'Weak Dependency Graph [60.0]'
------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, c(c(x1)) -> a(b(c(a(x1))))}
Details:
We have computed the following set of weak (innermost) dependency pairs:
{ a^#(x1) -> c_0()
, a^#(b(b(x1))) -> c_1(c^#(x1))
, c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
The usable rules are:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, c(c(x1)) -> a(b(c(a(x1))))}
The estimated dependency graph contains the following edges:
{a^#(b(b(x1))) -> c_1(c^#(x1))}
==> {c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
{c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
==> {a^#(b(b(x1))) -> c_1(c^#(x1))}
{c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
==> {a^#(x1) -> c_0()}
We consider the following path(s):
1) { a^#(b(b(x1))) -> c_1(c^#(x1))
, c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
The usable rules for this path are the following:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, c(c(x1)) -> a(b(c(a(x1))))}
We have applied the subprocessor on the union of usable rules and weak (innermost) dependency pairs.
'Weight Gap Principle'
----------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, c(c(x1)) -> a(b(c(a(x1))))
, a^#(b(b(x1))) -> c_1(c^#(x1))
, c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
Details:
We apply the weight gap principle, strictly orienting the rules
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)}
and weakly orienting the rules
{}
using the following strongly linear interpretation:
Processor 'Matrix Interpretation' oriented the following rules strictly:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)}
Details:
Interpretation Functions:
a(x1) = [1] x1 + [1]
b(x1) = [1] x1 + [0]
c(x1) = [1] x1 + [0]
a^#(x1) = [1] x1 + [1]
c_0() = [0]
c_1(x1) = [1] x1 + [1]
c^#(x1) = [1] x1 + [0]
c_2(x1) = [1] x1 + [1]
Finally we apply the subprocessor
We apply the weight gap principle, strictly orienting the rules
{c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
and weakly orienting the rules
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)}
using the following strongly linear interpretation:
Processor 'Matrix Interpretation' oriented the following rules strictly:
{c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
Details:
Interpretation Functions:
a(x1) = [1] x1 + [0]
b(x1) = [1] x1 + [0]
c(x1) = [1] x1 + [0]
a^#(x1) = [1] x1 + [1]
c_0() = [0]
c_1(x1) = [1] x1 + [1]
c^#(x1) = [1] x1 + [4]
c_2(x1) = [1] x1 + [0]
Finally we apply the subprocessor
'fastest of 'combine', 'Bounds with default enrichment', 'Bounds with default enrichment''
------------------------------------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost relative runtime-complexity with respect to
Strict Rules:
{ c(c(x1)) -> a(b(c(a(x1))))
, a^#(b(b(x1))) -> c_1(c^#(x1))}
Weak Rules:
{ c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))
, a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)}
Details:
The problem was solved by processor 'Bounds with default enrichment':
'Bounds with default enrichment'
--------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost relative runtime-complexity with respect to
Strict Rules:
{ c(c(x1)) -> a(b(c(a(x1))))
, a^#(b(b(x1))) -> c_1(c^#(x1))}
Weak Rules:
{ c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))
, a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)}
Details:
The problem is Match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ b_0(2) -> 2
, a^#_0(2) -> 4
, c_1_1(8) -> 4
, c^#_0(2) -> 7
, c^#_1(2) -> 8}
2) { a^#(b(b(x1))) -> c_1(c^#(x1))
, c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))
, a^#(x1) -> c_0()}
The usable rules for this path are the following:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, c(c(x1)) -> a(b(c(a(x1))))}
We have applied the subprocessor on the union of usable rules and weak (innermost) dependency pairs.
'Weight Gap Principle'
----------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, c(c(x1)) -> a(b(c(a(x1))))
, a^#(b(b(x1))) -> c_1(c^#(x1))
, c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))
, a^#(x1) -> c_0()}
Details:
We apply the weight gap principle, strictly orienting the rules
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, a^#(x1) -> c_0()}
and weakly orienting the rules
{}
using the following strongly linear interpretation:
Processor 'Matrix Interpretation' oriented the following rules strictly:
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, a^#(x1) -> c_0()}
Details:
Interpretation Functions:
a(x1) = [1] x1 + [1]
b(x1) = [1] x1 + [0]
c(x1) = [1] x1 + [0]
a^#(x1) = [1] x1 + [1]
c_0() = [0]
c_1(x1) = [1] x1 + [1]
c^#(x1) = [1] x1 + [0]
c_2(x1) = [1] x1 + [1]
Finally we apply the subprocessor
We apply the weight gap principle, strictly orienting the rules
{c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
and weakly orienting the rules
{ a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, a^#(x1) -> c_0()}
using the following strongly linear interpretation:
Processor 'Matrix Interpretation' oriented the following rules strictly:
{c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))}
Details:
Interpretation Functions:
a(x1) = [1] x1 + [0]
b(x1) = [1] x1 + [0]
c(x1) = [1] x1 + [0]
a^#(x1) = [1] x1 + [1]
c_0() = [0]
c_1(x1) = [1] x1 + [1]
c^#(x1) = [1] x1 + [8]
c_2(x1) = [1] x1 + [0]
Finally we apply the subprocessor
'fastest of 'combine', 'Bounds with default enrichment', 'Bounds with default enrichment''
------------------------------------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost relative runtime-complexity with respect to
Strict Rules:
{ c(c(x1)) -> a(b(c(a(x1))))
, a^#(b(b(x1))) -> c_1(c^#(x1))}
Weak Rules:
{ c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))
, a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, a^#(x1) -> c_0()}
Details:
The problem was solved by processor 'Bounds with default enrichment':
'Bounds with default enrichment'
--------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost relative runtime-complexity with respect to
Strict Rules:
{ c(c(x1)) -> a(b(c(a(x1))))
, a^#(b(b(x1))) -> c_1(c^#(x1))}
Weak Rules:
{ c^#(c(x1)) -> c_2(a^#(b(c(a(x1)))))
, a(x1) -> b(x1)
, a(b(b(x1))) -> c(x1)
, a^#(x1) -> c_0()}
Details:
The problem is Match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ b_0(2) -> 2
, a^#_0(2) -> 4
, c_0_0() -> 4
, c_1_1(8) -> 4
, c^#_0(2) -> 7
, c^#_1(2) -> 8}