the k tree

JohnE

01:05

there's an amazing amount of test "surface-area" to a K interpreter

Bubbler

That statement is probably true for any high-level Turing-complete language

jordancurve

03:09

@ngn I like backwards compatibility to the extent that I appreciate stability in a language. I don't want my knowledge of a language to go out of date too quickly.

Razetime

03:27

@ngn welp this came up during a project euler

Razetime

10:02

x(,/{@[x;y;]'(!10)^x*|/p[;y]=p,:,3/:-3!p:!9 9}')/&~*x apparently this is a sudoku solver

it takes an 81-length vector

and returns a result

not sure what it expects empty cells to be(probably 0)

ngn

11:23

@Razetime projecteuler.net/problem=96 ?

Razetime

@ngn nono, I took this from dfns.sudoku

It's Arthur Whitney's K5 solution

rak1507

huh cool ngn is number 1 in bulgaria

Razetime

indians are crazy

ngn

@Razetime mine is loosely based on arthur's

Razetime

there's a guy with pencil/paper at #147

rak1507

11:28

having a billion people probably helps :P

Razetime

eh yeah i guess

also the fact that everyone's obsessed with IT jobs and math

@ngn wait what

that's all your solutions?

is X some sort of defined adverb which runs each thing?

ngn

@Razetime i don't think what most people mention as their language matches what they use in reality. most of my solutions are in python and c but i'd still like to advertise apl/j/k

@Razetime that's the first 100

Razetime

yeah only 100 can be b=published right

ngn

@Razetime yes, defined near the top

Razetime

and what exactly is it doing

ngn

11:32

@Razetime from PE's "about" section: "Problems 1 to 100 provide a wealth of helpful introductory teaching material and if you are able to respect our requirements, then we give permission for those problems and their solutions to be discussed elsewhere."

rak1507

although people publish higher solutions all the time unfortunately

I looked one up once because I was certain I had the right answer, it was because I was pasting ¯1234.... instead of -1234...

wait, you can't even get negative answers right? must be misremembering

ngn

@rak1507 there are no restrictions. there's even at least one alphabetic answer.

@Razetime india is traditionally great at maths

rak1507

ah yeah it was 27, must have looked up a higher one for a different reason

Razetime

if i solve 26 problems I will be on the 35th page of the indian leadeboards

ngn

if it matters, you can always move to a small island nation and become #1 :)

Razetime

11:41

nah I want 69th place in india

and i want to maintain it

this requires > 200 solves

rak1507

or make your own micronation

Razetime

nah i have a decent goal

and it may be possible

Razetime

12:12

how do i assign at index?

ngn

@Razetime a[i]:b ?

Razetime

multiple idexes also work?

ngn

yes, they should

example

Razetime

12:33

what does mmap lmt mean?

ngn

you're trying to allocate too much memory

or maybe open too many files

Razetime

haha, open too many files

it was definitely allocating too much memory

Razetime

13:22

do defined verbs not work like infix operators?

ngn

@Razetime there are no user-defined verbs (except something+adverb). lambdas { } act as nouns syntactically.

Razetime

like nouns?

so the only correct calling syntax is []

ngn

@Razetime yes, a lambda can be an argument to another function

Razetime

oh ok functions are data kinda thing

ngn

@Razetime a[b] or a@b or a b - they are the same thing, application

for more than one arg you can use a[b;c] or a . (b;c) but not b a c

@Razetime note how "while" works, for instance: {a}{b}/c

{a} and c are the arguments of {b}/

{b}/ is a derived verb

so, thanks to {a} being a noun, it can become the left arg of {b}/

Razetime

13:29

interesting

ngn

this solves the old apl problem of static parsability, and i think it's one of the best design decisions in k

Razetime

what exactly is static parsability

ngn

being able to tell what the structure of the code is just by looking at the source locally, without any knowledge about runtime values

e.g. is a b c the same as (a b)c or a(b c)? in apl you can't tell. in k you can.

