/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 29 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 145 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A, B, C) -> Com_1(eval(A, B + 1, C + 1)) :|: A >= B + C + 1
start(A, B, C) -> Com_1(eval(A, B, C)) :|: TRUE

The start-symbols are:[start_3]


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(1) Koat Proof (FINISHED)
YES(?, Ar_0 + Ar_1 + Ar_2 + 1)



Initial complexity problem:

1:	T:

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

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

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

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

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

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(eval) = V_1 - V_2 - V_3

	Pol(start) = V_1 - V_2 - V_3

	Pol(koat_start) = V_1 - V_2 - V_3

orients all transitions weakly and the transition

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

strictly and produces the following problem:

3:	T:

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

		(Comp: 1, Cost: 1)                     start(Ar_0, Ar_1, Ar_2) -> Com_1(eval(Ar_0, Ar_1, Ar_2))

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

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0 + Ar_1 + Ar_2 + 1



Time: 0.027 sec (SMT: 0.024 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: start

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

      1: start -> eval : [], cost: 1



Checking for constant complexity:

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

      1: start -> eval : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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



   Accelerated rule 0 with metering function meter (where 2*meter==-C+A-B), yielding the new rule 2.

   Removing the simple loops: 0.



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

   Start location: start

      2: eval -> eval : B'=meter+B, C'=meter+C, [ A>=1+C+B && 2*meter==-C+A-B && meter>=1 ], cost: meter

      1: start -> eval : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      1: start -> eval : [], cost: 1

      3: start -> eval : B'=meter+B, C'=meter+C, [ A>=1+C+B && 2*meter==-C+A-B && meter>=1 ], cost: 1+meter



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

   Start location: start

      3: start -> eval : B'=meter+B, C'=meter+C, [ A>=1+C+B && 2*meter==-C+A-B && meter>=1 ], cost: 1+meter



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      3: start -> eval : B'=meter+B, C'=meter+C, [ A>=1+C+B && 2*meter==-C+A-B && meter>=1 ], cost: 1+meter



Computing asymptotic complexity for rule 3

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -C+A-B (+/+!), 1+meter (+), 1+2*meter+C-A+B (+/+!), 1-2*meter-C+A-B (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {meter==n,C==0,A==2*n,B==0}

      resulting limit problem:

      [solved]



   Solution:

      meter / n

      C / 0

      A / 2*n

      B / 0

   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+meter

   Rule guard:  [ A>=1+C+B && 2*meter==-C+A-B ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)