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Complexity_ITS 2019-03-21 04.46 pair #429990586
details
property
value
status
complete
benchmark
eric.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n089.star.cs.uiowa.edu
space
T2
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.82628 seconds
cpu usage
5.4208
user time
5.13392
system time
0.286886
max virtual memory
1.859044E7
max residence set size
213704.0
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), O(n^2))
output
5.32/2.79 WORST_CASE(Omega(n^1), O(n^2)) 5.32/2.80 proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat 5.32/2.80 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 5.32/2.80 5.32/2.80 5.32/2.80 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^2). 5.32/2.80 5.32/2.80 (0) CpxIntTrs 5.32/2.80 (1) Koat Proof [FINISHED, 233 ms] 5.32/2.80 (2) BOUNDS(1, n^2) 5.32/2.80 (3) Loat Proof [FINISHED, 1137 ms] 5.32/2.80 (4) BOUNDS(n^1, INF) 5.32/2.80 5.32/2.80 5.32/2.80 ---------------------------------------- 5.32/2.80 5.32/2.80 (0) 5.32/2.80 Obligation: 5.32/2.80 Complexity Int TRS consisting of the following rules: 5.32/2.80 f1(A, B, C) -> Com_1(f2(A, B, B + 1)) :|: A >= B && A >= 1 && B >= 1 5.32/2.80 f2(A, B, C) -> Com_1(f3(A, B, C)) :|: B >= C + 1 5.32/2.80 f2(A, B, C) -> Com_1(f3(A, B, C)) :|: C >= B + 1 5.32/2.80 f3(A, B, C) -> Com_1(f2(A, B, 0)) :|: A + 1 >= 0 && C >= 1 && C >= A + 1 5.32/2.80 f3(A, B, C) -> Com_1(f2(A, B, C + 1)) :|: A >= C && C + 1 >= 0 5.32/2.80 5.32/2.80 The start-symbols are:[f1_3] 5.32/2.80 5.32/2.80 5.32/2.80 ---------------------------------------- 5.32/2.80 5.32/2.80 (1) Koat Proof (FINISHED) 5.32/2.80 YES(?, 20*ar_0^2 + 32*ar_0*ar_1 + 280*ar_0 + 260*ar_1 + 12*ar_1^2 + 576) 5.32/2.80 5.32/2.80 5.32/2.80 5.32/2.80 Initial complexity problem: 5.32/2.80 5.32/2.80 1: T: 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f1(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_1 + 1)) [ ar_0 >= ar_1 /\ ar_0 >= 1 /\ ar_1 >= 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 0)) [ ar_0 + 1 >= 0 /\ ar_2 >= 1 /\ ar_2 >= ar_0 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_2 + 1)) [ ar_0 >= ar_2 /\ ar_2 + 1 >= 0 ] 5.32/2.80 5.32/2.80 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, ar_2)) [ 0 <= 0 ] 5.32/2.80 5.32/2.80 start location: koat_start 5.32/2.80 5.32/2.80 leaf cost: 0 5.32/2.80 5.32/2.80 5.32/2.80 5.32/2.80 Repeatedly propagating knowledge in problem 1 produces the following problem: 5.32/2.80 5.32/2.80 2: T: 5.32/2.80 5.32/2.80 (Comp: 1, Cost: 1) f1(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_1 + 1)) [ ar_0 >= ar_1 /\ ar_0 >= 1 /\ ar_1 >= 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 0)) [ ar_0 + 1 >= 0 /\ ar_2 >= 1 /\ ar_2 >= ar_0 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_2 + 1)) [ ar_0 >= ar_2 /\ ar_2 + 1 >= 0 ] 5.32/2.80 5.32/2.80 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, ar_2)) [ 0 <= 0 ] 5.32/2.80 5.32/2.80 start location: koat_start 5.32/2.80 5.32/2.80 leaf cost: 0 5.32/2.80 5.32/2.80 5.32/2.80 5.32/2.80 Applied AI with 'oct' on problem 2 to obtain the following invariants: 5.32/2.80 5.32/2.80 For symbol f2: X_1 - X_3 + 1 >= 0 /\ X_1 - X_2 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0 5.32/2.80 5.32/2.80 For symbol f3: X_1 - X_3 + 1 >= 0 /\ X_1 - X_2 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0 5.32/2.80 5.32/2.80 5.32/2.80 5.32/2.80 5.32/2.80 5.32/2.80 This yielded the following problem: 5.32/2.80 5.32/2.80 3: T: 5.32/2.80 5.32/2.80 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, ar_2)) [ 0 <= 0 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_2 + 1)) [ ar_0 - ar_2 + 1 >= 0 /\ ar_0 - ar_1 >= 0 /\ ar_1 - 1 >= 0 /\ ar_0 + ar_1 - 2 >= 0 /\ ar_0 - 1 >= 0 /\ ar_0 >= ar_2 /\ ar_2 + 1 >= 0 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 0)) [ ar_0 - ar_2 + 1 >= 0 /\ ar_0 - ar_1 >= 0 /\ ar_1 - 1 >= 0 /\ ar_0 + ar_1 - 2 >= 0 /\ ar_0 - 1 >= 0 /\ ar_0 + 1 >= 0 /\ ar_2 >= 1 /\ ar_2 >= ar_0 + 1 ] 5.32/2.80 5.32/2.80 (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 + 1 >= 0 /\ ar_0 - ar_1 >= 0 /\ ar_1 - 1 >= 0 /\ ar_0 + ar_1 - 2 >= 0 /\ ar_0 - 1 >= 0 /\ ar_2 >= ar_1 + 1 ]
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