To follow up on this thread, I have been playing with
psuedo liberties a bit, and here is my solution.
* I use 2 vectors of values. The first is used for
storing the pseudo liberty values. The second lists has
all 1*, 2*, 3*, and 4* the values, which are sorted.
Binary search will show
On Sat, 2007-11-24 at 08:38 -0700, [EMAIL PROTECTED] wrote:
To follow up on this thread, I have been playing with
psuedo liberties a bit, and here is my solution.
* I use 2 vectors of values. The first is used for
storing the pseudo liberty values. The second lists has
all 1*, 2*, 3*,
Jason House [EMAIL PROTECTED] said:
On Sat, 2007-11-24 at 08:38 -0700, [EMAIL PROTECTED] wrote:
To follow up on this thread, I have been playing with
psuedo liberties a bit, and here is my solution.
* I use 2 vectors of values. The first is used for
storing the pseudo liberty
On Sat, 2007-11-24 at 10:36 -0700, [EMAIL PROTECTED] wrote:
Since 160,000 2*(19*19)^2, a value well below the various possible
sums of squares, I have to ask what additional work you've done to prove
that the overlap doesn't cause problems?
160,000 is greater than (19 * 19 + 1)^2, so
The solution that John Tromp found, for 9x9 is this:
Keep an extended pseudo-liberty count that looks like this:
lower 8 bits: Standard pseudo-liberty count
next 12 bits: Encoding of x-coordinate information
next 12 bits: Encoding of y-coordinate information
For each block of 12 bits we need to
Jason,
I realize that I was looking at the IsOneSum() constructor,
which had the array lookups. My bad. I just didn't
understand how it worked.
-- M
On Sat, 2007-11-24 at 10:36 -0700, [EMAIL PROTECTED] wrote:
Since 160,000 2*(19*19)^2, a value well below the various possible
sums of
On Nov 24, 2007 10:38 AM, [EMAIL PROTECTED] wrote:
* For liberty values, I use the following equation:
values[i] = 25000 + (16 * (i+1)) + ((i+1) * (i+1));
Some bits are constant, some are linear, and some
are quadratic, which guarantees that the sum of
up to 4 values will
