Skip, Here is a little simpler version of fps1. I looked at forks in LJ and
what I am looking for is a dyadic fork. This doesn’t seem to have one either.
fps2=: 13 :'y#~0=(+/"1) 2|_ q:y'
(fps 2+i.1000)-:fps2 2+i.1000
1
fps2
┌─ ]
├─ ~ ─── #
│
──┤ ┌─ 0
│ ├─ =
│ │ ┌─ [:
└─────┤ │ ┌─ / ─── +
│ ├─ " ─┴─ 1
└───┤
│ ┌─ 2
│ ├─ |
└─────┤ ┌─ _
└─────┼─ q:
└─ ]
Linda
-----Original Message-----
From: Programming <[email protected]> On Behalf Of Linda
Alvord
Sent: Saturday, November 3, 2018 10:03 PM
To: [email protected]
Subject: Re: [Jprogramming] Square Roots and Extended Arithmetic
Skip, I can't find the fork in fps1
(fps 2+i.1000)-:fps1 2+i.1000
1
fps1
┌─ ]
├─ ~ ─── #
│
──┤ ┌─ 0
│ ├─ =
│ │ ┌─ [:
└─────┤ │ ┌─ / ─── +
│ ├─ " ─┴─ 1
└───┤
│ ┌─ [:
│ ├─ ]
└─────┤ ┌─ 2
│ ├─ |
└─────┤ ┌─ _
└───┼─ q:
└─ ]
fps
┌─ ]
├─ ~ ─── #
│
──┤ ┌─ [:
│ │ ┌─ =
└─────┼────┴─ >.
└─ %:
Linda
-----Original Message-----
From:S Programming <[email protected]> On Behalf Of Skip
Cave
Sent: Saturday, November 3, 2018 10:43 AM
To: [email protected]
Subject: Re: [Jprogramming] Square Roots and Extended Arithmetic
Also, there's a completely different way to attack the problem of finding
perfect squares using the dyadic form of q: (Prime Exponents). I discovered
this while reading NuVoc about q:
fps1 =: 13 :'y#~0=+/"1]2|_ q:y'
fps1 2+i.100
4 9 16 25 36 49 64 81 100
fps1 8200+i.1000
8281 8464 8649 8836 9025
%: 8281 8464 8649 8836 9025
91 92 93 94 95
fps1
] #~ 0 = [: +/"1 [: ] 2 | _ q: ]
The secret is: if all exponents of the prime factors of an integer are even,
the number is a perfect square.
Skip
Skip Cave
Cave Consulting LLC
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