The Maths Geek: Maths

Showing posts with label Maths. Show all posts

Wednesday, 1 June 2011

Two Ladders Puzzle: Answer

Previously I posted the Two Ladders Puzzle in which two ladders lean against opposite walls. The length of the ladders is given and so is the height of the point where they intersect. The task was to find the distance between the walls. The problem can be solved using Pythagoras and the law of similar triangles, so really it can classify as an easy problem. Nonetheless, the answer requires the solution of a quartic equation. We begin by naming the lengths of the ladders $a$ and $b$ and the, as yet unknown heights at which the ladders lean against the walls by $x$ and $y$ . Also let's divide the distance between the two walls into the parts $m$ and $n$ . We now have 5 unknowns, so we need 5 equations. The easy one is

$m + n = d$

Then we have Pythagoras for the two ladders

$a 2 = d 2 + x 2$

$b 2 = d 2 + y 2$

Finally we have the similar triangles

$\frac{d}{x} = \frac{n}{h}$

$\frac{d}{y} = \frac{m}{h}$

If we add these two equations, we get the sum $(m + n) / h$ on the right hand side. But $m + n = d$ and thus we can eliminate d to obtain the nice intermediate result

$\frac{1}{h} = \frac{1}{x} + \frac{1}{y}$

This means that $h$ is the harmonic mean of $x$ and $y$ , independent of the values of $a$ , $b$ or $d$ .

The rest of the solution will become a tiny bit messy. Instead of looking for a solution of $d$ , we can look first for the values of $x$ or $y$ . Then we can use Pythagoras to find $d$ . To do this we subtract the two Pythagorean equations to eliminate $d$

$x 2 - y 2 = a 2 - b 2$

Solving the harmonic equation for $y$ we get

$y = \frac{hx}{x-h}$

With this we can eliminate $y$ from the above difference of squares

$x^2-\frac{h^2 x^2}{(x-h)^2} = a^2-b^2$

Obviously this is a fourth order equation for $x$ . I know this is not really a nice result so we will have to get some help for solving it. So let's insert the values and stick it into WolframAlpha. Of course there are four possible solutions. We get one positive, one negative and two complex solutions. Only the positive solution of $x \approx 3.03692$ is relevant. Now we can use the first Pythagoras to find the value for $d$

$d \approx 2.60329$

Wednesday, 25 May 2011

Two Ladders Puzzle

This puzzle is an old favourite of mine. It looks quite straightforward but it's not as easy as one might expect.

Two ladders are leaning against opposite walls in an alleyway. The bottom of each ladder is placed on the ground in the opposite corner. One ladder is 4m long, the other is 3m long. The height at which the two ladders meet is 1m.

Question: How far are the walls apart?

Monday, 16 May 2011

Collatz: Bringing some order into the chaos

I recently wrote that I was rekindling my interest in the Collatz problem. I still haven't had the time to read any literature but I came up with an interesting transformation to the problem that seems to unearth some structure which is maybe not directly obvious. For regular readers of this blog, you might see the similarities with this transformation and the formula I posted earlier on calculating the points for interval halving using a closed formula. It really was the Collatz problem that inspired that formula. To the uninitiated, the Collatz problem is based on the sequence
$x_{n+1} = \begin{cases} x_n/2 & x_n \text{ even}\\ 3x_n + 1 & x_n \text{ odd} \end{cases}$

The question is whether, for any positive integer starting value, the sequence will always reach the value 1.

One can reverse the problem in the following way. Define the set C in the following way

$\begin{alignat}{2} &1 \in C\\ &\text{if } x \in C \quad\text{ then }\quad2x \in C\\ &\text{if } x \in C \quad\text{ and }\quad x \text{ mod } 3 = 1 \quad\text{ then }\quad\frac{x-1}{3} \in C \end{alignat}$

The problem then transforms to the question, whether $C$ contains all positive integers, in other words: is $C=\mathbb{N}$ ? In this form the problem can be represented as a tree with 1 at its root, every element $x$ has at most two children. One child is always $2 x$ , the other child is $(x - 1) / 3$ if that number is an integer.

Every odd number $x$ trivially has an infinite number of descendants of the form $2 k x$ . I will call the set of these numbers the tower of $x$ .

Definition: The tower $T x$ of an odd number $x$ is the set $T_x=\{2^k x | k=0,1,2,\ldots\}$ . $x$ is called the base of the tower.

We also define the children of $T$ in the natural way, as a set of towers which are based on those children of the elements of $T$ which are not elements of $T$ themselves.

