/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


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WORST_CASE(Omega(n^2), O(n^2))
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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^2, n^2).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 43 ms]
(2) BOUNDS(1, n^2)
(3) Loat Proof [FINISHED, 349 ms]
(4) BOUNDS(n^2, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f1(A, B) -> Com_1(f3(A, A)) :|: A >= 1
f3(A, B) -> Com_1(f3(A, B - 1)) :|: B >= 1
f3(A, B) -> Com_1(f3(A - 1, A - 1)) :|: 0 >= B && A >= 2

The start-symbols are:[f1_2]


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



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_0)) [ Ar_0 >= 1 ]

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

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(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: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_0)) [ Ar_0 >= 1 ]

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

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f1) = V_1

	Pol(f3) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transition

	f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\ Ar_0 >= 2 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)       f1(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_0)) [ Ar_0 >= 1 ]

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

		(Comp: Ar_0, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f3) = V_2

and size complexities

	S("koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-0) = Ar_0

	S("koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-1) = Ar_1

	S("f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\\ Ar_0 >= 2 ]", 0-0) = Ar_0

	S("f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\\ Ar_0 >= 2 ]", 0-1) = Ar_0

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

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

	S("f1(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_0)) [ Ar_0 >= 1 ]", 0-0) = Ar_0

	S("f1(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_0)) [ Ar_0 >= 1 ]", 0-1) = Ar_0

orients the transitions

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

weakly and the transition

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

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)                f1(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_0)) [ Ar_0 >= 1 ]

		(Comp: Ar_0^2 + Ar_0, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ]

		(Comp: Ar_0, Cost: 1)             f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 - 1, Ar_0 - 1)) [ 0 >= Ar_1 /\ Ar_0 >= 2 ]

		(Comp: 1, Cost: 0)                koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0^2 + 2*Ar_0 + 1



Time: 0.059 sec (SMT: 0.051 sec)


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(2)
BOUNDS(1, n^2)

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f1

      0: f1 -> f3 : B'=A, [ A>=1 ], cost: 1

      1: f3 -> f3 : B'=-1+B, [ B>=1 ], cost: 1

      2: f3 -> f3 : A'=-1+A, B'=-1+A, [ 0>=B && A>=2 ], cost: 1



Checking for constant complexity:

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

      0: f1 -> f3 : B'=A, [ A>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f3 -> f3 : B'=-1+B, [ B>=1 ], cost: 1

      2: f3 -> f3 : A'=-1+A, B'=-1+A, [ 0>=B && A>=2 ], cost: 1



   Accelerated rule 1 with metering function B, yielding the new rule 3.

   Found no metering function for rule 2.

   Nested simple loops 2 (outer loop) and 3 (inner loop) with metering function -1+A, resulting in the new rules: 4, 5.

   Removing the simple loops: 1 2.



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

   Start location: f1

      0: f1 -> f3 : B'=A, [ A>=1 ], cost: 1

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

      4: f3 -> f3 : A'=1, B'=1, [ B>=1 && A>=2 ], cost: -3/2-1/2*(-1+A)^2+(-1+A)*A+3/2*A

      5: f3 -> f3 : A'=1, B'=1, [ 0>=B && -1+A>=2 ], cost: -2+(-1+A)*(-2+A)+3/2*A-1/2*(-2+A)^2



Chained accelerated rules (with incoming rules):

   Start location: f1

      0: f1 -> f3 : B'=A, [ A>=1 ], cost: 1

      6: f1 -> f3 : B'=0, [ A>=1 ], cost: 1+A

      7: f1 -> f3 : A'=1, B'=1, [ A>=2 ], cost: -1/2-1/2*(-1+A)^2+(-1+A)*A+3/2*A



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

   Start location: f1

      6: f1 -> f3 : B'=0, [ A>=1 ], cost: 1+A

      7: f1 -> f3 : A'=1, B'=1, [ A>=2 ], cost: -1/2-1/2*(-1+A)^2+(-1+A)*A+3/2*A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f1

      6: f1 -> f3 : B'=0, [ A>=1 ], cost: 1+A

      7: f1 -> f3 : A'=1, B'=1, [ A>=2 ], cost: -1/2-1/2*(-1+A)^2+(-1+A)*A+3/2*A



Computing asymptotic complexity for rule 6

   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).



Computing asymptotic complexity for rule 7

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -1+3/2*A+1/2*A^2 (+), -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+1/2*n^2+3/2*n has complexity: Poly(n^2)



Found new complexity Poly(n^2).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^2)

   Cpx degree:  2

   Solved cost: -1+1/2*n^2+3/2*n

   Rule cost:   -1/2-1/2*(-1+A)^2+(-1+A)*A+3/2*A

   Rule guard:  [ A>=2 ]



WORST_CASE(Omega(n^2),?)


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(4)
BOUNDS(n^2, INF)