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Nov 5, 2013, 5:15:31 PM11/5/13

to

Dear All,

Question 1: I just wondered a little bit

about the following. Can I use some CLP(*)

for unit calculations?

Here is an example expectation:

?- X ~ 5*M, Y ~ 10*S, Z ~ X/Y

Z ~ 1/2*M/S.

The (~)/2 would be the equality of the

corresponding CLP(*).

Question 2: What CHR framework would

one use to implement such a CLP(*), and

how complex would it be.

Bye

P.S.: I know a couple of reasons why

CLP(FD) cannot do it:

Hurdle 1: 1/2 is not an integer.

Hurdle 2: CLP(FD) usually cannot deal

with X1^A1*...*Xn^An, although it can

often represent A*X1+..+An*Xn.

Hurdle 3: CLP(FD) usually does not

substitute expression, it just keeps

them in the constraint store.

P.P.S.: Puzzle inspired by:

http://perso.ens-lyon.fr/guillaume.allais/?en/main/blog/read/dimension-aware-computations

Question 1: I just wondered a little bit

about the following. Can I use some CLP(*)

for unit calculations?

Here is an example expectation:

?- X ~ 5*M, Y ~ 10*S, Z ~ X/Y

Z ~ 1/2*M/S.

The (~)/2 would be the equality of the

corresponding CLP(*).

Question 2: What CHR framework would

one use to implement such a CLP(*), and

how complex would it be.

Bye

P.S.: I know a couple of reasons why

CLP(FD) cannot do it:

Hurdle 1: 1/2 is not an integer.

Hurdle 2: CLP(FD) usually cannot deal

with X1^A1*...*Xn^An, although it can

often represent A*X1+..+An*Xn.

Hurdle 3: CLP(FD) usually does not

substitute expression, it just keeps

them in the constraint store.

P.P.S.: Puzzle inspired by:

http://perso.ens-lyon.fr/guillaume.allais/?en/main/blog/read/dimension-aware-computations

Nov 5, 2013, 5:37:23 PM11/5/13

to

"Jan Burse" <janb...@fastmail.fm> wrote in message

news:l5bqm2$d61$1...@news.albasani.net...

value, unit of measure }, i.e. a data type with its own operations. E.g.

(a; A)/''(b; B) := (a/b; A/'B).

Julio

news:l5bqm2$d61$1...@news.albasani.net...

> Dear All,

>

> Question 1: I just wondered a little bit

> about the following. Can I use some CLP(*)

> for unit calculations?

>

> Here is an example expectation:

>

> ?- X ~ 5*M, Y ~ 10*S, Z ~ X/Y

> Z ~ 1/2*M/S.

Why/how CLP? That rather makes me think of an ad-hoc compound structure {
>

> Question 1: I just wondered a little bit

> about the following. Can I use some CLP(*)

> for unit calculations?

>

> Here is an example expectation:

>

> ?- X ~ 5*M, Y ~ 10*S, Z ~ X/Y

> Z ~ 1/2*M/S.

value, unit of measure }, i.e. a data type with its own operations. E.g.

(a; A)/''(b; B) := (a/b; A/'B).

Julio

Nov 5, 2013, 5:41:32 PM11/5/13

to

Julio Di Egidio schrieb:

expressions that undergo some simplification, and we would

like to have results such as:

?- X ~ 5*KG, Y ~ 10*G, Z ~ X/Y

Z ~ 500.

You get the pair (a;A) for free in CLP(*) if you don't

think in units, just assume the M, S, etc.. are some

variables.

Maybe we need some predefinitions, such as KG ~ 1000*G.

Right?

> Why/how CLP? That rather makes me think of an ad-hoc compound structure

> { value, unit of measure }, i.e. a data type with its own operations.

> E.g. (a; A)/''(b; B) := (a/b; A/'B).

>

> Julio

The unit itself would be also structured, respectively
> { value, unit of measure }, i.e. a data type with its own operations.

> E.g. (a; A)/''(b; B) := (a/b; A/'B).

>

> Julio

expressions that undergo some simplification, and we would

like to have results such as:

?- X ~ 5*KG, Y ~ 10*G, Z ~ X/Y

Z ~ 500.

