Monday Mazes

Cumulative Maze

2006-03-29T00:56:49Z

Yeah, okay, it's Tuesday. I know. I'm late again. Sorry.

Anyway, I'd said said last week I had two more 45° mazes made that I'd save for this week. But I think instead I'll put up another new kind of maze this week, and save the new 45° mazes for some other week. So...this week's new maze type is called the Cumulative Maze.

A Cumulative Maze consists of a grid, some of the cells of which contain symbols. Each turn, you must move a number of cells equal to the number of times you've been on a cell with the same symbol as the cell you're currently in. The current cell counts, and landing on the same cell more than once adds up. For now, you can only move orthogonally--though the concept could certainly be generalized to allow diagonal directions, or other grid types, such as hexagonal. You can jump over a blank cell, but you can't land on it.

Here's an example:

So. We start in the upper left corner. Well, to begin with, let's try moving south. Since this is our first time on a red circle (obviously, since this is the first cell we've been on), we only move one square. This puts us on another red circle. This is now our second time on a red circle, so we have to move two cells. Well, moving two cells north or west would put us outside the maze, and moving two cells east would land us on a blank cell, so our only choice is to move south.

Now we're on a blue square. This is our first time on a blue square, so we have to move one cell away. Again, there's only one direction we can move without ending up either outside the maze or on a blank square, and in this case it's north. That puts us on a green triangle. This is, once again, our first time on this symbol, so we have to move one cell away--but we can move either north or south.

Let's try north. That lands us back on a red circle. This is now our third time on a red circle, so we have to move three cells. Well, the only way to do that without going outside the maze is to head east. Doing that lands us on a yellow star--another new symbol. So we move one square away, either north or south (west would take us to a blank cell, and east outside the maze). Hm...okay, let's try north.

Now we're back on a blue square. This is our second time on a blue square, so we have to move two cells. We can do that either west or south; let's try south. This lands us on a green triangle. It's our second time on a green triangle, so again we have to move two squares. This time, the only way we can do that is by going back north. This lands us back on the blue square we just left; it's now our third time on a blue square, so we can move three squares away. If we went west, we'd land back on a red circle, and we'd be at a dead end--because it would be our fourth time on a red circle, and moving four cells in any direction would take outside the maze! But in any case, it would be kind of stupid to move west, since the finish is three cells away to the south, so we can just go south and complete the maze!

Which...uh...to be honest, wasn't the intended solution; in fact, when I started writing this example, I'd expected to end up at a dead end. Oh, well. Actually, though, the solution I intended is one move shorter than this nine-move example. So see if you can find an eight-move solution...

This type of maze turned out to be much more difficult to design than I expected. (At least, to make small mazes...maybe if I'd used a bigger grid and given myself more room it would have been easier.) I had a couple of 5x5 cumulative mazes made up for this week...and they both turned out to have some drastic short cuts I couldn't get rid of. I finally managed to come up with one that didn't, though I'm still a bit unhappy with it, because I'd tried for a symmetrical arrangement of non-blank cells...but it turned out that whatever symbol I put in the cell east of Start (including a fifth symbol not found elsewhere in the maze), it led to a shorter alternate solution. Ah, well. Here's the maze, anyway:

The shortest solution is twelve moves long.

As an added bonus, here's another LAR maze; refer to this entry for the rules.

The shortest solution is 78 moves long.

I'd had another one of these for this week, too, but I was having a little trouble getting it to work out right, and I'm running late as it is...so I guess that, too, will be saved for some other week.

45° Mazes

2006-03-21T10:42:27Z

After two consecutive types of "mazes" that kind of stretch the definition, this week's new type of maze is of a more traditional sort--though it still has a set of rules I haven't quite seen used before. Here's a simple example:

And here are the instructions:

You start out heading in the direction of the arrow (southeast for all the mazes this week), and continue in a straight line until you hit a square with an octagonal ring in it. At that point, you must make a 45° turn--though it's your choice whether to turn left or right. Then you continue on in a straight line from there until you hit another octagonal ring, and so forth. You're not allowed to leave the grid--if you do, you can consider yourself to have hit a dead end.

Here are some illustrations of the rules, using the above sample maze:

Beginning in the upper left, we hit an octagonal ring pretty soon, which means we're forced to turn 45° either left or right:

Suppose we choose to turn left--so we're heading east. We immediately hit another ring, and have to make another turn:

Suppose we choose to turn right this time, so we're heading southeast. Well, again we immediately hit another ring...but as it turns out, no matter which way we go, we're going to end up going off the edge of the grid in no more than four moves. There are nine different ways we can go, but they're all dead ends:

Going off the edge of the grid isn't the only way to effectively reach a dead end, though. Suppose instead of starting out by turning left, we'd started out by heading right. This time it is possible to avoid going off the edge of the grid...but if we do, we don't end up much better off:

Sure, we haven't gone off the edge of the grid, but we've just gone around in a loop and ended up back where we started! This may not technically be a dead end, but it amounts to the same thing.

Of course, this maze does have a solution...and it lies in going the one way we haven't tried yet. We'll go left at the beginning, and then left again at the next turn...and it's not hard to figure out where to go from there:

Note that we've passed through one ring--the one in the square outlined in blue--twice. That's not unusual; in most 45° mazes you'll end up passing through many rings two or three times in the course of the solution. (Well, passing through a ring three times is rare...but it could happen!)

Now that you've seen an example, here are some mazes for you to solve. To start you off, here's an easy one: the shortest solution is 26 moves, but given that there are very few lengthy dead ends or false paths, you shouldn't have much trouble finding it:

The next maze, however, should prove a more difficult challenge:

The shortest solution to this one is 48 moves long.

