/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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WORST_CASE(?, O(1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, max(29 + -108 * Arg_0 + 27 * Arg_1, 2 + -54 * Arg_0 + 54 * Arg_1, 29, 29 + -108 * Arg_0 + 54 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 124 ms]
(2) BOUNDS(1, max(29 + -108 * Arg_0 + 27 * Arg_1, 2 + -54 * Arg_0 + 54 * Arg_1, 29, 29 + -108 * Arg_0 + 54 * Arg_1))
(3) Loat Proof [FINISHED, 234 ms]
(4) BOUNDS(1, INF)
(5) Koat Proof [FINISHED, 317 ms]
(6) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
l0(A, B) -> Com_1(l1(A, B)) :|: TRUE
l1(A, B) -> Com_1(l1(A + A, B + B + B)) :|: B + B + B + B >= A && A >= B && A >= 1 && B >= 1

The start-symbols are:[l0_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, 2+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]) {O(n)})



Initial Complexity Problem:

  Start:  l0  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  l0, l1  

  Transitions:

  0: l0->l1

  1: l1->l1  



Timebounds: 

  Overall timebound: 2+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]) {O(n)}

  0: l0->l1: 1 {O(1)}

  1: l1->l1: 1+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]) {O(n)}



Costbounds:

  Overall costbound: 2+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]) {O(n)}

  0: l0->l1: 1 {O(1)}

  1: l1->l1: 1+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]) {O(n)}



Sizebounds:

`Lower:

  0: l0->l1, Arg_0: Arg_0 {O(n)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  1: l1->l1, Arg_0: 3 {O(1)}

  1: l1->l1, Arg_1: 2 {O(1)}

`Upper:

  0: l0->l1, Arg_0: Arg_0 {O(n)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  1: l1->l1, Arg_0: 3^(1+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]))*max([0, Arg_0]) {O(EXP)}

  1: l1->l1, Arg_1: 2^(1+27*max([1, max([1+Arg_1+-4*Arg_0, max([1+-4*Arg_0+2*Arg_1, -2*Arg_0+2*Arg_1])])]))*max([0, Arg_1]) {O(EXP)}


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(2)
BOUNDS(1, max(29 + -108 * Arg_0 + 27 * Arg_1, 2 + -54 * Arg_0 + 54 * Arg_1, 29, 29 + -108 * Arg_0 + 54 * Arg_1))

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(3) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      1: l1 -> l1 : A'=3*A, B'=2*B, [ 4*A>=B && B>=A && B>=1 && A>=1 ], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      0: l0 -> l1 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: l1 -> l1 : A'=3*A, B'=2*B, [ 4*A>=B && B>=A && B>=1 ], cost: 1



   Found no metering function for rule 1.

   Removing the simple loops:.



Accelerated all simple loops using metering functions (where possible):

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      1: l1 -> l1 : A'=3*A, B'=2*B, [ 4*A>=B && B>=A && B>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      2: l0 -> l1 : A'=3*A, B'=2*B, [ 4*A>=B && B>=A && B>=1 ], cost: 2



Removed unreachable locations (and leaf rules with constant cost):

   Start location: l0

     <empty>



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: l0

     <empty>



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Constant

   Cpx degree:  0

   Solved cost: 1

   Rule cost:   1

   Rule guard:  []



WORST_CASE(Omega(1),?)


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(4)
BOUNDS(1, INF)

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(5) Koat Proof (FINISHED)
YES(?, 5)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(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)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1)) with all transitions in problem 2, the following new transition is obtained:

	l0(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

We thus obtain the following problem:

3:	T:

		(Comp: 1, Cost: 2)    l0(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition l0(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ] with all transitions in problem 3, the following new transition is obtained:

	l0(Ar_0, Ar_1) -> Com_1(l1(9*Ar_0, 4*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 ]

We thus obtain the following problem:

4:	T:

		(Comp: 1, Cost: 3)    l0(Ar_0, Ar_1) -> Com_1(l1(9*Ar_0, 4*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition l0(Ar_0, Ar_1) -> Com_1(l1(9*Ar_0, 4*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 ] with all transitions in problem 4, the following new transition is obtained:

	l0(Ar_0, Ar_1) -> Com_1(l1(27*Ar_0, 8*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 /\ 36*Ar_0 >= 4*Ar_1 /\ 4*Ar_1 >= 9*Ar_0 /\ 4*Ar_1 >= 1 /\ 9*Ar_0 >= 1 ]

We thus obtain the following problem:

5:	T:

		(Comp: 1, Cost: 4)    l0(Ar_0, Ar_1) -> Com_1(l1(27*Ar_0, 8*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 /\ 36*Ar_0 >= 4*Ar_1 /\ 4*Ar_1 >= 9*Ar_0 /\ 4*Ar_1 >= 1 /\ 9*Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition l0(Ar_0, Ar_1) -> Com_1(l1(27*Ar_0, 8*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 /\ 36*Ar_0 >= 4*Ar_1 /\ 4*Ar_1 >= 9*Ar_0 /\ 4*Ar_1 >= 1 /\ 9*Ar_0 >= 1 ] with all transitions in problem 5, the following new transition is obtained:

	l0(Ar_0, Ar_1) -> Com_1(l1(81*Ar_0, 16*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 /\ 36*Ar_0 >= 4*Ar_1 /\ 4*Ar_1 >= 9*Ar_0 /\ 4*Ar_1 >= 1 /\ 9*Ar_0 >= 1 /\ 108*Ar_0 >= 8*Ar_1 /\ 8*Ar_1 >= 27*Ar_0 /\ 8*Ar_1 >= 1 /\ 27*Ar_0 >= 1 ]

We thus obtain the following problem:

6:	T:

		(Comp: 1, Cost: 5)    l0(Ar_0, Ar_1) -> Com_1(l1(81*Ar_0, 16*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 /\ 36*Ar_0 >= 4*Ar_1 /\ 4*Ar_1 >= 9*Ar_0 /\ 4*Ar_1 >= 1 /\ 9*Ar_0 >= 1 /\ 108*Ar_0 >= 8*Ar_1 /\ 8*Ar_1 >= 27*Ar_0 /\ 8*Ar_1 >= 1 /\ 27*Ar_0 >= 1 ]

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transition from problem 6:

	l1(Ar_0, Ar_1) -> Com_1(l1(3*Ar_0, 2*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 ]

We thus obtain the following problem:

7:	T:

		(Comp: 1, Cost: 5)    l0(Ar_0, Ar_1) -> Com_1(l1(81*Ar_0, 16*Ar_1)) [ 4*Ar_0 >= Ar_1 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 1 /\ Ar_0 >= 1 /\ 12*Ar_0 >= 2*Ar_1 /\ 2*Ar_1 >= 3*Ar_0 /\ 2*Ar_1 >= 1 /\ 3*Ar_0 >= 1 /\ 36*Ar_0 >= 4*Ar_1 /\ 4*Ar_1 >= 9*Ar_0 /\ 4*Ar_1 >= 1 /\ 9*Ar_0 >= 1 /\ 108*Ar_0 >= 8*Ar_1 /\ 8*Ar_1 >= 27*Ar_0 /\ 8*Ar_1 >= 1 /\ 27*Ar_0 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 5



Time: 0.372 sec (SMT: 0.346 sec)


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(6)
BOUNDS(1, 1)