Problem Set 5

Due Friday, November 3 at 1:00 pm Pacific

This problem set – the last one purely on discrete mathematics – is designed as a cumulative review of the topics we’ve covered so far and a proving ground to try out your newfound skills with mathematical induction. The problems here span all sorts of topics and we hope that it serves as a fitting coda to our whirlwind tour of discrete math!

We recommend that you read the Guide to Induction before starting this problem set. It contains a lot of useful advice about how to approach problems inductively, how to structure inductive proofs, and how to not fall into common inductive traps. Additionally, before submitting, be sure to read over the Induction Proofwriting Checklist for a list of specific things to watch for in your solutions before submitting.

As a note on this problem set – normally, you're welcome to use any proof technique you'd like to prove results in this course. On this problem set, we've specifically asked on some problems that you prove a result inductively. For those problems, you should prove those results using induction or complete induction, even if there is another way to prove the result. (If you'd like to use induction in conjunction with other techniques like proof by contradiction or proof by contrapositive, that's perfectly fine.)

Good luck, and have fun!

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Problem One: Induction Proof Critiques

Below are a collection of proofs by induction. For each proof, do the following:

• Determine whether the "theorem" being proved is actually true. If it's true, just tell us that it's true. If it's false, give a counterexample.
• Critique the proofwriting style in regards to the Induction Proofwriting Checklist and Guide to Induction.
• Identify any logic errors in the execution of the proofs.

You do not need to rewrite these proofs – just give us your feedback on the items we mentioned.

1. Critique the following proof about sums of natural numbers.

   Theorem: The sum of the first $n$ natural numbers is $\frac{n(n - 1)}{2}$.

   Proof: Let $P(n) = \frac{n(n - 1)}{2}$. We will prove by induction on $n$ that $P(n)$ holds for all $n \in \naturals$, from which theorem follows.

   As our base cases, we prove $P(0)$ and $P(1)$. First we’ll prove $P(0)$, that the sum of the first zero natural numbers is $\frac{0(0 - 1)}{2}$. The sum of no numbers is the empty sum $(0)\text,$ and we see that $\frac{0(0 - 1)}{2} = 0$, so $P(0)$ is true. Next, we’ll prove $P(1)$. The sum of the first natural number is $0$ (since $0$ is the smallest natural number), and we note that $\frac{1(1 - 1)}{2} = 0$, so $P(1)$ is true.

   For our inductive step, assume for all natural numbers $k$ that $P(k)$ is true. This means

   \[0 + 1 + 2 + \dots + (k - 1) \quad = \quad \frac{k(k - 1)}{2}\text.\]

   We will prove $P(k+1)$, that the sum of the first $k+1$ natural numbers is equal to $\frac{(k + 1)k}{2}$. Starting with the sum of the first $k+1$ natural numbers, we see that

   \[\begin{array}{lcl} 0 + 1 + 2 + \dots + (k - 1) + k & = & \frac{(k+1)k}{2} \\ 0 + 1 + 2 + \dots + (k - 1) & = & \frac{(k + 1)k}{2} - k \\ 0 + 1 + 2 + \dots + (k - 1) & = & \frac{(k + 1)k}{2} - \frac{2k}{2}\\ 0 + 1 + 2 + \dots + (k - 1) & = & \frac{(k + 1 - 2)k}{2} \\ 0 + 1 + 2 + \dots + (k - 1) & = & \frac{k(k - 1)}{2} \text. \end{array}\]

   We've reached our inductive hypothesis, so the equation is true. Thus $P(k+1)$ holds, completing the induction. $\qed$

2. Critique the following proof about directed graphs.

   Theorem: No directed graphs have any cycles.

   Proof: Let $P(n)$ be the statement “for any $n \in \naturals$, no directed graph with $n$ nodes contains a cycle.” We will prove by induction that $P(n)$ holds for all $n \in \naturals$, from which the theorem follows.

