# Monthly Archives: October 2011

## Cut the rope!

This is a real problem that was sent around my former maths department. The inquirer had a boat, and he wanted ropes of various lengths to knot together to moor his canal barge (I think each rope may have somehow used an eye splice for knotting). This is more or less the e-mail as I received it:

“I have 5 pieces of rope of length:

• 1 x 10 metres
• 2 x 12 metres
• 2 x 40 metres

I want to be able to cut the ropes into pieces of different lengths and to be able to tie combinations of these together to make longer lengths.

Is there a formula to obtain the optimum number and lengths of pieces ropes (i.e. the minimum number of pieces of ropes to give the most possible combinations of lengths of rope!). The minimum length of rope I need is 6m.”

Being more or less a combinatorial problem, I doubted whether any nice formula existed (or that it would be any more useful than a specific solution!). So, wanting to help, I cheated slightly, and chatted to the guy to get a bit more info. After our discussion, he decided that he wanted ropes with at most two knots (ie. three pieces), and to be able to make lengths at intervals of one or two metres. He also was quite keen on having a 20m rope.

I’ve given some additional assumptions and my solution below. But please have a go first: you may come up with a better way of doing it!

Filed under Accessible, Applications

## Tie Fighters

Noughts and crosses (tic-tac-toe) is quite a boring game. The two main reasons for this are:

1. Most games end in a draw or tie.
2. The optimal strategy is too obvious (the first player wants to start in the middle).
With an ulterior mathematical motive in mind, I’d like to introduce you to two games that avoid the first complaint. Whether they address the second is up to you: I’d argue that only Game B, called Sim, does.
Game A
Let’s call the game triangle-tac-toe. Draw a grid in the shape of a right-angled isosceles triangle whose sides are five squares across (giving 15 squares in total). A board-drawing hint: draw five rectangles.
Starting grid
Take turns to add noughts or crosses as in traditional tic-tac-toe. A player wins when three of their marks form the corners of a right-angled isosceles triangle of any size in the same orientation as the board (the corners may be touching or spread out).
Some of the possible ways for Crosses to win.
Game B (Sim)
Draw six points in a hexagonal arrangement. Optionally, lightly join each possible pair of points (a total of 15 lines) with dotted lines—don’t worry whether three cross lines in the middle or not, but double check that each point has five lines coming out of it.
Sim starting layout
Players pick their favourite colours (we’ll use the traditional Red and Blue), and take turns drawing a straight line in their chosen hue between two points (that haven’t already been connected). A player loses if they form a triangle of their colour between any three of the hexagon’s vertices.
End of a sample game—Red loses due to the highlighted triangle
Animation of the sample game above (click if not playing)

Filed under Accessible

## Percentages for sceptics

Let’s suppose you see a newspaper article or an advert on television that claims “73% of women reported healthier looking hair” or “88% of cats prefer Meonards to other leading feline food brands”, but they don’t give the number of respondents to their survey (perhaps you also have to suppose it happened in the past—the Advertising Standards Agency in Britain seems to have clamped down on this behaviour). What is the minimum number of people they could have asked?

Clearly, we can get any integer percentage by asking 100 people (or pets with strong brand preferences), and to get, say, 75% we only need to have asked four (with three responding positively). But if we assume that they rounded to zero decimal places in the standard manner, the number of respondents may be far fewer than expected (or at least fewer than I expected).  Continue reading

Filed under Accessible, Maths in Life

## Cars overtaking

While cycling during one of the unseasonably sunny October days, I passed lots of cars on the opposite side of the road, but none seemed to be overtaking me. I wondered: how many more I should expect on one side compared to the other?

Making some unrealistic assumptions: I ride my bicycle at a constant speed of $b$ mph in traffic that is constant, with evenly spaced cars (a poor assumption) travelling at a constant speed $c$ mph (not slowing when they overtake—a realistic assumption, in my experience), the answer is easily calculated.

The unslowing cars approach from behind at a merciless relative speed of $c-b$ mph, and those safely across the central double line zoom past at $c+b$ mph. So, for every car I pass on the opposite side of the road, I would expect to be overtaken $\frac{c-b}{c+b}$ times.  Continue reading

Filed under Accessible, Maths in Life

## An inequality for the Consumer and Retail Price Indices

Since 1996, Britain has had two major ways of measuring inflation: but, when explaining the difference between the Retail Price Index (RPI) and Consumer Price Index (CPI), British newspapers typically mention that CPI is (generally) lower because it excludes housing costs, whereas RPI includes them. However, in 2013, the CPI is due to be updated, and may then also take these housing costs into account. This would cause CPI to rise closer to the level of RPI, but you would still expect inflation rates as given by RPI to be higher—why would this remain the case?

Let’s begin with some background. Both indices try to measure the rise in cost of an ‘average’ basket of goods bought by households or consumers across a year, and neither is an attempt to measure the cost of maintaining a minimum standard of living, which depends on how those minimum standards are set. Other methods exist: The Economist uses its partially tongue-in-cheek Big Mac index to double-check consumer inflation measures around the world—here the Big Mac burger is the physical basket of goods.

The Retail Price Index has the longer history—its predecessor is associated with price increases suffered by workers in World War 1—RPI officially began in 1956 (though an interim version started in 1947, after WW2). RPI tries to reflect the spending of the ‘average’ private household: it excludes the top 4% of households by income, and pensioners whose state pensions and benefits make up more than 3/4 of their incomes. It also excludes spending by overseas visitors (for instance university tuition fees paid by foreigners) and those living in institutions such as university accommodation or nursing homes. It also excludes, for instance, stockbroker fees.

The Consumer Price Index, on the other hand, was introduced in 1996 to harmonise inflationary measures across the European Union. For now, it excludes many housing costs such as mortgages, estate agent fees, council tax, as well as costs such as TV licences and trade union subscriptions. It also differs from its sister index in how it deals with car costs: whereas RPI imputes new car prices from those of second hand cars, CPI is obliged to use real data.

However, the most significant difference between the two, known as the formula effect, arises during the early stages of the calculation. The formula effect has contributed at least 0.4 percentage points difference each year since its inception (measured by recalculating each index with the other’s variables). In 2010, it contributed to a difference of 0.8 percentage points, compared with the 0.6 percentage point difference associated with housing costs.

The inequality

Essentially, this difference arises because the CPI uses a geometric mean, while RPI uses the more well-known average, the arithmetic mean. A famous inequality that links these two means, the AMGM inequality, tells us that the arithmetic mean is always greater than the geometric mean (but they will be equal if, and only if, all the numbers averaged are the same).

For a collection of non-negative numbers $x_1, x_2, \ldots, x_n$ we have:

$\frac{1}{n}(x_1 + x_2 + \cdots + x_n) \geq \sqrt[n]{x_1 \cdot x_2 \cdot \cdots \cdot x_n}$ ,

or in more succinct notation:

$\frac{1}{n} \sum_{i=1}^n x_i \geq (\prod_{i=1}^n x_i)^{1/n}$.