/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 64 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 164 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, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: 0 >= A && B >= A && B <= A && C >= D && C <= D && E >= F && E <= F
start(A, B, C, D, E, F) -> Com_1(lbl71(A, B - 1, C - 1, D, 1 + E, F)) :|: A >= 1 && B >= A && B <= A && C >= D && C <= D && E >= F && E <= F
lbl71(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: D >= C + 1 && B >= 0 && B <= 0 && E + C >= F + D && E + C <= F + D && A + C >= D && A + C <= D
lbl71(A, B, C, D, E, F) -> Com_1(lbl71(A, B - 1, C - 1, D, 1 + E, F)) :|: A + C >= D + 1 && D >= C + 1 && A + C >= D && E + C >= D + F && E + C <= D + F && B + D >= A + C && B + D <= A + C
start0(A, B, C, D, E, F) -> Com_1(start(A, A, D, D, F, F)) :|: TRUE

The start-symbols are:[start0_6]


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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, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: ?, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: ?, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_3 >= Ar_2 + 1 /\ Ar_1 = 0 /\ Ar_4 + Ar_2 = Ar_5 + Ar_3 /\ Ar_0 + Ar_2 = Ar_3 ]

		(Comp: ?, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 + Ar_2 >= Ar_3 + 1 /\ Ar_3 >= Ar_2 + 1 /\ Ar_0 + Ar_2 >= Ar_3 /\ Ar_4 + Ar_2 = Ar_3 + Ar_5 /\ Ar_1 + Ar_3 = Ar_0 + Ar_2 ]

		(Comp: ?, Cost: 1)    start0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start(Ar_0, Ar_0, Ar_3, Ar_3, Ar_5, Ar_5))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start0(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: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: ?, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_3 >= Ar_2 + 1 /\ Ar_1 = 0 /\ Ar_4 + Ar_2 = Ar_5 + Ar_3 /\ Ar_0 + Ar_2 = Ar_3 ]

		(Comp: ?, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 + Ar_2 >= Ar_3 + 1 /\ Ar_3 >= Ar_2 + 1 /\ Ar_0 + Ar_2 >= Ar_3 /\ Ar_4 + Ar_2 = Ar_3 + Ar_5 /\ Ar_1 + Ar_3 = Ar_0 + Ar_2 ]

		(Comp: 1, Cost: 1)    start0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start(Ar_0, Ar_0, Ar_3, Ar_3, Ar_5, Ar_5))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start0(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(start) = 1

	Pol(stop) = 0

	Pol(lbl71) = 1

	Pol(start0) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_3 >= Ar_2 + 1 /\ Ar_1 = 0 /\ Ar_4 + Ar_2 = Ar_5 + Ar_3 /\ Ar_0 + Ar_2 = Ar_3 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: 1, Cost: 1)    start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: 1, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_3 >= Ar_2 + 1 /\ Ar_1 = 0 /\ Ar_4 + Ar_2 = Ar_5 + Ar_3 /\ Ar_0 + Ar_2 = Ar_3 ]

		(Comp: ?, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 + Ar_2 >= Ar_3 + 1 /\ Ar_3 >= Ar_2 + 1 /\ Ar_0 + Ar_2 >= Ar_3 /\ Ar_4 + Ar_2 = Ar_3 + Ar_5 /\ Ar_1 + Ar_3 = Ar_0 + Ar_2 ]

		(Comp: 1, Cost: 1)    start0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start(Ar_0, Ar_0, Ar_3, Ar_3, Ar_5, Ar_5))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start0(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(start) = V_1

	Pol(stop) = V_2

	Pol(lbl71) = V_2

	Pol(start0) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transition

	lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 + Ar_2 >= Ar_3 + 1 /\ Ar_3 >= Ar_2 + 1 /\ Ar_0 + Ar_2 >= Ar_3 /\ Ar_4 + Ar_2 = Ar_3 + Ar_5 /\ Ar_1 + Ar_3 = Ar_0 + Ar_2 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)       start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: 1, Cost: 1)       start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 = Ar_0 /\ Ar_2 = Ar_3 /\ Ar_4 = Ar_5 ]

