3.63/1.90	WORST_CASE(Omega(n^1), O(n^1))
3.63/1.91	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
3.63/1.91	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.63/1.91	
3.63/1.91	
3.63/1.91	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).
3.63/1.91	
3.63/1.91	(0) CpxIntTrs
3.63/1.91	(1) Koat Proof [FINISHED, 21 ms]
3.63/1.91	(2) BOUNDS(1, n^1)
3.63/1.91	(3) Loat Proof [FINISHED, 117 ms]
3.63/1.91	(4) BOUNDS(n^1, INF)
3.63/1.91	
3.63/1.91	
3.63/1.91	----------------------------------------
3.63/1.91	
3.63/1.91	(0)
3.63/1.91	Obligation:
3.63/1.91	Complexity Int TRS consisting of the following rules:
3.63/1.91	f4(A, B, C) -> Com_1(f5(A, B, 1)) :|: 0 >= A && 0 >= B
3.63/1.91	f0(A, B, C) -> Com_1(f2(A, B, 1)) :|: A >= 1
3.63/1.91	f4(A, B, C) -> Com_1(f4(A, B - 1, C)) :|: B >= 1
3.63/1.91	f0(A, B, C) -> Com_1(f4(A, B, 0)) :|: 0 >= A
3.63/1.91	
3.63/1.91	The start-symbols are:[f0_3]
3.63/1.91	
3.63/1.91	
3.63/1.91	----------------------------------------
3.63/1.91	
3.63/1.91	(1) Koat Proof (FINISHED)
3.63/1.91	YES(?, ar_1 + 3)
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Initial complexity problem:
3.63/1.91	
3.63/1.91	1:	T:
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, 1)) [ 0 >= ar_0 /\ 0 >= ar_1 ]
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 1)) [ ar_0 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1, 0)) [ 0 >= ar_0 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.63/1.91	
3.63/1.91		start location:	koat_start
3.63/1.91	
3.63/1.91		leaf cost:	0
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Repeatedly propagating knowledge in problem 1 produces the following problem:
3.63/1.91	
3.63/1.91	2:	T:
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, 1)) [ 0 >= ar_0 /\ 0 >= ar_1 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 1)) [ ar_0 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1, 0)) [ 0 >= ar_0 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.63/1.91	
3.63/1.91		start location:	koat_start
3.63/1.91	
3.63/1.91		leaf cost:	0
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	A polynomial rank function with
3.63/1.91	
3.63/1.91		Pol(f4) = 1
3.63/1.91	
3.63/1.91		Pol(f5) = 0
3.63/1.91	
3.63/1.91		Pol(f0) = 1
3.63/1.91	
3.63/1.91		Pol(f2) = 0
3.63/1.91	
3.63/1.91		Pol(koat_start) = 1
3.63/1.91	
3.63/1.91	orients all transitions weakly and the transition
3.63/1.91	
3.63/1.91		f4(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, 1)) [ 0 >= ar_0 /\ 0 >= ar_1 ]
3.63/1.91	
3.63/1.91	strictly and produces the following problem:
3.63/1.91	
3.63/1.91	3:	T:
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, 1)) [ 0 >= ar_0 /\ 0 >= ar_1 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 1)) [ ar_0 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: ?, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1, 0)) [ 0 >= ar_0 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.63/1.91	
3.63/1.91		start location:	koat_start
3.63/1.91	
3.63/1.91		leaf cost:	0
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	A polynomial rank function with
3.63/1.91	
3.63/1.91		Pol(f4) = V_2
3.63/1.91	
3.63/1.91		Pol(f5) = V_2
3.63/1.91	
3.63/1.91		Pol(f0) = V_2
3.63/1.91	
3.63/1.91		Pol(f2) = V_2
3.63/1.91	
3.63/1.91		Pol(koat_start) = V_2
3.63/1.91	
3.63/1.91	orients all transitions weakly and the transition
3.63/1.91	
3.63/1.91		f4(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ]
3.63/1.91	
3.63/1.91	strictly and produces the following problem:
3.63/1.91	
3.63/1.91	4:	T:
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)       f4(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, 1)) [ 0 >= ar_0 /\ 0 >= ar_1 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)       f0(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 1)) [ ar_0 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: ar_1, Cost: 1)    f4(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 1)       f0(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1, 0)) [ 0 >= ar_0 ]
3.63/1.91	
3.63/1.91			(Comp: 1, Cost: 0)       koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ]
3.63/1.91	
3.63/1.91		start location:	koat_start
3.63/1.91	
3.63/1.91		leaf cost:	0
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Complexity upper bound ar_1 + 3
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Time: 0.048 sec (SMT: 0.044 sec)
3.63/1.91	
3.63/1.91	
3.63/1.91	----------------------------------------
3.63/1.91	
3.63/1.91	(2)
3.63/1.91	BOUNDS(1, n^1)
3.63/1.91	
3.63/1.91	----------------------------------------
3.63/1.91	
3.63/1.91	(3) Loat Proof (FINISHED)
3.63/1.91	
3.63/1.91	
3.63/1.91	### Pre-processing the ITS problem ###
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Initial linear ITS problem
3.63/1.91	
3.63/1.91	   Start location: f0
3.63/1.91	
3.63/1.91	      0: f4 -> f5 : C'=1, [ 0>=A && 0>=B ], cost: 1
3.63/1.91	
3.63/1.91	      2: f4 -> f4 : B'=-1+B, [ B>=1 ], cost: 1
3.63/1.91	
3.63/1.91	      1: f0 -> f2 : C'=1, [ A>=1 ], cost: 1
3.63/1.91	
3.63/1.91	      3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Removed unreachable and leaf rules:
3.63/1.91	
3.63/1.91	   Start location: f0
3.63/1.91	
3.63/1.91	      2: f4 -> f4 : B'=-1+B, [ B>=1 ], cost: 1
3.63/1.91	
3.63/1.91	      3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	### Simplification by acceleration and chaining ###
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Accelerating simple loops of location 0.