   As a base case, we prove $P(0)$, that no directed graphs with $0$ nodes contain a cycle. To see this, consider any directed graph $G$ with no nodes. Since $G$ has no nodes, it has no closed walks, since all closed walks contain at least one node. Therefore, $G$ has no cycles, as required.

   For our inductive step, assume for some arbitrary $k \in \naturals$ that P(k) is true and no directed graph with $k$ nodes has any cycles. We will prove that $P(k+1)$ is true, namely, that no directed graphs with $k+1$ nodes have any cycles.

   Consider a directed graph $G$ with $k$ nodes. By our inductive hypothesis, we know that $G$ has no cycles. Now, consider the graph $G'$ formed by adding a new node $v$ to $G$, then adding edges from $v$ to each node in $G$. We claim that this directed graph has no cycles. To see this, note that any cycle in $G'$ must involve $v$, since $G$ has no cycles on its own. But this is impossible, since any walk that leaves $v$ can never return to it and therefore can't be a closed walk. Therefore, $G'$ has no cycles, so $P(k+1)$ holds, completing the induction. $\qed$

Problem Two: Recurrence Relations

A recurrence relation is a way of defining an infinitely long sequence (usually, a sequence of numbers). A recurrence relation specifies the value of the first term or terms of the sequence, then defines the remaining entries from the previous terms. For example, here’s a simple recurrence relation:

The first terms of this sequence are given as follows:

• $a_0 = 1$, since that’s what the first rule says.
• $a_1 = 2$, since the second rule says that $a_1 = 2a_0 = 2 \cdot 1 = 2\text.$
• $a_2 = 4$, since the second rule says that $a_2 = 2a_1 = 2 \cdot 2 = 4\text.$
• $a_3 = 8$, since the second rule says that $a_3 = 2a_2 = 2 \cdot 4 = 8\text.$

Extending further, this sequence starts off with the numbers

which all happen to be powers of two. It turns out that this isn't a coincidence – this recurrence relation perfectly describes the powers of two.

Fill in the blanks below to complete the proof by induction.

int a(int n) {
    if (n == 0) return 1;
    return 2 * a(n - 1);
}

Taking a recurrence relation like the relation $a_n$ here and finding a simple, non-recursive expression for it is called solving the recurrence. It's common in theoretical CS to need to solve recurrences. For example, when determining the efficiency of a recursive algorithm, you might first start by writing a recurrence relation for its runtime, then solve the recurrence to get a simpler, direct way of calculating its efficiency. Take CS161 to learn how to do this!

Problem Three: Stacking Cans

On a recent trip to grocery store, I saw a bunch of cans stacked up in a pyramid like the one shown here. Each layer in the pyramid was made from a hexagon of cans, with each hexagon slightly bigger than the hexagon below it.

I went home to try making something like this from the cans in my pantry and quickly figured out that I didn't have enough cans to do so. Turns out, you need a lot of cans to make a pyramid like this! In this question, you'll figure out just how many cans are required.

The first step in this process is to figure out how many cans are in each hexagon. Here are cross-sectional views of the hexagons from the pyramid:

Notice that in moving from the first hexagon to the second we've surrounded the center can in six cans. In moving from the second hexagon to the third we've surrounded the seven previous cans with twelve new cans. And in moving from the third hexagon to the fourth we've surrounded the previous nineteen cans with eighteen new cans. And more generally, it seems like we go from one hexagon to the next by surrounding the previous hexagon with a number of cans equal to the next multiple of six. That lets us write a recurrence relation for how many cans are in the $n$th hexagon:

Pause for a minute to make sure you see why this is.

1. Prove by induction that $h_n = 3n(n - 1) + 1$ for all natural numbers $n \ge 1 \text.$

   As in Problem Two, you begin here with the recursive definition of $h$ given above, then need to show that the terms of the series are also given by the simpler formula $3n(n - 1) + 1$.

