/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^2), O(n^2))
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 22 ms]
(2) BOUNDS(1, n^2)
(3) Loat Proof [FINISHED, 263 ms]
(4) BOUNDS(n^2, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval1(A, B) -> Com_1(eval2(A, 0)) :|: A >= 1
eval2(A, B) -> Com_1(eval2(A, B + 1)) :|: A >= 1 && B >= 0 && A >= B + 1
eval2(A, B) -> Com_1(eval1(A - 1, B)) :|: A >= 1 && B >= 0 && B >= A
start(A, B) -> Com_1(eval1(A, B)) :|: TRUE

The start-symbols are:[start_2]


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(1) Koat Proof (FINISHED)
YES(?, 4*ar_0 + 2*ar_0^2 + 1)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_0 >= ar_1 + 1 ]

		(Comp: ?, Cost: 1)    eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_1 >= ar_0 ]

		(Comp: ?, Cost: 1)    start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: ?, Cost: 1)    eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_0 >= ar_1 + 1 ]

		(Comp: ?, Cost: 1)    eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_1 >= ar_0 ]

		(Comp: 1, Cost: 1)    start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(eval1) = 2*V_1

	Pol(eval2) = 2*V_1 - 1

	Pol(start) = 2*V_1

	Pol(koat_start) = 2*V_1

orients all transitions weakly and the transitions

	eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_1 >= ar_0 ]

	eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]

strictly and produces the following problem:

3:	T:

		(Comp: 2*ar_0, Cost: 1)    eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]

		(Comp: ?, Cost: 1)         eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_0 >= ar_1 + 1 ]

		(Comp: 2*ar_0, Cost: 1)    eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_1 >= ar_0 ]

		(Comp: 1, Cost: 1)         start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))

		(Comp: 1, Cost: 0)         koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(eval2) = V_1 - V_2

and size complexities

	S("koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]", 0-0) = ar_0

	S("koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]", 0-1) = ar_1

	S("start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))", 0-0) = ar_0

	S("start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))", 0-1) = ar_1

	S("eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\\ ar_1 >= 0 /\\ ar_1 >= ar_0 ]", 0-0) = ar_0

	S("eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\\ ar_1 >= 0 /\\ ar_1 >= ar_0 ]", 0-1) = ar_0

	S("eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\\ ar_1 >= 0 /\\ ar_0 >= ar_1 + 1 ]", 0-0) = ar_0

	S("eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\\ ar_1 >= 0 /\\ ar_0 >= ar_1 + 1 ]", 0-1) = ar_0

	S("eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]", 0-0) = ar_0

	S("eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]", 0-1) = 0

orients the transitions

	eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_0 >= ar_1 + 1 ]

weakly and the transition

	eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_0 >= ar_1 + 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 2*ar_0, Cost: 1)      eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, 0)) [ ar_0 >= 1 ]

		(Comp: 2*ar_0^2, Cost: 1)    eval2(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1 + 1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_0 >= ar_1 + 1 ]

		(Comp: 2*ar_0, Cost: 1)      eval2(ar_0, ar_1) -> Com_1(eval1(ar_0 - 1, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 0 /\ ar_1 >= ar_0 ]

		(Comp: 1, Cost: 1)           start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))

		(Comp: 1, Cost: 0)           koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 4*ar_0 + 2*ar_0^2 + 1



Time: 0.068 sec (SMT: 0.062 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: start

      0: eval1 -> eval2 : B'=0, [ A>=1 ], cost: 1

      1: eval2 -> eval2 : B'=1+B, [ A>=1 && B>=0 && A>=1+B ], cost: 1

      2: eval2 -> eval1 : A'=-1+A, [ A>=1 && B>=0 && B>=A ], cost: 1

      3: start -> eval1 : [], cost: 1



Simplified all rules, resulting in:

   Start location: start

      0: eval1 -> eval2 : B'=0, [ A>=1 ], cost: 1

      1: eval2 -> eval2 : B'=1+B, [ B>=0 && A>=1+B ], cost: 1

      2: eval2 -> eval1 : A'=-1+A, [ A>=1 && B>=0 && B>=A ], cost: 1

      3: start -> eval1 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: eval2 -> eval2 : B'=1+B, [ B>=0 && A>=1+B ], cost: 1



   Accelerated rule 1 with metering function A-B, yielding the new rule 4.

   Removing the simple loops: 1.



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

   Start location: start

      0: eval1 -> eval2 : B'=0, [ A>=1 ], cost: 1

      2: eval2 -> eval1 : A'=-1+A, [ A>=1 && B>=0 && B>=A ], cost: 1

      4: eval2 -> eval2 : B'=A, [ B>=0 && A>=1+B ], cost: A-B

      3: start -> eval1 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      0: eval1 -> eval2 : B'=0, [ A>=1 ], cost: 1

      5: eval1 -> eval2 : B'=A, [ A>=1 ], cost: 1+A

      2: eval2 -> eval1 : A'=-1+A, [ A>=1 && B>=0 && B>=A ], cost: 1

      3: start -> eval1 : [], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: start

      6: eval1 -> eval1 : A'=-1+A, B'=A, [ A>=1 ], cost: 2+A

      3: start -> eval1 : [], cost: 1



Accelerating simple loops of location 0.

   Accelerating the following rules:

      6: eval1 -> eval1 : A'=-1+A, B'=A, [ A>=1 ], cost: 2+A



   Accelerated rule 6 with metering function A, yielding the new rule 7.

   Removing the simple loops: 6.



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

   Start location: start

      7: eval1 -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 5/2*A+1/2*A^2

      3: start -> eval1 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      3: start -> eval1 : [], cost: 1

      8: start -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 1+5/2*A+1/2*A^2



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

   Start location: start

      8: start -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 1+5/2*A+1/2*A^2



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      8: start -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 1+5/2*A+1/2*A^2



Computing asymptotic complexity for rule 8

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      1+5/2*A+1/2*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 1+1/2*n^2+5/2*n 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+1/2*n^2+5/2*n

   Rule cost:   1+5/2*A+1/2*A^2

   Rule guard:  [ A>=1 ]



WORST_CASE(Omega(n^2),?)


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(4)
BOUNDS(n^2, INF)