, 194 occurs 3 times, 86 occurs 2 times, all others occur
once),
and since 37 > 26 (letters of the alphabet), this cannot be a simple
number-to-letter
encoding. Also, the message seems too short to dependably allow a
decryption.
Does the encoding depend on mathematical properties of the numbers, such as
their prime factorization? This problem does not seem interesting, because
too
much is unknown about it. More constraints are needed to make this an
interesting problem.
-- Mark Spahn
Mike
so i am not giving anything away too early. If people are genuinely
struggling after a while i will post some clues.
"Mark Spahn" <msp...@localnet.com> wrote in message
news:10eqtee...@corp.supernews.com...
Mike
will oblige.
"Mark Spahn" <msp...@localnet.com> wrote in message
news:10eqtee...@corp.supernews.com...
>
harry
"Mike Mc" <apa...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote in
message news:40ed2...@mk-nntp-2.news.uk.tiscali.com...
Mike
"harry" <ha...@hrsnospamx.fsnet.co.uk> wrote in message
news:ccjv9g$kt2$1...@news8.svr.pol.co.uk...
harry
"Mike" <to...@warts.com> wrote in message
news:hngHc.1460$pB1....@newsfe4-gui.ntli.net...
> ???
>
Sorry - I simply pressed the wrong keys while reading the post, and a null
reply got sent. I've seen others do the same thing, and now understand how
it happens.
Mike
I thought you posted the answer in invisible ink.
"harry" <ha...@hrsnospamx.fsnet.co.uk> wrote in message
news:cck467$coo$1...@newsg2.svr.pol.co.uk...
Bill McCray
I tried converting the numbers with "Frac(n/26)*26 + 1" and using the
letter at the resulting position in the alphabet. That gave TH for
the first two, but L for the third.
Adding the digits in each group also gives a strange starting string.
It can't be ASCII because of the 258.
No more ideas right now.
Bill
Swap first and last parts of username and ISP for address.
Mike
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:20hre09bvioc2usdj...@4ax.com...
Andrew B. Park
Mike,
Tried a few things but none of them really gave anything concrete to
grab onto.
Two interesting things about this string of numbers though:
1) The string has a quite consistent pattern of large-small-large
alternation. The pattern is broken here and there, but maybe that
breakage in the pattern in used to encode a message? (At least this
pecularity should suggest something about the method of encoding.... I
think.) (Tried to match that with Morse code--the message is too
short, and... it can't really work, as... unlike Morse code, there's
no gap in time to mark end of a letter or word.)
2) Not too few of the numbers are actually prime numbers. 227 is a
prime, 193 is a prime (er, I think...) and quite a few others... at
least more than would be suggested by chance alone, given the range of
numbers. Maybe this can also hint at the method of coding....
A little longer (twice... maybe three times as long?) string would
help in finding a pattern.
Best wishes,
Andrew
Andrew B. Park
Oh, one more pattern... all the numbers have three digits at maximum.
So, it is possible to take each number to be one syllable, where the
first digit stands for onset, the second digit, the vowel, and the
third, the coda. (Of course... when there's only one digit, it must
stand for the vowel, and if there are two digits, then it's ambiguous
which should stand for what.) The only problems with this idea is....
well, there's probably more than 9 possible consonant sound that can
go in the first or third place... and as can be seen, for the first
digit, we have only two variations...
