Short 30 second video of the AI in action here:

How to Play Minesweeper

If you’re an experienced minesweeper player, you can probably skip this section. Otherwise, I’ll just give a quick overview of some basic strategies that we can use to solve an easy minesweeper game.

We start with a 10×10 Beginner’s grid, and click on a square in the middle:

We can quickly identify some of the mines. When the number 1 has exactly one empty square around it, then we know there’s a mine there.

Let’s go ahead and mark the mines:

Now the next strategy: if a 1 has a mine around it, then we know that all the other squares around the 1 cannot be mines.

So let’s go ahead and click on the squares that we know are not mines:

Keep doing this. In this case, it turns out that these two simple strategies are enough to solve the Beginner’s grid:

Roadmap to an AI

All this seems easy enough. Here’s what we’ll need to do:

1. Read the board. If we use a screenshot function, we can get a bitmap of all the pixels on the board. We just need to ‘read’ the numbers on the screen. Luckily for us, the numbers tend to have different colors: 1 is blue, 2 is green, 3 is red, and so on.
2. Compute.  Run the calculations, figure out where the mines are. Enough said.
3. Click the board. This step is easy. In Java, we can use the Robot class in the standard library to send mouse clicks to the screen.

Reading the Field

There’s not a whole lot to this step, so I’m going to skim over it quickly.

At the beginning of the run, while we have a completely empty grid, we invoke a calibration routine – which takes a screenshot and looks for something that looks like a Minesweeper grid. Using heuristics, it determines the location of the grid, the size of a grid square, the dimensions of the board, and things like that.

Now that we know where the squares are, if we want to read a square, we crop a small section of the screenshot and pass it to a detection routine, which looks at a few pixels and figures out what’s in the square.

A few complications came up in the detection routine:

• The color for the number 1 is very close to the color of an unopened square: both are a dark-blue color. To separate them apart, I compared the ‘variance’ of the patch from the average color for the patch.
• The color for 3 is identical to that for 7. Here, I used a simple edge-detection heuristic.

Straightforward Algorithm

The trivially straightforward algorithm is actually good enough to solve the beginner and intermediate versions of the game a good percent of the time. Occasionally, if we’re lucky, it even manages to solve an advanced grid!

When humans play minesweeper, we compete for the fastest possible time to solve a grid of minesweeper. So it doesn’t matter if we lose 20 games for every game we win: only the wins count.

This is clearly a silly metric when we’re a robot that can click as fast as we want to. Instead, we’ll challenge ourselves with a more interesting metric:

Win as many games as possible.

Consider the following scenario:

Using the straightforward method, we seem to be stuck.

Up until now, whenever we mark a square as having a mine or safe, we’ve only had to look at a single 3×3 chunk at a time. This strategy fails us here: the trick is to employ a multisquare algorithm – look at multiple different squares at once.

From the lower 2, we know that one of the two circled squares has a mine, while the other doesn’t. We just don’t know which one has the mine:

Although this doesn’t tell us anything right now, we can combine this information with the next 2: we can deduce that the two yellowed squares are empty:

Let’s click them to be sure.

And voilà. They’re empty. The rest of the puzzle can be solved easily, after we’ve made the deduction that those two squares were empty.

The Tank Solver Algorithm

It’s difficult to make the computer think deductively like we just did. But there is a way to achieve the same results, without deductive thinking.

The idea for the Tank algorithm is to enumerate all possible configurations of mines for a position, and see what’s in common between these configurations.

In the example, there are two possible configurations:

You can check for yourself that no other configuration could work here. We’ve deduced that the one square with a cross must contain a mine, and the three squares shaded white below must not contain a mine:

This works even better than human deduction!

We always try to apply the simple algorithm first, and only if that gets us stuck, then we bring in the Tank algorithm.

To implement the Tank algorithm, we first make a list of border tiles: all the tiles we aren’t sure about but have some partial information.

Now we have a list of  border tiles. If we’re considering every possible configuration, there are of them. With backtracking, this number is cut down enough for this algorithm to be practical, but we can make one important optimization.

The optimization is segregating the border tiles into several disjoint regions:

If you look carefully, whatever happens in the green area has no effect on what happens in the pink area – we can effectively consider them separately.

