4.02/1.88	WORST_CASE(Omega(n^2), O(n^2))
4.02/1.89	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
4.02/1.89	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.02/1.89	
4.02/1.89	
4.02/1.89	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, n^2).
4.02/1.89	
4.02/1.89	(0) CpxIntTrs
4.02/1.89	(1) Koat Proof [FINISHED, 118 ms]
4.02/1.89	(2) BOUNDS(1, n^2)
4.02/1.89	(3) Loat Proof [FINISHED, 324 ms]
4.02/1.89	(4) BOUNDS(n^2, INF)
4.02/1.89	
4.02/1.89	
4.02/1.89	----------------------------------------
4.02/1.89	
4.02/1.89	(0)
4.02/1.89	Obligation:
4.02/1.89	Complexity Int TRS consisting of the following rules:
4.02/1.89	l0(A, B, C, D) -> Com_1(l1(0, B, C, D)) :|: TRUE
4.02/1.89	l1(A, B, C, D) -> Com_1(l2(A, B, 0, 0)) :|: B >= 1
4.02/1.89	l2(A, B, C, D) -> Com_1(l2(A, B, C + 1, D + C)) :|: B >= C + 1
4.02/1.89	l2(A, B, C, D) -> Com_1(l1(A + D, B - 1, C, D)) :|: C >= B
4.02/1.89	
4.02/1.89	The start-symbols are:[l0_4]
4.02/1.89	
4.02/1.89	
4.02/1.89	----------------------------------------
4.02/1.89	
4.02/1.89	(1) Koat Proof (FINISHED)
4.02/1.89	YES(?, 8*ar_1 + 2*ar_1^2 + 7)
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Initial complexity problem:
4.02/1.89	
4.02/1.89	1:	T:
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.02/1.89	
4.02/1.89		start location:	koat_start
4.02/1.89	
4.02/1.89		leaf cost:	0
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Repeatedly propagating knowledge in problem 1 produces the following problem:
4.02/1.89	
4.02/1.89	2:	T:
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 1)    l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)    l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.02/1.89	
4.02/1.89		start location:	koat_start
4.02/1.89	
4.02/1.89		leaf cost:	0
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	A polynomial rank function with
4.02/1.89	
4.02/1.89		Pol(l0) = V_2 + 1
4.02/1.89	
4.02/1.89		Pol(l1) = V_2 + 1
4.02/1.89	
4.02/1.89		Pol(l2) = V_2
4.02/1.89	
4.02/1.89		Pol(koat_start) = V_2 + 1
4.02/1.89	
4.02/1.89	orients all transitions weakly and the transition
4.02/1.89	
4.02/1.89		l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]
4.02/1.89	
4.02/1.89	strictly and produces the following problem:
4.02/1.89	
4.02/1.89	3:	T:
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 1)           l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))
4.02/1.89	
4.02/1.89			(Comp: ar_1 + 1, Cost: 1)    l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)           l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)           l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 0)           koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.02/1.89	
4.02/1.89		start location:	koat_start
4.02/1.89	
4.02/1.89		leaf cost:	0
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	A polynomial rank function with
4.02/1.89	
4.02/1.89		Pol(l2) = 1
4.02/1.89	
4.02/1.89		Pol(l1) = 0
4.02/1.89	
4.02/1.89	and size complexities
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-0) = ar_0
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-1) = ar_1
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-2) = ar_2
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-3) = ar_3
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-0) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-1) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-2) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-3) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-0) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-1) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-2) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-3) = ?
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-0) = ?
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-1) = ?
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-2) = 0
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-3) = 0
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-0) = 0
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-1) = ar_1
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-2) = ar_2
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-3) = ar_3
4.02/1.89	
4.02/1.89	orients the transitions
4.02/1.89	
4.02/1.89		l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89		l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89	weakly and the transition
4.02/1.89	
4.02/1.89		l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89	strictly and produces the following problem:
4.02/1.89	
4.02/1.89	4:	T:
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 1)           l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))
4.02/1.89	
4.02/1.89			(Comp: ar_1 + 1, Cost: 1)    l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]
4.02/1.89	
4.02/1.89			(Comp: ?, Cost: 1)           l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89			(Comp: ar_1 + 1, Cost: 1)    l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 0)           koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.02/1.89	
4.02/1.89		start location:	koat_start
4.02/1.89	
4.02/1.89		leaf cost:	0
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	A polynomial rank function with
4.02/1.89	
4.02/1.89		Pol(l2) = V_2 - V_3
4.02/1.89	
4.02/1.89	and size complexities
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-0) = ar_0
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-1) = ar_1
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-2) = ar_2
4.02/1.89	
4.02/1.89		S("koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]", 0-3) = ar_3
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-0) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-1) = 2*ar_1 + 4
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-2) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]", 0-3) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-0) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-1) = 2*ar_1 + 4
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-2) = ?
