I think it's easier to post my code than to try answering your question.
I've run successfully this in a new session, so I think it's
self-contained.
Apologies for the verbose code - it's as I wrote it, and I haven't
attempted
to polish it for presentation here. The function "part2" , not shown here,
was the one I wrote before I remembered there were three throws for
each turn! I would have preferred to deal with both players in a double-
go, but gave up. I do of course realise that there's a lot of
reduplicated
code for player 2 - it would have been neater to move the code to an
ecternal function, but I didn't!
Here it is - I hope it's not too opaque.
m10 =: 10&|&.<: NB. modulo function for origin 1
die3 =: ~. /:~,(1 2 3 +/])^:2 ]1 2 3
ctdie3 =: #/.~/:~,(1 2 3 +/])^:2 ]1 2 3
part2a =: 3 : 0
21 part2a y
:
cts =. 1 $ 1
'pos1 pos2' =. 2 1 $ y
'scr1 scr2' =. 2 1 $ 0
tgt =. x
wins =. 0 0
turns=. 0
ndie3=. # die3
while. #cts do.
NB. player 1
turns =. >: turns
pos1 =. m10 , pos1 +/ die3
cts =. ,cts <.@*/ ctdie3
scr1 =. pos1 + ndie3 # scr1
pscr =. pos1,. scr1,. ndie3#pos2,. scr2
cts =. pscr +//. cts
'pos1 scr1 pos2 scr2' =. |: pscr =. ~. pscr
if. +/ win1=. tgt <: scr1 do.
i1 =. I. win1
wins =. wins + 2 {. +/ i1{cts
echo 'new 1 wins at turn '; turns; wins
cts =. cts#~ -. win1
pscr =. pscr#~-. win1
'pos1 scr1 pos2 scr2' =. |: ~. pscr
end.
NB. player 2
turns =. >: turns
pos2 =. m10 , pos2 +/ die3
cts =. ,cts <.@*/ ctdie3
scr2 =. pos2 + ndie3 # scr2
pscr =. (pos2,. scr2),.~ ndie3#pos1,. scr1
cts =. pscr +//. cts
'pos1 scr1 pos2 scr2' =. |: pscr =. ~. pscr
if. +/ win2=. tgt <: scr2 do.
i2 =. I. win2
wins =. wins + _2 {. +/ i2{cts
echo 'new 2 wins at turn '; turns; wins
cts =. cts#~ -. win2
pscr =. pscr#~-. win2
'pos1 scr1 pos2 scr2' =. |: ~. pscr
end.
end.
turns; wins
)
Cheers,
Mike
On 11/01/2022 00:17, Raul Miller wrote:
Your 10&|&.<: is probably more comprehensible than my 11 |&.<:
But for part B, how did you handle the smearing of scores that
resulted from those new positions and counts?
Thanks,
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