Alloy :Scope of Integers - alloy

I'm novice at Alloy (4.2) Sat Solver
I have a problem with the scope of Integers
I want to guarentee that the age of a given person is always more than 18 years
Exemple:
sig Customer {age : Int }
fact f { all c:Customer | c.age > 18 }
pred NewRule { all c:Customer | c.age = 15 }
run NewRule for exactly 1 Customer
The problem is that the Integers in Alloy take only values between [-8,+7]
If I change the fact & mak c.age >0
and change the pred & make c.age =7
so the system is concistant
thinks

Related

How to sum more than two numbers in Alloy Analyzer?

I am trying to sum all the numbers in a set in Alloy.
For instance, in the signature abc, I want the value to be the sum of a.value + b.value + c.value, which is 4+1+3=8.
However, if I use "+", it gives me the union set and not the sum.
PS. I know there is the "plus" (as I used it in sig sumab), but this only allows me to sum two values.
Thanks
open util/integer
sig a{value: Int}
{
value = 4
}
sig b{value: Int}
{
value = 1
}
sig c{value: Int}
{
value = 3
}
sig abc{value: set Int}
{
value = a.value + b.value + c.value
}
sig sumab{
value : Int
}
{
value = plus[a.value, b.value]
}
pred add{}
run add for 4 int, exactly 1 sumab, exactly 1 a, exactly 1 b, exactly 1 c, exactly 1 abc
Note: I wrote this in pseudo-code, it may help to get to an answer:
fun plusN [setInt : set de Int] : Int { // function "plusN" should take a set of integers "setInt", return an integer
if #setInt = 2 //if only two numbers in set, sum them
then plus[max setInt , min setInt]
else // if more than 2, use recursion
plusN [max setInt , plusN[{setInt - max setInt}]]
}
Note 2. The function sum may seem to be a good idea, but if I sum 1+1+1=1, the result will be 1 intead of 3 as the only number in the set is 1.
Alloy comes with a built-in sum function, so just do sum[value].
As mentioned in this answer, Alloy has a sum keyword (different than the sum function) which can be used.
The general format is:
sum e: <set> | <expression involving e>
Variable sig values
Here's a simple example that sums the number of sales during a day for a restaurant's three meals. In particular, note the use of the sum keyword in fun sum_sales.
one sig Restaurant { total_sales: Int }
abstract sig Meal { sales: Int }
one sig Breakfast, Lunch, Dinner in Meal {}
// Each meal only falls in one category
fact { disj[Breakfast, Lunch, Dinner] }
// Every meal is in some category
fact { Breakfast + Lunch + Dinner = Meal }
// Keep the numbers small because the max alloy int is 7
fact { all m: Meal | m.sales <= 4 }
// Negative sales don't make sense
fact { all m: Meal | m.sales >= 0 }
fun sum_sales: Int { sum m: Meal | m.sales }
fact { Restaurant.total_sales = sum_sales }
This works even when all meals have the same number of sales (1 + 1 + 1 = 3), as shown in this check.
check { (all m: Meal | m.sales = 1) => Restaurant.total_sales = 3 }
Here are some other ways to play around with the example.
check {
{
// One meal with three sales
#{ m: Meal | m.sales = 3 } = 1
// Two meals with one sale
#{ m: Meal | m.sales = 1 } = 2
} => Restaurant.total_sales = 5
}
run { Restaurant.total_sales = 5 }
Constant sig values
Another way you might want to use this is to have the value associated with each type of Meal be constant, but allow the number of Meals to vary. You can model this with a relation mapping each meal type to a number of sales as follows.
one sig Restaurant { total_sales: Int }
abstract sig Meal {}
lone sig Breakfast, Lunch, Dinner in Meal {}
// Each meal only falls in one category
fact { disj[Breakfast, Lunch, Dinner] }
// Every meal is in some category
fact { Breakfast + Lunch + Dinner = Meal }
fun meal_sales: Meal -> Int {
Breakfast -> 2
+ Lunch -> 2
+ Dinner -> 3
}
fun sum_sales: Int { sum m: Meal | m.meal_sales }
fact { Restaurant.total_sales = sum_sales }
check { #Meal = 3 => Restaurant.total_sales = 6 }
run { Restaurant.total_sales = 5 }

