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


--------------------------------------------------------------------------------


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 36 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 526 ms]
(4) BOUNDS(n^1, INF)


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C) -> Com_1(f3(-(A), B, C)) :|: TRUE
f3(A, B, C) -> Com_1(f7(0, D, C)) :|: A >= 0 && A <= 0
f4(A, B, C) -> Com_1(f7(0, D, C)) :|: A >= 0 && A <= 0
f6(A, B, C) -> Com_1(f4(A, B, 1)) :|: A >= 1
f3(A, B, C) -> Com_1(f4(-(1) - A, B, 1)) :|: 0 >= A + 1 && 0 >= C
f3(A, B, C) -> Com_1(f4(-(1) - A, B, 1)) :|: 0 >= A + 1 && C >= 2
f3(A, B, C) -> Com_1(f4(-(1) - A, B, 1)) :|: A >= 1 && 0 >= C
f3(A, B, C) -> Com_1(f4(-(1) - A, B, 1)) :|: A >= 1 && C >= 2
f4(A, B, C) -> Com_1(f3(1 - A, B, 0)) :|: 0 >= A + 1 && C >= 1 && C <= 1
f4(A, B, C) -> Com_1(f3(1 - A, B, 0)) :|: A >= 1 && C >= 1 && C <= 1
f5(A, B, C) -> Com_1(f3(1 - A, B, 0)) :|: 0 >= A + 1 && C >= 1 && C <= 1
f5(A, B, C) -> Com_1(f3(1 - A, B, 0)) :|: A >= 1 && C >= 1 && C <= 1

The start-symbols are:[f6_3]


----------------------------------------

(1) Koat Proof (FINISHED)
YES(?, 2*Ar_0 + 3)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0, Ar_1, Ar_2))

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ Ar_2 >= 2 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ Ar_2 >= 2 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]

		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transitions from problem 1:

	f0(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0, Ar_1, Ar_2))

	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ Ar_2 >= 2 ]

	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ 0 >= Ar_2 ]

	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ Ar_2 >= 2 ]

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]

	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]

	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

We thus obtain the following problem:

2:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f4) = 1

	Pol(f7) = 0

	Pol(f3) = 1

	Pol(f6) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transitions

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]

	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]

		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f4) = V_1

	Pol(f7) = 0

	Pol(f3) = -V_1

	Pol(f6) = V_1

	Pol(koat_start) = V_1

orients all transitions weakly and the transitions

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]

		(Comp: Ar_0, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]

		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]

		(Comp: Ar_0, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]

		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 2*Ar_0 + 3



Time: 0.081 sec (SMT: 0.068 sec)


----------------------------------------

(2)
BOUNDS(1, n^1)

----------------------------------------

(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f6

      0: f0 -> f3 : A'=-A, [], cost: 1

      1: f3 -> f7 : A'=0, B'=free, [ A==0 ], cost: 1

      4: f3 -> f4 : A'=-1-A, C'=1, [ 0>=1+A && 0>=C ], cost: 1

      5: f3 -> f4 : A'=-1-A, C'=1, [ 0>=1+A && C>=2 ], cost: 1

      6: f3 -> f4 : A'=-1-A, C'=1, [ A>=1 && 0>=C ], cost: 1

      7: f3 -> f4 : A'=-1-A, C'=1, [ A>=1 && C>=2 ], cost: 1

      2: f4 -> f7 : A'=0, B'=free_1, [ A==0 ], cost: 1

      8: f4 -> f3 : A'=1-A, C'=0, [ 0>=1+A && C==1 ], cost: 1

      9: f4 -> f3 : A'=1-A, C'=0, [ A>=1 && C==1 ], cost: 1

      3: f6 -> f4 : C'=1, [ A>=1 ], cost: 1

     10: f5 -> f3 : A'=1-A, C'=0, [ 0>=1+A && C==1 ], cost: 1

     11: f5 -> f3 : A'=1-A, C'=0, [ A>=1 && C==1 ], cost: 1



Checking for constant complexity:

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

      3: f6 -> f4 : C'=1, [ A>=1 ], cost: 1



Removed unreachable and leaf rules:

   Start location: f6

      4: f3 -> f4 : A'=-1-A, C'=1, [ 0>=1+A && 0>=C ], cost: 1

      5: f3 -> f4 : A'=-1-A, C'=1, [ 0>=1+A && C>=2 ], cost: 1

      6: f3 -> f4 : A'=-1-A, C'=1, [ A>=1 && 0>=C ], cost: 1

      7: f3 -> f4 : A'=-1-A, C'=1, [ A>=1 && C>=2 ], cost: 1

      8: f4 -> f3 : A'=1-A, C'=0, [ 0>=1+A && C==1 ], cost: 1

      9: f4 -> f3 : A'=1-A, C'=0, [ A>=1 && C==1 ], cost: 1

      3: f6 -> f4 : C'=1, [ A>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Eliminated locations (on tree-shaped paths):

   Start location: f6

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

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

      3: f6 -> f4 : C'=1, [ A>=1 ], cost: 1



Accelerating simple loops of location 2.

   Accelerating the following rules:

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

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



   Accelerated rule 12 with NONTERM, yielding the new rule 14.

   Accelerated rule 13 with metering function meter (where 2*meter==-1+A), yielding the new rule 15.

   Removing the simple loops: 12 13.



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

   Start location: f6

     14: f4 -> [6] : [ 0>=1+A && C==1 ], cost: NONTERM

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

      3: f6 -> f4 : C'=1, [ A>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f6

      3: f6 -> f4 : C'=1, [ A>=1 ], cost: 1

     16: f6 -> f4 : A'=A-2*meter, C'=1, [ 0>=2-A && 2*meter==-1+A && meter>=1 ], cost: 1+2*meter



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

   Start location: f6

     16: f6 -> f4 : A'=A-2*meter, C'=1, [ 0>=2-A && 2*meter==-1+A && meter>=1 ], cost: 1+2*meter



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f6

     16: f6 -> f4 : A'=A-2*meter, C'=1, [ 0>=2-A && 2*meter==-1+A && meter>=1 ], cost: 1+2*meter



Computing asymptotic complexity for rule 16

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {A==1+2*n,meter==n}

      resulting limit problem:

      [solved]



   Solution:

      A / 1+2*n

      meter / n

   Resulting cost 1+2*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+2*n

   Rule cost:   1+2*meter

   Rule guard:  [ 0>=2-A && 2*meter==-1+A ]



WORST_CASE(Omega(n^1),?)


----------------------------------------

(4)
BOUNDS(n^1, INF)