7:26 how does your later solution not have an 8 when the only 1's in the last column are positions 1 and 8? If you need to add a different row, and you already finished with the first row, none of the other rows would affect 8, unless I'm misunderstanding how this row reduction works. edit: after watching another 10 times I realized you said swap ROWS 2 and 3, not columns. I misunderstood how row reduction worked. I guess the 8 could move up if it got swapped. Is it important which rows you choose to swap? I see row 2 could swap with 1, 3, 4 or 5. Also, if you do the row reduction and keep track of when the last column needs to change then you can use that and solve any puzzle of that size without having to use matricies.
If the individual columns were true vectors they could be 'added' in any order to produce the same result. That doesn't seem to be happening here however. ((X+ 5) + 3) + 2 for instance yields the result [0,0,0,0,1,0,0,0,0] I think rather than the [0,0,0,0,0,0,0,0,0] we want.
7:26 how does your later solution not have an 8 when the only 1's in the last column are positions 1 and 8? If you need to add a different row, and you already finished with the first row, none of the other rows would affect 8, unless I'm misunderstanding how this row reduction works.
edit: after watching another 10 times I realized you said swap ROWS 2 and 3, not columns. I misunderstood how row reduction worked. I guess the 8 could move up if it got swapped. Is it important which rows you choose to swap? I see row 2 could swap with 1, 3, 4 or 5. Also, if you do the row reduction and keep track of when the last column needs to change then you can use that and solve any puzzle of that size without having to use matricies.
Interesting. I don't know much about mathematics, but it amazes me when something seemingly complex has a relatively straightforward solution.
Thank you for your comment. There is more to come! :)
how to solve with brute force?
If the individual columns were true vectors they could be 'added' in any order to produce the same result. That doesn't seem to be happening here however. ((X+ 5) + 3) + 2 for instance yields the result [0,0,0,0,1,0,0,0,0] I think rather than the [0,0,0,0,0,0,0,0,0] we want.
Square 5 gets toggled 2 times (by 5 and by 2), so remains the same, black or 0.
اللي من جامعة سطام بالخرج ✋✋
Great video thanks
Thanks a lot! This is very useful for me.
Thank you for the solution. Thank you more for the explanation. Not many can do the same as clear as you.
I solved in less than a second