User Defined Functions Can Be Used Now! Also PRINT2 Added!

Conditionals and (finite) loops

In drocta ~ATH, the control flow (what part of the program is run when) is almost entirely determined by ~ATH loops.

The only case where it isn't is when the object initially pointed to by THIS dies. That ends the program immediately. 

This post will show in more detail the basics of its usage.(also introduces the variable NULL)

Is there anything in this post that needs to be clarified?

In a program, you might want something to only happen if something else is true, or you might want it to happen provided that something had not happened.

In many programming languages you would use a command called if.

drocta ~ATH does not have an if statement.

Instead, you create a loop on an object, and inside the loop kill the object.

A little morbid, but isn't ~ATH always?

for example:

SOMECODEHERE ~ATH(SOMEVAR){ SOMEVAR.DIE(); SOMEOTHERSTUFF } SOMEOTHERSTUFF 

However, this isn't the only way. There is another way that is likely preferable in most situations. I just thought this way was the most obvious, and would be an easy way to explain the other method, which is quite similar.

The loop doesn't exactly wait for the object to die, but rather repeats so long as the variable points to an object that is alive. The variable can be made to point to an object that is not alive, either by make the object it points to DIE(), or by making it so it points to a different object that is ALREADY DEAD.

(How can you expect to kill it, when it is ALREADY DEAD?!? haha)

So if the object you wanted the part of the script to run if it was alive was important, and you didn't want to kill it, you could just do this:

BIFURCATE [IMPORTANTOBJECTVAR,IMPORTANTOBJECTVAR]BLAH; BIFURCATE BLAH[V,V]; ~ATH(V){ BIFURCATE [V,NULL]V; BIFURCATE V[JUNK,V]; OTHER CODE TO ONLY EXECUTE IF IMPORTHATOBJECT IS ALIVE } 

Thats nice to be able to do, isn't it.

In fact, it leads nicely into how to make loops that go around a fixed number of times.

suppose if you had

BIFURCATE [BLAH,NULL]V; ~ATH(V){ BIFURCATE V[BLAH,V]; DO OTHER STUFF, POSSIBLY WITH BLAH  }  MORE STUFF 

V would initially be alive, so it would go into the loop, but then V would be made to point to the NULL object, which is dead, so it wouldn't loop the second time.(it would skip to after the loop

that would do the stuff in the loop once.

now what if you wanted to do it twice?

Just make it so it has to bifurcate the thing twice before moving on! Like so:

BIFURCATE [BLAH,NULL]V; BIFURCATE [BLAH,V]V; ~ATH(V){ BIFURCATE V[BLAH,V]; DO OTHER STUFF, POSSIBLY WITH BLAH  }  MORE STUFF 

 this way it will go through the loop, pop off the one side of V, doing the other stuff in the loop. V will still point to something alive after this, so it will do it again, but this time, when it pops off one side of V, the result will be the NULL object, so it will stop.

Thats a loop that goes around twice!

This can be extended to as many number of repetitions as follows:

BIFURCATE [BLAH,NULL]V; BIFURCATE [BLAH,V]V; BIFURCATE [BLAH,V]V; BIFURCATE [BLAH,V]V; ETCETERA BIFURCATE [BLAH,V]V; ~ATH(V){ BIFURCATE V[BLAH,V]; DO OTHER STUFF, POSSIBLY WITH BLAH  }  MORE STUFF 

where the loop will go around however many blahs there are attached to the NULL, because it will keep popping the BLAHs of until it reaches NULL.

The variable V is being used to store how many more loops need to be executed, Which is a number.

V is being used to store a number. 

You might be thinking something along the lines of

"Well this is all very nice, but that seems to only allow for loops that loop a predetermined amount of times."

NOT SO!

You can have another loop that makes a variable point to a number, which would then be in another loop! For example, the following will add two numbers in A and B, store the result in C, copy that result to CTEMP, and then print "some text" the ammount of times in CTEMP!

