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Complexity_ITS 2019-03-21 04.46 pair #429990986
details
property
value
status
complete
benchmark
random1d.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n123.star.cs.uiowa.edu
space
SAS10
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.22996 seconds
cpu usage
4.70929
user time
4.43672
system time
0.272567
max virtual memory
1.8524832E7
max residence set size
218688.0
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), O(n^1))
output
4.56/2.20 WORST_CASE(Omega(n^1), O(n^1)) 4.56/2.21 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 4.56/2.21 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.56/2.21 4.56/2.21 4.56/2.21 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). 4.56/2.21 4.56/2.21 (0) CpxIntTrs 4.56/2.21 (1) Koat Proof [FINISHED, 201 ms] 4.56/2.21 (2) BOUNDS(1, n^1) 4.56/2.21 (3) Loat Proof [FINISHED, 531 ms] 4.56/2.21 (4) BOUNDS(n^1, INF) 4.56/2.21 4.56/2.21 4.56/2.21 ---------------------------------------- 4.56/2.21 4.56/2.21 (0) 4.56/2.21 Obligation: 4.56/2.21 Complexity Int TRS consisting of the following rules: 4.56/2.21 start(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: 0 >= A && B >= C && B <= C && D >= A && D <= A && E >= F && E <= F 4.56/2.21 start(A, B, C, D, E, F) -> Com_1(lbl101(A, 2, C, D, 1, F)) :|: A >= 1 && B >= C && B <= C && D >= A && D <= A && E >= F && E <= F 4.56/2.21 start(A, B, C, D, E, F) -> Com_1(lbl101(A, 2, C, D, -(1), F)) :|: A >= 1 && B >= C && B <= C && D >= A && D <= A && E >= F && E <= F 4.56/2.21 lbl101(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: E + A >= 0 && A >= 1 && A >= E && B >= A + 1 && B <= A + 1 && D >= A && D <= A 4.56/2.21 lbl101(A, B, C, D, E, F) -> Com_1(lbl101(A, 1 + B, C, D, 1 + E, F)) :|: A >= B && E + B >= 1 && A + 1 >= B && B >= 2 && B >= E + 1 && D >= A && D <= A 4.56/2.21 lbl101(A, B, C, D, E, F) -> Com_1(lbl101(A, 1 + B, C, D, E - 1, F)) :|: A >= B && E + B >= 1 && A + 1 >= B && B >= 2 && B >= E + 1 && D >= A && D <= A 4.56/2.21 start0(A, B, C, D, E, F) -> Com_1(start(A, C, C, A, F, F)) :|: TRUE 4.56/2.21 4.56/2.21 The start-symbols are:[start0_6] 4.56/2.21 4.56/2.21 4.56/2.21 ---------------------------------------- 4.56/2.21 4.56/2.21 (1) Koat Proof (FINISHED) 4.56/2.21 YES(?, 4*ar_0 + 17) 4.56/2.21 4.56/2.21 4.56/2.21 4.56/2.21 Initial complexity problem: 4.56/2.21 4.56/2.21 1: T: 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, 1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, -1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ ar_4 + ar_0 >= 0 /\ ar_0 >= 1 /\ ar_0 >= ar_4 /\ ar_1 = ar_0 + 1 /\ ar_3 = ar_0 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 + 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 - 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start(ar_0, ar_2, ar_2, ar_0, ar_5, ar_5)) 4.56/2.21 4.56/2.21 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ] 4.56/2.21 4.56/2.21 start location: koat_start 4.56/2.21 4.56/2.21 leaf cost: 0 4.56/2.21 4.56/2.21 4.56/2.21 4.56/2.21 Repeatedly propagating knowledge in problem 1 produces the following problem: 4.56/2.21 4.56/2.21 2: T: 4.56/2.21 4.56/2.21 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] 4.56/2.21 4.56/2.21 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, 1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] 4.56/2.21 4.56/2.21 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, -1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ ar_4 + ar_0 >= 0 /\ ar_0 >= 1 /\ ar_0 >= ar_4 /\ ar_1 = ar_0 + 1 /\ ar_3 = ar_0 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 + 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] 4.56/2.21 4.56/2.21 (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 - 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] 4.56/2.21 4.56/2.21 (Comp: 1, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start(ar_0, ar_2, ar_2, ar_0, ar_5, ar_5)) 4.56/2.21 4.56/2.21 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ] 4.56/2.21 4.56/2.21 start location: koat_start 4.56/2.21 4.56/2.21 leaf cost: 0 4.56/2.21 4.56/2.21 4.56/2.21 4.56/2.21 A polynomial rank function with 4.56/2.21 4.56/2.21 Pol(start) = 1 4.56/2.21 4.56/2.21 Pol(stop) = 0 4.56/2.21 4.56/2.21 Pol(lbl101) = 1 4.56/2.21 4.56/2.21 Pol(start0) = 1 4.56/2.21 4.56/2.21 Pol(koat_start) = 1
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