Nabble - Alloy Analyzer

new community website / IMPORTANT

2008-07-16T10:48:51Z

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How to detect whether maxium/minimum bound of integer is exceeded?

2008-07-12T15:00:51Z

Hi,

I want to use some integer functions in Alloy, but only within the
limited boundary. For example, if x+y is less than the maximum atom of
the current bitwidth, than the function will compute the result,
otherwise I will use some other signature to hold the value of x+y (as
some symbolic value). But as mentioned in the document, the integer
functions will implicitly truncate the result if it overflows. So I can
not use "add[x, y] > max", because if will always return true. Is there
any way to tell whether an integer function produce a overflow result?
Thank you!

Re: How to define transitive closure of tree-like relations for reachability, transitive closure over a formula

2008-07-06T11:53:46Z

Hi Eunsuk,
Many Thanks!!! This creates a random tree,
For the benefit of other users I've put here the corrected code. Indeed different solvers produce different solutions.
Users: play with the run command - time increases as parameter of int increases, and the Domain size.
the parameter for int is bit width, default is 4, Domain size should not exceed (2^(int_size-1))-1, see note 6 on
http://alloy.mit.edu/alloy4/quickguide/a4.html
Experiment with the reach function in the evaluator, press Instance, then Evaluator button.
module startFormula
sig Domain {}
fact { #Domain > 6 } // <============= Setting domain size
// for predicate root[v] - whether v is root of tree.
sig root_id12'Sig in Domain { }
pred root_id12(v1:Domain) {
v1 in root_id12'Sig
}
//True iff node is reachable from root node.
sig r'root'_id17'Sig in Domain { }
pred r'root'_id17(v1:Domain) {
v1 in r'root'_id17'Sig
}
//True iff v2 is imediate right child of v1.
sig right_id15'Sig in Domain { right_id15'f: set Domain }
pred right_id15(v1:right_id15'Sig,v2:Domain) {
v2 in v1.right_id15'f
}
//True iff v2 is imediate left child of v1.
sig left_id16'Sig in Domain { left_id16'f: set Domain }
pred left_id16(v1:left_id16'Sig,v2:Domain) {
v2 in v1.left_id16'f
}
pred solveFormula {
//start forcing a Tree
// some node is root
(some v1:Domain|root_id12[v1] ) &&
// every node is either a root or pointed by a left or right of another node
(all v1:Domain | root_id12[v1] || (some v2:Domain |left_id16[v2,v1] || right_id15[v2,v1])) &&
//no two nodes connected by both left and right !!XXX Only if remarked there's solution
//(all v1:Domain,v2:Domain| left_id16[v1,v2] =>!(right_id15[v1,v2])) &&
// no self loops
(all v1:Domain,v2:Domain| left_id16[v1,v2] => !(v1==v2)) &&
(all v1:Domain,v2:Domain| right_id15[v1,v2] => !(v1==v2)) &&
//there's a fork, it's a real tree, not a list
(some v1:Domain,v2:Domain,v3:Domain| left_id16[v1,v2] && right_id15[v1,v3] ) &&
//no node points to root
(all v1:Domain,v2:Domain|!(root_id12[v1] && (left_id16[v2,v1] || right_id15[v2,v1]))) &&
// no mutual pointing
(all v1:Domain,v2:Domain| left_id16[v1,v2] => ! left_id16[v2,v1]) &&
(all v1:Domain,v2:Domain| right_id15[v1,v2] => ! right_id15[v2,v1]) &&
// a node may be pointed only once.
(all v1:Domain,v2:Domain,v3:Domain| (left_id16[v1,v2] && left_id16[v3,v2]) => v1==v3) &&
(all v1:Domain,v2:Domain,v3:Domain| (right_id15[v1,v2] && right_id15[v3,v2]) =>v1==v3) &&
(all v1:Domain,v2:Domain,v3:Domain| left_id16[v1,v2] => ! right_id15[v3,v2]) &&
(all v1:Domain,v2:Domain,v3:Domain| right_id15[v1,v2] =>! left_id16[v3,v2]) &&
// all nodes are in the same tree
(all v1:Domain|some v2:Domain| root_id12[v2] && v1 in reach[v2]) &&
// no cycles
( no v1:Domain| v1 in v1.^n'ftc0) &&
//end forcing a Tree
//root[v] holds for at most a single node.
(
all v2:Domain,v1:Domain|((root_id12[v1] && root_id12[v2]) => (v1 == v2)))
&& (
all v2:Domain|(
( some v1:Domain|(root_id12[v1] && ( ! (v1 == v2)))) => ( ! root_id12[v2])))
&&
// right is a pointer.
(
all v2:Domain,v1:Domain|(
( some v:Domain|(right_id15[v,v1] && right_id15[v,v2])) => (v1 == v2))) && (
all v2:Domain,v:Domain|(
( some v1:Domain|(right_id15[v,v1] && ( ! (v1 == v2)))) => ( ! right_id15[v,v2]))) && (
//left is a pointer.
all v2:Domain,v1:Domain|(
( some v:Domain|(left_id16[v,v1] && left_id16[v,v2])) => (v1 == v2))) && (
all v2:Domain,v:Domain|(
( some v1:Domain|(left_id16[v,v1] && ( ! (v1 == v2)))) => ( ! left_id16[v,v2]))) &&
//definition of r'root
( all v:Domain|((root_id12[v] ||
( some z___v2:Domain|(root_id12[z___v2] && tcFormula'TC[z___v2,v]))) => r'root'_id17[v])) &&
( all v:Domain|(( ! (root_id12[v] ||
( some z___v2:Domain|(root_id12[z___v2] && tcFormula'TC[z___v2,v]))))) => ( ! r'root'_id17[v]))
}
sig tcFormula'Sig in Domain { n'ftc0: set Domain }
fact tcFact { all v1:Domain,v2:Domain| (left_id16[v1,v2] || right_id15[v1,v2]) <=> (v1 in tcFormula'Sig && v2 in v1. n'ftc0) }
// is there a path between v1 and v2 on tree.
pred tcFormula'TC(v1:tcFormula'Sig, v2:Domain) {
v2 in v1.^n'ftc0
}
fun reach(v1:Domain): Domain {
v1 + v1.^n'ftc0
}
run solveFormula for 10 Domain, 4 int expect 1
Sharon

----- Original Message ----
From: Eunsuk Kang <eskang@...>
To: alloy-discuss@...
Sent: Sunday, July 6, 2008 2:13:02 AM
Subject: Re: [alloy-discuss] How to define transitive closure of tree-like relations for reachability, transitive closure over a formula

Hi Sharon,

The following constraint in pred solveFormula was causing a
contradiction in your model:

// right is a pointer.
....
( some v1:Domain|(right_ id15[v,v1] && ( ! (v1 == v2)))) => ( !
left_id16[v, v2]))) && ....

If you replace :left_id16[v, v2]" with "right_id15[ v,v2]", Alloy
generates an instance:

( some v1:Domain|(right_ id15[v,v1] && ( ! (v1 == v2)))) => ( !
right_id15[v, v2]))) && (

In general, if Alloy does not produce any instance, unsat core can be
very useful in determining
the set of constraints that led to a contradiction. To enable unsat
core, select Menu -> Options ->
Sat Solver: MiniSat with Unsat Core.

In your example, an unsat core returned from Alloy highlighted a small
subset of constraints in
pred solveFormula, from which I was able to locate the above overconstraint.

