22nd August 2011 - How Infinity Messes Up Stuff

I just love this video. It's caused me much amusement for me in tormenting my friends so I thought maybe I should write about it.

On Minute Physics, Henry Reich 'proves' that

 1+ 2 + 4 + 8 + 16+... = -1.

You can see how in the video so I'll just let you watch it before proceeding.

Now, according to the mathematics you're taught at school, every step is correct yet something seems terribly wrong here. How can you be adding positive numbers and end up with a negative number?

Let's look at what he does. 

1. He multiplies by 1 which doesn't change anything.
2. He rewrites 1 as (2 - 1) which is also fine since 2 - 1 = 1.
3. He applies the distributive law. The distributive law says that (a + b)c = ac + bc and a(b + c) = ab + bc). For example, (3+5)x2 = 3x2 + 5x2. If you check this, you will see that (3 + 5)2 = 8 x 2 =16 and also 3x2 + 5x2 = 6 + 10 = 16. To apply this to the example, he does 2(1+2+4+8+...) - 1(1+2+4+8+...) and then applies it again to multiply each number inside the bracket by the number outside to get 2 + 4 + 8 + 16 + ... - 1 - 2 - 4 - 8 - 16 - ...
4. He cancels all the terms and is left with -1.

So where is the flaw? It's in step 4 because when we deal with infinity strange things happen, and the distributive law doesn't work. The method in itself is a useful mathematical tool, as we'll see below, but unfortunately it doesn't work when we are dealing with infinite series.

Let's look at how it should work. What we will do is use another mathematical trick, and by trick I don't mean something bad or deceitful, but something clever. We will look at a finite series which has N+1 terms, and then we will let N get as big as we want, infinite in fact, and see what happens. This will show us how infinity is confusing us in the video above. We're going to follow the Minute Physics method 

Now, as we're doubling each time, so we have

1st Term: 1 = 2⁰

2nd Term: 2 = 2¹

3rd Term: 4 = 2x2=2²

4th Term: 8 = 2x2x2=2³

5th Term: 16 =2x2x2x2=2⁴

Notice that the power of 2 for each term is one less than its position in the series. This means for the N th term, where N is some number (we don't care what number it is at the moment, just some large positive whole number.)

N th Term: 2x2x2x..x2=2^N-1.

Right, back to the method given in the video, but this time for a finite series.

1 + 2 + 4 + 8 + ...+2^N-1 = 1(1 + 2 + 4 + 8 + ...+2^N-1)

= (2-1)(1 + 2 + 4 + 8 + ...+2^N-1)

= 2(1 + 2 + 4 + 8 + ...+2^N-1)-1(1 + 2 + 4 + 8 + ...+2^N-1)

= 2 + 4 + 8 + ...+ 2^N-1+2^N - 1 - 2 - 4 - 8 - ...- 2^N-1

= 2^N - 1.

Well, that means that 1 + 2 + 4 + 8 + ...+2^N-1= 2^N - 1, which is quite a nice formula but how does it help us with the infinite series in the video?

What we'll do now we've got this nicely written down is make N get bigger and bigger so it tends to infinity. That means the left hand side becomes 1 + 2 + 4+ 8 + ... where the dots mean carry on forever. What happens to the right hand side? Since N tends to infinite, then 2^N tends to infinity even more quickly, so the right hand side tends to infinity. This means that the sum 1+2+ 4+ 8 + ... tends to infinity too, and most definitely not -1.

15 August 2011 - 1 Multiplying Numbers When the First Digits are the Same & the Last Sum to Ten

Continuing the theme of multiplying various numbers, we come to some which satisfy a particular condition as explained in the examples below.

Example 1

We'll multiply 26 by 24. The method we'll use requires the first digits to be the same, here 2, and the last digit from each number sums to ten, here 6 + 4 = 10.