How much of a speedup do we get? In this case, the green region has 10 tiles, the pink has 7. Taken together, we need to search through combinations. With segregation, we only have : about a 100x speedup.

Practically, the optimization brought the algorithm from stopping for several seconds (sometimes minutes) to think, to giving the solution instantly.

Probability: Making the Best Guess

Are we done now? Can our AI dutifully solve any minesweeper grid we throw at it, with 100% accuracy?

Unsurprisingly, no:

One of the two squares has a mine. It could be in either, with equal probability. No matter how cleverly we program our AI, we can’t do better than a 50-50 guess. Sorry.

The Tank solver fails here, no surprise. Under exactly what circumstances does the Tank algorithm fail?

If it failed, it means that for every border tile, there exists some configuration that this tile has a mine, and some configuration that this tile is empty. Otherwise the Tank solver would have ‘solved’ this particular tile.

In other words, if it failed, we are forced to guess. But before we put in a random guess, we can do some more analysis, just to make sure that we’re making the best guess we could make.

Try this. What do we do here:

From the 3 in the middle, we know that three of them are mines, as marked. But marking mines doesn’t give us any new information about the grid: in order to gain information, we have to uncover some square. Out of the 13 possible squares to uncover, it’s not at all clear which one is the best.

The Tank solver finds 11 possible configurations. Here they are:

Each of these 11 configurations should be equally likely to be the actual position – so we can assign each square a probability that it contains a mine, by counting how many (of the 11) configurations does it contain a mine:

Our best guess would be to click on any of the squares marked ‘2’: in all these cases, we stand an 82% chance of being correct!

Two Endgame Tactics

Up until now, we haven’t utilized this guy:

The mine counter. Normally, this information isn’t of too much use for us, but in many endgame cases it saves us from guessing.

For example:

Here, we would have a 50-50 guess, where two possibilities are equally likely.

But what if the mine counter reads 1? The 2-mine configuration is eliminated, leaving just one possibility left. We can safely open the three tiles on the perimeter.

Now on to our final tactic.

So far we have assumed that we only have information on a tile if there’s a number next to it. For the most part, that’s true. If you pick a tile in some distant unexplored corner, who knows if there’s a mine there?

Exceptions can arise in the endgame:

The mine counter reads 2. Each of the two circled regions gives us a 50-50 chance – and the Tank algorithm stops here.

Of course, the middle square is safe!

To modify the algorithm to solve these cases, when there aren’t that many tiles left, do the recursion on all the remaining tiles, not just the border tiles.

The two tricks here have the shared property that they rely on the mine counter. Reading the mine counter, however, is a non-trivial task that I won’t attempt; instead, the program is coded in with the total number of mines in the grid, and keeps track of the mines left internally.

Conclusion, Results, and Source Code

At this point, I’m convinced that there isn’t much more we could do to improve the win rate. The algorithm uses every last piece of information available, and only fails when it’s provably certain that guessing is needed.

How well does it work? We’ll use the success rate for the advanced grid as a benchmark.

• The naïve algorithm could not solve it, unless we get very lucky.
• Tank Solver with probabilistic guessing solves it about 20% of the time.
• Adding the two endgame tricks bumps it up to a 50% success rate.

Here’s proof:

I’m done for now; the source code for the project is available on Github if anyone is inclined to look at it / tinker with it:

https://github.com/luckytoilet/MSolver

In case you’re not, the idea is that you’re given some rational function with an awful denominator that you want to integrate, like:

And you break it up into smaller, simpler fractions:

This is the idea. If we get into the details, it gets fairly ugly — in a typical calculus textbook, you’ll find a plethora of rules regarding what to do in all sorts of cases: what to do when there are repeated linear factors, quadratic factors, repeated quadratic factors, and so on.

Since the textbooks generously cover this for us, we’ll assume that we know what to do with a rational polynomial with some polynomial as the numerator, and some number of linear or quadratic factors in the denominator. We can do partial fraction decomposition on this. If we like, we could integrate it too. I’m talking about anything of this form:

Although we won’t prove this, this seems fairly believable. We’ll assume that once we get a fraction into this form, we’re done and we can let existing partial fraction methods take care of the rest.

