## Saturday, August 20, 2011

### Su Doku Solver

Cold Hard Logic

Let's be frank, Su Doku solvers are dime a dozen. My favourite is the one in Google Goggles - you point your camera at a puzzle and it uses OCR to interpret it, then solves it for you. None of that faffing with inputting the puzzle by hand. Anyway.

At the most basic level, computers run on maths and logic. So logic puzzles should be easy for computers to solve - it's just following a collection of logical rules. No creative thinking required.

The problem are,

1) Inputting a problem in a form that the computer can parse (understand)
2) Translating the logical rules into a form the computer can apply
3) Applying said rules to a given input problem to find its solution.

Here's the main puzzle I'll be using as the example for this post [source]. Feel free to try and solve it yourself before continuing (it should be easy).

The Rules

So what are the rules? In the traditional version you're presented with a 9x9 grid, with some numbers already filled in. The idea is to enter into the grid the numbers 1-9, so that each number appears only once in each row, column, or box (3x3 square).

So that's fairly straight forward. But in that particular form, the rules are really only good for checking if your solution is right.

Brute Force

This can be use in a brute-force approach to solving, e.g.

1) Stick numbers in empty squares at random
2) If any of the numbers break the rules, try again
3) Repeat until the right solution is found.

But that's incredibly inefficient and inelegant, and certainly not the way it's solved by people in the real world.

As an example, say our placement technique is this - take the numbers missing from each row, and just shuffle them around within their given row.

[There are probably better brute-force methods.]

If we use this approach on the puzzle above, there are ~4.95E26 possible number placements (permutations), of which only one is correct. Say our computer can check 1 billion solutions per second, it would take 16 billion years to solve. Which is about 2 billion years longer than the current age of the universe.

As a ball park estimate, the general formula for the number of permutations for a grid of side n is given by -> ((n-a)!)^n  - Where a is the average number of numbers already placed in each row.

If, for example, we try to solve a 16-grid with a=8 (half the numbers already placed), then the number of possible permutations is a mind-meltingly inconceivable 4.88x10^73. At 1 billion checks per second, that would take 1.5x10^57 years to solve - 1 billionth the lifespan of a black hole with the mass of the sun (according to wikipedia).

This is an example of a non-polynomial time (NP) algorithm. More on that later.

Rewriting the Rules

The rules can be re-interpreted and applied in two fundamental ways

1) Which is the only square this number can appear in?

That is, a particular number has to go in one particular square, because if you try to put it in any other square in that row (or column, or box) it will break the rules.

This is the most common approach in real world solving - that is, "three can't go in that row, or those columns because they already have threes in them, so it must go there" sort of thing.

2) What is the only number that can go in this square?

That is, if every other number already appears in the same row, column, or box as a given square, then that square can only be one particular number.

This is most obvious in the case where every other number in a given row is already filled in, but can also be used by considering all the rows, columns and boxes a square belongs to.
This tends to be required more in the harder puzzles; you could probably solve the above puzzle by Rule One alone.

Setting Up The Problem

First of all, we create our grid. We can refer to each square in the grid by a sort of coordinate - (i,j) - with the square in the top left corner being (0,0) and the square in the bottom right being (8,8)
We reference Row i as the set {(i,j) : j in [0,8]} and Column j as {(j,i) : j in [0,8]}

For boxes, we note that the index for the top-left square of each box are multiples of 3 - (0,0) (0,3) (0,6) (3,0) etc. So, Box (m,n) is given as {(3m+i,3n+j) : i,j in [0,2]} for m,n in [0,2]

So from this we can, for example, check the correctness of a puzzle (by rows) like this:
for i in range(0,9):
for j in range(0,8):
for k in range(j+1,9):
if Grid[i][j] == Grid[i][k]:
return "incorrect puzzle"
Checking columns and boxes works similarly.

If we have a fancy GUI, then inputting and parsing the grid is pretty straight forward. But we don't. So instead, we input the puzzle as one long string -> puzzle = "???36?9???2?.."

