3.89/1.87	WORST_CASE(Omega(n^1), O(n^1))
3.89/1.88	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
3.89/1.88	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.89/1.88	
3.89/1.88	
3.89/1.88	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).
3.89/1.88	
3.89/1.88	(0) CpxIntTrs
3.89/1.88	(1) Koat Proof [FINISHED, 17 ms]
3.89/1.88	(2) BOUNDS(1, n^1)
3.89/1.88	(3) Loat Proof [FINISHED, 217 ms]
3.89/1.88	(4) BOUNDS(n^1, INF)
3.89/1.88	
3.89/1.88	
3.89/1.88	----------------------------------------
3.89/1.88	
3.89/1.88	(0)
3.89/1.88	Obligation:
3.89/1.88	Complexity Int TRS consisting of the following rules:
3.89/1.88	evalSimpleSinglestart(A, B) -> Com_1(evalSimpleSingleentryin(A, B)) :|: TRUE
3.89/1.88	evalSimpleSingleentryin(A, B) -> Com_1(evalSimpleSinglebb3in(0, B)) :|: TRUE
3.89/1.88	evalSimpleSinglebb3in(A, B) -> Com_1(evalSimpleSinglebbin(A, B)) :|: B >= A + 1
3.89/1.88	evalSimpleSinglebb3in(A, B) -> Com_1(evalSimpleSinglereturnin(A, B)) :|: A >= B
3.89/1.88	evalSimpleSinglebbin(A, B) -> Com_1(evalSimpleSinglebb3in(A + 1, B)) :|: TRUE
3.89/1.88	evalSimpleSinglereturnin(A, B) -> Com_1(evalSimpleSinglestop(A, B)) :|: TRUE
3.89/1.88	
3.89/1.88	The start-symbols are:[evalSimpleSinglestart_2]
3.89/1.88	
3.89/1.88	
3.89/1.88	----------------------------------------
3.89/1.89	
3.89/1.89	(1) Koat Proof (FINISHED)
3.89/1.89	YES(?, 2*ar_1 + 6)
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Initial complexity problem:
3.89/1.89	
3.89/1.89	1:	T:
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglestart(ar_0, ar_1) -> Com_1(evalSimpleSingleentryin(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSingleentryin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglebbin(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglereturnin(ar_0, ar_1)) [ ar_0 >= ar_1 ]
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebbin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(ar_0 + 1, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglereturnin(ar_0, ar_1) -> Com_1(evalSimpleSinglestop(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(evalSimpleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]
3.89/1.89	
3.89/1.89		start location:	koat_start
3.89/1.89	
3.89/1.89		leaf cost:	0
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Repeatedly propagating knowledge in problem 1 produces the following problem:
3.89/1.89	
3.89/1.89	2:	T:
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)    evalSimpleSinglestart(ar_0, ar_1) -> Com_1(evalSimpleSingleentryin(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)    evalSimpleSingleentryin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglebbin(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglereturnin(ar_0, ar_1)) [ ar_0 >= ar_1 ]
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebbin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(ar_0 + 1, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglereturnin(ar_0, ar_1) -> Com_1(evalSimpleSinglestop(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(evalSimpleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]
3.89/1.89	
3.89/1.89		start location:	koat_start
3.89/1.89	
3.89/1.89		leaf cost:	0
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	A polynomial rank function with
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglestart) = 2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSingleentryin) = 2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglebb3in) = 2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglebbin) = 2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglereturnin) = 1
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglestop) = 0
3.89/1.89	
3.89/1.89		Pol(koat_start) = 2
3.89/1.89	
3.89/1.89	orients all transitions weakly and the transitions
3.89/1.89	
3.89/1.89		evalSimpleSinglereturnin(ar_0, ar_1) -> Com_1(evalSimpleSinglestop(ar_0, ar_1))
3.89/1.89	
3.89/1.89		evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglereturnin(ar_0, ar_1)) [ ar_0 >= ar_1 ]
3.89/1.89	
3.89/1.89	strictly and produces the following problem:
3.89/1.89	
3.89/1.89	3:	T:
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)    evalSimpleSinglestart(ar_0, ar_1) -> Com_1(evalSimpleSingleentryin(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)    evalSimpleSingleentryin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglebbin(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]
3.89/1.89	
3.89/1.89			(Comp: 2, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglereturnin(ar_0, ar_1)) [ ar_0 >= ar_1 ]
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)    evalSimpleSinglebbin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(ar_0 + 1, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 2, Cost: 1)    evalSimpleSinglereturnin(ar_0, ar_1) -> Com_1(evalSimpleSinglestop(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(evalSimpleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]
3.89/1.89	
3.89/1.89		start location:	koat_start
3.89/1.89	
3.89/1.89		leaf cost:	0
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	A polynomial rank function with
