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Apr 22, 2020 at 4:51 vote accept Andeanlll
Apr 22, 2020 at 3:10 comment added J. M.'s missing motivation Just to add to the answers below: you can do strict filtering with Alternatives (|) and Repeated[] e.g. Count[{{1, 1, 1, 1}, {1, 1, 0, 1}, {1, 0, 0, 1}, {0, 1, 0, 0}, {1, 0, 0, 0}, {1, 0, 1, 0}, {1, 1, 1, 1}, {0, 1, 0, 0}}, {1, Repeated[0 | 1, {2}], 1}]
Apr 22, 2020 at 3:04 answer added Nasser timeline score: 1
Apr 22, 2020 at 3:03 answer added Rohit Namjoshi timeline score: 1
Apr 22, 2020 at 2:55 history asked Andeanlll CC BY-SA 4.0