4.03/1.89	WORST_CASE(Omega(n^1), O(n^1))
4.03/1.90	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
4.03/1.90	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.03/1.90	
4.03/1.90	
4.03/1.90	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).
4.03/1.90	
4.03/1.90	(0) CpxIntTrs
4.03/1.90	(1) Koat Proof [FINISHED, 79 ms]
4.03/1.90	(2) BOUNDS(1, n^1)
4.03/1.90	(3) Loat Proof [FINISHED, 205 ms]
4.03/1.90	(4) BOUNDS(n^1, INF)
4.03/1.90	
4.03/1.90	
4.03/1.90	----------------------------------------
4.03/1.90	
4.03/1.90	(0)
4.03/1.90	Obligation:
4.03/1.90	Complexity Int TRS consisting of the following rules:
4.03/1.90	f3(A, B, C, D) -> Com_1(f1(0, B, C, D)) :|: TRUE
4.03/1.90	f1(A, B, C, D) -> Com_1(f2(A, B, C, E)) :|: B >= C
4.03/1.90	f1(A, B, C, D) -> Com_1(f2(1, 1 + B, C, E)) :|: 0 >= A && B + 1 >= C && B + 1 <= C
4.03/1.90	f1(A, B, C, D) -> Com_1(f1(0, 1 + B, -(1) + C, D)) :|: C >= 1 + B && 0 >= A && C >= 2 + B
4.03/1.90	
4.03/1.90	The start-symbols are:[f3_4]
4.03/1.90	
4.03/1.90	
4.03/1.90	----------------------------------------
4.03/1.90	
4.03/1.90	(1) Koat Proof (FINISHED)
4.03/1.90	YES(?, ar_1 + ar_2 + 3)
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Initial complexity problem:
4.03/1.90	
4.03/1.90	1:	T:
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f3(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1, ar_2, ar_3))
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(1, ar_1 + 1, ar_2, e)) [ 0 >= ar_0 /\ ar_1 + 1 = ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1 + 1, ar_2 - 1, ar_3)) [ ar_2 >= ar_1 + 1 /\ 0 >= ar_0 /\ ar_2 >= ar_1 + 2 ]
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.03/1.90	
4.03/1.90		start location:	koat_start
4.03/1.90	
4.03/1.90		leaf cost:	0
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Repeatedly propagating knowledge in problem 1 produces the following problem:
4.03/1.90	
4.03/1.90	2:	T:
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)    f3(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1, ar_2, ar_3))
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(1, ar_1 + 1, ar_2, e)) [ 0 >= ar_0 /\ ar_1 + 1 = ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1 + 1, ar_2 - 1, ar_3)) [ ar_2 >= ar_1 + 1 /\ 0 >= ar_0 /\ ar_2 >= ar_1 + 2 ]
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.03/1.90	
4.03/1.90		start location:	koat_start
4.03/1.90	
4.03/1.90		leaf cost:	0
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	A polynomial rank function with
4.03/1.90	
4.03/1.90		Pol(f3) = 1
4.03/1.90	
4.03/1.90		Pol(f1) = 1
4.03/1.90	
4.03/1.90		Pol(f2) = 0
4.03/1.90	
4.03/1.90		Pol(koat_start) = 1
4.03/1.90	
4.03/1.90	orients all transitions weakly and the transitions
4.03/1.90	
4.03/1.90		f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(1, ar_1 + 1, ar_2, e)) [ 0 >= ar_0 /\ ar_1 + 1 = ar_2 ]
4.03/1.90	
4.03/1.90		f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_2 ]
4.03/1.90	
4.03/1.90	strictly and produces the following problem:
4.03/1.90	
4.03/1.90	3:	T:
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)    f3(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1, ar_2, ar_3))
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(1, ar_1 + 1, ar_2, e)) [ 0 >= ar_0 /\ ar_1 + 1 = ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1 + 1, ar_2 - 1, ar_3)) [ ar_2 >= ar_1 + 1 /\ 0 >= ar_0 /\ ar_2 >= ar_1 + 2 ]
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.03/1.90	
4.03/1.90		start location:	koat_start
4.03/1.90	
4.03/1.90		leaf cost:	0
