# Concrete Mathematics Chapter 1 Exam Problems

It took me longer than I thought, and the outcome is slightly disappointing: I failed to solve two of the problems, and I solved the remaining ones way too slowly, so in a real exam conditions I probably would have solved just one or two…

## Exam Problems

### 4 Pegs Tower of Hanoi

First, it helps to see that the indices of the recurrence are actually $S_n$:

\begin{aligned} W_{n(n+1)/2}&= W_{S_n}\\ W_{n(n-1)/2}&= W_{S_{n-1}} \end{aligned}

And of course, $S_n = S_{n-1} + n$.

Setting $m=S_{n-1}$, we try to show:

\begin{aligned} W_{m+n} & \le 2W_{m} + T_n\\ \end{aligned}

Now, obviously, if we have $m+n$ discs, we can move the $m$ top ones from $A$ to $C$ using $B$ and $D$ as transfer pegs, then move the bottom $n$ ones from $A$ to $B$ using $D$ as transfer peg, and finally move the top $m$ ones from $C$ to $B$.

The first step takes $W_m$ moves, the second one is the classic Tower of Hanoi problem (as we can no longer use peg $C$, we only have three pegs), so it takes $T_n$ moves, and the last step takes $W_m$ moves again.

This is only one possible solution; the optimal one must be equal or better, so we have

\begin{aligned} W_{m+n} & \le 2W_m + T_n\\ \end{aligned}

This is true for any $m+n$ discs, and in particular for $S_n = S_{n-1} + n$ ones.

### Specific Zigs

I could not solve this problem. I had found that the half-lines did intersect, but then I failed to show that their intersections were all distinct.

Even with the solution from the book, it took me a while before I finally had a complete understanding.

One problem I had was that lines in a graph are basic college level mathematics, but college was a long, long time ago. I pretty much had to work from first principles.

Following the book in writing the positions as $(x_j, 0)$ and $(x_j - a_j, 1)$, I need to find $\alpha$ and $\beta$ such that $y=\alpha x + \beta$ is true for both points above.

\begin{aligned} 0 & = \alpha x_j + \beta \\ \beta & = - \alpha x_j\\ 1 & = \alpha (x_j - a_j) - \alpha x_j\\ & = \alpha x_j - \alpha a_j - \alpha x_j\\ & = - \alpha a_j\\ \alpha & = \frac{-1}{a_j}\\ y & = \frac{x_j - x}{a_j}\\ \end{aligned}

With this given, I can try to find the intersection of lines from different zigs, $j$ and $k$:

\begin{aligned} \frac{x_j - x}{a_j} & = \frac{x_k - x}{a_k}\\ a_k (x_j - x) & = a_j (x_k - x)\\ a_k x_j - a_k x & = a_j x_k - a_j x\\ a_k x_j - a_j x_k & = (a_k - a_j) x\\ \end{aligned}

Now, still following the book, I replace $x$ by $t$ with $x=x_j - t a_j$:

\begin{aligned} a_k x_j - a_j x_k & = (a_k - a_j) (x_j - t a_j)\\ a_k x_j - a_j x_k & = a_k x_j - a_j x_j - t a_j a_k + t a_j^2\\ - a_j x_k & = t a_j^ 2 - a_j x_j - t a_j a_k\\ - x_k & = t a_j - x_j -t a_k&&\text{dividing by } a_j\\ x_j - x_k & = t (a_j - a_k)\\ t & = \frac{x_j - x_k}{a_j - a_k}\\ \end{aligned}

Somehow, I have a faint memory of such a result; I need to check a college math book.

To complete, I need to show that $y = t$:

\begin{aligned} y & = \frac{x_j - x}{a_j}\\ & = \frac{x_j - x_j + t a_j}{a_j}\\ & = \frac{t a_j}{a_j}\\ & = t\\ \end{aligned}

So the intersection of any two pair of half-lines from different zigs is $(x_j - t a_j, t)$. Note that $t$ has the same value whether $j \gt k$ or $k \gt j$. To simplify further computations, I set $j \gt k$.

