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Complexity_C_Integer 2019-03-21 04.38 pair #429989012
details
property
value
status
complete
benchmark
svcomp_java_Continue1.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n065.star.cs.uiowa.edu
space
Adapted_from_Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c_complexity
runtime (wallclock)
1.25799 seconds
cpu usage
2.23021
user time
2.07222
system time
0.157991
max virtual memory
1.8363176E7
max residence set size
183704.0
stage attributes
key
value
starexec-result
WORST_CASE(?, O(1))
output
2.13/1.22 WORST_CASE(?, O(1)) 2.13/1.23 proof of /export/starexec/sandbox/output/output_files/bench.koat 2.13/1.23 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 2.13/1.23 2.13/1.23 2.13/1.23 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). 2.13/1.23 2.13/1.23 (0) CpxIntTrs 2.13/1.23 (1) Koat Proof [FINISHED, 74 ms] 2.13/1.23 (2) BOUNDS(1, 1) 2.13/1.23 2.13/1.23 2.13/1.23 ---------------------------------------- 2.13/1.23 2.13/1.23 (0) 2.13/1.23 Obligation: 2.13/1.23 Complexity Int TRS consisting of the following rules: 2.13/1.23 eval_foo_start(v_.0, v_c, v_i) -> Com_1(eval_foo_bb0_in(v_.0, v_c, v_i)) :|: TRUE 2.13/1.23 eval_foo_bb0_in(v_.0, v_c, v_i) -> Com_1(eval_foo_bb1_in(0, v_c, v_i)) :|: TRUE 2.13/1.23 eval_foo_bb1_in(v_.0, v_c, v_i) -> Com_1(eval_foo_bb2_in(v_.0, v_c, v_i)) :|: v_.0 < 20 2.13/1.23 eval_foo_bb1_in(v_.0, v_c, v_i) -> Com_1(eval_foo_bb3_in(v_.0, v_c, v_i)) :|: v_.0 >= 20 2.13/1.23 eval_foo_bb2_in(v_.0, v_c, v_i) -> Com_1(eval_foo_bb1_in(v_.0 + 1, v_c, v_i)) :|: v_.0 + 1 <= 10 2.13/1.23 eval_foo_bb2_in(v_.0, v_c, v_i) -> Com_1(eval_foo_bb1_in(v_.0 + 1, v_c, v_i)) :|: v_.0 + 1 > 10 2.13/1.23 eval_foo_bb3_in(v_.0, v_c, v_i) -> Com_1(eval_foo_stop(v_.0, v_c, v_i)) :|: TRUE 2.13/1.23 2.13/1.23 The start-symbols are:[eval_foo_start_3] 2.13/1.23 2.13/1.23 2.13/1.23 ---------------------------------------- 2.13/1.23 2.13/1.23 (1) Koat Proof (FINISHED) 2.13/1.23 YES(?, 123) 2.13/1.23 2.13/1.23 2.13/1.23 2.13/1.23 Initial complexity problem: 2.13/1.23 2.13/1.23 1: T: 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoostart(ar_0) -> Com_1(evalfoobb0in(ar_0)) 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb0in(ar_0) -> Com_1(evalfoobb1in(0)) 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb1in(ar_0) -> Com_1(evalfoobb2in(ar_0)) [ 19 >= ar_0 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb1in(ar_0) -> Com_1(evalfoobb3in(ar_0)) [ ar_0 >= 20 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb2in(ar_0) -> Com_1(evalfoobb1in(ar_0 + 1)) [ 9 >= ar_0 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb2in(ar_0) -> Com_1(evalfoobb1in(ar_0 + 1)) [ ar_0 >= 10 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb3in(ar_0) -> Com_1(evalfoostop(ar_0)) 2.13/1.23 2.13/1.23 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(evalfoostart(ar_0)) [ 0 <= 0 ] 2.13/1.23 2.13/1.23 start location: koat_start 2.13/1.23 2.13/1.23 leaf cost: 0 2.13/1.23 2.13/1.23 2.13/1.23 2.13/1.23 Repeatedly propagating knowledge in problem 1 produces the following problem: 2.13/1.23 2.13/1.23 2: T: 2.13/1.23 2.13/1.23 (Comp: 1, Cost: 1) evalfoostart(ar_0) -> Com_1(evalfoobb0in(ar_0)) 2.13/1.23 2.13/1.23 (Comp: 1, Cost: 1) evalfoobb0in(ar_0) -> Com_1(evalfoobb1in(0)) 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb1in(ar_0) -> Com_1(evalfoobb2in(ar_0)) [ 19 >= ar_0 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb1in(ar_0) -> Com_1(evalfoobb3in(ar_0)) [ ar_0 >= 20 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb2in(ar_0) -> Com_1(evalfoobb1in(ar_0 + 1)) [ 9 >= ar_0 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb2in(ar_0) -> Com_1(evalfoobb1in(ar_0 + 1)) [ ar_0 >= 10 ] 2.13/1.23 2.13/1.23 (Comp: ?, Cost: 1) evalfoobb3in(ar_0) -> Com_1(evalfoostop(ar_0)) 2.13/1.23 2.13/1.23 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(evalfoostart(ar_0)) [ 0 <= 0 ] 2.13/1.23 2.13/1.23 start location: koat_start 2.13/1.23 2.13/1.23 leaf cost: 0 2.13/1.23 2.13/1.23 2.13/1.23 2.13/1.23 A polynomial rank function with 2.13/1.23 2.13/1.23 Pol(evalfoostart) = 2 2.13/1.23 2.13/1.23 Pol(evalfoobb0in) = 2 2.13/1.23 2.13/1.23 Pol(evalfoobb1in) = 2 2.13/1.23 2.13/1.23 Pol(evalfoobb2in) = 2 2.13/1.23 2.13/1.23 Pol(evalfoobb3in) = 1 2.13/1.23 2.13/1.23 Pol(evalfoostop) = 0
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