Skip to content

ajnirp/j-scripts

Repository files navigation

About

A collection of some of my J scripts.

To future me

.deb files provided by J Software for versions 8 and above usually install the normal Unix way (so look for stuff in /usr/share/j/8.01/...). To run sentences like install 'all' you need to open j console as follows:

sudo ijconsole

Explanations

gray.ijs

Start with

0
1

next takes such a pattern and produces the next level, which in the above case is

0 0 0 1 1 0 1 1

Just append 0 to each cell, then reverse the pattern and append 1 to each cell. Then join to the two with ,

identity.ijs

id1: To get an identity matrix of order 4, just take 1 followed by four zeroes and cycle it in a 4 by 4 shape.

id2: i. 4 is 0 1 2 3. Do an =/ on itself. Actually, any list will do, not just i. 4

id3: The key observation here is that i. 4 4 has diagonal elements which are all multiples of 4+1.

josephus.ijs

I'm literally just simulating the problem. Kill the 2nd guy from the left, then rotate the list one to the left. Then repeat. Stop when the length of the list is 1.

remdup.ijs

Grab the last char with {: y

Get the next char by finding the index of the last char (a. i. {: y) and then adding one to that (>: a. i. {: y), then indexing into a. (a. {~ >: a. i. {: y)

Pad y with this, then rotate backward by 1 and check where equality isn't occurring. Drop the last item of this list of 1s and 0s, since we don't care about matching the first char with the padding char.

The resulting list of 1s and 0s is the parts we need to preserve from y. So we index with y #~ (...stuff we did earlier...)

numfact.ijs

q: y finds the prime factorization of y. We assign this into a temp variable b, and find its length. We then generate all binary strings of that length by simply doing i. 2 ^ length, then applying the antibase operation to each element of i. 2 ^ length. This results in the following (for length = 2)

0 0
0 1
1 0
1 1

We use the above matrix to index into our earlier variable b. This gives us a matrix in which each row has some nonzero primes and zero or more 0s. If we replace all the 0s by 1s and multiply the elements of each row together in the above matrix, we get a list of divisors of the number. Use ~. (nub) to filter out any repetitions, and count the number of remaining elements. This gives us the number of divisors of y.

About

Collection of J scripts, along with some explanations. Updated sporadically.

Resources

Stars

Watchers

Forks

Releases

No releases published

Packages

No packages published

Languages