The first row is given by the characters (numbers) numbered [0,..,8], the second row is [9,..,17], and so on. In general, row i is the set of characters with index [9i,..,9i+j,...,9i+8].

Parsing our puzzle into the Grid works like this then:

for i in range(0, 9):
for j in range(0, 9):
Grid[i][j] = puzzle[9i+j]
And so on..

Applying the Rules

Now, if you've done Su Dokus, you might be familiar with the technique of filling each square with the numbers one to nine and eliminating numbers as appropriate. That's basically the bulk of how this solver works:

1) Assign to each empty square ("?") the set [1,...,9]
2) For each number in a given square's set:
a) Check to see if that number is already fixed in another square in the same row, column, or box as the square in question. If so, eliminate the number from that square's set.

3) Repeat for all squares in the grid
From there, we can apply Rule Two:

4) For each square:
a) if that square's set contains only one number, fix that square's value accordingly.
5) Go back to (2). Repeat until no more squares can be fixed.

You can actually created a shortcut for this - create a collection of sets for each row (etc.), which contain the values already fixed in that row. That way, instead of checking every square along each row for fixed values, you just check in that row's set. (Trust me, it's easier.)

Applying Rule One is easy as looking along each row (or column, or box) to see if a given number appears in only one of the squares.
Or more technically,

1) For a given row (etc), for each number 1-9:
a) count how many times that number appears in square sets in that row
b) if a number's count equals 1, fix the value of the square it appears in accordingly.

2) If you fix any new squares this way, go back to (2) to see if that will allow you to set any more squares.

You then just loop through rules One and Two until the puzzle is (hopefully) solved.

One More 'Rule'

In fact, this is just Rule Two, but with an added 'quirk'.

Here's what happens if you apply the method, so far, to a more difficult puzzle
This is where the program gets stuck. What do you do from here? If we focus on the bottom-right corner box
Notice there are no squares that can take only one value, and no values that only appear in one of the squares.

But look on that bottom row - the circled boxes can only take the values 3 or 5. What this means is that 3 and 5 can't go in any other of the squares in that box. So we eliminate 3s and 5s from the other squares in that box. Similarly for the 1/6 squares in the middle.
We now have the square in the top-right of the box that can only be 8, and the middle-right that can only be 9. So we fix those values. And from there, it's pretty straight forward to solve the rest - going back to looping through Rules Two and One.

Not only is it tricky to spot in real world solving, it's also tricky to translate it into computer logic. It mostly involves looking for pairs of squares in a box (etc.) which have matching sets of size two (or triplets, size 3, or etc.). Fortunately, this sort of thing doesn't come up often.

A Note on P versus NP

So as mentioned above, it's easy enough to apply the standard rules (as states) to check if a solution is correct - you just check to make sure no number appears more than once in any given row, column, or box.
It turns out, checking the correctness of a puzzle can be done in polynomial time (fast). For a grid of side n, the time required to check a solution* is given by -> 3(n^3 - n)/2

..Multiplied by the amount of time it takes a computer to work out if 2 numbers are equal or not (call this time t). So for a 9-grid, that time is 972t.

For a grid of side n=100, the time is ~1.5million t. Which on a modern computer is peanuts. For example, if our hypothetical computer can check 1 billion 9-grids in one second, then it can check 650,000 100-grids in a second.

By comparison, actually solving the puzzle - by the method outlined about, and by all other known methods - is done in non-polynomial time.

What this means, is that if you apply this algorithm to a larger grid - say, a 25x25, or a 36x36 grid - then the amount of time it will take to solve the puzzle increases rapidly (exponentially), to the point where a puzzle couldn't be solve in the lifetime of the universe.

But, since checking the puzzle for correctness is polynomial, even if the puzzle took longer than the life of the universe to solve, you could still check to see if the solution is correct (that no number appears more than once in each row, etc) in a reasonable amount of time.

