Raul's explained a lot overnight but, now I'm awake,
I'll try to expand on the working of my effort - appended
below your (Bruno's) note in case anyone is still trying
to do problem 36.
Mike
Bruno wrote:
Thank you for the awnsers.
I'm having some trouble understanding the codes tough.... In Roger's code
what does the I. do? Here's how I read the code:
+/ I. ((-:|.)@#: *. (-:|.)@":)"0 i.1e6
from 1 to 1e6, check if the number is palyndrome in base 2 and in base
10.
I. the bolean list // don't understand this part
sum the elements
In Mike's code, there's a nice speed boost, but I can't understand the
code
flow... How do you prevent from doing a test if the other test was
false? I
beleive it has to do with the * primitive...
And if we wanted to do as Raul told and only test odd numbers, how do
we go
from 1 to 999999 jumping 2 numbers? That would cut the code execution
time
in half I beleive.
#######################
I tried going back to problem 1 and adapting your solutions to it, but
failed =(
+/ 1e3 #~ ((3|"0)) +. (5|"0)) i. 1e3
NB. how do I access the current number from inside the ( ) ?
Bruno
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(I've tried to anticipate line-unwraps, probably missed some)
My first offering was
+/(* ((-:|.)@#:"0)) (#~((-:|.)@":"0)) i. 1e6
You can construe this as 3 operations in series, given
the right argument as i. 1e6
Each operation is a tacit J expression, but the whole
is an explicit J expression.
(Raul's useful remark about odds only can be incorporated
by restating the right argument as >: +: i. 0.5e6)
Let's use a small set of numbers for exposition:
n =: 12 110 111 112 313
The first operation is
(#~((-:|.)@":"0))
Consider these fragments:
[sn =: ":"0 n NB. express each number as a string
12
110
111
112
313
[b =: (-: |.)"1 sn NB. are strings palindromes?
0 0 1 0 1
NB. this is a hook; see J dictionary
[n10 =: b#n NB. yield base-10 palindromes
111 313
NB. join these fragments together - we can now drop
NB. the "1 from the hook:
[n10 =: (#~(-:|.)@":"0)n NB.can drop a pair of brackets
111 313
NB. The second similar operation applies only to the results
NB. of the first. So in this approach you don't need to
NB. worry about "doing a test if the other test was false"
[n2 =: (* (-:|.)@#:"0) n10 NB.again drop a pair of brackets
0 313
NB. This is another hook f g where f is * and
NB. g is ((-:|.)@#:"0)
((-:|.)@#:"0) n10 NB. right monad side of hook
0 1
NB. The * just multiplies 0 1 by 111 313
NB. Alternatively we could have had
[n2 =: (#~ ((-:|.)@#:"0)) n10
313
NB. I think the latter (with #~) involves more
NB. restructuring while the former (with *) needs
NB. more multiplies. However, they time roughly equally
NB. on this problem.
NB. The last, third, operation +/ merely adds up the
NB. remaining numbers
My alternative solution just swaps the base-10 and
base-2 operations.
I hope that helps.
Mike
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