SOME CODE TO GET A AND B HERE import bluh BLAH; BIFURCATE [BLAH,A]ATEMP; BIFURCATE [BLAH,B]BTEMP; BIFURCATE ATEMP[JUNK,ATEMP]; BIFURCATE BTEMP[JUNK,BTEMP]; BIFURCATE [BLAH,NULL]C; BIFURCATE C[JUNK,C]; ~ATH(ATEMP){ BIFURCATE ATEMP[JUNK,ATEMP]; BIFURCATE [BLAH,C]C; } ~ATH(BTEMP){ BIFURCATE BTEMP[JUNK,BTEMP]; BIFURCATE [BLAH,C]C; } BIFURCATE [BLAH,C]CTEMP; BIFURCATE CTEMP[JUNK,CTEMP]; ~ATH(CTEMP){ BIFURCATE CTEMP[JUNK,CTEMP]; print some text; } print DONE!; 

YAY! We just added two numbers together! In similar ways, we can also subtract, multiply, divide, etcetera! 

if some of this was unclear I would appreciate requests for what needs to be clarified.

Is there anything in this post that needs clarification?

unary to binary (back again, an objects journey) (part 1)

ok, so last time I didn't actually get to converting from unary to binary again.

So I guess I will do that here. (HAHA I JUST GOT IT TO DIVIDE WITH REMAINDER. APPLYING THAT TO THE CONVERSION IN PART 2)

so to do this we have to find the largest power of two not greater than the number,  then the largest power of two not greater then what is remaining of the number, and repeat until we reach zero.

OR we can

find the largest power of two not greater than the number, then for each power of two less than the number, if it is not greater than it, use a 1 and subtract it from the remaining number.

and then for both of these, of course, reverse the result at the end so it is in the right order. 

I think so far the second one of these sounds better.

but come to think of it, repeated integer division by two might work well.

that is, instead of repeated doubling, and then checking which one is larger, it might be faster to do repeated halving. (with a remained of either zero or one)

this could be then used to make a integer base 2 logarithm, with a remainder thing.

and instead of checking each power of two less than it, we could just find the integer base 2 log, and the base 2 log of the "remainder" and the base 2 log of THAT remainder and so on.

that seems like its the best way to me so far, but how do we take the integer logarithm AND get the log remainder thing?

getting the logarithm is fairly simple, just keep dividing by the number until you reach 1.

but how do you get the remainder part? and also how do you do the integer division in the first place with these number like things that we have created?

well lets answer the first of those questions first.

It seems like the remainder of the logarithm is probably related to the remainders of the divisions that make it up.

so lets look at taking the integer base 2 logarithm of 9:

9 4 (rem 1) 2 1

that took 3 halvings, and there was one remainder of one.

what if we try on 10 or something though?

10 5 2 (rem 1) 1

that also took 3 steps and one remainder, but the remainder was on a different step.

now lets try 11

11

5 (rem 1) 2 (rem 1) 1 (rem 0)

now wait, theres a pattern here, which might allow for converting it even faster.

the remainders of the divisions are the remainders of the logs expressed in binary!

we COULD turn them back into unary, so we could take the log of it again,

but what we are doing it converting it into binary anyway! (so that would be a silly step)

so if we look at what we were doing with the repeated division again, but looking at the numbers in binary, it will be pretty evident how to convert it in a better way.:

lets try 9 again (which is 1001)

1001

100 (rem 1) 10 (rem 0) 1 (rem 0) 0 (rem 1)

whats that? the remainders are the number we want in binary?

and now that I think of that it was kind of obvious?

yes. I for some reason did not think of that earlier.

so yeah, thats what we will do, we will repeatedly integer divide the number by 2 until we reach zero, and the remainders will be the number in binary.

but wait, we haven't even written out how to divide the number by two in the first place!

well turns out its not actually that hard to do.

in a loop (that keeps looping provided the number is not zero) subtract one, and if its still not zero, subtract one again. if it is zero after the first subtracting, the remainder of that division is 1. (only the last loop will cause the remainder to be 1) count the number of times where 2 was subtracted. 

so like for 5 that would be

5 3 1 1 2 0 2 (remainder 1)

so that gives the correct answer (2 with remainder zero)

ok, now to implement this integer division by 2 in drocta ~ATH (wow! already? that was fast! /sarcasm)