Can Partial Fractions Fail?

What if we have a polynomial greater than a quadratic in the denominator? So let’s say:

Fortunately, here the denominator can be factored, giving us a form we can deal with:

But we were lucky that time. After all, not all polynomials can be factored, right? What if we have this:

We can’t factor this. What can we do?

It turns out that this isn’t a huge problem. We never required the coefficients of the factors to be integers! Although the factorization is awkward, it can still be factored:

Other than making the next step somewhat algebraically tedious, this decomposition is perfectly valid. The coefficients need not be integers, or even be expressed with radicals. As long as every coefficient is real, partial fraction decomposition will work fine.

Universality of Partial Fractions

The logical next question would be, can all radical functions be written in the previous partial fraction decomposition-suitable form? Looking through my calculus textbooks, none seemed to provide a proof of this — and failing to find a proof on the internet, I’ll give the proof here.

We need to prove that any polynomial that might appear in the denominator of a rational function, say , can be broken down into linear or quadratic factors with real coefficients.

In order to prove this, we’ll need the following two theorems:

• Fundamental Theorem of Algebra — any polynomial of degree n can be written as a product of n linear complex factors:
• Complex Conjugate Root Theorem — if some complex number is a root of some polynomial with real coefficients, then its conjugate is also a root.

Starting with the denominator polynomial , we break it down using the Fundamental Theorem of Algebra into complex factors. Of these factors, some will be real, while others will be complex.

Consider the complex factors of . By the complex conjugate root theorem, for every complex factor we have, its conjugate is also a factor. Hence we can take all of the complex factors and pair them up with their conjugates. Why? If we multiply a complex root by its complex conjugate root: — we always end up with a quadratic with real coefficients. (you can check this for yourself if you want)

Before, we were left with real linear factors and pairs of complex factors. The pairs of complex factors multiply to form quadratic polynomials with real coefficients, so we are done.

At least in theory — partial fraction decomposition always works. The problem is just that we relied on the Fundamental Theorem of Algebra to hand us the roots of our polynomial. Often, these roots aren’t simple integers or radicals — often they can’t really be expressed exactly at all. So we should say — partial fraction decomposition always works, if you’re fine with having infinitely long decimals in the decomposed product.

We have a square, and we put a point on each side of the square. Then we connect the four points to create a quadrilateral. How can we make this quadrilateral have the smallest possible perimeter?

Intuitively, you may believe that this natural, obvious configuration should produce the least perimeter:

Attempt with Calculus

How can we prove that this indeed gives us the smallest possible perimeter?

A first attempt might be to give variables to the side lengths, and somehow find the minimum perimeter using algebra and calculus tools. So there are four independent points — let’s parameterize them with four variables, and assume the side length of the square is 1:

Then we want to minimize this expression:

At this point, it isn’t clear how to proceed — there doesn’t seem to be any way to minimize this expression of four variables.

Proof by Net

We’ll have to try something different. It’s hard to make sense of anything when there are four independent variables. Instead, if we expand things out a bit, things start to become more manageable:

What we did was reflect the square three times, and each time the square is reflected, the inscribed quadrilateral goes with it. By taking only the relevant parts of the quadrilateral, we get the green path.

Now we might have a solution. If we had a different green path, can we reverse the steps and get the original quadrilateral back? Basically, the following requirements have to be met:

• The path has to cross all three of the internal lines BC, BA, and DA.
• The path’s position on the bottom-most line, DC must be the same when reflected onto the top-most line DC.

With these requirements in mind, the shortest green path that satisfies these requirements is a straight line connecting a point on the bottom left to its reflected point on the top right:

Our intuition at the start was well-founded.

Now notice that this isn’t the only possible shortest path. If we move the entire green line to the left or right, we get a different path of the same length!

For instance, the degenerate ‘quadrilateral’ formed by connecting two opposite corners has the same perimeter as the one we get by connecting the midpoints. Neat, huh?

After a few days, I was able to solve all of the problems except one — the second part of the seventh problem:

Seven people participate in a tournament, in which each pair of players play one game, and one player is declared the winner and the other the loser. A triplet ABC is considered cyclic if A beats B, B beats C, and C beats A.