You get the pair (a;A) for free in CLP(*) if you don't

think in units, just assume the M, S, etc.. are some

variables.

Maybe we need some predefinitions, such as KG ~ 1000*G.

Right?

Nov 5, 2013, 5:59:27 PM11/5/13

to

"Jan Burse" <janb...@fastmail.fm> wrote in message

news:l5bs6r$g88$1...@news.albasani.net...
> Julio Di Egidio schrieb:

>> Why/how CLP? That rather makes me think of an ad-hoc compound structure

>> { value, unit of measure }, i.e. a data type with its own operations.

>> E.g. (a; A)/''(b; B) := (a/b; A/'B).

>

>> Why/how CLP? That rather makes me think of an ad-hoc compound structure

>> { value, unit of measure }, i.e. a data type with its own operations.

>> E.g. (a; A)/''(b; B) := (a/b; A/'B).

>

> The unit itself would be also structured, respectively

> expressions that undergo some simplification, and we would

> like to have results such as:

>

> ?- X ~ 5*KG, Y ~ 10*G, Z ~ X/Y

> Z ~ 500.

>

> You get the pair (a;A) for free in CLP(*) if you don't

> think in units, just assume the M, S, etc.. are some

> variables.

But that would not be satisfactory: 5 Kg / 10 g ==> 0.5 Kg/g.
> expressions that undergo some simplification, and we would

> like to have results such as:

>

> ?- X ~ 5*KG, Y ~ 10*G, Z ~ X/Y

> Z ~ 500.

>

> You get the pair (a;A) for free in CLP(*) if you don't

> think in units, just assume the M, S, etc.. are some

> variables.

> Maybe we need some predefinitions, such as KG ~ 1000*G.

>

> Right?

CS competencies are limited anyway), I'd have only "normalised" units in the

core, and maybe do some normalisation-denormalisation before and after the

actual calculations. E.g. 5 Kg / 10 g ==>(normalised) 5000 g / 10 g ==> 500

g/g

==> 500 ==> (denormalised, if we like) 0.5 K.

Julio

Nov 6, 2013, 2:52:50 AM11/6/13

to

Julio Di Egidio schrieb:

For example we are expecting units to be independent.

So that the following query should not succeed:

?- M ~ S.

But what should it do? Fail? Throw an error? And

how to I teach the CLP(*) that they are independent.

Should I give some predefined inequality:

M \~ S

Or rather something like:

forall A, B in Number (A*M + B*S \~ 0)

Then there is another issue. Units are not only

independent, but I guess they are also incompatible.

Namely one wants the following not to succeed:

?- Z ~ KG + M.

Now assume we use some type inference to handle

the above cases. Thus adding a different layer

to our CLP(*). What would be the type inference

of the following:

?- A*M ~ B*S

I can immagine as a solution A ~ S and B ~ M,

but the following works also A ~ M^-1 and B ~ S^-1.

So the type of A and B can be anything along

X*S and X*M, where X can be any Unit.

Bye

>> You get the pair (a;A) for free in CLP(*) if you don't

>> think in units, just assume the M, S, etc.. are some

>> variables.

>

> But that would not be satisfactory: 5 Kg / 10 g ==> 0.5 Kg/g.

Yeah, probably there are more problems lurking.
>> think in units, just assume the M, S, etc.. are some

>> variables.

>

> But that would not be satisfactory: 5 Kg / 10 g ==> 0.5 Kg/g.

For example we are expecting units to be independent.

So that the following query should not succeed:

?- M ~ S.

But what should it do? Fail? Throw an error? And

how to I teach the CLP(*) that they are independent.

Should I give some predefined inequality:

M \~ S

Or rather something like:

forall A, B in Number (A*M + B*S \~ 0)

Then there is another issue. Units are not only

independent, but I guess they are also incompatible.

Namely one wants the following not to succeed:

?- Z ~ KG + M.

Now assume we use some type inference to handle

the above cases. Thus adding a different layer

to our CLP(*). What would be the type inference

of the following:

?- A*M ~ B*S

I can immagine as a solution A ~ S and B ~ M,

but the following works also A ~ M^-1 and B ~ S^-1.