Now that we know how many cans are in each hexagon, we can start working out how many cans it takes to make a pyramid with $n$ layers.

2. Fill in the following blanks. No justification is required.

   • A 0-layer tower has $\blank$ cans in it.
   • A 1-layer tower has $1$ can in it.
   • A 2-layer tower has $8$ cans in it.
   • A 3-layer tower has $\blank$ cans in it.
   • A 4-layer tower has $\blank$ cans in it.
   • A 5-layer tower has $\blank$ cans in it.
   • A 6-layer tower has $\blank$ cans in it.
   • A 7-layer tower has $\blank$ cans in it.
   • A 8-layer tower has $\blank$ cans in it.
   • A 9-layer tower has $\blank$ cans in it.
   • A 10-layer tower has $\blank$ cans in it.
     

   Just to make sure you didn't miss it, we want the total number of cans in the full tower made of $n$ layers, not just the number of cans in layer $n\text.$

3. Fill in the following blank with a simple mathematical expression (e.g. $n^2 + 19n + 1$, $\sqrt{n^2 + 5}$, $\floor{\frac{n}{2}}^2$, etc, but not a recurrence relation like $t_{n+1} = t_n + h_{n+1}$). No justification is required.

   An $n$-layer tower has $\blank$ cans in it.

4. Prove by induction on $n$ that your formula from part (iii) holds for all natural numbers $n$.

   We've asked you to prove this is true for all natural numbers $n$, including when $n = 0\text.$

Your answer to part (iv) explains why I wasn't able to build any reasonable pyramid from the cans I have at home. It's worth taking a minute or two to ponder how much it would cost to build a pyramid twenty layers deep!

A purely optional but very rewarding question to ponder: the formula you proved correct in part (iv) suggests there's a completely different way to arrange the cans in the pyramid into an appealing geometrical shape. What is that other shape? Can you find a way to explain why that shape - which superficially has nothing to do with hexagons - relates back to hexagonal layers?

Problem Four: The Circle Game

You have a circle with $2n$ arbitrarily-chosen points on its circumference for some natural number $n \ge 0$. Of the $2n$ points, $n$ are labeled $+1$, and the remaining $n$ are labeled $-1$. One sample circle with eight points, of which four are labeled $+1$ and four are labeled $-1$, is shown below.

Here's a game you can play. Pick some position on the circle as your starting point - it can either be one of the points with a number on it, or an unnumbered point that just happens to be on the circle. Now, move clockwise around the circle. You lose the game if at any point on you pass through more $-1$ points than $+1$ points. You win the game if you get all the way back to your starting point without losing. For example, if you start at point $A$, the game would go like this:

• Start at $A$: $+1$.
• Pass through $B$: $+2$.
• Pass through $C$: $+1$.
• Pass through $D$: $0$.
• Pass through $E$: $-1$. (You lose.)

If you started at point $G$, the game would go like this:

• Start at $G$: $-1$ (You lose.)

However, if you started at point $F$, the game would go like this:

• Start at $F$: $+1$.
• Pass through $G$: $0$.
• Pass through $H$: $+1$.
• Pass through $A$: $+2$.
• Pass through $B$: $+3$.
• Pass through $C$: $+2$.
• Pass through $D$: $+1$.
• Pass through $E$: $+0$.
• Return to $F$. (You win!)

There’s a remarkable theorem about this game:

Theorem: For any natural number $n$, if $n$ points labeled $+1$ are placed on the boundary of the circle and $n$ points labeled $-1$ are placed on the boundary of the circle, there’s always some starting position on the circle from which you can start and win the circle game.

We’re going to ask you to prove this using a proof by induction with this predicate $P(n)$:

$P(n)$ is the predicate “for any circle with $n$ points labeled $+1$ and $n$ points labeled $-1$ on its boundary, there is a starting position on the circle from which you can win the circle game.”

Before proceeding, let’s confirm you see the general structure of the proof.

