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


--------------------------------------------------------------------------------


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 11 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 144 ms]
(4) BOUNDS(n^1, INF)


----------------------------------------

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

The start-symbols are:[f2_3]


----------------------------------------

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



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(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)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f300) = 1

	Pol(f1) = 0

	Pol(f2) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 ]

strictly and produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f300) = V_1 - V_2

	Pol(f1) = V_1 - V_2

	Pol(f2) = V_1 - V_2

	Pol(koat_start) = V_1 - V_2

orients all transitions weakly and the transition

	f300(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: Ar_0 + Ar_1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)              f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 1)              f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))

		(Comp: 1, Cost: 0)              koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0 + Ar_1 + 2



Time: 0.026 sec (SMT: 0.023 sec)


----------------------------------------

(2)
BOUNDS(1, n^1)

----------------------------------------

(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f2

      0: f300 -> f300 : B'=1+B, [ A>=1+B ], cost: 1

      1: f300 -> f1 : C'=free, [ B>=A ], cost: 1

      2: f2 -> f300 : [], cost: 1



Checking for constant complexity:

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

      2: f2 -> f300 : [], cost: 1



Removed unreachable and leaf rules:

   Start location: f2

      0: f300 -> f300 : B'=1+B, [ A>=1+B ], cost: 1

      2: f2 -> f300 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f300 -> f300 : B'=1+B, [ A>=1+B ], cost: 1



   Accelerated rule 0 with metering function A-B, yielding the new rule 3.

   Removing the simple loops: 0.



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

   Start location: f2

      3: f300 -> f300 : B'=A, [ A>=1+B ], cost: A-B

      2: f2 -> f300 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f2

      2: f2 -> f300 : [], cost: 1

      4: f2 -> f300 : B'=A, [ A>=1+B ], cost: 1+A-B



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

   Start location: f2

      4: f2 -> f300 : B'=A, [ A>=1+B ], cost: 1+A-B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f2

      4: f2 -> f300 : B'=A, [ A>=1+B ], cost: 1+A-B



Computing asymptotic complexity for rule 4

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {A==0,B==-n}

      resulting limit problem:

      [solved]



   Solution:

      A / 0

      B / -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-B

   Rule guard:  [ A>=1+B ]



WORST_CASE(Omega(n^1),?)


----------------------------------------

(4)
BOUNDS(n^1, INF)