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Compl Integ Trans Syste 26843 pair #381745538
details
property
value
status
complete
benchmark
sas1.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n014.star.cs.uiowa.edu
space
T2
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
3.19792079926 seconds
cpu usage
7.13406897
max memory
3.60341504E8
stage attributes
key
value
output-size
21935
starexec-result
WORST_CASE(Omega(n^1), O(n^1))
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^1), O(n^1)) 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 312 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 1518 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C, D) -> Com_1(f4(0, B, C, D)) :|: TRUE f4(A, B, C, D) -> Com_1(f8(A + 1, B, 0, D)) :|: B >= A + 1 f8(A, B, C, D) -> Com_1(f16(A, B, C, 0)) :|: B >= A + 1 f8(A, B, C, D) -> Com_1(f16(A, B, C, D)) :|: A >= B f8(A, B, C, D) -> Com_1(f8(A + 1, B, C + 1, E)) :|: B >= A + 1 && 0 >= E + 1 f8(A, B, C, D) -> Com_1(f8(A + 1, B, C + 1, E)) :|: B >= A + 1 && E >= 1 f16(A, B, C, D) -> Com_1(f4(A, B, C, D)) :|: 0 >= C f16(A, B, C, D) -> Com_1(f4(A - 1, B, C, D)) :|: C >= 1 f4(A, B, C, D) -> Com_1(f20(A, B, C, D)) :|: A >= B The start-symbols are:[f0_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 17*ar_1 + 2) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2, ar_3) -> Com_1(f4(0, ar_1, ar_2, ar_3)) (Comp: ?, Cost: 1) f4(ar_0, ar_1, ar_2, ar_3) -> Com_1(f8(ar_0 + 1, ar_1, 0, ar_3)) [ ar_1 >= ar_0 + 1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2, ar_3) -> Com_1(f16(ar_0, ar_1, ar_2, 0)) [ ar_1 >= ar_0 + 1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2, ar_3) -> Com_1(f16(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2, ar_3) -> Com_1(f8(ar_0 + 1, ar_1, ar_2 + 1, e)) [ ar_1 >= ar_0 + 1 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2, ar_3) -> Com_1(f8(ar_0 + 1, ar_1, ar_2 + 1, e)) [ ar_1 >= ar_0 + 1 /\ e >= 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3) -> Com_1(f4(ar_0, ar_1, ar_2, ar_3)) [ 0 >= ar_2 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3) -> Com_1(f4(ar_0 - 1, ar_1, ar_2, ar_3)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f4(ar_0, ar_1, ar_2, ar_3) -> Com_1(f20(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f0(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0, ar_1, ar_2]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f4(ar_0, ar_1, ar_2) -> Com_1(f20(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0 - 1, ar_1, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2) -> Com_1(f4(ar_0, ar_1, ar_2)) [ 0 >= ar_2 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, ar_2 + 1)) [ ar_1 >= ar_0 + 1 /\ e >= 1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, ar_2 + 1)) [ ar_1 >= ar_0 + 1 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f16(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f16(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 + 1 ] (Comp: ?, Cost: 1) f4(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, 0)) [ ar_1 >= ar_0 + 1 ] (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f4(0, ar_1, ar_2))
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