/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 44 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 229 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, F) -> Com_1(f2(-(1) + A, -(1) + B, A, B, -(2) + A, F)) :|: A >= 1 && B >= 1
f3(A, B, C, D, E, F) -> Com_1(f2(A, B, C, D, E, F)) :|: TRUE
f2(A, B, C, D, E, F) -> Com_1(f4(A, G, C, D, E, H)) :|: 0 >= B && 0 >= G
f2(A, B, C, D, E, F) -> Com_1(f4(A, B, C, D, E, H)) :|: B >= 1 && 0 >= A

The start-symbols are:[f3_6]


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(1) Koat Proof (FINISHED)
YES(?, ar_0 + 3)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0 - 1, ar_1 - 1, ar_0, ar_1, ar_0 - 2, ar_5)) [ ar_0 >= 1 /\ ar_1 >= 1 ]

		(Comp: ?, Cost: 1)    f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5))

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, g, ar_2, ar_3, ar_4, h)) [ 0 >= ar_1 /\ 0 >= g ]

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, ar_1, ar_2, ar_3, ar_4, g)) [ ar_1 >= 1 /\ 0 >= ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 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, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0 - 1, ar_1 - 1, ar_0, ar_1, ar_0 - 2, ar_5)) [ ar_0 >= 1 /\ ar_1 >= 1 ]

		(Comp: 1, Cost: 1)    f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5))

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, g, ar_2, ar_3, ar_4, h)) [ 0 >= ar_1 /\ 0 >= g ]

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, ar_1, ar_2, ar_3, ar_4, g)) [ ar_1 >= 1 /\ 0 >= ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = 1

	Pol(f3) = 1

	Pol(f4) = 0

	Pol(koat_start) = 1

orients all transitions weakly and the transitions

	f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, g, ar_2, ar_3, ar_4, h)) [ 0 >= ar_1 /\ 0 >= g ]

	f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, ar_1, ar_2, ar_3, ar_4, g)) [ ar_1 >= 1 /\ 0 >= ar_0 ]

strictly and produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0 - 1, ar_1 - 1, ar_0, ar_1, ar_0 - 2, ar_5)) [ ar_0 >= 1 /\ ar_1 >= 1 ]

		(Comp: 1, Cost: 1)    f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5))

		(Comp: 1, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, g, ar_2, ar_3, ar_4, h)) [ 0 >= ar_1 /\ 0 >= g ]

		(Comp: 1, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, ar_1, ar_2, ar_3, ar_4, g)) [ ar_1 >= 1 /\ 0 >= ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f2) = V_1

	Pol(f3) = V_1

	Pol(f4) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transition

	f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0 - 1, ar_1 - 1, ar_0, ar_1, ar_0 - 2, ar_5)) [ ar_0 >= 1 /\ ar_1 >= 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: ar_0, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0 - 1, ar_1 - 1, ar_0, ar_1, ar_0 - 2, ar_5)) [ ar_0 >= 1 /\ ar_1 >= 1 ]

		(Comp: 1, Cost: 1)       f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5))

		(Comp: 1, Cost: 1)       f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, g, ar_2, ar_3, ar_4, h)) [ 0 >= ar_1 /\ 0 >= g ]

		(Comp: 1, Cost: 1)       f2(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f4(ar_0, ar_1, ar_2, ar_3, ar_4, g)) [ ar_1 >= 1 /\ 0 >= ar_0 ]

		(Comp: 1, Cost: 0)       koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound ar_0 + 3



Time: 0.105 sec (SMT: 0.097 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: f3

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

      2: f2 -> f4 : B'=free, F'=free_1, [ 0>=B && 0>=free ], cost: 1

      3: f2 -> f4 : F'=free_2, [ B>=1 && 0>=A ], cost: 1

      1: f3 -> f2 : [], cost: 1



Removed unreachable and leaf rules:

   Start location: f3

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

      1: f3 -> 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, C'=A, D'=B, E'=-2+A, [ A>=1 && B>=1 ], cost: 1



   Accelerated rule 0 with metering function B (after adding A>=B), yielding the new rule 4.

   Accelerated rule 0 with metering function A (after adding A<=B), yielding the new rule 5.

   Removing the simple loops: 0.



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

   Start location: f3

      4: f2 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: B

      5: f2 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: A

      1: f3 -> f2 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f3

      1: f3 -> f2 : [], cost: 1

      6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: 1+B

      7: f3 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: 1+A



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

   Start location: f3

      6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: 1+B

      7: f3 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: 1+A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f3

      6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: 1+B

      7: f3 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: 1+A



Computing asymptotic complexity for rule 6

   Simplified the guard:

      6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ B>=1 && A>=B ], cost: 1+B

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / n

      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+B

   Rule guard:  [ B>=1 && A>=B ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)