I have two more 45° mazes made, but I think I'll save them for next week (particularly since it's late, I still have to run them through the computer to double-check them, and I don't feel like doing that right now). So, instead of posting those right now, in my continued attempt to post extra mazes to somewhat compensate for the months of non-updating I give you two more Jigsaw Maze grids; refer to last week for the rules. (These two mazes use the same fourteen path pieces as last weeks'.)

Jigsaw Mazes (Also, I'm Back!)

2006-03-14T10:15:11Z

Okay, yikes. It's been since November that I haven't posted anything in this community, and October was the last actual maze. Sorry about that; the end of last year was a very difficult time for me, and, well, it took some time to get things back in gear.

Anyway, I figure it's well past time I ought to get this community going again. I have a lot of partly finished mazes, or finished but untested, so it's not like I'm out of ideas. Unfortunately, I don't have much time, but I did manage to come up with a new kind of maze for this week. Like the Foursquare Maze a while back, this maybe stretches the definition of a "maze", but, well, the goal is to find a particular path from Start to Finish, so it could qualify...

The thing is that in these mazes pieces of the solution are already given. Your goal is to fit them into the grid in such a way that they form a continuous path from start to finish. There are a few restrictions:

All the pieces must be used.
Path pieces may be rotated, but not reflected.
The lines on the path pieces must form a single continuous path (so, for example, you can't use some of them to make a path from Start to Finish and just make a loop with the others).
Path pieces cannot cross or overlap (though they can occupy adjacent cells).
A path piece cannot be placed on a black square.
A path piece cannot be placed wholly or partly outside the boundaries of the grid.

Here's an example: Suppose you had these path pieces:

and this grid:

Here's a solution:

Now, for the mazes to be solved, you're going to have a somewhat larger set of pieces to play around with. To be specific, both the mazes below will use the following set of fourteen pieces:

Now, here are the grids to put them in:

Speaking of Foursquare Mazes, to start to try to make up for my long absence I want to throw in some extra mazes, so here are a few more of those, too:

Refer back to this entry for the rules.

See you next week!

More Mazelessness

2005-11-29T07:56:17Z

Whoops...I said I was going to finally upload some more mazes last week, didn't I?

Sorry. I'm just going through some very hard times right now financially, and I've got a lot on my mind, and unfortunately this community, while I'm still definitely interested in it, isn't exactly my highest priority when I'm worried about where next month's rent is coming from. But I'll try to finally update it next week with some new mazes; sorry for the long delay.

A Month of Mazeless Mondays

2005-11-15T04:28:15Z

Actually, it's been a little over a month by now, but hey, I wanted to keep the alliteration.

Anyway, it's been long enough since I've posted a maze that I thought I should make a post here just to clarify that I have not, in fact, abandoned this community. I've just been really, really busy.

I do in fact have a few mazes I am going to post. But, again, I want to run them through a computer test first to make sure I haven't overlooked any drastic short cuts. And I haven't had time to do that.

But I will. Probably not tonight, alas, but...well, I'll shoot for next Monday. Yes, that makes four consecutive mazeless weeks, and I'm not happy about that. But, well, I'll try to post five mazes next Monday to make up for it.

The type of maze I will post next Monday, incidentally, I have not decided on a name for yet. Perhaps "unison maze". Or maybe by next Monday I'll have come up with a better name for it.

Anyway, sorry for the lack of posting. If all goes well, in the near future I should be in a position to have rather more free time, and this shouldn't happen again.

LAR Maze

2005-10-11T09:03:01Z

Hm. I think I yet again made the mistake of making too long and complicated a solution path, therefore leaving too little room for dead ends and false paths and actually making the maze rather easier than I intended. So I tried to make a second maze with the same rules to correct that, and, um, did the same thing with the second maze too. So, what the hey; these mazes are pretty easy, but maybe later I'll make some other, more difficult mazes using the same rules.

Anyway, here are this week's mazes:

And here are the rules:

Each space in the maze--except the Start and Finish spaces--either is blank or contains either an L, an R, an A. These letters stand for "Left", "Right", and "straight Ahead" (I used "A" instead of "S" to avoid confusion with the Start space), and direct which direction (if any) you have to turn when you get to it. Note that these are relative directions to the way you're facing--so, for example, if you enter a space with an "L" on it heading north, you have to leave it heading west; if you enter an "L" space heading west, you leave it heading south; and so on. You can leave a blank space in any direction, except the direction you just came from. If you end up outside the boundaries of the maze, you've hit a dead end--you have to start over. You can leave the Start space in any direction (well, any direction that doesn't take you outside the maze).

For example: In the maze on the left, suppose we leave the start space by heading east. This takes us to an R space, so we have to turn right, which means moving south. This takes us to an A space, so we move straight ahead--still south. This brings us to another A space, so it's south once again. Now we're at an R space, so we turn right--which means we move west. Now we're at a blank space--so it's our choice of directions! Unfortunately, no matter which way we go from here (except east, which is illegal because that's the way we just came from), we end up outside of the maze either immediately or after the next move--so it looks like this is a dead end. So apparently on the left-hand maze we have to start out by moving south. You can take it from there.

The solutions to the two mazes have 94 and 88 moves, respectively.

Foursquare Maze

2005-10-04T11:55:47Z

Hm...calling this a "maze" may be kind of pushing it. Still, the goal is to find a specific path from Start to Finish, so I guess it could qualify.

