/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 28 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 330 ms]
(4) BOUNDS(n^1, INF)


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

The start-symbols are:[start_3]


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



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



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

	eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ]

We thus obtain the following problem:

2:	T:

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(start(Ar_0, Ar_1, Ar_2)) [ 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, Ar_2) -> Com_1(eval(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

		(Comp: ?, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(eval) = V_1 + V_2

	Pol(start) = V_1 + V_2

	Pol(koat_start) = V_1 + V_2

orients all transitions weakly and the transitions

	eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

	eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: Ar_0 + Ar_1, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ 0 >= Ar_0 /\ Ar_1 >= 1 ]

		(Comp: Ar_0 + Ar_1, Cost: 1)    eval(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 + Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 1)              start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)              koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 2*Ar_0 + 2*Ar_1 + 1



Time: 0.053 sec (SMT: 0.045 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+B>=1+C && C>=0 && A>=1 ], cost: 1

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

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

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



Checking for constant complexity:

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

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



Removed rules with unsatisfiable guard:

   Start location: start

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

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

      3: 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+B>=1+C && C>=0 && A>=1 ], cost: 1

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



   Found no metering function for rule 0.

   Accelerated rule 1 with metering function -C+A+B, yielding the new rule 4.

   Removing the simple loops: 1.



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

   Start location: start

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

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

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

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



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

   Start location: start

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

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



Computing asymptotic complexity for rule 6

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==n,A==-n,B==3*n}

      resulting limit problem:

      [solved]



   Solution:

      C / n

      A / -n

      B / 3*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-C+A+B

   Rule guard:  [ A+B>=1+C && C>=0 && 0>=A ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)