Recurrences, Part 3 Troy Vasiga Centre for Education in Mathematics - - PowerPoint PPT Presentation

Recurrences, Part 3

Troy Vasiga Centre for Education in Mathematics and Computing Faculty of Mathematics, University of Waterloo cemc.uwaterloo.ca

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Outline

• Selected solutions to problems from last week
• L-systems
• Examples of L-systems
• Prouhet-Thue-Morse sequence
• Other definitions of PTM
• Morphisms
• Squares, cubes and overlaps
• Summary

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Solution to Problem 5

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n. Solution: We have a = 4, b = 2 and f (n) = n.

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Solution to Problem 5

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n. Solution: We have a = 4, b = 2 and f (n) = n. We calculate nlogb a = nlog2 4 = n2.

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Solution to Problem 5

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n. Solution: We have a = 4, b = 2 and f (n) = n. We calculate nlogb a = nlog2 4 = n2. Then, f (n) = n ∈ O(n2−ǫ for ǫ = 0.9 (for instance).

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Solution to Problem 5

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n. Solution: We have a = 4, b = 2 and f (n) = n. We calculate nlogb a = nlog2 4 = n2. Then, f (n) = n ∈ O(n2−ǫ for ǫ = 0.9 (for instance). Thus, we are in case 1 of the Master Theorem, and thus T(n) = Θ(nlogb a) = Θ(n2).

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Solution to Problem 7

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n3. Solution: We have a = 4, b = 2 and f (n) = n3. We calculate nlogb a = nlog2 4 = n2.

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Solution to Problem 7

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n3. Solution: We have a = 4, b = 2 and f (n) = n3. We calculate nlogb a = nlog2 4 = n2. Then, f (n) = n3 ∈ Ω(n3−ǫ) for ǫ = 1, for instance.

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Solution to Problem 7

Use the Master Theorem to find the asymptotic running time for the recurrence T(n) = 4T(n/2) + n3. Solution: We have a = 4, b = 2 and f (n) = n3. We calculate nlogb a = nlog2 4 = n2. Then, f (n) = n3 ∈ Ω(n3−ǫ) for ǫ = 1, for instance. Thus, we are in case 3 of the Master Theorem. We just need to verify that the second condition holds: that is, we need to show that af n b

• ≤ cf (n)

for some fixed c for all n sufficiently large.

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Solution to Problem 7 (continued)

Plugging in our constants and f (n) = n3, we have to show: 4 n 2 3 ≤ cn3.

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Solution to Problem 7 (continued)

Plugging in our constants and f (n) = n3, we have to show: 4 n 2 3 ≤ cn3. Pick c = 1 and then for n > 1, we have: n3 2 ≤ cn3.

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Solution to Problem 7 (continued)

Plugging in our constants and f (n) = n3, we have to show: 4 n 2 3 ≤ cn3. Pick c = 1 and then for n > 1, we have: n3 2 ≤ cn3. Thus, by case 3 of the Master Theorem, T(n) = Θ(n3).

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Definition of an L-system

An L-system or Lindenmayer system is a parallel rewriting system. By parallel, we mean we each step involves replacing every possible

• ccurrence at the same time.

We also need to specify the starting condition/symbol.

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Rewriting Rules

We use the rule X → Y to mean “replace every occurrence of X with Y .” For example, if we have the rule X → ABX and we have the current word: ABXBAX then the next word will be

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Rewriting Rules

We use the rule X → Y to mean “replace every occurrence of X with Y .” For example, if we have the rule X → ABX and we have the current word: ABXBAX then the next word will be ABABXBAABX

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What does this sequence do?

• 1. A → AB
• 2. B → ǫ

(Here ǫ means “this disappears.”) Start with A

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB.

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB. This indicates that AB is a fixed point of this recurrence.

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB. This indicates that AB is a fixed point of this recurrence. Definition: A fixed point of a function f is a value t such that f (t) = t.

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB. This indicates that AB is a fixed point of this recurrence. Definition: A fixed point of a function f is a value t such that f (t) = t. Notice: if f (t) = t, then f (f (t))

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB. This indicates that AB is a fixed point of this recurrence. Definition: A fixed point of a function f is a value t such that f (t) = t. Notice: if f (t) = t, then f (f (t)) = f (t) =

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB. This indicates that AB is a fixed point of this recurrence. Definition: A fixed point of a function f is a value t such that f (t) = t. Notice: if f (t) = t, then f (f (t)) = f (t) = t,

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Not much

This just generates AB at the first step, which becomes AB after the second step, and so on. So, this is just the word AB. This indicates that AB is a fixed point of this recurrence. Definition: A fixed point of a function f is a value t such that f (t) = t. Notice: if f (t) = t, then f (f (t)) = f (t) = t, which means f n(t) = f (f (· · · f (t)) · · · ) = t.

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What does this sequence do?

• 1. A → AB
• 2. B → B

Start with A

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Infinitely more

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Infinitely more

After n iterations of the rewriting rules, we have the word ABn = ABB · · · B.