Razetime

oh! that is very nice

ngn

yes, it offloads some of the work from runtime to compile time

Razetime

13:33

I am thinking of making a k tutorial series

sort of like a video introduction

ngn

cool! if there's anything we in this chat can help with, do tell us

Razetime

sure, will ask

Razetime

13:50

why isn't global assignment working here? program

b is supposed to be 2

ngn

@Razetime i+::1 - why not just i+:1?

Razetime

i need it to globally update?

is modified assignment global

ngn

@Razetime yes (as long as there is no local with that name)

Razetime

i can't believe that was the problem

so modified assignment with :: returns a value?

that is, if there is no other local with the same name

ok one more thing

{~2=#b}{i+::1;t:b;b::.',/$|dec/|b+(((#b)-#a)#0),a;a::t}/1

I'm trying to run this

but it goes on forever

nvm it's the same problem

Razetime

14:19

this program stops on the first iteration in my pc(repl): program

is there a way to load k program outside the repl?

also what is the maximum number limit in K?

ngn

@Razetime i think it parses as i+: :1 which makes the second : act like "return"

Razetime

i guess so

ngn

@Razetime \l file.k should work when part of a .k file too

Razetime

@ngn like, can i do ./k file.k?

ngn

@Razetime signed 64bit, i.e. 2^63 - 1 = 9223372036854775807

@Razetime yes

Razetime

14:27

will it show return values

or does it need to explicitly output

ngn

@Razetime it works the same way as if you've typed it in the repl

Razetime

also is there a way to get an executable?

ngn

@Razetime no.. but that could be a good idea

Razetime

alright

ngn

i was thinking, what if i make the k executable self-aware of its own size, and if it detects that it has a .k program concatented to its end, it executes it

this way turning a .k script into an executable program would be just: cat k file.k > myprogram

not sure if the elf format would allow that

Razetime

14:31

that is a very interesting idea

is there an equivalent to ⍷ in K?

ngn

@Razetime maybe something like *(3'"abcdefgh")?,"cde"

(or (3'"abcdefgh")?"cde" in k dialects with "rank-sensitivity")

Razetime

oh, there are k dialects with rank

ngn

rank-sensitivity

Razetime

ok

so this is basically taking windows

and then finding first index

ngn

yeah

in 2+ dimensions it gets more complicated..

Razetime

14:37

i can see why

i just need to find a recurring decimal so this 1D version works well

JohnE

@ngn might also be able to do something with permitting a shebang line and then having the interpreter executable follow a \\

ngn

@JohnE #! is allowed. i'm not sure what @Razetime needs this for but yeah, if the interpreter can be in a separate file, #! is the better solution

@JohnE i don't think it would be a valid elf file if the elf header is not at the beginning of the file

JohnE

it'd probably be deeply nasty to try doing something like lopping off the end of a shell script, storing it as a temp file, and then executing the temp file

ngn

@JohnE ah.. so your idea is that it should start like a shell script, not like a k file

JohnE

14:53

right

Razetime

i'm trying to find the first 1 in a boolean array

is *& better or ?,1 better?

ngn

i think x?1 because it stops when it finds the 1

JohnE

yes, I'd guess x?,1

Razetime

15:19

why am i getting 'val here?

dzaima

@Razetime no lexical scope

Razetime

oh hm

ngn

yep. shame on k.

ktye

@ngn you could append zip files (even recursively) to the end. they have their index at the end, so they know how much to read. but then you need unzip available or within k.

coltim

@coltim one example of how nuanced testing overloads/all different types of args can get is this - n?dict (deal) only works intuitively if the keys are 0..#dict. am I clear on what the desired behavior even is here? not really, ha

ktye

15:27

@ngn *& could be implemented as an optimized idiom.

Razetime

{a::x;p::0;fr::x#0;{(~x=0) &(0=fr x)}{fr[x]::p;p+::1;_(x*10)%a}/1}13 I'm not sure why it's running forever

but without the :: next to fr[x] and p+ it just halts immediately

ngn

@ktye yes, of course

@Razetime +::1 intentional?

Razetime

let me try again

Razetime

15:42

ngn/k, 74 bytes: {a::x;p::0;fr::x#0;v:{(~x=0) &(0=fr x)}{fr[x]::p;p+:1;a!x*10}/1;p-fr v}983 this snippet is giving me 4 in the repl

i have no clue why

i'm not sure why i'm having so many problems with the repl

how close is K3 to K6?

also, which dialect does kdb+ use?

ngn

just a few minutes to make a cup of coffee and i'll be back..

