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 <[email protected]> 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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>>
>
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