I did not know fill on |. worked that way. I see it now. neat.
----- Original Message -----
From: Joe Bogner <joebog...@gmail.com>
To: programm...@jsoftware.com
Sent: Thursday, December 10, 2015 9:40 AM
Subject: Re: [Jprogramming] advent 10
Thanks, the fill is needed otherwise it would return the wrong result
if the last number is the same as the first. I don't think my version
is susceptible to that but would be interested to find an example if
it is
correct:
(,@((#,{.);.1~ _1&(|.!._) ~: ])) (1 1 1 3 2 2 2 1 1 1)
3 1 1 3 3 2 3 1
vs (no fill)
(,@((#,{.);.1~ _1&(|.) ~: ])) (1 1 1 3 2 2 2 1 1 1)
1 3 3 2 3 1
You can see this in:
Correct:
(<;.1~ _1&(|.!._) ~: ]) (1 1 1 3 2 2 2 1 1 1)
┌─────┬─┬─────┬─────┐
│1 1 1│3│2 2 2│1 1 1│
└─────┴─┴─────┴─────┘
Incorrect:
(<;.1~ _1&(|.) ~: ]) (1 1 1 3 2 2 2 1 1 1)
┌─┬─────┬─────┐
│3│2 2 2│1 1 1│
└─┴─────┴─────┘
vs
On Thu, Dec 10, 2015 at 9:05 AM, 'Pascal Jasmin' via Programming
<programm...@jsoftware.com> wrote:
> 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 <joebog...@gmail.com>
> To: programm...@jsoftware.com
> 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 <ec...@cim.mcgill.ca> 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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