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglestart) = V_2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSingleentryin) = V_2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglebb3in) = -V_1 + V_2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglebbin) = -V_1 + V_2 - 1
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglereturnin) = -V_1 + V_2
3.89/1.89	
3.89/1.89		Pol(evalSimpleSinglestop) = -V_1 + V_2
3.89/1.89	
3.89/1.89		Pol(koat_start) = V_2
3.89/1.89	
3.89/1.89	orients all transitions weakly and the transition
3.89/1.89	
3.89/1.89		evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglebbin(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]
3.89/1.89	
3.89/1.89	strictly and produces the following problem:
3.89/1.89	
3.89/1.89	4:	T:
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)       evalSimpleSinglestart(ar_0, ar_1) -> Com_1(evalSimpleSingleentryin(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)       evalSimpleSingleentryin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ar_1, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglebbin(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]
3.89/1.89	
3.89/1.89			(Comp: 2, Cost: 1)       evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglereturnin(ar_0, ar_1)) [ ar_0 >= ar_1 ]
3.89/1.89	
3.89/1.89			(Comp: ?, Cost: 1)       evalSimpleSinglebbin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(ar_0 + 1, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 2, Cost: 1)       evalSimpleSinglereturnin(ar_0, ar_1) -> Com_1(evalSimpleSinglestop(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 0)       koat_start(ar_0, ar_1) -> Com_1(evalSimpleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]
3.89/1.89	
3.89/1.89		start location:	koat_start
3.89/1.89	
3.89/1.89		leaf cost:	0
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Repeatedly propagating knowledge in problem 4 produces the following problem:
3.89/1.89	
3.89/1.89	5:	T:
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)       evalSimpleSinglestart(ar_0, ar_1) -> Com_1(evalSimpleSingleentryin(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 1)       evalSimpleSingleentryin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: ar_1, Cost: 1)    evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglebbin(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]
3.89/1.89	
3.89/1.89			(Comp: 2, Cost: 1)       evalSimpleSinglebb3in(ar_0, ar_1) -> Com_1(evalSimpleSinglereturnin(ar_0, ar_1)) [ ar_0 >= ar_1 ]
3.89/1.89	
3.89/1.89			(Comp: ar_1, Cost: 1)    evalSimpleSinglebbin(ar_0, ar_1) -> Com_1(evalSimpleSinglebb3in(ar_0 + 1, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 2, Cost: 1)       evalSimpleSinglereturnin(ar_0, ar_1) -> Com_1(evalSimpleSinglestop(ar_0, ar_1))
3.89/1.89	
3.89/1.89			(Comp: 1, Cost: 0)       koat_start(ar_0, ar_1) -> Com_1(evalSimpleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]
3.89/1.89	
3.89/1.89		start location:	koat_start
3.89/1.89	
3.89/1.89		leaf cost:	0
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Complexity upper bound 2*ar_1 + 6
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Time: 0.040 sec (SMT: 0.036 sec)
3.89/1.89	
3.89/1.89	
3.89/1.89	----------------------------------------
3.89/1.89	
3.89/1.89	(2)
3.89/1.89	BOUNDS(1, n^1)
3.89/1.89	
3.89/1.89	----------------------------------------
3.89/1.89	
3.89/1.89	(3) Loat Proof (FINISHED)
3.89/1.89	
3.89/1.89	
3.89/1.89	### Pre-processing the ITS problem ###
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Initial linear ITS problem
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      0: evalSimpleSinglestart -> evalSimpleSingleentryin : [], cost: 1
3.89/1.89	
3.89/1.89	      1: evalSimpleSingleentryin -> evalSimpleSinglebb3in : A'=0, [], cost: 1
3.89/1.89	
3.89/1.89	      2: evalSimpleSinglebb3in -> evalSimpleSinglebbin : [ B>=1+A ], cost: 1
3.89/1.89	
3.89/1.89	      3: evalSimpleSinglebb3in -> evalSimpleSinglereturnin : [ A>=B ], cost: 1
3.89/1.89	
3.89/1.89	      4: evalSimpleSinglebbin -> evalSimpleSinglebb3in : A'=1+A, [], cost: 1
3.89/1.89	
3.89/1.89	      5: evalSimpleSinglereturnin -> evalSimpleSinglestop : [], cost: 1
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Removed unreachable and leaf rules:
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      0: evalSimpleSinglestart -> evalSimpleSingleentryin : [], cost: 1
3.89/1.89	
3.89/1.89	      1: evalSimpleSingleentryin -> evalSimpleSinglebb3in : A'=0, [], cost: 1
3.89/1.89	
3.89/1.89	      2: evalSimpleSinglebb3in -> evalSimpleSinglebbin : [ B>=1+A ], cost: 1
3.89/1.89	
3.89/1.89	      4: evalSimpleSinglebbin -> evalSimpleSinglebb3in : A'=1+A, [], cost: 1
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	### Simplification by acceleration and chaining ###
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Eliminated locations (on linear paths):
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      6: evalSimpleSinglestart -> evalSimpleSinglebb3in : A'=0, [], cost: 2
3.89/1.89	
3.89/1.89	      7: evalSimpleSinglebb3in -> evalSimpleSinglebb3in : A'=1+A, [ B>=1+A ], cost: 2
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Accelerating simple loops of location 2.