Anyway, here's the maze:

And here are the instructions:

Your goal is, as I already mentioned, to find a path from Start to Finish. However, there are some specific criteria the correct path must meet. First, it must follow the dotted blue lines. It may cross the solid black lines, but not retrace them. Likewise, it may cross itself, but not retrace itself. Finally, the completed path, along with the solid black lines, must divide the grid into separate cells each of which contains exactly four squares.

In case that last requirement isn't clear, here's an example:

Take this maze, subject to the same rules:

In this case, the unique solution is the following:

Note that the blue path is a continuous path from Start to Finish that crosses itself at one point and crosses a solid black line at another, but never retraces either, and that, as required, the grid is divided into nine cells each containing exactly four squares.

Of course, the first maze above is substantially larger, so there will be more cells (twenty-five, to be exact), but the same principles apply.

Equation Mazes

2005-09-27T08:54:34Z

Okay, I confess: I didn't get to verify this one by computer yet. I actually started to, but...the danged program took way too long to run. There probably was a way I could have simplified it, but, well, anyway, it's almost midnight as it is, and I don't want to be late with the maze this time, so...here it is. There may be a shorter alternate solution. If you find one, congratulations.

Well, anyway, I'm throwing in two mazes this week, so hopefully at least one of them doesn't have a short-cut solution. So here are the mazes:

Each turn you can jump to another square any distance in any orthogonal direction except the direction you just came from. Your goal is to spell out a series of valid equations; each equation must have two terms on the left-hand side and one term on the right. Thus "15+3=18" and "128*1=128" would be valid equations, for the purposes of solving the maze. "1=1", "17-1+6=22", and "18/3=40-34" would not. When an equation is complete, the result (the number on the right-hand side) becomes the first term (on the left-hand side) of the next equation. To finish the maze, you must end an equation on the finish square. Passing through the finish square in the middle of an equation is permissible, but does not solve the maze.

For example, you could get started on the first maze like so: From the 1, you can jump east to the 8, then north to any of the division signs, then north again to the 2, west to the equal sign, and south to the 9. You've just spelled out the valid equation 18/2=9. From the 9, you can now go east to the multiplication sign, south to the 3, west to the 7, north to the equal sign, right to the 3 (note that this is the finish square, but we're not at the end of an equation, so the maze isn't solved), south to another 3, and west to yet another, forming 9×37=333, another valid equation. That 333 would then become the first term in the third equation--except that we can't get to any operators from here, so this turns out to be a dead end.

A few nitpicky rules, just to close some possible loopholes: There are no leading zeroes allowed on the numbers. So that means that while you can get to the finish of the first maze quickly by spelling out "18/6=03", the leading zero on the "03" means this isn't a valid solution. Also, the operators count only as operators; you can't use the minus sign to signify a negative number, or the division sign to create a fraction. All the terms in the equations must be positive integers.

The intended solutions to both mazes are seven equations long. However, as I said, I didn't get a chance to thoroughly test these mazes for alternate solutions, so I won't promise no shorter solutions (very possibly much shorter solutions) exist.

Common Ground Maze #1

2005-09-21T00:06:26Z

Okay. I didn't have this week's maze up yesterday like I was supposed to. Sorry. The good news is, I already have a maze or two done for next week, so I'm a bit ahead now, and this shouldn't happen again.

The maze below is not the maze I had originally planned to have up this week. The maze I had originally planned to have up followed the same rules, but was rather larger, but I was having so much trouble getting it to work that I finally gave up for now and designed a smaller maze with those rules, figuring that would be easier. And it was, but it still wasn't easy. I had actually thought a maze like this would be quite simple and quick to design; I could hardly have been more wrong.

Well, anyway, here's the maze:

The rules are as follows: Each move, you can move to an adjacent square in any direction, except the direction you just came from. (Actually, I should clarify that: any orthogonal direction. You can't move diagonally.) On your first move, the square you move to must share at least one symbol with the square you started on. On your second move, the square you move to must share a different symbol with each of the last two squares you were on. And on every move thereafter, the square you move to must share a different symbol with each of the last three squares you were on.

So, here's an example: Suppose you start out by moving "east"--that square shares a stick-figure symbol with the start square, so that's legal. Now, the square east of here shares a spiral with the square you were just on and a stick-figure symbol with the start square, so you can move east again. But we can't move east again from here--the next square does not share three different symbols with the last three squares. Nor can we move south, for the same reason. So this is a dead end.

The shortest solution to this maze has twenty-six moves.

Another Monday (and Tuesday and Wednesday...) Gone...

2005-09-16T05:29:32Z

All right, I said yesterday that the maze that wasn't up that Monday would be up that night. Clearly, it wasn't. When I said that, I thought the maze was almost done, and that it should be no problem getting it up that night; however...

Okay, wait, stop. Actually, even if it had been, I still wouldn't have gotten it up that night, technically; that day ended up being busier than I had expected, and I didn't get time to work on it. However, I would have had it up this morning. Except that when I tried to check it on the computer, I found a major short-cut I'd missed. And when I eliminated that short-cut, I found I'd accidentally eliminated the actual solution instead. And when I patched that I created another short cut...repeat ad nauseam.

I think I've really almost got it now. But...at this point I'm thinking rather than put it up four days late, I might as well save it till next Monday.

But next week I'll do my best to finish up at least two new mazes...not so I can post them both the following Monday, but so I can have a maze ahead so if I have a busy week and fall behind I have a bit of a buffer.

Also, I've been planning to make a Monday Mazes home page where I can put the old mazes in a more convenient place than having to look through all the entries here. I probably won't have that done by this week, but I figured I might as well mention it so there's some relatively good news in this post.

