details

property	value
status	complete
benchmark	p-7b.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n039.star.cs.uiowa.edu
space	T2

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.92018 seconds
cpu usage	4.22272
user time	3.94819
system time	0.274534
max virtual memory	5.4667692E7
max residence set size	221508.0

stage attributes

key	value
starexec-result	WORST_CASE(Omega(n^1), O(n^1))

output

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4.03/1.90 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.03/1.90 4.03/1.90 4.03/1.90 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). 4.03/1.90 4.03/1.90 (0) CpxIntTrs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 popout

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all output return to Complexity_ITS 2019-03-21 04.46