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DeamonMonkey
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icon Could a Computer "Learn" to Play DROD? (0)  
Having just re-read '2001 A Space Odyssey', and '2010 Odyssey Two', I got to thinking about computers and learning. And DROD, like usual.

One of the points about DROD is that you don't need to bring any outside knowledge to it, all the information you need is present in the room your in. I think I saw that somewhere, but couldn't find it again... But anyway I've found that's true, and it sounds real nice.

So. My point is, well I don't really have a point, but I was wondering if a program could be created that would teach itself to play DROD.

It could learn just like you and I did. Through experimentation and remembering the actions or monsters and other objects. I'm thinking of when I first met a Goblin and had to figure out how to manipulate and kill it.

Or, what about a study that studied (because that's what studies do!) how a person learned to play DROD. You could put micro chips in someone's brain like they did to that chimp who learned to manipulate a mechanical arm with just it's brain. That was cool, but DROD is much cooler.

Now I'm rambling.
05-30-2005 at 05:54 AM
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Mattcrampy
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It's entirely possible. Building systems that learn is certainly doable, and I'd expect that a computer could learn the moves and come up with an efficient move balancing and pruning algorithm from that.

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05-30-2005 at 09:19 AM
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schep
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Possible, sure, but extremely hard. I would bet the problem of whether a given DROD room is conquerable is NP-complete, if we imagine that room size is variable. See some Wikipedia comments on machine learning, and another DROD.net topic where similar questions have been discussed.
05-30-2005 at 01:45 PM
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AlefBet
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It's definitely in NP. I'm not sure whether it's NP complete or not, but it certainly could be at least NP intermediate. It's not at all obvious to me that it would be in P, so it could be quite difficult to solve.

Writing a program that could "learn" to solve DROD rooms would certainly be possible (pretty straightforward, actually) but the way I'm thinking of it, it would probably take a ridiculously long time to solve each room.

This is made trickier by the fact that a lot of DROD rooms have some trick or theme to them or something you have to notice or catch (like tar's going to grow into an unsolvable shape after 120 turns or you need to leave a trapdoor path to get out). These can be tricky for humans, but we're pretty good at at least spotting there's a trick. Computer learners are often quite oblivious to such.

Edit: I just realized that I'm not entirely certain it's in NP since I'm not sure I can prove that the solution can't be exponential in the size of the room. Someone may be able to prove that, though, and then I'm pretty sure we could prove it's in NP.

[Edited by AlefBet at Local Time:05-30-2005 at 04:44 PM]

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05-30-2005 at 04:40 PM
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schep
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AlefBet wrote:
I'm not sure I can prove that the solution can't be exponential in the size of the room.
What does this mean? The Longest Room contest thread has several hold concepts for which the shortest solution is exponential in room size. See especially the "Eternity" series and my own "Eternal Script".
05-30-2005 at 06:40 PM
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silver
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Nevertheless, if we hooked each square of a room up to a neuron and added an intermediate layer of neurons, gave a small positive feedback for killing things, dropping red or black doors, dropping black doors, and a BIG positive feedback for exiting the room with everything dead, and BIG negative feedback for getting trapped or dying or having the room state be unchanged for a longish time... who knows?

EDIT: actually, a neural network might not do too well because the outputs are local based on relative position, but the inputs I suggested would be global. and I suspect it would EXPLODE on a multi-room puzzle.


[Edited by silver at Local Time:05-30-2005 at 06:52 PM]

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05-30-2005 at 06:43 PM
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schep
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Oh, here we go. The Satisfiability Problem is known to be NP-complete. It's rather simple to represent literals using orbs with toggle connections, and a conjunctive normal form in a grid of walls and doors. The total area used would be O(the length of the logical statement). I get a bit confused with some of the terminology in this topic, but that has to prove something interesting, right?
05-30-2005 at 06:51 PM
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AlefBet
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schep wrote:
Oh, here we go. The Satisfiability Problem is known to be NP-complete. It's rather simple to represent literals using orbs with toggle connections, and a conjunctive normal form in a grid of walls and doors.
That's convincing to me. A SAT construction seems pretty straightforward to me, so DROD is NP-Hard (NP complete or worse).

