Comment by merelysounds

Comment by merelysounds 4 days ago

3 replies

View on Hacker News

Spoiler warning, this comment contains a solution, this is your chance to stop reading, especially if you didn’t have a chance to play yet.

With 8 moves and rows only: 2->1, 1->2, 2->1, 3->2, 2->3, 4->3, 4->1, 1->4.

A more efficient solution should be possible; did anyone find any?

PinkRidingHood 4 days 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.

Reply View 2 replies
  • dandanua 4 days ago

    You 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).

    Reply View 0 replies