A1, Q2B Solution

[Brian]

Prof. Brudno, did you say that the solution to this question was that
we create a partition line of slope 1 through the centre of the
matrix, like so?

http://tinyurl.com/2btgcp6

Where the gray box should be n x n in in dimensions.

[Michael Brudno]

Yes, I said this. But then I thought about it (just now) and realized
that won't be balanced, as, for example, an alignment going along the
top row to the partition will not lead to balanced alignments, as one
will have size 0 and the other won't. So have to use a line of non-one
slope instead.

Sorry for the confusion.

-M

[Brian]

Oh okay, that's what I thought. I did some calculations and got that
the partition sizes were [j + (m-n)/2] x j and [(m + n)/2 - j] x (n-j)
where (i, j) is the cell on the partition line through which the path
of the optimal global alignment goes, but those didn't seem balanced.

On Nov 2, 1:43 am, Michael Brudno <bru...@gmail.com> wrote:
> Yes, I said this. But then I thought about it (just now) and realized
> that won't be balanced, as, for example, an alignment going along the
> top row to the partition will not lead to balanced alignments, as one
> will have size 0 and the other won't. So have to use a line of non-one
> slope instead.
>
> Sorry for the confusion.
>
> -M
>