4.02/1.89	
4.02/1.89		S("l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]", 0-3) = ?
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-0) = ?
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-1) = 2*ar_1 + 4
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-2) = 0
4.02/1.89	
4.02/1.89		S("l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]", 0-3) = 0
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-0) = 0
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-1) = ar_1
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-2) = ar_2
4.02/1.89	
4.02/1.89		S("l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))", 0-3) = ar_3
4.02/1.89	
4.02/1.89	orients the transitions
4.02/1.89	
4.02/1.89		l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89	weakly and the transition
4.02/1.89	
4.02/1.89		l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89	strictly and produces the following problem:
4.02/1.89	
4.02/1.89	5:	T:
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 1)                        l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3))
4.02/1.89	
4.02/1.89			(Comp: ar_1 + 1, Cost: 1)                 l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ]
4.02/1.89	
4.02/1.89			(Comp: 2*ar_1^2 + 6*ar_1 + 4, Cost: 1)    l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ]
4.02/1.89	
4.02/1.89			(Comp: ar_1 + 1, Cost: 1)                 l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ]
4.02/1.89	
4.02/1.89			(Comp: 1, Cost: 0)                        koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]
4.02/1.89	
4.02/1.89		start location:	koat_start
4.02/1.89	
4.02/1.89		leaf cost:	0
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Complexity upper bound 8*ar_1 + 2*ar_1^2 + 7
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Time: 0.132 sec (SMT: 0.124 sec)
4.02/1.89	
4.02/1.89	
4.02/1.89	----------------------------------------
4.02/1.89	
4.02/1.89	(2)
4.02/1.89	BOUNDS(1, n^2)
4.02/1.89	
4.02/1.89	----------------------------------------
4.02/1.89	
4.02/1.89	(3) Loat Proof (FINISHED)
4.02/1.89	
4.02/1.89	
4.02/1.89	### Pre-processing the ITS problem ###
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Initial linear ITS problem
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      0: l0 -> l1 : A'=0, [], cost: 1
4.02/1.89	
4.02/1.89	      1: l1 -> l2 : C'=0, D'=0, [ B>=1 ], cost: 1
4.02/1.89	
4.02/1.89	      2: l2 -> l2 : C'=1+C, D'=C+D, [ B>=1+C ], cost: 1
4.02/1.89	
4.02/1.89	      3: l2 -> l1 : A'=D+A, B'=-1+B, [ C>=B ], cost: 1
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	### Simplification by acceleration and chaining ###
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Accelerating simple loops of location 2.
4.02/1.89	
4.02/1.89	   Accelerating the following rules:
4.02/1.89	
4.02/1.89	      2: l2 -> l2 : C'=1+C, D'=C+D, [ B>=1+C ], cost: 1
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	   Accelerated rule 2 with metering function -C+B, yielding the new rule 4.
4.02/1.89	
4.02/1.89	   Removing the simple loops: 2.
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Accelerated all simple loops using metering functions (where possible):
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      0: l0 -> l1 : A'=0, [], cost: 1
4.02/1.89	
4.02/1.89	      1: l1 -> l2 : C'=0, D'=0, [ B>=1 ], cost: 1
4.02/1.89	
4.02/1.89	      3: l2 -> l1 : A'=D+A, B'=-1+B, [ C>=B ], cost: 1
4.02/1.89	
4.02/1.89	      4: l2 -> l2 : C'=B, D'=-1-1/2*C-C*(C-B)+1/2*(C-B)^2+D+1/2*B, [ B>=1+C ], cost: -C+B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Chained accelerated rules (with incoming rules):
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      0: l0 -> l1 : A'=0, [], cost: 1
4.02/1.89	
4.02/1.89	      1: l1 -> l2 : C'=0, D'=0, [ B>=1 ], cost: 1
4.02/1.89	
4.02/1.89	      5: l1 -> l2 : C'=B, D'=-1+1/2*B^2+1/2*B, [ B>=1 ], cost: 1+B
4.02/1.89	
4.02/1.89	      3: l2 -> l1 : A'=D+A, B'=-1+B, [ C>=B ], cost: 1
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Eliminated locations (on tree-shaped paths):
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      0: l0 -> l1 : A'=0, [], cost: 1
4.02/1.89	
4.02/1.89	      6: l1 -> l1 : A'=-1+1/2*B^2+A+1/2*B, B'=-1+B, C'=B, D'=-1+1/2*B^2+1/2*B, [ B>=1 ], cost: 2+B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Accelerating simple loops of location 1.
