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Compl Integ Trans Syste 26843 pair #381744162
details
property
value
status
complete
benchmark
sect2.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n023.star.cs.uiowa.edu
space
KoAT-2013
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.12145996094 seconds
cpu usage
4.660032134
max memory
3.00527616E8
stage attributes
key
value
output-size
18292
starexec-result
WORST_CASE(Omega(n^2), O(n^2))
output
/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^2), O(n^2)) proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, n^2). (0) CpxIntTrs (1) Koat Proof [FINISHED, 135 ms] (2) BOUNDS(1, n^2) (3) Loat Proof [FINISHED, 436 ms] (4) BOUNDS(n^2, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: l0(A, B, C, D) -> Com_1(l1(0, B, C, D)) :|: TRUE l1(A, B, C, D) -> Com_1(l1(A + 1, B - 1, C, D)) :|: B >= 1 l1(A, B, C, D) -> Com_1(l2(A, B, A, D)) :|: 0 >= B l2(A, B, C, D) -> Com_1(l3(A, B, C, C)) :|: C >= 1 l3(A, B, C, D) -> Com_1(l3(A, B, C, D - 1)) :|: D >= 1 && C >= 1 l3(A, B, C, D) -> Com_1(l2(A, B, C - 1, D)) :|: 0 >= D && C >= 1 The start-symbols are:[l0_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 5*ar_1 + 2*ar_1^2 + 2) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3)) (Comp: ?, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + 1, ar_1 - 1, ar_2, ar_3)) [ ar_1 >= 1 ] (Comp: ?, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_0, ar_3)) [ 0 >= ar_1 ] (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l3(ar_0, ar_1, ar_2, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) l3(ar_0, ar_1, ar_2, ar_3) -> Com_1(l3(ar_0, ar_1, ar_2, ar_3 - 1)) [ ar_3 >= 1 /\ ar_2 >= 1 ] (Comp: ?, Cost: 1) l3(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 - 1, ar_3)) [ 0 >= ar_3 /\ ar_2 >= 1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3)) (Comp: ?, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + 1, ar_1 - 1, ar_2, ar_3)) [ ar_1 >= 1 ] (Comp: ?, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_0, ar_3)) [ 0 >= ar_1 ] (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l3(ar_0, ar_1, ar_2, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) l3(ar_0, ar_1, ar_2, ar_3) -> Com_1(l3(ar_0, ar_1, ar_2, ar_3 - 1)) [ ar_3 >= 1 /\ ar_2 >= 1 ] (Comp: ?, Cost: 1) l3(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 - 1, ar_3)) [ 0 >= ar_3 /\ ar_2 >= 1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(l0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(l0) = 1 Pol(l1) = 1 Pol(l2) = 0 Pol(l3) = 0
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