1. Is $P(n)$ a universally-quantified statement or an existentially-quantified statement? Based on that, will you “induct up,” or will you “induct down?”

   There are multiple quantifiers in $P(n)$. We care about the first one.

2. Prove by induction that $P(n)$ is true for all $n \in \naturals$.

Problem Five: Regular Graphs

A graph $G = (V, E)$ is called $d$-regular if every node in $G$ has degree exactly $d$. (As a reminder, the degree of a node $v$ is the number of edges that have $v$ as an endpoint, or, equivalently, the number of nodes that $v$ is adjacent to.)

For example, here's a 3-regular graph with 6 nodes and a 4-regular graph with 7 nodes:

Prove by induction that for all natural numbers $n$ there exists an $n$-regular graph containing exactly $2^n$ nodes.

Regular graphs often make appearances in combinatorics and theoretical computer science. Graphs with the specific property you're exploring in this problem - where there are a huge number of nodes of modest degree - have applications in parallel computing, network design, and error-correcting codes. Take CS144 (Computer Networks), CS149 (Parallel Computing) or CS250 (Algebraic Error-Correcting Codes) to learn more!

Problem Six: Contractions

A function $f : \naturals \to \naturals$ is called a contraction if $f(0) = 0$ and the following is true about $f$:

Before we move on, let's confirm this definition makes sense.

1. Which of the following are contractions? No justification is necessary.

   1. $f : \naturals \to \naturals$ defined as $f(n) = n - 1\text.$

   2. $f : \naturals \to \naturals$ defined as follows:

      \[f(n) = \left\lbrace \begin{array}{cl} 0 & \text{if } n \le 1 \\ n - 1 & \text{otherwise} \end{array} \right.\]
   3. $f : \naturals \to \naturals$ defined as $f(n) = 0 \text.$
   4. $f : \naturals \to \naturals$ defined as $f(n) = \floor{\frac{n}{2}} \text.$
   5. $f : \naturals \to \naturals$ defined as $f(n) = \ceil{\frac{n}{2}} \text.$

   6. $f : \naturals \to \naturals$ defined as follows:

      \[f(n) = \begin{cases} 0 & \text{if } n \text{ is even} \\ n-1 & \text{if } n \text{ is odd}\end{cases}\]

Contractions have the following important property that makes them extremely useful in the analysis of algorithms and data structures.

2. Let $f$ be a contraction. Prove by induction that for all natural numbers $n\text,$ there is a natural number $r$ where $f^r(n) = 0\text.$

   Here, $f^r(n)$ means "$f$ applied to $n$ a total of $r$ times" rather than "$f(n)$ raised to the $r\text{th}$ power." Refer back to PS3 for the definition of $f^n\text.$

(The rest of this section is just for context on why what you just proved is important.)

Let $f$ be a contraction. We can define a new function $f^\star : \naturals \to \naturals$ so that $f^\star(n)$ is the smallest natural number $r$ for which $f^r(n) = 0\text.$ In other words, $f^\star(n)$ is the number of times we have to apply $f$ to $n$ before it drops to zero.

These functions $f^\star$ have applications in algorithmic analysis. If $f$ is a contraction function that grows "slowly," for some definition of "slowly," then $f^\star$ grows even more slowly than $f$. For example, if $f(n)$ is a function that sends $n$ to $n - 2$ (or zero, if that's too small), then $f^\star$ roughly halves its inputs, $f^{\star\star}$ roughly computes the base-2 logarithm of its input, and $f^{\star\star\star}$ grows so slowly that if $n$ is the number of atoms in the known universe, then $f^{\star\star\star}(n) = 6$. (Wow!)

Want to learn more about iterated functions? Take CS166!

Problem Seven: Hereditary Sets

A hereditary set is a special type of set that is defined as follows:

This is a recursive definition that defines hereditary sets in terms of themselves. Accordingly, it's not surprising to see induction make an appearance when reasoning about them.

