/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 12 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 623 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A, B) -> Com_1(eval(A - 1, B)) :|: A >= 1
eval(A, B) -> Com_1(eval(A - 1, B)) :|: B >= 1 && A >= 1
eval(A, B) -> Com_1(eval(A, B - 1)) :|: A >= 1 && 0 >= A && B >= 1
eval(A, B) -> Com_1(eval(A, B - 1)) :|: B >= 1 && 0 >= A
eval(A, B) -> Com_1(eval(A, B)) :|: A >= 1 && 0 >= A && 0 >= B
eval(A, B) -> Com_1(eval(A, B)) :|: B >= 1 && 0 >= A && 0 >= B
start(A, B) -> Com_1(eval(A, B)) :|: TRUE

The start-symbols are:[start_2]


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(1) Koat Proof (FINISHED)
YES(?, Ar_1 + 2*Ar_0 + 1)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_0 >= 1 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1)) [ Ar_0 >= 1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(start(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transitions from problem 1:

	eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_0 >= 1 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

	eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1)) [ Ar_0 >= 1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ]

	eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ]

We thus obtain the following problem:

2:	T:

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(start(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(start(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(eval) = V_2

	Pol(start) = V_2

	Pol(koat_start) = V_2

orients all transitions weakly and the transition

	eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ]

strictly and produces the following problem:

4:	T:

		(Comp: Ar_1, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ]

		(Comp: ?, Cost: 1)       eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)       eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 1)       start(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(start(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(eval) = V_1

	Pol(start) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transitions

	eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_1 >= 1 /\ Ar_0 >= 1 ]

	eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ]

strictly and produces the following problem:

5:	T:

		(Comp: Ar_1, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ]

		(Comp: Ar_0, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: Ar_0, Cost: 1)    eval(Ar_0, Ar_1) -> Com_1(eval(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 1)       start(Ar_0, Ar_1) -> Com_1(eval(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(start(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_1 + 2*Ar_0 + 1



Time: 0.085 sec (SMT: 0.076 sec)


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(2)
BOUNDS(1, n^1)

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(3) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: start

      0: eval -> eval : A'=-1+A, [ A>=1 ], cost: 1

      1: eval -> eval : A'=-1+A, [ B>=1 && A>=1 ], cost: 1

      2: eval -> eval : B'=-1+B, [ A>=1 && 0>=A && B>=1 ], cost: 1

      3: eval -> eval : B'=-1+B, [ B>=1 && 0>=A ], cost: 1

      4: eval -> eval : [ A>=1 && 0>=A && 0>=B ], cost: 1

      5: eval -> eval : [ B>=1 && 0>=A && 0>=B ], cost: 1

      6: start -> eval : [], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      6: start -> eval : [], cost: 1



Removed rules with unsatisfiable guard:

   Start location: start

      0: eval -> eval : A'=-1+A, [ A>=1 ], cost: 1

      1: eval -> eval : A'=-1+A, [ B>=1 && A>=1 ], cost: 1

      3: eval -> eval : B'=-1+B, [ B>=1 && 0>=A ], cost: 1

      6: start -> eval : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: eval -> eval : A'=-1+A, [ A>=1 ], cost: 1

      1: eval -> eval : A'=-1+A, [ B>=1 && A>=1 ], cost: 1

      3: eval -> eval : B'=-1+B, [ B>=1 && 0>=A ], cost: 1



   Accelerated rule 0 with metering function A, yielding the new rule 7.

   Accelerated rule 1 with metering function A, yielding the new rule 8.

   Accelerated rule 3 with metering function B, yielding the new rule 9.

   Removing the simple loops: 0 1 3.



Accelerated all simple loops using metering functions (where possible):

   Start location: start

      7: eval -> eval : A'=0, [ A>=1 ], cost: A

      8: eval -> eval : A'=0, [ B>=1 && A>=1 ], cost: A

      9: eval -> eval : B'=0, [ B>=1 && 0>=A ], cost: B

      6: start -> eval : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      6: start -> eval : [], cost: 1

     10: start -> eval : A'=0, [ A>=1 ], cost: 1+A

     11: start -> eval : A'=0, [ B>=1 && A>=1 ], cost: 1+A

     12: start -> eval : B'=0, [ B>=1 && 0>=A ], cost: 1+B



Removed unreachable locations (and leaf rules with constant cost):

   Start location: start

     10: start -> eval : A'=0, [ A>=1 ], cost: 1+A

     11: start -> eval : A'=0, [ B>=1 && A>=1 ], cost: 1+A

     12: start -> eval : B'=0, [ B>=1 && 0>=A ], cost: 1+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

     10: start -> eval : A'=0, [ A>=1 ], cost: 1+A

     11: start -> eval : A'=0, [ B>=1 && A>=1 ], cost: 1+A

     12: start -> eval : B'=0, [ B>=1 && 0>=A ], cost: 1+B



Computing asymptotic complexity for rule 10

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      A (+/+!), 1+A (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {A==n}

      resulting limit problem:

      [solved]



   Solution:

      A / n

   Resulting cost 1+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^1)

   Cpx degree:  1

   Solved cost: 1+n

   Rule cost:   1+A

   Rule guard:  [ A>=1 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)