Definition: The children $C T$ of a tower $T$ is the set

$C_T = \{T_y | y = (x-1)/3 \quad\text{ where }\quad x \in T\quad\text{ and }\quad x \text{ mod } 3 = 1 \}$

It's easy to show that every tower either has no children, if the base is a multiple of 3, or an infinite number of children otherwise.

I would like to find a way to map each tower onto a single number, i.e. every element $x$ of $T x$ should be mapped onto the same value. To do this I take the largest power $k x$ of 2 that is smaller or equal to $x$ : $2^{k_x} \le x < 2^{k_x+1}$ . We now define

$t_x := 2^{-k_x} x - 1$

The values of $t x$ lie between 0 and 1 and are rational numbers of the form

$t_x = \frac{p}{2^k}$ .

It is clear that $k 2 x = k x + 1$ and therefore $t 2 x = t x$ . This means, all the elements of a tower are uniquely mapped onto a single value. We can, therefore, identify these values with the towers they originate from and we will refer to the towers $T$ by their $t_T := t_x, x \in T$ .

The next question that arises is, given an tower $t T$ what do the children of a tower look like. As an example, the children of $t 7 = 3 / 4$ are plotted in the figure on the right. The values of the first few of these children are given in the following table.

Children of t = 3/4
Fraction	Decimal
1 / 8	0.125
5 / 32	0.15625
21 / 128	0.16406
85 / 512	0.16602
341 / 2048	0.16650

It can be seen, and easily proved, that if $p / 2 k$ is a child of any $t$ , then so is $(4 p + 1) / 2 k + 2$ . This results in an geometric series that converges to

$(3p+1)/(3\times 2^k)$

The figure above plots the children of the first 1000 towers against the values of the towers themselves. Interestingly a clear structure emerges. The sequences of the children all seem to converge against two straight line segments. These line segments can be represented by the function

$y = \begin{cases} \frac{1}{3}( 4x + 1) & \quad \text{for } x< \frac{1}{2}\\ \frac{1}{3}( 2x - 1) & \quad \text{for } x> \frac{1}{2} \end{cases}$

When I first plotted this graph and discovered this very simple structure hidden in the Collatz problem, I was utterly amazed. Of course I don't know enough to decide if this structure could hold the key to unravelling the problem but it certainly gives me hope.

Finally, an apology to all those who know more about the problem. These things probably have been discovered already. I have not supplied any references and most likely got the terminology wrong. If so, I will be happy to be corrected.

Thursday, 28 April 2011

An Old Problem in Disguise

Here is a well known mathematical problem stated in a slightly different way than usual. See if you can find out which problem I am talking about.

Start with any binary number ending in a 1, e.g.

100101101

Now shift this number one place to the left, filling the rightmost digit with a 1

1001011011

then add this number to the original number

Finally eliminate all trailing zeros

Now repeat the whole process until you reach the number 1. The question is, will you always end up at the number 1, irrespective of the number you started with?

You probably guessed it by now that this is the well known Collatz problem. In its standard form the Collatz conjecture reads

Given a starting number x₀ and the recursion

$x_{n+1} = \begin{cases} x_n/2 & x_n \text{ even}\\ 3x_n + w & x_n \text{ odd} \end{cases}$

the sequence x_n will always reach the number 1.

I have always been fascinated by this problem, but never so seriously that I would have started reading up on it and following the newest results. Only recently I started pondering over it and came up with the representation in the beginning. Naive that I was, I thought I could see some patterns in the binary representation of the problem. Of course I was in no way the first person to think of this an some further reading on Wikipedia revealed that this is one of the standard alternative representations of the problem. Some more extended Google search revealed this online script for calculating the Collatz number sequence in binary form.

I like the binary representation because it somehow reminds me of cellular automata. In a cellular automaton the state of a cell in the next generation is determined by the states of the cells in a given neighbourhood in the previous generation. Although it looks similar, the binary Collatz problem doesn't map in a straightforward way onto a cellular automaton. This is because the addition of the two binary numbers can cause non-local changes due to the carry over. It turns out, however, that there is a way of translating the Collatz problem into a cellular automaton.

There are a few patterns that one can identify in the binary representation. If a number x ends on k ones then there will be k successive steps where the 3n+1 operation just adds one zero at the end. This means that the number will grow from its original value x to
$\left(\frac{3n+1}{2}\right)^k x$

On the other hand, if x ends in k alternating ones and zeros, then the next 3n+1 operation will turn this sequence into zeros and the result will be
$\frac{3n+1}{2^k}$

These quick investigations have certainly sparked my interest in the Collatz conjecture again, and I am now looking forward to reading a bit more in depth on the problem.