As mentioned above, the brute-force method outlined is non-polynomial (NP). Notice that it takes catastrophically longer - 70 orders of magnitude - to solve a 16-grid by brute-force, than it takes to check the solution to a, much larger, 100-grid.

Admittedly, I don't know what the time function is for the method outlined above. But suffice it to say, it's more efficient than the brute-force, but still non-polynomial.

What P versus NP asks, then, is

if the solution to a problem can be checked in polynomial time (quickly), then does there exist an algorithm that can solve the problem in polynomial time?

At present, this is one of the great unsolved questions of mathematics and computer science, and a proof of whether or not P=NP will earn its discoverer \$1million...

Where's The Code?

You'll probably have noticed, I've given a general, hand-wavy outline of how a program would work, and only given odd snippets of code.

I did write a solver in Python a few years back. I'd taken a Java programming course in my first year at university, and when the lecturer discussed a tic-tac-toe (naughts and crosses) playing program, it gave me the idea of how to represent a grid in a way a computer can parse. It also made me realise that there is actually a point to object-oriented programming.

Anyway, long story short, I wrote the solver 3 years ago. It's kind of sloppy and inelegant. And it can't handle the 'tricky' puzzles - although, oddly enough, according to my notebook, I had worked out a way of implementing it, but apparently never did.

Or to put it another way, it works, but compared to the myriad others out there, it's nothing special.

If you're interested, as I mentioned at the beginning there are plenty of other solvers out there. They might not work according to the methods discussed in this post. But you should at least have an idea of how one might work, now.

And as a final note, these methods can also be used for hand-solving. It's generally not necessary to be so rigidly systematic, though. It's easier for humans to pick out number placements at a glace, without having to go through the whole set-and-elimination approach.

In Case You Were Interested

Here's the solution to the Su Doku at the top of the page
You're on your own with finishing the other ('difficult') one..

Oatzy.

## Sunday, August 07, 2011

No, this isn't a Doctor Who thing (sorry). No, this is based on another lost article I read a while back:

How many photos do you have to take in order to get at least one where no one's blinking?

First things first, what's the probability of one person blinking when you take a photo of them?

Well for one thing, that's going to vary depending on environment, lighting, etc. And also on the shutter speed of the camera being used to take the picture.. But for simplicity, I'm ignoring all that.

The average person blinks around 10 times per minute, with an average blink lasting 300-400 milliseconds (call it 350ms). So in any given minute, the average person's eyes will be closed for a total of 3.5 seconds.

We'll say the shutter is open for less than the duration of a blink. So the probability of a persons eyes being closed while the shutter is open -> p = 0.058.

[I'll be honest, I'm not not entirely convinced that's right. But I'll go with it anyway.]

So the probability they don't blink -> (1-p) = 0.942

If you're taking a picture of one person, that's only a 5.8% chance of the subject ruining a photo by blinking. So your odds of a good shot are pretty good.

But if we have a much larger group of people - n = 30 - the probability that none of those people blink while a photo is being taken:

p1 = (1-p)^n = 0.165

That's an 83.5% chance that at least one person will blink. Those odds aren't so good.

So if we take S number of photos, what is the probability that at least one of those is 'perfect'?

This goes back to the methods use in the Law Of Truly Large Numbers post - the probability of at least one perfect photo, big P, is one minus the probability that none of the S photos are perfect:

P = 1 - (1-p1)^S

So if we were to take, say, S = 5 photos -> P = 0.594

Bearing in mind that you have to make these 30 people stand around while you take your however many photos, an almost 60% chance of the perfect shot from 5 tries it pretty reasonable.

But let's say you're the panicky sort, and you want to be 90% certain that you have at least one perfect shot..?

Without going into the nitty-gritty, we can find S for P = 0.9, using some logarithms and algebra thus:

S = ln(1-P)/ln(1-p1) = 12.8 shots

And if you can get a group of 30 people to stand still for 13 photos, knowing that there's still a 10% chance you won't get that perfect shot, then more power to you.

