Where does the multiplicity arise?
For convenience I transpose the matrix.
The first state is1000 00000000 . . .
It is made into (N=1)
000011000000 . . .
Now there are two possibilities for the next move: (N=2)
000001001100 . . .
and
000010000110 . . .
So C(2)=2
See what I mean?
/Bo.
Den søndag den 5. september 2021 14.24.42 CEST skrev Raul Miller
<rauldmil...@gmail.com>:
If I understand correctly C(1) through C(9) would look like this (each
generation occupying only one column), if every amoeba divided at each
opportunity (so, technically, each column should be surrounded by
other columns consisting only of zeros .. what I am showing here is a
compacted representation of that):
|:(+ _1&|.)^:(i.9) 1 0 0 0
1 1 1 1 2 6 16 36 72
0 1 2 3 4 6 12 28 64
0 0 1 3 6 10 16 28 56
0 0 0 1 4 10 20 36 64
But from the description, I imagine that instead, exactly one amoeba
divides at each relevant point in time.
I hope this helps...
--
Raul
On Sun, Sep 5, 2021 at 7:12 AM 'Michael Day' via Chat
<c...@jsoftware.com> wrote:
>
> The good news: 2 new problems yesterday after a lull for several months.
> The bad news (for me): I don't understand them!
>
> ... which is often the case, but here, I don't even get the Mickey
> Mouse case.
>
> It's not done to "spoil" these problems for others, but I don't think
> it's cheating to
> invite an explanation for how C(2) = 2 for problem 762. Once I
> understand what they
> want I can go on and (probably not) be able to solve it for myself.
>
> As I mis-understand it, the starting position is (0,0) and there's only
> one state at each generation.
> Where does the multiplicity arise?
>
> I should post my query in their Clarifications Forum, but don't fancy
> being trolled there - much
> better to be teased by fellow J(oker)s.
>
> The original of problem 762 can be found here:
> https://projecteuler.net/problem=762
>
> Here's a copy, modified to non-graphics, roughly as pasted into my
> proto-script:
>
> ... better in a fixed width font:
> NB. Problem 762
> NB. Consider a two dimensional grid of squares. The grid has 4 rows but
> infinitely many columns.
>
> NB. An amoeba in square (x, y) can divide itself into two amoebas to
> occupy the squares (x+1,y) and
> NB. (x+1,4|y+1), provided these squares are empty.
>
> NB. The following diagrams show two cases of an amoeba placed in square
> A of each grid. When it divides, it is
> NB. replaced with two amoebas, one at each of the squares marked with B:
>
> NB. (origin at J matrix index 3 0)
> NB. ('abb' (3 2 3, each 0 1 1) } 4 6 $ '.'),.(4 4$' '),. 'abb' (0 0
> 3, each 3 4 4) } 4 6 $ '.'
> NB. ...... ...ab.
> NB. ...... ......
> NB. .b.... ......
> NB. ab.... ....b.
>
> NB. Originally there is only one amoeba in the square (0, 0). After N
> divisions there will be N+1 amoebas
> NB. arranged in the grid. An arrangement may be reached in several
> different ways but it is only counted once.
> NB. Let C(N) be the number of different possible arrangements after N
> divisions.
>
> NB. For example, C(2) = 2, C(10) = 1301, C(20)=5895236 and the last nine
> digits of C(100) are 125923036.
> NB. Find C(100,000), enter the last nine digits as your answer.
>
> One for Joseph Turco, perhaps; and thanks for any, non-spoiler, tips!
>
> Mike
>
> --
> This email has been checked for viruses by Avast antivirus software.
> https://www.avast.com/antivirus
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
