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Compl Integ Trans Syste 26843 pair #381744154
details
property
value
status
complete
benchmark
speedpldi3.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n056.star.cs.uiowa.edu
space
WTC
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.41932702065 seconds
cpu usage
5.580921393
max memory
3.22637824E8
stage attributes
key
value
output-size
60246
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, n^2).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, 321 ms] (2) BOUNDS(1, n^2) (3) Loat Proof [FINISHED, 832 ms] (4) BOUNDS(n^2, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalspeedpldi3start(A, B, C, D) -> Com_1(evalspeedpldi3entryin(A, B, C, D)) :|: TRUE evalspeedpldi3entryin(A, B, C, D) -> Com_1(evalspeedpldi3returnin(A, B, C, D)) :|: 0 >= A evalspeedpldi3entryin(A, B, C, D) -> Com_1(evalspeedpldi3returnin(A, B, C, D)) :|: A >= B evalspeedpldi3entryin(A, B, C, D) -> Com_1(evalspeedpldi3bb5in(A, B, 0, 0)) :|: A >= 1 && B >= A + 1 evalspeedpldi3bb5in(A, B, C, D) -> Com_1(evalspeedpldi3bb2in(A, B, C, D)) :|: B >= D + 1 evalspeedpldi3bb5in(A, B, C, D) -> Com_1(evalspeedpldi3returnin(A, B, C, D)) :|: D >= B evalspeedpldi3bb2in(A, B, C, D) -> Com_1(evalspeedpldi3bb3in(A, B, C, D)) :|: A >= C + 1 evalspeedpldi3bb2in(A, B, C, D) -> Com_1(evalspeedpldi3bb4in(A, B, C, D)) :|: C >= A evalspeedpldi3bb3in(A, B, C, D) -> Com_1(evalspeedpldi3bb5in(A, B, C + 1, D)) :|: TRUE evalspeedpldi3bb4in(A, B, C, D) -> Com_1(evalspeedpldi3bb5in(A, B, 0, D + 1)) :|: TRUE evalspeedpldi3returnin(A, B, C, D) -> Com_1(evalspeedpldi3stop(A, B, C, D)) :|: TRUE The start-symbols are:[evalspeedpldi3start_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 6*ar_1 + 6*ar_0 + 12*ar_0*ar_1 + 9) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalspeedpldi3start(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3)) (Comp: ?, Cost: 1) evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_1 ] (Comp: ?, Cost: 1) evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb5in(ar_0, ar_1, 0, 0)) [ ar_0 >= 1 /\ ar_1 >= ar_0 + 1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb5in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb2in(ar_0, ar_1, ar_2, ar_3)) [ ar_1 >= ar_3 + 1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb5in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3)) [ ar_3 >= ar_1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb2in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb3in(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb2in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb4in(ar_0, ar_1, ar_2, ar_3)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalspeedpldi3bb3in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb5in(ar_0, ar_1, ar_2 + 1, ar_3)) (Comp: ?, Cost: 1) evalspeedpldi3bb4in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb5in(ar_0, ar_1, 0, ar_3 + 1)) (Comp: ?, Cost: 1) evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3stop(ar_0, ar_1, ar_2, ar_3)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3start(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) evalspeedpldi3start(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3)) (Comp: 1, Cost: 1) evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3)) [ 0 >= ar_0 ] (Comp: 1, Cost: 1) evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_1 ] (Comp: 1, Cost: 1) evalspeedpldi3entryin(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb5in(ar_0, ar_1, 0, 0)) [ ar_0 >= 1 /\ ar_1 >= ar_0 + 1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb5in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb2in(ar_0, ar_1, ar_2, ar_3)) [ ar_1 >= ar_3 + 1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb5in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3returnin(ar_0, ar_1, ar_2, ar_3)) [ ar_3 >= ar_1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb2in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb3in(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalspeedpldi3bb2in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalspeedpldi3bb4in(ar_0, ar_1, ar_2, ar_3)) [ ar_2 >= ar_0 ]
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