Predicate not consistent

I have to realize an application which permits to reserve a seat in a store. There are no sintax error, but I don't understand why terminal reply me "predicate not consistent". Can u help me?
This is my code:
open util/integer
sig Email{}
sig CF{}
sig Time{
hour: one Int,
minute:one Int,
second: one Int
}
abstract sig RegisteredUser{
email: one Email
}
sig User extends RegisteredUser{
cf: one CF
}
sig SM extends RegisteredUser{
cf: one CF,
store: one Store
}
sig Location{}
sig Date{}
abstract sig Status {}
//only one of this can be true
one sig Waiting extends Status {}
one sig Expired extends Status{}
one sig Pending extends Status{}
sig Ticket{
owner:one User,
date: one Date,
time: one Time,
status : one Status,
}
sig Visit{
owner:one User,
date: one Date,
time: one Time,
duration:one Time,
status : one Status,
products: some Product,
category: some Category,
}
//tickets are assumed with a default duration time of 15 min and entrance one for time, so 4*8=32 visit per day
sig TicketQueue {
max_visit:one Int,
ticket: some Ticket,
manager: one SM
}
sig VisitQueue{
category: some Category,
visit: some Visit,
manager: one SM
}
sig Store{
max_simultaneous: one Int,
location: one Location,
visitqueue: one VisitQueue,
ticketqueue: one TicketQueue,
product_category: some Category
}
sig Category{
simultaneous_seats: one Int,
}
sig Product{
category: one Category
}
// Constraints
// Registration data for the system are unique(Unique username and Fiscal Code)
fact registrationDataUniqueness {
no disjoint u1, u2: RegisteredUser | u1.email = u2.email
no disjoint u1,u2: User | u1.cf = u2.cf
no disjoint s1,s2: SM | s1.cf = s2.cf
no disjoint s:SM, u:User | s.cf=u.cf
}
//Ticket and Visit can have only one of status' values defined before
fact requestConsistency {
all s: Status | (s = Waiting && s != Expired && s !=Pending ) || (s != Waiting && s = Expired && s != Pending) || (s !=Waiting && s != Expired && s = Pending)
}
//SM can manage only one Store
fact StoreUniqueness{
no disjoint s1,s2: SM |s1.store=s2.store
}
// the same ticket or visit cannot be of two or more different user
fact UserUniqueness{
no disjoint t1,t2: Ticket | t1.owner=t2.owner
no disjoint v1,v2: Visit | v1.owner=v2.owner
}
//different products can not be of the same category
fact CategoryUniqueness{
no disjoint p1,p2: Product | p1.category=p2.category
}
//different stores can't have the same ticket or visit queue
fact QueueUniqueness{
no disjoint s1,s2: Store | s1.ticketqueue=s2.ticketqueue
no disjoint s1,s2: Store | s1.visitqueue=s2.visitqueue
}
//different tickets cannot have the same time
fact TimeUniqueness{
no disjoint t1,t2: Ticket | t1.time = t2.time
}
//for semplicity,in the visit/ticket queue, we consider "active" only visit/ticket with Waiting status
fact StatusQueue{
all tq:TicketQueue | tq.ticket.status=Waiting
all vq:VisitQueue | vq.visit.status=Waiting
}
//one ticket/visit can not belong to two or more different queue
fact ReservationUniqueness{
no disjoint tq1,tq2:TicketQueue | tq1.ticket=tq2.ticket
no disjoint vq1,vq2:VisitQueue | vq1.visit=vq2.visit
}
//SM is unique for a specific store, and its related ticket/visit queue
fact SMUniqueness{
no disjoint s1,s2: Store | s1.ticketqueue.manager=s2.ticketqueue.manager
and s1.visitqueue.manager=s2.visitqueue.manager
and s1.ticketqueue.manager=s2.visitqueue.manager
and s1.visitqueue.manager=s2.ticketqueue.manager
}
//different tickets/visits associated with the same user have different time
fact UniqueTimeUser {
all u: User, t1, t2: Ticket, v1,v2:Visit | ((u in t1.owner) and (u in t2.owner) and (u in v1.owner) and (u in v2.owner)
