details

property	value
status	complete
benchmark	ndecr.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n016.star.cs.uiowa.edu
space	WTC_V2

run statistics

property	value
solver	AProVE
configuration	c_complexity
runtime (wallclock)	1.1846139431 seconds
cpu usage	2.112584565
max memory	1.92708608E8

stage attributes

key	value
output-size	6125
starexec-result	WORST_CASE(?, O(n^1))

output

/export/starexec/sandbox/solver/bin/starexec_run_c_complexity /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(?, O(n^1)) proof of /export/starexec/sandbox/output/output_files/bench.koat # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 81 ms] (2) BOUNDS(1, n^1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_ndecr_start(v_i.0, v_n) -> Com_1(eval_ndecr_bb0_in(v_i.0, v_n)) :|: TRUE eval_ndecr_bb0_in(v_i.0, v_n) -> Com_1(eval_ndecr_bb1_in(v_n - 1, v_n)) :|: TRUE eval_ndecr_bb1_in(v_i.0, v_n) -> Com_1(eval_ndecr_bb2_in(v_i.0, v_n)) :|: v_i.0 > 1 eval_ndecr_bb1_in(v_i.0, v_n) -> Com_1(eval_ndecr_bb3_in(v_i.0, v_n)) :|: v_i.0 <= 1 eval_ndecr_bb2_in(v_i.0, v_n) -> Com_1(eval_ndecr_bb1_in(v_i.0 - 1, v_n)) :|: TRUE eval_ndecr_bb3_in(v_i.0, v_n) -> Com_1(eval_ndecr_stop(v_i.0, v_n)) :|: TRUE The start-symbols are:[eval_ndecr_start_2] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*ar_1 + 6) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalndecrstart(ar_0, ar_1) -> Com_1(evalndecrbb0in(ar_0, ar_1)) (Comp: ?, Cost: 1) evalndecrbb0in(ar_0, ar_1) -> Com_1(evalndecrbb1in(ar_1 - 1, ar_1)) (Comp: ?, Cost: 1) evalndecrbb1in(ar_0, ar_1) -> Com_1(evalndecrbb2in(ar_0, ar_1)) [ ar_0 >= 2 ] (Comp: ?, Cost: 1) evalndecrbb1in(ar_0, ar_1) -> Com_1(evalndecrbb3in(ar_0, ar_1)) [ 1 >= ar_0 ] (Comp: ?, Cost: 1) evalndecrbb2in(ar_0, ar_1) -> Com_1(evalndecrbb1in(ar_0 - 1, ar_1)) (Comp: ?, Cost: 1) evalndecrbb3in(ar_0, ar_1) -> Com_1(evalndecrstop(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(evalndecrstart(ar_0, ar_1)) [ 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) evalndecrstart(ar_0, ar_1) -> Com_1(evalndecrbb0in(ar_0, ar_1)) (Comp: 1, Cost: 1) evalndecrbb0in(ar_0, ar_1) -> Com_1(evalndecrbb1in(ar_1 - 1, ar_1)) (Comp: ?, Cost: 1) evalndecrbb1in(ar_0, ar_1) -> Com_1(evalndecrbb2in(ar_0, ar_1)) [ ar_0 >= 2 ] (Comp: ?, Cost: 1) evalndecrbb1in(ar_0, ar_1) -> Com_1(evalndecrbb3in(ar_0, ar_1)) [ 1 >= ar_0 ] (Comp: ?, Cost: 1) evalndecrbb2in(ar_0, ar_1) -> Com_1(evalndecrbb1in(ar_0 - 1, ar_1)) (Comp: ?, Cost: 1) evalndecrbb3in(ar_0, ar_1) -> Com_1(evalndecrstop(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(evalndecrstart(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalndecrstart) = 2 Pol(evalndecrbb0in) = 2 Pol(evalndecrbb1in) = 2 Pol(evalndecrbb2in) = 2 Pol(evalndecrbb3in) = 1 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Compl C Integ Progr 85445