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	A polynomial rank function with
4.03/1.90	
4.03/1.90		Pol(f3) = -V_2 + V_3
4.03/1.90	
4.03/1.90		Pol(f1) = -V_2 + V_3
4.03/1.90	
4.03/1.90		Pol(f2) = -V_2 + V_3
4.03/1.90	
4.03/1.90		Pol(koat_start) = -V_2 + V_3
4.03/1.90	
4.03/1.90	orients all transitions weakly and the transition
4.03/1.90	
4.03/1.90		f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1 + 1, ar_2 - 1, ar_3)) [ ar_2 >= ar_1 + 1 /\ 0 >= ar_0 /\ ar_2 >= ar_1 + 2 ]
4.03/1.90	
4.03/1.90	strictly and produces the following problem:
4.03/1.90	
4.03/1.90	4:	T:
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)              f3(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1, ar_2, ar_3))
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)              f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 1)              f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(1, ar_1 + 1, ar_2, e)) [ 0 >= ar_0 /\ ar_1 + 1 = ar_2 ]
4.03/1.90	
4.03/1.90			(Comp: ar_1 + ar_2, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(0, ar_1 + 1, ar_2 - 1, ar_3)) [ ar_2 >= ar_1 + 1 /\ 0 >= ar_0 /\ ar_2 >= ar_1 + 2 ]
4.03/1.90	
4.03/1.90			(Comp: 1, Cost: 0)              koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f3(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.03/1.90	
4.03/1.90		start location:	koat_start
4.03/1.90	
4.03/1.90		leaf cost:	0
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Complexity upper bound ar_1 + ar_2 + 3
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Time: 0.096 sec (SMT: 0.076 sec)
4.03/1.90	
4.03/1.90	
4.03/1.90	----------------------------------------
4.03/1.90	
4.03/1.90	(2)
4.03/1.90	BOUNDS(1, n^1)
4.03/1.90	
4.03/1.90	----------------------------------------
4.03/1.90	
4.03/1.90	(3) Loat Proof (FINISHED)
4.03/1.90	
4.03/1.90	
4.03/1.90	### Pre-processing the ITS problem ###
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Initial linear ITS problem
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      0: f3 -> f1 : A'=0, [], cost: 1
4.03/1.90	
4.03/1.90	      1: f1 -> f2 : D'=free, [ B>=C ], cost: 1
4.03/1.90	
4.03/1.90	      2: f1 -> f2 : A'=1, B'=1+B, D'=free_1, [ 0>=A && 1+B==C ], cost: 1
4.03/1.90	
4.03/1.90	      3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ C>=1+B && 0>=A && C>=2+B ], cost: 1
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Removed unreachable and leaf rules:
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      0: f3 -> f1 : A'=0, [], cost: 1
4.03/1.90	
4.03/1.90	      3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ C>=1+B && 0>=A && C>=2+B ], cost: 1
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Simplified all rules, resulting in:
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      0: f3 -> f1 : A'=0, [], cost: 1
4.03/1.90	
4.03/1.90	      3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ 0>=A && C>=2+B ], cost: 1
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	### Simplification by acceleration and chaining ###
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Accelerating simple loops of location 1.
4.03/1.90	
4.03/1.90	   Accelerating the following rules:
4.03/1.90	
4.03/1.90	      3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ 0>=A && C>=2+B ], cost: 1
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	   Accelerated rule 3 with metering function meter (where 2*meter==-1+C-B), yielding the new rule 4.
4.03/1.90	
4.03/1.90	   Removing the simple loops: 3.
