Hi all!
When you do these kind of set operations it is common to represent the
set as a bitmap with ones for each item. A k-set {0 1 2} of the set {0 1
2 3 4} would then be represented like this.
1 1 1 0 0
This is called GRAY CODES.
One advantage is that you can move an item in only one operation. You
don't have to remove it somewhere and add it somewhere else. The move
can be made with exclusive or.
1 1 ~: 0 1
1 0
0 0 ~: 0 1
0 1
You can switch both zeroes and ones with ones. Zeros do nothing.
The first example shows a move of a one one step left.
Two bitmaps of 64 bits can be handled in one assembler instruction. All
that is needed to create the next set.
Here is an example of the generation of a 3-set from a 5-set with gray
codes.
[q=:10 5 $1 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 1 1 1
0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 1 0 0 0 1 0 0 1
1 1 1 0 0
0 0 1 1 0
0 0 0 1 1
0 1 1 1 1
0 0 0 1 1
1 1 0 0 0
0 0 1 1 0
1 1 0 0 0
1 0 1 0 0
0 1 0 0 1
~:/\ q
1 1 1 0 0
1 1 0 1 0
1 1 0 0 1
1 0 1 1 0
1 0 1 0 1
0 1 1 0 1
0 1 0 1 1
1 0 0 1 1
0 0 1 1 1
0 1 1 1 0
/:~>(~:/\ q) <@#"1 i.5
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
Cheers,
Erling Hellenäs
On 2017-11-03 18:16, Linda Alvord wrote:
It seems that you mighg want to use a diferent name for your combE
load 'stats'
(3 comb 4);3 combE 4
┌─────┬───────┐
│0 1 2│0 0 0 1│
│0 1 3│0 0 0 2│
│0 2 3│0 0 1 1│
│1 2 3│0 0 1 2│
│ │0 0 2 1│
│ │0 0 2 2│
│ │0 1 0 1│
│ │0 1 0 2│
│ │0 1 1 1│
│ │0 1 1 2│
│ │0 1 2 1│
│ │0 1 2 2│
│ │0 2 0 1│
│ │0 2 0 2│
│ │0 2 1 1│
│ │0 2 1 2│
│ │0 2 2 1│
│ │0 2 2 2│
│ │1 0 0 1│
│ │1 0 0 2│
│ │1 0 1 1│
│ │1 0 1 2│
│ │1 0 2 1│
│ │1 0 2 2│
│ │1 1 0 1│
│ │1 1 0 2│
│ │1 1 1 1│
│ │1 1 1 2│
│ │1 1 2 1│
│ │1 1 2 2│
│ │1 2 0 1│
│ │1 2 0 2│
│ │1 2 1 1│
│ │1 2 1 2│
│ │1 2 2 1│
│ │1 2 2 2│
└─────┴───────┘
Linda
-----Original Message-----
From: Programming [mailto:[email protected]] On Behalf
Of Erling Hellenäs
Sent: Friday, November 3, 2017 12:44 PM
To: [email protected]
Subject: Re: [Jprogramming] Partitions
Hi all!
I did not correctly understand the e. verb
24 e. r4
0
24 e. ,r4
1
So 358358 surely was there and was double-counted.
Cheers,
Erling Hellenäs
On 2017-11-03 17:07, Erling Hellenäs wrote:
No, its not. It seems it should have bin there. Maybe something is
wrong. /Erling
On 2017-11-03 16:56, Erling Hellenäs wrote:
Lol. This root is double-counted. 1 1 1 1 358358 /Erling
On 2017-11-03 16:42, Erling Hellenäs wrote:
Hi all!
My take:
v=:1 1 1 1 2 7 11 13 179
r=:5 parRuskeyE 9
r2=: >r (*/)@:{&.>"1 0 < v
r3=:/:~"1 r2
r4=: ~.r3
NB. One root is 1 1 1 1 358358
NB. All permutations
(!5)*1+#r4
6360
Cheers,
Erling Hellenäs
On 2017-11-03 15:03, Erling Hellenäs wrote:
Sorry. Add ones. /Erling
On 2017-11-03 15:00, Erling Hellenäs wrote:
OK. The five prime factors are needed if you want to multiply 5
numbers and get 358358, except that you can possibly add zeros and
358358 itself? /Erling
On 2017-11-03 14:35, 'Mike Day' via Programming wrote:
Language differences?
Personally, I tend to be rather careless, using the terms
“factor” and “divisor” for the same thing, when “divisor” is
perhaps to be preferred. However, Pari GP provides a library
function “factor” to deliver the prime factorisation of its
argument, in a similar fashion to J’s q: .
So, strictly speaking, perhaps, the factors of 358358 are 2 7
11 13 179 and its divisors are 1 2 7 11 13 14 22 etc...
You’re both right!
Mike
Please reply to [email protected].
Sent from my iPad
On 3 Nov 2017, at 11:25, Erling Hellenäs
<[email protected]> wrote:
Hi all!
Raul:
"Hmm... actually, thinking about it, the par approach here is
not efficient enough for this example. 5 parRuskeyE 32 is too
big of a result, I think. (358358 has 5 distinct prime factors
and, thus, 32 integer factors.)"
However there are only 5 integer factors:
q: 358358
2 7 11 13 179
*/q:358358
358358
Now you ask me for 44 5 integer factorizations.
As far as I understand there is only one 5 integer factorization
of 358358 unless you count 1 and 358358 as factors.
In any case it can be handled by parRuskeyE.
Cheers,
Erling Hellenäs
Den 2017-11-03 kl. 10:58, skrev Raul Miller:
I'm not sure where you showed the 44 different 5 integer
factorizations of 358358?
Thanks,
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