and (t1 != t2) and (v1 !=v2))
implies (t1.time != t2.time)
and (v1.time != v2.time)
}
//different ticket and visit associated with the same user have different time
fact UniqueTimeUser2 {
all u: User, t: Ticket, v:Visit | ((u in t.owner) and (u in v.owner) )
implies (v.time != t.time)
}
//Maximum number of visit with the same time for category
fact MaxTime{
all v:Visit, c: Category |(( c in v.category)) implies #v.time< c.simultaneous_seats
}
//Considering meaningfull integer value
fact PossibleValues{
all c: Category | c.simultaneous_seats>1 and c.simultaneous_seats<5
all s: Store | s.max_simultaneous >0 and s.max_simultaneous<50
all t: Ticket | t.time.hour>7 and t.time.hour<21 and ((t.time.minute=15) || (t.time.minute=30) || (t.time.minute=45)) and t.time.second=0
all v: Visit | v.time.hour>7 and v.time.hour<21 and ((v.time.minute=15) || (v.time.minute=30) || (v.time.minute=45)) and v.time.second=0 and v.duration.hour=0 and v.duration.minute>0 and v.duration.minute=<30 and v.duration.second=0
all tq: TicketQueue | tq.max_visit=32
}
pred addTicket[ t :Ticket,ti,ti':Time, tq, tq':TicketQueue]{
//precondition
ti' not in Ticket.time
#tq.ticket< tq.max_visit //seats available
//postconditions
tq'.manager= tq.manager
tq'.ticket= tq.ticket+t
t in tq'.ticket
#tq'.ticket< tq.max_visit
all t': Ticket | t' in tq.ticket implies t' in tq'.ticket
}
pred show{
#User = 4
#Store = 2
#SM = 2
#Category = 3
#TicketQueue = 2
#VisitQueue= 2
#Category = 5
some v, v': Visit | v.time != v'.time
#Ticket = 5
#Visit=5
}
run show for 4
This is the complete response of Alloy terminal:
Executing "Run show for 4"
Solver=sat4j Bitwidth=4 MaxSeq=4 SkolemDepth=1 Symmetry=20
15098 vars. 840 primary vars. 17697 clauses. 28ms.
No instance found. Predicate may be inconsistent. 1ms.
Have you run unsat core (using the MiniSAT with Unsat Core solver)? That might give you a hint. Another thought: I would simplify the model first and then only add features as you need them. I'm not sure that you really need to split the times into hours, minutes and seconds. That adds a lot of solver complexity, perhaps needlessly.
General observations:
Some inconsistences in your show predicate:
#Category = 3
#Category = 5
Your scope is definitely too small. You can't expect 5 Visit elements if you have a general scope of 4.
I'd rewrite your show predicate and run command as follows:
pred show{
some v, v': Visit | v.time != v'.time
}
run show for 10 but 7 Int, exactly 4 User, exactly 2 Store, exactly 2 SM, exactly 3 Category, exactly 2 TicketQueue, exactly 2 VisitQueue, exactly 5 Ticket, exactly 5 Visit
Notice the 7 Int.
It means that your model will contain int type atoms that represent all the integers you can express with a bitwidth of 7 (interval is [-64,63]).
As Daniel wrote, i'd also advise to stay as abstract as possible when you model in Alloy and thus to prefer concepts over quantitative values.
I concur with the previous advice of simplifying and would add to use check statements to test as you go. I'm not an expert with Alloy, so I take it slow and test early and often.
One thing that sticks out to me immediately is that your "show" predicate seems to be specifying multiplicities for some signatures that are greater than those allowed by the run statement ("for 4"). I've never specified multiplicity in a predicate that way, so I don't know how Alloy handles that. Offhand, those seem like contradictory constraints.
Also, I figured this out the hard way, but Alloy counts instances (atoms) of extended signatures as instances (atoms) of the parent(s). So, be aware of that.