So the type of A and B can be anything along

X*S and X*M, where X can be any Unit.

Bye

Nov 8, 2013, 8:20:52 PM11/8/13

to

"Jan Burse" <janb...@fastmail.fm> wrote in message

news:l5csgj$1ht$1...@news.albasani.net...
> Julio Di Egidio schrieb:

>>

>>> You get the pair (a;A) for free in CLP(*) if you don't

>>> think in units, just assume the M, S, etc.. are some

>>> variables.

>>

>> But that would not be satisfactory: 5 Kg / 10 g ==> 0.5 Kg/g.

>

> Yeah, probably there are more problems lurking.

> For example we are expecting units to be independent.

> So that the following query should not succeed:

>

> ?- M ~ S.

>

> But what should it do? Fail? Throw an error? And

> how to I teach the CLP(*) that they are independent.

I would expect it to fail, 2 = 1 fails.
>>

>>> You get the pair (a;A) for free in CLP(*) if you don't

>>> think in units, just assume the M, S, etc.. are some

>>> variables.

>>

>> But that would not be satisfactory: 5 Kg / 10 g ==> 0.5 Kg/g.

>

> Yeah, probably there are more problems lurking.

> For example we are expecting units to be independent.

> So that the following query should not succeed:

>

> ?- M ~ S.

>

> But what should it do? Fail? Throw an error? And

> how to I teach the CLP(*) that they are independent.

Anyway, units of measure never come alone, so I'd rather think about:

(V1, U1) ~ (V2, U2)

where, say, Vs are real numbers.

Or even:

(V1, M1, U1) ~ (V2, M2, U2)

to consider the multipliers.

Then I'd imagine an arithmetic on these be developed, finally a CLP that

leverages that arithmetic.

Julio

Nov 16, 2013, 7:34:55 PM11/16/13

to

Who needs Go, Dart, etc.. if you have Frink! Disclaimer:

I didn't use it much yet, only tried the Android app once.

But somehow feel sympathetic for a language far away from the

JavaScript, WebGL, Concurrency, etc.. frenzy.

Frink for Android

https://play.google.com/store/apps/developer?id=Alan+Eliasen

Frink Samples

http://futureboy.us/frinkdocs/#SampleCalculations

The bet is still open. Any small Frink kernel in Prolog?

Jan Burse schrieb:

I didn't use it much yet, only tried the Android app once.

But somehow feel sympathetic for a language far away from the

JavaScript, WebGL, Concurrency, etc.. frenzy.

Frink for Android

https://play.google.com/store/apps/developer?id=Alan+Eliasen

Frink Samples

http://futureboy.us/frinkdocs/#SampleCalculations

The bet is still open. Any small Frink kernel in Prolog?

Jan Burse schrieb:

Nov 20, 2013, 3:33:46 PM11/20/13

to

For what it's worth, I've written an evaluator for expressions with units,

though not a proper constraint solver. Nevertheless, it solves the sample

problems you posed, and can also do some of the Frink stuff as well.

The main predicate is isu/2 (corresponding to is/2):

% Jan's original example

?- X isu 5m, Y isu 10s, Z isu X/Y.

X = 5 m

Y = 10 s

Z = 0.5 m/(s)

% mixing units (result is always in SI units)

?- X isu 3m + 4ft.

X = 4.2192 m

% units with higher exponents (Note: multiply sign in 4*m^2 is required!)

?- X isu 3m * 4*m^2.

X = 12 * (m)^3

% How much energy is left after you eat a bar of chocolate,

% then spend 1 hour on the treadmill?

?- Y isu 600kcal - 1h*200'W'.

Y = 1790400.0 'J'

% Ohm's law

?- X isu 230'V' / 882'Ohm'.

X = 0.26077097505668934 'A'

% Frink example (http://futureboy.us/frinkdocs/#SampleCalculations)

?- Water isu 1kg/l, X isu 10ft*12ft*8ft * Water / lb.

