the power of symmetry

The power of symmetry

People love symmetry

They find symmetrical faces more attractive

Mathematicians are just like people :)

They too love symmetry

This is Johann Carl Friedrich Gauss (1777-1855)

He used symmetry to sum the numbers from 1 to 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Suppose you have to sum these guys

How would you do that?

You could do this…

Is this a smart solution?

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100

Well… it are 99 plusses, so 99 points of effort (and possibly mistake)

Could we do this with less operations(plusses)?

Remember Gauss?

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100

Remember Gauss?

That’s me!

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100

Remember Gauss?

He could do it with only 2 operations, using symmetry

That’s me!

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100

1+2+3+4+5+ … +95+96+97+98+99+100

Let’s position the numbers a bit differently

Now we can ‘see’ the symmetry

1+2+3+4+5+ … +95+96+97+98+99+100

It’s like there’s a mirror in the middle!

This way we can make 50 groups op two

1+100 = 1012+99 = 1013+98 = 101

1+2+3+4+5+ … +95+96+97+98+99+100

50 groups of 101 -> 50 * (100 + 1) = 5050

With only 2 operations!

1+100 = 1012+99 = 1013+98 = 101

Start loving symmetry already?

Start loving symmetry already?

I do!

Let’s have a look at the next problem

Concerning a shepherd, a sheep, a wolf and a cabbage

And a river, which they have to cross

And a boat. But, the boat can only carry the shepherd plus one item

So he takes either the sheep with him, or the wolf or the cabbage

Ow, and one more thing

When left alone (without the shepherd present)

The wolf will eat the sheep, and the sheep will eat the cabbage

Can we get the entire party safely to the other side?

Lets make a diagram representing the shepherds choices

Initially, all are at the right side of the river

|HSCW

So let’s place the whole party right of the |

The shepherd (his name is Hank), the Sheep, the Wolf and the Cabbage

|HSCW

Now what can Hank do?

He can take either one of the three items with him

|HSCW

HC|SW HS | CW HW | SC

However, remember

Wolf eats sheep, sheep eats cabbage

|HSCW


So these two options are no good!

|HSCW


Lets go back to the original problem:

Wolf eats sheep, sheep eats cabbage

If we were to model this problem

We might come up with something this

But is this necessary?

All that matters is sheep can’t be with wolfs and cabbages

Who eats who…

Is irrelevant!

So let’s try modeling the problem again!

We have a shepherd, a sheep,

Plus two things that we shouldn’t combine with sheep

Lets call them anti-sheep (A)

It we construct the diagram now, it is way more simple

So next time when you are modeling something, be aware of antisheep

Don’t overmodel

The symmetric solution was invented by this guy

Edsger W. Dijkstra, a Dutch computer scientist

Edsger W. Dijkstra – Pruning the search tree http://www.cs.utexas.edu/users/EWD/transcriptions/EWD12xx/EWD1255.html

http://www.cs.utexas.edu/users/EWD/transcriptions/EWD12xx/EWD1255.html

http://www.cs.utexas.edu/users/EWD/transcriptions/EWD12xx/EWD1255.html

So far we’ve see how symmetry can help in solving problems

Of course seeing this symmetry is not always easy

But if you practice a little, you start seeing it everywhere

Let’s try another problem!

Finding the greatest common divisor (gcd) of two numbers

Given two numbers A en B

Determine the maximum number x, that divides both A en B

612

For instance 6 and 12

2, 3 and 6 all divide both numbers, but 6 is the largest

612

But how do we calculate the gcd of larger numbers?

We could try all numbers up to 105

105252

Can we do better?

Using symmetry?

105252

So let’s search for a problem that is smaller

But can be solved in the same way

105252

Rephrasing our problem definition

We search for an x that ‘fits’ into both A and B

105 = x * z 252 = x * y

But we don’t really care about how often x fit in A and B

Remember: be aware of antisheep!

105 = x * z 252 = x * y

So how can we reduce this problem?

While keeping it similar?

105 = x * z 252 = x * y

Well, we could subtract

Now we again have two numbers in which x fits nicely, but smaller!

105 = x * z - 252 = x * y

147 = x * (y-z)

Now we can continue with 105 and 147

And again have two numbers in which x fits nicely, but smaller!

105 = x * z - 147 = x * y

42 = x * (y-z)

You can check yourself that within 5 steps we reach the answer

That’s way better that 105 steps. Thanks to…

105 = 5 * 21

252 = 21 * 21

Symmetry!

Wanna try it with a new problem?

Suppose you’re going on a backpacking trip

And you need to decide what items to bring with you

For example: Food, a towel, a first aid kit and entertainment

Each item has a certain weight and a value

But we cannot take all of them, we can only carry three kilos

Weight

Value

1 2

2 1

1 3

3 4

How to decide on the maximum value we can bring?

Weight

Value

1 2

2 1

1 3

3 4

Like before, we could make a decision tree

For each item, we can either take it (left arrow) or leave it (right arrow)

Start

W:1 V:2 W:0 V:0

W:3 V:3 W:1 V:2 W:2 V:1 W:0 V:0

Weight

Value

1 2

2 1

1 3

3 4

Is that a smart move?

Well for four items it is okay, we have to create 16 path (2 ^ 4)

Start

W:1 V:2 W:0 V:0

W:3 V:3 W:1 V:2 W:2 V:1 W:0 V:0

Weight

Value

1 2

2 1

1 3

3 4

However: 2^n is not a sweet formula!!

