/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 74 ms]
(2) BOUNDS(1, n^1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval_speedSingleSingle_start(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_bb0_in(v_n, v_x.0)) :|: TRUE
eval_speedSingleSingle_bb0_in(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_bb1_in(v_n, 0)) :|: TRUE
eval_speedSingleSingle_bb1_in(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_bb2_in(v_n, v_x.0)) :|: v_x.0 < v_n
eval_speedSingleSingle_bb1_in(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_bb3_in(v_n, v_x.0)) :|: v_x.0 >= v_n
eval_speedSingleSingle_bb2_in(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_0(v_n, v_x.0)) :|: TRUE
eval_speedSingleSingle_0(v_n, v_x.0) -> Com_2(eval_nondet_start(v_n, v_x.0), eval_speedSingleSingle_1(v_n, v_x.0)) :|: TRUE
eval_speedSingleSingle_1(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_bb1_in(v_n, v_x.0 + 1)) :|: TRUE
eval_speedSingleSingle_bb3_in(v_n, v_x.0) -> Com_1(eval_speedSingleSingle_stop(v_n, v_x.0)) :|: TRUE

The start-symbols are:[eval_speedSingleSingle_start_2]


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(1) Koat Proof (FINISHED)
YES(?, 16*ar_1 + 6)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    evalspeedSingleSinglestart(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb0in(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb0in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb2in(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb3in(ar_0, ar_1)) [ ar_0 >= ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb2in(ar_0, ar_1) -> Com_1(evalspeedSingleSingle0(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSingle0(ar_0, ar_1) -> Com_2(evalnondetstart(ar_0, ar_1), evalspeedSingleSingle1(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSingle1(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(ar_0 + 1, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb3in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestop(ar_0, ar_1))

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestart(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)    evalspeedSingleSinglestart(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb0in(ar_0, ar_1))

		(Comp: 1, Cost: 1)    evalspeedSingleSinglebb0in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb2in(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb3in(ar_0, ar_1)) [ ar_0 >= ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb2in(ar_0, ar_1) -> Com_1(evalspeedSingleSingle0(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSingle0(ar_0, ar_1) -> Com_2(evalnondetstart(ar_0, ar_1), evalspeedSingleSingle1(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSingle1(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(ar_0 + 1, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb3in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestop(ar_0, ar_1))

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalspeedSingleSinglestart) = 2

	Pol(evalspeedSingleSinglebb0in) = 2

	Pol(evalspeedSingleSinglebb1in) = 2

	Pol(evalspeedSingleSinglebb2in) = 2

	Pol(evalspeedSingleSinglebb3in) = 1

	Pol(evalspeedSingleSingle0) = 2

	Pol(evalnondetstart) = 0

	Pol(evalspeedSingleSingle1) = 2

	Pol(evalspeedSingleSinglestop) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	evalspeedSingleSinglebb3in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestop(ar_0, ar_1))

	evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb3in(ar_0, ar_1)) [ ar_0 >= ar_1 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    evalspeedSingleSinglestart(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb0in(ar_0, ar_1))

		(Comp: 1, Cost: 1)    evalspeedSingleSinglebb0in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb2in(ar_0, ar_1)) [ ar_1 >= ar_0 + 1 ]

		(Comp: 2, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb3in(ar_0, ar_1)) [ ar_0 >= ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb2in(ar_0, ar_1) -> Com_1(evalspeedSingleSingle0(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSingle0(ar_0, ar_1) -> Com_2(evalnondetstart(ar_0, ar_1), evalspeedSingleSingle1(ar_0, ar_1))

		(Comp: ?, Cost: 1)    evalspeedSingleSingle1(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(ar_0 + 1, ar_1))

		(Comp: 2, Cost: 1)    evalspeedSingleSinglebb3in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestop(ar_0, ar_1))

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestart(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 evalspeedSingleSingle0: X_2 - 1 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ X_1 >= 0

  For symbol evalspeedSingleSingle1: X_2 - 1 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ X_1 >= 0

  For symbol evalspeedSingleSinglebb1in: X_1 >= 0

  For symbol evalspeedSingleSinglebb2in: X_2 - 1 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ X_1 >= 0

