Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/4.18)

The rewrite relation of the following TRS is considered.

gcd(x,0)	→	x	(1)
gcd(0,y)	→	y	(2)
gcd(s(x),s(y))	→	if(<(x,y),gcd(s(x),-(y,x)),gcd(-(x,y),s(y)))	(3)

The evaluation strategy is innermost.

Property / Task

Determine bounds on the runtime complexity.

Answer / Result

An upperbound for the complexity is O(n²).

Proof (by AProVE @ termCOMP 2023)

1 Dependency Tuples

We get the following set of dependency tuples:

gcd^#(z0,0)

→

(5)

originates from

gcd(z0,0)

→

(4)

gcd^#(0,z0)

→

(7)

originates from

gcd(0,z0)

→

(6)

gcd^#(s(z0),s(z1))

→

c2(gcd^#(s(z0),-(z1,z0)),gcd^#(-(z0,z1),s(z1)))

(9)

originates from

gcd(s(z0),s(z1))

→

if(<(z0,z1),gcd(s(z0),-(z1,z0)),gcd(-(z0,z1),s(z1)))

(8)

Moreover, we add the following terms to the innermost strategy.

gcd^#(z0,0)

gcd^#(0,z0)

gcd^#(s(z0),s(z1))

1.1 Usable Rules

We remove the following rules since they are not usable.

gcd(z0,0)	→	z0	(4)
gcd(0,z0)	→	z0	(6)
gcd(s(z0),s(z1))	→	if(<(z0,z1),gcd(s(z0),-(z1,z0)),gcd(-(z0,z1),s(z1)))	(8)

1.1.1 Rule Shifting

The rules

gcd^#(z0,0)	→	c	(5)
gcd^#(0,z0)	→	c1	(7)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c]	=	0
[c1]	=	0
[c2(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[gcd^#(x₁, x₂)]	=	3 · x₁ + 0 + 2 · x₂
[0]	=	3
[s(x₁)]	=	0
[-(x₁, x₂)]	=	0

which has the intended complexity. Here, only the following usable rules have been considered:

gcd^#(z0,0)	→	c	(5)
gcd^#(0,z0)	→	c1	(7)
gcd^#(s(z0),s(z1))	→	c2(gcd^#(s(z0),-(z1,z0)),gcd^#(-(z0,z1),s(z1)))	(9)

1.1.1.1 Rule Shifting

The rules

gcd^#(s(z0),s(z1))

→

c2(gcd^#(s(z0),-(z1,z0)),gcd^#(-(z0,z1),s(z1)))

(9)

are strictly oriented by the following non-linear polynomial interpretation over the naturals

[c]	=	0
[c1]	=	0
[c2(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[gcd^#(x₁, x₂)]	=	2 · x₁ · x₂ + 0
[0]	=	0
[s(x₁)]	=	2
[-(x₁, x₂)]	=	0

which has the intended complexity. Here, only the following usable rules have been considered:

gcd^#(z0,0)	→	c	(5)
gcd^#(0,z0)	→	c1	(7)
gcd^#(s(z0),s(z1))	→	c2(gcd^#(s(z0),-(z1,z0)),gcd^#(-(z0,z1),s(z1)))	(9)

1.1.1.1.1 R is empty

There are no rules in the TRS R. Hence, R/S has complexity O(1).