//given that NUM is the variable that has the name we are halving BIFURCATE [NUM,NUM]2NUM; BIFURCATE 2NUM[NUMCPY,JUNK]; BIFURCATE [BLAH,BLAH]2BLAH; BIFURCATE 2NULL[SUBCOUNT,REMAINDER]; ~ATH(NUMCPY){ BIFURCATE 2BLAH[UNEVEN,JUNK]; BIFURCATE NUMCPY[JUNK,NUMCPY]; BIFURCATE [NUMCPY,NUMCPY]2NUMCPY; BIFURCATE 2NUMCPY[NUMCPYCPY,JUNK]; ~ATH(NUMCPYCPY){ BIFURCATE [BLAH,SUBCOUNT]SUBCOUNT; BIFURCATE NUMCPY[JUNK,NUMCPY]; BIFURCATE 2NULL[UNEVEN,NUMCPYCPY]; } ~ATH(UNEVEN){ BIFURCATE [BLAH,REMAINDER]REMAINDER; BIFURCATE 2NULL[UNEVEN,JUNK]; } } BIFURCATE [NUMCPY,NUMCPY]2NUMCPY; BIFURCATE 2NUMCPY[NUMDIV2,JUNK]; 

ok, that should divide the number by 2, put the result in NUMDIV2, and put the remainder in REMAINDER.

this takes linear time based on the size of the unary number. (provided I didn't make a mistake)

so now it is time to make sure it works, hold on a second while I check it. (I mean, im not going to post this until after I check it, so it doesn't really make sense to tell you to wait, because you just keep reading, but as I am writing this, I am about to test it.)

ok, I tested it. and it didn't seem to work, but then I realized that I made a mistake in making the test. (I checked the wrong variable)

but yeah, turns out that works...

ok, so now we need to repeatedly divide the number by two to get the remainders.

and we need to store these remainders in a list or something.

reading binary numbers from the user, continued.

when we left off, we had code to interpret one binary number from user input, and output it as a unary number. In this post we will add the second user input, and maybe make a thing that converts from unary to binary, and then outputs it.

The code for that was the following:

import blah BLAH; print please enter whatever character you will be using for binary zero.; INPUT ZEROCHAR; BIFURCATE ZEROCHAR[ZEROCHAR,JUNK]; import chrcmp CMP0; BIFURCATE [CMP0,ZEROCHAR]D; D.DIE(); print please enter whatever character you will be using for binary one.; INPUT ONECHAR; BIFURCATE ONECHAR[ONECHAR,JUNK]; import chrcmp CMP1; BIFURCATE [CMP1,ONECHAR]D; D.DIE(); BIFURCATE [NULL,NULL]2NULL; BIFURCATE 2NULL[OUTNUM,JUNK]; print please input the binary number you want.(it will be converted to unary); INPUT BINNUM; ~ATH(BINNUM){ BIFURCATE [OUTNUM,OUTNUM]G; BIFURCATE G[NCOPY,OUTNUM]; ~ATH(NCOPY){ BIFURCATE NCOPY[JUNK,NCOPY]; BIFURCATE [BLAH,OUTNUM]OUTNUM; }  BIFURCATE BINNUM[CHAR,BINNUM]; BIFURCATE [CMP0,CHAR]NEQ0; ~ATH(NEQ0){ BIFURCATE [BLAH,OUTNUM]OUTNUM; BIFURCATE 2NULL[NEQ0,JUNK]; } } print ok, going to print it out in unary, with each digit on one line. If the number you entered was large you might want to close the program instead of hitting enter.; INPUT JUNK; BIFURCATE [OUTNUM,OUTNUM]GOUTNUM; BIFURCATE GOUTNUM[OUTNUMCOPY,JUNK]; ~ATH(OUTNUMCOPY){ BIFURCATE OUTNUMCOPY[JUNK,OUTNUMCOPY]; print 1; }

ok, so what do we need to do after that?

Well we need to make it get two numbers, right?

Yes. Yes we do.

But getting one number was kind of long, wasn't it.