Water = 1000.0 kg / (m)^3

X = 59930.842153098834

The code is a bit long for posting, but you can download it from

http://eclipseclp.org/wiki/Examples/ComputationWithNumbersAndUnits

It's written for ECLiPSe, for other systems you may have to replace

the do-loops with recursions.

Enjoy,

Joachim

though not a proper constraint solver. Nevertheless, it solves the sample

problems you posed, and can also do some of the Frink stuff as well.

The main predicate is isu/2 (corresponding to is/2):

% Jan's original example

?- X isu 5m, Y isu 10s, Z isu X/Y.

X = 5 m

Y = 10 s

Z = 0.5 m/(s)

% mixing units (result is always in SI units)

?- X isu 3m + 4ft.

X = 4.2192 m

% units with higher exponents (Note: multiply sign in 4*m^2 is required!)

?- X isu 3m * 4*m^2.

X = 12 * (m)^3

% How much energy is left after you eat a bar of chocolate,

% then spend 1 hour on the treadmill?

?- Y isu 600kcal - 1h*200'W'.

Y = 1790400.0 'J'

% Ohm's law

?- X isu 230'V' / 882'Ohm'.

X = 0.26077097505668934 'A'

% Frink example (http://futureboy.us/frinkdocs/#SampleCalculations)

?- Water isu 1kg/l, X isu 10ft*12ft*8ft * Water / lb.

Water = 1000.0 kg / (m)^3

X = 59930.842153098834

The code is a bit long for posting, but you can download it from

http://eclipseclp.org/wiki/Examples/ComputationWithNumbersAndUnits

It's written for ECLiPSe, for other systems you may have to replace

the do-loops with recursions.

Enjoy,

Joachim

Nov 20, 2013, 6:40:39 PM11/20/13

to

Joachim Schimpf schrieb:

> For what it's worth, I've written an evaluator for expressions with units,

Thats gr8. Thank you for sharing.
Nov 20, 2013, 6:51:03 PM11/20/13

to

And for those puzzlers that want to compare

notes of their solution or another solution

with what CAS, computer algebra systems, have

in store, two links:

51. ezunits

http://maxima.sourceforge.net/docs/manual/en/maxima_51.html#SEC257

79. unit

http://maxima.sourceforge.net/docs/manual/en/maxima_79.html#SEC369

By way of hardmath / scicomp.stackexchange.com.

Can we push toward CLP(*)?

notes of their solution or another solution

with what CAS, computer algebra systems, have

in store, two links:

51. ezunits

http://maxima.sourceforge.net/docs/manual/en/maxima_51.html#SEC257

79. unit

http://maxima.sourceforge.net/docs/manual/en/maxima_79.html#SEC369

By way of hardmath / scicomp.stackexchange.com.

Can we push toward CLP(*)?

Nov 24, 2013, 10:13:02 AM11/24/13

to

Hi,

I just read:

?- op(300,yf,m).

Yes

?- X = 3m.

X = 3 m

?- X = 3m^4.

Error: Parenthesis ("(") before operator missing.

X = 3m^4.

^

But I guess the dark forces are also strong with

these Prolog systems that are more tolerant. For

example SWI-Prolog since version 7 cannot handle

unquoted operators by default anymore:

> % How much energy is left after you eat a bar of chocolate,

> % then spend 1 hour on the treadmill?

> ?- Y isu 600kcal - 1h*200'W'.

> Y = 1790400.0 'J'

?- op(300,yf,'W').

ERROR: op/3: Domain error: `unquoted_atom' expected, found `W'

To solve the exponentiation problem, I got the idea

that for example the following operators could

be used:

m2 for m^2

m3 for m^3

I see this even sometimes in real world text, so it

would come quite natural.

Bye

I just read:

> % units with higher exponents (Note: multiply sign in

> 4*m^2 is required!)

> ?- X isu 3m * 4*m^2.

> X = 12 * (m)^3

Yes, I get an error on my side:
> 4*m^2 is required!)

> ?- X isu 3m * 4*m^2.

> X = 12 * (m)^3

?- op(300,yf,m).

Yes

?- X = 3m.

X = 3 m

?- X = 3m^4.

Error: Parenthesis ("(") before operator missing.

X = 3m^4.

