我是合金新手。我正在尝试为具有 512 个状态的模型找到解决方案。但是它的内存不足。我将内存和堆栈设置为最大级别,但这还不够。有没有其他方法可以用来增加合金使用的内存?感谢您的时间和帮助。非常感谢,法蒂耶
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2 回答
1
很难知道从哪里开始。看起来好像您正在编写合金模型,就好像您期望它成为模型检查器一样。但是 Alloy 的重点是允许您分析状态具有复杂结构的系统,并以关系逻辑编写约束。将低级模型直接编码到合金中不会走得太远;对于那种事情,你最好使用模型检查器。
于 2013-10-26T23:39:30.247 回答
0
module threeprocesses
abstract sig boolean {
}
one sig true,false extends boolean{}
sig state {
e1: boolean,
t1: boolean,
ready1: boolean,
e2: boolean,
t2: boolean,
ready2: boolean,
e3: boolean,
t3: boolean,
ready3: boolean
}
sig relation {
lambda : state -> one Int,
next1 : state -> state
}
pred LS (s : state) {
(((s.t1 =s.t3) and (s.t2 =s.t1) and (s.t3 =s.t2))
or ((s.t1 != s.t3) and (s.t2 !=s.t1) and (s.t3 =s.t2))
or ((s.t1 != s.t3) and (s.t2 =s.t1) and (s.t3 !=s.t2))) and
((s.e1 =s.e3) or (s.e2 !=s.e1) or (s.e3 !=s.e2))
}
pred show (r:relation) {
all s : state |
LS [s] implies LS [s.(r.next1)]
all s : state |
(not (LS [s])) implies not (s =s.(r.next1))
all s : state |
(not (LS [s])) implies (all s2 : (s.(r.next1)) | s2. (r.lambda) > s.(r.lambda))
all s : state,s2 : state |
((s.t1 = s2.t1) and (s.e1 = s2.e1) and (s.ready1 = s2.ready1) and (s.e3 = s2.e3)
and (s.t3 = s2.t3)) implies
( (((s2.(r.next1)).ready1)= ((s.(r.next1)).ready1)) and (((s2.(r.next1)).e1)= ((s.
(r.next1)).e1)) and
(((s2.(r.next1)).t1)= ((s.(r.next1)).t1)) )
all s : state,s2 : state |
((s.t2 = s2.t2) and (s.e2 = s2.e2) and (s.ready2 = s2.ready2) and (s.e1 = s2.e1)
and (s.t1 = s2.t1)) implies
( (((s2.(r.next1)).ready2)= ((s.(r.next1)).ready2)) and (((s2.(r.next1)).e2)= ((s.
(r.next1)).e2)) and
(((s2.(r.next1)).t2)= ((s.(r.next1)).t2)) )
all s : state,s2 : state |
((s.t3 = s2.t3) and (s.e3 = s2.e3) and (s.ready3 = s2.ready3) and (s.e2 = s2.e2)
and (s.t2 = s2.t2)) implies
( (((s2.(r.next1)).ready3)= ((s.(r.next1)).ready3)) and (((s2.(r.next1)).e3)= ((s.
(r.next1)).e3)) and
(((s2.(r.next1)).t3)= ((s.(r.next1)).t3)) )
all s : state |
(not ( (s.e1 = ((s.(r.next1)).e1)) and (s.t1 = ((s.(r.next1)).t1)) and (s.ready1
= ((s.(r.next1)).ready1)) ) ) implies
(s.e1 = s.e3)
all s : state |
(not ( (s.e2 = ((s.(r.next1)).e2)) and (s.t2 = ((s.(r.next1)).t2)) and (s.ready2
= ((s.(r.next1)).ready2)) ) ) implies
(not (s.e2 = s.e1))
all s : state |
(not ( (s.e3 = ((s.(r.next1)).e3)) and (s.t3 = ((s.(r.next1)).t3)) and (s.ready3
= ((s.(r.next1)).ready3)) ) ) implies
(not (s.e3 = s.e2))
all s : state ,s2:state |
(s != s2) implies (not ((s.e1 = s2.e1) and (s.e2 = s2.e2) and (s.e3 = s2.e3) and
(s.t1 = s2.t1) and (s.t2 = s2.t2) and (s.t3 = s2.t3) and
(s.ready1 = s2.ready1) and (s.ready2 = s2.ready2) and (s.ready3 = s2.ready3)))
}
run show for 3 but 1 relation, exactly 512 state
于 2013-10-24T17:55:28.893 回答