Inferring Expected Runtimes Using Sizes

KoAT2 Proof WORST_CASE( ?, 6/5 {O(1)})

Initial Complexity Problem (after preprocessing)

Start:f
Program_Vars:Arg_0
Temp_Vars:
Locations:f, g
Transitions:
f(Arg_0) -{0}> g(1)
g(Arg_0) -> 3/4:g(0) :+: 1/12:g(0) :+: 1/6:g(1) :|: 0<Arg_0 && Arg_0<=1 && 0<=Arg_0

G f f g g f->g t₀ ∈ g₀ η (Arg_0) = 1 {0} g->g t₁ ∈ g₁ p = 3/4 η (Arg_0) = 0 τ = 0<Arg_0 g->g t₂ ∈ g₁ p = 1/12 η (Arg_0) = 0 τ = 0<Arg_0 g->g t₃ ∈ g₁ p = 1/6 η (Arg_0) = 1 τ = 0<Arg_0

Timebounds:

Overall timebound:inf {Infinity}
0,0: f->g: 1 {O(1)}
1,1: g->g: 1 {O(1)}
2,1: g->g: 1 {O(1)}
3,1: g->g: inf {Infinity}

Expected Timebounds:

Overall expected timebound: 11/5 {O(1)}
0: f->[1:g]: 1 {O(1)}
1: g->[3/4:g; 1/12:g; 1/6:g]: 6/5 {O(1)}

Costbounds:

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

Expected Costbounds:

Overall expected costbound: 6/5 {O(1)}
0: f->[1:g]: 0 {O(1)}
1: g->[3/4:g; 1/12:g; 1/6:g]: 6/5 {O(1)}

Sizebounds:

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

ExpSizeBounds:

(0: f->[1:g], g), Arg_0: 1 {O(1)}
(1: g->[3/4:g; 1/12:g; 1/6:g], g), Arg_0: 1 {O(1)}