Tool CaT
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^2))
Problem:
average(s(x),y) -> average(x,s(y))
average(x,s(s(s(y)))) -> s(average(s(x),y))
average(0(),0()) -> 0()
average(0(),s(0())) -> 0()
average(0(),s(s(0()))) -> s(0())
Proof:
Complexity Transformation Processor:
strict:
average(s(x),y) -> average(x,s(y))
average(x,s(s(s(y)))) -> s(average(s(x),y))
average(0(),0()) -> 0()
average(0(),s(0())) -> 0()
average(0(),s(s(0()))) -> s(0())
weak:
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {3}
transitions:
01() -> 3*
average0(1,2) -> 3*
average0(2,1) -> 3*
average0(1,1) -> 3*
average0(2,2) -> 3*
s0(2) -> 3,1
s0(1) -> 1*
s0(3) -> 3*
00() -> 2*
problem:
strict:
average(s(x),y) -> average(x,s(y))
average(x,s(s(s(y)))) -> s(average(s(x),y))
average(0(),s(0())) -> 0()
average(0(),s(s(0()))) -> s(0())
weak:
average(0(),0()) -> 0()
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {3}
transitions:
01() -> 3*
average0(1,2) -> 3*
average0(2,1) -> 3*
average0(1,1) -> 3*
average0(2,2) -> 3*
s0(2) -> 3,1
s0(1) -> 1*
s0(3) -> 3*
00() -> 2*
problem:
strict:
average(s(x),y) -> average(x,s(y))
average(x,s(s(s(y)))) -> s(average(s(x),y))
average(0(),s(s(0()))) -> s(0())
weak:
average(0(),s(0())) -> 0()
average(0(),0()) -> 0()
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {3}
transitions:
s1(7) -> 3*
01() -> 7*
average0(1,2) -> 3*
average0(2,1) -> 3*
average0(1,1) -> 3*
average0(2,2) -> 3*
s0(2) -> 1*
s0(1) -> 1*
s0(3) -> 3*
00() -> 3,2
problem:
strict:
average(s(x),y) -> average(x,s(y))
average(x,s(s(s(y)))) -> s(average(s(x),y))
weak:
average(0(),s(s(0()))) -> s(0())
average(0(),s(0())) -> 0()
average(0(),0()) -> 0()
Matrix Interpretation Processor:
dimension: 4
max_matrix:
[1 0 0 0]
[0 0 0 0]
[0 0 0 1]
[0 0 0 0]
interpretation:
[0]
[0]
[0] = [0]
[3],
[1 0 0 0] [1 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [3]
[average](x0, x1) = [0 0 0 0]x0 + [0 0 0 0]x1 + [3]
[0 0 0 0] [0 0 0 0] [3],
[1 0 0 0] [1]
[0 0 0 0] [0]
[s](x0) = [0 0 0 1]x0 + [0]
[0 0 0 0] [0]
orientation:
[1 0 0 0] [1 0 0 0] [1] [1 0 0 0] [1 0 0 0] [1]
[0 0 0 0] [0 0 0 0] [3] [0 0 0 0] [0 0 0 0] [3]
average(s(x),y) = [0 0 0 0]x + [0 0 0 0]y + [3] >= [0 0 0 0]x + [0 0 0 0]y + [3] = average(x,s(y))
[0 0 0 0] [0 0 0 0] [3] [0 0 0 0] [0 0 0 0] [3]
[1 0 0 0] [1 0 0 0] [3] [1 0 0 0] [1 0 0 0] [2]
[0 0 0 0] [0 0 0 0] [3] [0 0 0 0] [0 0 0 0] [0]
average(x,s(s(s(y)))) = [0 0 0 0]x + [0 0 0 0]y + [3] >= [0 0 0 0]x + [0 0 0 0]y + [3] = s(average(s(x),y))
[0 0 0 0] [0 0 0 0] [3] [0 0 0 0] [0 0 0 0] [0]
[2] [1]
[3] [0]
average(0(),s(s(0()))) = [3] >= [3] = s(0())
[3] [0]
[1] [0]
[3] [0]
average(0(),s(0())) = [3] >= [0] = 0()
[3] [3]
[0] [0]
[3] [0]
average(0(),0()) = [3] >= [0] = 0()
[3] [3]
problem:
strict:
average(s(x),y) -> average(x,s(y))
weak:
average(x,s(s(s(y)))) -> s(average(s(x),y))
average(0(),s(s(0()))) -> s(0())
average(0(),s(0())) -> 0()
average(0(),0()) -> 0()
Matrix Interpretation Processor:
dimension: 2
max_matrix:
[1 1]
[0 1]
interpretation:
[0]
[0] = [0],
[1 1]
[average](x0, x1) = [0 1]x0 + x1,
[1]
[s](x0) = x0 + [1]
orientation:
[1 1] [2] [1 1] [1]
average(s(x),y) = [0 1]x + y + [1] >= [0 1]x + y + [1] = average(x,s(y))
[1 1] [3] [1 1] [3]
average(x,s(s(s(y)))) = [0 1]x + y + [3] >= [0 1]x + y + [2] = s(average(s(x),y))
[2] [1]
average(0(),s(s(0()))) = [2] >= [1] = s(0())
[1] [0]
