(I see RE Boss has just posted, while I was drafting - sorry if it's the
same!)
I thought to exploit the property that
S3(n) = sum i^2, i <: n, is n(n+1)(2n+1)%6
but solving S(n) - S(m) = 2018 isn't that easy!
Anyway, we could define
S3 =: 6 <.@%~ (* (>: * >:@+:))
but as we need a string of squares it's overkill.
Either way, it's useful to be able to identify the indices of ones in a
boolean
matrix:
ixbin =: $#.inv I.@,
So we get, accprding to taste:
-/"1 ixbin 2018 = -/~ +/\ *: i. >: 20
12
-/"1 ixbin 2018 = -/~ S3 i. 20 NB. not as efficient as it looks!
12
ts'-/"1 ixbin 2018 = -/~ S3 i. 20'
1.53e_5 8768
ts'-/"1 ixbin 2018 = -/~ +/\ *: i. >: 20'
8.7e_6 9856
MIke
On 24/10/2019 07:36, Skip Cave 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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