of 23 /23
Non-trivial Decidable Nested Recurrence Relations Sahand Saba University of Victoria June 5, 2013

# Non-trivial Decidable Nested Recurrence Relationscanadam.math.ca/2013/program/slides/Saba.Sahand.Decidable.Nest… · I Discussed in \Godel, Escher, Bach: an Eternal Golden Braid"

• Author
others

• View
2

0

Embed Size (px)

### Text of Non-trivial Decidable Nested Recurrence...

• Non-trivial Decidable Nested RecurrenceRelations

Sahand Saba

University of Victoria

[email protected]

June 5, 2013

• Overview

Background

Decidable Nested Recurrences - Bounded Example

Decidable Nested Recurrences - Unbounded Example

Open Problems

• Nested Recurrence Relations

I Expression of the form . . .F (. . .F (. . .) . . .) . . . appears in therecursive definition for F .

I In other words, previous values are used to determine how farto look back for values, which are in turn used to generate thenew number in the sequence.

I Discussed in “Godel, Escher, Bach: an Eternal Golden Braid”by Douglas R. Hofstadter, where Q sequence, in particular, isintroduced.

I Also studied by Golomb, Ruskey, Tanny, etc.

I Some display chaotic behaviour, highly dependent on initialconditions.

• Decidability of Nested Recurrence Relations

I Celaya and Ruskey, in their 2012 paper, study nestedrecurrence relations from a computational perspective.

I The idea is to consider a recurrence relation, such asG (n) = G (n − G (n − 1)), as a an abstract computingmachine.

I The recurrence takes in as input a set of initial conditions,such as F (1) = 1 and F (2) = 2.

I The recurrence then produces an output in the form of asequence of numbers, in the above example 1, 2, 1, 1, 1, . . ..

• Decision Problems and Nested Recurrence Relations

I Similar to tag systems’ halting condition, a nested recurrencerelation can reach points in the sequence where it can nolonger generate a new number based on the recurrence.

I Simple example: let G (n) = G (n − G (n − 1)) and G (1) = 2,then G (2) = G (2− 2) = G (0) which is undefined.

I This is referred to as being “well-defined”, and also as “livingor dying”.

I We can formulate decision problems using this condition.

• Undecidable Example

I Celaya and Ruskey provide an example of a universal nestedrecurrence relation in their 2012 paper.

I The recurrence is

A(n) = A(n − 4− A(A(n − 4))) + 4A(A(n − 4)) +A(2A(n − 4− A(n − 2)) + A(n − 2)).

I Universality is proved by showing that the recurrence cansimulate reverse tag systems, which can in turn simulate tagsystems.

I Question of decidable examples is brought up.

• A Decidable Example

I First example is G (n) = G (n − G (n − 1)) discussed in UnjengCheng’s 1981 doctoral dissertation.

I Example graph given here.

0 5 10 15 20 25 30 35 40 45 502

4

6

8 G

Figure : Chart of values of G (n) = G (n − G (n − 1)) with initial values9, 3, 6, 5, 2, 2, 8, 5, 8, 6.

• Two observations about G (n − G (n − 1))

I Finiteness: Only a finite set of values are ever part of thesequence, namely the values given in the initial conditions.

I Boundedness: The furthest the recurrence ever goes back isdetermined by the largest number in the initial conditions.

I Example: In the above graph, the initial conditions are9, 3, 6, 5, 2, 2, 8, 5, 8, 6 so only values in {2, 3, 5, 6, 8, 9} aregoing to be in the sequence. Similarly, the furthest therecurrence rule will ever look back is K = 9 numbers back.

• Simplified Decidability Result for G (n − G (n − 1))

I Let K be the largest number in the initial conditions, and Rthe set of numbers in the initial condition.

I Construct a graph with vertices RK ∪ {⊥}. Connect(s1, s2, . . . , sK ) to (s2, s3, . . . , sK , sK+1−sK ) if sK ≤ K andlabel the edge with sK+1−sK , otherwise connect it to ⊥ andlabel the edge with ⊥.

I Connect ⊥ to ⊥ and label the edge ⊥.

• Simplified Decidability Result - Continued

I Starting with a set of K initial conditions, G (1) = s1, . . . ,G (K ) = sK , find the vertex (s1, . . . , sK ) on the graph, andfollow its infinite traversal.

I The edge labels then give the sequence G . If the sequencebecomes undefined at any point, then we get ⊥ at that pointonwards.

I Since the graph is finite by construction, either ⊥ or anothercycle is reached. This implies that either the sequence dies, orit becomes periodic at some point.

• Example Functional Di-Graph

Figure : Example directed graph constructed for initial conditions 3, 2.

