Somewhat clumsy, as uses each for the loop.
The key idiom is to use subarrays to iterate a window over the vector …
x u;.3 y
So to see subarrays of various lengths … here length 8 in i.10 (there are 3)
and then length 9 in i.10 (there are 2) … the 2 results (for length 8 and
length 9 are shown vertically for clarity)
,.((*:i.10)&(] <;. _3 [)) each 8 9
NB. Note here the u function is just < so we can see
the lists
┌────────────────────────────────────────────────────────────────┐
│┌───────────────────┬────────────────────┬─────────────────────┐│
││0 1 4 9 16 25 36 49│1 4 9 16 25 36 49 64│4 9 16 25 36 49 64 81││
│└───────────────────┴────────────────────┴─────────────────────┘│
├────────────────────────────────────────────────────────────────┤
│┌──────────────────────┬───────────────────────┐ │
││0 1 4 9 16 25 36 49 64│1 4 9 16 25 36 49 64 81│ │
│└──────────────────────┴───────────────────────┘ │
└────────────────────────────────────────────────────────────────┘
So now change the input vector to (*:i.100) and the length of the subarrays to
the list 9 10 11 12 13 to show any sets that match the rule …
,.((*:i.100)&(] ([: < ([: 2018&= +/) # ] );. _3 [)) each 9 10 11 12 13
NB. Change the u function to a tacit expression to return the box of
the subarray if its sum matches 2018
┌──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
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├──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┤
│┌┬┬┬┬┬┬┬────────────────────────────────────────────┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┬┐│
│││││││││49 64 81 100 121 144 169 196 225 256 289
324│││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││││
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└──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┘
> On 24 Oct 2019, at 5:36 pm, Skip Cave <[email protected]> wrote:
>
> I occasionally try to solve Quora math problems to polish my J skills.
> Here's a problem that
>
> * "The integer 2018 can be written uniquely as the sum
> of consecutive squared integers. How many squares does the sum contain?"*
>
> a=.2^~i.100
>
> NB. Find how many consecutive squared integers it takes to sum to 2018. Try
> each length until I get a hit.
>
> +/2018=+/"1]5,\a
>
> 0
>
> +/2018=+/"1]6,\a
>
> 0
>
> +/2018=+/"1]7,\a
>
> 0
>
> +/2018=+/"1]8,\a
>
> 0
>
> +/2018=+/"1]9,\a
>
> 0
>
> +/2018=+/"1]10,\a
>
> 0
>
> +/2018=+/"1]11,\a
>
> 0
>
> +/2018=+/"1]12,\a
>
> 1 NB. A hit!
>
> +/2018=+/"1]13,\a
>
> 0
>
>
> So the answer is 12
>
> List them:
>
> b#~2018=+/"1 b=.12,\a
>
> 49 64 81 100 121 144 169 196 225 256 289 324
>
>
> Is there a way to do this search without explicit looping?
>
>
> Skip
>
>
> Skip Cave
> Cave Consulting LLC
> ----------------------------------------------------------------------
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