3 ms·
I found one using a program: [('row', 0, 1), ('row', 1, 2), ('row', 2, 0), ('row', 0, 3), ('col', 2, 3), ('col', 1, 2), ('col', 0, 1)]. It says it's the optimal
by PinkRidingHood 1y ago
I found one using a program: [('row', 0, 1), ('row', 1, 2), ('row', 2, 0), ('row', 0, 3), ('col', 2, 3), ('col', 1, 2), ('col', 0, 1)]. It says it's the optimal.
- merelysounds 1y agoThanks for sharing and congrats!
- dandanua 1y agoYou could use 7 row operations. row and col ops commute, and your last 3 col ops are equivalent to ('row', 1, 0), ('row', 2, 1), ('row', 3, 2) if acted on identity matrix. So, use them at first, and then your four row ops. Alternatively, you could use 7 col. Your 4 row ops are equivalent to ('col', 3, 0), ('col', 0, 2), ('col', 2, 1), ('col', 1, 0).