		(Comp: 1, Cost: 1)       lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_3 >= Ar_2 + 1 /\ Ar_1 = 0 /\ Ar_4 + Ar_2 = Ar_5 + Ar_3 /\ Ar_0 + Ar_2 = Ar_3 ]

		(Comp: Ar_0, Cost: 1)    lbl71(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(lbl71(Ar_0, Ar_1 - 1, Ar_2 - 1, Ar_3, Ar_4 + 1, Ar_5)) [ Ar_0 + Ar_2 >= Ar_3 + 1 /\ Ar_3 >= Ar_2 + 1 /\ Ar_0 + Ar_2 >= Ar_3 /\ Ar_4 + Ar_2 = Ar_3 + Ar_5 /\ Ar_1 + Ar_3 = Ar_0 + Ar_2 ]

		(Comp: 1, Cost: 1)       start0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start(Ar_0, Ar_0, Ar_3, Ar_3, Ar_5, Ar_5))

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(start0(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 + 4



Time: 0.138 sec (SMT: 0.116 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 : [ 0>=A && B==A && C==D && E==F ], cost: 1

      1: start -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ A>=1 && B==A && C==D && E==F ], cost: 1

      2: lbl71 -> stop : [ D>=1+C && B==0 && C+E==F+D && C+A==D ], cost: 1

      3: lbl71 -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ C+A>=1+D && D>=1+C && C+A>=D && C+E==F+D && D+B==C+A ], cost: 1

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



Checking for constant complexity:

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

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



Removed unreachable and leaf rules:

   Start location: start0

      1: start -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ A>=1 && B==A && C==D && E==F ], cost: 1

      3: lbl71 -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ C+A>=1+D && D>=1+C && C+A>=D && C+E==F+D && D+B==C+A ], cost: 1

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



Simplified all rules, resulting in:

   Start location: start0

      1: start -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ A>=1 && B==A && C==D && E==F ], cost: 1

      3: lbl71 -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ C+A>=1+D && D>=1+C && C+E==F+D && D+B==C+A ], cost: 1

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      3: lbl71 -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ C+A>=1+D && D>=1+C && C+E==F+D && D+B==C+A ], cost: 1



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

   Removing the simple loops: 3.



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

   Start location: start0

      1: start -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ A>=1 && B==A && C==D && E==F ], cost: 1

      5: lbl71 -> lbl71 : B'=-C+D-A+B, C'=D-A, E'=C-D+A+E, [ C+A>=1+D && D>=1+C && C+E==F+D && D+B==C+A ], cost: C-D+A

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



Chained accelerated rules (with incoming rules):

   Start location: start0

      1: start -> lbl71 : B'=-1+B, C'=-1+C, E'=1+E, [ A>=1 && B==A && C==D && E==F ], cost: 1

      6: start -> lbl71 : B'=-C+D-A+B, C'=D-A, E'=C-D+A+E, [ A>=1 && B==A && C==D && E==F && -1+C+A>=1+D && C+E==F+D && -1+D+B==-1+C+A ], cost: C-D+A

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



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

   Start location: start0

      6: start -> lbl71 : B'=-C+D-A+B, C'=D-A, E'=C-D+A+E, [ A>=1 && B==A && C==D && E==F && -1+C+A>=1+D && C+E==F+D && -1+D+B==-1+C+A ], cost: C-D+A

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



Eliminated locations (on linear paths):

   Start location: start0

      7: start0 -> lbl71 : B'=0, C'=D-A, E'=F+A, [ -1+D+A>=1+D ], cost: 1+A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start0

      7: start0 -> lbl71 : B'=0, C'=D-A, E'=F+A, [ -1+D+A>=1+D ], cost: 1+A



Computing asymptotic complexity for rule 7

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -1+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:

      D / 0

      A / 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+A

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)