1. Take the first digit	2
2. Add one	2 + 1 = 3
3. Multiply the numbers from step 1 and 2.	2 x 3 = 6
4. Multiply the last 2 digits of the original numbers	6 x 4 = 24
5. Write the answer to step 4 after step 3.	624

Hence 26 x 24 = 624. The method also works for larger numbers.

Example 2

We'll multiply 113 by 117. We'll check the numbers satisfy the conditions for the method. The first digits are the same, both 11, and the last two digits sum to ten: 3 + 7 = 10.

1. Take the first digit	11
2. Add one	11 + 1 = 12
3. Multiply the numbers from step 1 and 2.	11 x 12 = 132
4. Multiply the last 2 digits of the original numbers	3x7=21
5. Write the answer to step 4 after step 3.	13221

Hence 113 x 117 = 13221.

Proof

Again, let's check that the examples above weren't just flukes. Let's multiply two numbers a and b which both start with c which could be a single digit or several digits, and end in d and (10 - d) respectively. Hence

a = 10c + d and b = 10c + (10-d).

Multiplying a and b, we get

ab = (10c + d)(10c + (10-d)) 
= 100c²+ 100c -10cd + 10cd + 10d - d²

= 100c²+ 100c + 10d - d².

We'll now check to see if the method gives the same result.

1. Take the first digit	c
2. Add one	c + 1
3. Multiply the numbers from step 1 and 2.	c(c + 1) = c2 + c
4. Multiply the last 2 digits of the original numbers	d(d-10) = d2 - 10d
5. Write the answer to step 4 after step 3. This is the equivalent to multiplying the answer to step 3 by 10 and then adding the answer to step 4.	10(c2 + c)+d2 - 10d = 10c2 + 10c + d2 - 10d

Hence the method gives 10c² + 10c + d² - 10d which is ab so it works for numbers of the prescribed form.

Source: http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp

13th August 2011 -1 Squaring Two Digit Numbers Ending in 5

I saw this method in a GCSE foundation paper, as well as here. Squaring a number is simply multiplying it by itself, so 3 squared, written as 3² means 3 x 3 = 9. That is 3²=9.

Method

This method works for numbers ending in 5. As an example, we'll square 35.

Example	Method
3	1) Write down the tens figure.
3 + 1 = 4	2) Add 1 to it.
3 x 4 = 12	3) Multiply the numbers from steps 1 and 2.
1225	4) Write 25 after the number obtained in step 3. This is the original number squared.

Does this method really work?

Proof

Let the number to be squared be a where a is a number ending in 5. Let the tens digit be b. Then we can write

a = 10b + 5. 

Squaring a gives

a² = (10b + 5)²

= 100b² + 100b + 25. (1)

Now we'll apply the method to a and see if we get the same result. First we write down the tens figure b and then add one to it, giving b + 1. These are then multiplied to give

b(b + 1) = b² + b.

Since, we write 25 immediately after this number, we are actually multiplying by 100, which shifts the 12 in the example above to 1200,  and then adding 25.

100(b² + b) + 25 = 100b² + 100b + 25,

which is what we obtained when we multiplied a by itself in (1). Hence the method gives us a², that is, it works.

Note that this proof can be adapted for any number ending in 5 as we never use that b is a single digit. For instance 115² is found as follows:

11 x 12 = 132     (since 12 = 11+1)
so 115² = 13225 (putting 25 after the number above).

10th August 2011 - 1 Multiplying Numbers just over 100

Today, we're going to look at extending the methods discussed earlier this week to numbers just above 100. Again, we'll look first at the method, then its feasibility and finally a proof.

Method

We'll multiply 105 by 107. How does the method differ from earlier methods? Instead of subtracting the numbers from 100, we subtract 100 from then, and instead of subtracting diagonally, we add. (See table below.)