For 10 items, we already have more than 1000 (1024) paths

Weight

Value

1 2

2 1

1 3

3 4

Like before, we want to find a smaller problem

That we can solve similar to the ‘big’ problem

Suppose this rectangle is an empty backpack

If we put one item is, what we in fact have left is…

If we put one item is, what we in fact have left is…

A smaller backpack!

With this idea we are going to solve the problem:

We put an item in the backpack, and continue with a smaller one

But we are cheating a little bit,

Since we don’t know beforehand which items to take

Therefore, we need a table listing all values for smaller backpack

From those values, we can calculate the end result

We calculate the maximum value to be obtained

For each weight (horizontal) choosing from a certain list of items (vertical)

W V

1 2

2 1

1 3

3 4

Max. value Max weight:0

Max weight:1

Max weight:2

Max weight:3

Items up to 0

Items up to 1

Items up to 2

Items up to 3

Items up to 4

This way, the final solution can be found in the lower right corner

Since there we can choose from all items, carrying weight 3

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0

Items up to 1

Items up to 2

Items up to 3

Items up to 4

So how to fill this nice table?

Some values are very easy to figure out. Do you know some?

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0

Items up to 1

Items up to 2

Items up to 3

Items up to 4

Well, if you cannot carry any weight…

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0

Items up to 1

Items up to 2

Items up to 3

Items up to 4

Well, if you cannot carry any weight…

You will not carry anything of value anyway

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0Items up to 1 0Items up to 2 0Items up to 3 0Items up to 4 0

And if you cannot choose from any item…

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0Items up to 1 0Items up to 2 0Items up to 3 0Items up to 4 0

And if you cannot choose from any item…

Nothing is to be gained either

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0 0 0 0Items up to 1 0Items up to 2 0Items up to 3 0Items up to 4 0

What else is easy?

Lets try the second row…

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3



Max weight:1

Max weight:2

Max weight:3


With the burger only, we can never do better than value 2

Which is the value of the burger

W V

1 2

2 1

1 3

3 4

Now for the remainder of the second row

In the yellow box we have: items up to 2, max weight 1

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0 0 0 0Items up to 1 0 2 2 2Items up to 2 0Items up to 3 0Items up to 4 0

What can we do? We can’t take the towel, since it weights 2

So there’s no option, but just sticking with the burger (value 2)

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0 0 0 0Items up to 1 0 2 2 2Items up to 2 0 2Items up to 3 0Items up to 4 0

Let’s do one more!

Now we can choose from the first three items

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


So now we could also take the mushroom.

It has weight 1, so at least it fits

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


But is this better?

Well… What we have now is value 2, and the mushroom is worth 3

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


So let’s ditch the burger, and go for the mushrooms

Than our maximum value is 3, and that is obviously better than 2.

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0 0 0 0Items up to 1 0 2 2 2Items up to 2 0 2Items up to 3 0 3Items up to 4 0

This way, we can fill the entire table

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3

Items up to 0 0 0 0 0Items up to 1 0 2 2 2Items up to 2 0 2 2 3Items up to 3 0 3 3 3Items up to 4 0 3 3 5

To calculate the values, we secretly used a recurrence relation

This means describing a value in the table with other values in the table

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


Let’s call the table T. T has two dimensions, i (items) and w(weight)

Here we highlight T[2,1], indicating items up to 2, maximum weight 1

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


How can we calculate this value T[2,1]?

In stead of reasoning about it as we did before?

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


Well what did we do before?

We said: Since the towel does not fit, we stick with the burger

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


if wi < w (the new item is too heavy)

then T[i,w] = T[i-1,w] (we stick with what we already have)

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


Now, let’s have a look at what we did at T[3,1]

We said: It fits, and now we have value 2, and the mushrooms is worth 3

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


if wi <= w (this new item fits)

then T[i,w] = Max (T[i-1,w], T[i-1-w-wi]+vi) (we check if we can do better)

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


To perform this calculation, we only need n2 operations

In stead of 2n Okay, in this case both are 16, but you get the idea :)

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


Again, we reduced complexity by seeing the symmetry!

W V

1 2

2 1

1 3

3 4


Max weight:1

Max weight:2

Max weight:3


So let’s do one little more problem

Calculating all prime numbers

Did you know they were named after this guy?

Okay they weren’t, but he does makes a nice background

Prime numbers are numbers only divisible by 1 and itself

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

So how do we know what’s a prime number and what’s not?

We could check all divisors, per number.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

But that could take up to the root of each number

Lets use symmetry, and find a smaller problem like this one.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Suppose we already have a number that’s prime.

Say 2. What do we know now?

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

A hint: It is something about every multiple of two…

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Exactly, they are not primes!

Since they have 2 as a divisor

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

We have eliminated half of the numbers

(Okay we have infinitely many left, but still…)

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

We can do the same trick with 3

Identify it as prime, and remove all numbers divisible by it

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

This way we can quickly determine all prime numbers

Thanks to…

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100


Thanks for your attention!


And remember:

Try to find symmetry everywhere2n is never funAnd beware of antisheep


This presentation was created by Felienne Hermans, PhD student and entrepreneur

And presented at the yearly Devnology Community Day 6 november 2010

@Felienne

http://www.devnology.nl/

http://www.devnology.nl/

If you liked this presentation, you might also like these books

The author of the first two is nice to follow on Twitter: @marcusdusautoy

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