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Accelerated all simple loops using metering functions (where possible):
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      0: f3 -> f1 : A'=0, [], cost: 1
4.03/1.90	
4.03/1.90	      4: f1 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ 0>=A && C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: meter
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Chained accelerated rules (with incoming rules):
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      0: f3 -> f1 : A'=0, [], cost: 1
4.03/1.90	
4.03/1.90	      5: f3 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: 1+meter
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Removed unreachable locations (and leaf rules with constant cost):
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      5: f3 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: 1+meter
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	### Computing asymptotic complexity ###
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Fully simplified ITS problem
4.03/1.90	
4.03/1.90	   Start location: f3
4.03/1.90	
4.03/1.90	      5: f3 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: 1+meter
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Computing asymptotic complexity for rule 5
4.03/1.90	
4.03/1.90	   Solved the limit problem by the following transformations:
4.03/1.90	
4.03/1.90	      Created initial limit problem:
4.03/1.90	
4.03/1.90	      C-2*meter-B (+/+!), 2-C+2*meter+B (+/+!), 1+meter (+), -1+C-B (+/+!) [not solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      applying transformation rule (C) using substitution {C==1+2*meter+B}
4.03/1.90	
4.03/1.90	      resulting limit problem:
4.03/1.90	
4.03/1.90	      1 (+/+!), 1+meter (+), 2*meter (+/+!) [not solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      applying transformation rule (B), deleting 1 (+/+!)
4.03/1.90	
4.03/1.90	      resulting limit problem:
4.03/1.90	
4.03/1.90	      1+meter (+), 2*meter (+/+!) [not solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      removing all constraints (solved by SMT)
4.03/1.90	
4.03/1.90	      resulting limit problem: [solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      applying transformation rule (C) using substitution {meter==n}
4.03/1.90	
4.03/1.90	      resulting limit problem:
4.03/1.90	
4.03/1.90	      [solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	   Solved the limit problem by the following transformations:
4.03/1.90	
4.03/1.90	      Created initial limit problem:
4.03/1.90	
4.03/1.90	      C-2*meter-B (+/+!), 2-C+2*meter+B (+/+!), 1+meter (+), -1+C-B (+/+!) [not solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      applying transformation rule (C) using substitution {C==1+2*meter+B}
4.03/1.90	
4.03/1.90	      resulting limit problem:
4.03/1.90	
4.03/1.90	      1 (+/+!), 1+meter (+), 2*meter (+/+!) [not solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      applying transformation rule (B), deleting 1 (+/+!)
4.03/1.90	
4.03/1.90	      resulting limit problem:
4.03/1.90	
4.03/1.90	      1+meter (+), 2*meter (+/+!) [not solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      removing all constraints (solved by SMT)
4.03/1.90	
4.03/1.90	      resulting limit problem: [solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	      applying transformation rule (C) using substitution {meter==n}
4.03/1.90	
4.03/1.90	      resulting limit problem:
4.03/1.90	
4.03/1.90	      [solved]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	   Solution:
4.03/1.90	
4.03/1.90	      C / 1+2*n
4.03/1.90	
4.03/1.90	      meter / n
4.03/1.90	
4.03/1.90	      B / 0
4.03/1.90	
4.03/1.90	   Resulting cost 1+n has complexity: Poly(n^1)
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Found new complexity Poly(n^1).
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	Obtained the following overall complexity (w.r.t. the length of the input n):
4.03/1.90	
4.03/1.90	   Complexity:  Poly(n^1)
4.03/1.90	
4.03/1.90	   Cpx degree:  1
4.03/1.90	
4.03/1.90	   Solved cost: 1+n
4.03/1.90	
4.03/1.90	   Rule cost:   1+meter
4.03/1.90	
4.03/1.90	   Rule guard:  [ C>=2+B && 2*meter==-1+C-B ]
4.03/1.90	
4.03/1.90	
4.03/1.90	
4.03/1.90	WORST_CASE(Omega(n^1),?)
4.03/1.90	
4.03/1.90	
4.03/1.90	----------------------------------------
4.03/1.90	
4.03/1.90	(4)
4.03/1.90	BOUNDS(n^1, INF)
4.03/1.92	EOF