Brachylog

Leaky Nun

4:09 AM

brachylog.tryitonline.net/…

@Fatalize If empty strings are allowed then why doesn't it extend to infinity?

Inconsistent behaviour

Leaky Nun

4:26 AM

How to reverse predicate names, i.e. what is the reverse of 2&?

Never mind

brachylog.tryitonline.net/…

Why doesn't this terminate? Remove the ,.h'0 and it terminates

brachylog.tryitonline.net/…

Leaky Nun

5:09 AM

You might want to add some constraints to modulo so that it doesn't go on forever

(e.g. finding solutions of 81:x%2)

for example (in Brachylog syntax): A:B%C,A:BohM,0<=C<m

actually, since idk whether A:BohM does not backtrack indefinitely:

A:B%C,0<=C<=A should work

negative modulo though

Nope, B is still not constrained

A:B%C,0<=C<=A,0<=C<B,{A<B,AC} (I have no idea how to do pred1 and (pred2->pred3)

oh wait... A:B%C,A:C-$ps*B

http://brachylog.tryitonline.net/#code=MWdmLgpBODEsQzEsQTpCJUMsQTpDLSRwcypCLEIuLC49&input=&args=Wg

works but with duplicates

Leaky Nun

6:05 AM

http://brachylog.tryitonline.net/#code=MWdmLgpBODEsQzEsQTpCJUMsMDw9Qzw9QSwwPD1DPEIsKEE-QyxBOkMtPj1CPjApLEIuLC49&input=&args=Wg

this is much better

Leaky Nun

6:25 AM

So basically:
if A and C are fixed:
if A<C: return false
if A==C: return all B>A
if A>C: set 0<B<=A-C, find all A%B == C

basically there are infinitely many solutions if A==C

Leaky Nun

6:50 AM

if B and C are fixed:
please, just return all N*B+C

Leaky Nun

8:54 AM

if A and B are fixed:
come on

Fatalize

9:56 AM

@LeakyNun Here?

Leaky Nun

@Fatalize yes

Fatalize

I kind of glanced at your posts this morning because I don't really have much time right now

Leaky Nun

Alright

So summary:

1. reversed concatenation (~c) yields empty strings, but not indefinitely

2. please fix modulo

that's all

@Fatalize

Fatalize

@LeakyNun Noted

I think it limits itself to the number of chars of the string

Leaky Nun

Which is completely arbitrary

brachylog_concatenate_limit_length(Max,H) :-
    Length #>= 0,
    Length #=< Max,
    indomain(Length),
brachylog_length(H,'integer':Length).

Fatalize

10:05 AM

well not quite, after that point all bigger sets will only be bigger because of empty strings

Leaky Nun

so either make it infinite with empty strings or finite without empty strings

Fatalize

the problem is things like this

brachylog.tryitonline.net/…

if I prevent empty strings then you would only get b

which I guess still makes sense but can be a bit annoying

Leaky Nun

@Fatalize which makes sense

then why limit itself to the length

Fatalize

@LeakyNun What's even the problem with it?

Leaky Nun

@Fatalize Solving 81:.%1 backtracks indefinitely

I detailed my solution above

Why doesn't this work?

binary brachylog.tryitonline.net/…

Fatalize

11:12 AM

@LeakyNun $p doesn't work when its input is a free variable, might be fixable

Leaky Nun

Imperative binary brachylog.tryitonline.net/…

@Fatalize I see, thanks

http://brachylog.tryitonline.net/#code=fisuIyMjZDp7MXwkcGR0Mjw9P31h&input=Mw&args=Wg
Indefinite backtracking again

Somehow indefinite backtracking is the biggest problem in declarative languages

Fatalize

infinite*

what's it trying to do?

Leaky Nun

come up with a list of distinct powers of 2 which sums to the input

Fatalize

11:45 AM

brachylog.tryitonline.net/…

Not quite sure why this fails infinitely on the +, the constraints in the debugger seem like it should succeed

Leaky Nun

@Fatalize How do you detect and do things based on which variable is free?

in Prolog I mean

Fatalize

12:03 PM

@LeakyNun If you code your relations well it shouldn't matter if they are free or not

obviously it's not always possible, so in that case there is var and nonvar to know that

Leaky Nun

So could you implement modulo as I have stated above?

Fatalize

brachylog_modulo(['integer':I1,'integer':I2],'integer':Rem) :-
Rem #= I1 mod I2.

There shouldn't be a problem with it because it's basic CLPFD…

so the problem might come from elsewhere

Leaky Nun

I saw the code

Does 1 #= 81 mod X backtrack infinitely?

Fatalize

?- 1 #= 81 mod X.
X in inf.. -1\/1..sup,
81 mod X#=1.