There are two remaining steps: show that $t$ is different for different pairs of $j$, $k$ (with $j \ne k$); and then show that the four intersections for a pair $j$, $k$ are also distinct.

$a_j$ can be of two forms: $n^j$ and $n^j + n^{-n}$. So $a_j - a_k$ can be one of

\begin{aligned} & n^j - n^k\\ & n^j + n^{-n} - n^k\\ & n^j - n^k - n^{-n}\\ n^j + n^{-n} - n^k - n^{-n} = & n^j - n^k\\ \end{aligned}

So there are three different forms for $a_j - a_k$, which I will simply write $n^j - n^k + \epsilon$ where $|\epsilon| \lt 1$.

\begin{aligned} t & = \frac{n^{2j} - n^{2k}}{n^j - n^k + \epsilon}\\ & = \frac{(n^j - n^k)(n^j + n^k)}{n^j - n^k + \epsilon}\\ \end{aligned}

Let’s show that $n^j+n^k - 1 \lt t \lt n^j+n^k + 1$: multiply the whole inequality by $n^j - n^k + \epsilon$. As

\begin{aligned} n^j - n^k & \ge n\\ & \ge 2\\ & \gt |\epsilon|\\ \end{aligned}

so $n^j - n^k + \epsilon \gt 0$. Defining

\begin{aligned} N_{jk} & = n^j + n^k\\ N'_{jk} & = n^j - n^k\\ \end{aligned}

the left and right inequalities become

\begin{aligned} (N_{jk} - 1) (N'_{jk} + \epsilon) & = N_{jk}N'_{jk} - N'_{jk} + \epsilon N_{jk} - \epsilon\\ (N_{jk} + 1) (N'_{jk} + \epsilon) & = N_{jk}N'_{jk} + N'_{jk} + \epsilon N_{jk} + \epsilon\\ \end{aligned}

Subtracting $N_{jk}N'_{jk} = (n^j-n^k)(n^j+n^k)$ from the original inequality:

\begin{aligned} -N'_{jk}+\epsilon N_jk - \epsilon \lt 0 \lt N'_{jk} + \epsilon N_{jk} + \epsilon\\ \end{aligned}

I need to prove the following inequality

\begin{aligned} (n^j - n^k) & \gt |\epsilon| + |\epsilon| (n^j - n^k)\\ \end{aligned}

We already know $|\epsilon| \lt 1$, so looking at the second term (and assuming $\epsilon \ne 0$, as this case is trivial)

\begin{aligned} |\epsilon| (n^j-n^k) & = n^{-n} (n^j - n^k)\\ & = n^{j-n} - n^{k-n}\\ &\lt 1\\ \end{aligned}

and we have

\begin{aligned} n^j - n^k & \ge 2 & \gt |\epsilon| + |\epsilon (n^j - n^k)|\\ \end{aligned}

So the inequalities are established. $N_{jk}$ can be seen as a number in based $n$ where the digits are all zeroes except the $j$ and $k$ ones, $N_{jk} = N_{j'k'} \implies j=j', k=k'$, and therefore $t$ uniquely defines $j$ and $k$ or, two pairs of zigs must have different $t$.

I still need to show that for a given pair, when $t$ is the same, the intersections are different. There are three different values of $t$, so two intersections points have the same height. This happens for

\begin{aligned} t & = \frac{n^{2j} - n^{2k}}{n^j - n^k}\\ \end{aligned}

which happens when $a_j = n^j$, $a_k = n^k$ and $a_j = n^j + n^{-n}$, $a_k = n^k + n^{-n}$. But the $x = x_j - t a_j$ value for intersections is different: $t n^j$ and $t (n^j + n^{-n})$, so there are indeed four distinct intersection points.

### 30 degrees Zigs

I could not solve this problem. Once again, my lack of intuition with geometry was to blame.