Eunsuk

Sharon G. wrote:

> Hi,
>
> How do I define reachability of a binary tree, each node having a
> left, right fields. Please see example below. Eventually I'd like
> Alloy to generate a tree structure, and have predicate
> tcFormula'TC( v1,v2) ==True iff there is a path on tree between v1 and
> v2 (assuming v1!=v2).
> I'm concerned with how to define the fact tcFact - is it correct ?
> Also no solution to this unless I remark the requirement labeled with
> !!XXX forcing a tree structure.
> How can i force tree structure, and still have tcFormula'TC( v1,v2)
> correct ?
> I must have a general solution for transitive closure of a formula,
> since I'm developing a static analysis tool using Alloy as solver.
> In this example the formula like TC(left_id16[ v1,v2] ||
> right_id15[v1, v2])(w1,w2) .
>
> module startFormula
> sig Domain {}
> fact { #Domain > 4 }
> // for predicate root[v] - whether v is root of tree.
> sig root_id12'Sig in Domain { }
> pred root_id12(v1: Domain) {
> v1 in root_id12'Sig
> }
> //True iff node is reachable from root node.
> sig r'root'_id17' Sig in Domain { }
> pred r'root'_id17( v1:Domain) {
> v1 in r'root'_id17' Sig
> }
> //True iff v2 is imediate right child of v1.
> sig right_id15'Sig in Domain { right_id15'f: set Domain }
> pred right_id15(v1: right_id15' Sig,v2:Domain) {
> v2 in v1.right_id15' f
> }
> //True iff v2 is imediate left child of v1.
> sig left_id16'Sig in Domain { left_id16'f: set Domain }
> pred left_id16(v1: left_id16' Sig,v2:Domain) {
> v2 in v1.left_id16' f
> }
> pred solveFormula {
> //start forcing a Tree
> // some node is root
> (some v1:Domain|root_ id12[v1] ) &&
> // every node is either a root or pointed by a left or right of
> another node
> (all v1:Domain | root_id12[v1] || (some v2:Domain |left_id16[v2, v1]
> || right_id15[v2, v1])) &&
> //no two nodes connected by both left and right !!XXX Only if
> remarked there's solution
> (all v1:Domain,v2: Domain| left_id16[v1, v2] =>!(right_id15[ v1,v2])) &&
> // no self loops
> (all v1:Domain,v2: Domain| left_id16[v1, v2] => !(v1==v2)) &&
> (all v1:Domain,v2: Domain| right_id15[v1, v2] => !(v1==v2)) &&
> //there's a fork, it's a real tree, not a list
> (some v1:Domain,v2: Domain,v3: Domain| left_id16[v1, v2] &&
> right_id15[v1, v3] ) &&
> //no node points to root
> (all v1:Domain,v2: Domain|!( root_id12[ v1] && (left_id16[v2, v1] ||
> right_id15[v2, v1]))) &&
> // no mutual pointing
> (all v1:Domain,v2: Domain| left_id16[v1, v2] => ! left_id16[v2, v1]) &&
> (all v1:Domain,v2: Domain| right_id15[v1, v2] => ! right_id15[v2, v1]) &&
> //end forcing a Tree
> //root[v] holds for at most a single node.
> (
> all v2:Domain,v1: Domain|(( root_id12[ v1] && root_id12[v2] ) => (v1 ==
> v2)))
> && (
> all v2:Domain|(
> ( some v1:Domain|(root_ id12[v1] && ( ! (v1 == v2)))) => ( !
> root_id12[v2] )))
> &&
> // right is a pointer.
> (
> all v2:Domain,v1: Domain|(
> ( some v:Domain|(right_ id15[v,v1] && right_id15[v, v2])) => (v1 ==
> v2))) && (
> all v2:Domain,v: Domain|(
> ( some v1:Domain|(right_ id15[v,v1] && ( ! (v1 == v2)))) => ( !
> left_id16[v, v2]))) && (
> //left is a pointer.
> all v2:Domain,v1: Domain|(
> ( some v:Domain|(left_ id16[v,v1] && left_id16[v, v2])) => (v1 == v2)))
> && (
> all v2:Domain,v: Domain|(
> ( some v1:Domain|(left_ id16[v,v1] && ( ! (v1 == v2)))) => ( !
> left_id16[v, v2]))) &&
> //definition of r'root
> ( all v:Domain|((root_ id12[v] ||
> ( some z___v2:Domain| (root_id12[ z___v2] && tcFormula'TC[ z___v2,v] )))
> => r'root'_id17[ v])) &&
> ( all v:Domain|(( ! (root_id12[v] ||
> ( some z___v2:Domain| (root_id12[ z___v2] &&
> tcFormula'TC[ z___v2,v] ))))) => ( ! r'root'_id17[ v]))
> }
> sig tcFormula'Sig in Domain { n'ftc0: set Domain }
> fact tcFact { all v1:Domain,v2: Domain| (left_id16[v1, v2] ||
> right_id15[v1, v2]) <=> (v1 in tcFormula'Sig && v2 in v1. n'ftc0) }
> // no cycles
> fact { no v1:Domain| v1 in v1.^n'ftc0}
> // is there a path between v1 and v2 on tree.
> pred tcFormula'TC( v1:tcFormula' Sig, v2:Domain) {
> v2 in v1.^n'ftc0
> }
> //reflexive transitive reachability
> fun reach(v1:Domain) : Domain {
> v1 + v1.^n'ftc0
> }
> run solveFormula for 7 Domain expect 1
>
> Many Thanks,
> Sharon
>
>

Re: How to define transitive closure of tree-like relations for reachability, transitive closure over a formula

2008-07-05T17:13:02Z

Hi Sharon,

The following constraint in pred solveFormula was causing a
contradiction in your model:

// right is a pointer.
....
( some v1:Domain|(right_id15[v,v1] && ( ! (v1 == v2)))) => ( !
left_id16[v,v2]))) && ....

If you replace :left_id16[v, v2]" with "right_id15[v,v2]", Alloy
generates an instance:

( some v1:Domain|(right_id15[v,v1] && ( ! (v1 == v2)))) => ( !
right_id15[v,v2]))) && (

In general, if Alloy does not produce any instance, unsat core can be
very useful in determining
the set of constraints that led to a contradiction. To enable unsat
core, select Menu -> Options ->
Sat Solver: MiniSat with Unsat Core.

In your example, an unsat core returned from Alloy highlighted a small
subset of constraints in
pred solveFormula, from which I was able to locate the above overconstraint.

Eunsuk

Sharon G. wrote:

> Hi,
>
> How do I define reachability of a binary tree, each node having a
> left, right fields. Please see example below. Eventually I'd like
> Alloy to generate a tree structure, and have predicate
> tcFormula'TC(v1,v2) ==True iff there is a path on tree between v1 and
> v2 (assuming v1!=v2).
> I'm concerned with how to define the fact tcFact - is it correct ?
> Also no solution to this unless I remark the requirement labeled with
> !!XXX forcing a tree structure.
> How can i force tree structure, and still have tcFormula'TC(v1,v2)
> correct ?
> I must have a general solution for transitive closure of a formula,
> since I'm developing a static analysis tool using Alloy as solver.
> In this example the formula like TC(left_id16[v1,v2] ||
> right_id15[v1,v2])(w1,w2).
>
> module startFormula
> sig Domain {}
> fact { #Domain > 4 }
> // for predicate root[v] - whether v is root of tree.
> sig root_id12'Sig in Domain { }
> pred root_id12(v1:Domain) {
> v1 in root_id12'Sig
> }
> //True iff node is reachable from root node.
> sig r'root'_id17'Sig in Domain { }
> pred r'root'_id17(v1:Domain) {
> v1 in r'root'_id17'Sig
> }
> //True iff v2 is imediate right child of v1.
> sig right_id15'Sig in Domain { right_id15'f: set Domain }
> pred right_id15(v1:right_id15'Sig,v2:Domain) {
> v2 in v1.right_id15'f
> }
> //True iff v2 is imediate left child of v1.
> sig left_id16'Sig in Domain { left_id16'f: set Domain }
> pred left_id16(v1:left_id16'Sig,v2:Domain) {
> v2 in v1.left_id16'f
> }
> pred solveFormula {
> //start forcing a Tree
> // some node is root
> (some v1:Domain|root_id12[v1] ) &&
> // every node is either a root or pointed by a left or right of
> another node
> (all v1:Domain | root_id12[v1] || (some v2:Domain |left_id16[v2,v1]
> || right_id15[v2,v1])) &&
> //no two nodes connected by both left and right !!XXX Only if
> remarked there's solution
> (all v1:Domain,v2:Domain| left_id16[v1,v2] =>!(right_id15[v1,v2])) &&
> // no self loops
> (all v1:Domain,v2:Domain| left_id16[v1,v2] => !(v1==v2)) &&
> (all v1:Domain,v2:Domain| right_id15[v1,v2] => !(v1==v2)) &&
> //there's a fork, it's a real tree, not a list
> (some v1:Domain,v2:Domain,v3:Domain| left_id16[v1,v2] &&
> right_id15[v1,v3] ) &&
> //no node points to root
> (all v1:Domain,v2:Domain|!(root_id12[v1] && (left_id16[v2,v1] ||
> right_id15[v2,v1]))) &&
> // no mutual pointing
> (all v1:Domain,v2:Domain| left_id16[v1,v2] => ! left_id16[v2,v1]) &&
> (all v1:Domain,v2:Domain| right_id15[v1,v2] => ! right_id15[v2,v1]) &&
> //end forcing a Tree
> //root[v] holds for at most a single node.
> (
> all v2:Domain,v1:Domain|((root_id12[v1] && root_id12[v2]) => (v1 ==
> v2)))
> && (
> all v2:Domain|(
> ( some v1:Domain|(root_id12[v1] && ( ! (v1 == v2)))) => ( !
> root_id12[v2])))
> &&
> // right is a pointer.
> (
> all v2:Domain,v1:Domain|(
> ( some v:Domain|(right_id15[v,v1] && right_id15[v,v2])) => (v1 ==
> v2))) && (
> all v2:Domain,v:Domain|(
> ( some v1:Domain|(right_id15[v,v1] && ( ! (v1 == v2)))) => ( !
> left_id16[v,v2]))) && (
> //left is a pointer.
> all v2:Domain,v1:Domain|(
> ( some v:Domain|(left_id16[v,v1] && left_id16[v,v2])) => (v1 == v2)))
> && (
> all v2:Domain,v:Domain|(
> ( some v1:Domain|(left_id16[v,v1] && ( ! (v1 == v2)))) => ( !
> left_id16[v,v2]))) &&
> //definition of r'root
> ( all v:Domain|((root_id12[v] ||
> ( some z___v2:Domain|(root_id12[z___v2] && tcFormula'TC[z___v2,v])))
> => r'root'_id17[v])) &&
> ( all v:Domain|(( ! (root_id12[v] ||
> ( some z___v2:Domain|(root_id12[z___v2] &&
> tcFormula'TC[z___v2,v]))))) => ( ! r'root'_id17[v]))
> }
> sig tcFormula'Sig in Domain { n'ftc0: set Domain }
> fact tcFact { all v1:Domain,v2:Domain| (left_id16[v1,v2] ||
> right_id15[v1,v2]) <=> (v1 in tcFormula'Sig && v2 in v1. n'ftc0) }
> // no cycles
> fact { no v1:Domain| v1 in v1.^n'ftc0}
> // is there a path between v1 and v2 on tree.
> pred tcFormula'TC(v1:tcFormula'Sig, v2:Domain) {
> v2 in v1.^n'ftc0
> }
> //reflexive transitive reachability
> fun reach(v1:Domain): Domain {
> v1 + v1.^n'ftc0
> }
> run solveFormula for 7 Domain expect 1
>
> Many Thanks,
> Sharon
>
>

Displaying Alloy optimizations impact

2008-07-05T14:28:18Z

Hi,
I'd like to show Alloy indeed helps in solving problems faster.
How can I see the number of variables and clauses before and after Alloy optimizations ?
Many Thanks,
Sharon

How to define transitive closure of tree-like relations for reachability, transitive closure over a formula

2008-07-05T14:23:01Z

Hi,
How do I define reachability of a binary tree, each node having a left, right fields. Please see example below. Eventually I'd like Alloy to generate a tree structure, and have predicate tcFormula'TC(v1,v2) ==True iff there is a path on tree between v1 and v2 (assuming v1!=v2).
I'm concerned with how to define the fact tcFact - is it correct ?
Also no solution to this unless I remark the requirement labeled with !!XXX forcing a tree structure.
How can i force tree structure, and still have tcFormula'TC(v1,v2) correct ?
I must have a general solution for transitive closure of a formula, since I'm developing a static analysis tool using Alloy as solver.
In this example the formula like TC(left_id16[v1,v2] || right_id15[v1,v2])(w1,w2).
module startFormula
sig Domain {}
fact { #Domain > 4 }
// for predicate root[v] - whether v is root of tree.
sig root_id12'Sig in Domain { }
pred root_id12(v1:Domain) {
v1 in root_id12'Sig
}
//True iff node is reachable from root node.
sig r'root'_id17'Sig in Domain { }
pred r'root'_id17(v1:Domain) {
v1 in r'root'_id17'Sig
}
//True iff v2 is imediate right child of v1.
sig right_id15'Sig in Domain { right_id15'f: set Domain }
pred right_id15(v1:right_id15'Sig,v2:Domain) {
v2 in v1.right_id15'f
}
//True iff v2 is imediate left child of v1.
sig left_id16'Sig in Domain { left_id16'f: set Domain }
pred left_id16(v1:left_id16'Sig,v2:Domain) {
v2 in v1.left_id16'f
}
pred solveFormula {
//start forcing a Tree
// some node is root
(some v1:Domain|root_id12[v1] ) &&
// every node is either a root or pointed by a left or right of another node
(all v1:Domain | root_id12[v1] || (some v2:Domain |left_id16[v2,v1] || right_id15[v2,v1])) &&
//no two nodes connected by both left and right !!XXX Only if remarked there's solution
(all v1:Domain,v2:Domain| left_id16[v1,v2] =>!(right_id15[v1,v2])) &&
// no self loops
(all v1:Domain,v2:Domain| left_id16[v1,v2] => !(v1==v2)) &&
(all v1:Domain,v2:Domain| right_id15[v1,v2] => !(v1==v2)) &&
//there's a fork, it's a real tree, not a list
(some v1:Domain,v2:Domain,v3:Domain| left_id16[v1,v2] && right_id15[v1,v3] ) &&
//no node points to root
(all v1:Domain,v2:Domain|!(root_id12[v1] && (left_id16[v2,v1] || right_id15[v2,v1]))) &&
// no mutual pointing
(all v1:Domain,v2:Domain| left_id16[v1,v2] => ! left_id16[v2,v1]) &&
(all v1:Domain,v2:Domain| right_id15[v1,v2] => ! right_id15[v2,v1]) &&
//end forcing a Tree
//root[v] holds for at most a single node.
(
all v2:Domain,v1:Domain|((root_id12[v1] && root_id12[v2]) => (v1 == v2)))
&& (
all v2:Domain|(
( some v1:Domain|(root_id12[v1] && ( ! (v1 == v2)))) => ( ! root_id12[v2])))
&&
// right is a pointer.
(
all v2:Domain,v1:Domain|(
( some v:Domain|(right_id15[v,v1] && right_id15[v,v2])) => (v1 == v2))) && (
all v2:Domain,v:Domain|(
( some v1:Domain|(right_id15[v,v1] && ( ! (v1 == v2)))) => ( ! left_id16[v,v2]))) && (
//left is a pointer.
all v2:Domain,v1:Domain|(
( some v:Domain|(left_id16[v,v1] && left_id16[v,v2])) => (v1 == v2))) && (
all v2:Domain,v:Domain|(
( some v1:Domain|(left_id16[v,v1] && ( ! (v1 == v2)))) => ( ! left_id16[v,v2]))) &&
//definition of r'root
( all v:Domain|((root_id12[v] ||
( some z___v2:Domain|(root_id12[z___v2] && tcFormula'TC[z___v2,v]))) => r'root'_id17[v])) &&
( all v:Domain|(( ! (root_id12[v] ||
( some z___v2:Domain|(root_id12[z___v2] && tcFormula'TC[z___v2,v]))))) => ( ! r'root'_id17[v]))
}
sig tcFormula'Sig in Domain { n'ftc0: set Domain }
fact tcFact { all v1:Domain,v2:Domain| (left_id16[v1,v2] || right_id15[v1,v2]) <=> (v1 in tcFormula'Sig && v2 in v1. n'ftc0) }
// no cycles
fact { no v1:Domain| v1 in v1.^n'ftc0}
// is there a path between v1 and v2 on tree.
pred tcFormula'TC(v1:tcFormula'Sig, v2:Domain) {
   v2 in v1.^n'ftc0
}
//reflexive transitive reachability
fun reach(v1:Domain): Domain {
   v1 + v1.^n'ftc0
}
run solveFormula for 7 Domain expect 1
Many Thanks,
Sharon