X's domain is infinite

Leaky Nun

exactly

So please implement what I described

Fatalize

12:11 PM

?- 1 #= 81 mod X, brachylog_equals(integer:X,integer:X).
X = 2 ;
X = 4 ;
X = 5 ;
X = 8 ;
X = 10 ;
X = 16 ;
X = 20 ;
X = 40 ;
X = 80 ;
infinite loop

Leaky Nun

So please implement ce que je decrivais

Fatalize

Seems more to be a bug with CLPFD no?

1 #= 81 mod X.
X in inf.. -1\/1..sup,
81 mod X#=1.

Leaky Nun

So please fix this bug by implementing ce que je decrivais

Fatalize

Can't we say for sure that X domain is at least smaller in absolute value than 81?

we can can't we?

Leaky Nun

I haven't dealt with negative modulo yet

How is it defined?

X is just the factors of A-C where A%X=C

4%3 = 1
-4%3 = ?
4%-3 = ?
-4%-3 = ?

Fatalize

12:24 PM

?- 1 + X*K #= 81.
X in -80.. -1\/1..80,
X*K#=80,
K in -80.. -1\/1..80.

I think the problem is in how mod is defined in Prolog

Leaky Nun

How is it defined?

Also, can you help me find out how negative number is treated ?

Fatalize

-4 mod 3 = 2

because -4 = 3*-2 + 2

it's the same but you want a positive remainder

at least that's what it does in Prolog

Leaky Nun

So i think you can implement mod as:
mod(A,B,C):
A=B*K+C,0<=C<B

(completely shitty prolog grammar)

what about -4mod-3?

Fatalize

13 ?- X is -4 mod 3.
X = 2.

14 ?- X is 4 mod -3.
X = -2.

15 ?- X is -4 mod -3.
X = -1.

pretty stupid if you ask me

Leaky Nun

ok, so branch on the sign of B

Fatalize

12:31 PM

I'll ask a question on SO about that mod CLPFD propagation

Leaky Nun

I believe javascript also uses that format

Fatalize

we're lucky that the creator of SWI-Prolog's CLPFD seems pretty interested in Brachylog and is active on SO ; should get a quick answer

Leaky Nun

whaaaat

Can you invite them to this room?

Fatalize

I'd rather post a question on SO, more useful for others and I won't bother him if he doesn't have time :p

@LeakyNun swi-prolog.org/pldoc/man?section=clpfd-applications

Leaky Nun

Alright

Fatalize

12:35 PM

I'm famous mom!

Leaky Nun

Damn, nice

Fatalize

0

Q: Modulo function and variable domain

Here is a simple CLPFD relation: 1 #= 81 mod X. This returns: X in inf.. -1\/1..sup, 81 mod X#=1. Unless my math is completely incorrect, shouldn't the domain of X be -80.. -1\/1..80?

Leaky Nun

Have an upvote

So you're right, the range should be -80 to 80

that is if 81 > 1

Fatalize

I think so

I'm just unsure if a similar argument applies to any triplet

Leaky Nun

12:53 PM

Just refer to what I wrote above lol

Leaky Nun

1:18 PM

@Fatalize If you said que that format is ridiculous, what do you have in mind?

Fatalize

@LeakyNun What format?

Leaky Nun

@Fatalize what should -4%3 give?

Fatalize

2

Leaky Nun

1:33 PM

You said it is ridiculous

Wtf, in Prolog you can make strings without quotes?

Fatalize

they are atoms

in Prolog there is no difference between code and data

so reverse can be both a predicate name but also just a string of data

and so you can query data if it corresponds to code, or store code as data

Leaky Nun

2:03 PM

Alright

@Fatalize Can I implicitly declare functions in Prolog?

Fatalize

you mean arithmetic functions or predicates ?

you can do both

Leaky Nun

Could you give an example?

Fatalize

wait implicitely

not really then

Leaky Nun

Can I define f(x)=y such that x*x=y

Alright...

Fatalize

well sure

Leaky Nun

2:09 PM

How?

Fatalize

square(X,Y) :- Y is X*X.

Leaky Nun

Oh, nice

then what were you thinking about when you said "not really"?

Fatalize

square(X,Y) :- Y #= X*X.

if you have clpfd library loaded

well I don't really understand your "implicitely"

Leaky Nun

Or, in other words, what is not possible?

What did you think it means?

Fatalize

some sort of unnamed function

Leaky Nun

2:11 PM

Oh...

Fatalize

I mean you could but that would simply just be writing code normally so the questions didn't make sense to me

Leaky Nun

Can I define f(x) such that f(f(x))= -x?

Fatalize

a(a(X),-X).

that was easy :p

Leaky Nun

What is the arity of a?

You made it both 1 and 2

Fatalize

?- a(a(3),Z).
Z = - 3.

Leaky Nun

2:15 PM

I think you defined two functions...

Fatalize

a/2 and a/1 are different things anyway so it's not a problem that they don't have the same arity

Leaky Nun

Exactly

I want only one function

Fatalize

But I didn't even define a

that's the thing :P

it's kind of a trick because a/1 doesn't event exist here as a predicaet

Leaky Nun

Can I only use a/2?