Whoops...Late Maze Again

2005-09-14T18:19:14Z

So...uh...last week I said it shouldn't come down again to having to choose between finishing my comic strip or finishing a maze. Well, it kinda did, again. Only this time it's because, well, between Sunday and Monday I ended up doing five strips, four of them in color. Which...took a while. Hence no maze. (And Tuesday...well, there's a reason I planned to post new mazes on Mondays, not Tuesdays or Wednesdays. Though, in retrospect, choosing the same day as my comic site updates probably wasn't the best of ideas. Well, it's alliterative...)

I should have this week's maze up tonight, though. It's basically done (has been for well over a week); I just want to double-check it for alternate solutions. And I think I'm going to try to get another maze done, while I'm at it, so I can get a week ahead. That way even if something does come up one week that prevents me from getting a maze done in time, I'll have a bit of a buffer so I can still get one up.

Oh...this has nothing to do with mazes, but as for the reason I was doing those four extra strips? They were for the Webcomic Hurricane Relief Telethon. Check it out, if you haven't already.

EPR Maze #1

2005-09-06T07:53:09Z

All right, so, despite my assurances to the contrary, there was no maze up last week. Sorry. This has just been an extremely busy week (and, to tell the truth, after making that post on Wednesday I, um, kept forgetting to finish and post the maze). Hey, at least I managed to get my comics up on time. Apparently when it comes to a choice between the Monday Mazes and the comics, the comics take priority. However, the worst is over as far as the things that have been keeping me busy of late, so hopefully this choice shouldn't come up again.

Anyway, I suppose I could put up two mazes this week...but, uh, I won't. I do have one maze for this week, though (the one I'd intended to put up last week), and this time I took the time to double-check it by computer to make sure it had a unique solution.

The main thing that requires some explanation is the meaning of the squares with numbers in them. Each of these squares can be replaced with either a horizontal path, like so:

or a vertical path, like so:

It's your choice which, except that the like-numbered squares must have different paths in them; if one contains a horizontal path, the other must contain a vertical path, and vice versa. Once you have the right paths in the right squares, finding the path through the maze is easy--but there's only one way to fill the numbered squares that leaves a valid path through the maze!

One other thing that's probably self-explanatory, but that I'll explicitly go over just in case: Where the paths cross like so: , you can go over or under the crossing, but you can't "jump" it. That is, either of the following are valid moves:

This, however, is not:

Yeah, this maze is relatively simple; there are only 256 possible choices for how to fill in the squares. I may make a more complicated maze of this type later.

As you've probably already seen from the subject, I call this an "EPR maze". Extra points for those who realize why. ;)

Late Maze

2005-09-01T01:27:26Z

Well, there was no "Monday Maze" up on Monday this week.

Nor was there one up on Tuesday.

I didn't forget this time; I've simply had no time. Between the 17th and the 29th, I was home a grand total of about twelve hours.

Part of the problem is I've decided (largely due to 11011110's having found short-cuts in a number of past mazes) that I really need to start double-checking my mazes more carefully again. So, I do have some new mazes in progress, of some new types that I think are fairly interesting...but I want to carefully double-check them and make sure they're free of shorter alternate solutions before I post them. Anyway, combining the time I need to double-check my mazes with the time I've been away and unable to do so the last few weeks...well, it's made it quite difficult to get a new maze done (and checked).

So...it is now Wednesday, and there may be a maze up later today. But don't necessarily count on it. If not, though, there'll be a maze up tomorrow for sure.

Inconstant Knight Maze

2005-08-24T06:58:51Z

Yes, this week's maze is up on Tuesday again--and very late on Tuesday, at that. No, I didn't forget again. This week, the maze is late for the simple reason that I wasn't home on Monday. I was supposed to be--but I overslept and missed my flight, and got to spend Monday in an airport. Yay.

Anyway, here's this week's maze:

It probably goes without saying by this point that you start on the square with the S and your goal is to reach the square with the F. In fact, that so probably goes without saying that from this point on I'm not going to bother to say it anymore.

However, there are a few rules this week that are worth saying. For one, you can't land on a square with a red tint. You're free to pass through red-tinted squares during your move; you just can't end your move on one. You can only end your move on a square that is pure black or pure white. Yes, that does pretty drastically narrow your choices.

So what constitutes a legal move? Well, on your first move, you move like a chess knight: two squares in one direction, and one square in a perpendicular direction. (In this particular maze, that only leaves you one choice for your first move--two squares south and one square east.) Thereafter, for each move you move X squares in one direction and Y squares in a perpendicular direction, where either X or Y is changed by one from its value in the previous move, while the other stays the same. For instance, since on the first move X=2 and Y=1, on the second move you can choose either X=1 and Y=1, or X=3 and Y=1, or X=2 and Y=0, or X=2 and Y=2. In this particular maze, X=3 and Y=1 won't get you anywhere on your second move--every square you could reach that way is tinted red--but any of the other three choices works. Suppose you choose X=2 and Y=2. The only square you can reach that way is two squares north and two squares east. From there, you can now choose X=1 and Y=2 or X=3 and Y=2. (Or X=2 and Y=1 or X=2 and Y=3, technically, but these are, of course, equivalent to the other two choices; switching the values of X and Y has no effect.) However, neither of these choices can get you to any square that isn't tinted red. So apparently X=2 and Y=2 was a bad choice as your second move.

Hope this makes sense. The intended solution is twenty-eight moves long.