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05-30-2005 at 07:18 PM
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DeamonMonkey
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There must be a few Math Majors here...

You guys lost me, can you explain in layman terms, NP, NP-Complete, and Polynomial Time?
05-31-2005 at 05:33 AM
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eytanz
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Well, I'm neither a math or a computer science major, but I'll try to explain how I understand these terms and then the more knowledgable people can correct my mistakes:

Polynomal Time essentially means "reasonably fast". If a problem can be solved in polynomal time, that means its worth writing a program to do so.

NP problems are problems for which, given a putative solution, you can check whether it's right in polynomal time. Note that this doesn't mean the problem can be solved in polynomal time.

NP complete means that it's an NP problem that probably (though not certainly, there's no proof of this yet) can't be solved in polynomal time.

Solving a DROD room is NP complete because it's easy to write a program that verifies whether a given solution works or not (e.g., the demo integrity checker included in DROD), but it's probably impossible to write a program that can solve arbitrary DROD rooms (because, as Schep pointed out, it's possible to write DROD rooms that include problems that it's already been proven are NP complete)

[Edited by eytanz at Local Time:05-31-2005 at 06:33 AM]

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05-31-2005 at 06:33 AM
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AlefBet
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Eytan, that's very correct for a non-computer scientist/non-mathematician. It's actually pretty correct for a computer scientist, too. I'm impressed. There are only two fine points where it's not quite accurate, but they're minor nits and many CS people probably wouldn't notice them either. I'll mention them now to both try to be complete and to try to look smarter than I actually am :blush.

First, NP complete isn't exactly synonymous with "can't be solved in polynomial time." (I'm assuming P!=NP here, because everybody does anyway.) There are some problems that are NP-intermediate, which aren't NP complete (since NPC problems can't be reduced to them) but which also aren't in P. I think that integer factoring is believed to be one of these, but don't quote me on that.

Second, DROD is probably harder than NP complete since apparently the solution can be exponential in the size of the problem, and to be in NP the solution verification has to be a polynomial process in the size of the problem. This means that the reduction can't go both ways, since DROD couldn't be polynomial-reduced to Satisfiability.

Hmm, now that I've written this out, you could have just as well not have mentioned them because they're confusing and not important. Oh, well. I've typed it all out anyway.

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05-31-2005 at 07:08 AM
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DeamonMonkey
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Thanks eytanz.

You've probably already been talking about this, and I just missed it. But what I envisioned was a program that would "guess" a move, and record Data from the movements of monsters. Then "guess" the next most logical move determined by the data it already possessed.

05-31-2005 at 05:46 PM
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DeamonMonkey wrote:
Thanks eytanz.

You've probably already been talking about this, and I just missed it. But what I envisioned was a program that would "guess" a move, and record Data from the movements of monsters. Then "guess" the next most logical move determined by the data it already possessed.
Sure, that's a straightforward way to go about solving a room. The problem arises when you've guessed a large sequence of moves and the room hasn't been solved yet. Do you keep the program running? Has the room already become impossible to solve and the program just doesn't know it? Or, even assuming the program could always know when a room is no longer solvable, what should it do when it reaches that dead end? Back up a move and try something different? Sure, but that's where it becomes impractical to try to solve rooms. You'll have, in the immortal words of Carl Sagan, billions and billions of different paths to try out, and maybe none of them will work. How many days, weeks, months, or years do you let the program keep searching for new ways to solve a room before you let it give up?

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05-31-2005 at 05:57 PM
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Malarame
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Following this discussion, I'm reminded why I chose to become an engineer rather than a computer scientist or math major. I don't think I've been this lost since Numerical Analysis last semester.

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05-31-2005 at 06:52 PM
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Hm... if the average room would require 500 moves to solve, it could be brute-forced roughly in less than 0.48 e30 (less than since a lot of moves are instantly invalid). Let's assume half that, or 0.24 e30. Given a 10 GHz computer and assuming 1k cycles per move (which is probably overdoing it, except when brains are involved) that gives us 0.24 e24 seconds, or about 7.5 trillion years.

