/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 92 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 637 ms]
(4) BOUNDS(n^1, INF)


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

The start-symbols are:[f2_4]


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(1) Koat Proof (FINISHED)
YES(?, 3*Ar_0 + 3*Ar_1 + 2)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3))

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_2, Ar_3)) [ Fresh_2 >= 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_1, Ar_3)) [ 0 >= Fresh_1 + 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0 - 1, Ar_1, 0, Ar_3)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3))

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_2, Ar_3)) [ Fresh_2 >= 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_1, Ar_3)) [ 0 >= Fresh_1 + 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0 - 1, Ar_1, 0, Ar_3)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = 1

	Pol(f300) = 1

	Pol(f1) = 0

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3))

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_2, Ar_3)) [ Fresh_2 >= 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_1, Ar_3)) [ 0 >= Fresh_1 + 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0 - 1, Ar_1, 0, Ar_3)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = V_1 - V_2

	Pol(f300) = V_1 - V_2

	Pol(f1) = V_1 - V_2

	Pol(koat_start) = V_1 - V_2

orients all transitions weakly and the transitions

	f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_2, Ar_3)) [ Fresh_2 >= 1 /\ Ar_0 >= Ar_1 + 1 ]

	f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_1, Ar_3)) [ 0 >= Fresh_1 + 1 /\ Ar_0 >= Ar_1 + 1 ]

	f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0 - 1, Ar_1, 0, Ar_3)) [ Ar_0 >= Ar_1 + 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)              f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3))

		(Comp: Ar_0 + Ar_1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_2, Ar_3)) [ Fresh_2 >= 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: Ar_0 + Ar_1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0, Ar_1 + 1, Fresh_1, Ar_3)) [ 0 >= Fresh_1 + 1 /\ Ar_0 >= Ar_1 + 1 ]

		(Comp: Ar_0 + Ar_1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f300(Ar_0 - 1, Ar_1, 0, Ar_3)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)              f300(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 0)              koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 3*Ar_0 + 3*Ar_1 + 2



Time: 0.112 sec (SMT: 0.096 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: f2

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

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

      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1

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

      4: f300 -> f1 : D'=free_2, [ B>=A ], cost: 1



Checking for constant complexity:

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

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



Removed unreachable and leaf rules:

   Start location: f2

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

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

      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

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

      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1

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



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

   Accelerated rule 2 with metering function A-B, yielding the new rule 6.

   Accelerated rule 3 with metering function A-B, yielding the new rule 7.

   Removing the simple loops: 1 2 3.



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

   Start location: f2

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

      5: f300 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: A-B

      6: f300 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: A-B

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



Chained accelerated rules (with incoming rules):

   Start location: f2

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

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

      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B

     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B



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

   Start location: f2

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

      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B

     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f2

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

      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B

     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B



Computing asymptotic complexity for rule 8

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {free==1,A==0,B==-n}

      resulting limit problem:

      [solved]



   Solution:

      free / 1

      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:  [ free>=1 && A>=1+B ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)