details

property	value
status	complete
benchmark	practical1.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n109.star.cs.uiowa.edu
space	CAV02

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.11424 seconds
cpu usage	4.1841
user time	3.94669
system time	0.237408
max virtual memory	1.8525456E7
max residence set size	218324.0

stage attributes

key	value
starexec-result	WORST_CASE(Omega(n^2), O(n^2))

output

4.10/2.07 WORST_CASE(Omega(n^2), O(n^2)) 4.10/2.08 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 4.10/2.08 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.10/2.08 4.10/2.08 4.10/2.08 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, max(1, 3 + 2 * Arg_0) + nat(3 * Arg_0) + nat(Arg_0 * max(3 * Arg_0, -3) + max(3 * Arg_0, -3))). 4.10/2.08 4.10/2.08 (0) CpxIntTrs 4.10/2.08 (1) Koat2 Proof [FINISHED, 255 ms] 4.10/2.08 (2) BOUNDS(1, max(1, 3 + 2 * Arg_0) + nat(3 * Arg_0) + nat(Arg_0 * max(3 * Arg_0, -3) + max(3 * Arg_0, -3))) 4.10/2.08 (3) Loat Proof [FINISHED, 426 ms] 4.10/2.08 (4) BOUNDS(n^2, INF) 4.10/2.08 4.10/2.08 4.10/2.08 ---------------------------------------- 4.10/2.08 4.10/2.08 (0) 4.10/2.08 Obligation: 4.10/2.08 Complexity Int TRS consisting of the following rules: 4.10/2.08 eval1(A, B) -> Com_1(eval2(A, 0)) :|: A >= 0 4.10/2.08 eval2(A, B) -> Com_1(eval2(A, B + 1)) :|: A >= 1 + B 4.10/2.08 eval2(A, B) -> Com_1(eval1(A - 1, B)) :|: B >= A 4.10/2.08 start(A, B) -> Com_1(eval1(A, B)) :|: TRUE 4.10/2.08 4.10/2.08 The start-symbols are:[start_2] 4.10/2.08 4.10/2.08 4.10/2.08 ---------------------------------------- 4.10/2.08 4.10/2.08 (1) Koat2 Proof (FINISHED) 4.10/2.08 YES( ?, 1+2*max([0, 1+Arg_0])+max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)}) 4.10/2.08 4.10/2.08 4.10/2.08 4.10/2.08 Initial Complexity Problem: 4.10/2.08 4.10/2.08 Start: start 4.10/2.08 4.10/2.08 Program_Vars: Arg_0, Arg_1 4.10/2.08 4.10/2.08 Temp_Vars: 4.10/2.08 4.10/2.08 Locations: eval1, eval2, start 4.10/2.08 4.10/2.08 Transitions: 4.10/2.08 4.10/2.08 eval1(Arg_0,Arg_1) -> eval2(Arg_0,0):|:0 <= Arg_0 4.10/2.08 4.10/2.08 eval2(Arg_0,Arg_1) -> eval1(Arg_0-1,Arg_1):|:Arg_1 <= Arg_0 && 0 <= Arg_1 && 0 <= Arg_0+Arg_1 && 0 <= Arg_0 && Arg_0 <= Arg_1 4.10/2.08 4.10/2.08 eval2(Arg_0,Arg_1) -> eval2(Arg_0,Arg_1+1):|:Arg_1 <= Arg_0 && 0 <= Arg_1 && 0 <= Arg_0+Arg_1 && 0 <= Arg_0 && 1+Arg_1 <= Arg_0 4.10/2.08 4.10/2.08 start(Arg_0,Arg_1) -> eval1(Arg_0,Arg_1):|: 4.10/2.08 4.10/2.08 4.10/2.08 4.10/2.08 Timebounds: 4.10/2.08 4.10/2.08 Overall timebound: 1+2*max([0, 1+Arg_0])+max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)} 4.10/2.08 4.10/2.08 0: eval1->eval2: max([0, 1+Arg_0]) {O(n)} 4.10/2.08 4.10/2.08 1: eval2->eval2: max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)} 4.10/2.08 4.10/2.08 2: eval2->eval1: max([0, 1+Arg_0]) {O(n)} 4.10/2.08 4.10/2.08 3: start->eval1: 1 {O(1)} 4.10/2.08 4.10/2.08 4.10/2.08 4.10/2.08 Costbounds: 4.10/2.08 4.10/2.08 Overall costbound: 1+2*max([0, 1+Arg_0])+max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)} 4.10/2.08 4.10/2.08 0: eval1->eval2: max([0, 1+Arg_0]) {O(n)} 4.10/2.08 4.10/2.08 1: eval2->eval2: max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)} 4.10/2.08 4.10/2.08 2: eval2->eval1: max([0, 1+Arg_0]) {O(n)} 4.10/2.08 4.10/2.08 3: start->eval1: 1 {O(1)} 4.10/2.08 4.10/2.08 4.10/2.08 4.10/2.08 Sizebounds: 4.10/2.08 4.10/2.08 `Lower: 4.10/2.08 4.10/2.08 0: eval1->eval2, Arg_0: 0 {O(1)} 4.10/2.08 4.10/2.08 0: eval1->eval2, Arg_1: 0 {O(1)} 4.10/2.08 4.10/2.08 1: eval2->eval2, Arg_0: 1 {O(1)} 4.10/2.08 4.10/2.08 1: eval2->eval2, Arg_1: 1 {O(1)} 4.10/2.08 4.10/2.08 2: eval2->eval1, Arg_0: -1 {O(1)} 4.10/2.08 4.10/2.08 2: eval2->eval1, Arg_1: 0 {O(1)} 4.10/2.08 popout

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all output return to Complexity_ITS 2019-03-21 04.46