Which goes to show that brute forcing is a patently bad idea :)

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05-31-2005 at 09:32 PM
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DeamonMonkey
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Malarame wrote:
Following this discussion, I'm reminded why I chose to become an engineer rather than a computer scientist or math major. I don't think I've been this lost since Numerical Analysis last semester.

It's also making me glad that I'm going to be an engineer.
06-01-2005 at 07:23 AM
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ArchMageOmega
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Lol. I'm double majoring in computer science and math. (But then again, I enjoy this kind of stuff.)
06-02-2005 at 08:20 AM
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eytanz
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AlefBet wrote:
Eytan, that's very correct for a non-computer scientist/non-mathematician. It's actually pretty correct for a computer scientist, too. I'm impressed. There are only two fine points where it's not quite accurate, but they're minor nits and many CS people probably wouldn't notice them either. I'll mention them now to both try to be complete and to try to look smarter than I actually am :blush.

Gee, thanks :blush

The hardest part actually was figuring out that "polynomal time" means fast. It doesn't sound fast, it sounds like a lot of time. It's only when I realized that the alternative must be exponential time or something like that that I figured out how this scale works. I assume there's also "linear time" which probably means "too fast, add a talking paperclip to slow the program down" or something.

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06-02-2005 at 02:49 PM
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masonjason
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Linear time is polynomial time. A linear expression is a polynomial of degree one. Infact linear time can be longer than any polynomial you like if your linear polynomial is suitable... for example, 1000000x grows faster than x^2 for quite a while.

(disclaimer: I'm a mathematician, not a computer scientist, so I could be wrong about some computery details.

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06-02-2005 at 08:33 PM
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eytanz
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But by that logic, can't you say that polynomal time is exponential time with an exponential degree of 1 as well?

x^2 = 1^x * x^2.

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06-02-2005 at 08:49 PM
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AlefBet
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Linear problems are interesting, too. All "regular problems" (problems that can be solved with a finite state automaton) are linear in time (and constant in space), although there are many linearly bound problems that are not "regular". If you can show that a problem is linear, that's significant. However, many interesting problems are provably harder than linear. Sorting (without assumptions) is O(n*log(n)), many graph based problems are in the neighborhood of O(n^2) or O(n^3), and dense matrix problems are practically O(n^3). (Technically, there are matrix operation algorithms better than O(n^3), but their overhead typically makes them impractical.)

For practical algorithms, generally people draw the line around O(n^2) or O(n^3), although there are some machine learning algorithms that are in the neighborhood of O(n^5). The great thing about polynomials, though, is that there are many ways you can combine them and still end up with polynomials. This is convenient in complexity theory (the computer science study of what is hard to compute), because combining two or more algorithms to make a new algorithm solve a more complicated problem often corresponds with combining their computation time. Combining two linear algorithms can often give you a quadratic one (or another linear one, depending on how they're combined). And, even moderately high polynomials still have a somewhat palatable rate of growth, whereas exponential problems get ridiculus fast.

So, computer theory generally draws the line of "tractable" at "polynomial complexity". (In case you were curious.)

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06-02-2005 at 08:55 PM
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AlefBet
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eytanz wrote:
But by that logic, can't you say that polynomal time is exponential time with an exponential degree of 1 as well?

x^2 = 1^x * x^2.
This isn't exponential complexity, since the base of the exponent isn't greater than 1. A truly exponential function can't be bounded by a polynomial as x becomes large, but this one is bounded by 2*x^2, which is a polynomial.

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06-02-2005 at 08:59 PM
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Jacob
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schep wrote:
Oh, here we go. The Satisfiability Problem is known to be NP-complete. It's rather simple to represent literals using orbs with toggle connections, and a conjunctive normal form in a grid of walls and doors. The total area used would be O(the length of the logical statement). I get a bit confused with some of the terminology in this topic, but that has to prove something interesting, right?

I'm a little confused. I started working through how you'd do this in my head and thought k=number of doors an orb can toggle, m=number of doors and n=number of orbs (using the k,m,n of the satisfiability problem) but then I'm kind of stuck. Cos what constitutes a solution. Something along the lines of: there's a path of open doors through the grid, or what? If someone could make an example hold with small k,m,n or just explain briefly, that would be great. Sorry if my explanation here is nonsense, I'm not a mathematician, I was only discussing it with one who doesn't play drod.