average(0(),s(0())) = [1] >= [0] = 0()
[0] [0]
average(0(),0()) = [0] >= [0] = 0()
problem:
strict:
weak:
average(s(x),y) -> average(x,s(y))
average(x,s(s(s(y)))) -> s(average(s(x),y))
average(0(),s(s(0()))) -> s(0())
average(0(),s(0())) -> 0()
average(0(),0()) -> 0()
Qed
Tool IRC1
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^1))
Tool IRC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
Proof Output:
'matrix-interpretation of dimension 1' proved the best result:
Details:
--------
'matrix-interpretation of dimension 1' succeeded with the following output:
'matrix-interpretation of dimension 1'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
Proof Output:
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
average(x1, x2) = [3] x1 + [2] x2 + [0]
s(x1) = [1] x1 + [1]
0() = [1]Tool RC1
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^1))
Tool RC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
Proof Output:
'matrix-interpretation of dimension 1' proved the best result:
Details:
--------
'matrix-interpretation of dimension 1' succeeded with the following output:
'matrix-interpretation of dimension 1'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
Proof Output:
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
average(x1, x2) = [3] x1 + [2] x2 + [0]
s(x1) = [1] x1 + [1]
0() = [1]Tool pair1rc
| Execution Time | Unknown |
|---|
| Answer | TIMEOUT |
|---|
| Input | AG01 3.15 |
|---|
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
Certificate: TIMEOUT
Application of 'pair1 (timeout of 60.0 seconds)':
-------------------------------------------------
Computation stopped due to timeout after 60.0 seconds
Arrrr..Tool pair2rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^2))
Application of 'pair2 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
average(x1, x2) = [1 2 1] x1 + [2 0 2] x2 + [0]
[2 2 2] [1 1 1] [0]
[0 0 0] [0 0 0] [1]
s(x1) = [1 0 1] x1 + [0]
[0 1 0] [1]
[0 0 0] [1]
0() = [0]
[0]
[1]
Hurray, we answered YES(?,O(n^2))Tool pair3irc
| Execution Time | Unknown |
|---|
| Answer | TIMEOUT |
|---|
| Input | AG01 3.15 |
|---|
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: innermost
Certificate: TIMEOUT
Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
Computation stopped due to timeout after 60.0 seconds
Arrrr..Tool pair3rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^2))
Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
average(x1, x2) = [1 2 1] x1 + [2 0 2] x2 + [0]
[2 2 2] [1 1 1] [0]
[0 0 0] [0 0 0] [1]
s(x1) = [1 0 1] x1 + [0]
[0 1 0] [1]
[0 0 0] [1]
0() = [0]
[0]
[1]
Hurray, we answered YES(?,O(n^2))Tool rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'rc (timeout of 60.0 seconds)':
----------------------------------------------
'dp' proved the goal fastest:
We have computed the following dependency pairs
Strict Dependency Pairs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, average^#(0(), 0()) -> c_3()
, average^#(0(), s(0())) -> c_4()
, average^#(0(), s(s(0()))) -> c_5()}
We consider the following Problem:
Strict DPs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, average^#(0(), 0()) -> c_3()
, average^#(0(), s(0())) -> c_4()
, average^#(0(), s(s(0()))) -> c_5()}
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'usablerules':
-----------------------------
No rule is usable.