Best wishes,
Andrew
Mike Mc
technique :-
27 144 257 180 136 148 74 77 198 3 194 208 181 11 105 220 202 95 246 223 14
243 93 134 196 13 151 119 76 159 165 67 71 212 211 150 252 233 58 177 250 62
136 27 255 173 4 147 25 26 204 72 11 75 97 77 242 250 102 71 41 41 165 104
105 182 83 162 53 47 149 20 117 231 66 201 96 74 235 161 160 109 25 143 177
233 213 5 139 234 110 173 128 131 118 90 103 69 14 62 250 15 99 93 125 39
121 65 160 138 40 240 167 129 99 27 200 117 192 152 213 4 53 18 26 84 32 0
137 168 139 211 20 49 173 116 91 38 180 237 135 22 193 102 118 125 53 84 24
150 43 237 25 113 69 223 192 69 172 65 23 7 82 202 76 60 123 65 87 11 123 52
221 150 193 237 84 138 20 162 104 245 29 236 158 89 38 122 56 254 33 7 88
140 236 137 104 216 211 172 184 255 86 126 105 177 45 108 139 208 138 109
130 58 158 248 231 251 33 36 168 100 226 178 212 104 2 252 190 48 214 190 98
159 125 242 172 108 227 87 36 254 214 99 40 183 176 227 142 155 95 213 218
74 231 59 146 200 36 224 112 116 175 45 43 63 9 44 134 218 3 49 76 71 209 14
28 218 115 198 84 86 185 52 60 199 35 164 64 57 136 192 210 14 124 227 200
201 180 103 231 143 78 217 239 163 20 63 11 44 107 72 159 47 10 51 93 83 17
36 128 248 77 145 203 232 23 98 232 26 166 184 184 125 21 115 126 34 69 110
253 221 249 182 103 31 141 32 51 67 205 205 240 115 179 3 86 9 71 33 116 239
11 158 118 81 50 176 174 33 201 51 113 114 62 234 56 84 26 53 119 110 13 81
179 182 175 130 35 170 93 130 9 152 61 142 251 46 79 22 92 155 83 98 60 172
155 225 13 182 243 229 152 139 62 23 7 218 13 3 237 240 254 5 231 87 176 190
159 7
"Mike Mc" <apa...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote in
message news:40ed2...@mk-nntp-2.news.uk.tiscali.com...
Bill McCray
There are 430 values and 209 distinct values extending from zero to
257. The largest frequency count is five for the values 11, 84, and
231.
In a simple substitution cipher in English, the expectation is that
about 1/8 of the characters will be E and we would expect no more than
maybe 42 distinct values (allowing for numbers, space, and some
punctuation). Perhaps there are several values that translate to a
given letter.
Although the 257 and the 258 in the previous cipher rule out simple
ASCII, the fact that the numbers don't go much above 255 leads me to
think that there's connection to ASCII or to values that can be
contained in one byte each.
Mike
You seem to be getting on well without any help from me so I will leave you
to it. I am on vacation until 19th July so I hope to return to a solution ;)
Here is a (not very helpful) hint - This message is the first two lines of a
classic book ;)
If I return and you still haven't solved it I will give you another hint.
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:tk6te0provltanc1e...@4ax.com...
Andrew B. Park
were, I guess we would be impossible but for someone who knows which
book was used for the code to crack it--and thus it will have nothing
to do with math... or code breaking in general...
Bill McCray
the puzzle here, because there would be nothing mathematical about it.
Therefore, there must be a way for us to attack it.
Bill
On 10 Jul 2004 02:33:21 -0700, "Andrew B. Park" <nov...@yahoo.com>
wrote:
Andrew B. Park
number might be representing two consecutive letters or one (or, say,
one letter plus a space). It doesn't seem too unlikely from what we
have so far.
Bill McCray wrote:
> On Fri, 9 Jul 2004 12:39:28 +0100, "Mike Mc"
> <apa...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote:
>
> > OK guys - here is a whole new longer message encoded using the same
> > technique :-
(a whole new longer message snipped out)
>
> There are 430 values and 209 distinct values extending from zero to
> 257. The largest frequency count is five for the values 11, 84, and
> 231.
>
> In a simple substitution cipher in English, the expectation is that
> about 1/8 of the characters will be E and we would expect no more
than
> maybe 42 distinct values (allowing for numbers, space, and some
> punctuation). Perhaps there are several values that translate to a
> given letter.
If each number represents two consecutive letters, it would solve this
problem... as, although 'e' might appear as often as 1/8 of the time,
when it can be put together with other letters in such combinations as
'ea', 'ae', 'ex', 'ed', etc. etc. then none of those combinations would
occur very frequently.
>
> Although the 257 and the 258 in the previous cipher rule out simple
> ASCII, the fact that the numbers don't go much above 255 leads me to
> think that there's connection to ASCII or to values that can be
> contained in one byte each.
It may be possible to represent 'one letter combinations' + 'all the
naturally-occuring two-letter combinations' in one byte. Granted, at
the minimum, 26*26 is much greater than 258, but not all possible
combinations (like, say, xs, or sz) occur naturally. One good way to
test this idea would be to feed a dictionary into an algorithm that
counts all distinct occurances of two-letter combinations (and the
result should be somewhat smaller than 258, if we are to allow for
things like punctuations, one-letter combination, possible
capitalization, etc.). Unfortunately, I don't have an access to a
computer with a compiler, so I can't do that :( (nor do I have an
easily usable dictionary entries in text file format).
