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# Re: [tlaplus] Re: Simpler method of making multiple changes to a function in a single step

 Apologies for the typo – of course it should have been [N -> SUBSET R] and I should really have checked the definitions before sending them out. For the record, the spec below works as intended. I obtain 42 distinct states (for 80053 states computed overall) when declaring N and R as symmetry sets of 3, resp. 4 elements.Regards,Stephan----------------------------- MODULE Allocation -----------------------------EXTENDS Integers, FiniteSets, TLCCONSTANT R, NVARIABLE registerIsBalanced(reg) == \A m,n \in N : Cardinality(reg[m]) - Cardinality(reg[n]) \in {-1,0,1}BalancedAllocations == { reg \in [N -> SUBSET R] : IsBalanced(reg) }Init ==  register = [n \in N |-> {}]Rebalance ==  /\ register' \in BalancedAllocations  /\ PrintT(register')Spec == Init /\ [][Rebalance]_register=============================================================================On 16 Jan 2019, at 20:57, Jack Vanlightly wrote:Hi,Sets are fine and I have modified the spec accordingly, also I removed the LET \intersect as you suggested and it works. About the CHOOSE, in theory it shouldn't matter that it is deterministic, the important requirement is even distribution. But, I need to consider that point more deeply.Regarding your much scaled down version, it is almost there but not quite.BalancedAllocations == { reg \in [N -> R] : IsBalanced(reg) }fails with Attempted to apply the operator overridden by the Java methodpublic static tlc2.value.IntValue tlc2.module.FiniteSets.Cardinality(tlc2.value.Value),but it produced the following error:Attemtped to compute cardinality of the value"r1"If I modify it to be:BalancedAllocations == { reg \in [N -> SUBSET R] : IsBalanced(reg) }Then it doesn't fail but happily makes no assignments.If I modify it to be:BalancedAllocations == { reg \in [N -> {R}] : IsBalanced(reg) }then it assigns all resources to all nodes.So it needs a tweak to make sure it assigns all resources, but without assigning the same resource twice which is an invariant of the algorithm.I will have a crack at that tomorrow.ThanksJackOn Wednesday, January 16, 2019 at 5:35:42 PM UTC+1, Jack Vanlightly wrote:Hi,I have a set of nodes N, a set of resources R and a function "register" that maps N to R. The algorithm is a resource allocation algorithm that must assign the resources of R evenly across the nodes N.So if N is { "n1", "n2", "n3"} and R is {"r1", "r2", "r3", "r4" } then once allocation has taken place a valid value for register would be:[n1 |-> <<"r4", "r1">>, n2 |-> <<"r2">>, n3 |-> <<"r3">>]I want to set the values of register in a single step and I have managed it, though the formula seems overly complex and I wonder if there is a simpler way of doing that would help me also gain more insight into TLA+.I have isolated the allocation logic into a toy spec as follows:EXTENDS Integers, FiniteSets, Sequences, TLCCONSTANT R, NVARIABLE registerInit ==  register = [n \in N |-> << >>]HasMinResources(counts, nd) ==  \A x \in N : counts[nd] <= counts[x] Allocate ==   LET newRegister == [n \in N |-> << >>]       counts == [n \in N |-> 0]       al[pendingRes \in SUBSET R, assignCount \in [N -> Nat]] ==            LET n == CHOOSE nd \in N : HasMinResources(assignCount, nd)                r == LET int == R \intersect pendingRes                      IN CHOOSE x \in int : TRUE             IN                IF Cardinality(pendingRes) = 0 THEN newRegister                ELSE                     LET remaining == pendingRes \ { r }                        newAssignCount == [assignCount EXCEPT ![n] = @ + 1]                    IN [al[remaining, newAssignCount] EXCEPT ![n] = Append(@, r)]   IN al[R, counts]   Rebalance ==  /\ register' = Allocate  /\ PrintT(register')(* ignore the spec, I just wanted to run the Rebalance action once *)Spec == Init /\ [][Rebalance]_registerNotes:- I made Allocate recursive as that is the only way I could figure out making all the changes to register in a single step.- I did the intersect so that I could use CHOOSE. Else it complained that is was an unbounded CHOOSE so I figured if I did an intersect with R then it would be interpretted as bounded.Any insights or suggestions would be great.ThanksJack -- You received this message because you are subscribed to the Google Groups "tlaplus" group. To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+u...@xxxxxxxxxxxxxxxx. To post to this group, send email to tla...@xxxxxxxxxxxxxxxx. Visit this group at https://groups.google.com/group/tlaplus. For more options, visit https://groups.google.com/d/optout.