details

property	value
status	complete
benchmark	p-37.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n044.star.cs.uiowa.edu
space	T2

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.75373 seconds
cpu usage	3.63621
user time	3.40591
system time	0.230293
max virtual memory	1.8341456E7
max residence set size	217284.0

stage attributes

key	value
starexec-result	WORST_CASE(?, O(1))

output

3.35/1.72 WORST_CASE(?, O(1)) 3.35/1.73 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 3.35/1.73 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.35/1.73 3.35/1.73 3.35/1.73 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). 3.35/1.73 3.35/1.73 (0) CpxIntTrs 3.35/1.73 (1) Koat Proof [FINISHED, 22 ms] 3.35/1.73 (2) BOUNDS(1, 1) 3.35/1.73 3.35/1.73 3.35/1.73 ---------------------------------------- 3.35/1.73 3.35/1.73 (0) 3.35/1.73 Obligation: 3.35/1.73 Complexity Int TRS consisting of the following rules: 3.35/1.73 f12(A, B, C, D, E, F) -> Com_1(f5(4, 0, 0, D, E, F)) :|: TRUE 3.35/1.73 f5(A, B, C, D, E, F) -> Com_1(f11(A, B, 0, 0, 0, F)) :|: B >= A && C >= 0 && C <= 0 3.35/1.73 f5(A, B, C, D, E, F) -> Com_1(f10(A, B, C, C, C, F)) :|: C >= 1 3.35/1.73 f5(A, B, C, D, E, F) -> Com_1(f10(A, B, C, C, C, F)) :|: 0 >= C + 1 3.35/1.73 f7(A, B, C, D, E, F) -> Com_1(f5(A, 1 + B, G, H, E, G)) :|: TRUE 3.35/1.73 f8(A, B, C, D, E, F) -> Com_1(f5(A, 1 + B, G, H, E, G)) :|: TRUE 3.35/1.73 f5(A, B, C, D, E, F) -> Com_1(f5(A, 1 + B, G, H, E, G)) :|: A >= 1 + B && C >= 0 && C <= 0 3.35/1.73 3.35/1.73 The start-symbols are:[f12_6] 3.35/1.73 3.35/1.73 3.35/1.73 ---------------------------------------- 3.35/1.73 3.35/1.73 (1) Koat Proof (FINISHED) 3.35/1.73 YES(?, 8) 3.35/1.73 3.35/1.73 3.35/1.73 3.35/1.73 Initial complexity problem: 3.35/1.73 3.35/1.73 1: T: 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f12(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(4, 0, 0, ar_3, ar_4, ar_5)) 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f11(ar_0, ar_1, 0, 0, 0, ar_5)) [ ar_1 >= ar_0 /\ ar_2 = 0 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_2, ar_5)) [ ar_2 >= 1 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_2, ar_5)) [ 0 >= ar_2 + 1 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(ar_0, ar_1 + 1, g, h, ar_4, g)) 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(ar_0, ar_1 + 1, g, h, ar_4, g)) 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f5(ar_0, ar_1 + 1, g, h, ar_4, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ] 3.35/1.73 3.35/1.73 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(f12(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ] 3.35/1.73 3.35/1.73 start location: koat_start 3.35/1.73 3.35/1.73 leaf cost: 0 3.35/1.73 3.35/1.73 3.35/1.73 3.35/1.73 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0, ar_1, ar_2]. 3.35/1.73 3.35/1.73 We thus obtain the following problem: 3.35/1.73 3.35/1.73 2: T: 3.35/1.73 3.35/1.73 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, ar_1, ar_2)) [ 0 <= 0 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) [ ar_0 >= ar_1 + 1 /\ ar_2 = 0 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ 0 >= ar_2 + 1 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2) -> Com_1(f10(ar_0, ar_1, ar_2)) [ ar_2 >= 1 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ] 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f12(ar_0, ar_1, ar_2) -> Com_1(f5(4, 0, 0)) 3.35/1.73 3.35/1.73 start location: koat_start 3.35/1.73 3.35/1.73 leaf cost: 0 3.35/1.73 3.35/1.73 3.35/1.73 3.35/1.73 Testing for reachability in the complexity graph removes the following transitions from problem 2: 3.35/1.73 3.35/1.73 f8(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) 3.35/1.73 3.35/1.73 f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1 + 1, g)) 3.35/1.73 3.35/1.73 We thus obtain the following problem: 3.35/1.73 3.35/1.73 3: T: 3.35/1.73 3.35/1.73 (Comp: ?, Cost: 1) f5(ar_0, ar_1, ar_2) -> Com_1(f11(ar_0, ar_1, 0)) [ ar_1 >= ar_0 /\ ar_2 = 0 ] popout

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all output return to Complexity_ITS 2019-03-21 04.46