Two-Turn Maze

2005-08-17T01:14:59Z

Yes, it's Tuesday. And the thing is, I had this week's maze done yesterday--actually, I had it done Sunday morning--but I, um, kind of forgot to upload it. Anyway, here it is now:

As usual, you start on the node with the S and the goal is to reach the node with the F. The rules this time are as follows: You must go straight every third node. On the other nodes, you must turn. (It's up to you where to start the counting, though...so you can, for example, go straight through the first node you hit, and then turn the next two nodes and go straight after that; or you can turn on the first node, go straight the second, then turn on the next two, etc.) Also, see those triangles over some of the paths? Those are arrows, and they indicate one-way paths; you have to go the direction the arrows are pointing.

So, for example: say we start out by heading toward the center of the maze. Now, on our first node (besides the Start node) we have a choice whether we want to turn or not; let's decide to keep going straight. This means at the next nodes we hit we have to turn. Well, we can't turn right at the next node, because that would be going against an arrow, so we have to turn left. And now we have a choice. We can either make a sharp left or a sharp right. (A wide left would take us against another arrow.) Let's take a sharp left. Now, at the node after this, we have to go straight...and this takes us one node away from Start. You should be able to take it from here. (Or take a different path, of course.)

The intended solution is thirty-five moves long.

Three-in-One Directional Maze

2005-08-09T05:53:33Z

Whoops...I've had this week's maze done since yesterday morning, but I almost forgot to post it. Anyway, remember what I said last week about making a maze solvable through either last week's ruleset or the one from the week before? Well, I decided to go ahead and make such a maze, and here it is:

I call this a "three-in-one directional maze" for a reason. (Um...as opposed to the completely random and arbitrary names I usually give the mazes?) The goal is, of course, to get from one red circle to the other. But you can do so using either the 1-2-3 maze rules from last week or the lag maze rules from the week before. So where does the third path come in? Well, for the 1-2-3 rules, the solution is different depending on which red circle you start on. (For the lag maze rules, it isn't; you can solve it by the same path regardless of which direction you go.)

The thing I'm most dissatisfied about with this maze is the asymmetrical arrangement of the "notches" in the border. I probably could have avoided that had I planned ahead better. But I suppose that's just an aesthetic matter.

The intended solutions are 36, 45, and 74 turns long. (I won't bother to say which is which--well, except that you know the 74-move solution has to correspond to the lag maze rules, since the number of turns in any solution to the maze under the 1-2-3 maze rules has to be divisible by three (I leave the proof of this as an exercise to the reader ;) ).) But, as usual, there are quite possibly unintended shorter alternate solutions.

Directional 1-2-3 Maze

2005-08-01T21:12:22Z

This maze may look very similar to last week's maze, but the rules are entirely different. It's actually another 1-2-3 maze, except that instead of going by color we're going by direction. That is, the rules are as follows:

On the first turn, you may move one node in any direction. On the second turn, you must move two nodes in a straight line. The third turn must be three nodes long in a straight line, and then repeat: the fourth turn is one move, the fifth two, and so forth. You can move in the same direction on two successive turns, but you can't go back the way you just came. Of course, you must follow the gray paths; if there aren't gray paths connecting the nodes you want to move across, you can't move there. And, of course, you start at the node with the S, and the goal is to reach the node with the F.

Here's an example: Suppose we start by moving one move to the east. Now, we can move east or south; let's pick east again. Since this is our second turn, we have to take two moves (in a straight line). Now, from here we can only move east; we can't move south because we have to take three moves, and we can only move two nodes to the south before running out of gray path. So, all right, we'll move east. From here we can move (one node) either east or south; let's pick east, which takes us to the northeast corner of the maze. Now we can only move south, so let's do it...and, since we took one move last turn, this turn we have to take two moves. Again, we can only move south from where we end up, and this turn we have to take three moves. We're just two nodes away from the finish node now! But this turn we can only make one move. We can move either west or south; let's choose south, toward the finish node. Oops! The next turn, we have to make two moves, but we can't...so we've hit a dead end, just one node from the finish.

The intended solution takes thirty-nine turns. However, as usual, I won't guarantee there aren't shorter solutions I didn't intend.

Hm...regarding the fact that this maze looks so much like last week's: it should be possible to build a single maze that can be solved using either set of rules (with different solutions, naturally). But I guess that's a project for another time...

[EDIT: Actually, it turns out this maze can be solved using last week's rules, but the solution isn't very interesting. And last week's maze, alas, has no solution using this week's rules...as is actually quite easy to prove without going to the trouble of trying to solve it (though alas I didn't hit on the proof till after I'd already tried the hard way). [EDIT 2: Last week's maze does, however, have a solution using this week's rules if you start from the Finish and work back to Start...though it's a very short and, again, not very interesting one.]]

Directional Lag Maze

2005-07-26T06:05:13Z

Here are the rules for this week's maze: Start at the S; you're trying to get to the F (as usual). Each turn, you move along a grey path from the node you're currently on to another node. However, there is a restriction (obviously; otherwise it would be a really easy maze). You may not move in the same direction as you did the turn before last. (It's okay to move in the same direction as you did last turn...just not the same direction as the turn before that.) Also, no backtracking; you can't immediately return back along the same path you just took (though you can return to it at a later point).

For example: Suppose you start off by moving south on your first move. Well, that means you'll have to go south on your second move, too. Now, you can't go south on your third move, because you went south two turns ago (on your first move), so you have to go east. Again, you can't go south on your fourth move, because you went south on your second move, but you can go east or north from here. Let's choose east. Whoops! We can't go east again, because we went east two moves ago, but there's no other direction we can go. Looks like we've hit a dead end.

The intended solution has fifty moves. However, as usual, I won't swear that a shorter solution doesn't exist.

Phase Maze #1

2005-07-19T05:18:09Z

I'm not sure what I think of this week's maze. The rules may be more trouble than they're worth. Then again, they may not. Hm. I may experiment more with this later. (Actually, this may be another maze type that lends itself to a Flash implementation--not that I've done any mazes in Flash yet, but I've been meaning to get around to it; in fact, that's the "something different" I've been saying for a while now that I was going to do (but haven't yet gotten around to doing).)

Here's the maze:

Navigating the maze is fairly straightforward. Your path through the maze--in a spate of unusual terminology, we'll call it a "beam", for reasons that may or may not become apparent later--proceeds in the direction of the last arrow it passed over until it reaches another arrow, at which point it proceeds in the direction of that arrow. Where there are arrows pointing in multiple directions in one square, you can choose which one to have the beam follow. As usual, you start on the square marked S, and you're trying to get to the square marked F.

However, you see those white arrows with red outlines? Those are special. We'll call them "beam splitters". There, you don't choose which arrow to follow. Instead, you follow both of them--the "beam" splits in two! The beams rejoin if they both trace the same path, in the same direction. However, there's a caveat here: in order to rejoin, the two beams must have, before rejoining, arrived at the point of juncture in the same number of moves since the beginning of the "maze". Otherwise, they're "out of phase", and trying to rejoin two out-of-phase beams, well, let's say it makes the maze blow up, or something. Anyway, you can't do it. And you can't just rationalize that, well, they're following the same path, but they're not rejoining. Two beams can only go along the same path, in the same direction, if they're rejoining. Otherwise, the maze explodes. Or something. (This includes the same beam retracing its path in the same direction--that's illegal, too.) Also, all split beams must have rejoined before reaching the finish square.

This is probably kinda confusing, so let's see an example. Take this simple maze:

Here's a solution:

This solution as presented may still not be terribly clear, so allow me to walk you through it. The beam starts on a splitter, so it proceeds both "south" and "east". The east beam continues east to the northeast corner of the maze, then turns south, proceeds two squares in that direction, then turns west--and hits another splitter, splitting again into two smaller beams. (Well, two more beams; they're drawn narrower here to attempt to clarify what's going on, but the "size" of the beam has no real effect on solving the maze.) Meanwhile, the south beam goes only one square before turning east, after which it continues east all the way to the edge of the maze, then turns north, travels north one square, turns west, travels west two squares, turns south...and then after traveling south one square rejoins one of the small beams that split off from the east beam, which had gone north and then west. If you count up the moves taken to reach this point along each path, you'll find that it's eight moves for both beams: the east beam traveled six moves before splitting, and the small beam that split off from it traveled two moves before the join, for eight moves total, while the south beam, well, it traveled eight moves too. If these numbers of moves hadn't matched, then having the beams join here would have been illegal.

Meanwhile, the other small beam split off from the east beam at the second splitter goes west two squares, turns south, moves south one square, turns east, turns east one square...and rejoins the joined beam we've already discussed above. Again, if you count the numbers of moves, they match: we've already seen that the thick beam traveled eight moves before the second join, and it traveled two moves south in the meantime, making ten moves total; meanwhile, we've already said that the east beam traveled six moves before splitting, and this small beam traveled a further four moves before rejoining, making, again, ten moves total. Anyway, all the beams are rejoined by now, so the joined beam travels two more squares east to finish and...voilà.

Note, incidentally, that in the upper right part of the maze there are two beams traveling along the same path without rejoining--but they're traveling in opposite directions, so that's okay.

Got all that? Hope so, though I realize it may still be a tetch confusing.

There is, incidentally, one rule I'm still undecided on: whether or not to impose a requirement that the solution path go over all splitters. In the small sample maze, leaving out this requirement allows a much simpler alternate solution:

In the "main" maze, however, as far as I know there's no such simpler solution, and this rule is unnecessary. I could be wrong, though...so if you find a solution that doesn't pass through all four splitters, congratulations; you've found an alternate route. Now try to find a solution that does. ;)

One-Two-Three Maze #2

2005-07-12T07:33:39Z

All right, I still haven't had a chance to finish any of the more time-consuming mazes I'm working on, so here's another one-two-three maze. Actually, this could benefit from my spending more time on it, too, I suppose; again, I haven't really had a chance to do any extensive checking, and it wouldn't surprise me at all to find out there are some drastic short-cuts through the maze. But...well...this is all I had time to do this week, and I don't think it's too bad, anyway.

Rules are exactly the same as last week's maze, although we have three colors of paths to deal with now instead of just two. The intended solution is thirty-nine moves long, but as I said there may very well be short-cuts that drastically reduce that number. (Whoops...in fact, I just found a twenty-move solution. And it's quite possible that yet shorter solutions exist. I've got to start double-checking these mazes again...)

One-Two-Three Maze #1

2005-07-05T08:01:51Z

Okay, this maze was admittedly thrown together fairly quickly; I have a number of more complicated mazes I've been working on for some time, but...well, I didn't get any of them done by today. So you get this instead. Still, I think the concept is interesting, even if the maze itself might be a bit on the easy side--as I seem to so often do when creating a maze of a type I haven't worked with before, I overdid it on making a long, winding solution path at the expense of not leaving enough room to put in loops and false paths. Ah, well. I do like the concept behind the maze, as I said, and I'll probably revisit it later with a little more care.

So, here are the rules behind this maze: You start on the intersection with the S, and your goal, of course, is to land on the intersection with the F. On your first turn, you may make one move in any direction. (A "move" consists of passing along a segment of colored path from one intersection to another.) On your second turn, you make two moves, both of which must be along paths of the same color, which must be a different color from the path you moved on during your first move. On your third turn, you make three successive moves, again all along paths of the same color, which must be a different color from the path you moved on during your second move. Your fourth turn, you're back to one move again, along a path of a different color from the ones you took in your third turn.