Can you always separate the seven players into two rooms, so that neither room contains a cyclic triplet?

(Note: the first half of the problem asked the same question for six people — and it’s not too difficult to prove that no matter what, we can put them into two rooms so that neither the first nor the second room contains a cyclic triplet.)

But what happens when we add another person? Can we still put four people in one room, and three people in the other, so that neither rooms contain a cyclic triplet?

There are two possibilities here:

• One, it’s always possible. No matter what combinations of wins and losses have occurred, we can always separate them into two rooms in such a way. To prove this, we’ll need to systematically consider all possible combinations, and one by one, verify that the statement is possible for each of the cases.
• Two, it’s not always possible. Then there is some counterexample — some combination of wins and losses so that no matter how we separate them, one of the rooms has a cyclic triplet. This is easier to prove: provided that we have the counterexample, we just have to verify that indeed, this case is a counterexample to the statement.

But there’s a problem. Which of the cases does the solution fall into? That is, should we look for a quick solution by counterexample, or look for some mathematical invariant that no counterexample can exist?

Brute Force?

It would be really helpful if we knew the counterexample, or knew for sure what the counterexample was. What if we wrote a computer program to check all the cases? After all, there are only 7 people in the problem, and 7 choose 2 or 21 games played. Then since each game is either won by one player or the other, there are only 2^21 combinations overall (although that does count some duplicates). And 2^21 is slightly over two million cases to check — completely within the bounds of brute force.

So I coded up a possibility-checker. Generate all 2^21 possible arrangements, then for each one, check all possible ways to separate them into two rooms. If it turns out that no matter how we arrange them, a cyclic triplet persists, then display the counterexample. Simple.

I ran the program. It quickly cycled through every possible arrangement, three seconds later exiting without producing a counterexample.

Alright. So there’s no counterexample. I would have to find some nice mathematical invariant, showing that no matter what, there is always some way to group the players so that neither room has a cyclic triplet.

But no such invariant came. I tried several things, but in each attempt couldn’t quite show that the statement held for every case. I knew that there was no counterexample, but I couldn’t prove it. But why? There must be some tricky way to show that no counterexample existed; whatever it was, I couldn’t find it.

Brute Force poorly implemented

Reluctantly, as the deadline came and passed, I submitted my set of solutions without solving the problem. When the solutions came out a week later, the solution to this problem did not contain any tricky way to disprove the counterexample. Instead, what I found was this:

Let be seven players. Let beat when the difference .

Huh? A counterexample, really? Let’s look at it.

Everything is symmetric — we can ‘cycle’ the players around without changing anything. Also, if we take four players, two of them are consecutive. Let them be and .

At this point everything falls into place: in any subset of four players, three of them are cyclic.

But wait … my program had not found any counterexamples! And right here is a counterexample! The culprit was obvious (the reader may have foreseen this by now) — of course, there had to be a problem with my program.

Running my code through a debugger, I found a logic error in the routine converting binary numbers to array configurations, meaning that not all possible configurations were tried. As a result, the counterexample slipped through the hole.

After fixing the code, the program found not one, but a total of 7520 (although not necessarily distinct) counterexamples. Most of them had no elegant structure, but the solution’s configuration was among them.

For the interested, here is the fixed code.

When to Start Over?

It is true that the program could have been better written, better debugged. But how could you know whether a counterexample existed and your program didn’t find it, or if no counterexample existed at all?

In hindsight, it seems that writing the brute force program made me worse off than if I hadn’t written it at all. After the program ran without finding a single counterexample, I was confident that no counterexample existed, and set out about proving that, instead of looking for counterexamples or symmetry.

When you are stuck on such a math problem — that is, after making a bit of progress you get stuck — it might be profitable to start over. More often than I would like, I prove a series of neat things, without being able to prove the desired result. Then a look at the solutions manual reveals that a very short solution — one or two steps — lay in the opposite direction.

I’ll put an end to my philosophical musings of the day. Fortunately, the cutoff for the CMOQR was low enough that even without solving every single problem, I was still able to meet the cutoff.