Re: How does the Analyzer slice Alloy models?

2008-07-05T04:55:00Z

On Sat, 5 Jul 2008, ikarus1983de wrote:
> I'm looking for some information (e.g. technical report) on how an
> Alloy model is reduced / sliced before the analysis.
>
> I know about the work on the Kato-Tool by Ilya Shlyakhter et. al. but
> up to my knowledge this isn't integrated into Alloy (or Kodkod) yet.
> Furthermore, I'd like to know if Alloy does stuff like "signature x is
> not used for check y, thus remove x for the analysis of y".

Dear Christian:

There is currently no technical report on Alloy4's translation yet,
but I will write it soon as part of my thesis.

To answer your question, since Alloy translates the user problem into
a single CNF problem, that means "each predicate call" adds to
the final CNF, and unused predicates are completely omitted.

Several optimizations include: 1) calls to the same predicates
with the same actual arguments are merged, so that if you call
the same predicate from multiple places in your model with the
same arguments, then the cost is borne only once. 2) Even between
two calls with different arguments, identical subformulas and
subexpressions are merged ("shared") to further reduce the cost.

Sincerely,
Felix Chang
Alloy4 Developer

How does the Analyzer slice Alloy models?

2008-07-05T02:56:28Z

Hi,

I'm looking for some information (e.g. technical report) on how an
Alloy model is reduced / sliced before the analysis.

I know about the work on the Kato-Tool by Ilya Shlyakhter et. al. but
up to my knowledge this isn't integrated into Alloy (or Kodkod) yet.
Furthermore, I'd like to know if Alloy does stuff like "signature x is
not used for check y, thus remove x for the analysis of y".

Any help or hint is appreciated.

Best regards,
Christian

How does the Analyzer slice Alloy models?

2008-07-05T02:53:18Z

Hi,

I'm looking for some information (e.g. technical report) on how an
Alloy model is reduced / sliced before the analysis.

I know about the work on the Kato-Tool by Ilya Shlyakhter et. al. but
up to my knowledge this isn't integrated into Alloy (or Kodkod) yet.
Furthermore, I'd like to know if Alloy does stuff like "predicate x is
not used for check y, thus remove x for the analysis of y".

Any help or hint is appreciated.

Best regards,
Christian

Re: haw to declare integers

2008-07-04T09:00:21Z

Felix,

You do a fantastic job in replying to queries. But seeing some of the
replies to the last discussion, it occurs to me that you might want
to hold off on some more basic questions to encourage other people to
answer.

--Daniel

On Jun 25, 2008, at 5:31 PM, Felix Chang wrote:

> On Wed, 25 Jun 2008, ylayouni wrote:
>> please i want to manipulate integer, for example :
>> sig user { check_ : set machine}
>> sig machine { temperature: int} // this is wrong !
>> sig machine1 extends machine{}
>>
>> fact {
>> all u: user | machine.temperature >10 => u.check_= machine
>> }
>
> The lowercase "int" refers to a "2's complement" integer.
> So if your model is configured to use a bitwidth of 4, let's say,
> then your model would have integers ranging
> from -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7
>
> (You can set the bitwidth for each "run" and each "check" separately.
> If you write "check someAssertion for 5 int", that means
> you want to check that assertion with a bitwidth of 5)
>
> The uppercase "Int" refers to a set of special atoms
> that correspond to each integer.
>
> Only atoms can belong in relations. So you would write
> "sig machine { temperature: Int }" if you want
> each machine to have an Int atom associated with it.
>
> Note that the conversion between Int and int is very subtle.
> Please search the mailinglist messages for previous discussions,
> and please consult the Alloy3 user manual and/or Prof. Jackson's book.
>
> Sincerely,
> Felix Chang
> Alloy4 Developer
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>

Re: haw to declare integers

2008-06-25T14:31:43Z

On Wed, 25 Jun 2008, ylayouni wrote:
> please i want to manipulate integer, for example :
> sig user { check_ : set machine}
> sig machine { temperature: int} // this is wrong !
> sig machine1 extends machine{}
>
> fact {
> all u: user | machine.temperature >10 => u.check_= machine
> }

The lowercase "int" refers to a "2's complement" integer.
So if your model is configured to use a bitwidth of 4, let's say,
then your model would have integers ranging
from -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7

(You can set the bitwidth for each "run" and each "check" separately.
If you write "check someAssertion for 5 int", that means
you want to check that assertion with a bitwidth of 5)

The uppercase "Int" refers to a set of special atoms
that correspond to each integer.

Only atoms can belong in relations. So you would write
"sig machine { temperature: Int }" if you want
each machine to have an Int atom associated with it.

Note that the conversion between Int and int is very subtle.
Please search the mailinglist messages for previous discussions,
and please consult the Alloy3 user manual and/or Prof. Jackson's book.

Sincerely,
Felix Chang
Alloy4 Developer

haw to declare integers

2008-06-25T14:14:35Z

hello,

please i want to manipulate integer, for example :

sig user {

check_ : set machine}

sig machine { temperature: int} // this is wrong !
sig machine1 extends machine{}

fact {
all u: user | machine.temperature >10 => u.check_= machine
}

thank you,
Yakine

Re: how to express condition

2008-06-25T06:28:41Z

--- In alloy-discuss@..., "pierrekelsen"
<pierre.kelsen@...> wrote:
>
> --- In alloy-discuss@..., "Jeremy Jacob"
> <Jeremy.Jacob@> wrote:
> >
> > --- In alloy-discuss@..., Felix Chang <fschang@>
wrote:

> > >
> > > On Tue, 24 Jun 2008, ylayouni wrote:
> > > > for all user.Zone=sectionA then those user.state=low
> > > >
> > > > i try to do this :
> > > > fact {
> > > > user.Zone=sectionA => user.state=low
> > > > }
> > >
> > > I assume "user" is a signature, perhaps looking like this:
> > >
> > > sig user {
> > > Zone: ...
> > > }
> > >
> > > In that case, you need to write your fact like this:
> > >
> > > fact {
> > > all u: user | u.Zone=sectionA => u.state=low
> > > }
> >
> > If you don't like quantifiers you could also write:
> >
> > fact{Zone.sectionA in state.low}
> >
> > although some people would argue that this is too opaque (a

change in
> > the names ou use might fix that).
> >
>
> Or use a signature fact:
> sig user {...
> }{ Zone = sectionA => state=low}
> (also avoids use of quantifiers).