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08-07-2005 at 10:50 AM
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Jacob wrote:
If someone could make an example hold with small k,m,n or just explain briefly, that would be great.

Here's approximately what I was thinking. The attached hold represents the expression
(A v !B) ^ (B v !C) ^ (!A v C) ^ (!A v !B v !C)

Since the expression is not satisfiable, the exit stairs cannot be reached. I used more area than was strictly necessary to make my procedure more obvious, for area equal to O((number of literals) * (number of disjunctions)), but I think my previous claim works out if you compact the grid vertically. I only mentioned expression length because I wasn't sure what's normally meant by calling satisfiability non-polynomial (in what variable?)
08-08-2005 at 03:13 PM
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AlefBet
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schep wrote:
I only mentioned expression length because I wasn't sure what's normally meant by calling satisfiability non-polynomial (in what variable?)
If you're talking about whether a problem is polynomial solvable, that typically means that the computation needed to find the solution is proportional to (actually, bounded by) a polynomial in terms of the length of the problem's formulation (often called the "input"). The length of the input depends on how it's encoded, but when you're talking abut polynomial versus non-polynomial there are usually a wide variety of encodings that more or less give you the same answer. So for satisfiability, you could roughly state that it's verifiable in polynomial time in terms of the number of terms in the expression, but not solvable in polynomial time. (The "solvable" part is not strictly proven, but is widely believed to be true.)

As a point of trivia, in computer theory the satisfiability problem is typically limited to three variables per term. This isn't a serious limitation, though, since a problem with more variables in a term can be converted to one of the proper form that's roughly a comparable size. This proper form is called 3-SAT, and that is used instead of 2-SAT because the same conversion is not necessarily possible for 2-SAT. (In fact, 2-SAT is known to be polynomially solvable.)

Edit: Incidentally, the expression (A or not B) and (B or not C) and (not A or C) and (not A or not B or not C) is in fact satisfiable (by setting all variables to false) and your example hold shows the solution at level start. But the hold is a good representation of the expression, and gives a good idea of how it works.

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08-08-2005 at 05:17 PM
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AlefBet wrote:
Edit: Incidentally, the expression (A or not B) and (B or not C) and (not A or C) and (not A or not B or not C) is in fact satisfiable (by setting all variables to false) and your example hold shows the solution at level start.
Ha ha, good catch. :blush Oh well.
08-09-2005 at 01:11 AM
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I know this was a while ago, but thanks for that hold. I'd kind of got the wrong end of the stick in how to implement the problem.
I was discussing DROD with my dad, who's a mathematician, and he found it quite interesting, especially how it can operate as a kind of universal turing machine, i.e. other problems can be expressed in drod, such as this one, and ones in holds such as Palace of Puzzles etc. He wanted to know if: (a) Drod offered an alternative way of presenting existing problems, allowing an interesting way to look at/solve them. (b) DROD itself offered up *new* mathematical problems, not reducible to existing ones (I suggested optimisation of simple rooms, solvability of orb problems). Any ideas on these? Has anyone actually done any work into the mathematics of drod? (sorry if this is kind of off-topic)

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08-24-2005 at 02:36 PM
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Jacob wrote:
Has anyone actually done any work into the mathematics of drod? (sorry if this is kind of off-topic)
Well, there's always the article Invariants of Tar Cutting by Watcher.

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08-24-2005 at 03:40 PM
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I suppose it depends on what you're looking for. On a theoretical level, any computable problem can be reduced to the halting problem. Drod is definitely computable (although it may be exponential).

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08-24-2005 at 04:14 PM
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One of the puzzles in the Palace of Puzzles is a Sokoban problem. It has apparently been shown that Sokoban is PSPACE-complete (although I don't understand the proof), so it would follow that DROD is also PSPACE-complete, even using only door/orb puzzles. (It is obviously in PSPACE, by Savitch's theorem.)

I've written a mini-hold with around three different NP-complete problems implemented in DROD, but it's on my other computer, so I can't attach it.
09-27-2005 at 03:45 AM
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