Any thoughts?
Best wishes,
Andrew
Bill McCray
Association) asking for ideas for attacking it (the cipher, not the
association). Nothing dramatic has come from there, but one person
noted that Mike's hint that the text is the first two lines from a
book is interesting. A line of text in a book generally contains
something like 50-75 characters, so the plain text of the cipher
should have 100-150 characters. The cipher text we have been given
has 430 characters (number groups), so it appears that maybe three
number groups taken together give us a single character of the cipher
text or maybe every third group gives a character and the other two
must be ignored.
Another thought I had was that perhaps the numbers indicate bit
patterns of the vertical lines of a display of the plain text, but
expanding the values with neither the "1" bit nor the "128" bit on top
looks like anything recognizable.
Bill
On 12 Jul 2004 13:01:48 -0700, "Andrew B. Park" <nov...@yahoo.com>
wrote:
> I just thought of an idea of what the coding method might be: each
Swap first and last parts of username and ISP for address.
Mike
defiantely mathematical and very easy maths at that !!
Let me know if you want any clues and let me know how your getting on so
far.
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:he36f0d0btm831dol...@4ax.com...
Mike
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:he36f0d0btm831dol...@4ax.com...
Prai Jei
<ccuqnc$j...@odah37.prod.google.com>:
I remember devising a cipher along these lines when I was a kid. Since there
are 26 letters in the alphabet, an encoding of two letters would require
numbers up to 26^2 = 676. After several experiments, I eventually got it to
where the position of one letter in the alphabet got multiplied by 30
(easier to do mentally than multiply by 26) and the position of the second
letter got added on, thus giving ciphergroups from 1 (a single A) to 806
(ZZ) with some values skipped. Within a word the ciphergroups are simply
run together.
585 241665 1 225 699608 608289 579406365 13169571215.
A longer sample:
284 1 255365 284 20245 228471424 20245545 12292154 1 255062290. 14470 1
14049625, 4288625, 23170 255365, 189372154 699608 20245 164139 456 23468409
1424 44 465805 19395372, 14468 25170 1 4565, 61545, 19044145 255365 699608
14470249427 284 290 615 19290 135704 464 468 615 5050: 290 23049 1
255062290-255365, 1424 408050 13151439 3463195560.
--
Paul Townsend
I put it down there, and when I went back to it, there it was GONE!
Interchange the alphabetic elements to reply
Mike
"Prai Jei" <pvsto...@zyx-abc.fsnet.co.uk> wrote in message
news:cdeni6$jv9$1...@newsg4.svr.pol.co.uk...
Bill McCray
<pvsto...@zyx-abc.fsnet.co.uk> wrote:
> I remember devising a cipher along these lines when I was a kid. Since there
> are 26 letters in the alphabet, an encoding of two letters would require
> numbers up to 26^2 = 676. After several experiments, I eventually got it to
> where the position of one letter in the alphabet got multiplied by 30
> (easier to do mentally than multiply by 26) and the position of the second
> letter got added on, thus giving ciphergroups from 1 (a single A) to 806
> (ZZ) with some values skipped. Within a word the ciphergroups are simply
> run together.
>
> 585 241665 1 225 699608 608289 579406365 13169571215.
There are eleven digits in the last group, 13169571215. Thus, it
could be taken as 131 695 712 15, 131 695 71 215, 131 69 571 215, or
13 169 571 215. Did your rules of encryption allow this ambiguity or
would you have put a zero in any of the first three situations to
remove the ambiguity?
Bill
Bill McCray
characters and there are 430 integers in the list below. This implies
that there are more than one integer involved in determining each
character of the plaintext message. 430 is not divisible by either 3
or 4, leaving 2 and 5 as possible divisors.
Combining two seems to be the most likely, but how to combine them?
Perhaps each integer gives the value for half of a byte.
I tried adding the digits of each integer, but the ninth integer, 198,
adds to 18, which is too big (> 15) for four bits.
I have tried taking the integers MOD 16 and combining pairs of the
results to give byte values and intepreting with ASCII. I combined
them once with the first of a pair as the low four bits of the byte
and once with them as the high four bits of the byte. Neither made
any sense.
That's all for now. I'm waiting for another inspiration.
