4.99/2.39	WORST_CASE(Omega(n^1), O(n^1))
4.99/2.40	proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat
4.99/2.40	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.99/2.40	
4.99/2.40	
4.99/2.40	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, nat(2 * Arg_0 + -2 * Arg_1) + max(2, 2 + Arg_0 + -1 * Arg_1)).
4.99/2.40	
4.99/2.40	(0) CpxIntTrs
4.99/2.40	(1) Koat2 Proof [FINISHED, 640 ms]
4.99/2.40	(2) BOUNDS(1, nat(2 * Arg_0 + -2 * Arg_1) + max(2, 2 + Arg_0 + -1 * Arg_1))
4.99/2.40	(3) Loat Proof [FINISHED, 736 ms]
4.99/2.40	(4) BOUNDS(n^1, INF)
4.99/2.40	
4.99/2.40	
4.99/2.40	----------------------------------------
4.99/2.40	
4.99/2.40	(0)
4.99/2.40	Obligation:
4.99/2.40	Complexity Int TRS consisting of the following rules:
4.99/2.40	f2(A, B, C, D) -> Com_1(f300(A, B, C, D)) :|: TRUE
4.99/2.40	f300(A, B, C, D) -> Com_1(f300(1 + A, B, E, D)) :|: E >= 1 && B >= 1 + A
4.99/2.40	f300(A, B, C, D) -> Com_1(f300(1 + A, B, E, D)) :|: 0 >= E + 1 && B >= 1 + A
4.99/2.40	f300(A, B, C, D) -> Com_1(f300(A, -(1) + B, 0, D)) :|: B >= 1 + A
4.99/2.40	f300(A, B, C, D) -> Com_1(f1(A, B, C, E)) :|: A >= B
4.99/2.40	
4.99/2.40	The start-symbols are:[f2_4]
4.99/2.40	
4.99/2.40	
4.99/2.40	----------------------------------------
4.99/2.40	
4.99/2.40	(1) Koat2 Proof (FINISHED)
4.99/2.40	YES( ?, max([0, Arg_0-Arg_1])+max([2, 2+Arg_0-Arg_1])+max([0, Arg_0-Arg_1]) {O(n)})
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Initial Complexity Problem:
4.99/2.40	
4.99/2.40	  Start:  f2  
4.99/2.40	
4.99/2.40	  Program_Vars:  Arg_0, Arg_1, Arg_2, Arg_3  
4.99/2.40	
4.99/2.40	  Temp_Vars:  E  
4.99/2.40	
4.99/2.40	  Locations:  f1, f2, f300  
4.99/2.40	
4.99/2.40	  Transitions:
4.99/2.40	
4.99/2.40	  f2(Arg_0,Arg_1,Arg_2,Arg_3) -> f300(Arg_0,Arg_1,Arg_2,Arg_3):|:
4.99/2.40	
4.99/2.40	  f300(Arg_0,Arg_1,Arg_2,Arg_3) -> f1(Arg_0,Arg_1,Arg_2,E):|:Arg_0 <= Arg_1
4.99/2.40	
4.99/2.40	  f300(Arg_0,Arg_1,Arg_2,Arg_3) -> f300(Arg_0,1+Arg_1,E,Arg_3):|:1 <= E && 1+Arg_1 <= Arg_0
4.99/2.40	
4.99/2.40	  f300(Arg_0,Arg_1,Arg_2,Arg_3) -> f300(Arg_0,1+Arg_1,E,Arg_3):|:E+1 <= 0 && 1+Arg_1 <= Arg_0
4.99/2.40	
4.99/2.40	  f300(Arg_0,Arg_1,Arg_2,Arg_3) -> f300(-1+Arg_0,Arg_1,0,Arg_3):|:1+Arg_1 <= Arg_0  
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Timebounds: 
4.99/2.40	
4.99/2.40	  Overall timebound: max([0, Arg_0-Arg_1])+max([2, 2+Arg_0-Arg_1])+max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300: 1 {O(1)}
4.99/2.40	
4.99/2.40	  1: f300->f300: max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300: max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300: max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  4: f300->f1: 1 {O(1)}
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Costbounds:
4.99/2.40	
4.99/2.40	  Overall costbound: max([0, Arg_0-Arg_1])+max([2, 2+Arg_0-Arg_1])+max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300: 1 {O(1)}
4.99/2.40	
4.99/2.40	  1: f300->f300: max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300: max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300: max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  4: f300->f1: 1 {O(1)}
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Sizebounds:
4.99/2.40	
4.99/2.40	`Lower:
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_0: Arg_0 {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_1: Arg_1 {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_2: Arg_2 {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_0: Arg_0-max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_1: Arg_1 {O(n)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_2: 1 {O(1)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_0: Arg_0-max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_1: Arg_1 {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_0: Arg_0-max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_1: Arg_1 {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_2: 0 {O(1)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  4: f300->f1, Arg_0: min([Arg_0, -(-(Arg_0)+max([0, Arg_0-Arg_1]))]) {O(n)}
4.99/2.40	
4.99/2.40	  4: f300->f1, Arg_1: Arg_1 {O(n)}
4.99/2.40	
4.99/2.40	`Upper:
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_0: Arg_0 {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_1: Arg_1 {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_2: Arg_2 {O(n)}
4.99/2.40	
4.99/2.40	  0: f2->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_0: Arg_0 {O(n)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_1: Arg_1+2*max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  1: f300->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_0: Arg_0 {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_1: Arg_1+2*max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_2: -1 {O(1)}
4.99/2.40	
4.99/2.40	  2: f300->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_0: Arg_0 {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_1: Arg_1+2*max([0, Arg_0-Arg_1]) {O(n)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_2: 0 {O(1)}
4.99/2.40	
4.99/2.40	  3: f300->f300, Arg_3: Arg_3 {O(n)}
4.99/2.40	
4.99/2.40	  4: f300->f1, Arg_0: Arg_0 {O(n)}
4.99/2.40	
4.99/2.40	  4: f300->f1, Arg_1: max([Arg_1, Arg_1+2*max([0, Arg_0-Arg_1])]) {O(n)}
4.99/2.40	
4.99/2.40	
4.99/2.40	----------------------------------------
4.99/2.40	
4.99/2.40	(2)
4.99/2.40	BOUNDS(1, nat(2 * Arg_0 + -2 * Arg_1) + max(2, 2 + Arg_0 + -1 * Arg_1))
4.99/2.40	
4.99/2.40	----------------------------------------
4.99/2.40	
4.99/2.40	(3) Loat Proof (FINISHED)
4.99/2.40	
4.99/2.40	
4.99/2.40	### Pre-processing the ITS problem ###
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Initial linear ITS problem
4.99/2.40	
4.99/2.40	   Start location: f2
4.99/2.40	
4.99/2.40	      0: f2 -> f300 : [], cost: 1
4.99/2.40	
4.99/2.40	      1: f300 -> f300 : B'=1+B, C'=free, [ free>=1 && A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      3: f300 -> f300 : A'=-1+A, C'=0, [ A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      4: f300 -> f1 : D'=free_2, [ B>=A ], cost: 1
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Removed unreachable and leaf rules:
4.99/2.40	
4.99/2.40	   Start location: f2
4.99/2.40	
4.99/2.40	      0: f2 -> f300 : [], cost: 1
4.99/2.40	
4.99/2.40	      1: f300 -> f300 : B'=1+B, C'=free, [ free>=1 && A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      3: f300 -> f300 : A'=-1+A, C'=0, [ A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	### Simplification by acceleration and chaining ###
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Accelerating simple loops of location 1.
4.99/2.40	
4.99/2.40	   Accelerating the following rules:
4.99/2.40	
4.99/2.40	      1: f300 -> f300 : B'=1+B, C'=free, [ free>=1 && A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	      3: f300 -> f300 : A'=-1+A, C'=0, [ A>=1+B ], cost: 1
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	   Accelerated rule 1 with metering function A-B, yielding the new rule 5.
4.99/2.40	
4.99/2.40	   Accelerated rule 2 with metering function A-B, yielding the new rule 6.
4.99/2.40	
4.99/2.40	   Accelerated rule 3 with metering function A-B, yielding the new rule 7.
4.99/2.40	
4.99/2.40	   Removing the simple loops: 1 2 3.
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Accelerated all simple loops using metering functions (where possible):
4.99/2.40	
4.99/2.40	   Start location: f2
4.99/2.40	
4.99/2.40	      0: f2 -> f300 : [], cost: 1
4.99/2.40	
4.99/2.40	      5: f300 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: A-B
4.99/2.40	
4.99/2.40	      6: f300 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: A-B
4.99/2.40	
4.99/2.40	      7: f300 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: A-B
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Chained accelerated rules (with incoming rules):
4.99/2.40	
4.99/2.40	   Start location: f2
4.99/2.40	
4.99/2.40	      0: f2 -> f300 : [], cost: 1
4.99/2.40	
4.99/2.40	      8: f2 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Removed unreachable locations (and leaf rules with constant cost):
4.99/2.40	
4.99/2.40	   Start location: f2
4.99/2.40	
4.99/2.40	      8: f2 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	### Computing asymptotic complexity ###
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Fully simplified ITS problem
4.99/2.40	
4.99/2.40	   Start location: f2
4.99/2.40	
4.99/2.40	      8: f2 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Computing asymptotic complexity for rule 8
4.99/2.40	
4.99/2.40	   Solved the limit problem by the following transformations:
4.99/2.40	
4.99/2.40	      Created initial limit problem:
4.99/2.40	
4.99/2.40	      free (+/+!), 1+A-B (+), A-B (+/+!) [not solved]
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	      removing all constraints (solved by SMT)
4.99/2.40	
4.99/2.40	      resulting limit problem: [solved]
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	      applying transformation rule (C) using substitution {free==1,A==0,B==-n}
4.99/2.40	
4.99/2.40	      resulting limit problem:
4.99/2.40	
4.99/2.40	      [solved]
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	   Solution:
4.99/2.40	
4.99/2.40	      free / 1
4.99/2.40	
4.99/2.40	      A / 0
4.99/2.40	
4.99/2.40	      B / -n
4.99/2.40	
4.99/2.40	   Resulting cost 1+n has complexity: Poly(n^1)
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Found new complexity Poly(n^1).
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	Obtained the following overall complexity (w.r.t. the length of the input n):
4.99/2.40	
4.99/2.40	   Complexity:  Poly(n^1)
4.99/2.40	
4.99/2.40	   Cpx degree:  1
4.99/2.40	
4.99/2.40	   Solved cost: 1+n
4.99/2.40	
4.99/2.40	   Rule cost:   1+A-B
4.99/2.40	
4.99/2.40	   Rule guard:  [ free>=1 && A>=1+B ]
4.99/2.40	
4.99/2.40	
4.99/2.40	
4.99/2.40	WORST_CASE(Omega(n^1),?)
4.99/2.40	
4.99/2.40	
4.99/2.40	----------------------------------------
4.99/2.40	
4.99/2.40	(4)
4.99/2.40	BOUNDS(n^1, INF)
5.15/2.43	EOF