To put it more concisely, each turn must consist of a number of moves along paths of the same color, and the number of moves on a given turn changes in the sequence one-two-three-one-two-three-etc.

There are two more details that should be specified: first, at no point may you go back in the direction you just came from--you can't move "east" and immediately back "west", etc. This rule holds whether the moves in question are part of the same turn or span two successive turns. Second, to solve the maze you must land on the finish intersection at the end of a turn. It's perfectly legal to pass through the finish intersection (or the start intersection, for that matter) in the middle of a turn, but that doesn't count as solving the maze.

Here's an example of how to move through the maze: Suppose we start out by going "east". Since we took a red path on our first turn (not that in this particular maze we had a choice about that), our second turn must consist of two moves along blue paths. We can either go "south" and then "east", or "south" and then "south". Let's choose the latter option. Now, our third turn we have to make three moves along red paths. Unfortunately, that's impossible--there's no way to make three successive moves along red paths from our current position. So apparently we've hit a dead end.

By the way, just in case it's not clear, I should specify that the place where the two blue paths touch in the middle of the maze does count as an intersection; you can move through it.

The intended solution to this maze consists of twenty-seven moves. But, again, I didn't have time to do a thorough check for alternate solutions, and it's quite possible there's a shorter solution than the one I intended. (Actually, I did find one unintended alternate solution, but it's the same length as the intended solution, and only differs from it by taking a slightly path for two moves, so...oh well.)

Obviously, there are a number of ways this concept can be extended. There's nothing special about the sequence "one-two-three" in particular; a maze could be constructed using just "one-two", or "two-three", or just about anything else. (In fact, in retrospect, in some ways "one-two-three" is a less than ideal choice, because it turns out it leads to some odd parity issues; it's quite easy to prove that half the intersections you'll only ever land on at the end of a three-move turn (though of course you can pass through them at other times), and the other half you'll never land on at the end of a three-move turn.) Also, it's quite possible to use more than two colors (which is why in the instructions I said "a different color from the previous move" rather than "the other color"). And, of course, there's no reason the mazes have to be constrained to square grids. I'll probably work with some of these variations on this theme later.

Some Assembly Required Maze #2

2005-06-28T07:49:57Z

So, I said last week the previous "Some Assembly Required" maze might be a bit too difficult, given the number of possible combinations of the cells (assuming there aren't loads of alternate solutions, which there probably are). So, this week's maze is another "Some Assembly Required" maze, but with a few fewer combinations. Only 46,080 possible combinations, instead of 82,575,360.

Rules are the same as last time, with one exception: unlike in the previous "Some Assembly Required" maze, the cells including the Start and Finish square can be interchanged with other cells, just like any other cell. Note too that the rule that the solution path must pass through all the cells is still in effect.

Some Assembly Required Maze #1

2005-06-21T06:12:36Z

The rules for navigating this maze are pretty much what you'd probably guess. You start in the lower left, on the square with the S, and your goal is to reach the upper right, the square with the F. Each move, you move one square in the direction of the arrow in the current square. If there are two arrows in the current square (as in the start square), you can choose which one to follow. You may not leave the boundaries of the maze, and you may not move onto a gray square.

Which, however, presents a problem, because as it stands the maze is rather obviously unsolvable. No matter which way you choose to go from the start square, you hit a gray square almost immediately. That's where the rest of the rules come in. See, the rules stated above are the rules for navigating the maze. But there are also rules for assembling the maze...

You've probably noticed that some of the lines between the squares are darker and thicker than others. In fact, in addition to being divided into thirty-six small squares, the maze is divided by the thicker black lines into nine two-by-two cells. These cells can be rearranged. The start cell and the finish cell must remain fixed in their current positions, but the other seven can be freely rotated and interchanged, yielding in all--uh, let me do some quick math--82,575,360 possible combinations. (Yikes! Honestly, that's more than I expected.) At least one (and possibly more than one) of these combinations produces a solvable maze. Going through all the possible arrangements may seem an impossibly daunting dask, but of course some large classes of arrangements can be ruled out immediately.

(Though actually, given the huge number of combinations, I'm now uncertain whether this maze can realistically be solved (without computer aid)...if anyone actually does try it, let me know how it goes. Maybe I should have used a smaller number of cells...seven seemed like a small number, but eighty million possible arrangements isn't so small...)

[EDIT: D'oi...I just realized there exists a fairly trivial solution (or rather, a fairly trivial class of solutions). So, new rule to eliminate these trivial solutions: the path through the maze must pass through all nine cells. Maybe some other time I'll try to devise a Some Assembly Required maze that doesn't have such trivial solutions so that rule isn't necessary...]

Tetrahedron-on-Icosahedron Maze

2005-06-14T17:24:01Z

I've discussed before the difficulties in making "rolling mazes" using Euclidean solids other than the cube. The tetrahedron, in particular, poses two difficulties. First, if you're avoiding letting one face go down (or, equivalently, one corner go up), that limits you to just rolling the tetrahedron around in a circle on the other three faces. This can be gotten around by amending the usual rolling maze rules so that the requirement isn't just to prevent one face from being down. But this leaves the other, bigger problem: when you roll a tetrahedron around on a plane, whenever it comes to the same place it'll always be in the same orientation, which removes what makes rolling-cube mazes interesting in the first place.

Previously, I've gotten around this by having some cells rotate the tetrahedron--leading to the Four-In-One Tetrahedron Maze, which remains one of my favorite Monday Mazes to date, albeit mostly for its four-in-one nature. But there are other ways to avoid the problem--one of which is, well, not rolling the tetrahedron on a plane.