Bill
On Fri, 9 Jul 2004 12:39:28 +0100, "Mike Mc"
<apa...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote:
Swap first and last parts of username and ISP for address.
Mike Mc
"Mark Spahn" <msp...@localnet.com> wrote in message
news:10eqtee...@corp.supernews.com...
>
Mike Mc
Here is Clue No. 1 :-
The first and last 3 digits of the sequence are red herrings and can be
ignored.
;)
"Mike Mc" <ne...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote in message
news:YPyLc.241$sx5...@newsfe3-gui.ntli.net...
Bill McCray
<ne...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote:
> OK,
>
> Here is Clue No. 1 :-
>
> The first and last 3 digits of the sequence are red herrings and can be
> ignored.
Actually, that was number 2. The first was that the message is the
first two lines of a classic book (or something like that).
Some questions to clarify this clue: Is it the first digit and the
last three digits that are red herrings or the first three digits and
the last three digits? Assuming the latter, did you really mean that
many digits or did you mean the first and last integers (digit
groups), since the first integer is two digits and the last is a
single one.
Assuming you meant to ignore the first and last integers, I redid the
conversion of integer pairs to bytes starting with the second integer.
This led to nothing intelligible.
Bill
Prai Jei
<6tssf0549eq88nlij...@4ax.com>:
Only one of your options gives a sensible word that's a reasonable
conclusion of the sentence. In the longer passage, the cipherwords where
the number of digits is a multiple of three should enable you to recognise
the text, so that the other words may be filled in from one of the many
copies out there. The unambiguous rules by which my cipher operates can
then be discerned.
8046505 8644609427 !!
Mike Mc
!
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:37ivf0luh8t3eqfrg...@4ax.com...
Mike Mc
have another puzzle to bring to our attention can you start a seperate
thread please.
"Prai Jei" <pvsto...@zyx-abc.fsnet.co.uk> wrote in message
news:cdeni6$jv9$1...@newsg4.svr.pol.co.uk...
Jim Gillogly
> The first 3 numbers and the last 3 numbers in the sequence are red herrings
> !
At least that gets rid of the problem of the 257
being larger than 255. I'm guessing some kind of
autokey system, where subsequent characters depend
on the plaintext and/or ciphertext of previous ones.
Jim Gillogly
Mike Mc
;)
"Jim Gillogly" <j...@acm.org> wrote in message
news:pan.2004.07.23....@acm.org...
Mike Mc
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:6tssf0549eq88nlij...@4ax.com...
Bill McCray
will see that the last integer has eleven digits. It's not your
cipher I'm talking about. My comment is about Prai Jei' encoding
scheme.
I haven't had any fresh ideas on your cipher yet.
Bill
Mike Mc
message without the red herrings :-
180 136 148 74 77 198 3 194 208 181 11 105 220 202 95 246 223 14
243 93 134 196 13 151 119 76 159 165 67 71 212 211 150 252 233 58 177 250 62
136 27 255 173 4 147 25 26 204 72 11 75 97 77 242 250 102 71 41 41 165 104
105 182 83 162 53 47 149 20 117 231 66 201 96 74 235 161 160 109 25 143 177
233 213 5 139 234 110 173 128 131 118 90 103 69 14 62 250 15 99 93 125 39
121 65 160 138 40 240 167 129 99 27 200 117 192 152 213 4 53 18 26 84 32 0
137 168 139 211 20 49 173 116 91 38 180 237 135 22 193 102 118 125 53 84 24
150 43 237 25 113 69 223 192 69 172 65 23 7 82 202 76 60 123 65 87 11 123 52
221 150 193 237 84 138 20 162 104 245 29 236 158 89 38 122 56 254 33 7 88
140 236 137 104 216 211 172 184 255 86 126 105 177 45 108 139 208 138 109
130 58 158 248 231 251 33 36 168 100 226 178 212 104 2 252 190 48 214 190 98
159 125 242 172 108 227 87 36 254 214 99 40 183 176 227 142 155 95 213 218
74 231 59 146 200 36 224 112 116 175 45 43 63 9 44 134 218 3 49 76 71 209 14
28 218 115 198 84 86 185 52 60 199 35 164 64 57 136 192 210 14 124 227 200
201 180 103 231 143 78 217 239 163 20 63 11 44 107 72 159 47 10 51 93 83 17
36 128 248 77 145 203 232 23 98 232 26 166 184 184 125 21 115 126 34 69 110
253 221 249 182 103 31 141 32 51 67 205 205 240 115 179 3 86 9 71 33 116 239
11 158 118 81 50 176 174 33 201 51 113 114 62 234 56 84 26 53 119 110 13 81
179 182 175 130 35 170 93 130 9 152 61 142 251 46 79 22 92 155 83 98 60 172
155 225 13 182 243 229 152 139 62 23 7 218 13 3 237 240 254 5 231 87 176
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:37ivf0luh8t3eqfrg...@4ax.com...