Luckily we don't have to have the entire thing there twice, much of it we can jusst have it in there once.

we COULD define a function, but I haven't added that to the interpreter yet, so yeah...

so a significant portion of that was initialization stuff that doesn't need to be duplicated. the first 14 lines in fact!

so pretty much we just duplicate everything but the first 14 lines.

gee, now I sound lazy for not writing this part earlier.

anyway, here goes:

import blah BLAH;

print please enter whatever character you will be using for binary zero.;

INPUT ZEROCHAR;

BIFURCATE ZEROCHAR[ZEROCHAR,JUNK];

import chrcmp CMP0;

BIFURCATE [CMP0,ZEROCHAR]D;

D.DIE();

print please enter whatever character you will be using for binary one.;

INPUT ONECHAR;

BIFURCATE ONECHAR[ONECHAR,JUNK];

import chrcmp CMP1;

BIFURCATE [CMP1,ONECHAR]D;

D.DIE();

BIFURCATE [NULL,NULL]2NULL;

BIFURCATE 2NULL[OUTNUM,JUNK];

print please input the first binary number you want.(it will be converted to unary);

INPUT BINNUM;

~ATH(BINNUM){

BIFURCATE [OUTNUM,OUTNUM]G;

BIFURCATE G[NCOPY,OUTNUM];

~ATH(NCOPY){

BIFURCATE NCOPY[JUNK,NCOPY];

BIFURCATE [BLAH,OUTNUM]OUTNUM;

} 

BIFURCATE BINNUM[CHAR,BINNUM];

BIFURCATE [CMP0,CHAR]NEQ0;

~ATH(NEQ0){

BIFURCATE [BLAH,OUTNUM]OUTNUM;

BIFURCATE 2NULL[NEQ0,JUNK];

}

}

BIFURCATE [BLAH,OUTNUM]UINNUM1;

BIFURCATE UINNUM1[JUNK,UINNUM1];

BIFURCATE 2NULL[OUTNUM,JUNK];

print input the second binary number:;

INPUT BINNUM;

~ATH(BINNUM){

BIFURCATE [OUTNUM,OUTNUM]G;

BIFURCATE G[NCOPY,OUTNUM];

~ATH(NCOPY){

BIFURCATE NCOPY[JUNK,NCOPY];

BIFURCATE [BLAH,OUTNUM]OUTNUM;

} 

BIFURCATE BINNUM[CHAR,BINNUM];

BIFURCATE [CMP0,CHAR]NEQ0;

~ATH(NEQ0){

BIFURCATE [BLAH,OUTNUM]OUTNUM;

BIFURCATE 2NULL[NEQ0,JUNK];

}

}

BIFURCATE [BLAH,OUTNUM]UINNUM2;

BIFURCATE UINNUM2[JUNK,UINNUM2];

BIFURCATE [UINNUM1,UINNUM1]CUINNUM1;

BIFURCATE CUINNUM1[SUM,JUNK];//haha, some junk

BIFURCATE [UINNUM2,UINNUM2]CUINNUM2;

BIFURCATE CUINNUM2[UINNUM2CPY,JUNK];

~ATH(UINNUM2CPY){

BIFURCATE UINNUM2CPY[JUNK,UINNUM2CPY];

BIFURCATE [BLAH,SUM]SUM;

}

print ok, going to print the sum out in unary, with each digit on one line. If the numbers you entered were large you might want to close the program instead of hitting enter.;

INPUT JUNK;

BIFURCATE [SUM,SUM]GSUM;

BIFURCATE GSUM[SUMCOPY,JUNK];

~ATH(SUMCOPY){

BIFURCATE SUMCOPY[JUNK,SUMCOPY];

print 1;

}

ok. That worked. I haven't made it convert the output to binary yet. I started this the day of the other post, but there was a bug I didnt get around to fixing in it that I didn't find the fix for until today.

that bug was that in the part where it changes the binary number to a unary number, the first time around, the unary number starts at zero and increases to what it should be...

...but I forgot to include the line to reset it back to zero for the second number.

but in works now.

I might make the part where it converts from unary to binary now instead of later, but I have to do stuff.

Sorry for having such a slow post rate, it is partly because I am busy and partly because I am lazy. mostly the [FORMER,LATTER].

hey, it adds the numbers that were entered in binary...

So a future improvement is to make it so it outputs it in binary as well.

Another possible future post is something that shouldn't work but the interpreter has a certain bug that might be amusing, and will not negatively affect many things.

specifically, if you have a "}" inside of a print statement, and you are jumping to the end of a loop, it will jump to the inside of the print statement, and execute the text as code. This is a bug, not a feature, so you should not rely on it when writing stuff. It might be fun to mess with though.

I might add user defined functions to the interpreter soon...

I don't feel that this post really explained much of anything, but I have to do stuff now, so I am going to post this, and possibly modify it later with clarifications.

