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Title
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1966_Sackson_352_November 28.jpg
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Creator
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Sid Sackson
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Date
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1966
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Type
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image
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Format
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.jpg
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Source
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Box 1, Object 4, Sid Sackson collection
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Language
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English
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Coverage
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1966
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Rights
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The Strong, Rochester, New York.
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transcription
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English
Monday 28 November 1966
332nd day - 33 days to come
Called Jules Cooper to tell him of the idea of playing
OPERATION SEARCH with three hidden cards. Also suggested
using for three players using passing of ordinary
requests, but only giving the number of
bonus requests.
He says some of his men have played it but find it is
inferior to CLUE in chances for deduction. I read over
the rules of the latter and feel that the game is
played simply be keeping the positive and negative information
on the detective sheets. It seems almost automatic.
Claude called. He was at Arthur's. Nothing new.
Solution of puzzle on 11/29
1) A has 2-SU and C has 1-SU, so all SU are accounted for.
2) B has 2-UL which must be NUL & WUL since EUL & SUL are
already accounted for.
3) B has 1-UM but 0-EM. Since NUM and SUM are already accounted
for, B must have WUM.
4) B has 1-WR. B also has 3-W. Since B has already been shown
to have WUL and WUM and the 3rd W is in the Rural area,
B cannot have WIL or WIM.
5) A has 1-NM. C has 3-N but 0-NF. Since NUL and NUM have
already been accounted for, C's 3-N must be NIL, NRL, and
either NUM or NRM. (It can be proven that C has NIM
rather than NRM, but this not necessary for the solution.)
6) C has 1-SI. This must be SIL since SIF and SIM are
accounted for.
7) C has 2-IL. Since C has already been shown to have NIL
and SIL, C cannot have WIL or EIL.
8) A, as stated before, has 2-SU. A also has 2-EI and 4-R.
This limits the location of all 8 of A's cards to the SU, EI,
and R areas. Among the cards that A cannot have is WIL,
which has already been proven not to be in the possession
of B or C. WIL is the solution.
The statement, A had 2-W was thrown in as a small red
herring. They must both be in the Rural area.
