/export/starexec/sandbox2/solver/bin/starexec_run_c_complexity /export/starexec/sandbox2/benchmark/theBenchmark.c /export/starexec/sandbox2/output/output_files


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WORST_CASE(?, O(n^1))
proof of /export/starexec/sandbox2/output/output_files/bench.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, n^1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 84 ms]
(2) BOUNDS(1, n^1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval_speed_popl10_simple_single_start(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_bb0_in(v_n, v_x.0)) :|: TRUE
eval_speed_popl10_simple_single_bb0_in(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_bb1_in(v_n, 0)) :|: TRUE
eval_speed_popl10_simple_single_bb1_in(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_bb2_in(v_n, v_x.0)) :|: v_x.0 < v_n
eval_speed_popl10_simple_single_bb1_in(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_bb3_in(v_n, v_x.0)) :|: v_x.0 >= v_n
eval_speed_popl10_simple_single_bb2_in(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_0(v_n, v_x.0)) :|: TRUE
eval_speed_popl10_simple_single_0(v_n, v_x.0) -> Com_2(eval_nondet_start(v_n, v_x.0), eval_speed_popl10_simple_single_1(v_n, v_x.0)) :|: TRUE
eval_speed_popl10_simple_single_1(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_bb1_in(v_n, v_x.0 + 1)) :|: TRUE
eval_speed_popl10_simple_single_bb3_in(v_n, v_x.0) -> Com_1(eval_speed_popl10_simple_single_stop(v_n, v_x.0)) :|: TRUE

The start-symbols are:[eval_speed_popl10_simple_single_start_2]


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(1) Koat Proof (FINISHED)
YES(?, 16*Ar_1 + 6)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglestart(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesingle0(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle0(Ar_0, Ar_1) -> Com_2(evalnondetstart(Ar_0, Ar_1), evalspeedpopl10simplesingle1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle1(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(Ar_0 + 1, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestart(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)    evalspeedpopl10simplesinglestart(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesingle0(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle0(Ar_0, Ar_1) -> Com_2(evalnondetstart(Ar_0, Ar_1), evalspeedpopl10simplesingle1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle1(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(Ar_0 + 1, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalspeedpopl10simplesinglestart) = 2

	Pol(evalspeedpopl10simplesinglebb0in) = 2

	Pol(evalspeedpopl10simplesinglebb1in) = 2

	Pol(evalspeedpopl10simplesinglebb2in) = 2

	Pol(evalspeedpopl10simplesinglebb3in) = 1

	Pol(evalspeedpopl10simplesingle0) = 2

	Pol(evalnondetstart) = 0

	Pol(evalspeedpopl10simplesingle1) = 2

	Pol(evalspeedpopl10simplesinglestop) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestop(Ar_0, Ar_1))

	evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    evalspeedpopl10simplesinglestart(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1))

		(Comp: 1, Cost: 1)    evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: 2, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesingle0(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle0(Ar_0, Ar_1) -> Com_2(evalnondetstart(Ar_0, Ar_1), evalspeedpopl10simplesingle1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle1(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(Ar_0 + 1, Ar_1))

		(Comp: 2, Cost: 1)    evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestop(Ar_0, Ar_1))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestart(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Applied AI with 'oct' on problem 3 to obtain the following invariants:

  For symbol evalspeedpopl10simplesingle0: X_2 - 1 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ X_1 >= 0

  For symbol evalspeedpopl10simplesingle1: X_2 - 1 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ X_1 >= 0

  For symbol evalspeedpopl10simplesinglebb1in: X_1 >= 0

  For symbol evalspeedpopl10simplesinglebb2in: X_2 - 1 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ X_1 >= 0

  For symbol evalspeedpopl10simplesinglebb3in: X_1 - X_2 >= 0 /\ X_1 >= 0





This yielded the following problem:

4:	T:

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestart(Ar_0, Ar_1)) [ 0 <= 0 ]

		(Comp: 2, Cost: 1)    evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestop(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle1(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(Ar_0 + 1, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesingle0(Ar_0, Ar_1) -> Com_2(evalnondetstart(Ar_0, Ar_1), evalspeedpopl10simplesingle1(Ar_0, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesingle0(Ar_0, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1)) [ Ar_0 >= 0 /\ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)    evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(0, Ar_1))

		(Comp: 1, Cost: 1)    evalspeedpopl10simplesinglestart(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1))

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = 4*V_2

	Pol(evalspeedpopl10simplesinglestart) = 4*V_2

	Pol(evalspeedpopl10simplesinglebb3in) = -4*V_1 + 4*V_2

	Pol(evalspeedpopl10simplesinglestop) = -4*V_1 + 4*V_2

	Pol(evalspeedpopl10simplesingle1) = -4*V_1 + 4*V_2 - 3

	Pol(evalspeedpopl10simplesinglebb1in) = -4*V_1 + 4*V_2

	Pol(evalspeedpopl10simplesingle0) = -4*V_1 + 4*V_2 - 2

	Pol(evalnondetstart) = -4*V_1

	Pol(evalspeedpopl10simplesinglebb2in) = -4*V_1 + 4*V_2 - 1

	Pol(evalspeedpopl10simplesinglebb0in) = 4*V_2

orients all transitions weakly and the transitions

	evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesingle0(Ar_0, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

	evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

	evalspeedpopl10simplesingle1(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(Ar_0 + 1, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

	evalspeedpopl10simplesingle0(Ar_0, Ar_1) -> Com_2(evalnondetstart(Ar_0, Ar_1), evalspeedpopl10simplesingle1(Ar_0, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 0)         koat_start(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestart(Ar_0, Ar_1)) [ 0 <= 0 ]

		(Comp: 2, Cost: 1)         evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglestop(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: 4*Ar_1, Cost: 1)    evalspeedpopl10simplesingle1(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(Ar_0 + 1, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: 4*Ar_1, Cost: 1)    evalspeedpopl10simplesingle0(Ar_0, Ar_1) -> Com_2(evalnondetstart(Ar_0, Ar_1), evalspeedpopl10simplesingle1(Ar_0, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: 4*Ar_1, Cost: 1)    evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesingle0(Ar_0, Ar_1)) [ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

		(Comp: 2, Cost: 1)         evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb3in(Ar_0, Ar_1)) [ Ar_0 >= 0 /\ Ar_0 >= Ar_1 ]

		(Comp: 4*Ar_1, Cost: 1)    evalspeedpopl10simplesinglebb1in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb2in(Ar_0, Ar_1)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)         evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb1in(0, Ar_1))

		(Comp: 1, Cost: 1)         evalspeedpopl10simplesinglestart(Ar_0, Ar_1) -> Com_1(evalspeedpopl10simplesinglebb0in(Ar_0, Ar_1))

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 16*Ar_1 + 6



Time: 0.101 sec (SMT: 0.083 sec)


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(2)
BOUNDS(1, n^1)