Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Complexity_ITS 2019-03-21 04.46 pair #429990928
details
property
value
status
complete
benchmark
sect5-sumSum.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n001.star.cs.uiowa.edu
space
KoAT-2013
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
1.91559 seconds
cpu usage
4.15148
user time
3.90145
system time
0.250037
max virtual memory
1.8576724E7
max residence set size
214596.0
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^2), O(n^2))
output
4.02/1.88 WORST_CASE(Omega(n^2), O(n^2)) 4.02/1.89 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 4.02/1.89 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.02/1.89 4.02/1.89 4.02/1.89 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, n^2). 4.02/1.89 4.02/1.89 (0) CpxIntTrs 4.02/1.89 (1) Koat Proof [FINISHED, 118 ms] 4.02/1.89 (2) BOUNDS(1, n^2) 4.02/1.89 (3) Loat Proof [FINISHED, 324 ms] 4.02/1.89 (4) BOUNDS(n^2, INF) 4.02/1.89 4.02/1.89 4.02/1.89 ---------------------------------------- 4.02/1.89 4.02/1.89 (0) 4.02/1.89 Obligation: 4.02/1.89 Complexity Int TRS consisting of the following rules: 4.02/1.89 l0(A, B, C, D) -> Com_1(l1(0, B, C, D)) :|: TRUE 4.02/1.89 l1(A, B, C, D) -> Com_1(l2(A, B, 0, 0)) :|: B >= 1 4.02/1.89 l2(A, B, C, D) -> Com_1(l2(A, B, C + 1, D + C)) :|: B >= C + 1 4.02/1.89 l2(A, B, C, D) -> Com_1(l1(A + D, B - 1, C, D)) :|: C >= B 4.02/1.89 4.02/1.89 The start-symbols are:[l0_4] 4.02/1.89 4.02/1.89 4.02/1.89 ---------------------------------------- 4.02/1.89 4.02/1.89 (1) Koat Proof (FINISHED) 4.02/1.89 YES(?, 8*ar_1 + 2*ar_1^2 + 7) 4.02/1.89 4.02/1.89 4.02/1.89 4.02/1.89 Initial complexity problem: 4.02/1.89 4.02/1.89 1: T: 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3)) 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ] 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ] 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ] 4.02/1.89 4.02/1.89 (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 ] 4.02/1.89 4.02/1.89 start location: koat_start 4.02/1.89 4.02/1.89 leaf cost: 0 4.02/1.89 4.02/1.89 4.02/1.89 4.02/1.89 Repeatedly propagating knowledge in problem 1 produces the following problem: 4.02/1.89 4.02/1.89 2: T: 4.02/1.89 4.02/1.89 (Comp: 1, Cost: 1) l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3)) 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ] 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ] 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ] 4.02/1.89 4.02/1.89 (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 ] 4.02/1.89 4.02/1.89 start location: koat_start 4.02/1.89 4.02/1.89 leaf cost: 0 4.02/1.89 4.02/1.89 4.02/1.89 4.02/1.89 A polynomial rank function with 4.02/1.89 4.02/1.89 Pol(l0) = V_2 + 1 4.02/1.89 4.02/1.89 Pol(l1) = V_2 + 1 4.02/1.89 4.02/1.89 Pol(l2) = V_2 4.02/1.89 4.02/1.89 Pol(koat_start) = V_2 + 1 4.02/1.89 4.02/1.89 orients all transitions weakly and the transition 4.02/1.89 4.02/1.89 l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ] 4.02/1.89 4.02/1.89 strictly and produces the following problem: 4.02/1.89 4.02/1.89 3: T: 4.02/1.89 4.02/1.89 (Comp: 1, Cost: 1) l0(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(0, ar_1, ar_2, ar_3)) 4.02/1.89 4.02/1.89 (Comp: ar_1 + 1, Cost: 1) l1(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, 0, 0)) [ ar_1 >= 1 ] 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l2(ar_0, ar_1, ar_2 + 1, ar_3 + ar_2)) [ ar_1 >= ar_2 + 1 ] 4.02/1.89 4.02/1.89 (Comp: ?, Cost: 1) l2(ar_0, ar_1, ar_2, ar_3) -> Com_1(l1(ar_0 + ar_3, ar_1 - 1, ar_2, ar_3)) [ ar_2 >= ar_1 ] 4.02/1.89
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Complexity_ITS 2019-03-21 04.46