Runtime Complexity proof of /tmp/tmpb1UKrn/pl8.4.1.pl

Runtime Complexity of the query pattern
even(g)
w.r.t. the given Prolog program could be shown to be BOUNDS(O(1), O(n^1)).

0 Prolog

↳1 PrologToCdtProblemTransformerProof (UPPER BOUND (ID), 52 ms)

↳2 CdtProblem

    ↳3 CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID), 0 ms)

    ↳4 CdtProblem

    ↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 109 ms)

    ↳6 CdtProblem

    ↳7 SIsEmptyProof (⇔, 0 ms)

    ↳8 BOUNDS(O(1), O(1))

↳9 PrologToCdtProblemTransformerProof (UPPER BOUND (ID), 76 ms)

↳10 CdtProblem

    ↳11 CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID), 0 ms)

    ↳12 CdtProblem

    ↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 124 ms)

    ↳14 CdtProblem

(0) Obligation:

Clauses:

even(0).
even(s(X)) :- odd(X).
odd(s(X)) :- even(X).

Query: even(g)

(1) PrologToCdtProblemTransformerProof (UPPER BOUND (ID) transformation)

Built complexity over-approximating cdt problems from derivation graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: even(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: even(T1), SCOPE: 1, CLAUSE: 0 | even(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: true | even(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: even(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\neven(T1) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: even(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: odd(T4)\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: odd(T4), SCOPE: 2, CLAUSE: 2\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: even(T8)\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 23 [arrowhead = none ];
23 -> 4;

24 [label="EVAL with clause\neven(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 6;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 25 [arrowhead = none ];
25 -> 8;

26 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 26 [arrowhead = none ];
26 -> 9;

27 [label="EVAL with clause\neven(s(X4)) :- odd(X4).\nand substitutionX4 -> T4,\nT1 -> s(T4)", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 27 [arrowhead = none ];
27 -> 18;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 28 [arrowhead = none ];
28 -> 19;

29 [label="BACKTRACK\nfor clause: even(s(X)) :- odd(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 29 [arrowhead = none ];
29 -> 11;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 30 [arrowhead = none ];
30 -> 20;

31 [label="EVAL with clause\nodd(s(X8)) :- even(X8).\nand substitutionX8 -> T8,\nT4 -> s(T8)", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 31 [arrowhead = none ];
31 -> 21;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 32 [arrowhead = none ];
32 -> 22;

33 [label="INSTANCE with matching:\nT1 -> T8", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 33 [arrowhead = none , style=dashed];
33 -> 1[style=dashed];

}

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in(0) → f1_out1
f1_in(s(s(z0))) → U1(f1_in(z0), s(s(z0)))
U1(f1_out1, s(s(z0))) → f1_out1

Tuples:

F1_IN(s(s(z0))) → c1(U1'(f1_in(z0), s(s(z0))), F1_IN(z0))

S tuples:

F1_IN(s(s(z0))) → c1(U1'(f1_in(z0), s(s(z0))), F1_IN(z0))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c1

(3) CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in(0) → f1_out1
f1_in(s(s(z0))) → U1(f1_in(z0), s(s(z0)))
U1(f1_out1, s(s(z0))) → f1_out1

Tuples:

F1_IN(s(s(z0))) → c1(F1_IN(z0))

S tuples:

F1_IN(s(s(z0))) → c1(F1_IN(z0))

K tuples:none
Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c1

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F1_IN(s(s(z0))) → c1(F1_IN(z0))

We considered the (Usable) Rules:none
And the Tuples:

F1_IN(s(s(z0))) → c1(F1_IN(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F1_IN(x₁)) = [2]x₁
POL(c1(x₁)) = x₁
POL(s(x₁)) = [1] + x₁

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f1_in(0) → f1_out1
f1_in(s(s(z0))) → U1(f1_in(z0), s(s(z0)))
U1(f1_out1, s(s(z0))) → f1_out1

Tuples:

F1_IN(s(s(z0))) → c1(F1_IN(z0))

S tuples:none
K tuples:

F1_IN(s(s(z0))) → c1(F1_IN(z0))

Defined Rule Symbols:

f1_in, U1

Defined Pair Symbols:

F1_IN

Compound Symbols:

c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))

(9) PrologToCdtProblemTransformerProof (UPPER BOUND (ID) transformation)

Built complexity over-approximating cdt problems from derivation graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="2: even(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
3 [label="3: even(T1), SCOPE: 1, CLAUSE: 0 | even(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true | even(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: even(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\neven(T1) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: even(0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: odd(T3)\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: odd(T3), SCOPE: 2, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: even(T6)\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="EVAL with clause\neven(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 20 [arrowhead = none ];
20 -> 7;

21 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 21 [arrowhead = none ];
21 -> 10;

22 [label="EVAL with clause\neven(s(X3)) :- odd(X3).\nand substitutionX3 -> T3,\nT1 -> s(T3)", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 22 [arrowhead = none ];
22 -> 13;

23 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 23 [arrowhead = none ];
23 -> 14;

24 [label="BACKTRACK\nfor clause: even(s(X)) :- odd(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 24 [arrowhead = none ];
24 -> 12;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 25 [arrowhead = none ];
25 -> 15;

26 [label="EVAL with clause\nodd(s(X6)) :- even(X6).\nand substitutionX6 -> T6,\nT3 -> s(T6)", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 26 [arrowhead = none ];
26 -> 16;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 27 [arrowhead = none ];
27 -> 17;

28 [label="INSTANCE with matching:\nT1 -> T6", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 28 [arrowhead = none , style=dashed];
28 -> 2[style=dashed];

}

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in(0) → f2_out1
f2_in(s(s(z0))) → U1(f2_in(z0), s(s(z0)))
U1(f2_out1, s(s(z0))) → f2_out1

Tuples:

F2_IN(s(s(z0))) → c1(U1'(f2_in(z0), s(s(z0))), F2_IN(z0))

S tuples:

F2_IN(s(s(z0))) → c1(U1'(f2_in(z0), s(s(z0))), F2_IN(z0))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1

(11) CdtGraphRemoveTrailingTuplepartsProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in(0) → f2_out1
f2_in(s(s(z0))) → U1(f2_in(z0), s(s(z0)))
U1(f2_out1, s(s(z0))) → f2_out1

Tuples:

F2_IN(s(s(z0))) → c1(F2_IN(z0))

S tuples:

F2_IN(s(s(z0))) → c1(F2_IN(z0))

K tuples:none
Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F2_IN(s(s(z0))) → c1(F2_IN(z0))

We considered the (Usable) Rules:none
And the Tuples:

F2_IN(s(s(z0))) → c1(F2_IN(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F2_IN(x₁)) = [2]x₁
POL(c1(x₁)) = x₁
POL(s(x₁)) = [1] + x₁

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f2_in(0) → f2_out1
f2_in(s(s(z0))) → U1(f2_in(z0), s(s(z0)))
U1(f2_out1, s(s(z0))) → f2_out1

Tuples:

F2_IN(s(s(z0))) → c1(F2_IN(z0))

S tuples:none
K tuples:

F2_IN(s(s(z0))) → c1(F2_IN(z0))

Defined Rule Symbols:

f2_in, U1

Defined Pair Symbols:

F2_IN

Compound Symbols:

c1