Alloy Analyzer element comparision from set

Some background: my project is to make a compiler that compiles from a c-like language to Alloy. The input language, that has c-like syntax, must support contracts. For now, I am trying to implement if statements that support pre and post condition statements, similar to the following:
int x=2
if_preCondition(x>0)
if(x == 2){
x = x + 1
}
if_postCondtion(x>0)
The problem is that I am a bit confused with the results of Alloy.
sig Number{
arg1: Int,
}
fun addOneConditional (x : Int) : Number{
{ v : Number |
v.arg1 = ((x = 2 ) => x.add[1] else x)
}
}
assert conditionalSome {
all n: Number| (n.arg1 = 2 ) => (some field: addOneConditional[n.arg1] | { field.arg1 = n.arg1.add[1] })
}
assert conditionalAll {
all n: Number| (n.arg1 = 2 ) => (all field: addOneConditional[n.arg1] | { field.arg1 = n.arg1.add[1] })
}
check conditionalSome
check conditionalAll
In the above example, conditionalAll does not generate any Counterexample. However, conditionalSomegenerates Counterexamples. If I understand all and some quantifiers correctly then there is a mistake. Because from mathematical logic we have Ɐx expr(x) => ∃x expr(x) ( i.e. If expression expr(x) is true for all values of x then there exist a single x for which expr(x) is true)
The first thing is that you need to model your pre-, post- and operations. Functions are terrible for that because they cannot not return something that indicates failure. You, therefore, need to model the operation as a predicate. The value of a predicate indicates if the pre/post are satisfied, the arguments and return values can be modeled as parameters.
This is as far as I can understand your operation:
pred add[ x : Int, x' : Int ] {
x > 0 -- pre condition
x = 2 =>
x'=x.plus[1]
else
x'=x
x' > 0 -- post condition
}
Alloy has no writable variables (Electrum has) so you need to model the before and after state with a prime (') character.
We can now use this predicate to calculate the set of solutions to your problem:
fun solutions : set Int {
{ x' : Int | some x : Int | add[ x,x' ] }
}
We create a set with integers for which we have a result. The prime character is nothing special in Alloy, it is only a convention for the post-state. I am abusing it slightly here.
This is more than enough Alloy source to make mistakes so let's test this.
run { some solutions }
If you run this then you'll see in the Txt view:
skolem $solutions={1, 3, 4, 5, 6, 7}
This is as expected. The add operation only works for positive numbers. Check. If the input is 2, the result is 3. Ergo, 2 can never be a solution. Check.
I admit, I am slight confused by what you're doing in your asserts. I've tried to replicate them faithfully, although I've removed unnecessary things, at least I think we're unnecessary. First your some case. Your code was doing an all but then selecting on 2. So removed the outer quantification and hardcoded 2.
check somen {
some x' : solutions | 2.plus[1] = x'
}
This indeed does not give us any counterexample. Since solutions was {1, 3, 4, 5, 6, 7}, 2+1=3 is in the set, i.e. the some condition is satisfied.
check alln {
all x' : solutions | 2.plus[1] = x'
}
However, not all solutions have 3 as the answer. If you check this, I get the following counter-example:
skolem $alln_x'={7}
skolem $solutions={1, 3, 4, 5, 6, 7}
Conclusion. Daniel Jackson advises not to learn Alloy with Ints. Looking at your Number class you took him literally: you still base your problem on Ints. What he meant is not use Int, don't hide them under the carpet in a field. I understand where Daniel is coming from but Ints are very attractive since we're so familiar with them. However, if you use Ints, let them at least use their full glory and don't hide them.
Hope this helps.
And the whole model:
pred add[ x : Int, x' : Int ] {
x > 0 -- pre condition
x = 2 =>
x'=x.plus[1]
else
x'=x
x' > 0 -- post condition
}
fun solutions : set Int { { x' : Int | some x : Int | add[ x,x' ] } }
run { some solutions }
check somen { some x' : solutions | x' = 3 }
check alln { all x' : solutions | x' = 3 }

