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?

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
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
Well… it are 99 plusses, so 99 points of effort (and possibly mistake)
Could we do this with less operations(plusses)?

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?

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
That’s me!
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
That’s me!
Remember Gauss?
He could do it with only 2 operations, using symmetry

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+100 = 101
2+99 = 101
3+98 = 101
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 = 101
2+99 = 101
3+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!

Start loving symmetry already?

I do!
Start loving symmetry already?

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?

|HSCW
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
HC|SW
HS | CW
HW | SC
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
HC|SW
HS | CW
HW | SC
So these two options are no good!

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

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

12
6
Given two numbers A en B
Determine the maximum number x, that divides both A en B

12
6
For instance 6 and 12
2, 3 and 6 all divide both numbers, but 6 is the largest

252
105
But how do we calculate the gcd of larger numbers?
We could try all numbers up to 105

252
105
Can we do better?
Using symmetry?

252
105
So let’s search for a problem that is smaller
But can be solved in the same way

252 = x * y
105 = x * z
Rephrasing our problem definition
We search for an x that ‘fits’ into both A and B

252 = x * y
105 = x * z
But we don’t really care about how often x fit in A and B
Remember: be aware of antisheep!

252 = x * y
105 = x * z
So how can we reduce this problem?
While keeping it similar?

252 = x * y
105 = x * z -
147 = x * (y-z)
Well, we could subtract
Now we again have two numbers in which x fits nicely, but smaller!

147 = x * y
105 = x * z -
42 = x * (y-z)
Now we can continue with 105 and 147
And again have two numbers in which x fits nicely, but smaller!

252 = 21 * 21
105 = 5 * 21
You can check yourself that within 5 steps we reach the answer
That’s way better that 105 steps. Thanks to…

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

How to decide on the maximum value we can bring?

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
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
Is that a smart move?
Well for four items it is okay, we have to create 16 path (2 ^ 4)

However: 2^n is not a sweet formula!!
For 10 items, we already have more than 1000 (1024) paths

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)

This way, the final solution can be found in the lower right corner
Since there we can choose from all items, carrying weight 3

So how to fill this nice table?
Some values are very easy to figure out. Do you know some?

Well, if you cannot carry any weight…

Well, if you cannot carry any weight…
You will not carry anything of value anyway

And if you cannot choose from any item…

And if you cannot choose from any item…
Nothing is to be gained either

What else is easy?
Lets try the second row…

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

Now for the remainder of the second row
In the yellow box we have: items up to 2, max weight 1

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)

Let’s do one more!
Now we can choose from the first three items

So now we could also take the mushroom.
It has weight 1, so at least it fits

But is this better?
Well… What we have now is value 2, and the mushroom is worth 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.

This way, we can fill the entire table

To calculate the values, we secretly used a recurrence relation
This means describing a value in the table with other values in the table

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

How can we calculate this value T[2,1]?
In stead of reasoning about it as we did before?

Well what did we do before?
We said: Since the towel does not fit, we stick with the burger

if wi < w (the new item is too heavy)
then T[i,w] = T[i-1,w] (we stick with what we already have)

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

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)

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 :)

Again, we reduced complexity by seeing the symmetry!

So let’s do one little more problem

Calculating all prime numbers
Did you know they were named after this guy?

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
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…)

234 5 6 7 8910 11 12 13 141516 17 18 19 202122 23 24 25 262728 29 30 31 323334 35 36 37 383940 41 42 43 444546 47 48 49 505152 53 54 55 565758 59 60 61 626364 65 66 67 686970 71 72 73 747576 77 78 79 808182 83 84 85 868788 89 90 91 929394 95 96 97 9899100
We can do the same trick with 3
Identify it as prime, and remove all numbers divisible by it

234 5 6 7 8910 11 12 13 141516 17 18 19 202122 23 24 25 262728 29 30 31 323334 35 36 37 383940 41 42 43 444546 47 48 49 505152 53 54 55 565758 59 60 61 626364 65 66 67 686970 71 72 73 747576 77 78 79 808182 83 84 85 868788 89 90 91 929394 95 96 97 9899100
This way we can quickly determine all prime numbers
Thanks to…

Thanks for your attention!

And remember:
Try to find symmetry everywhere
2n is never fun
And beware of antisheep

@Felienne
This presentation was created by FelienneHermans,
PhD student and entrepreneur
And presented at the yearly DevnologyCommunity Day 6 november 2010

Books on symmetry
(in increasing order of difficulty)
General mathematical
awesomeness
More cool programming
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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Symmetry

by Felienne Hermans on Nov 15, 2010

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