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What does this sequence do?

• 1. A → B
• 2. B → AB

Start with A. Hint: write out the first few terms and look at them.

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Fibonacci

Notice the length. Notice the ratio |A| |B|

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What does this sequence do?

• 1. A → ABA
• 2. B → BBB

Start with A. Hint: draw a straight line.

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Cantor Dust

How much of the line is removed? What about the number 1

4 for instance?

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What does this sequence do?

• 1. X → X + YF
• 2. Y → FX − Y

Start with FX.

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Dragon Curve

Do the following things:

• F means move forward
• + means turn clockwise 90o
• − means turn counterclockwise 90o

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Dragon Curve

Here are some steps:

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Dragon Curve

Here are some steps:

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Dragon Curve

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Another L-system

Start with: F − −F − −F One rule: F → F + F − −F + F

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Another L-system

Start with: F − −F − −F One rule: F → F + F − −F + F Let:

• F: move forward
• +: move clockwise by π

3

• −: move counterclockwise by π

3

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Koch Snowflake

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What does this sequence do?

• 1. 0 → 01
• 2. 1 → 10

Start 0.

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Prouhet-Thue-Morse Sequence

It does lots of things. Call the sequence t = t0t1t2... = 0 1 1 0 1 0 0 1 ...

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Draw a picture

• If tn = 0, turn by π
• If tn = 1, move ahead one unit and then rotate

counterclockwise by an angle of π

3 .

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Koch Curve

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Another definition of the PTM Sequence

Write out all integers in base-2 (i.e., binary). Define s2(x) to the sum of the digits in the base-2 representation

• f x.

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Yet another definition of the PTM Sequence

Write the number, then flip the bits and rewrite it. Define: X0 = Xn+1 = Xn ¯ Xn where ¯ x means change all 0’s to 1’s, and 1’s to 0’s.

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Equivalent definitions

Prove that the last two definitions are equivalent. In other words, prove: tn = s2(n) mod 2 Proof: Use induction. Base case: n = 0. Assume true for all n < n′. Let k be the integer such that 2k ≤ n < 2k+1.

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Morphism

A morphism is a map h on strings that satisfies the identity h(xy) = h(x)h(y) for all strings x, y.

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PTM morphism

Define the PTM morphism µ(0) = 01, µ(1) = 10. Then µ(0) = 01

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PTM morphism

Define the PTM morphism µ(0) = 01, µ(1) = 10. Then µ(0) = 01 µ2(0) = µ(µ(0)) = 0110

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PTM morphism

Define the PTM morphism µ(0) = 01, µ(1) = 10. Then µ(0) = 01 µ2(0) = µ(µ(0)) = 0110 µ3(0) = 01101001 µ4(0) = 0110100110010110

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Fixed point

The fixed point of a morphism h beginning with a is: hω(a) = lim

m→∞ hm(a).

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction.

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0.

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk.

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) =

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0))

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) =

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01)

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01) =

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01) = µk(0)µk(1)

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01) = µk(0)µk(1) =

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01) = µk(0)µk(1) = Xk ¯ Xk

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01) = µk(0)µk(1) = Xk ¯ Xk =

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Equivalent definition

Let’s prove µn(0) = Xn. Use mathematical induction. Base case: n = 0. µ0(0) = 0 = X0. Assume the result holds for n = k. That is, µk(0) = Xk. Prove true for n = k + 1. µk+1(0) = µk(µ(0)) = µk(01) = µk(0)µk(1) = Xk ¯ Xk = Xk+1

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Overlap-free

A square is a word of the form xx. A cube is a word of the form xxx. An overlap is a word axaxa, with a single letter a and x being any word. English word examples?

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Overlap-free

A square is a word of the form xx. A cube is a word of the form xxx. An overlap is a word axaxa, with a single letter a and x being any word. English word examples? How does this relate to the Thue-Morse sequence?

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What does this sequence converge to?

Consider the sequence: 1 2, 1/2 3/4,

1/2 3/4 5/6 7/8

,

1/2 3/4 5/6 7/8 9/10 11/12 13/14 15/16

What does this converge to?

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Proof

First, observe the limit is: Πn≥0 2n + 1 2n + 2 (−1)tn Let: P = Πn≥0 2n + 1 2n + 2 (−1)tn Q = Πn≥0

• 2n

2n + 1 (−1)tn Notice that PQ = 1 2Πn≥1

• n

n + 1 (−1)tn

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Proof

PQ = 1 2Πn≥1

• n

n + 1 (−1)tn Rewrite, breaking over odd and even indices: PQ = 1 2Πn≥0 2n + 1 2n + 2 (−1)t2n+1 Πn≥1

• 2n

2n + 1 (−1)tn = 1 2P−1Q. Thus, P2 = 1

2.

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Summary

• recursion is very natural
• proving results using recursion is powerful
• recursion has a self-contained, succinct form
• recursion is beautiful

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