4.02/1.89	
4.02/1.89	   Accelerating the following rules:
4.02/1.89	
4.02/1.89	      6: l1 -> l1 : A'=-1+1/2*B^2+A+1/2*B, B'=-1+B, C'=B, D'=-1+1/2*B^2+1/2*B, [ B>=1 ], cost: 2+B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	   Accelerated rule 6 with metering function B, yielding the new rule 7.
4.02/1.89	
4.02/1.89	   Removing the simple loops: 6.
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Accelerated all simple loops using metering functions (where possible):
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      0: l0 -> l1 : A'=0, [], cost: 1
4.02/1.89	
4.02/1.89	      7: l1 -> l1 : A'=1+1/6*B^3+A-7/6*B, B'=0, C'=1, D'=0, [ B>=1 ], cost: 1/2*B^2+5/2*B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Chained accelerated rules (with incoming rules):
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      0: l0 -> l1 : A'=0, [], cost: 1
4.02/1.89	
4.02/1.89	      8: l0 -> l1 : A'=1+1/6*B^3-7/6*B, B'=0, C'=1, D'=0, [ B>=1 ], cost: 1+1/2*B^2+5/2*B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Removed unreachable locations (and leaf rules with constant cost):
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      8: l0 -> l1 : A'=1+1/6*B^3-7/6*B, B'=0, C'=1, D'=0, [ B>=1 ], cost: 1+1/2*B^2+5/2*B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	### Computing asymptotic complexity ###
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Fully simplified ITS problem
4.02/1.89	
4.02/1.89	   Start location: l0
4.02/1.89	
4.02/1.89	      8: l0 -> l1 : A'=1+1/6*B^3-7/6*B, B'=0, C'=1, D'=0, [ B>=1 ], cost: 1+1/2*B^2+5/2*B
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Computing asymptotic complexity for rule 8
4.02/1.89	
4.02/1.89	   Solved the limit problem by the following transformations:
4.02/1.89	
4.02/1.89	      Created initial limit problem:
4.02/1.89	
4.02/1.89	      1+1/2*B^2+5/2*B (+), B (+/+!) [not solved]
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	      removing all constraints (solved by SMT)
4.02/1.89	
4.02/1.89	      resulting limit problem: [solved]
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	      applying transformation rule (C) using substitution {B==n}
4.02/1.89	
4.02/1.89	      resulting limit problem:
4.02/1.89	
4.02/1.89	      [solved]
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	   Solution:
4.02/1.89	
4.02/1.89	      B / n
4.02/1.89	
4.02/1.89	   Resulting cost 1+5/2*n+1/2*n^2 has complexity: Poly(n^2)
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Found new complexity Poly(n^2).
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	Obtained the following overall complexity (w.r.t. the length of the input n):
4.02/1.89	
4.02/1.89	   Complexity:  Poly(n^2)
4.02/1.89	
4.02/1.89	   Cpx degree:  2
4.02/1.89	
4.02/1.89	   Solved cost: 1+5/2*n+1/2*n^2
4.02/1.89	
4.02/1.89	   Rule cost:   1+1/2*B^2+5/2*B
4.02/1.89	
4.02/1.89	   Rule guard:  [ B>=1 ]
4.02/1.89	
4.02/1.89	
4.02/1.89	
4.02/1.89	WORST_CASE(Omega(n^2),?)
4.02/1.89	
4.02/1.89	
4.02/1.89	----------------------------------------
4.02/1.89	
4.02/1.89	(4)
4.02/1.89	BOUNDS(n^2, INF)
4.02/1.91	EOF