• Generalization

Generalizes to any nested recurrence relation that satisfies thefiniteness and boundedness conditions. Examples:

I G (n) = G (n − G (n − 1)− G (n − 2))I G (n) = G (n − 2G (n − 1) + 3G (n − 2))I G (n) = G (n − G (G (n − 1) + G (n − 2)))I etc.

• Nested Recurrences as Discrete Dynamical Systems

It is to interesting point out that some recurrences that satisfy thefiniteness and boundedness conditions can be embedded indiscrete dynamical systems on reals. A brief construction of thedynamical system corresponding to G (n) = G (n − G (n − 1)) isgiven here:

I Assume that b is a number that is larger than the largestgiven initial condition.

I Represent initial segments of the sequence in reverse as base bdecimals. For example, letting b = 10, represent(1, 3, 2, 4, 2, 5) with x = 0.524231.

I Then G (n − k) corresponds to bbkxc.I After some algebra and simplification we get the following

function acting on [0, 1] as the dynamical system:

Ξ(x) = 1b⌊bbbxcx

⌋−⌊bbbxc−1x

⌋+ xb

• Graph of Ξ(x)

Figure : Graph of Ξ(x) with b = 10 and the line y = x .

• Non-homogenous case: G (n) = 1 + G (n − G (n − 1))

Now let us consider G (n) = 1 + G (n − G (n − 1)).I Sequences given by this recurrence relation are unbounded

provided they are well-defined.

I Therefore further and further previous numbers are needed togenerate the next number.

I Meaning both above observations fail.

I Need a new approach to prove decidability.

I However, looking at the graphs, an observable patterndefinitely exists. See below for a few examples.

• G (n) = 1 + G (n − G (n − 1)) Example 1

Figure : Graph of (n) = 1 + G (n − G (n − 1)) with initial conditions(2, 1, 3).

• G (n) = 1 + G (n − G (n − 1)) Example 2

Figure : Graph of (n) = 1 + G (n − G (n − 1)) with initial conditions(1, 3, 2, 4).

• G (n) = 1 + G (n − G (n − 1)) Example 3

Figure : Graph of (n) = 1 + G (n − G (n − 1)) with initial conditions(4, 4, 8, 7, 2, 1, 2, 9, 1, 7).

• Sketch of the Decidability Proof

Decidability can be proved by showing the following:

1. Let M(n) = maxni=1 G (i) and K the number of initialconditions. Then M(n) is slow-growing for n ≥ K .

2. If for some n0 we have M(n0) ≤ n0 then the sequence lives.3. M(n) does not grow in two consecutive positions.

4. Therefore G (n) ≤ C + n2 for n > N for some constants N,Cboth of which are computable from the initial conditions.

5. This means n0 required in condition 2 above can be found tobe less than 2M(K ).

6. End result: if the sequence dies, it dies before 2M(K )numbers.

• Open Problems

I M(n) = Θ(√n) provided the sequence lives.

I Complete characterization of sequences generated byG (n) = 1 + G (n − G (n − 1)). See next slide.

I Generalize to G (n) = v + G (n − G (n − 1)) for v ≥ 2.I Is Q(n) = Q(n − Q(n − 1)) + Q(n − Q(n − 2)) decidable or

undecidable?

I Can the decision problem forQ(n) = Q(n − Q(n − 1)) + Q(n − Q(n − 2)) be reduced tothat of other similar “meta-Fibonacci” recurrences, such asC (n) = C (n − C (n − 1)) + C (n − 1− C (n − 2))?

• Conjecture About G (n) = 1 + G (n − G (n − 1))

I Let Ak be the sequence of indices where M(n) grows.

I Define generations gk for k ≥ 0 by letting gk be the word(G (Ak+1),G (Ak+1 + 1),G (Ak+1 + 2), . . . ,G (Ak+2 − 1)).

I Then assuming G (n) lives, as an infinite word we haveG = g0g1g2g3g4 . . .

I Let ĝk = gk − k where subtraction is done component wise.I Conjecture: There exists K and p ≥ 1 such that for i > K we

haveĝip+r = xry

i−1zr

for 0 ≤ r < p, where |y | = p.

• Example

I Take initial conditions (1, 3, 2, 4). Graph is given in Example 2in previous slides. Omitting brackets and commas we have:

ĝ0 = 13242

ĝ1 = 43442

ĝ2 = 4434424

ĝ3 = 443442

ĝ4 = 43443442

. . .

I We get p = 3, y = 443 and the following:

ĝ3k = 443(443)k−1442

ĝ3k+1 = 43443(443)k−1442

ĝ3k+2 = 443(443)k−14424

• The End

BackgroundDecidable Nested Recurrences - Bounded ExampleDecidable Nested Recurrences - Unbounded ExampleOpen Problems

Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents