NB. No history in this comment - side-effect of using iPad away from home WiFi
My farthingsworth:
Suppose you’ve got one canonical solution in the form of an array of m-hands
compatible (in some way) with one canonical n-hand. (Devon asked for all
5-hands given 1 (or 6 permutations of 1) 4-hand, while Mr Boss has also looked
at all 4-hands given a 5-hand.)
If, say, we require all 5-hands given a 4-hand, I suggest it’s sufficient to
solve for
0 1 2 3. All solutions for other 4-hands are equivalent after mapping.
eg, if we require the solution for h =. 3 5 8 13, the to-vector is
to =. 3 5 8 13 0 1 2 4 6 7 9 10 11 12 14 .... 51,
ie h, h-.~ i.52
If the canonical solution for i.4 is S, the solution for 3 5 8 13 is
to { S .
I assume the order of the 4-hand is partially significant, in that the first
pair prescribes required cards, and the second pair excluded cards. If desired,
we could
set, or perhaps require:
h =. (/:~ 2{.h), /:~ _2{.h .
The mapped solution, (to { S) , may, if required, be re-sorted within hands
and between hands to exactly reproduce a solution generated by other mechanisms
discussed in this thread.
Apologies if this is so painfully obvious that it didn’t need stating. Even
more if it’s wrong!
Mike
Sent from my iPad
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