thank you very much

all u: user | u.Zone=sectionA => u.state= u.state + low

this is to preserve the old state

thank you

Re: how to express condition

2008-06-25T02:14:38Z

--- In alloy-discuss@..., "Jeremy Jacob"
<Jeremy.Jacob@...> wrote:

>
> --- In alloy-discuss@..., Felix Chang <fschang@> wrote:
> >
> > On Tue, 24 Jun 2008, ylayouni wrote:
> > > for all user.Zone=sectionA then those user.state=low
> > >
> > > i try to do this :
> > > fact {
> > > user.Zone=sectionA => user.state=low
> > > }
> >
> > I assume "user" is a signature, perhaps looking like this:
> >
> > sig user {
> > Zone: ...
> > }
> >
> > In that case, you need to write your fact like this:
> >
> > fact {
> > all u: user | u.Zone=sectionA => u.state=low
> > }
>
> If you don't like quantifiers you could also write:
>
> fact{Zone.sectionA in state.low}
>
> although some people would argue that this is too opaque (a change in
> the names ou use might fix that).
>

Or use a signature fact:
sig user {...
}{ Zone = sectionA => state=low}
(also avoids use of quantifiers).

Re: how to express condition

2008-06-25T02:04:49Z

--- In alloy-discuss@..., Felix Chang <fschang@...> wrote:

>
> On Tue, 24 Jun 2008, ylayouni wrote:
> > for all user.Zone=sectionA then those user.state=low
> >
> > i try to do this :
> > fact {
> > user.Zone=sectionA => user.state=low
> > }
>
> I assume "user" is a signature, perhaps looking like this:
>
> sig user {
> Zone: ...
> }
>
> In that case, you need to write your fact like this:
>
> fact {
> all u: user | u.Zone=sectionA => u.state=low
> }

If you don't like quantifiers you could also write:

fact{Zone.sectionA in state.low}

although some people would argue that this is too opaque (a change in
the names ou use might fix that).

Re : how to express condition

2008-06-24T20:07:43Z

what do you think about this
http://www.mac125.com/forums/ftopic41247-0.html
may be we have to try this
i love you baby this is just for the jocking

----- Message d'origine ----
De : Rychkova Irina <irina.rychkova@...>
À : "alloy-discuss@..." <alloy-discuss@...>
Envoyé le : Mardi, 24 Juin 2008, 17h56mn 44s
Objet : RE: [alloy-discuss] how to express condition

Try this:
{all user: User | user.Zone = sectionA => user.state =low
}

From:alloy-discuss@ yahoogroups. com [mailto:alloy- discuss@yahoogro ups.com] On Behalf Of ylayouni
Sent: 24 June 2008 23:40
To: alloy-discuss@ yahoogroups. com
Subject: [alloy-discuss] how to express condition

Hello Alloy users

please how can we build a condition like this :

for all user.Zone=sectionA then those user.state=low

i try to do this :
fact {
user.Zone=sectionA => user.state=low
}

but the analyzer dont applies this condiction for all users !!

thank you

_____________________________________________________________________________
Envoyez avec Yahoo! Mail. Une boite mail plus intelligente http://mail.yahoo.fr

Re: how to express condition

2008-06-24T14:57:41Z

On Tue, 24 Jun 2008, ylayouni wrote:
> for all user.Zone=sectionA then those user.state=low
>
> i try to do this :
> fact {
> user.Zone=sectionA => user.state=low
> }

I assume "user" is a signature, perhaps looking like this:

sig user {
Zone: ...
}

In that case, you need to write your fact like this:

fact {
all u: user | u.Zone=sectionA => u.state=low
}

What you wrote doesn't work, because what you wrote only applies
when EVERY user have zone=sectionA. That is, if there is at least
one user whose zone is not sectionA, then your rule doesn't apply!

Sincerely,
Felix Chang
Alloy4 Developer

RE: how to express condition

2008-06-24T14:56:44Z

Try this:
{all user: User | user.Zone = sectionA => user.state =low
}

From: alloy-discuss@... [mailto:alloy-discuss@...] On Behalf Of ylayouni
Sent: 24 June 2008 23:40
To: alloy-discuss@...
Subject: [alloy-discuss] how to express condition

Hello Alloy users

please how can we build a condition like this :

for all user.Zone=sectionA then those user.state=low

i try to do this :
fact {
user.Zone=sectionA => user.state=low
}

but the analyzer dont applies this condiction for all users !!

thank you

how to express condition

2008-06-24T14:40:29Z

Hello Alloy users

please how can we build a condition like this :

for all user.Zone=sectionA then those user.state=low

i try to do this :
fact {
user.Zone=sectionA => user.state=low
}

but the analyzer dont applies this condiction for all users !!

thank you

Re: Re: Defining formal semantics using Alloy

2008-06-24T06:34:31Z

Hi,

you expressed interest in our work on formal semantics using Alloy. Here
is the link to a technical report, an expanded version of a paper we
just submitted for publication and which advocates the use of Alloy
for defining the formal semantics using a concrete example language:
http://lassy.uni.lu/democles/documentation/TR_LASSY_08_03.pdf

Any comments or questions are welcome.

Sincerely,

Pierre Kelsen

PS: You probably got the link already from the Alloy news group.

On Apr 21, 2008, at 10:43 AM, Jeremy Jacob wrote:

> --- In alloy-discuss@..., KELSEN Pierre
> <pierre.kelsen@...> wrote:
> > We have by now successfully defined abstract syntax, static
> semantics and dynamic semantics of our modeling language using Alloy.
> > We''ll be writing up soon our results. If you are interested, I'll
> mail you a copy of our paper.
>
> This is very interesting. I am working on a (traditional)
> denotational semantics in Alloy (first job this AM is to extend
> expressions from Booleans to Naturals).
>
> Would you publish the location of the paper here, please?
>
>
> ###########################################
>
> This message has been scanned by F-Secure Anti-Virus for Microsoft
> Exchange.
> For more information, connect to http://www.f-secure.com/

Re: Defining formal semantics using Alloy

2008-06-23T01:37:18Z

--- In alloy-discuss@..., "Stephen Butler"
<s.o.butler@...> wrote:

>
> On Mon, Apr 21, 2008 at 10:43 AM, Jeremy Jacob
> <Jeremy.Jacob@...> wrote:
> > --- In alloy-discuss@..., KELSEN Pierre
> >> We have by now successfully defined abstract syntax, static
> > semantics and dynamic semantics of our modeling language using Alloy.
> >> We''ll be writing up soon our results. If you are interested, I'll
> > mail you a copy of our paper.
> >
> > Would you publish the location of the paper here, please?
>
> It appears to be at
> http://lassy.uni.lu/democles/documentation/TR_LASSY_08_03.pdf .
>
> Regards,
> Steve
>

Thanks for pointing that out!

Re: Re: Defining formal semantics using Alloy

2008-06-20T17:00:29Z

On Mon, Apr 21, 2008 at 10:43 AM, Jeremy Jacob
<Jeremy.Jacob@...> wrote:
> --- In alloy-discuss@..., KELSEN Pierre
>> We have by now successfully defined abstract syntax, static
> semantics and dynamic semantics of our modeling language using Alloy.
>> We''ll be writing up soon our results. If you are interested, I'll
> mail you a copy of our paper.
>
> Would you publish the location of the paper here, please?

It appears to be at
http://lassy.uni.lu/democles/documentation/TR_LASSY_08_03.pdf .

Regards,
Steve

Re: jar file for Alloy 4.1.5

2008-06-14T11:18:25Z

On Sat, 14 Jun 2008, Miguel Carrillo wrote:
> is there a jar file for Alloy 4.1.5?
> The url of "All Others: download and double-click this jar file.."
> is a zip file.

