Non-trivial Decidable Nested RecurrenceRelations

Sahand Saba

University of Victoria

saba@uvic.ca

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

⌋+ x

b

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 unboundedprovided 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,C

both 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 orundecidable?

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 gk = gk − k where subtraction is done component wise.

I Conjecture: There exists K and p ≥ 1 such that for i > K wehave

gip+r = xryi−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:

g0 = 13242

g1 = 43442

g2 = 4434424

g3 = 443442

g4 = 43443442

. . .

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

g3k = 443(443)k−1442

g3k+1 = 43443(443)k−1442

g3k+2 = 443(443)k−14424

The End