Numbers to Multiply	Left hand column minus 100
105	5
107	7
ADD the number diagonally opposite. It doesn't matter which pair we use as we get the same answer. The pairs are shown in red and blue.	Multiply the numbers in this column
105 + 7 or 107  + 5 = 112	5 x 7 = 35
Multiply this by 100
112 x 100 = 11200	35
Finally, we add the two numbers.
11200 + 35 = 11235

When is it useful?

If you can multiply the two digits after 100 by each other, then this method works.

Example 1
For instance, 102 x 145 since 2 x 45 = 90 is easy if you're comfortable with doubling.  Once you've multiplied those, all that is left to do is add 102 and 45 which is 147, put a couple of zeroes on the end and add the 90 you got earlier. That is,

147 x 100 = 14700,

14700 + 90 = 14790,

and so

102 x 145 = 14790.

Example 2
Similarly, 120 x 145 is as follows. Do 20 x 45, which is just 2 x 45 = 90 with a zero on the end, that is

20 x 45 = 900.

Then

120 + 45 = 165,

165 x 100 = 16500,

16500 + 900 = 17400,

and so

120 x 145 = 17400.

Proof

The proof is similar to those I've given for the last three posts on this topic.

 We will multiply a and b, which are numbers greater than 100. Next we subtract 100 from each of them to give a - 100 and b - 100, and then multiply them together.

 (a - 100)(b - 100) = ab - 100a - 100b + 10000. (1) 

 Moving on to the other part of the calculation, we add a - 100 to b to get a + b - 100. Next we multiply by 100 which gives

100(a + b - 100) = 100a + 100b - 10000. (2) 

 Adding (1) and (2), we obtain

ab - 100a - 100b + 10000 + 100a + 100b - 10000 = ab, 

again showing that our method does indeed multiply a and b together.

 Source: http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp

8th August 2011 - 1 Multiplying Numbers Just Below 100

The underlying method I talked about two days ago can be extended so you can multiply numbers just below 100. What do we need to do to adjust it? Wherever we used 10, we replace it by 100. Let's try 96 x 91.

Method

Numbers to Multiply	100 minus left hand column
96	4
91	9
Subtract the number diagonally opposite. It doesn't matter which pair we use as we get the same answer. The pairs are shown in red and blue.	Multiply the numbers in this column
96 - 9 or 91 - 4 = 87	4 x 9 = 36
Multiply this by 100
87 x 100 = 8700	36
Finally, we add the two numbers.
8700 + 36 = 8736

When is it Useful?

So let's consider when this method might be useful. Personally, I think the method for small numbers is interesting but too time consuming compared with the benefit of learning tables up to at least 10 x 10. When the maths gets more advanced, to have to resort to fingers to do 8 x 7 is slow compared with recalling from memory. Small numbers are multiplied fairly often so the investment of time is worth it, even if you struggle to learn things by rote, like I do.

 If you know your tables up to 10 x 10, then any two numbers in the 90s will be easy to multiply together because the differences will be at most 10.

Multiplying 98 by any number is going to be fairly straightforwards if doubling is something you can do easily. For example 98 x 46.

Numbers to Multiply	100 minus left hand column
98	2
46	54
Subtract the number diagonally opposite. It doesn't matter which pair we use as we get the same answer. The pairs are shown in red and blue.	Multiply the numbers in this column
98-54  or I prefer 46 - 2 = 44	2 x 54 = 108
Multiply this by 100
44 x 100 = 4400	108
Finally, we add the two numbers.
4400 + 108 = 4508

If you're happy to double and double again, you'll find you can multiply by 96 as that is 4 less than 100.  Similarly, 80 and 60 aren't too bad since then you're just multiplying by 20 (double and put a 0 on the end) or 40 (double, double again and put a 0 on the end).

And just to make sure, let's check the proof still works out ok.

Proof

Let's apply this method to a and b which are two whole numbers less than 100.

