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Complexity_ITS 2019-03-21 04.46 pair #429989752
details
property
value
status
complete
benchmark
complete2.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
VMCAI04
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
1.81927 seconds
cpu usage
3.94034
user time
3.68657
system time
0.253763
max virtual memory
1.8505492E7
max residence set size
216748.0
stage attributes
key
value
starexec-result
WORST_CASE(?, O(1))
output
3.87/1.78 WORST_CASE(?, O(1)) 3.87/1.79 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 3.87/1.79 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.87/1.79 3.87/1.79 3.87/1.79 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). 3.87/1.79 3.87/1.79 (0) CpxIntTrs 3.87/1.79 (1) Koat Proof [FINISHED, 84 ms] 3.87/1.79 (2) BOUNDS(1, 1) 3.87/1.79 3.87/1.79 3.87/1.79 ---------------------------------------- 3.87/1.79 3.87/1.79 (0) 3.87/1.79 Obligation: 3.87/1.79 Complexity Int TRS consisting of the following rules: 3.87/1.79 eval(A) -> Com_1(eval(B)) :|: A >= 0 && B + 2 * A >= 10 && 10 >= 2 * A + B 3.87/1.79 start(A) -> Com_1(eval(A)) :|: TRUE 3.87/1.79 3.87/1.79 The start-symbols are:[start_1] 3.87/1.79 3.87/1.79 3.87/1.79 ---------------------------------------- 3.87/1.79 3.87/1.79 (1) Koat Proof (FINISHED) 3.87/1.79 YES(?, 5) 3.87/1.79 3.87/1.79 3.87/1.79 3.87/1.79 Initial complexity problem: 3.87/1.79 3.87/1.79 1: T: 3.87/1.79 3.87/1.79 (Comp: ?, Cost: 1) eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] 3.87/1.79 3.87/1.79 (Comp: ?, Cost: 1) start(ar_0) -> Com_1(eval(ar_0)) 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ] 3.87/1.79 3.87/1.79 start location: koat_start 3.87/1.79 3.87/1.79 leaf cost: 0 3.87/1.79 3.87/1.79 3.87/1.79 3.87/1.79 Repeatedly propagating knowledge in problem 1 produces the following problem: 3.87/1.79 3.87/1.79 2: T: 3.87/1.79 3.87/1.79 (Comp: ?, Cost: 1) eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 1) start(ar_0) -> Com_1(eval(ar_0)) 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ] 3.87/1.79 3.87/1.79 start location: koat_start 3.87/1.79 3.87/1.79 leaf cost: 0 3.87/1.79 3.87/1.79 3.87/1.79 3.87/1.79 By chaining the transition start(ar_0) -> Com_1(eval(ar_0)) with all transitions in problem 2, the following new transition is obtained: 3.87/1.79 3.87/1.79 start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] 3.87/1.79 3.87/1.79 We thus obtain the following problem: 3.87/1.79 3.87/1.79 3: T: 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 2) start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] 3.87/1.79 3.87/1.79 (Comp: ?, Cost: 1) eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ] 3.87/1.79 3.87/1.79 start location: koat_start 3.87/1.79 3.87/1.79 leaf cost: 0 3.87/1.79 3.87/1.79 3.87/1.79 3.87/1.79 By chaining the transition start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] with all transitions in problem 3, the following new transition is obtained: 3.87/1.79 3.87/1.79 start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ] 3.87/1.79 3.87/1.79 We thus obtain the following problem: 3.87/1.79 3.87/1.79 4: T: 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 3) start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ] 3.87/1.79 3.87/1.79 (Comp: ?, Cost: 1) eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] 3.87/1.79 3.87/1.79 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ] 3.87/1.79 3.87/1.79 start location: koat_start 3.87/1.79 3.87/1.79 leaf cost: 0 3.87/1.79
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