There are other shapes with equilateral triangular faces, after all^*. The octahedron and icosahedron come to mind, but both of those have too few faces for an interesting maze. (Well, it could possibly be pulled off with the icosahedron, but it would still be a very simple maze.) There's the cumulated dodecahedron, with sixty faces. Or cumulated versions of various Archimedean solids. Or judicious placement of square-based pyramids on a plane could lead to a surface with equilateral triangular faces on which a tetrahedron could roll so that it didn't always arrive at the same place in the same orientation. But for this maze, I decided to just use an icosahedron with each face subdivided into four triangles^**

So here's the maze:

As for how the faces join up: well, basically, the left side wraps around and joins the right side in sort of a cylinder, and then all five of the top triangles fold in and meet, as do all five of the botton triangles. So the tips of the five top triangles all meet at a single vertex, and so do the tips of the five bottom ones. It may be a bit hard to keep track at first of the tetrahedron's orientation as it rolls around, but hopefully not too much.

The rules are similar to those of the previous tetrahedron mazes, except for the omission of the rotation cells. You have a tetrahedron with one corner colored red, one blue, one yellow, and one green. Start with the tetrahedron on the S cell, with the red corner pointing "outward", that is, away from the maze. (You can choose which way it's rotated otherwise, though only one way should lead to a solution. I'd intended to specify that the blue corner should be pointing "up" (i.e. toward the top of the maze as it's displayed on the screen), but...uh...looking at the maze now, I just realized that's not how I designed it. Oops. Maybe next time.) Each turn, you can roll the tetrahedron into any adjacent cell (including cells that are not adjacent on the maze as it's displayed on the screen but would be adjacent in the "folded" icosahedron), provided that the cell in question is either white or the same color as the corner of the tetrahedron currently pointing "out". (So for your first move, you can roll it onto any of the three adjacent cells, since they're all red and so is the corner pointing "out".) Your goal, of course, is to reach the cell with the F.

There are, of course, two non-equivalent (mirror-image) ways the corners of the tetrahedron can be colored. The maze is solvable with the corners colored either way, although of course the solutions will be different. So you can try to find both of them, if you're so inclined. The two tetrahedra are reproduced below, so you can print them out and make paper tetrahedra to roll about the maze if you want to (though I'm guessing you probably won't):

I'm not wholly happy with how this maze turned out; that long streak of white in the lower part I admit is something of a last-minute desperate measure to take one of the two solutions to completion, and I wouldn't be terribly surprised to find out it leads to a big unintended short cut or two. (Both the intended solutions have more than sixty moves, but I won't guarantee those are really the shortest solutions.) Plus, there's that unintentional issue of the arbitrary initial rotation of the tetrahedron. Eh. Oh well.

Of course, these same three-dimensional surfaces can be used for a rolling octahedron maze as well as a tetrahedron. In fact, maybe I'll make one of those later...

^*A polyhedron with all faces equilateral triangles is known as a "deltahedron". There are only eight convex deltahedra, none with more faces than an icosahedron, so the cumulated dodecahedron and the other shapes mentioned here wouldn't actually be convex--but that's not really a problem as long none of the concave angles are narrow enough to prevent the tetrahedron from "fitting" on each of its faces.

^**I did contemplate using a cumulated dodecahedron, but decided against it because I figured it might be too hard to visualize how the faces joined each other. Even with the icosahedron, I fear this may be, like the tesseract mazes, more a curiosity than a maze most people would actually be interested in trying to solve, but at least the icosahedron stands a better chance of being comprehensible enough to be solvable than I think the cumulated dodecahedron would. The ideal would be an interactive version of such a maze, done probably in Shockwave since it has 3D capabilities (though it could be done in simulated 3D in Flash), but that's currently somewhat beyond my capabilities. (Well...actually, I probably could do it in Flash, with a lot of effort, but I don't have the time to put into that. Maybe later once I'm better with Flash and have more tools. Or maybe not.)

One-Way Turn Maze

2005-06-07T07:10:29Z

I'm not particularly happy with this week's maze. Not just because the maze itself isn't all that great (though it isn't), but because it's not really an original concept. Not that no one had ever done mazes before like those I've posted here previous weeks, but at least those were new concepts to me. This one I thought was at first, but then I realized it wasn't.

See, I've had a number of ideas for mazes, but haven't had time to implement them--it's been a busy week (again). But then this week I had an idea for a maze that I figured I could make relatively quickly. How about a maze, I thought, in which the solver was constrained to move according to a cyclic pattern--first right, then down, then left, then up, and so forth? It wasn't until after I'd already finished making such a maze that I realized this was precisely equivalent to just forbidding left turns--and I'd already seen mazes that used that rule. So, eh, okay, it's been done before. But it's almost midnight; I don't have time to make a new maze from scratch; and better post this than nothing at all.

Anyway, it's not really that bad, I suppose. It's kind of easy, but I guess that's okay.

The rules should be fairly simple to understand, but I'll explain them in a little more detail anyway. The goal, as usual, is to get from Start, the space marked with an S, to Finish, the space marked with an F. You do so by moving from one square to another, with the following provisos: You can't move out of the maze; you can't move onto a black square, and, as mentioned above, you can't ever turn left. For example, take the following maze:

The solution is as follows:

And here's the maze itself:

Like I said, it's kind of easy, but if you're left wanting more to do, here's a little extra something: Both the sample maze and the "official" maze can also be solved if, instead of being forbidden to turn left, you instead aren't allowed to turn right. So you can try to find those solutions too...