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 <[email protected]> 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