Usually this is not much of a problem — the missing steps become obvious when you sit down and think about what’s going on with a pencil and some paper. But sometimes, this is not the case.

The problem

One of the worst examples I’ve seen comes in the book Inequalities, A Mathematical Olympiad Approach. By all means, this is an excellent book. Anyways, here’s one of its easier problems — and you’re expected to solve it using the triangle inequality:

Prove that for all real numbers a and b,

Attempt 1: Intuitive solution

It isn’t clear how the triangle inequality fits. If I weren’t required to use the triangle inequality, I might be tempted to do an intuitive, case-by-case argument.

Let’s visualize the absolute value of as the difference between the two numbers on a number line. Now we compare this distance with the distance after you take the absolute value of both of them, . If one of the numbers is positive and the other negative, we clearly have a smaller distance if we ‘reflect’ the negative one over. Of course, if they’re both positive, or they’re both negative, then nothing happens and the distances remain equal.

There, a simple, fairly clear argument. Now let’s see what the book says.

The book’s solution

Flip to the end of the book, and find

Consider and , and apply the triangle inequality.

Huh. Perhaps if you are better versed than I am in the art of solving inequalities, you’ll understand what this solution is saying. But I, of course, had no idea.

Maybe try the substitution they suggest. I only see one place to possibly substitute for anything — and substituting gives . Now what? I don’t think I did it right — this doesn’t make any sense.

To be fair, I cheated a little bit in the first attempt: I didn’t use the triangle inequality. Fair enough — let’s solve it with the triangle inequality then and come back to see if the solution makes any sense now.

Attempt 2: Triangle inequality solution

A standard corollary to the triangle inequality of two variables is the following:

Combine this with the two variables switched around:

Combine the two inequalities and we get the desired

Now let’s look at the solution again. Does it make sense? No, at no point here  did we do any substitution. Clearly the authors were thinking of a different solution that happened to also use the triangle inequality. Whatever it was, I had no idea what the solution meant.

The book’s solution, decrypted

Out of ideas and hardly apt to let the issue rest, I consulted help online at a math forum. And look — it turns out that my solution was without a doubt the same solution as the book’s intended solution!

What the author meant was this: considering that , we have from the triangle inequality. Then, moving the over we get .

After that, the steps I took above are left to the reader.

Perhaps I’m a bit thick-headed, but your solution can’t possibly be very clear if a reader has the exact same solution yet can’t even recognize your solution as the same solution. Come to think of it, if I couldn’t even recognize the solution, what chance is there of anybody being able to follow the solution — especially if they’re new to inequalities?

Almost every one of the one-sentence phrasings of this solution I could think of would be clearer and less puzzling than the solution the book gives me.

If you use Digsby with Facebook, you might have noticed that things behave strangely — the program pops up a window looking like this when it tries to connect to Facebook:

Then after you give it your credentials, Digsby still thinks you’re not logged in, and so on.

If you found this page via a google search, there’s a simple hack / workaround you can use to patch up this problem. Basically, instead of using the Facebook protocol to connect, we let Digsby use the Jabber protocol as a ‘proxy’ to connect to Facebook:

1. Go to Digsby -> My Accounts and in the Add Accounts section at the top, select the Jabber icon.
2. You should get a window that looks like this:
   
3. In the Jabber ID box, put your.id@chat.facebook.com, and in the password field, put your facebook password. For example, if your facebook page is at facebook.com/yourname, your Jabber id is yourname@chat.facebook.com.
4. Remove the facebook account from Digsby

At this point, you’re done: Digsby should give you no more problems about Facebook.

Warning: the following is unnecessary and experimental! It might screw up the entire Digsby installation, forcing you to reinstall!

However, you can replace the Jabber icon with the Facebook one (this is for purely cosmetic purposes):

1. Go to C:\Program Files (x86)\Digsby\res\skins\default\serviceicons (that’s the default installation path on my machine, yours may be different)
2. Delete jabber.png, duplicate facebook.png, and rename it jabber.png
3. Restart Digsby

There you have it — hack accomplished:

AB and CD are harmonic conjugates if this equation holds:

If you’re like me, you’re thinking along the lines of “But why? Why is this defined this way? Why would we spend so much time proving things about this weird concept? What’s the point, what’s the use?”

