/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^1), O(n^1))
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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^1, n^1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 28 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 215 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
start(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: 1 >= A && B >= C && B <= C && D >= A && D <= A
start(A, B, C, D) -> Com_1(lbl32(A, B, C, D - 1)) :|: A >= 2 && B >= C && B <= C && D >= A && D <= A
lbl32(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: A >= 2 && D >= 1 && D <= 1 && B >= C && B <= C
lbl32(A, B, C, D) -> Com_1(lbl32(A, B, C, D - 1)) :|: D >= 2 && D >= 1 && A >= D + 1 && B >= C && B <= C
start0(A, B, C, D) -> Com_1(start(A, C, C, A)) :|: TRUE

The start-symbols are:[start0_4]


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(1) Koat Proof (FINISHED)
YES(?, Ar_0 + 4)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 1 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_0 >= 2 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: ?, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 2 /\ Ar_3 = 1 /\ Ar_1 = Ar_2 ]

		(Comp: ?, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_3 >= 2 /\ Ar_3 >= 1 /\ Ar_0 >= Ar_3 + 1 /\ Ar_1 = Ar_2 ]

		(Comp: ?, Cost: 1)    start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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)    start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 1 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_0 >= 2 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: ?, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 2 /\ Ar_3 = 1 /\ Ar_1 = Ar_2 ]

		(Comp: ?, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_3 >= 2 /\ Ar_3 >= 1 /\ Ar_0 >= Ar_3 + 1 /\ Ar_1 = Ar_2 ]

		(Comp: 1, Cost: 1)    start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(start) = 1

	Pol(stop) = 0

	Pol(lbl32) = 1

	Pol(start0) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 2 /\ Ar_3 = 1 /\ Ar_1 = Ar_2 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 1 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_0 >= 2 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: 1, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 2 /\ Ar_3 = 1 /\ Ar_1 = Ar_2 ]

		(Comp: ?, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_3 >= 2 /\ Ar_3 >= 1 /\ Ar_0 >= Ar_3 + 1 /\ Ar_1 = Ar_2 ]

		(Comp: 1, Cost: 1)    start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(start) = V_1

	Pol(stop) = V_4

	Pol(lbl32) = V_4

	Pol(start0) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transition

	lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_3 >= 2 /\ Ar_3 >= 1 /\ Ar_0 >= Ar_3 + 1 /\ Ar_1 = Ar_2 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)       start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 1 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: 1, Cost: 1)       start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_0 >= 2 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ]

		(Comp: 1, Cost: 1)       lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 2 /\ Ar_3 = 1 /\ Ar_1 = Ar_2 ]

		(Comp: Ar_0, Cost: 1)    lbl32(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl32(Ar_0, Ar_1, Ar_2, Ar_3 - 1)) [ Ar_3 >= 2 /\ Ar_3 >= 1 /\ Ar_0 >= Ar_3 + 1 /\ Ar_1 = Ar_2 ]

		(Comp: 1, Cost: 1)       start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0))

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0 + 4



Time: 0.078 sec (SMT: 0.066 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: start0

      0: start -> stop : [ 1>=A && B==C && D==A ], cost: 1

      1: start -> lbl32 : D'=-1+D, [ A>=2 && B==C && D==A ], cost: 1

      2: lbl32 -> stop : [ A>=2 && D==1 && B==C ], cost: 1

      3: lbl32 -> lbl32 : D'=-1+D, [ D>=2 && D>=1 && A>=1+D && B==C ], cost: 1

      4: start0 -> start : B'=C, D'=A, [], cost: 1



Checking for constant complexity:

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

      4: start0 -> start : B'=C, D'=A, [], cost: 1



Removed unreachable and leaf rules:

   Start location: start0

      1: start -> lbl32 : D'=-1+D, [ A>=2 && B==C && D==A ], cost: 1

      3: lbl32 -> lbl32 : D'=-1+D, [ D>=2 && D>=1 && A>=1+D && B==C ], cost: 1

      4: start0 -> start : B'=C, D'=A, [], cost: 1



Simplified all rules, resulting in:

   Start location: start0

      1: start -> lbl32 : D'=-1+D, [ A>=2 && B==C && D==A ], cost: 1

      3: lbl32 -> lbl32 : D'=-1+D, [ D>=2 && A>=1+D && B==C ], cost: 1

      4: start0 -> start : B'=C, D'=A, [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      3: lbl32 -> lbl32 : D'=-1+D, [ D>=2 && A>=1+D && B==C ], cost: 1



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

   Removing the simple loops: 3.



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

   Start location: start0

      1: start -> lbl32 : D'=-1+D, [ A>=2 && B==C && D==A ], cost: 1

      5: lbl32 -> lbl32 : D'=1, [ D>=2 && A>=1+D && B==C ], cost: -1+D

      4: start0 -> start : B'=C, D'=A, [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start0

      1: start -> lbl32 : D'=-1+D, [ A>=2 && B==C && D==A ], cost: 1

      6: start -> lbl32 : D'=1, [ A>=2 && B==C && D==A && -1+D>=2 ], cost: -1+D

      4: start0 -> start : B'=C, D'=A, [], cost: 1



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

   Start location: start0

      6: start -> lbl32 : D'=1, [ A>=2 && B==C && D==A && -1+D>=2 ], cost: -1+D

      4: start0 -> start : B'=C, D'=A, [], cost: 1



Eliminated locations (on linear paths):

   Start location: start0

      7: start0 -> lbl32 : B'=C, D'=1, [ -1+A>=2 ], cost: A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start0

      7: start0 -> lbl32 : B'=C, D'=1, [ -1+A>=2 ], cost: A



Computing asymptotic complexity for rule 7

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      A (+), -2+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 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:   A

   Rule guard:  [ -1+A>=2 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)