/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 7 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 137 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B) -> Com_1(f3(A, 1)) :|: 0 >= A
f0(A, B) -> Com_1(f2(A, 0)) :|: TRUE
f2(A, B) -> Com_1(f2(A - 1, B)) :|: A >= 1

The start-symbols are:[f0_2]


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



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, 1)) [ 0 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Ar_0, 0))

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

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 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) -> Com_1(f3(Ar_0, 1)) [ 0 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Ar_0, 0))

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

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = 1

	Pol(f3) = 0

	Pol(f0) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, 1)) [ 0 >= Ar_0 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, 1)) [ 0 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Ar_0, 0))

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

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = V_1

	Pol(f3) = V_1

	Pol(f0) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transition

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

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)       f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, 1)) [ 0 >= Ar_0 ]

		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1) -> Com_1(f2(Ar_0, 0))

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

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0 + 2



Time: 0.033 sec (SMT: 0.030 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: f0

      0: f2 -> f3 : B'=1, [ 0>=A ], cost: 1

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

      1: f0 -> f2 : B'=0, [], cost: 1



Checking for constant complexity:

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

      1: f0 -> f2 : B'=0, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

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

      1: f0 -> f2 : B'=0, [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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



   Accelerated rule 2 with metering function A, yielding the new rule 3.

   Removing the simple loops: 2.



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

   Start location: f0

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

      1: f0 -> f2 : B'=0, [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      1: f0 -> f2 : B'=0, [], cost: 1

      4: f0 -> f2 : A'=0, B'=0, [ A>=1 ], cost: 1+A



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

   Start location: f0

      4: f0 -> f2 : A'=0, B'=0, [ A>=1 ], cost: 1+A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      4: f0 -> f2 : A'=0, B'=0, [ A>=1 ], cost: 1+A



Computing asymptotic complexity for rule 4

   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)