details

property	value
status	complete
benchmark	ndecr.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n041.star.cs.uiowa.edu
space	SAS10

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.22431 seconds
cpu usage	4.00929
user time	3.73506
system time	0.274236
max virtual memory	1.8552604E7
max residence set size	219344.0

stage attributes

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

output

3.87/1.85 WORST_CASE(Omega(n^1), O(n^1)) 3.95/1.86 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 3.95/1.86 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.95/1.86 3.95/1.86 3.95/1.86 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). 3.95/1.86 3.95/1.86 (0) CpxIntTrs 3.95/1.86 (1) Koat Proof [FINISHED, 118 ms] 3.95/1.86 (2) BOUNDS(1, n^1) 3.95/1.86 (3) Loat Proof [FINISHED, 229 ms] 3.95/1.86 (4) BOUNDS(n^1, INF) 3.95/1.86 3.95/1.86 3.95/1.86 ---------------------------------------- 3.95/1.86 3.95/1.86 (0) 3.95/1.86 Obligation: 3.95/1.86 Complexity Int TRS consisting of the following rules: 3.95/1.86 start(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: 1 >= A && B >= C && B <= C && D >= A && D <= A 3.95/1.86 start(A, B, C, D) -> Com_1(lbl32(A, B, C, D - 1)) :|: A >= 2 && B >= C && B <= C && D >= A && D <= A 3.95/1.86 lbl32(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: A >= 2 && D >= 1 && D <= 1 && B >= C && B <= C 3.95/1.86 lbl32(A, B, C, D) -> Com_1(lbl32(A, B, C, D - 1)) :|: D >= 2 && D >= 1 && A >= D + 1 && B >= C && B <= C 3.95/1.86 start0(A, B, C, D) -> Com_1(start(A, C, C, A)) :|: TRUE 3.95/1.86 3.95/1.86 The start-symbols are:[start0_4] 3.95/1.86 3.95/1.86 3.95/1.86 ---------------------------------------- 3.95/1.86 3.95/1.86 (1) Koat Proof (FINISHED) 3.95/1.86 YES(?, ar_0 + 4) 3.95/1.86 3.95/1.86 3.95/1.86 3.95/1.86 Initial complexity problem: 3.95/1.86 3.95/1.86 1: T: 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3)) [ 1 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 ] 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3) -> Com_1(lbl32(ar_0, ar_1, ar_2, ar_3 - 1)) [ ar_0 >= 2 /\ ar_1 = ar_2 /\ ar_3 = ar_0 ] 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) lbl32(ar_0, ar_1, ar_2, ar_3) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= 2 /\ ar_3 = 1 /\ ar_1 = ar_2 ] 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) lbl32(ar_0, ar_1, ar_2, ar_3) -> Com_1(lbl32(ar_0, ar_1, ar_2, ar_3 - 1)) [ ar_3 >= 2 /\ ar_3 >= 1 /\ ar_0 >= ar_3 + 1 /\ ar_1 = ar_2 ] 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3) -> Com_1(start(ar_0, ar_2, ar_2, ar_0)) 3.95/1.86 3.95/1.86 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(start0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ] 3.95/1.86 3.95/1.86 start location: koat_start 3.95/1.86 3.95/1.86 leaf cost: 0 3.95/1.86 3.95/1.86 3.95/1.86 3.95/1.86 Repeatedly propagating knowledge in problem 1 produces the following problem: 3.95/1.86 3.95/1.86 2: T: 3.95/1.86 3.95/1.86 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3)) [ 1 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 ] 3.95/1.86 3.95/1.86 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3) -> Com_1(lbl32(ar_0, ar_1, ar_2, ar_3 - 1)) [ ar_0 >= 2 /\ ar_1 = ar_2 /\ ar_3 = ar_0 ] 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) lbl32(ar_0, ar_1, ar_2, ar_3) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= 2 /\ ar_3 = 1 /\ ar_1 = ar_2 ] 3.95/1.86 3.95/1.86 (Comp: ?, Cost: 1) lbl32(ar_0, ar_1, ar_2, ar_3) -> Com_1(lbl32(ar_0, ar_1, ar_2, ar_3 - 1)) [ ar_3 >= 2 /\ ar_3 >= 1 /\ ar_0 >= ar_3 + 1 /\ ar_1 = ar_2 ] 3.95/1.86 3.95/1.86 (Comp: 1, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3) -> Com_1(start(ar_0, ar_2, ar_2, ar_0)) 3.95/1.86 3.95/1.86 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(start0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ] 3.95/1.86 3.95/1.86 start location: koat_start 3.95/1.86 3.95/1.86 leaf cost: 0 3.95/1.86 3.95/1.86 3.95/1.86 3.95/1.86 A polynomial rank function with 3.95/1.86 3.95/1.86 Pol(start) = 1 3.95/1.86 3.95/1.86 Pol(stop) = 0 3.95/1.86 3.95/1.86 Pol(lbl32) = 1 3.95/1.86 3.95/1.86 Pol(start0) = 1 3.95/1.86 3.95/1.86 Pol(koat_start) = 1 3.95/1.86 3.95/1.86 orients all transitions weakly and the transition 3.95/1.86 3.95/1.86 lbl32(ar_0, ar_1, ar_2, ar_3) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= 2 /\ ar_3 = 1 /\ ar_1 = ar_2 ] 3.95/1.86 3.95/1.86 strictly and produces the following problem: 3.95/1.86 3.95/1.86 3: T: 3.95/1.86 3.95/1.86 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3)) [ 1 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 ] popout

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all output return to Complexity_ITS 2019-03-21 04.46