  For symbol evalspeedSingleSinglebb3in: 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(evalspeedSingleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]

		(Comp: 2, Cost: 1)    evalspeedSingleSinglebb3in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestop(ar_0, ar_1)) [ ar_0 - ar_1 >= 0 /\ ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSingle1(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(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)    evalspeedSingleSingle0(ar_0, ar_1) -> Com_2(evalnondetstart(ar_0, ar_1), evalspeedSingleSingle1(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)    evalspeedSingleSinglebb2in(ar_0, ar_1) -> Com_1(evalspeedSingleSingle0(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)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb3in(ar_0, ar_1)) [ ar_0 >= 0 /\ ar_0 >= ar_1 ]

		(Comp: ?, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb2in(ar_0, ar_1)) [ ar_0 >= 0 /\ ar_1 >= ar_0 + 1 ]

		(Comp: 1, Cost: 1)    evalspeedSingleSinglebb0in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(0, ar_1))

		(Comp: 1, Cost: 1)    evalspeedSingleSinglestart(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb0in(ar_0, ar_1))

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = 4*V_2

	Pol(evalspeedSingleSinglestart) = 4*V_2

	Pol(evalspeedSingleSinglebb3in) = -4*V_1 + 4*V_2

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

	Pol(evalspeedSingleSingle1) = -4*V_1 + 4*V_2 - 3

	Pol(evalspeedSingleSinglebb1in) = -4*V_1 + 4*V_2

	Pol(evalspeedSingleSingle0) = -4*V_1 + 4*V_2 - 2

	Pol(evalnondetstart) = -4*V_1

	Pol(evalspeedSingleSinglebb2in) = -4*V_1 + 4*V_2 - 1

	Pol(evalspeedSingleSinglebb0in) = 4*V_2

orients all transitions weakly and the transitions

	evalspeedSingleSinglebb2in(ar_0, ar_1) -> Com_1(evalspeedSingleSingle0(ar_0, ar_1)) [ ar_1 - 1 >= 0 /\ ar_0 + ar_1 - 1 >= 0 /\ -ar_0 + ar_1 - 1 >= 0 /\ ar_0 >= 0 ]

	evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb2in(ar_0, ar_1)) [ ar_0 >= 0 /\ ar_1 >= ar_0 + 1 ]

	evalspeedSingleSingle1(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(ar_0 + 1, ar_1)) [ ar_1 - 1 >= 0 /\ ar_0 + ar_1 - 1 >= 0 /\ -ar_0 + ar_1 - 1 >= 0 /\ ar_0 >= 0 ]

	evalspeedSingleSingle0(ar_0, ar_1) -> Com_2(evalnondetstart(ar_0, ar_1), evalspeedSingleSingle1(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(evalspeedSingleSinglestart(ar_0, ar_1)) [ 0 <= 0 ]

		(Comp: 2, Cost: 1)         evalspeedSingleSinglebb3in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglestop(ar_0, ar_1)) [ ar_0 - ar_1 >= 0 /\ ar_0 >= 0 ]

		(Comp: 4*ar_1, Cost: 1)    evalspeedSingleSingle1(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(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)    evalspeedSingleSingle0(ar_0, ar_1) -> Com_2(evalnondetstart(ar_0, ar_1), evalspeedSingleSingle1(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)    evalspeedSingleSinglebb2in(ar_0, ar_1) -> Com_1(evalspeedSingleSingle0(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)         evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb3in(ar_0, ar_1)) [ ar_0 >= 0 /\ ar_0 >= ar_1 ]

		(Comp: 4*ar_1, Cost: 1)    evalspeedSingleSinglebb1in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb2in(ar_0, ar_1)) [ ar_0 >= 0 /\ ar_1 >= ar_0 + 1 ]

		(Comp: 1, Cost: 1)         evalspeedSingleSinglebb0in(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb1in(0, ar_1))

		(Comp: 1, Cost: 1)         evalspeedSingleSinglestart(ar_0, ar_1) -> Com_1(evalspeedSingleSinglebb0in(ar_0, ar_1))

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 16*ar_1 + 6



Time: 0.130 sec (SMT: 0.115 sec)


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(2)
BOUNDS(1, n^1)