details

property	value
status	complete
benchmark	Bangalore_true-termination.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n090.star.cs.uiowa.edu
space	Adapted_from_Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c_complexity
runtime (wallclock)	1.36624598503 seconds
cpu usage	2.482672815
max memory	2.39353856E8

stage attributes

key	value
output-size	11458
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, n^1).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, 183 ms] (2) BOUNDS(1, n^1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_foo_start(v_.0, v_x, v_y) -> Com_1(eval_foo_bb0_in(v_.0, v_x, v_y)) :|: TRUE eval_foo_bb0_in(v_.0, v_x, v_y) -> Com_1(eval_foo_bb1_in(v_x, v_x, v_y)) :|: v_y >= 1 eval_foo_bb0_in(v_.0, v_x, v_y) -> Com_1(eval_foo_bb3_in(v_.0, v_x, v_y)) :|: v_y < 1 eval_foo_bb1_in(v_.0, v_x, v_y) -> Com_1(eval_foo_bb2_in(v_.0, v_x, v_y)) :|: v_.0 >= 0 eval_foo_bb1_in(v_.0, v_x, v_y) -> Com_1(eval_foo_bb3_in(v_.0, v_x, v_y)) :|: v_.0 < 0 eval_foo_bb2_in(v_.0, v_x, v_y) -> Com_1(eval_foo_bb1_in(v_.0 - v_y, v_x, v_y)) :|: TRUE eval_foo_bb3_in(v_.0, v_x, v_y) -> Com_1(eval_foo_stop(v_.0, v_x, v_y)) :|: TRUE The start-symbols are:[eval_foo_start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 4*ar_2 + 11) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalfoostart(ar_0, ar_1, ar_2) -> Com_1(evalfoobb0in(ar_0, ar_1, ar_2)) (Comp: ?, Cost: 1) evalfoobb0in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb1in(ar_0, ar_2, ar_2)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) evalfoobb0in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb3in(ar_0, ar_1, ar_2)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb2in(ar_0, ar_1, ar_2)) [ ar_1 >= 0 ] (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb3in(ar_0, ar_1, ar_2)) [ 0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalfoobb2in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb1in(ar_0, ar_1 - ar_0, ar_2)) (Comp: ?, Cost: 1) evalfoobb3in(ar_0, ar_1, ar_2) -> Com_1(evalfoostop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalfoostart(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) evalfoostart(ar_0, ar_1, ar_2) -> Com_1(evalfoobb0in(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 1) evalfoobb0in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb1in(ar_0, ar_2, ar_2)) [ ar_0 >= 1 ] (Comp: 1, Cost: 1) evalfoobb0in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb3in(ar_0, ar_1, ar_2)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb2in(ar_0, ar_1, ar_2)) [ ar_1 >= 0 ] (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb3in(ar_0, ar_1, ar_2)) [ 0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalfoobb2in(ar_0, ar_1, ar_2) -> Com_1(evalfoobb1in(ar_0, ar_1 - ar_0, ar_2)) (Comp: ?, Cost: 1) evalfoobb3in(ar_0, ar_1, ar_2) -> Com_1(evalfoostop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalfoostart(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalfoostart) = 2 Pol(evalfoobb0in) = 2 Pol(evalfoobb1in) = 2 popout

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job log

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actions

all output return to Compl C Integ Progr 85445