Feel free to send me messages asking about how some part works.

reading the contents of user input, and calculator improvement

This post will cover how to actually determine WHAT the user has typed, instead of just how long it is. It will also include how to interpret what the user enters as a binary number, so that its easier to type.

An Essential part of making it interpret binary numbers is making it double numbers repeatedly.

This actually has a few ways that can be done, so this is one of the first situations where coding style for this problem might differ from person to person. Because of this, I will say more than one way to do it.

The first way to do this it to copy the number twice, and then start from zero and add both of the copies. This is relatively inefficient, and would take

a copy thing, consisting of two bifurcates (which would take a little time)

where the size of the initial number is N, 2N normal bifurcates, 2N reverse bifurcates, and 4N lines relating to the actual loop

assuming each command takes the same amount of time (which is an oversimplification) this would take 9N+C line times. (C is a constant) This might be acceptable, but there is a more efficient and nicer looking way.

The second way is nicer looking, but still not the most effecient. However, when multiplying by a larger number(such as 3, or 4, or even large numbers), this method is part of what would be used.

The second method is essentially copying the number (using a reverse bifurcate and a normal bifurcate), and then adding the number to zero, except instead of each loop increasing the new number by one, it increases it by two. This is shorter, and it looks nicer. It also only takes half as many normal bifurcates. As a result, the number of steps it would take (again assuming each step is the same length) is 8N+C, instead of 9N+C 

this one I will write out, but it is still not the best way:

//N is the number initially BIFURCATE [NULL,NULL]2NULL; BIFURCATE [N,N]G; BIFURCATE G[NCOPY,JUNK]; BIFURCATE 2NULL[RESULT,JUNK]; ~ATH(NCOPY){ BIFURCATE NCOPY[JUNK,NCOPY]; BIFURCATE [BLAH,RESULT]RESULT; BIFURCATE [BLAH,RESULT]RESULT; } 

ok, so yeah. that takes N, and puts twice N into RESULT, but it is still inefficient.

A more efficient version is to copy the initial number, and add the number to itself. This way you only have to do half the number of reverse BIFURCATE statements. This is much more efficient, taking instead the steps:

a copy thing, consisting of two bifurcates (which would take a little time)

where the size of the initial number is N, N normal bifurcates, N reverse bifurcates, and 2N lines relating to the actual loop

This has 7N+C steps, which is a significant improvement. I think it is the fastest way to double a number in drocta ~ATH.

It is as follows (N is the number)

BIFURCATE [N,N]G; BIFURCATE G[NCOPY,RESULT]; ~ATH(NCOPY){ BIFURCATE NCOPY[JUNK,NCOPY]; BIFURCATE [BLAH,RESULT]RESULT; } 

This is shortest and fastest solution I have found. If you find a shorter or faster method, please tell me.

Ok. Now we can double numbers. That is good. That is an important step. But we still haven't gotten user input to be read in any reasonable way.

Hang on, I'm GETTING TO THAT. GEEZ. (I'm kidding, no one has been complaining about my taking so long, other than myself)

Ok, so here goes:

To interpret the binary number input and convert it to a "number", we can follow the following algorithm:

start with zero.(this is before the loop)

If there are any characters left, double the number that is being created.

remove the first character from the remaining characters. If it is "1" or whatever symbol (or alternatively if it is not "0"), add one to the number that is being created. Otherwise, continue onto step 4 without doing anything first.

go back to the start of the loop (step 2)

Ok. thats the algorithm we are going to use. But I STILL haven't explained how to recognize what the next character is. Seriously what is up with that?

What you do is you bifurcate the rest of the input into [the next character,the rest of the input].

Now you have the next character. Then what you do is you reverse bifurcate it with some other object, and then you check whether that object is already dead or not.

But how do you make it so the combination is already dead? How do you get the object for the character before the user has even inputed it?