We subtract them from 100 to get 100-a and 100-b. We multiply these (right hand column in the table) to get

 (100 - a)(100 - b) = 10000 - 100a - 100b + ab. (1)

Next we subtract to get a - (100 - b) = a + b - 100 and then multiply  by 100,

100(a + b - 100) = 100a + 100b -10000. (2)

Finally we add (1) and (2) to get


10000 - 100a - 100b + ab+100a + 100b -10000 = ab.

This means that applying the method does indeed multiply a and b as claimed.

Source: http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp

Edit: Corrected two typos where I put 10 instead of 100.

7th August 2011 - 1 Multiplying Single Digits Part 2 - Using Fingers

The Method

Using the principle covered in yesterday's post, here are two videos showing how to use your fingers to apply it. The first touches fingers together

and the second folds some down.

I think the videos explains the methods clearly, but I do have one minor issue with the second and that is the comment "To show it's no fluke, let's try another multiplication." Unfortunately, two examples do not show it is not a fluke. Checking all possibilities would, or a proof such as the one I gave yesterday. The proof needs a slight change from yesterday's to apply to this method.

The Proof

Let's look at what is happening with the numbers first, and then generalise to show how the method works for any two numbers a and b, between 5 and 10 inclusive.

The circled fingers in the first picture correspond to the fingers in both circles in the second picture.

Circled Fingers

On my left hand I have 3 fingers folded down because 8 - 5 = 3. On the right hand it's 7 - 5 = 2. These are added, 3 + 2 = 5. This corresponds to the number of fingers circled in the first picture. This number is multiplied by 10 to give 50. In general, on the left hand there are a - 5 fingers circled, and on the right b - 5. Summing these gives

(a - 5) + (b - 5) = a - 5 + b - 5 = a + b - 10.

This is multiplied by 10 to give

10(a + b - 10) = 10a + 10b - 100,

which is the same as we had yesterday. We need to work out the value obtained from the non-circled fingers and then add that to this value.

Non-circled Fingers

The number of non-circled fingers is 10 - 8 = 2 on the left hand and 10 - 7 = 3 on the right hand. We multiply these two numbers together to give 6 and add it to the answer from the circled fingers, calculated above.

On the left hand we have 10 - a non-circled fingers, and on the right 10 - b. Multiplying these together gives

(10 - a)(10 - b) = 100 -10a - 10b + ab.

Finally, adding this to the circled fingers answer, and simplifying, gives

10a + 10b - 100 + 100 -10a - 10b + ab = ab,

showing that both methods do correctly multiply a and b together.

6th August 2011 - 1 Multiplying Single Digits Part 1

There are many methods for multiplying numbers together.

The first method I'm going to look at is for single digits which are both 5 or above. If one of the digits is below 5, you gain nothing over just multiplying them together, or at best, very little. If both digits are 5 or above, all you need to know is how to subtract and times tables up to 5 x 5.

The Method

As an example, we'll multiply 8 and 6.

First we subtract both numbers from 10 to get 2 and 4, and write the answer in the corresponding right hand column below.

Numbers to Multiply	10 minus left hand column
8	2
6	4
Subtract the number diagonally opposite. It doesn't matter which pair we use as we get the same answer. The pairs are shown in red and blue.	Multiply the numbers in this column
8-4 or 6-2 = 4	2 x 4 = 8
Multiply this by 10
4 x 10 = 40	8
Finally, we add the two numbers.
40 + 8 = 48

So when we multiply 8 and 6 we get 48.

Sometimes when you multiply the two new numbers together, we get a number which is two digits, as shown in the example below where we multiply 6 by 6. This doesn't make any difference to the method.

Numbers to Multiply	10 minus left hand column
6	4
6	4
6 - 4 or 6 - 4 = 2	4 x 4 = 16
2 x 10 = 20	16
20 + 16 = 36

So 6 x 6 = 36.

How does it work?

With some algebra, we can see that the method above really does give the right answer.

Let the two numbers to be multiplied be a and b. 

Numbers to Multiply	10 minus left hand column
a	10-a
b	10-b
a - (10 - b) = a -10 + b = a + b -10	(10 - a)(10 - b)
Multiply by 10.	Multiply out the brackets.
10(a + b -10)=10a + 10b -100	(10 - a)(10 - b) = 100 - 10a - 10b + ab
We now add the two expression above.
10a + 10b -100 + 100 - 10a - 10b + ab =  (-100 + 100) + (10a - 10a) + (10b - 10b) +ab = ab
As we get ab, this means the method does indeed give us the correct answer when we multiply a and b.

Source: http://www.vedicmaths.org/introduction/tutorial/tutorial.asp#tutorial2

5th August 2011 - 3 Correlation but is it causation?

Ben Goldacre posted an interesting illustration of how data can mislead and needs to be analysed before posting. His example shows a correlation between lung cancer and drinking, that is, if you drink you are more likely to get lung cancer than if you don't drink. Does that really mean that drinking causes lung cancer or is there another factor in play?

When the figures are split into smokers and non-smokers, the smoker who drink and the smokers who don't drink have the same occurrence of lung cancer. Similarly, for non-smokers, drinker and non-drinkers have the same occurrence of lung cancer. This shows that the occurrence of lung cancer is explained by smoking versus not smoking rather than drinking versus not drinking.

5th August 2011 - 2 Bad Maths

I noticed as my twitter feed flew past today the following link which amused me in a strange sort of way. It's an image of a poster claiming to solve the world financial crisis along with some conspiracy claims. In the middle of the poster they make the following claim.

*An estimated 51% of retail prices are a result of interest on the National debt. This means that as of right now, everything in stores is 51% more than it would be if we used a national currency like the United States Note.

What amused me was that someone claiming to solve the monetary crisis doesn't understand basic percentages.

Suppose an item costs $10. From the first sentence, we have that 51% of this price is interest on the National Debt. That would be $10 x 0.51 = $5.10, so without this interest on the item, it would cost $10 - $5.10 = $4.90.

The claim in the second sentence claims that everything in stores is 51% than it would be without the National Debt, as the poster claims that the United States Note will solve this. This 51% is now referring to the 51% of $4.90 (the price without debt) which is $4.90 * 0.51 which is about $2.50. If the second sentence were true, the item would cost $4.90+2.50=$7.40 which it doesn't. So how much more is being paid as the 51% is clearly wrong and not enough.

As we want the percentage to be compared to the price without the debt, we write the extra paid, $5.10, as a fraction of the price without debt and multiply by 100 to turn it into a percentage.

5.10/4.90 * 100 = 104% (to nearest percent)

What this tells us is that instead of paying about half extra as the poster claims, people are paying more than double what the price would be without the debt payments!

This illustrates how important it is to know what any percentage is of so that mistakes like this do not occur.

4th August 2011 - 3 The Horizon

Assuming the Earth is a sphere leads to simple calculation, based on Pythagoras' Theorem, of the distance to the horizon given the height, h,  of the viewer's eye. The details are nicely explained by Phil Plait in his blog Bad Astronomy.

The distance, d (in km),  to the horizon is

d = square root of  h(h+2R)

where h is the height of the eye (in km) and R = 6365 km, an approximation of the radius of the Earth.

For instance, if you eye is at ground level, the horizon is 0 metres away, but if your eye is at 1.65 m (about 5'5"),  then the horizon is at 4.6 km (2.9 miles). At the top of Mount Everest, which is 8,848 m high, it would be 335.75 km (208 miles).

26th April 2011 Mersienne Primes

For which values of a and n is N=aⁿ-1 a prime?

Jason Rosenhouse at Scienceblogs gives a nice explanation of the possibility of finding primes for different values. He shows why the only possible value for a is 2, and that n must be a prime. If these conditions are not met, then there N is not a prime, but if they are N may be a prime.

Sources

Evolutionblog: Monday Math MersennePrimes

24th April 2011 Fractals

This video is a short introduction to fractals. It explains their repetitive nature, called self symmetry, and briefly introduces fractional dimension, that is that some things have dimensions other than 1,2 or 3.