When people know they're having their picture taken, they generally try harder not to blink. Especially if you use a count down. So the probability of blinking, and by extension, the number of photos you'd have to take, drops dramatically. The numbers worked out above are probably more applicable for candid shots.

On the other hand, maybe the flashing going off (if you use one) will cause some people to automatically blink. And, admittedly, I've taken pictures of myself that have still managed to get capture me mid-blink. Though that might be down to a delay between click and shutter.

Of course, in this day and age, of digital cameras with instant preview, you can just keep shooting until you get the photo you want. Not like the dark old days of film cameras and photo roulette...

Oatzy.

[The word 'blink' and its variants appear ~17 times in this post]

[As a random aside, working out the number of pokéball you need to throw to catch a given Pokémon is done in much the same.]

## Wednesday, August 03, 2011

### The Toilet Seat Conundrum

Gentlemen, do you leave to toilet seat up, or courteously put it down after use?

I read a (somewhat tongue-in-cheek) article a while back, I can't remember where, that explained the toilet seat conundrum in terms of game theory.  As best I can remember, it was quite clever. I was recently reminded of it when reading a Cracked article, and thought I'd try to recreate it, and - as is my wont - take it a step further.

Game theory, for those unfamiliar, is an area of maths/economics that studies 'competitive' interactions, "in which an individual's success in making choices depends on the choices of others".

Preamble

The (average) probability of the gentleman needing to 'sit down' when visiting the bathroom, we call p*. The probability of not is (1-p).

If the toilet seat is in the 'wrong position' for a given visit, we call the cost of this c1, and we assume that this cost is the same for both genders. This may not be strictly true.

The simplest cost would be in having to move the seat, typically in the form of mild inconvenience, and the potentially unpleasant experience of having to touch the underside of the seat. I'm also told that there are certain perils in visiting the toilet at the night, if the seat is in the upright position and is required to be otherwise. I can't say this is a cost I've ever experienced.

One might argue that the 'costs' are inconsequential; but for the sake of arguing, they aren't.

For the sheer hell of it, we'll call the woman Alice and the man Bob. Alice and Bob have been in a relationship/living together just long enough to quarrel over such matters. I suspect, for most people, this is a non-issue; but that's not the point of the post.

There is a third possible game, not discussed below, in which Bob can just leave the seat down at all times. In this case, we have c3, the cost of clean up if Bob's aim isn't quite up to scratch.

Oh, and there is a fourth game, where the default position of the the seat is upright. This is the worst possible game for Alice, and is only the best possible game for Bob if p<0.28. I can only imagine this game working in an all male household, and even then (a re-adjusted) Game One works out better.

Game One - Leave It As Is

Probability of the seat being down is the probability of Alice being the last to visit the lav plus the probability Bob was the last and left the seat down. Probability the seat is up is 1 minus the above.

Here's the cost matrix for this game
Where cost is c1 multiplied by the probability that the seat is in the wrong position

To get the total costs to Alice and Bob for this game, we work out

(Probability the seat is down x the cost if seat is down) +
(probability seat is up x cost if seat is up)
And we can work out the ratio of costs
B:A  =>  2p+1 : 1

In the extreme case, where p=0 (Bob never poops), their costs are equal. But in all other cases, Bob's cost is greater than Alice's.

Default meaning the seat is always returned to the downright position after use. The probability of the seat being up is always 0.

Here's what the matrix looks like
In this case, Alice incurs no cost. Bob, on the other hand, incurs double cost - when he needs to urinate, he has to move the toilet seat twice: up before use, and down afterwards.

It's obvious that Alice, once again, fairs better than Bob.

Game Three, mentioned in the preamble, works the same as this, but with 2c1 replaced with c3. Alice still comes out better though. Unless she doesn't like the thought of sitting on a toilet seat that's (potentially) been peed on - even if it is cleaned - in which case, there's some abstract cost to her.

If she doesn't mind, then which of games Two and Three Bob would prefer depends on which is smaller: 2c1 or c3.

