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Non-trivial Decidable Nested Recurrence Relations Sahand Saba University of Victoria [email protected] 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"

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  • 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