details

property	value
status	complete
benchmark	ex4.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)	1.8987 seconds
cpu usage	3.90436
user time	3.65731
system time	0.247052
max virtual memory	1.840904E7
max residence set size	223120.0

stage attributes

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

output

3.36/1.86 WORST_CASE(?, O(1)) 3.36/1.87 proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat 3.36/1.87 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.36/1.87 3.36/1.87 3.36/1.87 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). 3.36/1.87 3.36/1.87 (0) CpxIntTrs 3.36/1.87 (1) Koat Proof [FINISHED, 120 ms] 3.36/1.87 (2) BOUNDS(1, 1) 3.36/1.87 3.36/1.87 3.36/1.87 ---------------------------------------- 3.36/1.87 3.36/1.87 (0) 3.36/1.87 Obligation: 3.36/1.87 Complexity Int TRS consisting of the following rules: 3.36/1.87 f0(A, B, C, D, E, F, G) -> Com_1(f9(0, 0, H, D, E, F, G)) :|: TRUE 3.36/1.87 f9(A, B, C, D, E, F, G) -> Com_1(f10(A, B, C, C, E, F, G)) :|: 0 >= C + 1 3.36/1.87 f9(A, B, C, D, E, F, G) -> Com_1(f10(A, B, C, C, E, F, G)) :|: C >= 1 3.36/1.87 f10(A, B, C, D, E, F, G) -> Com_1(f9(A + 1, A + 1, H, D, E, F, G)) :|: 9 >= A 3.36/1.87 f16(A, B, C, D, E, F, G) -> Com_1(f28(A, B, C, D, E, F, G)) :|: A >= 10 3.36/1.87 f16(A, B, C, D, E, F, G) -> Com_1(f16(A + 1, B, C, D, A, H, H)) :|: 9 >= A && 0 >= H + 1 3.36/1.87 f16(A, B, C, D, E, F, G) -> Com_1(f16(A + 1, B, C, D, A, H, H)) :|: 9 >= A && H >= 1 3.36/1.87 f16(A, B, C, D, E, F, G) -> Com_1(f28(A, B, C, D, A, 0, 0)) :|: 9 >= A 3.36/1.87 f10(A, B, C, D, E, F, G) -> Com_1(f16(0, B, C, D, E, F, G)) :|: A >= 10 3.36/1.87 f9(A, B, C, D, E, F, G) -> Com_1(f16(0, B, 0, 0, E, F, G)) :|: C >= 0 && C <= 0 3.36/1.87 3.36/1.87 The start-symbols are:[f0_7] 3.36/1.87 3.36/1.87 3.36/1.87 ---------------------------------------- 3.36/1.87 3.36/1.87 (1) Koat Proof (FINISHED) 3.36/1.87 YES(?, 61) 3.36/1.87 3.36/1.87 3.36/1.87 3.36/1.87 Initial complexity problem: 3.36/1.87 3.36/1.87 1: T: 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f9(0, 0, h, ar_3, ar_4, ar_5, ar_6)) 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f9(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_4, ar_5, ar_6)) [ 0 >= ar_2 + 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f9(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_4, ar_5, ar_6)) [ ar_2 >= 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f10(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f9(ar_0 + 1, ar_0 + 1, h, ar_3, ar_4, ar_5, ar_6)) [ 9 >= ar_0 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f28(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6)) [ ar_0 >= 10 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(ar_0 + 1, ar_1, ar_2, ar_3, ar_0, h, h)) [ 9 >= ar_0 /\ 0 >= h + 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(ar_0 + 1, ar_1, ar_2, ar_3, ar_0, h, h)) [ 9 >= ar_0 /\ h >= 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f28(ar_0, ar_1, ar_2, ar_3, ar_0, 0, 0)) [ 9 >= ar_0 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f10(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6)) [ ar_0 >= 10 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f9(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(0, ar_1, 0, 0, ar_4, ar_5, ar_6)) [ ar_2 = 0 ] 3.36/1.87 3.36/1.87 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6)) [ 0 <= 0 ] 3.36/1.87 3.36/1.87 start location: koat_start 3.36/1.87 3.36/1.87 leaf cost: 0 3.36/1.87 3.36/1.87 3.36/1.87 3.36/1.87 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0, ar_2]. 3.36/1.87 3.36/1.87 We thus obtain the following problem: 3.36/1.87 3.36/1.87 2: T: 3.36/1.87 3.36/1.87 (Comp: 1, Cost: 0) koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] 3.36/1.87 3.36/1.87 (Comp: ?, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) 3.36/1.87 3.36/1.87 start location: koat_start 3.36/1.87 popout

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all output return to Complexity_ITS 2019-03-21 04.46