We consider the following Problem:
Strict DPs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, average^#(0(), 0()) -> c_3()
, average^#(0(), s(0())) -> c_4()
, average^#(0(), s(s(0()))) -> c_5()}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'Fastest':
-------------------------
'removetails >>> ... >>> ... >>> ...' proved the goal fastest:
We consider the the dependency-graph
1: average^#(s(x), y) -> c_1(average^#(x, s(y)))
--> average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y)): 2
--> average^#(0(), s(s(0()))) -> c_5(): 5
--> average^#(0(), s(0())) -> c_4(): 4
--> average^#(s(x), y) -> c_1(average^#(x, s(y))): 1
2: average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
--> average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y)): 2
--> average^#(s(x), y) -> c_1(average^#(x, s(y))): 1
3: average^#(0(), 0()) -> c_3()
4: average^#(0(), s(0())) -> c_4()
5: average^#(0(), s(s(0()))) -> c_5()
together with the congruence-graph
->{1,2}
|
|->{4} Noncyclic, trivial, SCC
|
`->{5} Noncyclic, trivial, SCC
->{3} Noncyclic, trivial, SCC
Here rules are as follows:
{ 1: average^#(s(x), y) -> c_1(average^#(x, s(y)))
, 2: average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, 3: average^#(0(), 0()) -> c_3()
, 4: average^#(0(), s(0())) -> c_4()
, 5: average^#(0(), s(s(0()))) -> c_5()}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 3: average^#(0(), 0()) -> c_3()
, 4: average^#(0(), s(0())) -> c_4()
, 5: average^#(0(), s(s(0()))) -> c_5()}
We consider the following Problem:
Strict DPs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'simpDPRHS >>> ...':
-----------------------------------
No rule was simplified
We apply the transformation 'usablerules' on the resulting sub-problems:
Sub-problem 1:
--------------
We consider the problem
Strict DPs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))}
StartTerms: basic terms
Strategy: none
No rule is usable.
We abort the transformation and continue with the subprocessor on the problem
Strict DPs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))}
StartTerms: basic terms
Strategy: none
1) The weightgap principle applies, where following rules are oriented strictly:
Dependency Pairs: {average^#(s(x), y) -> c_1(average^#(x, s(y)))}
Interpretation:
---------------
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {}, Uargs(average^#) = {},
Uargs(c_1) = {1}, Uargs(c_2) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
average(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
s(x1) = [0 0] x1 + [0]
[0 1] [2]
0() = [0]
[0]
average^#(x1, x2) = [0 2] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
c_1(x1) = [1 0] x1 + [3]
[0 1] [0]
c_2(x1) = [1 0] x1 + [3]
[0 1] [3]
c_3() = [0]
[0]
c_4() = [0]
[0]
c_5() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict DPs: {average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))}
Weak DPs: {average^#(s(x), y) -> c_1(average^#(x, s(y)))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'removetails >>> ... >>> ... >>> ...':
-----------------------------------------------------
No dependency-pair could be removed
We abort the transformation and continue with the subprocessor on the problem
Strict DPs: {average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))}
Weak DPs: {average^#(s(x), y) -> c_1(average^#(x, s(y)))}
StartTerms: basic terms
Strategy: none
1) The weightgap principle applies, where following rules are oriented strictly:
Dependency Pairs:
{average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))}
Interpretation:
---------------
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {}, Uargs(average^#) = {},
Uargs(c_1) = {1}, Uargs(c_2) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
average(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
s(x1) = [0 0] x1 + [0]
[0 1] [2]
0() = [0]
[0]
average^#(x1, x2) = [0 2] x1 + [0 2] x2 + [0]
[0 0] [0 0] [0]
c_1(x1) = [1 0] x1 + [0]
[0 1] [0]
c_2(x1) = [1 0] x1 + [1]
[0 1] [0]
c_3() = [0]
[0]
c_4() = [0]
[0]
c_5() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Weak DPs:
{ average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, average^#(s(x), y) -> c_1(average^#(x, s(y)))}
StartTerms: basic terms
Strategy: none
Certificate: YES(O(1),O(1))
Application of 'removetails >>> ... >>> ... >>> ...':
-----------------------------------------------------
We consider the the dependency-graph
1: average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
--> average^#(s(x), y) -> c_1(average^#(x, s(y))): 2
--> average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y)): 1
2: average^#(s(x), y) -> c_1(average^#(x, s(y)))
--> average^#(s(x), y) -> c_1(average^#(x, s(y))): 2
--> average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y)): 1
together with the congruence-graph
->{1,2} Weak SCC
Here rules are as follows:
{ 1: average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, 2: average^#(s(x), y) -> c_1(average^#(x, s(y)))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 1: average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, 2: average^#(s(x), y) -> c_1(average^#(x, s(y)))}
We consider the following Problem:
StartTerms: basic terms
Strategy: none
Certificate: YES(O(1),O(1))
Application of 'simpDPRHS >>> ...':
-----------------------------------
No rule was simplified
We apply the transformation 'usablerules' on the resulting sub-problems:
Sub-problem 1:
--------------
We consider the problem
StartTerms: basic terms
Strategy: none
The input problem is not a DP-problem.