Answer: You don't. Not in the current version of drocta ~ATH anyway. You will have to tell the user to enter all the characters they will be using ahead of time. Yes this is horrible and stupid. No its not exactly like that in the comic. Its ~ATH what do you expect? :P

that might change in future versions, but I will try to stay backwards compatible with that.

but anyway, back to comparing it:

so you say something along the lines of:

import comparingobject CMP1; othercodehere makeNEQ1besomethingalive BIFURCATE [CMP1,CHAR]EQ1; BIFURCATE [NULL,NULL]2NULL; ~ATH(EQ1){ print yep, they are equal; BIFURCATE 2NULL[EQ1,NEQ1]; } ~ATH(NEQ1){ print nope, they are not equal; BIFURCATE 2NULL[NEQ1,JUNK]; }  

in the othercodehere you get the character a head of time, and say BIFURCATE[CMP1,THECHARTHATMATCHESWITHCMP1]D; D.DIE();

That makes it so that it will go through the one section of code if the character is the right one, but something else if it is something else.

Which is what we want.

So to put it all together, and make the thing that interprets the input as a binary number(hold on tight(ok, what, why did I say that), this will be a long one(why am I talking like this?)):

import blah BLAH; print please enter whatever character you will be using for binary zero.; INPUT ZEROCHAR; BIFURCATE ZEROCHAR[ZEROCHAR,JUNK]; import chrcmp CMP0; BIFURCATE [CMP0,ZEROCHAR]D; D.DIE(); print please enter whatever character you will be using for binary one.; INPUT ONECHAR; BIFURCATE ONECHAR[ONECHAR,JUNK]; import chrcmp CMP1; BIFURCATE [CMP1,ONECHAR]D; D.DIE(); BIFURCATE [NULL,NULL]2NULL; BIFURCATE 2NULL[OUTNUM,JUNK]; print please input the binary number you want.(it will be converted to unary); INPUT BINNUM; ~ATH(BINNUM){ BIFURCATE [OUTNUM,OUTNUM]G; BIFURCATE G[NCOPY,OUTNUM]; ~ATH(NCOPY){ BIFURCATE NCOPY[JUNK,NCOPY]; BIFURCATE [BLAH,OUTNUM]OUTNUM; }  BIFURCATE BINNUM[CHAR,BINNUM]; BIFURCATE [CMP0,CHAR]NEQ0; ~ATH(NEQ0){ BIFURCATE [BLAH,OUTNUM]OUTNUM; BIFURCATE 2NULL[NEQ0,JUNK]; } } print ok, going to print it out in unary, with each digit on one line. If the number you entered was large you might want to close the program instead of hitting enter.; INPUT JUNK; BIFURCATE [OUTNUM,OUTNUM]GOUTNUM; BIFURCATE GOUTNUM[OUTNUMCOPY,JUNK]; ~ATH(OUTNUMCOPY){ BIFURCATE OUTNUMCOPY[JUNK,OUTNUMCOPY]; print 1; } print Am I a terrible person for writing this?; 

Oh gosh. I wish I could indent in tumblr. that is terrible to read. tumblr is a terrible source code editor.

One time someone called me a masochaist for writing this type of stuff.

And then we just have to put that together with the adding thing, and then maybe add a better way of outputting the number. maybe in binary.

HAHAHAHAH

ok, yeah, I'm going to put it together in the next post, not this one, because I have to homework now.(using the noun homework as a verb was intentional)

yeah. putting it together in the next post.

As always, if something was confusing, please ask for clarification.

okay, this might be a liiittle specific, and more HS oriented, but still. can you power a First Guardian using ~ATH? and if so, what are some ways?

This is a slightly off of the usual topic of this blog, which is more about a language which I made up and is based on and named after the language in the comic, and attempts to be as close to the comic while still being possible to implement and use, (the language in the comic is not particularly well documented), but I’ll answer it to the best of my abilities.

I’m not sure what you mean by “power”.

“Can I write a first guardian character that has some of their behavior be based on ~ATH in some ways?” : I don’t see any reason why not. If you can do ectobiology with a puppet and a cueball, why not with a computer running a program?

“Can I make a first guardian character who is a first guardian because of the continual execution of a ~ATH script?” : While you can of course write whatever you want, I’m not sure that this is consistent with what is said in acts 1-7 (I have chosen to disregard everything after act 7) about first guardians. It seems like an entity either is or is not a first guardian, and that this fact is established at the time of that entity’s creation. It seems to me like what happens with first guardians is: using ectobiology, the genetic code from some players’ subconscious-or-whatever is combined with some other source(s), and the result is a first guardian.