Koch Snowflake

This fractal is called the Koch Snowflake. Starting with an equilateral triangle, the snowflake is formed by splitting each edge into thirds and then, for each edge drawing another triangle with the centre third of the line as its base. This is repeated infinitely.

By António Miguel de Campos via Wikipedia

At any step, the length of the line is 4/3 times longer than on the previous iteration, since three sections are replaced by four. This means that the line formed enclosing the snowflake has infinite length. This infinite length surrounds a finite area.

Calculation of Area of Koch Snowflake

At iteration 1, there are 3 edges of length s.
At iteration 2, there are 4 * 3 edges since each edge is replaced by 4 edges of length 1/3 s.
At iteration 3, there are 4*(4*3) = 4² * 3 edges since each of the 4 * 3 edges is replaces by 4 edges of length 1/3 * (1/3 s) = 1/3²  s.
...
At iteration n, there are 4^n-1 * 3 edges of length 1/3^n-1 s.

If the area of the original triangle is A, then the area of one of the additional triangles is (1/3)² times the area of the original triangle because the length of its base is a third of the size of original.  There are 3 such triangles so the total area after one iteration is

A +3(1/3)²A = A(1 + 1/3) = 4/3 A.

At iteration n, there will be one new triangle for each edge in the previous iteration, iteration n-1 giving 4^n-2*3 triangles. The area of these triangles will each be (1/3^n-1)²A = (1/3²)^n-1A  = (1/9)^n-1A = 1/9^n-1A where A is the area of the original triangle. This gives an increase in area from iteration n-1 to iteration n of

4^n-2*3 * 1/9^n-1A = 3 * 4^n-2/9^n-1A

So, after n iterations, the total area is

4/3 A + 3*4/9²A + 3*4²/9³ +...+ 3*4^n-2/9^n-1.

Apart from the first term, 4/3 A, this is a geometric progression, with initial term   a=3*4/9²A and common ratio r=4/9. As n tends to infinite, we have the sum of an infinite geometric progression with 0<r<1. The sum of this is

a/(1-r) =   3*4/9²A / (1 - 4/9) = 3*4/9²A*9/5 =  4/15 A,

and hence the total area of the Koch snowflake is

   4/3 A + 4/15 A = 24/15 A = 8/5 A.

Hence, the area is 1.4 times the area of the original triangle.

Fractal Dimension

The fractal dimension of the Koch Snowflake can be thought of as too big to be 1 dimensional because every piece of it is made up of lots of smaller pieces joined at angles which means it is not like a line. It's also too small to be two-dimensional as it doesn't cover a plane.

To convert an increase in the length by a fixed factor of the sides of an object in Euclean space (everyday space) to the corresponding increase in area, we have to square the factor. For instance, if we triple the length of the sides of a rectangle we have to multiply the area of the rectangle by 3². If we increase the length of the sides of a three dimensional object, he have to cube the factor we increased the length by. In other words, we can consider dimension as the power we have to use to convert.

Another way of looking at it is to look at a decrease in length.  Suppose we decrease a length by a factor of  1/f. How many copies of the new object will it take to cover the new object?  For example, if we cut the sides of a cube in half, then we can fit 2³=8 smaller cubes inside the original cube.

Suppose we decrease the length of the side by a factor 1/f.  How many copies F of the new object will we need to fill the old object?

in 2-dimensions is F=f²,
in 3-dimensions is F=f³, and
in D-dimensions is F=f^D.

To extract D from this, we use logs.

log F= log (f^D)
log F = D log f
D = log F/log f

For the Koch Snowflake, at each iteration, 1 line is replaced by 4 lines, each of length 1/3 the original. Hence F=4 and f=3, and so its fractal dimension is log 4/log 3  which is about 1.26.

