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Compl Integ Trans Syste 26843 pair #381744125
details
property
value
status
complete
benchmark
a.04.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n003.star.cs.uiowa.edu
space
pasta
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
1.73045682907 seconds
cpu usage
3.722423679
max memory
2.35782144E8
stage attributes
key
value
output-size
4101
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(1, 1 + Arg_0 + -1 * Arg_1)). (0) CpxIntTrs (1) Koat2 Proof [FINISHED, 26 ms] (2) BOUNDS(1, max(1, 1 + Arg_0 + -1 * Arg_1)) (3) Loat Proof [FINISHED, 125 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval(A, B) -> Com_1(eval(A, B + 1)) :|: A >= B + 1 start(A, B) -> Com_1(eval(A, B)) :|: TRUE The start-symbols are:[start_2] ---------------------------------------- (1) Koat2 Proof (FINISHED) YES( ?, max([1, 1+Arg_0-Arg_1]) {O(n)}) Initial Complexity Problem: Start: start Program_Vars: Arg_0, Arg_1 Temp_Vars: Locations: eval, start Transitions: eval(Arg_0,Arg_1) -> eval(Arg_0,Arg_1+1):|:Arg_1+1 <= Arg_0 start(Arg_0,Arg_1) -> eval(Arg_0,Arg_1):|: Timebounds: Overall timebound: max([1, 1+Arg_0-Arg_1]) {O(n)} 0: eval->eval: max([0, Arg_0-Arg_1]) {O(n)} 1: start->eval: 1 {O(1)} Costbounds: Overall costbound: max([1, 1+Arg_0-Arg_1]) {O(n)} 0: eval->eval: max([0, Arg_0-Arg_1]) {O(n)} 1: start->eval: 1 {O(1)} Sizebounds: `Lower: 0: eval->eval, Arg_0: Arg_0 {O(n)} 0: eval->eval, Arg_1: Arg_1 {O(n)} 1: start->eval, Arg_0: Arg_0 {O(n)} 1: start->eval, Arg_1: Arg_1 {O(n)} `Upper: 0: eval->eval, Arg_0: Arg_0 {O(n)} 0: eval->eval, Arg_1: Arg_1+max([0, Arg_0-Arg_1]) {O(n)} 1: start->eval, Arg_0: Arg_0 {O(n)} 1: start->eval, Arg_1: Arg_1 {O(n)} ----------------------------------------
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