/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 19 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 327 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B, C) -> Com_1(f2(1 + A, 1 + B, C)) :|: 1 >= A
f2(A, B, C) -> Com_1(f2(1 + A, 1 + B, C)) :|: 2 >= B && A >= 2
f2(A, B, C) -> Com_1(f300(A, B, D)) :|: B >= 3 && A >= 2
f1(A, B, C) -> Com_1(f2(A, B, C)) :|: TRUE

The start-symbols are:[f1_3]


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(1) Koat Proof (FINISHED)
YES(?, Ar_0 + Ar_1 + 8)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 2 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 3 /\ Ar_0 >= 2 ]

		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 2 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 3 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = 1

	Pol(f300) = 0

	Pol(f1) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 3 /\ Ar_0 >= 2 ]

strictly and produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 2 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 3 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = -V_1 + 3

	Pol(f300) = -V_1

	Pol(f1) = -V_1 + 3

	Pol(koat_start) = -V_1 + 3

orients all transitions weakly and the transition

	f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_0 ]

strictly and produces the following problem:

4:	T:

		(Comp: Ar_0 + 3, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)           f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 2 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)           f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 3 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)           f1(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)           koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = -V_2 + 3

	Pol(f300) = -V_2

	Pol(f1) = -V_2 + 3

	Pol(koat_start) = -V_2 + 3

orients all transitions weakly and the transition

	f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 2 >= Ar_1 /\ Ar_0 >= 2 ]

strictly and produces the following problem:

5:	T:

		(Comp: Ar_0 + 3, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_0 ]

		(Comp: Ar_1 + 3, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 2 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)           f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 3 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 1)           f1(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)           koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0 + Ar_1 + 8



Time: 0.075 sec (SMT: 0.063 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: f1

      0: f2 -> f2 : A'=1+A, B'=1+B, [ 1>=A ], cost: 1

      1: f2 -> f2 : A'=1+A, B'=1+B, [ 2>=B && A>=2 ], cost: 1

      2: f2 -> f300 : C'=free, [ B>=3 && A>=2 ], cost: 1

      3: f1 -> f2 : [], cost: 1



Checking for constant complexity:

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

      3: f1 -> f2 : [], cost: 1



Removed unreachable and leaf rules:

   Start location: f1

      0: f2 -> f2 : A'=1+A, B'=1+B, [ 1>=A ], cost: 1

      1: f2 -> f2 : A'=1+A, B'=1+B, [ 2>=B && A>=2 ], cost: 1

      3: f1 -> f2 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f2 -> f2 : A'=1+A, B'=1+B, [ 1>=A ], cost: 1

      1: f2 -> f2 : A'=1+A, B'=1+B, [ 2>=B && A>=2 ], cost: 1



   Accelerated rule 0 with metering function 2-A, yielding the new rule 4.

   Accelerated rule 1 with metering function 3-B, yielding the new rule 5.

   Removing the simple loops: 0 1.



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

   Start location: f1

      4: f2 -> f2 : A'=2, B'=2-A+B, [ 1>=A ], cost: 2-A

      5: f2 -> f2 : A'=3+A-B, B'=3, [ 2>=B && A>=2 ], cost: 3-B

      3: f1 -> f2 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f1

      3: f1 -> f2 : [], cost: 1

      6: f1 -> f2 : A'=2, B'=2-A+B, [ 1>=A ], cost: 3-A

      7: f1 -> f2 : A'=3+A-B, B'=3, [ 2>=B && A>=2 ], cost: 4-B



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

   Start location: f1

      6: f1 -> f2 : A'=2, B'=2-A+B, [ 1>=A ], cost: 3-A

      7: f1 -> f2 : A'=3+A-B, B'=3, [ 2>=B && A>=2 ], cost: 4-B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f1

      6: f1 -> f2 : A'=2, B'=2-A+B, [ 1>=A ], cost: 3-A

      7: f1 -> f2 : A'=3+A-B, B'=3, [ 2>=B && A>=2 ], cost: 4-B



Computing asymptotic complexity for rule 6

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      2-A (+/+!), 3-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 3+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: 3+n

   Rule cost:   3-A

   Rule guard:  [ 1>=A ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)