Esoteric Programming Languages

8:16 AM

@Sp3000 do you have evil installed? would q be a valid submission to the self-mutilating program? (i.e. does it terminate without error or output?)

Sp3000

8:27 AM

Will try, gimme a sec

Terminates with no error and no output

(not even a newline)

Martin Büttner

sweet

too bad I'm losing the tie to /// (which I totally thought of, but couldn't be bothered to post from mobile :D)

0

A: Self-Mutilating Program

evil, 1 byte q evil has several memory stores - one is the wheel which is a circular queue which is initialised to a single zero. q swaps the source code and the wheel, so it replaces the source with a null-byte. However, only lower-case letters are actual operators in evil, so that character ...

thanks!

btw, my next tagging project is stack (why do we not have that?)

I already made a list of applicable challenges... what do you think about including "generalised stacks" like the pancake challenges?

i.e. all list-like data structures, where all operations are relative to one end of the list?

(otherwise, even most stack-based languages which include swap or rotate wouldn't count as stacks)

Sp3000

8:52 AM

We have stack-manipulation I guess

Or was that array

I think relative to one end sounds okay

Martin Büttner

yeah, only array-manipulation

counting the pancake ones, I found 26 applicable questions for the new tag

Sp3000

Although tbh I'm not sure how to count, say ><> "stacks" which has {} for shifting

Martin Büttner

oh, hm.

Sp3000

Pop/push is from one end, but shifting gives a method of random access

Martin Büttner

I'd still consider them stacks because all "real" operations take their operators form one end

not sure how to put that in a definition though

although for tags it's not a huge issue if it boils down to "use your best judgement"

Sp3000

9:00 AM

How do the pancake ones work? Pop multiple at once?

Martin Büttner

reverse a prefix (or suffix, depending on which end you're talking about)

Sp3000

I almost want to say: Challenges which involve a data structure whereby elements are removed and inserted from the same end

But that's a bit different for pancake :/

(Trying to exclude deques there)

Hate to say it, but I guess best judgement would probably be the easiest

Do we want stack-manipulation to be consistent, or just [stacks]?

Martin Büttner

I was thinking stack

I don't like array-manipulation

we don't have anything else like that

(and in most cases there will only be a single stack)

Sp3000

9:16 AM

k :) Any thoughts about queues?

Martin Büttner

I'll probably add that as well at some point, but I think there will be by far not as many questions

Sp3000

Makes sense

Martin Büttner

12:50 PM

@Sp3000 What esolangs can you think of that use only letters as commands? evil, Alphuck, AlphaBeta... what else?

Headsecks and Glypho I guess

Sp3000

1:04 PM

ETA

If you're counting Glypho you can count the other related ones, like Udage and Trigger

Er... I'll think of more later, doing something

Martin Büttner

no worries

Sp3000

Oh, ACIDIC

Martin Büttner

hm, all interesting, but I don't think any of them would work (I'm brainstorming languages for RTTTL)

my best bet so far is l33t

Sp3000

1:32 PM

Oh you're doing RTTTL, that changes things

Martin Büttner

well I'm not sure yet, but I thought about it

Sp3000

a-g makes things a bit harder - Headsecks probably gives the most flexibility of the ones listed

Martin Büttner

l33t sounds like a good choice actually (using the digits instead of the letters)

Sp3000

1:50 PM

Good luck then :)

Sp3000

2:17 PM

@MartinBüttner Pretty convinced cuddle is possible as long as log_10(a) is irrational now

For any positive integer n, there exists b such that a^b starts with n

Martin Büttner

2:32 PM

provided log_10(a) is irrational, you mean?

Sp3000

Yeah

Martin Büttner

and that is the case for all numbers which aren't 10^n, because 10 doesn't have any prime factor more than once?

Sp3000

So let's say you have a and you want a^b to start with n

a^b = 10^(log_10(a^b)) = 10^(b log_10(a)) = 10^floor(x) * 10^decimal_part(x) where x = b log_10(a)

is what I had before on the Nineteenth Byte

floor(x) is an int, so 10^floor(x) is irrelevant since it doesn't change the actual digits (i.e. dividing a^b by 10^floor(x) just normalises the power to between 1 and 10)

The important part is 10^decimal_part(x), so say we want a number to start with 1337, then 10^decimal_part(x) needs to be between 1.337 and 1.338

Which means that decimal_part(x) needs to be between log_10(1.337) and log_10(1.338)

And since x = b log_10(a), if log_10(a) is irrational, then there exists a b such that the decimal part being in the range holds

Martin Büttner

neat

I guess that means in other bases there might be more numbers (which aren't just powers of the base)

Sp3000

Hmm can log_x(y) be rational if x,y are integers and y is not a power of x? Not sure about that one

import math

a = 2
n = 123456

L = math.log(a, 10)

lower = n
while lower >= 10: lower /= 10
lower = math.log(lower, 10)

upper = n+1
while upper >= 10: upper /= 10
upper = math.log(upper, 10)

b = 0
x = 0

while not (lower <= x%1 < upper):
    b += 1
    x += L

print(b)

^^ Proof of concept, just for fun :)

(works for n small enough not to run into float issues)

Martin Büttner

2:47 PM

x = 8, y = 2

so yes, if the prime factor multiplicities of the base are all divisible by the same number, there will be rational (but not integer) logarithms for other numbers than b^n

Sp3000

Ah, right. Then i guess that's indeed the case then :)

I can already imagine a CJam answer happening: Ab:+2< for the initial check :P

Sp3000

3:13 PM

@MartinBüttner I think I have 34 for time echo - how did you check whether yours worked?

Martin Büttner

by hand

Sp3000

... hm damn, I don't think this works for the wait at the end

Here it is, dunno if it'll help: {esl_}g](es-:T+2/{~{_es-T>}g;oNo}/

Martin Büttner