* Step 1: Bounds WORST_CASE(?,O(n^1))
+ Considered Problem:
- Strict TRS:
activate(X) -> X
activate(n__f(X)) -> f(X)
f(X) -> n__f(X)
f(0()) -> cons(0(),n__f(s(0())))
f(s(0())) -> f(p(s(0())))
p(s(X)) -> X
- Signature:
{activate/1,f/1,p/1} / {0/0,cons/2,n__f/1,s/1}
- Obligation:
innermost runtime complexity wrt. defined symbols {activate,f,p} and constructors {0,cons,n__f,s}
+ Applied Processor:
Bounds {initialAutomaton = minimal, enrichment = match}
+ Details:
The problem is match-bounded by 2.
The enriched problem is compatible with follwoing automaton.
0_0() -> 1
0_0() -> 2
0_1() -> 1
0_1() -> 2
0_1() -> 3
0_1() -> 6
0_2() -> 7
0_2() -> 10
activate_0(2) -> 1
cons_0(2,2) -> 1
cons_0(2,2) -> 2
cons_1(3,4) -> 1
cons_2(7,8) -> 1
f_0(2) -> 1
f_1(2) -> 1
n__f_0(2) -> 1
n__f_0(2) -> 2
n__f_1(2) -> 1
n__f_1(5) -> 4
n__f_2(2) -> 1
n__f_2(9) -> 8
p_0(2) -> 1
p_1(5) -> 1
p_1(5) -> 2
s_0(2) -> 1
s_0(2) -> 2
s_1(6) -> 5
s_2(10) -> 9
2 -> 1
6 -> 1
6 -> 2
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
+ Considered Problem:
- Weak TRS:
activate(X) -> X
activate(n__f(X)) -> f(X)
f(X) -> n__f(X)
f(0()) -> cons(0(),n__f(s(0())))
f(s(0())) -> f(p(s(0())))
p(s(X)) -> X
- Signature:
{activate/1,f/1,p/1} / {0/0,cons/2,n__f/1,s/1}
- Obligation:
innermost runtime complexity wrt. defined symbols {activate,f,p} and constructors {0,cons,n__f,s}
+ Applied Processor:
EmptyProcessor
+ Details:
The problem is already closed. The intended complexity is O(1).
WORST_CASE(?,O(n^1))