^

But I guess the dark forces are also strong with

these Prolog systems that are more tolerant. For

example SWI-Prolog since version 7 cannot handle

unquoted operators by default anymore:

> % How much energy is left after you eat a bar of chocolate,

> % then spend 1 hour on the treadmill?

> ?- Y isu 600kcal - 1h*200'W'.

> Y = 1790400.0 'J'

ERROR: op/3: Domain error: `unquoted_atom' expected, found `W'

To solve the exponentiation problem, I got the idea

that for example the following operators could

be used:

m2 for m^2

m3 for m^3

I see this even sometimes in real world text, so it

would come quite natural.

Bye

Nov 24, 2013, 12:59:22 PM11/24/13

to

On 24/11/2013 15:13, Jan Burse wrote:> Hi,

to 3*(m^4) and you can't have that with Prolog syntax.

> For example SWI-Prolog since version 7 cannot handle

> unquoted operators by default anymore:

>

>> % How much energy is left after you eat a bar of chocolate,

>> % then spend 1 hour on the treadmill?

>> ?- Y isu 600kcal - 1h*200'W'.

>> Y = 1790400.0 'J'

>

> ?- op(300,yf,'W').

> ERROR: op/3: Domain error: `unquoted_atom' expected, found `W'

A new SWI feature, incompatible with all other Prologs and the Standard.

-- Joachim

>

> I just read:

>> % units with higher exponents (Note: multiply sign in

>> 4*m^2 is required!)

>> ?- X isu 3m * 4*m^2.

>> X = 12 * (m)^3

>

> Yes, I get an error on my side:

>

> ?- op(300,yf,m).

> Yes

> ?- X = 3m.

> X = 3 m

> ?- X = 3m^4.

> Error: Parenthesis ("(") before operator missing.

> X = 3m^4.

> ^

> But I guess the dark forces are also strong with

> these Prolog systems that are more tolerant.

Nothing to do with tolerance. You want 3m^4 to be equivalent
> I just read:

>> % units with higher exponents (Note: multiply sign in

>> 4*m^2 is required!)

>> ?- X isu 3m * 4*m^2.

>> X = 12 * (m)^3

>

> Yes, I get an error on my side:

>

> ?- op(300,yf,m).

> Yes

> ?- X = 3m.

> X = 3 m

> ?- X = 3m^4.

> Error: Parenthesis ("(") before operator missing.

> X = 3m^4.

> ^

> But I guess the dark forces are also strong with

> these Prolog systems that are more tolerant.

to 3*(m^4) and you can't have that with Prolog syntax.

> For example SWI-Prolog since version 7 cannot handle

> unquoted operators by default anymore:

>

>> % How much energy is left after you eat a bar of chocolate,

>> % then spend 1 hour on the treadmill?

>> ?- Y isu 600kcal - 1h*200'W'.

>> Y = 1790400.0 'J'

>

> ?- op(300,yf,'W').

> ERROR: op/3: Domain error: `unquoted_atom' expected, found `W'

-- Joachim

Nov 24, 2013, 2:03:42 PM11/24/13

to

Joachim Schimpf schrieb:

If I use a different operator level, I don't

get an error anymore on my side, but its still

not usable, since m refers to 3 and is not captured

by the (^)/2:

Jekejeke Prolog, Development Environment 0.9.12

(c) 1985-2013, XLOG Technologies GmbH, Switzerland

?- op(100,yf,m).

X = 3 m^7

?- X = 3m^7, write_canonical(X).

^(m(3),7)X = 3 m^7

Now I understand why you did use the operator level

300. Oki Doki.

If I try a Prolog system that I dub tolerant,

I get some strange result:

Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 7.1.0)

Copyright (c) 1990-2013 University of Amsterdam, VU Amsterdam

?- op(300,yf,m).

true.

?- X = 3m^4.

X = 3^4 m.

?- X = 3m^4, write_canonical(X).

m(^(3,4))

X = 3^4 m.

My expectation would be that m is postfix of 3,

in case the prolog system is tolerant. What

is going on here?

Bye

>> ?- X = 3m^4.

>> >Error: Parenthesis ("(") before operator missing.