Lowest Costs

First of all, we note that in both games Alice comes out better than Bob - incurring a lower cost in both cases. That said, Alice does better in the latter game, incurring no cost at all in that one. So Game Two is preferable to Alice.

If we take the cost ratio of game one to game two for Bob, this is what we get
B1:B2  =>  2p+1 : 4

In the extreme case of p=1 (Bob never urinates), Game Two incurs a greater cost for Bob (3:4) - and by extension, Game Two always incurs a greater cost to Bob.

THIS is where and why the conflict arises.

Alice prefers Game Two, Bob prefers Game One.

Tipping the Scales

So Alice would prefer to play Game Two, but she has to encourage Bob towards it. So Alice introduces a new penalty - c2 - for Bob leaving the toilet seat up.

The cost will typically be something along the lines of a bollocking, silent treatment, arguments, or whatever.

So what we do is this - the odds of Bob leaving the toilet seat up, and Alice being the next to use the bathroom -> (1-p)/2

Multiplied by the cost, c2, and added to the pre-existing total cost for game one
Now, we - or rather, Alice - wants the cost to be such that Game Two is preferable, i.e. B'1 > B2

Rearranging and simplifying, we get
c2 > 4c1(2p+1)

However, if Alice were feeling kind, she could introduce a 'reward' for putting the toilet seat down, instead. It works effectively the same - barring psychological, carrot/stick considerations.

For this, Alice would have to offer a reward, R, with
Arguably, Bob could introduce a new cost - or enticement - himself, to 'persuade' Alice towards Game One. But TV leads me to believe that this is seldom thought of, or executed.

This might be because Alice has more to gain/lose - in as much as, Alice can avoid any cost by 'playing' Game Two. Bob, on the other hand, incurs some cost in both games.

You can draw your own conclusions on that one.

In terms of a co-operative solution, if we add together Alice and Bob' costs in each game and compare, we find that Game One has a lower total cost than Game Two.

So one could argue that Game One is better overall. The challenge, though, is convincing Alice that that is the best solution for both of them, given that, from Alice' point of view, she does worse in Game One.

Casino Bathrooms

So this is all well and good, but it's kind of a specialised case - the situation of a house with one male, one female, and one toilet. In our house, for example, we have two males, two females and three toilets. What then?

There are a few other problems with the probability-based approach, as well. For one thing, it uses an average poop-probability for the gentleman. It also assumes both Alice and Bob use the bathroom about the same number of time during a given time period - whereas some people have more robust insides than others.

So for this, we create a Monte Carlo simulation.

[This is what we call excessive commitment to an idea.]

In the simulation, we create a 'person' object, and assign to them a gender, an average number of bathroom visits, and, for males, an average bladder to bowel movement ratio. To capture the day to day variability in number of visits to the bathroom, we use Poisson distributions.

We also create a 'toilets' set, representing however many toilets there are, and their current states -> 1 = toilet seat up, 0 = toilet seat down. Each toilet has an equal chance of being chosen for use by any given person at any given time.

Each person has a counter, which is incremented when the person in question has to move the seat. At the end, these counters are grouped by gender for comparison.

In the Middle of Our Street

So I created a 'house' of two males, two females, and three toilets (variables chosen arbitrarily). Then ran the simulation for 10,000 hypothetical days.

The Game One simulation gives a result of ~ 3.28 seat moves per male per day, and 2.27 seat moves per female per day. That's a male:female seat move ratio of 1.44.

The Game Two simulation gives a result of 9.11 seat moves per male per day - bearing in mind, men have to move the seat twice per standing visit, in this version - and women never have to move the seat.

Code here.

Fun fact: Without additional costs and rewards, Game One is always preferable to men, Game Two to women. Regardless of the balance of men and women in a house.

So now you know!

Of course, in some cultures the conflict never really arises, since it is 'the norm' for men to sit for all visits to the lavatory.

Oatzy.

*[inb4 shouldn't p be the probability of needing to urinate lol]