3 1(1) is the shape of the result block, which contains the result of(p
# n)&>t
[you can see this for yourself by setting tooltip detail to 'Tutorial'
and hovering over the components of 3 1(1)]
The (1) indicates that fill was added to the result, bring the shape up
to 1.
The crosshatching in the result indicates where the fill was added to
the result.
p and n don't show results yet because they are being applied to
multiple cells as part of (p # n) and there is no guarantee that the
results of the individual cells can be assembled into an array. If you
click on the result of #, that will select an individual cell, and you
can then see what p and n contributed to that result.
If you want to see what p is doing, select a single result-cell of p and
then right-click the name p, to get a dissection of that name.
Henry Rich
On 12/4/2016 1:18 PM, Moon S wrote:
OK, simple case
t NB. three boxed strings
┌──┬──┬──┐
│22│33│44│
└──┴──┴──┘
p NB. test for all two's
*./@('2' = ])
n NB. append zero and cvt to number
".@('0' ,~ ])
(p&>t)#(n&>t) NB. ok
220
(p # n)&>t NB. zero-padded
220
0
0
dissect doesn't show anything for p and n in the second case; and what's "3
1(1)" for '#'?
http://mas.orgfree.com/adventofcode2016/diss1.png
On Sun, Dec 4, 2016 at 6:14 PM, Henry Rich <henryhr...@gmail.com> wrote:
I have no time to help, other than to suggest you use dissect to see
what's happening.
Henry Rich
On 12/4/2016 11:02 AM, Moon S wrote:
I have this code:
t =: cutLF CR-.~fread '04.dat'
a =: a.{~97+i.26 NB. 'abc...z'
c =: 5{.a\:[:+/"1 a=/[:/:~_10}.'-'-.~] NB. calculate checksum string
v =: c -: _1}._6&{. NB. checksums match = good string
n =: [:".3{._10{.] NB. extract number (id)
s =: 4 :'((26|x+])&.(_97+a.i.]))"0 y' NB. shift cypher, n s 'str'
p =. +./@('north' E. n s ]) NB. if decrypted string contains
'north'
echo +/(n*v)&> t NB. sum of ids of good strings
echo +/(n*p)&> t NB. sum of ids of strings with
'north'
exit 0 NB. ^^^ we assume such
string
is only one
It works, but I planned to use '#' instead of +/...*... for the second
task.
Verb n takes a number from a string.
Verb p decrypts a string using the number and tells if it contains
'north'.
So far so good.
Noun t is the input data as boxed strings.
n&> t gives numbers (like 111 222 333 444)
p&> t gives a boolean vector (like 0 0 1 0)
The idea was to have something like "(p # n)&> t" but that didn't work.
(p&> t)#(n&> t) is ok
(p # n)&> t is not
Something wrong with ranks, I guess. Any idea how to use (p # n)?
Thank you.
Georgiy Pruss
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