/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 10 ms]
(2) BOUNDS(1, n^2)
(3) Loat Proof [FINISHED, 429 ms]
(4) BOUNDS(n^2, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f4(A, B, C) -> Com_1(f4(A, B + 1, C)) :|: A >= B + 1
f4(A, B, C) -> Com_1(f4(A + 1, 0, C)) :|: C >= A + 2 && B >= A
f0(A, B, C) -> Com_1(f4(0, 0, C)) :|: C >= 1

The start-symbols are:[f0_3]


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(1) Koat Proof (FINISHED)
YES(?, Ar_2^2 + Ar_2 + 1)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\ Ar_1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(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)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f4) = -V_1 + V_3

	Pol(f0) = V_3

	Pol(koat_start) = V_3

orients all transitions weakly and the transition

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\ Ar_1 >= Ar_0 ]

strictly and produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: Ar_2, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f4) = V_1 - V_2

and size complexities

	S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-0) = Ar_0

	S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-1) = Ar_1

	S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-2) = Ar_2

	S("f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]", 0-0) = 0

	S("f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]", 0-1) = 0

	S("f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]", 0-2) = Ar_2

	S("f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\\ Ar_1 >= Ar_0 ]", 0-0) = Ar_2

	S("f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\\ Ar_1 >= Ar_0 ]", 0-1) = 0

	S("f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\\ Ar_1 >= Ar_0 ]", 0-2) = Ar_2

	S("f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]", 0-0) = Ar_2

	S("f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]", 0-1) = ?

	S("f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]", 0-2) = Ar_2

orients the transitions

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

weakly and the transition

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: Ar_2^2, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ]

		(Comp: Ar_2, Cost: 1)      f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 + 1, 0, Ar_2)) [ Ar_2 >= Ar_0 + 2 /\ Ar_1 >= Ar_0 ]

		(Comp: 1, Cost: 1)         f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, 0, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: 1, Cost: 0)         koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_2^2 + Ar_2 + 1



Time: 0.065 sec (SMT: 0.057 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: f0

      0: f4 -> f4 : B'=1+B, [ A>=1+B ], cost: 1

      1: f4 -> f4 : A'=1+A, B'=0, [ C>=2+A && B>=A ], cost: 1

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1



Checking for constant complexity:

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

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f4 -> f4 : B'=1+B, [ A>=1+B ], cost: 1

      1: f4 -> f4 : A'=1+A, B'=0, [ C>=2+A && B>=A ], cost: 1



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

   Found no metering function for rule 1.

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

   Removing the simple loops: 0 1.



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

   Start location: f0

      3: f4 -> f4 : B'=A, [ A>=1+B ], cost: A-B

      4: f4 -> f4 : A'=-1+C, B'=-1+C, [ C>=2+A && B>=A && 1+A>=1 ], cost: -3/2+1/2*(-1+C-A)^2+3/2*C-3/2*A+A*(-1+C-A)

      5: f4 -> f4 : A'=-1+C, B'=-1+C, [ A>=1+B && C>=2+A && 1+A>=1 ], cost: -3/2+1/2*(-1+C-A)^2+3/2*C-1/2*A-B+A*(-1+C-A)

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1

      6: f0 -> f4 : A'=-1+C, B'=-1+C, [ C>=2 ], cost: -1/2+3/2*C+1/2*(-1+C)^2



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

   Start location: f0

      6: f0 -> f4 : A'=-1+C, B'=-1+C, [ C>=2 ], cost: -1/2+3/2*C+1/2*(-1+C)^2



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      6: f0 -> f4 : A'=-1+C, B'=-1+C, [ C>=2 ], cost: -1/2+3/2*C+1/2*(-1+C)^2



Computing asymptotic complexity for rule 6

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      1/2*C+1/2*C^2 (+), -1+C (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      C / n

   Resulting cost 1/2*n+1/2*n^2 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: 1/2*n+1/2*n^2

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

   Rule guard:  [ C>=2 ]



WORST_CASE(Omega(n^2),?)


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(4)
BOUNDS(n^2, INF)