Even now, I can’t really give you an intuitive explanation of why this equality is so important. On the other hand, I could certainly come up with a few problems in which the concept of the harmonic conjugate turns to be useful.

Apollonius and Fleeing Ships

Apollonius’s problem was this: you are in control of a ship (point A on diagram), and you are in pursuit of another ship (point B). The other ship is fleeing in a straight line in some direction:

Your speed is (obviously) faster than the speed of the other ship: say they’re going at 30 km/h and you’re going at 50 km/h. Additionally, your ship is required to go in a straight line.

In which direction should you set off in order to intercept the fleeing ship?

Solution with Harmonic Conjugates

The first step of the solution is to construct harmonic conjugates CD so that their ratio is the ratio of your speed to the other ship’s speed (we’ll prove later that this is actually possible; assume we can do this for now):

Next, draw a circle with diameter CD:

There is a point where the ray from B (their ship) intersects this circle. Now go to this point immediately, in a straight line: the ship will be there.

The Proof

In order to prove that this works, we’ll need to take a step back and look at how we constructed the points C and D. The solution turns out to be evident directly from the construction of the harmonic conjugates.

Again, let’s assume our desired ratio is 5/3. Starting with the points A and B, the first step is constructing some point P so that:

This is fairly easy to do. Draw a circle of radius 5 around A, and draw a circle of radius 3 around B — the intersection P of these two circles forms the correct ratio. (if the circles don’t intersect, just scale everything up and try again)

Next, dropping the internal and external angle bisectors of the new triangle gives the harmonic conjugates C and D:

Why angle bisectors? From the angle bisector theorems (which I won’t prove here):

Combining the two proves that C and D are indeed harmonic conjugates to AB.

As a corollary, notice that because of angle bisecting, the angle CPD is always a right angle — hence, the locus of all points P forms a circle with diameter CD.

Returning to the ship problem, since each point P is defined as a point so that , it follows that when both ships travel to such a point P, they will meet at the same time.

You have a 20×20 board and some pieces (which you choose yourself and get to set up wherever you want, with some restrictions). You’re trying to kill all of your opponent’s pieces, which you have to do in order to win.

Each piece has a certain move radius — so it can move to any square within that radius (diagonals only counting as a distance of one). Then, once it has moved, it is allowed to attack an enemy within its range — again varying from piece to piece.

That’s the gist of the game. Oh yes, at certain times you can also use some silly special attacks in addition to moving and attacking — for instance, freeze an enemy piece for one move or temporarily boost the attack of one of your own pieces. And these special moves can be used on any pieces during your move, and in any combination.

A Conquest AI… or not

So some months ago, I was playing Conquest. I was a decent player, but after a string of losses, I thought: maybe I can write an AI to play this game for me — perhaps better than a human can? Perhaps the same way you would write a chess AI?

Well, I scribbled some back of the envelope calculations, but I quickly realized that the number of possible moves at each position was far too large.

How large? A player has maybe 5 pieces, and if a piece’s average move radius is 4, then we already have 5*9*9 (remember 4 is the radius, not diameter) which is over 400. And that’s if we never attack or use special moves. If we take special moves into account, since special moves can be stacked on top of each other, assuming 5 valid targets per special move and 3 special moves we would multiply the 400 again by 5*5*5. For one move, we would need to consider 50000 possibilities.

In comparison, from any position in chess there are typically 30 or 40 legal moves. So the standard minimax algorithm would probably be no good, at least not without an incredible amount of heuristics. I had to try something different for this to work.

Random has a chance

It turns out that there exists a very different strategy that worked quite well for Go-playing AIs. Instead of constructing and evaluating a huge search tree to find the best move, we simulate thousands random games, with random moves from each side. The move that gives us the best win percentage is played. All else being equal, a position that gives you a 90% chance of winning given random play from that point on is usually better than one with only a 30% chance of winning.