Mike Mc
"Prai Jei" <pvsto...@zyx-abc.fsnet.co.uk> wrote in message
news:cdp9d6$ocd$2...@newsg4.svr.pol.co.uk...
Mike Mc
cracked it. I expected it to be cracked within a day of posting it because
the encoding method is extremely simple!!
"Mike Mc" <ne...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote in message
news:X4YLc.928$805...@newsfe5-gui.ntli.net...
Bill McCray
<ne...@REMOVETHISBITamateur-astronomer.ANDTHIScom> wrote:
> So - have you all given up on this one? I am very surprised someone hasn't
> cracked it. I expected it to be cracked within a day of posting it because
> the encoding method is extremely simple!!
My guess is that the method is extremely simple once you have
discovered it. There are a tremendous number of possible algorithms
only one of which is the right one. I haven't stumbled upon it yet.
Lots of problems are like that.
If I knew the problem was simple substitution, I have techniques and
skill to attack it. Same with a lot of other encrypting schemes. But
with an unknown encryption scheme, you can only hypothesize schemes
and try them. And I have lots of other projects going besides this
one.
Now that you have eliminated the value greater than 255, I have tried
converting each integer to a binary number and reversing the order of
the bits, subtracting each integer from 256, and subtracting each
integer from 255. I have tried ORing and XORing pairs together. It
took only a few characters to determine that these led nowhere useful.
This is in addition to the other things I have tried.
I don't think I have tried ANDing pairs yet. (Does) No, not that.
The first two pairs give C with a cedilla (or however it's spelling)
followed by hex zero.
Mike Mc
mathematical function on a pair of numbers. Those values are the message.
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:r0bqg0lfdes7gioda...@4ax.com...
Prai Jei
<VebOc.1086$tW2...@newsfe1-gui.ntli.net>:
> I give up!
OK, here's how it works.
>> >> 585 241665 1 225 699608 608289 579406365 13169571215.
>> >
>> > There are eleven digits in the last group, 13169571215. Thus, it
>> > could be taken as 131 695 712 15, 131 695 71 215, 131 69 571 215, or
>> > 13 169 571 215. Did your rules of encryption allow this ambiguity or
>> > would you have put a zero in any of the first three situations to
>> > remove the ambiguity?
Your last guess was correct.
13 -> M
169 = _5_ * 30 + _19_ -> E S
571 = _19_ * 30 + _1_ -> S A
215 = _7_ * 30 + _5_ -> G E
13169571215 decodes unambiguously as MESSAGE. The remaining words of my
original should now present no problems.
(It helps of course to know the series A=1, B=2, etc. right through the
alphabet. I learnt them straight off many years ago.)
Extra digits (over and above a multiple of three) are always taken at the
beginning. In the original version they were taken at the end, which made
the cipher look less "natural" and more easily cracked, e.g. MESSAGE =
395589037005, since in this version all numbers had a multiple of 3 for the
number of digits and extremely low groups always occurred at the end - you
couldn't have for example 005 in the middle of a number.
Accidental repeats confuse many would-be crackers - the cipherword
12169069044 does not repeat any letters when deciphered despite the repeat
of the group 690 and the digit 4 in the original.
Bill McCray
<ne...@REMOVETHISBITamateur-astronomer.ANDTHIScom> wrote:
> OK new clue - ASCII values are generated by carrying out a simple
> mathematical function on a pair of numbers. Those values are the message.
Not the difference between the pair.
Not the average of the pair.
Doesn't appear to be a linear combination of the pair.
k.Wallace
kW
"Mike Mc" <ne...@REMOVETHISBITamateur-astronomerANDTHIS.com> wrote in message
news:YO1Oc.102$o06...@newsfe6-gui.ntli.net...
Mike Mc
"k.Wallace" <wall...@engr.orst.edu> wrote in message
news:UP6dnfZwY7M...@comcast.com...
