A Survey of Program Termination: Practical and Theoretical - PowerPoint PPT Presentation

A Survey of Program Termination: Practical and Theoretical Challenges Jo¨ el Ouaknine Department of Computer Science, Oxford University VTSA 2014 Luxembourg, October 2014

Instructive Example Consider the following order-5 recurrence: u n +5 = − 19 25 u n +4 − 114 125 u n +3 + 114 125 u n +2 + 19 25 u n +1 + u n

Instructive Example Consider the following order-5 recurrence: u n +5 = − 19 25 u n +4 − 114 125 u n +3 + 114 125 u n +2 + 19 25 u n +1 + u n This is simple, with characteristic roots 1 , λ 1 , λ 1 , λ 2 , λ 2 , where λ 1 = − 3 + 4 i and λ 2 = − 7 + 24 i 5 25

Instructive Example Consider the following order-5 recurrence: u n +5 = − 19 25 u n +4 − 114 125 u n +3 + 114 125 u n +2 + 19 25 u n +1 + u n This is simple, with characteristic roots 1 , λ 1 , λ 1 , λ 2 , λ 2 , where λ 1 = − 3 + 4 i and λ 2 = − 7 + 24 i 5 25 For suitably chosen initial values we have u n = 33 8 + λ n 1 + λ n 1 + 2 λ n 2 + 2 λ n 2

Orbits of Characteristic Roots { λ n 1 : n ∈ N } and { λ n 2 : n ∈ N } are both dense in T .

Orbits of Characteristic Roots { λ n 1 : n ∈ N } and { λ n 2 : n ∈ N } are both dense in T . 2 ) : n ∈ N } not dense in T 2 due to relation λ 2 { ( λ n 1 , λ n 1 λ 2 = 1.

Orbits of Characteristic Roots { λ n 1 : n ∈ N } and { λ n 2 : n ∈ N } are both dense in T . 2 ) : n ∈ N } not dense in T 2 due to relation λ 2 { ( λ n 1 , λ n 1 λ 2 = 1. 2 ) : n ∈ N } dense in helix { ( z 1 , z 2 ) ∈ T 2 : z 2 { ( λ n 1 , λ n 1 z 2 = 1 } .

Orbits of Characteristic Roots { λ n 1 : n ∈ N } and { λ n 2 : n ∈ N } are both dense in T . 2 ) : n ∈ N } not dense in T 2 due to relation λ 2 { ( λ n 1 , λ n 1 λ 2 = 1. 2 ) : n ∈ N } dense in helix { ( z 1 , z 2 ) ∈ T 2 : z 2 { ( λ n 1 , λ n 1 z 2 = 1 } . Point ( − 1 , − 1) does not lie on helix.

Example √ √ Critical Point! ( − 1 63 i 8 , − 31 63 i 8 + 32 + 32 )

Example √ √ Critical Point! ( − 1 63 i 8 , − 31 63 i 8 + 32 + 32 ) For ( λ n 1 , λ n 2 ) near this point, u n := 33 8 + λ n 1 + λ n 1 + 2 λ n 2 + 2 λ n 2 is close to 0.

Example √ √ Critical Point! ( − 1 63 i 8 , − 31 63 i 8 + 32 + 32 ) For ( λ n 1 , λ n 2 ) near this point, u n := 33 8 + λ n 1 + λ n 1 + 2 λ n 2 + 2 λ n 2 is close to 0. � u n � is ultimately positive—just.

Example √ √ Critical Point! ( − 1 63 i 8 , − 31 63 i 8 + 32 + 32 ) For ( λ n 1 , λ n 2 ) near this point, u n := 33 8 + λ n 1 + λ n 1 + 2 λ n 2 + 2 λ n 2 is close to 0. � u n � is ultimately positive—just. But what about u n − 1 2 n ?