Here’s a screenshot of a connect-4 AI using this technique:

The Monte Carlo approach, as that’s what it’s called, seems absurd — there’s no way playing random moves can beat a good minimax! Indeed, Monte Carlo is never used to play chess, and I could sometimes beat the above AI. But where Monte Carlo really shines is when standard minimax is impractical — my scenario.

After finding and reading a few papers describing the method in detail, I was ready to start coding. However, there is still one more catch: to make it practical to set up the AI against other players over Runescape (and where the time limit for each move is usually 30 seconds), I would have to find a way to integrate the back end — the part that does the computations — with a front end that relayed the moves back and forth to the Runescape game client.

All is not lost, however: there were several freely available hacked clients for botting purposes: Powerbot allowed for users to write java scripts that automated tedious ingame actions.

That was about as far as I got though, unfortunately. Coding the game logic proved trickier than expected. A month or so after I started my account was banned for botting; later, Jagex (the company behind the game) made an update that rendered Powerbot (as well as every other similar hacked client) unusable. Without a game account nor a practical front-end, I called it quits.

• If some digit in Y coincides with a digit in X (in the same position), then the guesser scores an A.
• If some digit in Y exists in X but is in the wrong place, then the guesser scores a B.

For instance, if X is 1234 and we guess 2674, we get an A and a B, because the 4 is in the right place, and the 2 is one of the right numbers but isn’t in the right place.

This proceeds until the guesser gets the exact number.

When humans (or at least beginners) play the AB game, they usually do some form of uncoordinated trial and error, which gets the right answer after some large number of moves. This takes anywhere from about 8 to possibly 30 guesses. When I played this game with a friend, I didn’t have a very systematic strategy, but I wondered if a computer program could solve the game, always entering the optimal guess.

A Bruteforce Solver

My first approach happened to work fairly well. Simply, the computer keeps track of a list of all possible numbers that the answer can be. At random, the computer guesses one of the numbers, and upon receiving feedback, eliminates every number in its list that doesn’t match that feedback. Quickly it eliminates whole swaths of combinations and arrives at the answer.

A typical session might go like this:

Guess 1: 6297. Score: 5040
b
Guess 2: 8512. Score: 1440
abb
Guess 3: 2315. Score: 83
bb
Guess 4: 5842. Score: 29
bb
Guess 5: 9581. Score: 13
ab
Guess 6: 8021. Score: 1

It took only 5 guesses for the computer to narrow down the choices to the only possible answer (8021).

A variant that is usually much harder for humans to solve is to allow the number chosen to have repeats. Although significantly trickier for humans to guess by trial and error, brute force doesn’t seem to be affected too much by it:

Guess 1: 1796. Score: 10000
b
Guess 2: 5881. Score: 3048
a
Guess 3: 0131. Score: 531
abb
Guess 4: 3311. Score: 15
aa
Guess 5: 3201. Score: 9
abb
Guess 6: 2011. Score: 1

The computer’s average of five or six is much better than a human can normally do! (although I haven’t researched possible human algorithms).

Optimal or Not?

At this point you may begin to wonder if this strategy is the optimal one. Unfortunately, it is not — and I only need one counterexample to demonstrate that.

Suppose that instead of four numbers, you were allowed to choose 4 letters from A to Z. You choose the combination ABCW. Now suppose that the computer ‘knows’ that the first three letters are ABC — that is, it has eliminated all other combinations except for ABCD, ABCE, …, ABCZ.

By the random guessing algorithm, the computer is forced to guess ABCJ, ABCP, etc, until it eventually hits on ABCW at random. This may take a very high number of guesses.

A smarter strategy would be to guess combinations of four unknown letters, say DEFG, then HIJK, etc. Instead of eliminating one letter, you eliminate four letters at a time. Although some guesses have no chance of being correct, the number of guesses required is fewer in the long run.