Mike Mc
it out??
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:r0bqg0lfdes7gioda...@4ax.com...
k.Wallace
can email return to me. I haven't given it my full attention yet anyway,
that will be a good weekend project.
-kw
"Mike Mc" <ne...@REMOVETHISBITamateur-astronomer.ANDTHIScom> wrote in message
news:WatSc.203$rA4...@newsfe2-gui.ntli.net...
Nate Smith
> Have you all given up? Anyone want a further clue or are you still working
> it out??
has anyone tried to imagine these numbers as representing
a visual image of characters?
OOO O*O
OOO ==> O*O
OOO O*O for letter I, = 010010010 = 128 + 32 + 2 = 162
this would work as written, but i think there is a direction
there to explore....
- nate
Nate Smith
> OOO ==> O*O
> OOO O*O for letter I, = 010010010 = 128 + 32 + 2 = 162
ooops
128 + *16* + 2 = 146
obviously some other layout is needed to get the values in the cypher.
- nate
Jim Gillogly
I don't think this works: 3x3 is pretty sparse to do the whole
alphabet. It's worse than this, though: what you're suggesting
would give simple substitution. However, there are 207 different
symbols used, which is more than the 127 7-bit characters typically
used for English.
Taking the visual idea a little further, one can imagine a bit
map picture comprised of each of the 8-bit values turned into
ones and zeroes displayed on an m*n screen. I think this is
unlikely because the first sample he gave is :
227 111 193 149 189 194 46 157 23 227 5 254 86 204 27 188 227
194 77 229 193
245 135 109 220 67 176 68 170 25 130 79 244 166 144 12 176 258
86 194
is 34 bytes long after you remove the first and last three
(which the "258" suggests you need to do), or 2*17. The second
cryptogram is 424 bytes long, or 2^3 * 53. These don't seem
very suitable for consistent screen sizes. I guess it's possible
if the scan widths aren't even multiples of 8 bits, though.
So I still think it's a substitution cipher with some kind of
inter-number feedback involved, and not something visual as you
suggest.
As a point of order, since I'm already writing, Mike Mc asks:
> Have you all given up? Anyone want a further clue or are you
> still working it out??
This in itself leads to a challenging psychological puzzle.
He has said in the past:
> I think it is a fairly easy one to crack and was coded
> using a simple method so i am not giving anything away
> too early. If people are genuinely struggling after a
> while i will post some clues.
and
> ... and very easy maths at that !!
and
> I am very surprised someone hasn't cracked it. I expected
> it to be cracked within a day of posting it because
> the encoding method is extremely simple!!
The puzzle is this: if we admit we're still trying, then
we're stupid because it was so simple, but if we admit we
finally succeeded after a whole sheaf of non-clue clues
we're stupid because it we needed so many clues for such
a simple puzzle.
From a motivational point of view I'd like the reward of
solving a cipher to be that I can pat myself on the back
and tell myself how clever I was to solve it, but you took
this away from the very beginning!
Anyway, I'll confess that I'm stupid and also waiting for
another clue. A real one this time.
Cheers -
Jim Gillogly
Mike Mc
OK, it is not a visual puzzle. You are looking at ending up with some
numbers, which are ascii representationsof the letters in the book.
Two numbers in the puzzle need to have a mathematical function carried out
on them to give you a single number, which is an ascii character.
"Jim Gillogly" <j...@acm.org> wrote in message
news:pan.2004.08.11...@acm.org...
Bill McCray
<ne...@REMOVETHISBITamateur-astronomer.ANDTHIScom> wrote:
> Have you all given up? Anyone want a further clue or are you still working
> it out??
I have tried taking one hex digit from the first of a pair and one hex
digit from the second of a pair and combining them in various orders.
None of these trials gave anything that makes sense.
No other new ideas to try. I think I'm ready for another hint.
Mike Mc
"Bill McCray" <McCra...@SpringMind.com> wrote in message
news:i1vkh05mtvb3ribfm...@4ax.com...
Bill McCray
Bill
Swap first and last parts of username and ISP for address.
Bob Harris
hundred random numbers then continually prod people with meaningless
'clues'. See how many people respond. Whenever they haven't responded for
a while, post the same clues again. The goal is to see how long you can
keep it going.
Bob H
Mike Mc
genuine cipher.
OK, the 'mathematical function' you carry out on the two numbers is a
logical operation, i.e. AND, OR, NOT, etc.