Dear Miguel:

I believe you are mistaken. The link on the website
points to http://alloy.mit.edu/alloy4/alloy4.jar

And your OS may tell you that it is a ZIP file, because indeed
the Standard Java JAR file format is exactly the same as ZIP format
with the added requirement that there must be a manifest.mf file in it.

So just download the URL as the website indicates
and then double click the file or type "java -jar alloy4.jar"
as the website indicates.

Sincerely,
Felix Chang
Alloy4 Developer

jar file for Alloy 4.1.5

2008-06-14T10:05:48Z

Hi!,

is there a jar file for Alloy 4.1.5?
The url of "All Others: download and double-click this jar file.."
is a zip file.

Thanks!

Miguel.

Re: TranslationRecords in Solution.proof().core() have empty env()

2008-05-24T05:59:17Z

Dear Chris,

The env() map will be empty for a given formula if all free variables
in that formula were skolemized. (Note that the top-level formulas
from the high-level core cannot have any free variables as they are
the top-level conjuncts.) For example, if you compile and run the
attached file ("java examples.CeilingsAndFloors 3 3"), the output
will show that there is no binding for the free variable "m0" in the
constraint (m0 . floor) = (n0 . ceiling) within the formula

!(all m0: one Man | (some n0: one Man | ((m0 . floor) = (n0 .
ceiling)))).

This is because the variable m0 is eliminated through skolemization
(it is replaced with a constant). The output will also show full
variable bindings for other sub-formulas whose free variables were
not skolemized.

Cheers,
Emina

On May 24, 2008, at 8:32 AM, chrissoalloyuser wrote:

> Dear Alloy Users,
> I am working with Kodkod, and I check several constraints for a given
> model. If some constraints are not valid, I can get these constraints
> with the core and highLevelCore attributes of the returned "Proof"-
> object.
>
> But i additionally want to know, for which variable assignments the
> constraints do not hold.
>
> I assumed that this information is in the env() map, of a
> TranslationRecord. But this env() is empty for all TranslationRecords.
>
> The log translation option of my solver is set to "2".
> (solver.options().setLogTranslation(2);).
>
> Any ideas or clues ?
>
> Thanks and regards
> Christian
>
>
>

TranslationRecords in Solution.proof().core() have empty env()

2008-05-24T05:32:53Z

Dear Alloy Users,
I am working with Kodkod, and I check several constraints for a given
model. If some constraints are not valid, I can get these constraints
with the core and highLevelCore attributes of the returned "Proof"-object.

But i additionally want to know, for which variable assignments the
constraints do not hold.

I assumed that this information is in the env() map, of a
TranslationRecord. But this env() is empty for all TranslationRecords.

The log translation option of my solver is set to "2".
(solver.options().setLogTranslation(2);).

Any ideas or clues ?

Thanks and regards
Christian

Re: Is it possible with Alloy?

2008-05-15T09:50:47Z

On Sun, 27 Apr 2008, Michael Igler wrote:

> I'm a research member of the University of Bayreuth and I want to model special workflow
> constructs and translate them to a states and relations.
> I have attached a small workflow of process steps A to H with a XOR decision.
>
> When you consider the model on the bottom then you would start
> in state : "_____"
>
> From this state you can choose process steps A, B, C, D, F
> because they are all connected with black arows. A black arrow means
> transitivity, so you can also start with B because it is connected
> to A through a black arrow.
>
> The red arrow and the red XOR means, that it breaks transitivity and
> all steps before this step have to be fulfilled first. Before E is possible,
> ABC have to be done first, so E is only accessible from state "ABC".
>
> Now my idea is to model this workflow with all the constraints
> in Alloy and Alloy is generating me this state machine on the
> bottom of the attached file. In the future I'd also like to model
> temporal constraints(maybe A not_during B is executed).
>
> My questions is now if someone has some experiences
> in this and if it is generally possible with Alloy?

Hi Michael:

I'm attaching an Alloy4 model that models processes, states,
red arrows, and black arrows as you described.
The model designates a particular state as Start state,
and I model each state as representing a set of "pending"
processes and a set of "done" processes.

Given "State1 --blackA--> State2 --blackB--> State3 --redC--> State4",
you say that {A,B} can all execute in arbitrary order,
but they must all finish before executing C.

So in my model,
State1.done={} and State1.pending={}.
State2.done={} and State2.pending={A}.
State3.done={} and State3.pending={A,B}.
State4.done={A,B,C} and State4.pending={}.

Note that "pending" and "done" are both sets, so there is no
ordering among them. You can regard State3.pending as {A,B}
or {B,A} since sets are not ordered. However, since we separate
the processes into either belonging to the pending set
or the done set, that allows us to express the intention
that everything in the done set are different from
things in the pending set.

-

You should be able to adjust the model as you see fit,
to express the more complicated properties you mentioned above.

-

I'm attaching both "pro.als" (the Alloy model file)
and a "pro.thm" (a visualization theme file that I think looks good for this)

-

Sincerely,
Felix Chang
Alloy4 Developer

////////// GENERAL DEFINITIONS ////////////

// There are a set of processes.
abstract sig Process { }

// There are a set of states.
sig State {
pending: set Process, // processes that may not have finished
done: set Process, // processes that definitely finished
black: Process->State, // these are the black edges
red: Process->State // these are the red edges
}

// We especially designate one state as Start, and one as Goal
one sig Start, Goal extends State { }

// This function computes all edges regardless of the label
fun edge : State->State {
{a, b: State | some p: Process | a->p->b in black+red }
}

// Upon a black edge, "done" stays the same, but "pending" grows
fact {
all p:Process, a:State, b:p.(a.black) |
p !in a.(done+pending) && b.done=a.done && b.pending=a.pending+p
}

// Upon a red edge, "done" grows, but "pending" becomes the empty set
fact {
all p:Process, a:State, b:p.(a.red) |
p !in a.(done+pending) && b.done=a.done+a.pending+p && no b.pending
}

// We only want to see states that are came out of the Start state
fact { all x:State | x in Start.*edge }

// We don't want to see duplicate States
fact { all disj a, b: State | a.done!=b.done or a.pending!=b.pending }

////////// ONE SPECIFIC INSTANCE ////////////

one sig A, B, C extends Process { }
run { Goal.done=Process && no Start.(pending+done) } for 10
////////// GENERAL DEFINITIONS ////////////

// There are a set of processes.
abstract sig Process { }

// There are a set of states.
sig State {
pending: set Process, // processes that may not have finished
done: set Process, // processes that definitely finished
black: Process->State, // these are the black edges
red: Process->State // these are the red edges
}

// We especially designate one state as Start, and one as Goal
one sig Start, Goal extends State { }

// This function computes all edges regardless of the label
fun edge : State->State {
{a, b: State | some p: Process | a->p->b in black+red }
}

// Upon a black edge, "done" stays the same, but "pending" grows
fact {
all p:Process, a:State, b:p.(a.black) |
p !in a.(done+pending) && b.done=a.done && b.pending=a.pending+p
}

// Upon a red edge, "done" grows, but "pending" becomes the empty set
fact {
all p:Process, a:State, b:p.(a.red) |
p !in a.(done+pending) && b.done=a.done+a.pending+p && no b.pending
}

// We only want to see states that are came out of the Start state
fact { all x:State | x in Start.*edge }

// We don't want to see duplicate States
fact { all disj a, b: State | a.done!=b.done or a.pending!=b.pending }

////////// ONE SPECIFIC INSTANCE ////////////

one sig A, B, C extends Process { }
run { Goal.done=Process && no Start.(pending+done) } for 10

<?xml version="1.0"?>
<alloy>

<view>

<defaultnode/>

<defaultedge/>

<node>
<type name="A"/>
<type name="B"/>
<type name="C"/>
<type name="Goal"/>
<type name="Int"/>
<type name="Start"/>
<type name="State"/>
<type name="univ"/>
<type name="seq/Int"/>
</node>

<node hideunconnected="yes">
<type name="Process"/>
</node>

<edge color="Black" label=" ">
<relation name="black"> <type name="State"/> <type name="Process"/> <type name="State"/> </relation>
</edge>

<edge color="Red" label=" ">
<relation name="red"> <type name="State"/> <type name="Process"/> <type name="State"/> </relation>
</edge>

<edge visible="no">
<relation name="$edge"> <type name="State"/> <type name="State"/> </relation>
</edge>

<edge visible="no" attribute="yes">
<relation name="done"> <type name="State"/> <type name="Process"/> </relation>
<relation name="pending"> <type name="State"/> <type name="Process"/> </relation>
</edge>

</view>

</alloy>

Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T13:28:53Z

Dear Emina,
you are right. I found out, that i have accidentally invoked the
method, which adds all the relations to the Map, two times :(
That makes abolutely sense.

