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Compl Integ Trans Syste 26843 pair #381744255
details
property
value
status
complete
benchmark
scaling-doubly-exp-growth.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
KoAT-2014
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.2859108448 seconds
cpu usage
5.211574461
max memory
3.38706432E8
stage attributes
key
value
output-size
17331
starexec-result
WORST_CASE(NON_POLY, ?)
output
/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(EXP, EXP). (0) CpxIntTrs (1) Koat Proof [FINISHED, 153 ms] (2) BOUNDS(1, EXP) (3) Loat Proof [FINISHED, 687 ms] (4) BOUNDS(EXP, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f(A, B, C) -> Com_1(g(A, 1, 1)) :|: TRUE g(A, B, C) -> Com_1(g(A - 1, 2 * B, C)) :|: A > 0 g(A, B, C) -> Com_1(h(A, B, C)) :|: A <= 0 h(A, B, C) -> Com_1(h(A, B - 1, 2 * C)) :|: B > 0 h(A, B, C) -> Com_1(i(A, B, C)) :|: B <= 0 i(A, B, C) -> Com_1(i(A, B, C - 1)) :|: C > 0 The start-symbols are:[f_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, pow(2, ar_2) * 2 + pow(2, pow(2, ar_2) * 2) * 2 + ar_2 + 5) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f(ar_0, ar_1, ar_2) -> Com_1(g(1, 1, ar_2)) (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(g(2*ar_0, ar_1, ar_2 - 1)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(h(ar_0, ar_1, ar_2)) [ 0 >= ar_2 ] (Comp: ?, Cost: 1) h(ar_0, ar_1, ar_2) -> Com_1(h(ar_0 - 1, 2*ar_1, ar_2)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) h(ar_0, ar_1, ar_2) -> Com_1(i(ar_0, ar_1, ar_2)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) i(ar_0, ar_1, ar_2) -> Com_1(i(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f(ar_0, ar_1, ar_2)) [ 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) f(ar_0, ar_1, ar_2) -> Com_1(g(1, 1, ar_2)) (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(g(2*ar_0, ar_1, ar_2 - 1)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(h(ar_0, ar_1, ar_2)) [ 0 >= ar_2 ] (Comp: ?, Cost: 1) h(ar_0, ar_1, ar_2) -> Com_1(h(ar_0 - 1, 2*ar_1, ar_2)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) h(ar_0, ar_1, ar_2) -> Com_1(i(ar_0, ar_1, ar_2)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) i(ar_0, ar_1, ar_2) -> Com_1(i(ar_0, ar_1 - 1, ar_2)) [ ar_1 >= 1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f) = 2 Pol(g) = 2 Pol(h) = 1 Pol(i) = 0
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