"Bob Harris" <plastic...@wrappermindspring.com> wrote in message
news:BD418D59.41FE2%plastic...@wrappermindspring.com...
Jim Gillogly
> I take it from that you want another clue. No it is not a trick it is a
> genuine cipher.
>
> OK, the 'mathematical function' you carry out on the two numbers is a
> logical operation, i.e. AND, OR, NOT, etc.
Ah, that did it. Since it's a logical operation, it's sufficient to
look at just the top bits. If you look at it by pairs, you see that
there are examples where either or both members of the pair has a
1-bit in the top position; none of the 16 logical operations will
turn that into a 0 (to end up with 7-bit ascii) without trivially
destroying the rest of the bits... so the two numbers being
manipulated are not adjacent. Next I considered first and last
(which also doesn't work), but when I separated it in half and
ran the first half against the second, it looked like:
149 189 194 46 157 23 227 5 254 86 204 27 188 227 194 77 229
193 245 135 109 220 67 176 68 170 25 130 79 244 166 144 12 176
for the first cipher; in each case the high bit is the same in these
pairs, which narrows down the possible logical operations. The right
one turns out to be XOR. The first set of plaintext is:
THECATSATONTHERAU (sic)
The second set is:
Looking back to all that has occurred to me since that eventful
day, I am scarcely able to believe in the reality of my
adventures. They were truly so wonderful that even now I am
bewildered when I think of them.
Jim Gillogly
Prai Jei
<pan.2004.08.13....@acm.org>:
Has anybody figured out my numerical cipher yet? Nowhere near so
complicated.
Bill McCray
bit. NOT is a unary operator, not binary, so it can't be used by
itself to combine two operands, but could be used in combination with
other functions.
Following the first three "red herring" entries are these decimal
ones: 180 136 148 74 77 198
hex: B4 88 94 4A 4D C6
Bin: 10110100 10001000 10010100 01001010 01001101 11000110
Paired vertically:
10110100 10010100 01001101
10001000 01001010 11000110
AND: 10000000 00000000 01000100
OR: 10111100 11011110 11001111
XOR: 00111100 11011110 10001011
Since all English letters in ASCII have the high bit as a zero, your
function can't be any of these. Perhaps the NOT of the OR:
OR: 10111100 11011110 11001111
NOR: 01000011 00100001 00110000
hex: 43 21 30
decimal: 67 33 48
ASCII: C ! 0
That doesn't appear to be it, unless either Mike or I have made a
mistake on that second character. I'll try a few more to test that
possibility).
The next eight values:
Dec: 3 194 208 181 11 105 220 202
Hex: 03 C2 D0 B5 0B 69 DC CA
00000011 11010000 00001011 11011100
11000010 10110101 01101001 11001010
OR: 11000011 11110101 01101011 11011110
NOR: 00111100 00001010 10010100 00100001
The second resulting byte is among the control codes in ASCII and the
third is over 127, so the NOR function isn't the function in question.
Side topic: I have just discovered that the calculator that comes
with Windows 2000 can do logical operations on decimal values, so I
didn't have to do all those conversions. While I'm on the subject of
the calculator program: Anybody know what the "Lsh", "dms", and "F-E"
keys do?
Anyway, the function in question doesn't appear to be any of the
elementary logical functions.
Bill
On Fri, 13 Aug 2004 06:11:01 GMT, "Mike Mc"
Bill McCray
might not be adjacent. Now I can quit wasting time on this.
Bill
Swap first and last parts of username and ISP for address.
Bob Harris
dms probably stands for degrees minutes seconds. Lsh could be left shift.
No idea on F-E.
Bill McCray
Your guesses appear to be correct.
Lsh appears to be a binary function that shifts the first operand left
the number of places specified by the second operand, where the places
are by bit in the binary representation of the number. E.g.
4 Lsh 2 = 16
16 Lsh -2 = 4
Dms did appear to convert an operand in decimal into Degrees, minutes,
seconds. E.g.
8.321 dms = 8.19156
where 8.321 degrees = 8 degrees, 19 minutes, 15.6 seconds.
It dawned on me that F-E might convert from floating-point notation to
engineering notation, and experiment shows that this is apparently
what it does.
Thanks for the suggestions.
Bill
Mike Mc
"Jim Gillogly" <j...@acm.org> wrote in message
news:pan.2004.08.13....@acm.org...