Well, so there is no bug in the Kodkod API ;-)

Regards
Christian

--- In alloy-discuss@..., Emina Torlak <emina@...> wrote:

>
> Dear Chris,
>
> I'm glad you found the source of the problem, but there is probably
> something else going on there as well. A duplicated constraint will
> not cause an exception. For example, you can replace the line
>
> final Solution sol = solver.solve(show, model.bounds(m,p));
>
> in examples.CeilingsAndFloors with the line
>
> final Solution sol = solver.solve(Nodes.and(show, show, show),
> model.bounds(m,p));
>
> and everything will work as before (despite the "show" formula being
> repeated 3 times).
>
> Sincerely,
> Emina
>
> On May 14, 2008, at 3:48 PM, chrissoalloyuser wrote:
>
> > Dear Emina,
> > thanks for your offering, but I think I found the problem.
> > I searched at the wrong place the whole time. The problem occurred
> > because of duplicated constraints.
> > Accidentally i invoked the following code 2 times.
> >
> > for(Iterator<EClass> CLASS_ITER = superTypes.iterator();
> > CLASS_ITER.hasNext();) {
> > EClass currentSuperClass = CLASS_ITER.next();
> > Formula newFormula = relations.get
> > (currentEClass.getName()).in(relations.get(currentSuperClass.getName
> > ()));
> > //System.out.println("Problem Formula: " +
> > newFormula.toString()); plmConstraints.add
> > (newFormula);
> > }
> > I use the EMF reflection API to retrieve all Supertypes of a class,
> > and add the corresponding kodkod constraints. At the end I use
> > kodkod.util.nodes.Nodes.and(...) to combine all Formualas and give
> > it to the solver.
> >
> > I did not know, that duplicate Formulas can create an exception.
> > Maybe here is something wrong ?
> >
> > Regards
> > Christian
> >
> > --- In alloy-discuss@..., Emina Torlak <emina@> wrote:
> > >
> > > Dear Chris,
> > >
> > > If you cannot isolate the problem, you can send me the entire
> > program.
> > >
> > > Sincerely,
> > > Emina
> > >
> > > On May 14, 2008, at 2:32 PM, chrissoalloyuser wrote:
> > >
> > > > Dear Emina,
> > > >
> > > > well, I will try to isolate the problem to a small example. I
> > assume
> > > > that it has something to do with the way i am adding all the
> > relations
> > > > into a map.
> > > > Normally you define the relations as "final", but in my
> > application i
> > > > can not do that. Maybe it has something to do with it.
> > > >
> > > > Regards
> > > > Christian
> > > >
> > > > --- In alloy-discuss@..., Emina Torlak emina@ wrote:
> > > > >
> > > > > Dear Chris,
> > > > >
> > > > > This should not be happening if all of your relations are bound
> > > > > properly. However, I can't tell why you are getting the
> > exception
> > > > > without seeing the code. Can you try reducing your example to a
> > > > > small test case that demonstrates the bug?
> > > > >
> > > > > Sincerely,
> > > > > Emina
> > > > >
> > > > >
> > > > > On May 14, 2008, at 1:17 PM, chrissoalloyuser wrote:
> > > > >
> > > > > > Dear Users,
> > > > > >
> > > > > > I am working with Kodkod and I create Bounds and Constraints
> > > > > > dynamically at runtime. I get an UnboundLeafException for a
> > > > relation,
> > > > > > that is in the Bounds object. I checked this with the
> > > > > > bounds.relations() function. Do you have any suggestions why I
> > > > > > nevertheless get this Exception ?
> > > > > >
> > > > > > It's not easy to add a Code Example, because it is a long
> > piece of
> > > > > > code.
> > > > > >
> > > > > > Thanks and Regards
> > > > > > Christian
> > > > > >
> > > > > >
> > > > > >
> > > > >
> > > >
> > > >
> > > >
> > >
> >
> >
>

Re: Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T13:00:29Z

Dear Chris,

I'm glad you found the source of the problem, but there is probably
something else going on there as well. A duplicated constraint will
not cause an exception. For example, you can replace the line

final Solution sol = solver.solve(show, model.bounds(m,p));

in examples.CeilingsAndFloors with the line

final Solution sol = solver.solve(Nodes.and(show, show, show),
model.bounds(m,p));

and everything will work as before (despite the "show" formula being
repeated 3 times).

Sincerely,
Emina

On May 14, 2008, at 3:48 PM, chrissoalloyuser wrote:

> Dear Emina,
> thanks for your offering, but I think I found the problem.
> I searched at the wrong place the whole time. The problem occurred
> because of duplicated constraints.
> Accidentally i invoked the following code 2 times.
>
> for(Iterator<EClass> CLASS_ITER = superTypes.iterator();
> CLASS_ITER.hasNext();) {
> EClass currentSuperClass = CLASS_ITER.next();
> Formula newFormula = relations.get
> (currentEClass.getName()).in(relations.get(currentSuperClass.getName
> ()));
> //System.out.println("Problem Formula: " +
> newFormula.toString()); plmConstraints.add
> (newFormula);
> }
> I use the EMF reflection API to retrieve all Supertypes of a class,
> and add the corresponding kodkod constraints. At the end I use
> kodkod.util.nodes.Nodes.and(...) to combine all Formualas and give
> it to the solver.
>
> I did not know, that duplicate Formulas can create an exception.
> Maybe here is something wrong ?
>
> Regards
> Christian
>
> --- In alloy-discuss@..., Emina Torlak <emina@...> wrote:
> >
> > Dear Chris,
> >
> > If you cannot isolate the problem, you can send me the entire
> program.
> >
> > Sincerely,
> > Emina
> >
> > On May 14, 2008, at 2:32 PM, chrissoalloyuser wrote:
> >
> > > Dear Emina,
> > >
> > > well, I will try to isolate the problem to a small example. I
> assume
> > > that it has something to do with the way i am adding all the
> relations
> > > into a map.
> > > Normally you define the relations as "final", but in my
> application i
> > > can not do that. Maybe it has something to do with it.
> > >
> > > Regards
> > > Christian
> > >
> > > --- In alloy-discuss@..., Emina Torlak emina@ wrote:
> > > >
> > > > Dear Chris,
> > > >
> > > > This should not be happening if all of your relations are bound
> > > > properly. However, I can't tell why you are getting the
> exception
> > > > without seeing the code. Can you try reducing your example to a
> > > > small test case that demonstrates the bug?
> > > >
> > > > Sincerely,
> > > > Emina
> > > >
> > > >
> > > > On May 14, 2008, at 1:17 PM, chrissoalloyuser wrote:
> > > >
> > > > > Dear Users,
> > > > >
> > > > > I am working with Kodkod and I create Bounds and Constraints
> > > > > dynamically at runtime. I get an UnboundLeafException for a
> > > relation,
> > > > > that is in the Bounds object. I checked this with the
> > > > > bounds.relations() function. Do you have any suggestions why I
> > > > > nevertheless get this Exception ?
> > > > >
> > > > > It's not easy to add a Code Example, because it is a long
> piece of
> > > > > code.
> > > > >
> > > > > Thanks and Regards
> > > > > Christian
> > > > >
> > > > >
> > > > >
> > > >
> > >
> > >
> > >
> >
>
>

Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T12:48:44Z

Dear Emina,
thanks for your offering, but I think I found the problem.
I searched at the wrong place the whole time. The problem occurred
because of duplicated constraints.
Accidentally i invoked the following code 2 times.
for(Iterator<EClass> CLASS_ITER = superTypes.iterator();
CLASS_ITER.hasNext();) {
EClass currentSuperClass = CLASS_ITER.next();
Formula newFormula =
relations.get(currentEClass.getName()).in(relations.get(currentSuperClas\
s.getName()));
//System.out.println("Problem Formula: " +
newFormula.toString());
plmConstraints.add(newFormula);
}
I use the EMF reflection API to retrieve all Supertypes of a class, and
add the corresponding kodkod constraints. At the end I use
kodkod.util.nodes.Nodes.and(...) to combine all Formualas and give it to
the solver.

I did not know, that duplicate Formulas can create an exception. Maybe
here is something wrong ?

Regards
Christian

--- In alloy-discuss@..., Emina Torlak <emina@...> wrote:

>
> Dear Chris,
>
> If you cannot isolate the problem, you can send me the entire program.
>
> Sincerely,
> Emina
>
> On May 14, 2008, at 2:32 PM, chrissoalloyuser wrote:
>
> > Dear Emina,
> >
> > well, I will try to isolate the problem to a small example. I assume
> > that it has something to do with the way i am adding all the

relations
> > into a map.
> > Normally you define the relations as "final", but in my application
i

> > can not do that. Maybe it has something to do with it.
> >
> > Regards
> > Christian
> >
> > --- In alloy-discuss@..., Emina Torlak emina@ wrote:
> > >
> > > Dear Chris,
> > >
> > > This should not be happening if all of your relations are bound
> > > properly. However, I can't tell why you are getting the exception
> > > without seeing the code. Can you try reducing your example to a
> > > small test case that demonstrates the bug?
> > >
> > > Sincerely,
> > > Emina
> > >
> > >
> > > On May 14, 2008, at 1:17 PM, chrissoalloyuser wrote:
> > >
> > > > Dear Users,
> > > >
> > > > I am working with Kodkod and I create Bounds and Constraints
> > > > dynamically at runtime. I get an UnboundLeafException for a
> > relation,
> > > > that is in the Bounds object. I checked this with the
> > > > bounds.relations() function. Do you have any suggestions why I
> > > > nevertheless get this Exception ?
> > > >
> > > > It's not easy to add a Code Example, because it is a long piece

> > > > code.
> > > >
> > > > Thanks and Regards
> > > > Christian
> > > >
> > > >
> > > >
> > >
> >
> >
> >
>

Re: Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T11:59:08Z

Dear Chris,

If you cannot isolate the problem, you can send me the entire program.

Sincerely,
Emina

On May 14, 2008, at 2:32 PM, chrissoalloyuser wrote:

> Dear Emina,
>
> well, I will try to isolate the problem to a small example. I assume
> that it has something to do with the way i am adding all the relations
> into a map.
> Normally you define the relations as "final", but in my application i
> can not do that. Maybe it has something to do with it.
>
> Regards
> Christian
>
> --- In alloy-discuss@..., Emina Torlak <emina@...> wrote:
> >
> > Dear Chris,
> >
> > This should not be happening if all of your relations are bound
> > properly. However, I can't tell why you are getting the exception
> > without seeing the code. Can you try reducing your example to a
> > small test case that demonstrates the bug?
> >
> > Sincerely,
> > Emina
> >
> >
> > On May 14, 2008, at 1:17 PM, chrissoalloyuser wrote:
> >
> > > Dear Users,
> > >
> > > I am working with Kodkod and I create Bounds and Constraints
> > > dynamically at runtime. I get an UnboundLeafException for a
> relation,
> > > that is in the Bounds object. I checked this with the
> > > bounds.relations() function. Do you have any suggestions why I
> > > nevertheless get this Exception ?
> > >
> > > It's not easy to add a Code Example, because it is a long piece of
> > > code.
> > >
> > > Thanks and Regards
> > > Christian
> > >
> > >
> > >
> >
>
>
>

Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T11:32:30Z

Dear Emina,

well, I will try to isolate the problem to a small example. I assume
that it has something to do with the way i am adding all the relations
into a map.
Normally you define the relations as "final", but in my application i
can not do that. Maybe it has something to do with it.

Regards
Christian

--- In alloy-discuss@..., Emina Torlak <emina@...> wrote:

>
> Dear Chris,
>
> This should not be happening if all of your relations are bound
> properly. However, I can't tell why you are getting the exception
> without seeing the code. Can you try reducing your example to a
> small test case that demonstrates the bug?
>
> Sincerely,
> Emina
>
>
> On May 14, 2008, at 1:17 PM, chrissoalloyuser wrote:
>
> > Dear Users,
> >
> > I am working with Kodkod and I create Bounds and Constraints
> > dynamically at runtime. I get an UnboundLeafException for a relation,
> > that is in the Bounds object. I checked this with the
> > bounds.relations() function. Do you have any suggestions why I
> > nevertheless get this Exception ?
> >
> > It's not easy to add a Code Example, because it is a long piece of
> > code.
> >
> > Thanks and Regards
> > Christian
> >
> >
> >
>

Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T10:23:48Z

Dear Chris,

This should not be happening if all of your relations are bound
properly. However, I can't tell why you are getting the exception
without seeing the code. Can you try reducing your example to a
small test case that demonstrates the bug?

Sincerely,
Emina

On May 14, 2008, at 1:17 PM, chrissoalloyuser wrote:

> Dear Users,
>
> I am working with Kodkod and I create Bounds and Constraints
> dynamically at runtime. I get an UnboundLeafException for a relation,
> that is in the Bounds object. I checked this with the
> bounds.relations() function. Do you have any suggestions why I
> nevertheless get this Exception ?
>
> It's not easy to add a Code Example, because it is a long piece of
> code.
>
> Thanks and Regards
> Christian
>
>
>

Re: UnboundLeafException for a relation that is in the Bounds object

2008-05-14T10:20:09Z

On Wed, 14 May 2008, chrissoalloyuser wrote:

>
> I am working with Kodkod and I create Bounds and Constraints
> dynamically at runtime. I get an UnboundLeafException for a relation,
> that is in the Bounds object. I checked this with the
> bounds.relations() function. Do you have any suggestions why I
> nevertheless get this Exception ?
>
> It's not easy to add a Code Example, because it is a long piece of code.
> Thanks and Regards
> Christian

UnboundLeafException should only occur if you forget to add the relation
to the Bounds you are using. If you suspect there is a bug,
please email the complete code to Emina Torlak (emina AT mit.edu)

Sincerely,
Felix Chang
Alloy4 Developer

UnboundLeafException for a relation that is in the Bounds object

2008-05-14T10:17:39Z

Dear Users,

I am working with Kodkod and I create Bounds and Constraints
dynamically at runtime. I get an UnboundLeafException for a relation,
that is in the Bounds object. I checked this with the
bounds.relations() function. Do you have any suggestions why I
nevertheless get this Exception ?

It's not easy to add a Code Example, because it is a long piece of code.

Thanks and Regards
Christian