| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
My sister has a jig-saw puzzle from one of those mail order gift places.
It is some kind of cypher of a message. It says the cypher was used
long ago to save a would-be presedent. The numbers corrospond to
the english alphabet and something about a square?
Does anyone have any clues on how to decode this. Here is a sample
of the code text.
372,824,824,234,894
707,769,153,153,862
824,530,234,318,703
824,824,864,372,621,333,703,234
197,824,894,530,944,704,509,769
703,707,824,509,372,894,511,197
509,824,530,234,509,511,944,703
511,769,234,530,944,769,509,234
...
...7 more lines this long,
...2 more lines (2 shorter on the right than the above)
...1 last line (3 shorter on the right than the above)
*There is a picture of a mouse in the clearing
on the lower left*
??Any clues on what this code is??
Thanks, Jon
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1006.1 | DWOVAX::YOUNG | Sharing is what Digital does best. | Wed Jan 04 1989 22:19 | 8 | |
Cyphers (if this truly is one) are usually broken by statistical
analysis. If you include the whole text the we can take a pretty
good crack at it.
Oh yeah, are the commas (",") in the puzzle or did you add them
here?
-- Barry
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| 1006.2 | complete text | DIODE::CROWELL | Jon Crowell | Wed Jan 04 1989 22:43 | 22 |
Complete list, the commas are shown on the puzzle.
372,824,824,234,894
707,769,153,153,862
824,530,234,318,703
824,824,864,372,621,333,703,234
197,824,894,530,944,704,509,769
703,707,824,509,372,894,511,197
509,824,530,234,509,511,944,703
511,769,234,530,944,769,509,234
372,333,511,153,511,769,318,703
197,611,865,611,824,621,318,824
894,769,824,703,197,824,333,216
704,707,865,509,530,894,769,333
509,953,707,153,511,333,944,530
216,511,703,707,234,769,372,707
611,611,953,333,611,530,769,703
197,234,824,707,511,153
511,824,894,824,611,865
223,511,333,611,944 <--picture of mouse here
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| 1006.3 | make sure you have correct cipher before solving! | HANNAH::OSMAN | type hannah::hogan$:[osman]eric.vt240 | Thu Jan 05 1989 13:21 | 13 |
Caution: Before spending too much time on this cipher, keep in mind that it is probably in error in at least one place. One possible way to remove errors would be for the submitter to type it in a *second* time without reference to the first copy. Then use a computer to compare the two type-ins. By the way, since the submitter used the word "presedent", there may indeed be an error in the numbers. /Eric | |||||
| 1006.4 | SMURF::DIKE | Thu Jan 05 1989 14:52 | 6 | ||
It's possible that the numbers represent letters or words from some
existing document. 523 is the 523'rd word or letter, for example.
This is a fairly common way of encrypting a message, and has been
done commerically as a contest in the not-too-distant past.
Jeff
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| 1006.5 | you could code on the document itself | HANNAH::OSMAN | type hannah::hogan$:[osman]eric.vt240 | Thu Jan 05 1989 16:16 | 11 |
Or, perhaps the rank of the letter in the document ITSELF.
For example, to code this very reply so far, I would write:
1,2,3,4,5,2,7,8,4
I did this by saying the "o" is in position 1, the "r" is 2, the space
is 3, the p is 4, the e is 5, the r is the same as what's at 2, the h
is 7, the a is 8, the p is the same as what's in 4 etc.
/Eric
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| 1006.6 | SMURF::DIKE | Fri Jan 06 1989 09:13 | 6 | ||
If you assume that the decryption is meant to be easy for those
people who know the key, then encrypting on the document itself
is locking the key in the box it opens. The only way to decrypt
the document is to know what the document says. Also, there will
be an arbitrary number of possible decodings of this cipher.
Jeff
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| 1006.7 | exit | NIZIAK::YARBROUGH | Fri Jan 06 1989 10:00 | 13 | |
Another possible interpretation: each group of three digits describes
a line or page in a document and the letter or word; e.g. 372 =
line 3, 72nd letter; or page 3, 72nd word. From the repetitions
It appears the writer used the same symbols over and over (he might
have made it a lot harder) so I would guess that a simple substitution
was used (perhaps 824 = 'o' or 'e', so the message might begin with
'Look' or 'Keep') ...
If this message has historical significance it is probably discussed
in David Kahn's "The Codebreakers" which might be found in your local
public library.
Lynn Yarbrough
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| 1006.8 | Substitution. | RDVAX::COOPER | Topher Cooper | Fri Jan 06 1989 11:53 | 22 |
RE: .6 (Jeff)
> If you assume that the decryption is meant to be easy for those
> people who know the key ...
Although generally a good assumption for "real" ciphers there really
is no reason to assume that here -- this is a challenge cipher after
all, and the only intended recipients are those who "crack" it.
> Also, there will be an arbitrary number of possible decodings
> of this cipher
Statistically very unlikely. What is described is a simple
substitution cipher and they are unambiguously decodable (assuming
English text) after relatively few characters (about 50 I believe).
Its probably worth reminding people that the ","'s may simply be
conveniences for reading or may be red hearings to deliberately
mislead. There is a very good chance that they have no significance
to the cipher.
Topher
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| 1006.9 | 4GL::GILBERT | Ownership Obligates | Fri Jan 06 1989 17:05 | 6 | |
Notice that there are only 24 distinct 3-digit numbers in the message.
824 occurs 15 times
511 occurs 10 times
769 occurs 9 times
...
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| 1006.10 | They're at it again | WOODRO::BOTTOMS | Thu Jan 26 1989 15:53 | 13 | |
This is one of a series of puzzles put out by this company, usually
at Christmas. (Is it gold printing on black background)? I have
one
from an earlier date. The solutions to this series puzzles (if
it is) is found in a historical document. The one I have is an
offset word count into the Constitution. Cracking this one is not
a mathematical problem but one perhaps best suited for a librarian.
-jb
(I am a member of the American Cryptogram Association, pass member
of the New York Cypher Society (D.Kahn director) and past crypto
security officer -- advisor to S.Vietnamese on cryptosecurity).
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