3.63/1.91	
3.63/1.91	   Accelerating the following rules:
3.63/1.91	
3.63/1.91	      2: f4 -> f4 : B'=-1+B, [ B>=1 ], cost: 1
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	   Accelerated rule 2 with metering function B, yielding the new rule 4.
3.63/1.91	
3.63/1.91	   Removing the simple loops: 2.
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Accelerated all simple loops using metering functions (where possible):
3.63/1.91	
3.63/1.91	   Start location: f0
3.63/1.91	
3.63/1.91	      4: f4 -> f4 : B'=0, [ B>=1 ], cost: B
3.63/1.91	
3.63/1.91	      3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Chained accelerated rules (with incoming rules):
3.63/1.91	
3.63/1.91	   Start location: f0
3.63/1.91	
3.63/1.91	      3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1
3.63/1.91	
3.63/1.91	      5: f0 -> f4 : B'=0, C'=0, [ 0>=A && B>=1 ], cost: 1+B
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Removed unreachable locations (and leaf rules with constant cost):
3.63/1.91	
3.63/1.91	   Start location: f0
3.63/1.91	
3.63/1.91	      5: f0 -> f4 : B'=0, C'=0, [ 0>=A && B>=1 ], cost: 1+B
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	### Computing asymptotic complexity ###
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Fully simplified ITS problem
3.63/1.91	
3.63/1.91	   Start location: f0
3.63/1.91	
3.63/1.91	      5: f0 -> f4 : B'=0, C'=0, [ 0>=A && B>=1 ], cost: 1+B
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Computing asymptotic complexity for rule 5
3.63/1.91	
3.63/1.91	   Solved the limit problem by the following transformations:
3.63/1.91	
3.63/1.91	      Created initial limit problem:
3.63/1.91	
3.63/1.91	      1-A (+/+!), 1+B (+), B (+/+!) [not solved]
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	      removing all constraints (solved by SMT)
3.63/1.91	
3.63/1.91	      resulting limit problem: [solved]
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	      applying transformation rule (C) using substitution {A==0,B==n}
3.63/1.91	
3.63/1.91	      resulting limit problem:
3.63/1.91	
3.63/1.91	      [solved]
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	   Solution:
3.63/1.91	
3.63/1.91	      A / 0
3.63/1.91	
3.63/1.91	      B / n
3.63/1.91	
3.63/1.91	   Resulting cost 1+n has complexity: Poly(n^1)
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Found new complexity Poly(n^1).
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	Obtained the following overall complexity (w.r.t. the length of the input n):
3.63/1.91	
3.63/1.91	   Complexity:  Poly(n^1)
3.63/1.91	
3.63/1.91	   Cpx degree:  1
3.63/1.91	
3.63/1.91	   Solved cost: 1+n
3.63/1.91	
3.63/1.91	   Rule cost:   1+B
3.63/1.91	
3.63/1.91	   Rule guard:  [ 0>=A && B>=1 ]
3.63/1.91	
3.63/1.91	
3.63/1.91	
3.63/1.91	WORST_CASE(Omega(n^1),?)
3.63/1.91	
3.63/1.91	
3.63/1.91	----------------------------------------
3.63/1.91	
3.63/1.91	(4)
3.63/1.91	BOUNDS(n^1, INF)
3.63/1.93	EOF