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Compl C Integ Progr 85445 pair #381745946
details
property
value
status
complete
benchmark
svcomp_a.08.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n024.star.cs.uiowa.edu
space
Adapted_from_Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c_complexity
runtime (wallclock)
1.21031618118 seconds
cpu usage
2.153559238
max memory
1.98586368E8
stage attributes
key
value
output-size
7341
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, n^1).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, 72 ms] (2) BOUNDS(1, n^1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_foo_start(v_.01, v_.02, v_c, v_x, v_y) -> Com_1(eval_foo_bb0_in(v_.01, v_.02, v_c, v_x, v_y)) :|: TRUE eval_foo_bb0_in(v_.01, v_.02, v_c, v_x, v_y) -> Com_1(eval_foo_bb1_in(v_x, v_y, v_c, v_x, v_y)) :|: TRUE eval_foo_bb1_in(v_.01, v_.02, v_c, v_x, v_y) -> Com_1(eval_foo_bb2_in(v_.01, v_.02, v_c, v_x, v_y)) :|: v_.01 > v_.02 eval_foo_bb1_in(v_.01, v_.02, v_c, v_x, v_y) -> Com_1(eval_foo_bb3_in(v_.01, v_.02, v_c, v_x, v_y)) :|: v_.01 <= v_.02 eval_foo_bb2_in(v_.01, v_.02, v_c, v_x, v_y) -> Com_1(eval_foo_bb1_in(v_.01 + 1, v_.02 + 2, v_c, v_x, v_y)) :|: TRUE eval_foo_bb3_in(v_.01, v_.02, v_c, v_x, v_y) -> Com_1(eval_foo_stop(v_.01, v_.02, v_c, v_x, v_y)) :|: TRUE The start-symbols are:[eval_foo_start_5] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*ar_1 + 2*ar_3 + 8) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalfoostart(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb0in(ar_0, ar_1, ar_2, ar_3)) (Comp: ?, Cost: 1) evalfoobb0in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb1in(ar_1, ar_1, ar_3, ar_3)) (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb2in(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb3in(ar_0, ar_1, ar_2, ar_3)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalfoobb2in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb1in(ar_0 + 1, ar_1, ar_2 + 2, ar_3)) (Comp: ?, Cost: 1) evalfoobb3in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoostop(ar_0, ar_1, ar_2, ar_3)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoostart(ar_0, ar_1, ar_2, ar_3)) [ 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, ar_3) -> Com_1(evalfoobb0in(ar_0, ar_1, ar_2, ar_3)) (Comp: 1, Cost: 1) evalfoobb0in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb1in(ar_1, ar_1, ar_3, ar_3)) (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb2in(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalfoobb1in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb3in(ar_0, ar_1, ar_2, ar_3)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalfoobb2in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoobb1in(ar_0 + 1, ar_1, ar_2 + 2, ar_3)) (Comp: ?, Cost: 1) evalfoobb3in(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoostop(ar_0, ar_1, ar_2, ar_3)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(evalfoostart(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalfoostart) = 2 Pol(evalfoobb0in) = 2 Pol(evalfoobb1in) = 2 Pol(evalfoobb2in) = 2 Pol(evalfoobb3in) = 1
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