Fatalize

the a(3) in a/2 is just a data construct

a(test(X),-X).

is basically the same thing

so that's a bit cheating in a sense, I think what you want is a(X,Temp),a(Temp,-X).

Leaky Nun

2:18 PM

Full stop in clause-body?  Cannot redefine ,/2

procedure `a(A,B)' does not exist
Reachable from:
	  swish_trace:swish_call(a(4,A))
	  '$swish wrapper'(a(4,A),B)

Fatalize

show your code

Leaky Nun

a(X,Temp),a(Temp,-X).

?- a(4,What).

Fatalize

a(X,Temp),a(Temp,-X).

that's not a valid fact

or rule

Leaky Nun

but I just typed what you said

Fatalize

what I'm saying is

your f(f(x),-x) in prolog

you would QUERY that as a(X,Temp), a(Temp,-X).

now you need to define what a/2 is

so that this actually holds

and I don't see how you actually can

(without cheating)

Leaky Nun

2:21 PM

@Fatalize define cheating

I thought you can implicitly define functions

implicitly meaning declaratively

Fatalize

sure but what relation exists such that a(X,Temp), a(Temp,-X). would be true?

Leaky Nun

implicitly meaning declaratively

@Fatalize I don't know, that's what I'm asking Prolog

a(X,Y):-
    a(Y,-X).

?- a(4,-4).

Out of global stack

Fatalize

well of course it stack overflows

your a/2 says "a(X,Y) is true iff a(Y,-X) is true"

which will infinite loop

Leaky Nun

Alright

Thanks

How do I use it such that it does not stackoverflow?

Like can I query a(a(4)) directly?

Fatalize

there are no functions in prolog

a(4) doesn't return anything so you can't use it as an argument to another predicate as you would expect in imperative languages

Leaky Nun

2:25 PM

@Fatalize You know what I mean

Fatalize

No I don't

Leaky Nun

Is there a way to query for a(4,Temp),a(Temp,What) without finding solutions for Temp

Fatalize

a(X,a:X):- \+ (X = a:_).
a(a:X,Y):- Y is -X.

THis is what I would call cheating

Leaky Nun

How does it work?

Fatalize

but outside Prolog, what relation is there such that f(f(x)) = -x?

I can't think of any

Leaky Nun

2:28 PM

Neither can I, that's why I'm using Prolog's declarativeness

Fatalize

I'm pretty sure it's impossible without going outside arithmetic

@LeakyNun It cheats by appending the a: before X, then when calling it again on a:X is will only work with the second line, yielding -X

Leaky Nun

@Fatalize Alright

Is there a way to query for a(4,Temp),a(Temp,What) without finding solutions for Temp?

What I mean is, just leave Temp undefined

Fatalize

Well yes, if a doesn't unifies Temp in the process

Leaky Nun

How to do so?

Fatalize

Prolog just give values to things when needed so if it's not needed it doesnt

I don't understand your goal

Leaky Nun

2:30 PM

a(4,Temp),a(Temp,What).

Fatalize

if you don't want Temp to get a value from a then it's most likely useless in a isn't it?

Leaky Nun

I just want to carry it to What

Fatalize

a(4,_Temp),a(_Temp,What)

Leaky Nun

Whaaaat

Nope

Out of global stack

Fatalize

I told you your a is wrong

doesn't matter what you call it on

Leaky Nun

2:32 PM

How do I make it right?

Fatalize

I don't understand

f(f(x)) = -x isn't possible so you can't make it right if you only use arithmetic

and if you don't I just gave you a solution

Leaky Nun

I want to define a relation a(X,Y) such that a(Y,-X)

Fatalize

…? that's simply a(X,Y) :- Y is -X.

Leaky Nun

@Fatalize no, that's a(X,-X)

Fatalize

I mean with your definition a is just negation

Leaky Nun

2:34 PM

if a(X,Y) then a(Y,X) under your definition

Fatalize

right nevermind

but then we're back to what I said

« cheat » or you can't

Leaky Nun

Alright

Fatalize

Prolog is about writing relations between things, if you can't think of a relation it's not magically gonna come up with it for you

Leaky Nun

Thanks

By the way, source

Fatalize

f(n) = sign(n) - (-1)n * n

well just code this then if this relation works

Leaky Nun

2:37 PM

Alright

@Fatalize That's integer to integer

Fatalize

@LeakyNun ?

a(X,Y) :- Y is sign(X) - (-1)^X * X.

Leaky Nun

I'm thinking of mapping the rationals to two integers

Fatalize

wat

Who I was thinking of answered

0

A: Modulo function and variable domain

First things first: Is there a mistake in the constraint solver? No(t necessarily), because all admissible solutions are still contained in the domain that is deduced, and no wrong solutions are reported. The domain that you deduced is a proper subset of what the solver reports. Does th...