You may have gotten lucky with _1&|. (fill not needed afaiu)
an improvement to mine, after seeing yours.
# ,@( ((# , {. );. 1)~ 1 , 2 ~:/\ ])^:(40) 1 1 1 3 2 2 2 1 1 3
----- Original Message -----
From: Joe Bogner <[email protected]>
To: [email protected]
Sent: Thursday, December 10, 2015 8:47 AM
Subject: Re: [Jprogramming] advent 10
Clever use of the state machine. I enjoy seeing those...
Here's mine:
NB. part 1
# (,@((#,{.);.1~ _1&(|.!._) ~: ]))^:40 input=:(1 1 1 3 2 2 2 1 1 3)
252594
NB. part 2
# (,@((#,{.);.1~ _1&(|.!._) ~: ]))^:50 input=:(1 1 1 3 2 2 2 1 1 3)
3579328
To explain, we are going to use cut to split the array when the number
is different than the previous number
Return 1 if the number is different than the previous number, otherwise 0
(_1&(|.!._) ~: ]) (1 1 1 3 2 2 2 1 1 3)
1 0 0 1 1 0 0 1 0 1
Apply cut to the sequence to cut into intervals
(<;.1~ _1&(|.!._) ~: ]) (1 1 1 3 2 2 2 1 1 3)
┌─────┬─┬─────┬───┬─┐
│1 1 1│3│2 2 2│1 1│3│
└─────┴─┴─────┴───┴─┘
Instead of boxing, return count of the first item in each interval
((#,{.);.1~ _1&(|.!._) ~: ]) input=:(1 1 1 3 2 2 2 1 1 3)
3 1
1 3
3 2
2 1
1 3
from here, just ravel and count the numbers
On Thu, Dec 10, 2015 at 3:55 AM, Ryan Eckbo <[email protected]> wrote:
> Oops, typo in the title - this is for day 10, not 8.
>
> On 10 Dec 2015, at 19:52, Ryan Eckbo wrote:
>
>> A previous advent answer got me thinking about state machines, so I wrote
>> one for this problem. The extra initial state ruins the natural mapping
>> between states and numbers, but I think it's unavoidable? Also someone
>> could probably write a clever verb to generate the table.
>>
>>
>> NB. state machine -- can be confusing because rows/states 1,2,3,..
>> represent numbers 0,1,2,..
>> S=: 0 10#: 10* ". }. [;._2 noun define
>> 0 1 2 3 4 5 6 7 8 9
>> 1.1 2.1 3.1 4.1 5.1 6.1 7.1 8.1 9.1 10.1 NB. initial
>> 1.0 2.2 3.2 4.2 5.2 6.2 7.2 8.2 9.2 10.2 NB. 0
>> 1.2 2.0 3.2 4.2 5.2 6.2 7.2 8.2 9.2 10.2 NB. 1
>> 1.2 2.2 3.0 4.2 5.2 6.2 7.2 8.2 9.2 10.2 NB. 2
>> 1.2 2.2 3.2 4.0 5.2 6.2 7.2 8.2 9.2 10.2 NB. 3
>> 1.2 2.2 3.2 4.2 5.0 6.2 7.2 8.2 9.2 10.2 NB. 4
>> 1.2 2.2 3.2 4.2 5.2 6.0 7.2 8.2 9.2 10.2 NB. 5
>> 1.2 2.2 3.2 4.2 5.2 6.2 7.0 8.2 9.2 10.2 NB. 6
>> 1.2 2.2 3.2 4.2 5.2 6.2 7.2 8.0 9.2 10.2 NB. 7
>> 1.2 2.2 3.2 4.2 5.2 6.2 7.2 8.2 9.0 10.2 NB. 8
>> 1.2 2.2 3.2 4.2 5.2 6.2 7.2 8.2 9.2 10.0 NB. 9
>> )
>> v=:[: ,@:((#,{.)every) (0;S)&;:
>> smoutput $ v^:40 Input=: 1 1 1 3 2 2 2 1 1 3
>> smoutput $ v^:50 Input
>> ----------------------------------------------------------------------
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>
> ----------------------------------------------------------------------
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