Source Code

I’ll put my somewhat unwieldy java code here for all to see:

import java.util.*;

public class ABSolver{
  
  // Does cc1 and cc2 together match the pattern?
  // If repeats are allowed, cc2 is matched against cc1.
  static boolean fits(char[] cc1, char[] cc2, int A, int B){

    int a = 0;
    int b = 0;
    for(int i=0; i<4; i++){
      if(cc1[i] == cc2[i]) a++;
      else{
        for(int j=0; j<4; j++){
          if(i != j && cc1[j] == cc2[i] && cc1[j]!=cc2[j]){
            b++;
            break;
          }
        }
      }
    }

    return a==A && b==B;
  }

  public static void main(String[] args){
    
    Random rand = new Random();
    Scanner scan = new Scanner(System.in);

    // Combinations that haven't been eliminated yet
    List<char[]> ok = new ArrayList<char[]>(10000);

    // Generate all possible combinations
    for(int i = 0; i <= 9999; i++){
      
      String i_s = Integer.toString(i);
      while(i_s.length() != 4) i_s = "0" + i_s;

      // Check for sameness
      char a = i_s.charAt(0);
      char b = i_s.charAt(1);
      char c = i_s.charAt(2);
      char d = i_s.charAt(3);
      boolean same = a==b || a==c || a==d || b==c || b==d || c==d;

      // Comment this line out if we're allowing repeats
      if(same) continue;

      char[] i_ia = i_s.toCharArray();
      ok.add(i_ia);

    }

    // Pick the first guess randomly
    char[] firstg = ok.get(rand.nextInt(ok.size()));
    char[] guess = null;

    int nguesses = 1;

    // Question answer cycle until we get it right
    while(true){

      // Ask for user response
      if(nguesses > 1){
        String ans = scan.nextLine();
        int A = 0;
        int B = 0;
        for(char cc : ans.toCharArray()){
          if(cc == 'a') A++;
          if(cc == 'b') B++;
        }

        if(A==4) return;

        // For each one check to see if it still fits
        List<char[]> ok_ = new ArrayList<char[]>(ok.size());
        for(char[] zhe : ok){
          if(fits(zhe, guess, A, B))
            ok_.add(zhe);
        }
        ok = ok_;
      }

      char[] nextguess = null;
      if(nguesses == 1)
        nextguess = firstg;
      else nextguess = ok.get(rand.nextInt(ok.size()));

      System.out.println("Guess " + nguesses + ": " + new String(nextguess)
                       + ". Score: " + ok.size());
      
      // we win!
      if(ok.size() == 1)
        return;

      guess = nextguess;
      nguesses++;
    }
    
  }
}

The angle between any two edges of the tetrahedron is , so it’s easy to (falsely) conclude that the angle between two faces must also be .

But this isn’t the case: the plane angle is defined as the angle between two lines on the two planes that are both perpendicular to the edge. None of the edges in a tetrahedron is perpendicular to any other, so the answer of is invalid.

We can try to compute the angle using regular Euclidean solid geometry, but things tend to get messy. A different way to approach the problem is by using vector geometry: using vector methods we can easily calculate the plane angle of the tetrahedron (as well as the icosahedron and the dodecahedron).

Assume symmetry. We represent three concurrent edges of the polyhedron as three vectors beginning at the same point: , , and ; let and be the angles between the vectors (by symmetry we’re assuming that the two alpha’s are equal):
(in case my poor drawing skills does not make this apparent, vectors a and b form one face of the polyhedron and c and b form another face)

For simplicity, let’s also say the lengths of each of the three vectors is 1.

We want to compute the angle between the plane formed by and , and the plane formed by and . Hence let and be the perpendiculars to and respectively each ending at the same point as their respective vectors:

For any two vectors, the dot product is defined with being the angle between the vectors. Given that the lengths of the vectors are all 1, we have:

Also, by vector addition:

Hence and .

We want to find the angle between and — call this angle . Then

The dot product of vectors x and y is simply:

Additionally . Hence the cosine of the angle is:

We can now use this newly derived formula to calculate plane angles! For example…

Tetrahedron

In the tetrahedron, both and are 60:
So  and .

Icosahedron

In the icosahedron, but :
The top ‘cap’ is a regular pentagon, which has a vertex angle of 108; each side of the pentagon constitutes a side of an equilateral triangle. Since , , and .

Dodecahedron

Computing angles for the dodecahedron works a bit differently from the tetrahedron and icosahedron. Instead of using existing edges as vectors, we construct an equilateral triangle by connecting three vertices:

So (since it’s part of a regular pentagon) and . Then and .