Little explorations with HP calculators (no Prime)

12312018, 08:36 PM
Post: #342




RE: Little explorations with HP calculators (no Prime)
(12312018 08:20 PM)pier4r Wrote: Hmm, am I wrong when I think that 71 matches should be simply impossible? Here's a straightforward way to create 21 arrangements where all starting positions (171) result in a "match": 1) Find all unique arrangements of a combination of 5 blue and 2 white balls: (1, 1, 1, 1, 0, 0, 1) (0, 1, 1, 1, 0, 1, 1) (1, 1, 0, 0, 1, 1, 1) (1, 1, 1, 0, 1, 1, 0) (1, 0, 1, 0, 1, 1, 1) (0, 1, 1, 0, 1, 1, 1) (1, 1, 1, 1, 0, 1, 0) (1, 1, 0, 1, 1, 0, 1) (1, 0, 1, 1, 1, 0, 1) (1, 1, 1, 0, 0, 1, 1) (0, 1, 1, 1, 1, 0, 1) (1, 1, 0, 1, 1, 1, 0) (1, 0, 1, 1, 1, 1, 0) (1, 0, 0, 1, 1, 1, 1) (0, 1, 1, 1, 1, 1, 0) (0, 0, 1, 1, 1, 1, 1) (0, 1, 0, 1, 1, 1, 1) (1, 1, 1, 1, 1, 0, 0) (1, 1, 1, 0, 1, 0, 1) (1, 1, 0, 1, 0, 1, 1) (1, 0, 1, 1, 0, 1, 1) 2) For each of those groups, repeat the indicated pattern 20 times to form a particular arrangement of 100 1s (blue) and 40 0s (white) Each resulting 140ball arrangement has 71 "matches" for this problem. In other words, any contiguous subgroup of 70 elements from any of those lists will result in a group containing 50 1s and 20 0s. These aren't the only ones, though. I believe the same logic applies if you start with a set of 10 1s and 4 0s, stepping through each unique arrangement (this time repeated only 10 times). There are undoubtedly others, but I haven't taken the time to verify them. 

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