Oh, I suppose if you just mean “an entity has the powers of a first guardian as a result of a ~ATH script”, then, uh, I guess that doesn’t contradict anything. How would I depict such a script? Well, I don’t know that anything that we’ve seen of the language gives any particularly clear direction for what such a script would look like. If you want it to just have the keywords we’ve seen, possibly re-interpreted to mean something different, I guess one idea could be to like, say bifurcate an existing first guardian into parts with names that seem relevant, and also whatever entity is to have the abilities of a first guardian, and then like, un-bifurcate (unbifurcation is not something shown in the comic, but is something I’ve added to my language, in order to make it usable) the other entity, but with some component of the first guardian mixed in somehow? Though if you did that, you would probably want to justify why, if ~ATH can do that that easily, why it doesn’t happen all the time.

There is very little in terms of rules that the story gives for how ~ATH behaves. If you want to have a ~ATH script “power” a first guardian, just make up some stuff. If you want to depict the actual script, then look at all the pages of the comic that depict code from it, see what it says they do, and just make something up that seems to aesthetically fit.

The language that I have defined (”drocta ~ath”) does not provide any means of doing this, because the language I designed was designed to technically be usable, while also resembling the language in the comic.

I suppose that, as all first guardians get their power from the green sun, and the green sun was created by the circumstantially simultaneous destruction of, uh, either universes A and B, or specifically of A2 and B1, not sure which, and this was in some ways tied to the Möbius Double Reacharoud virus, you could argue that all first guardians get their powers from, something somewhat connected to a ~ATH script.

Why can’t I add tags to an answer to an ask?

Happy 10/25

Today is the 10 year anniversary of the destruction of the universe frog. Such an occasion is traditionally celebrated by destroying the universe.

I couldn't help but notice that out of some of the best things you could do, no link to the download in the sidebar? You may want to do that...

uh, yes. that would be good to do.

I put a link in the about section, but I don't actually know how to put links in the sidebar outside of that.

do you think you could explain that for me?(I am not particularly experienced with tumblr)

EDIT:Nevermind, my friend from school told me where the feature is.

Learn ~ATH turned 1 today!

yaaaay!

So I think that my explanations of things hasn't been very clear so far, and while I can't promise that they will improve, I do intend to teach programming in person at my school, which might help me to know how to make this tutorial better. Maybe, maybe not. 

May things go well!

Hello again all!

So recently I started kind of working on this again for a bit. I have fixed some bugs with the parser that I haven't pushed yet. I am also writing an improved interpreter that will use the parser instead of the hacky thing that just goes through strings.

However, for the time being, even after I release this version, I would recommend maybe using the older version for a while if anyone is using it, because this version is probably even more buggy.

However, you know how a few posts ago (but more than a year ago (wow) ) I posted that post where I said that I didn't think bifurcate can be used to split values into more than 2 values?

Well I still kind of think that, but on the map page for homestuck on act 6, it says split Act_6[Act_1,Act_2,Act_3,<etc>];

So this is something I intend to implement, and something I am implementing.

And like I said before I would like it to be done with repeated bifurcation, as a sort of syntactic sugar.

And I am thinking I want it to be like

[a,b,c] means the same thing as [a,[b,c]]

so split Z[A,B,C];

would be the same as

BIFURCATE Z[A,BCTEMP]; BIFURCATE BCTEMP[B,C];

and that split [A,B,C]Z;

would be the same as

BIFURCATE [B,C]BCTEMP; BIFURCATE [A,BCTEMP]Z;

But the way the splits would be done could also be backwards

so [a,b,c] could be the same as [[a,b],c]

I'm pretty sure I prefer the first way, but the second way is actually easier to implement.

or at least cleaner looking to implement.

Why doesn't my code look clean ever?

Anyway, my reason for this post is this:

Does anyone have any opinions about how split is implemented?

ok, made some small commits...

(no features added, but cleaned up some code and added example programs)

yup.

it has the basis for how I am going to make functions if I make them, and cleaned up the code for BIFURCATEion, and removed some lines that were commented out.

if you want the new version it is on the github.

no real difference when it runs though.

(if something you wrote doesn't work with the new version, try using BIFFURCATE instead of BIFURCATE to use the old code. You probably won't need to, though. It works fine for me. and you probably haven't written anything in it anyway, because you kind of have to uh, want to, I guess)