Razetime

i think I should clone the latest version

ngn

15:59

@Razetime i'm not sure but probably closer than k6 vs k9

@Razetime k4

@Razetime what does it do?

Razetime

it finds the length of recurring decimal

ngn

i get 982 everywhere i try

Razetime

I'm recloning

will check

ngn

shorter: fr:&a:983;p:0;p-fr{~(~x)|fr x}{fr[x]::p;p+:1;a!x*10}/1

what we usually do in such cases is drag all of the state in the loop as argument(s) and result

so here the "state" could be (x;p;fr)

i'll try to rewrite it like that..

Razetime

@ngn can we omit brackets when assigning a function to a variable?

ngn

16:12

@Razetime usually yes

Razetime

oh wow

ngn

f:+ /this works

Razetime

f:x+y ?

ngn

@Razetime no, not like that

Razetime

ok

ngn

16:16

a:983
(x;p;fr):{(x;p;fr):x;(~~x)&~fr x}{(x;p;fr):x;(a!x*10;p+1;@[fr;x;:;p])}/(1;0;&a)
p-fr x

not too pretty but avoids closures

Razetime

instead of ~~x we can use just x right

ngn

@Razetime only if we are sure that x is a boolean

Razetime

huh. what if x is an int?

and positive only

ngn

then x could be different from ~~x. but since this is the stop condition of a while loop, it probably doesn't matter, as the truthiness of x is preserved

ok, here it is:

a:983
(x;p;fr):{(x;p;fr):x;x&~fr x}{(x;p;fr):x;(a!x*10;p+1;@[fr;x;:;p])}/(1;0;&a)
p-fr x

tbh, i think this looked better:

fr:&a:983;p:0;p-fr{x&~fr x}{fr[x]:p;p+:1;a!x*10}/1

Razetime

yep it does

I'll add it in commented

ngn

16:26

@Razetime how big can be a?

Razetime

a can go upto 1000

it's euler 26 i think

ngn

ah, so not a lot. maybe we can generate all powers of 10 mod a and use a single amend for fr

Razetime

16:38

maybe, I didn't want to risk it

thanks a bunch for the help

good night

ngn

17:15

@Razetime here's another idea (tortoise&hare cycle detection): we start with 1 1 and keep multiplying by 10 100 (mod a) until they become equal again. if the last pair is 0 0, there is no recurring part in the decimal fraction. otherwise, the cycle length is the length of our sequence.

|/{r:(-/)(x!t*)\x!t:10 100;$[**|r;#r;0]}'1+!1000

the above only finds the max cycle length. to actually solve the problem:

*>{r:(-/)(x!t*)\x!t:10 100;$[**|r;#r;0]}'!1000

coltim

18:19

@ngn hmm this reminded me of a golf but the results are different? not sure I have the brainpower at the moment to figure out why

ngn

@coltim f doesn't look correct to me. it tries to find the first 1 in the sequence instead of detecting a cycle (which could consist of anything)

coltim

@ngn hmm the question's test examples could be wrong? there's this: "Given a positive integer n, (n>1 and n<10000), find the length of pattern in 1/n, if it's repeating. Otherwise, return any non-positive integer (e.g., cases: 1/5,1/94,1/22)."

ngn

@coltim 65 -> -1 seems wrong. 1%65. does look like it has a pattern

even if that pattern doesn't start immediately after the dot

coltim

@ngn ah, that would be the difference. 1/65 has that leading 0 but repeats after that

ngn

"Note that, the pattern should start just after the decimal point" (from the challenge)

coltim

18:32

ah it could be this

coltim

19:01

I wonder if it's possible to prove that they'd return the same value

rak1507

22:20

how come x\\:x doesn't work for split each left?

dzaima

because \ for splitting is still an adverb

rak1507

oh right

ngn

@rak1507 it's equivalent to (x\)\:x which is missing a left arg

rak1507

is there anything shorter than x{x\y}\:x

ngn

22:36

probably not

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