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Complexity_ITS 2019-03-21 04.46 pair #429990988
details
property
value
status
complete
benchmark
easy2.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n182.star.cs.uiowa.edu
space
SAS10
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
1.92333 seconds
cpu usage
4.23167
user time
3.95758
system time
0.274093
max virtual memory
1.8341456E7
max residence set size
225820.0
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), O(n^1))
output
3.44/1.89 WORST_CASE(Omega(n^1), O(n^1)) 3.44/1.90 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 3.44/1.90 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.44/1.90 3.44/1.90 3.44/1.90 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). 3.44/1.90 3.44/1.90 (0) CpxIntTrs 3.44/1.90 (1) Koat Proof [FINISHED, 119 ms] 3.44/1.90 (2) BOUNDS(1, n^1) 3.44/1.90 (3) Loat Proof [FINISHED, 216 ms] 3.44/1.90 (4) BOUNDS(n^1, INF) 3.44/1.90 3.44/1.90 3.44/1.90 ---------------------------------------- 3.44/1.90 3.44/1.90 (0) 3.44/1.90 Obligation: 3.44/1.90 Complexity Int TRS consisting of the following rules: 3.44/1.90 start(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: 0 >= A && B >= A && B <= A && C >= D && C <= D && E >= F && E <= F 3.44/1.90 start(A, B, C, D, E, F) -> Com_1(lbl71(A, B - 1, C - 1, D, 1 + E, F)) :|: A >= 1 && B >= A && B <= A && C >= D && C <= D && E >= F && E <= F 3.44/1.90 lbl71(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: D >= C + 1 && B >= 0 && B <= 0 && E + C >= F + D && E + C <= F + D && A + C >= D && A + C <= D 3.44/1.90 lbl71(A, B, C, D, E, F) -> Com_1(lbl71(A, B - 1, C - 1, D, 1 + E, F)) :|: A + C >= D + 1 && D >= C + 1 && A + C >= D && E + C >= D + F && E + C <= D + F && B + D >= A + C && B + D <= A + C 3.44/1.90 start0(A, B, C, D, E, F) -> Com_1(start(A, A, D, D, F, F)) :|: TRUE 3.44/1.90 3.44/1.90 The start-symbols are:[start0_6] 3.44/1.90 3.44/1.90 3.44/1.90 ---------------------------------------- 3.44/1.90 3.44/1.90 (1) Koat Proof (FINISHED) 3.44/1.90 YES(?, ar_0 + 4) 3.44/1.90 3.44/1.90 3.44/1.90 3.44/1.90 Initial complexity problem: 3.44/1.90 3.44/1.90 1: T: 3.44/1.90 3.44/1.90 (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_0 /\ ar_2 = ar_3 /\ ar_4 = ar_5 ] 3.44/1.90 3.44/1.90 (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl71(ar_0, ar_1 - 1, ar_2 - 1, ar_3, ar_4 + 1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_0 /\ ar_2 = ar_3 /\ ar_4 = ar_5 ] 3.44/1.90 3.44/1.90 (Comp: ?, Cost: 1) lbl71(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_3 >= ar_2 + 1 /\ ar_1 = 0 /\ ar_4 + ar_2 = ar_5 + ar_3 /\ ar_0 + ar_2 = ar_3 ] 3.44/1.90 3.44/1.90 (Comp: ?, Cost: 1) lbl71(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl71(ar_0, ar_1 - 1, ar_2 - 1, ar_3, ar_4 + 1, ar_5)) [ ar_0 + ar_2 >= ar_3 + 1 /\ ar_3 >= ar_2 + 1 /\ ar_0 + ar_2 >= ar_3 /\ ar_4 + ar_2 = ar_3 + ar_5 /\ ar_1 + ar_3 = ar_0 + ar_2 ] 3.44/1.90 3.44/1.90 (Comp: ?, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start(ar_0, ar_0, ar_3, ar_3, ar_5, ar_5)) 3.44/1.90 3.44/1.90 (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 ] 3.44/1.90 3.44/1.90 start location: koat_start 3.44/1.90 3.44/1.90 leaf cost: 0 3.44/1.90 3.44/1.90 3.44/1.90 3.44/1.90 Repeatedly propagating knowledge in problem 1 produces the following problem: 3.44/1.90 3.44/1.90 2: T: 3.44/1.90 3.44/1.90 (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_0 /\ ar_2 = ar_3 /\ ar_4 = ar_5 ] 3.44/1.90 3.44/1.90 (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl71(ar_0, ar_1 - 1, ar_2 - 1, ar_3, ar_4 + 1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_0 /\ ar_2 = ar_3 /\ ar_4 = ar_5 ] 3.44/1.90 3.44/1.90 (Comp: ?, Cost: 1) lbl71(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_3 >= ar_2 + 1 /\ ar_1 = 0 /\ ar_4 + ar_2 = ar_5 + ar_3 /\ ar_0 + ar_2 = ar_3 ] 3.44/1.90 3.44/1.90 (Comp: ?, Cost: 1) lbl71(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl71(ar_0, ar_1 - 1, ar_2 - 1, ar_3, ar_4 + 1, ar_5)) [ ar_0 + ar_2 >= ar_3 + 1 /\ ar_3 >= ar_2 + 1 /\ ar_0 + ar_2 >= ar_3 /\ ar_4 + ar_2 = ar_3 + ar_5 /\ ar_1 + ar_3 = ar_0 + ar_2 ] 3.44/1.90 3.44/1.90 (Comp: 1, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start(ar_0, ar_0, ar_3, ar_3, ar_5, ar_5)) 3.44/1.90 3.44/1.90 (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 ] 3.44/1.90 3.44/1.90 start location: koat_start 3.44/1.90 3.44/1.90 leaf cost: 0 3.44/1.90 3.44/1.90 3.44/1.90 3.44/1.90 A polynomial rank function with 3.44/1.90 3.44/1.90 Pol(start) = 1 3.44/1.90 3.44/1.90 Pol(stop) = 0 3.44/1.90 3.44/1.90 Pol(lbl71) = 1 3.44/1.90 3.44/1.90 Pol(start0) = 1 3.44/1.90 3.44/1.90 Pol(koat_start) = 1 3.44/1.90 3.44/1.90 orients all transitions weakly and the transition 3.44/1.90 3.44/1.90 lbl71(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_3 >= ar_2 + 1 /\ ar_1 = 0 /\ ar_4 + ar_2 = ar_5 + ar_3 /\ ar_0 + ar_2 = ar_3 ] 3.44/1.90 3.44/1.90 strictly and produces the following problem: 3.44/1.90 3.44/1.90 3: T: 3.44/1.90 3.44/1.90 (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_0 /\ ar_2 = ar_3 /\ ar_4 = ar_5 ]
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