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#Post#: 311--------------------------------------------------
Re: Math question about the knight and hawk.
By: HGMuller Date: January 31, 2018, 5:11 am
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I was a bit hasty in my initial posting, and now I realize that
there are a few squares near the corners of the reachable area
that can actually not be reached. I was also wrongly imagining
the shape of the boundary as if the distance between the
outermost points was 4 steps (which you would get for a piece
that could leap over distances 3 and 4). This does not change
the answer for how many steps you need to cover a million
squares, though, as this was overshoorting that target by more
than 4000 squares. The correct argument is this:
Using oly steps of length 3, a 'diluted grid' with spacing 3 an
be reached. The image below shows one quadrant of the reachable
area for 5 such moves, the Hawk starting at #, the * indicating
reachable squares.
[font=courier]* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
# . . * . . * . . * . . * . . *[/font]
Next we check how we could get to in-betwee squares. The
upper-right corner can only be reached by 5 diagonal leaps of
length 3. But by replacing oe of these by a leap of length 2, we
advanced one square less along the diagonal ('1' below), and
doing that twice even two squares less ('2'). To advance even
less along the diagonal (e.g. to 'C'), we can just take fewer
steps.
[font=courier]* . . * . . * . . * . . * . . *
. . . . . . . . . . a b c . 1 .
. . . . . . . . . . 2 2 . 2 . .
* . . * . . * . . B 3 x C . c *
. . . . . . . . 1 1 y 1 x 2 b .
. . . . . . . . + + + y 3 2 a .
* . . * . . * . + A + 1 B . . *
. . . . . . . . + + + 1 . . . .
. . . . . . . . . . . . . . . .
* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
* . . * . . * . . * . . * . . *
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
# . . * . . * . . * . . * . . *
[/font]
The diluted-grid squares on the outer rim next to the corner can
only be reached by 4 diagonal leaps, followed by one orthogonal
one. But this gives us the choice to replace only diagonal leaps
by a length of 2 (tracing a diagonal via 'b' and '2' to 'B'), or
also replace the orthogonal step by a shorter one (leaving us at
'c'). For any square you can reach you can also reache squares
diagonally inward by replacing one of the diagonal leaps by a
shorter one. Note that even to go 5 leaps straight up you can
always take half the steps diagonally up-right, and the other
half diagonally up-left, so there are always many diagonal steps
available in the path that you could make shorter.
Note that for reaching squares on the diluted grid two steps
inward from the outermost ring (e.g. 'A'), you need two steps
less than you have, so you can make the remaining two steps
3-out, 2-in to step, to step to the adjacent squares '+'. This
fills up the entire plane in the inner region. Square 'a' can be
reached by stepping to 'C', and then take a leap of length 2
perpendicularly.
So all squares just inside the outer periphery are reachable,
except those next to the '1'. And for each of those you can move
to the squares more diagonally inward. That leaves only the
diagonals adjacent to the main diagonal. C can be reached with 3
diagonal leaps, then one up and one right. By taking one of the
latter of length 2, we can reached the 'x' squares. (And thus
also the squares diagoally inward from those, such as 'y'.) So
from the (6*N-1) x (6*N-1) area only 4 x 4 squares near the
corners are not reachable.
On the outermost rim only 2*N squares are reachable on each edge
(counting the corners as 1/2, because they are on 2 edges). So
the total number of squares reachable in N moves is
(6*N-1)*(6*N-1) + 8*N - 16. For N=167 this is 1001*1001 + 1336 -
16 = 1,003,321 squares.
Note that this calculation is based on that the number of
allowed moves is large, so that you could always find enough
steps in the desired directions to replace by a shorter one. For
N <= 3 this would not always be possible, and the formula would
not apply.
#Post#: 314--------------------------------------------------
Re: Math question about the knight and hawk.
By: chilipepper Date: January 31, 2018, 9:58 am
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Thanks HGMuller, that is awesome work. It is interesting that
your initial calculation had a slight error (no critisism) and
yet the final result (number of moves) was correct, and the
total number of reachable squares very close to your initial
figure. (small errors in math and algebra aren't always so
inconsequential).
I now take it that 1,000,000 or more destination squares can be
reached by 167 moves of the hawk. (1,003,321 with N=167).
This problem would also lend itself to solution by a worksheet
macro or a simple computer program. But doing it with only
mathematical equations is way more elegant and impressive!
Nicely done. :)
To help visualize the problem, here is a diagram with all the
hawk's moves for N=2 (1 quadrant), and counting the ways each
square can be reached (from only within that quadrant). The
total squares that can be reached for N=2 is 105 (assuming I
made no errors).
[attachimg=1]
#Post#: 319--------------------------------------------------
Re: Math question about the knight and hawk.
By: HGMuller Date: January 31, 2018, 12:07 pm
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I think the counting isn't entirely correct. (E.g. the (2,5)
square should be reachable by (2,2) + (0,3).)
Anyway, I think diagrams like this are important for predicting
leaper values with more precision. Being able to reach the same
square in multiple ways is somewhat better than reaching it only
in one way. But not early as good as reaching multiple squares
that are all different. Ths explains why the Knight is one of
the best leapers with 8 targets. There is no overlap between its
one- and two-move targets. And the redundancy within the
two-move set is as small as you can expect from a fully
symmetric piece (each target reachable in 2 ways, except those 8
made by two identical moves). So it can reach 41 squares
(including its square of origin). A Commoner can only reach 25
squares in 2 steps. A range-2 Rook only 33.
#Post#: 330--------------------------------------------------
Re: Math question about the knight and hawk.
By: chilipepper Date: January 31, 2018, 5:20 pm
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Thanks for noticing the error in my diagram. I get that the hawk
in two jumps can land on 113 distinct squares. A corrected
diagram is here:
[attachimg=1]
Your result for the knight also matches what I found. In two
jumps it can land on 41 different squares. Diagram:
[attachimg=2]
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