/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 34 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 420 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, E) -> Com_1(f1(A, F, C, D, E)) :|: 1 >= A
f2(A, B, C, D, E) -> Com_1(f300(A, B, C, D, E)) :|: A >= 2 && C >= 2
f300(A, B, C, D, E) -> Com_1(f1(A, F, C, D, E)) :|: D + 1 >= 0
f300(A, B, C, D, E) -> Com_1(f300(A, B, C, 1 + D, 1 + E)) :|: E >= 0 && 0 >= 2 + D
f300(A, B, C, D, E) -> Com_1(f300(A, B, C, 1 + D, 1 + E)) :|: 0 >= 2 + E && 0 >= 2 + D

The start-symbols are:[f2_5]


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(1) Koat Proof (FINISHED)
YES(?, 2*Ar_3 + 3)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 2 /\ Ar_2 >= 2 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_0, Ar_2, Ar_3, Ar_4)) [ Ar_3 + 1 >= 0 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ Ar_4 >= 0 /\ 0 >= Ar_3 + 2 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ 0 >= Ar_4 + 2 /\ 0 >= Ar_3 + 2 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 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, Ar_4) -> Com_1(f1(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 2 /\ Ar_2 >= 2 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_0, Ar_2, Ar_3, Ar_4)) [ Ar_3 + 1 >= 0 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ Ar_4 >= 0 /\ 0 >= Ar_3 + 2 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ 0 >= Ar_4 + 2 /\ 0 >= Ar_3 + 2 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = 1

	Pol(f1) = 0

	Pol(f300) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_0, Ar_2, Ar_3, Ar_4)) [ Ar_3 + 1 >= 0 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 2 /\ Ar_2 >= 2 ]

		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_0, Ar_2, Ar_3, Ar_4)) [ Ar_3 + 1 >= 0 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ Ar_4 >= 0 /\ 0 >= Ar_3 + 2 ]

		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ 0 >= Ar_4 + 2 /\ 0 >= Ar_3 + 2 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = -V_4

	Pol(f1) = -V_4

	Pol(f300) = -V_4

	Pol(koat_start) = -V_4

orients all transitions weakly and the transitions

	f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ Ar_4 >= 0 /\ 0 >= Ar_3 + 2 ]

	f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ 0 >= Ar_4 + 2 /\ 0 >= Ar_3 + 2 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)       f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_0 ]

		(Comp: 1, Cost: 1)       f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 2 /\ Ar_2 >= 2 ]

		(Comp: 1, Cost: 1)       f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Fresh_0, Ar_2, Ar_3, Ar_4)) [ Ar_3 + 1 >= 0 ]

		(Comp: Ar_3, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ Ar_4 >= 0 /\ 0 >= Ar_3 + 2 ]

		(Comp: Ar_3, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4 + 1)) [ 0 >= Ar_4 + 2 /\ 0 >= Ar_3 + 2 ]

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 2*Ar_3 + 3



Time: 0.092 sec (SMT: 0.075 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 -> f1 : B'=free, [ 1>=A ], cost: 1

      1: f2 -> f300 : [ A>=2 && C>=2 ], cost: 1

      2: f300 -> f1 : B'=free_1, [ 1+D>=0 ], cost: 1

      3: f300 -> f300 : D'=1+D, E'=1+E, [ E>=0 && 0>=2+D ], cost: 1

      4: f300 -> f300 : D'=1+D, E'=1+E, [ 0>=2+E && 0>=2+D ], cost: 1



Checking for constant complexity:

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

      0: f2 -> f1 : B'=free, [ 1>=A ], cost: 1



Removed unreachable and leaf rules:

   Start location: f2

      1: f2 -> f300 : [ A>=2 && C>=2 ], cost: 1

      3: f300 -> f300 : D'=1+D, E'=1+E, [ E>=0 && 0>=2+D ], cost: 1

      4: f300 -> f300 : D'=1+D, E'=1+E, [ 0>=2+E && 0>=2+D ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      3: f300 -> f300 : D'=1+D, E'=1+E, [ E>=0 && 0>=2+D ], cost: 1

      4: f300 -> f300 : D'=1+D, E'=1+E, [ 0>=2+E && 0>=2+D ], cost: 1



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

   Accelerated rule 4 with metering function -1-D (after adding D>=E), yielding the new rule 6.

   Accelerated rule 4 with metering function -1-E (after adding D<=E), yielding the new rule 7.

   Removing the simple loops: 3 4.



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

   Start location: f2

      1: f2 -> f300 : [ A>=2 && C>=2 ], cost: 1

      5: f300 -> f300 : D'=-1, E'=-1-D+E, [ E>=0 && 0>=2+D ], cost: -1-D

      6: f300 -> f300 : D'=-1, E'=-1-D+E, [ 0>=2+E && 0>=2+D && D>=E ], cost: -1-D

      7: f300 -> f300 : D'=-1+D-E, E'=-1, [ 0>=2+E && 0>=2+D && D<=E ], cost: -1-E



Chained accelerated rules (with incoming rules):

   Start location: f2

      1: f2 -> f300 : [ A>=2 && C>=2 ], cost: 1

      8: f2 -> f300 : D'=-1, E'=-1-D+E, [ A>=2 && C>=2 && E>=0 && 0>=2+D ], cost: -D

      9: f2 -> f300 : D'=-1, E'=-1-D+E, [ A>=2 && C>=2 && 0>=2+E && 0>=2+D && D>=E ], cost: -D

     10: f2 -> f300 : D'=-1+D-E, E'=-1, [ A>=2 && C>=2 && 0>=2+E && 0>=2+D && D<=E ], cost: -E



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

   Start location: f2

      8: f2 -> f300 : D'=-1, E'=-1-D+E, [ A>=2 && C>=2 && E>=0 && 0>=2+D ], cost: -D

      9: f2 -> f300 : D'=-1, E'=-1-D+E, [ A>=2 && C>=2 && 0>=2+E && 0>=2+D && D>=E ], cost: -D

     10: f2 -> f300 : D'=-1+D-E, E'=-1, [ A>=2 && C>=2 && 0>=2+E && 0>=2+D && D<=E ], cost: -E



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f2

      8: f2 -> f300 : D'=-1, E'=-1-D+E, [ A>=2 && C>=2 && E>=0 && 0>=2+D ], cost: -D

      9: f2 -> f300 : D'=-1, E'=-1-D+E, [ A>=2 && C>=2 && 0>=2+E && 0>=2+D && D>=E ], cost: -D

     10: f2 -> f300 : D'=-1+D-E, E'=-1, [ A>=2 && C>=2 && 0>=2+E && 0>=2+D && D<=E ], cost: -E



Computing asymptotic complexity for rule 8

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==n,D==-n,A==n,E==n}

      resulting limit problem:

      [solved]



   Solution:

      C / n

      D / -n

      A / n

      E / n

   Resulting cost 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: n

   Rule cost:   -D

   Rule guard:  [ A>=2 && C>=2 && E>=0 && 0>=2+D ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)