Sources

Wikipedia: Koch Snowflake
Wikipedia: Fractal dimension

22nd April 2011 Password Security

I really really really hope this is a spoof but is it? Can American Express really be saying that a short letter only password of at most length 8 is better than a longer one using special characters?

It reminds me of my university days, back before we all had home computers. We were advised to pick a password that meant something to us and was easy to remember. My friend cracked mine in a matter of minutes and then found it hilarious to wreak havoc with what I was doing. It seems that just allowing letters is really encouraging people to use easy passwords.

Just how silly is this?

Letters only, length 8

With just upper and lower case letters, there are

52 + 52² + 52³ + 52⁴ + 52⁵ + 52⁶+ 52⁷ + 52⁸
= 52(52^8-1)/(52-1)
= 54 507 958 502 660,

which is about 5.5 x 10¹³ passwords.

Letters, numbers, special characters, length 8

Using all upper and lower characters, numbers and special characters (96 in all), the number of 8 character passwords jumps to

96 + 962 + 963 + 964 + 965 + 966+ 967 + 968
= 96(96^8-1)/(96-1)
=7 289 831 534 994 528,

that is about 7.2 x 10¹⁵, which is 100 times more than for the letter only passwords.

Letters, numbers, special characters, length 12

If we allow passwords of up to length 12, the number jumps to

96(96^12-1)/(96-1) which is about 6.2 x 10²³ passwords.

Cracking Times

So in the worse case it is going to take a brute force attack 10 billion times longer, to crack a password of up to length 12 using all special characters, letters and numbers than it is to crack a letter only password of length 8.

Using a brute force attack (and that is the slowest method of attack),  at 1 billion checks per second, (supercomputer in 2009 [1]), in the worse case scenario,  this would take 15 hours,  84 days, and 20 million years respectively.

In November 2010, using brute force, 14 passwords of length at most 6 using all characters were cracked in 49 minutes using Amazon EC2.[2]  There are 7.9 billion such passwords, which is 100 times fewer than for  letter only length 8 passwords, so just for comparison, we could say it would take about 49/14*100 minutes, that is 6 hours, to crack one of these type of passwords.

So what can we say? Length really does matter!

Sources

1. http://www.lockdown.co.uk/?pg=combi#classF 
2. http://www.esecurityplanet.com/headlines/article.php/3920306/Cracking-Wi-Fi-Password-Protection-with-Amazon-EC2.htm
   

14th April 2011 Password Security

A new method of password encryption uses an image to encrypt a longer piece of data. Using a short key, the process of generating the image from the original image is reversed. The longer piece of data can then be read off from the image. Since computers have a harder time in telling they have found something that makes sense, this is more secure than just using a text. The reversed image may not be identical to the original image because the system is chaotic.  A small error in an image can lead to completely missing the correct image.

http://plus.maths.org/content/captcha-chaos
Paper (unpublished in peer reviewed journal)

14th April 2011 Ants go Home

Ants use a combination of vector addition and root mean square to find their way home.

When they move they estimate the direction (from the sun) and their distance travelled (how many steps). They add these vectors to always have a "home that way" vector.  Unfortunately small errors in either the distance or direction could lead to the ant not being home when she thinks she is.  What happens then?

An ant learns the appearance of the environment around the home.  Her eyesight is low resolution so she can not easily recognise home in the same way as we can by high resolution appearance, plus her environment is mostly trees and more trees.What she does is processes each new image in comparison to these images she has of home.  She does the equivalent of a root mean square of the difference between the pixels in the new image compared with one of her stock images. She moves, and recalculates. If the value is tending towards zero, she continues in that direction. If not, she tries a different route. When the value is zero, she is home.

http://plus.maths.org/content/finding-way-home

5th April 2011 Finite Simple Groups

Finite Simple Groups can be completely classified into 18 families (abstract analogues of the rotation groups of our regular polyhedra.) and  26 individual groups, the so-called sporadic groups.

Enormous Theorem Classification Finite Simple Groups