Lock Challenge in Alloy

I would like to solve the following lock challenge using Alloy.
My main issue is how to model the integers representing the digit keys.
I created a quick draft:
sig Digit, Position{}
sig Lock {
d: Digit one -> lone Position
}
run {} for exactly 1 Lock, exactly 3 Position, 10 Digit
In this context, could you please:
tell me if Alloy seems to you suitable to solve this kind of problem?
give me some pointers regarding the way I could model the key digits (without using Ints)?
Thank you.
My frame of this puzzle is:
enum Digit { N0,N1,N2,N3,N4,N5,N6,N7,N8,N9 }
one sig Code {a,b,c:Digit}
pred hint(h1,h2,h3:Digit, matched,wellPlaced:Int) {
matched = #(XXXX) // fix XXXX
wellPlaced = #(XXXX) // fix XXXX
}
fact {
hint[N6,N8,N2, 1,1]
hint[N6,N1,N4, 1,0]
hint[N2,N0,N6, 2,0]
hint[N7,N3,N8, 0,0]
hint[N7,N8,N0, 1,0]
}
run {}
A simple way to get started, you do not always need sig's. The solution found is probably not the intended solution but that is because the requirements are ambiguous, took a shortcut.
pred lock[ a,b,c : Int ] {
a=6 || b=8 || c= 2
a in 1+4 || b in 6+4 || c in 6+1
a in 0+6 || b in 2+6 || c in 2+0
a != 7 && b != 3 && c != 8
a = 7 || b=8 || c=0
}
run lock for 6 int
Look in the Text view for the answer.
upate we had a discussion on the Alloy list and I'd like to amend my solution with a more readable version:
let sq[a,b,c] = 0->a + 1->b + 2->c
let digit = { n : Int | n>=0 and n <10 }
fun correct[ lck : seq digit, a, b, c : digit ] : Int { # (Int.lck & (a+b+c)) }
fun wellPlaced[ lck : seq digit, a, b, c : digit ] : Int { # (lck & sq[a,b,c]) }
pred lock[ a, b, c : digit ] {
let lck = sq[a,b,c] {
1 = correct[ lck, 6,8,2] and 1 = wellPlaced[ lck, 6,8,2]
1 = correct[ lck, 6,1,4] and 0 = wellPlaced[ lck, 6,1,4]
2 = correct[ lck, 2,0,6] and 0 = wellPlaced[ lck, 2,0,6]
0 = correct[ lck, 7,3,8]
1 = correct[ lck, 7,8,0] and 0 = wellPlaced[ lck, 7,8,0]
}
}
run lock for 6 Int
When you think solve complete, let's examine whether the solution is generic.
Here is another lock.
If you can’t solve this in same form, your solution may not enough.
Hint1: (1,2,3) - Nothing is correct.
Hint2: (4,5,6) - Nothing is correct.
Hint3: (7,8,9) - One number is correct but wrong placed.
Hint4: (9,0,0) - All numbers are correct, with one well placed.
Yes, I think Alloy is suitable for this kind of problem.
Regarding digits, you don't need integers at all: in fact, it is a bit irrelevant for this particular purpose if they are digits or any set of 10 different identifiers (no arithmetic is performed with them). You can use singleton signatures to declare the digits, all extending signature Digit, which should be marked as abstract. Something like:
abstract sig Digit {}
one sig Zero, One, ..., Nine extends Digit {}
A similar strategy can be used to declare the three different positions of the lock. And btw since you have exactly one lock you can also declare Lock as singleton signature.
I like the Nomura solution on this page. I made a slight modification of the predicate and the fact to solve.
enum Digit { N0,N1,N2,N3,N4,N5,N6,N7,N8,N9 }
one sig Code {a,b,c: Digit}
pred hint(code: Code, d1,d2,d3: Digit, correct, wellPlaced:Int) {
correct = #((code.a + code.b + code.c)&(d1 + d2 + d3))
wellPlaced = #((0->code.a + 1->code.b + 2->code.c)&(0->d1 + 1->d2 + 2->d3))
}
fact {
some code: Code |
hint[code, N6,N8,N2, 1,1] and
hint[code, N6,N1,N4, 1,0] and
hint[code, N2,N0,N6, 2,0] and
hint[code, N7,N3,N8, 0,0] and
hint[code, N7,N8,N0, 1,0]
}
run {}
Update (2020-12-29):
The new puzzle presented by Nomura (https://stackoverflow.com/a/61022419/5005552) demonstrates a weakness in the original solution: it does not account for multiple uses of a digit within a code. A modification to the expression for "correct" fixes this. Intersect each guessed digit with the union of the digits from the passed code and sum them for the true cardinality. I encapsulated the matching in a function, which will return 0 or 1 for each digit.
enum Digit {N0,N1,N2,N3,N4,N5,N6,N7,N8,N9}
let sequence[a,b,c] = 0->a + 1->b + 2->c
one sig Code {c1, c2, c3: Digit}
fun match[code: Code, d: Digit]: Int { #((code.c1 + code.c2 + code.c3) & d) }
pred hint(code: Code, d1,d2,d3: Digit, correct, wellPlaced:Int) {
// The intersection of each guessed digit with the code (unordered) tells us
// whether any of the digits match each other and how many
correct = match[code,d1].plus[match[code,d2]].plus[match[code,d3]]
// The intersection of the sequences of digits (ordered) tells us whether
// any of the digits are correct AND in the right place in the sequence
wellPlaced = #(sequence[code.c1,code.c2,code.c3] & sequence[d1, d2, d3])
}
pred originalLock {
some code: Code |
hint[code, N6,N8,N2, 1,1] and
hint[code, N6,N1,N4, 1,0] and
hint[code, N2,N0,N6, 2,0] and
hint[code, N7,N3,N8, 0,0] and
hint[code, N7,N8,N0, 1,0]
}
pred newLock {
some code: Code |
hint[code, N1,N2,N3, 0,0] and
hint[code, N4,N5,N6, 0,0] and
hint[code, N7,N8,N9, 1,0] and
hint[code, N9,N0,N0, 3,1]
}
run originalLock
run newLock
run test {some code: Code | hint[code, N9,N0,N0, 3,1]}

