* Step 1: Sum WORST_CASE(Omega(n^1),O(n^1))
+ Considered Problem:
- Strict TRS:
dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
revApp#2(Nil(),x16) -> x16
- Signature:
{dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
- Obligation:
innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
,Node}
+ Applied Processor:
Sum {left = someStrategy, right = someStrategy}
+ Details:
()
** Step 1.a:1: DecreasingLoops WORST_CASE(Omega(n^1),?)
+ Considered Problem:
- Strict TRS:
dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
revApp#2(Nil(),x16) -> x16
- Signature:
{dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
- Obligation:
innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
,Node}
+ Applied Processor:
DecreasingLoops {bound = AnyLoop, narrow = 10}
+ Details:
The system has following decreasing Loops:
dfsAcc#3(y,z){y -> Node(x,y)} =
dfsAcc#3(Node(x,y),z) ->^+ dfsAcc#3(y,dfsAcc#3(x,z))
= C[dfsAcc#3(y,dfsAcc#3(x,z)) = dfsAcc#3(y,z){z -> dfsAcc#3(x,z)}]
** Step 1.b:1: Bounds WORST_CASE(?,O(n^1))
+ Considered Problem:
- Strict TRS:
dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
revApp#2(Nil(),x16) -> x16
- Signature:
{dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
- Obligation:
innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
,Node}
+ Applied Processor:
Bounds {initialAutomaton = minimal, enrichment = match}
+ Details:
The problem is match-bounded by 2.
The enriched problem is compatible with follwoing automaton.
Cons_0(2,2) -> 1
Cons_0(2,2) -> 2
Cons_1(2,1) -> 1
Cons_1(2,2) -> 1
Cons_1(2,2) -> 3
Cons_1(2,3) -> 1
Cons_1(2,3) -> 3
Cons_1(2,4) -> 4
Cons_1(2,6) -> 4
Cons_2(2,5) -> 1
Cons_2(2,5) -> 7
Cons_2(2,7) -> 1
Cons_2(2,7) -> 7
Leaf_0(2) -> 1
Leaf_0(2) -> 2
Nil_0() -> 1
Nil_0() -> 2
Nil_1() -> 5
Nil_1() -> 6
Node_0(2,2) -> 1
Node_0(2,2) -> 2
dfsAcc#3_0(2,2) -> 1
dfsAcc#3_1(2,2) -> 3
dfsAcc#3_1(2,3) -> 1
dfsAcc#3_1(2,3) -> 3
dfsAcc#3_1(2,4) -> 4
dfsAcc#3_1(2,6) -> 4
main_0(2) -> 1
revApp#2_0(2,2) -> 1
revApp#2_1(2,1) -> 1
revApp#2_1(4,5) -> 1
revApp#2_2(4,7) -> 1
revApp#2_2(6,7) -> 1
2 -> 1
7 -> 1
** Step 1.b:2: EmptyProcessor WORST_CASE(?,O(1))
+ Considered Problem:
- Weak TRS:
dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
revApp#2(Nil(),x16) -> x16
- Signature:
{dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
- Obligation:
innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
,Node}
+ Applied Processor:
EmptyProcessor
+ Details:
The problem is already closed. The intended complexity is O(1).
WORST_CASE(Omega(n^1),O(n^1))