OK - I've got to C(10) = 1301, but my current, crude, method is, of
course, so inefficient that C(20) takes
quite a while before reporting it's out of memory. Early days.
Pierpaolo pointed out to Raul that the propagation rule specifies that
only one amoeba may occupy one cell.
Thanks again, everyone. I suggest we don't discuss further, in order
not to spoil the problem for others.
Mike
On 05/09/2021 14:59, 'Mike Day' via Chat wrote:
Aaaaah....
Thanks all. I suppose the remark, that there are N+1 amoebas after N
divisions, was the clue, as well as your various examples.
I’d thought they meant that all members of a generation spawn offspring;
evidently only one may reproduce at each stage. Onwards and downwards.
Thanks again,
Mike
Sent from my iPad
On 5 Sep 2021, at 12:12, '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
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