We abort the transformation and continue with the subprocessor on the problem
StartTerms: basic terms
Strategy: none
1) We fail transforming the problem using 'weightgap of dimension Nat 2, maximal degree 1, cbits 4'
All strict components are empty, nothing to further orient
We try instead 'weightgap of dimension Nat 3, maximal degree 3, cbits 4 <> ...' on the problem
StartTerms: basic terms
Strategy: none
We fail transforming the problem using 'weightgap of dimension Nat 3, maximal degree 3, cbits 4'
All strict components are empty, nothing to further orient
We try instead 'weightgap of dimension Nat 4, maximal degree 3, cbits 4' on the problem
StartTerms: basic terms
Strategy: none
All strict components are empty, nothing to further orient
We abort the transformation and continue with the subprocessor on the problem
StartTerms: basic terms
Strategy: none
1) No dependency-pair could be removed
We apply the transformation 'weightgap of dimension Nat 2, maximal degree 1, cbits 4 <> ...' on the resulting sub-problems:
Sub-problem 1:
--------------
We consider the problem
StartTerms: basic terms
Strategy: none
We fail transforming the problem using 'weightgap of dimension Nat 2, maximal degree 1, cbits 4'
All strict components are empty, nothing to further orient
We try instead 'weightgap of dimension Nat 3, maximal degree 3, cbits 4 <> ...' on the problem
StartTerms: basic terms
Strategy: none
We fail transforming the problem using 'weightgap of dimension Nat 3, maximal degree 3, cbits 4'
All strict components are empty, nothing to further orient
We try instead 'weightgap of dimension Nat 4, maximal degree 3, cbits 4' on the problem
StartTerms: basic terms
Strategy: none
All strict components are empty, nothing to further orient
We abort the transformation and continue with the subprocessor on the problem
StartTerms: basic terms
Strategy: none
1) 'Sequentially' proved the goal fastest:
'empty' succeeded:
Empty rules are trivially bounded
Hurray, we answered YES(?,O(n^1))Tool tup3irc
| Execution Time | 3.647048ms |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | AG01 3.15 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Application of 'tup3 (timeout of 60.0 seconds)':
------------------------------------------------
The input problem contains no overlaps that give rise to inapplicable rules.
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: innermost
1) 'dp' proved the goal fastest:
We have computed the following dependency pairs
Strict Dependency Pairs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, average^#(0(), 0()) -> c_3()
, average^#(0(), s(0())) -> c_4()
, average^#(0(), s(s(0()))) -> c_5()}
We consider the following Problem:
Strict DPs:
{ average^#(s(x), y) -> c_1(average^#(x, s(y)))
, average^#(x, s(s(s(y)))) -> c_2(average^#(s(x), y))
, average^#(0(), 0()) -> c_3()
, average^#(0(), s(0())) -> c_4()
, average^#(0(), s(s(0()))) -> c_5()}
Weak Trs:
{ average(s(x), y) -> average(x, s(y))
, average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), 0()) -> 0()
, average(0(), s(0())) -> 0()
, average(0(), s(s(0()))) -> s(0())}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Application of 'Fastest':
-------------------------
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(average) = {}, Uargs(s) = {}, Uargs(average^#) = {},
Uargs(c_1) = {1}, Uargs(c_2) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
average(x1, x2) = [2 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
s(x1) = [1 0] x1 + [1]
[0 0] [0]
0() = [0]
[0]
average^#(x1, x2) = [2 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [0]
c_1(x1) = [1 0] x1 + [0]
[0 0] [0]
c_2(x1) = [1 0] x1 + [0]
[0 0] [0]
c_3() = [0]
[0]
c_4() = [1]
[0]
c_5() = [0]
[0]
Hurray, we answered YES(?,O(n^1))