But if we have two zigs with half-lines angles $\phi$, $\phi + 30^{\circ}$ and $\theta$, $\theta + 30^{\circ}$, then for any two pairs of half-lines from the two zigs to intersect, their angles must be between $0^{\circ}$ and $180^{\circ}$. Taken together, these constraints give $30^{\circ} \lt |\phi - \theta| \lt 150^{\circ}$.

Update: The original version of this post had a lower bound of $0$. Thanks to Tailshot for pointing out the error

This means there cannot be more than $5$ such pairs (and to be honest, I would have said 4, but the book says it’s indeed 5).

### Recurrence Equations

Using the repertoire method, solve the recurrence equations

\begin{aligned} h(1) & = \alpha\\ h(2n+j) & = 4h(n) + \gamma_j n + \beta_j\\ \end{aligned}

The general form of $h(n)$ is

\begin{aligned} h(n) & = \alpha A(n) + \beta_0 B_0(n) + \beta_1 B_1(n) + \gamma_0 C_0(n) + \gamma_1 C_1(n)\\ \end{aligned}

We get three of these functions directly by solving

\begin{aligned} h(1) & = \alpha\\ h(2n+j) & = 4h(n) + \beta_j\\ h(2^m+b_m\cdots b_0) & = (1\beta_{b_m}\cdots\beta_{b_0})_4\\ \end{aligned}

So we have a solution for $A(n)$, $B_0(n)$ and $B_1(n)$.

Setting $h(n) = n$

\begin{aligned} \alpha & = 1\\ 2n+j & = 4n + \gamma_j n + \beta_j\\ \beta_j & = j\\ \gamma_j & = -2\\ \end{aligned}

which gives the equation $n = A(n) + B_1(n) -2(C_0(n) + C_1(n))$.

Setting $h(n) = n^2$

\begin{aligned} \alpha & = 1\\ 4n^2 + 4jn + j & = 4n^2 + \gamma_j n + \beta_j\\ \beta_j & = j\\ \gamma_j & = 4j\\ \end{aligned}

which gives the equation $n^2 = A(n) + B_1(n) + 4C_1(n)$

The latest gives us $C_1(n) = (n^2 - A(n) - B_1(n))/4$. To solve for $C_0$, one can either replace the value of $C_1$ in the equation for $h(n) = n$ above, or, equivalently, add twice that equation to the one for $h(n) = n^2$, which eliminates $C_1(n)$:

\begin{aligned} 2n + n^2 & = 3A(n) + 3B_1(n) -4C_0(n)\\ C_0(n) & = \frac{3A(n) + 3B_1(n) - n^2 - 2n}{4}\\ \end{aligned}

### Good and Bad Persons in Josephus Problem

It took me a while, as I was trying to find a recurrence equation of some sort which would help me with this problem and the bonus one (where Josephus’ position is fixed but he can pick $m$). Eventually I found one, which did not help me with the bonus problem, but led me to a solution for this problem.

Obviously, if we have $k$ persons and want to remove the last one in the first round, we can choose $m=k$ and that will work. Actually, any multiple $m=ak$ works as well.

This shows that at each round, if we have $k$ persons left, and we start counting on the first one, when $m=ak$ we will remove the $k^{th}$ person then start counting from the first one again.

Back to the original problem: there are $2n$ persons, and we want to get rid of the $n+1, \cdots, 2n$ first. If we take $m=lcm(n+1,\cdots, 2n)$, then for the first $n$ rounds the last (bad) person will be remove, leaving only the good ones at the end.

When first solving the problem, I picked $m=\prod_{i=1}^n (n+i)$, which has the same property as the least common multiple, but is larger. Perhaps a smaller number is better for the nerves of the participants.

### Bonus Problems

I tried to solve the bonus questions, but after repeatedly failing, I had a glimpse at the solutions: they obviously require either knowledge of later chapters, or other concepts I know nothing about, so I will get back to these bonus problems after I finish the book.

I am now working through Chapter 2. It is a much larger chapter than the first, so it will take me some time.