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Compl Integ Trans Syste 26843 pair #381744914
details
property
value
status
complete
benchmark
array_free.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)
1.75477194786 seconds
cpu usage
3.899271289
max memory
2.44862976E8
stage attributes
key
value
output-size
3359
starexec-result
WORST_CASE(?, O(1))
output
/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(?, O(1)) proof of /export/starexec/sandbox/benchmark/theBenchmark.koat # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 20 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A) -> Com_1(f3(0)) :|: TRUE f3(A) -> Com_1(f3(A + 1)) :|: 41 >= A f3(A) -> Com_1(f3(A + 1)) :|: 41 >= A && 0 >= B + 1 f3(A) -> Com_1(f13(A)) :|: A >= 42 The start-symbols are:[f0_1] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 86) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0) -> Com_1(f3(0)) (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f3(ar_0 + 1)) [ 41 >= ar_0 ] (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f3(ar_0 + 1)) [ 41 >= ar_0 /\ 0 >= b + 1 ] (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f13(ar_0)) [ ar_0 >= 42 ] (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 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) f0(ar_0) -> Com_1(f3(0)) (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f3(ar_0 + 1)) [ 41 >= ar_0 ] (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f3(ar_0 + 1)) [ 41 >= ar_0 /\ 0 >= b + 1 ] (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f13(ar_0)) [ ar_0 >= 42 ] (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f0) = 1 Pol(f3) = 1 Pol(f13) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition f3(ar_0) -> Com_1(f13(ar_0)) [ ar_0 >= 42 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f3(0)) (Comp: ?, Cost: 1) f3(ar_0) -> Com_1(f3(ar_0 + 1)) [ 41 >= ar_0 ]
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