>> >X = 3m^4.

>> > ^

>> >But I guess the dark forces are also strong with

>> >these Prolog systems that are more tolerant.

> Nothing to do with tolerance. You want 3m^4 to be equivalent

> to 3*(m^4) and you can't have that with Prolog syntax.

Oh, now I see, I was adopting your operator level.
>> >Error: Parenthesis ("(") before operator missing.

>> >X = 3m^4.

>> > ^

>> >But I guess the dark forces are also strong with

>> >these Prolog systems that are more tolerant.

> Nothing to do with tolerance. You want 3m^4 to be equivalent

> to 3*(m^4) and you can't have that with Prolog syntax.

If I use a different operator level, I don't

get an error anymore on my side, but its still

not usable, since m refers to 3 and is not captured

by the (^)/2:

Jekejeke Prolog, Development Environment 0.9.12

(c) 1985-2013, XLOG Technologies GmbH, Switzerland

?- op(100,yf,m).

Yes

?- X = 3m.

X = 3 m

?- X = 3m^7.
?- X = 3m.

X = 3 m

X = 3 m^7

?- X = 3m^7, write_canonical(X).

^(m(3),7)X = 3 m^7

Now I understand why you did use the operator level

300. Oki Doki.

If I try a Prolog system that I dub tolerant,

I get some strange result:

Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 7.1.0)

Copyright (c) 1990-2013 University of Amsterdam, VU Amsterdam

?- op(300,yf,m).

true.

?- X = 3m^4.

X = 3^4 m.

?- X = 3m^4, write_canonical(X).

m(^(3,4))

X = 3^4 m.

My expectation would be that m is postfix of 3,

in case the prolog system is tolerant. What

is going on here?

Bye

Nov 24, 2013, 2:12:49 PM11/24/13

to

Jan Burse schrieb:

as on my side:

ECLiPSe Constraint Logic Programming System [kernel]

Version 6.1 #163 (x86_64_nt), Tue Jul 9 22:11 2013

?- op(100,yf,m).

Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 7.1.0)

Copyright (c) 1990-2013 University of Amsterdam, VU Amsterdam

?- op(100,yf,m).

So SWI-Prolog only produces something strange, in

my opinion, for a case that I dub tolerant. i.e. when

it accepts the op(300,yf,m). See my previous post.

Interesting

> Joachim Schimpf schrieb:

>>> ?- X = 3m^4.

>>> >Error: Parenthesis ("(") before operator missing.

>>> >X = 3m^4.

>>> > ^

>>> >But I guess the dark forces are also strong with

>>> >these Prolog systems that are more tolerant.

>> Nothing to do with tolerance. You want 3m^4 to be equivalent

>> to 3*(m^4) and you can't have that with Prolog syntax.

>

> Oh, now I see, I was adopting your operator level.

> If I use a different operator level, I don't

More Testresults. ECLiPSe, same result
>>> ?- X = 3m^4.

>>> >Error: Parenthesis ("(") before operator missing.

>>> >X = 3m^4.

>>> > ^

>>> >But I guess the dark forces are also strong with

>>> >these Prolog systems that are more tolerant.

>> Nothing to do with tolerance. You want 3m^4 to be equivalent

>> to 3*(m^4) and you can't have that with Prolog syntax.

>

> Oh, now I see, I was adopting your operator level.

> If I use a different operator level, I don't

as on my side:

ECLiPSe Constraint Logic Programming System [kernel]

Version 6.1 #163 (x86_64_nt), Tue Jul 9 22:11 2013

?- op(100,yf,m).

?- X = 3m^7, write_canonical(X).

^(m(3), 7)

X = 3 m ^ 7

And another check:
^(m(3), 7)

X = 3 m ^ 7

Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 7.1.0)

Copyright (c) 1990-2013 University of Amsterdam, VU Amsterdam

?- X = 3m^7, write_canonical(X).

^(m(3),7)

X = 3 m^7.
^(m(3),7)

So SWI-Prolog only produces something strange, in

my opinion, for a case that I dub tolerant. i.e. when

it accepts the op(300,yf,m). See my previous post.

Interesting

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