Prove by induction that $\wp^n(\emptyset)$ is a hereditary set for every natural number $n\text.$

Two fun facts about what you just proved. First, the sets you just reasoned about are enormously huge. Specifically, it's possible to show that

where there are $n$ copies of $2$ in the tower. This operation of "raise $2$ to itself $n$ times" is called tetration, where tetration is to exponentation what exponentiation is to multiplication and what multiplication is to addition. These numbers grow incredibly quickly; in fact, $\abs{\wp^7(\emptyset)}$ is inconcievably huge: its number of digits is so big, we couldn't write it out using all the protons in the known universe! Yet surprisingly tetration sometimes appears in the analysis of algorithms and data structures - take CS166 for details.

Second, hereditary sets have applications in foundational mathematics, a field that studies how to build all of mathematics from first principles. In a sense, hereditary sets represent the sets you can make if you only allow sets to contain other sets. You sometimes hear the class of all sets that can be made this way referred to as the von Neumann universe. To learn more, take Math 161 (Set Theory). Don't let the course number scare you; just because $161 \gt 3 \cdot 51$ doesn't mean it's three times harder than Math 51. In fact, Math 161 is an excellent follow-up course to CS103!

Optional Fun Problem: Egyptian Fractions

Leonardo Fibonacci, an eleventh-century Italian mathematician, is credited with introducing Hindu-Arabic numerals (the number system we use today) to Europe in his book Liber Abaci. The book Liber Abaci is also the source of the Fibonacci sequence (a sequence that begins 0, 1 and where each successive term is the sum of the two previous terms).

Relevant for this problem, Liber Abaci also described a method of writing out fractions called Egyptian fractions, which has been employed since ancient times; the Rhind Mathematical Papyrus, composed about 3,500 years ago in Thebes, includes several tables of fractions written out this way.

An Egyptian fraction is a sum of distinct fractions whose numerators are all 1 (these fractions are called unit fractions). For example, here are some sample Egyptian fraction representations:

Egyptian fractions are useful for divvying up objects fairly. For example, suppose you have two cakes to distribute to fifteen people – that is, everyone should get a $\frac{2}{15}$ fraction of those cakes. Begin by slicing each cake into tenths and giving each person one ($\frac{1}{10}$). Now, take the remaining tenths you haven’t distributed and cut them into thirds, giving thirtieths of the original cake. Each person then takes one of those ($\frac{1}{30}$). Because $\frac{1}{10} + \frac{1}{30} = \frac{2}{15}$, everyone gets their fair share. Pretty cool, isn’t it?

It's not immediately obvious that every fraction between 0 and 1 can be written this way, but, surprisingly, that is indeed the case. One way of finding an Egyptian fraction representation of a number is to use a greedy algorithm that works by finding the largest unit fraction at any point that can be subtracted out from the rational number. For example, to compute an Egyptian fraction representation for $\frac{42}{137}$, we would start off by noting that $\frac{1}{4}$ is the largest unit fraction less than $\frac{42}{137}$. We then say that

We then repeat this process by finding the largest unit fraction less than $\frac{31}{548}$ and subtracting it out. This number is $\frac{1}{18}$, so we get

The largest unit fraction we can subtract from $\frac{5}{4932}$ is $\frac{1}{987}$:

And at this point we're done, because the leftover fraction is itself a unit fraction.

Prove that the greedy algorithm for Egyptian fractions always terminates for any rational number $r$ in the range $0 \lt r \lt 1$ and always produces a valid Egyptian fraction. (A rational number is a real number that can be written as $r = \frac{p}{q}$ for some integers $p$ and $q$ where $q \ne 0$.) That is, the sum of the unit fractions should be the original number, there should only be finitely many fractions, and no unit fraction should be repeated. This shows that every rational number in the range $0 \lt r \lt 1$ has at least one Egyptian fraction representation.