3.89/1.89	
3.89/1.89	   Accelerating the following rules:
3.89/1.89	
3.89/1.89	      7: evalSimpleSinglebb3in -> evalSimpleSinglebb3in : A'=1+A, [ B>=1+A ], cost: 2
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	   Accelerated rule 7 with metering function -A+B, yielding the new rule 8.
3.89/1.89	
3.89/1.89	   Removing the simple loops: 7.
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Accelerated all simple loops using metering functions (where possible):
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      6: evalSimpleSinglestart -> evalSimpleSinglebb3in : A'=0, [], cost: 2
3.89/1.89	
3.89/1.89	      8: evalSimpleSinglebb3in -> evalSimpleSinglebb3in : A'=B, [ B>=1+A ], cost: -2*A+2*B
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Chained accelerated rules (with incoming rules):
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      6: evalSimpleSinglestart -> evalSimpleSinglebb3in : A'=0, [], cost: 2
3.89/1.89	
3.89/1.89	      9: evalSimpleSinglestart -> evalSimpleSinglebb3in : A'=B, [ B>=1 ], cost: 2+2*B
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Removed unreachable locations (and leaf rules with constant cost):
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      9: evalSimpleSinglestart -> evalSimpleSinglebb3in : A'=B, [ B>=1 ], cost: 2+2*B
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	### Computing asymptotic complexity ###
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Fully simplified ITS problem
3.89/1.89	
3.89/1.89	   Start location: evalSimpleSinglestart
3.89/1.89	
3.89/1.89	      9: evalSimpleSinglestart -> evalSimpleSinglebb3in : A'=B, [ B>=1 ], cost: 2+2*B
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Computing asymptotic complexity for rule 9
3.89/1.89	
3.89/1.89	   Solved the limit problem by the following transformations:
3.89/1.89	
3.89/1.89	      Created initial limit problem:
3.89/1.89	
3.89/1.89	      2+2*B (+), B (+/+!) [not solved]
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	      removing all constraints (solved by SMT)
3.89/1.89	
3.89/1.89	      resulting limit problem: [solved]
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	      applying transformation rule (C) using substitution {B==n}
3.89/1.89	
3.89/1.89	      resulting limit problem:
3.89/1.89	
3.89/1.89	      [solved]
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	   Solution:
3.89/1.89	
3.89/1.89	      B / n
3.89/1.89	
3.89/1.89	   Resulting cost 2+2*n has complexity: Poly(n^1)
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Found new complexity Poly(n^1).
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	Obtained the following overall complexity (w.r.t. the length of the input n):
3.89/1.89	
3.89/1.89	   Complexity:  Poly(n^1)
3.89/1.89	
3.89/1.89	   Cpx degree:  1
3.89/1.89	
3.89/1.89	   Solved cost: 2+2*n
3.89/1.89	
3.89/1.89	   Rule cost:   2+2*B
3.89/1.89	
3.89/1.89	   Rule guard:  [ B>=1 ]
3.89/1.89	
3.89/1.89	
3.89/1.89	
3.89/1.89	WORST_CASE(Omega(n^1),?)
3.89/1.89	
3.89/1.89	
3.89/1.89	----------------------------------------
3.89/1.89	
3.89/1.89	(4)
3.89/1.89	BOUNDS(n^1, INF)
4.00/1.91	EOF