Inferring Expected Runtimes Using Sizes

KoAT2 Proof WORST_CASE( ?, 1+17/3*Arg_0 {O(n)})

Initial Complexity Problem (after preprocessing)

Start:f
Program_Vars:Arg_0
Temp_Vars:
Locations:f, g, h
Transitions:
f(Arg_0) -> g(Arg_0)
g(Arg_0) -> 1/4:h(Arg_0+1) :+: 3/4:h(Arg_0-1) :|: 0<Arg_0
h(Arg_0) -> 1/5:h(Arg_0+1) :+: 4/5:h(Arg_0-1) :|: 0<Arg_0 && 0<=Arg_0
h(Arg_0) -> g(Arg_0) :|: Arg_0<1 && 0<=Arg_0

G f f g g f->g t₀ ∈ g₀ h h g->h t₁ ∈ g₁ p = 1/4 η (Arg_0) = Arg_0+1 τ = 0<Arg_0 g->h t₂ ∈ g₁ p = 3/4 η (Arg_0) = Arg_0-1 τ = 0<Arg_0 h->g t₅ ∈ g₃ τ = Arg_0<1 h->h t₃ ∈ g₂ p = 1/5 η (Arg_0) = Arg_0+1 τ = 0<Arg_0 h->h t₄ ∈ g₂ p = 4/5 η (Arg_0) = Arg_0-1 τ = 0<Arg_0

Timebounds:

Overall timebound:inf {Infinity}
0,0: f->g: 1 {O(1)}
1,1: g->h: max([0, Arg_0]) {O(n)}
2,1: g->h: max([0, Arg_0]) {O(n)}
3,2: h->h: inf {Infinity}
4,2: h->h: inf {Infinity}
5,3: h->g: max([0, 2*Arg_0]) {O(n)}

Expected Timebounds:

Overall expected timebound: 1+17/3*Arg_0 {O(n)}
0: f->[1:g]: 1 {O(1)}
1: g->[1/4:h; 3/4:h]: 2*Arg_0 {O(n)}
2: h->[1/5:h; 4/5:h]: 5/3*Arg_0 {O(n)}
3: h->[1:g]: 2*Arg_0 {O(n)}

Costbounds:

Overall costbound: inf {Infinity}
0,0: f->g: inf {Infinity}
1,1: g->h: inf {Infinity}
2,1: g->h: inf {Infinity}
3,2: h->h: inf {Infinity}
4,2: h->h: inf {Infinity}
5,3: h->g: inf {Infinity}

Expected Costbounds:

Overall expected costbound: 1+17/3*Arg_0 {O(n)}
0: f->[1:g]: 1 {O(1)}
1: g->[1/4:h; 3/4:h]: 2*Arg_0 {O(n)}
2: h->[1/5:h; 4/5:h]: 5/3*Arg_0 {O(n)}
3: h->[1:g]: 2*Arg_0 {O(n)}

Sizebounds:

0,0: f->g, Arg_0: Arg_0 {O(n)}
1,1: g->h, Arg_0: 1+Arg_0 {O(n)}
2,1: g->h, Arg_0: (-1)+Arg_0 {O(n)}
5,3: h->g, Arg_0: 0 {O(1)}

ExpSizeBounds:

(0: f->[1:g], g), Arg_0: Arg_0 {O(n)}
(1: g->[1/4:h; 3/4:h], h), Arg_0: 1+Arg_0 {O(n)}
(3: h->[1:g], g), Arg_0: 0 {O(1)}