How to express Nothing else in the set?

[Update] I spent a lot of time studying #Hovercouch's fantastic solution (see his solution below). I took his solution, along with Peter Krien's insights and wrote up a summary: 3 ways to model the set of non-negative even numbers. I welcome your comments.
I am trying to create an Alloy model that defines a set of integers. I want to constrain the set to the integers 0, 2, 4, ...
I want to use a "generative" approach to defining the set:
0 is in the set.
If i is in the set, then i+2 is in the set.
Nothing else is in the set.
I am struggling with the last one - nothing else is in the set. How do I express that?
Below is the Alloy model that I created.
one sig PositiveEven {
elements: set Int
}
pred generate_set_members {
0 in PositiveEven.elements
all i: Int | i in PositiveEven.elements => i.plus[2] in PositiveEven.elements
// Nothing else is in the set - How to express this?
}
The simplest way to do this would be to create a relationship that maps each number N to N+2, and then take the reflexive-transitive closure of that relationship over 0.
one sig PositiveEven {
elements: set Int
}
one sig Generator {
rel: Int -> Int
} {
all i: Int | i.rel = i.next.next
}
pred generate_set_members {
PositiveEven.elements = 0.*(Generator.rel)
}
assert only_positive_elements {
generate_set_members =>
all i: Int | i in PositiveEven.elements <=> i >= 0 and i.rem[2] = 0
}
Note that you cannot use i.plus[2] instead of i.next.next, because Alloy integers overflow to negative.
What do you think of this:
let iset[min,max,step] = { i : Int |
i>= min
and i<max
and i.minus[min].div[step].mul[step]
= i.minus[min] }
pred show[ s : set Int ] {
iset[ 0, 10, 2 ] = s
}
run show for 0 but 8 int
The visualiser does not show the Int types so look in the Tree or Text view.

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