This article may as well be part of a series: 12 (read this first)

A note on notation: We use the abbreviation “simple^3” in this post to refer to simple simple simple sliding block puzzles.

Warning: This article contains original research! (It’s also more papery than most of the other articles on this site.)

In the previous post I mentioned several methods that could be used to speed up the finding and exhaustive searching of sliding block puzzles, as well as a candidate for the ‘hardest’ one-piece-goal diagonal-traverse 4×4 sliding block puzzle. I am pleased to say that the various bugs have been worked out of the sliding block puzzle searcher, and that 132 has been confirmed to be the maximum number of moves for a simple simple simple 4×4 sliding block puzzle with no internal walls!

(that’s not a lot of qualifiers at all)

As the reader may expect, this blog post will report the main results from the 4×4 search. Additionally, I’ll go over precisely how the algorithms very briefly mentioned in the previous post work, and some estimated running times for various methods. Following that, various metrics for determining the difficulty of a SBP will be discussed, and a new metric will be introduced which, while it is slow and tricky to implement, is expected to give results actually corresponding to the difficulty as measured by a human.

Search Results

Running SBPSearcher on all the 4x4s takes about 3 hours (processing 12295564 puzzles), which was quite useful for fixing hidden bugs in the search algorithm. Here are the best simple^3 4×4 n-piece SBPs, where n goes from 1 to 15:

And for comparison, the move counts vs. the previously reported move counts:

Notice that all the entries in the top row of the table are either less than or equal to their respective entry in the bottom row of the table, some (such as in the case of p=7 or p=4) being much less. This is because the previous search (run, in fact, as a subfeature in the sliding block puzzle evolver) started with all possible positions and defined the goal piece to be the first piece encountered going left to right, top to bottom, across the board. As such, the search included both all the 4×4 simple^3 sliding block puzzles as well as a peculiar subclass of sliding block puzzles where the upper-left square is empty but the goal piece is often a single square away! This accounts for the 6-piece case and the 8-piece case (in which the first move is to move the goal piece to the upper-left), but what about the other two cases?

For the 4-piece case, the originally reported puzzle (see here for the whole list) wasn’t simple^3, and it isn’t possible to convert it into a simple^3 SBP by shifting blocks in less than 5 moves. Interestingly, the new ‘best’ puzzle for 4 pieces is just James Stephens’ Simplicity with a few blocks shifted, a different goal piece, and a different goal position! (There could be copyright issues, so unless stated, the puzzles in this article are public domain except for the one in the upper-right corner of this image) Additionally, the 7-piece 68-move puzzle in the previous article is impossible to solve! The upper-left triomino should be flipped vertically in place. I expect this to have been a typing error, but the question still stands: Why is there a 6-move difference?

As mentioned before, the actual stating of what constitutes a simple simple simple puzzle is as such: “A sliding block puzzle where the piece to be moved to the goal is in the upper-left corner, and the goal is to move the goal piece to the lower-right corner of the board”. Unfortunately, there’s some ambiguity as to when a piece is in the lower-right corner of the board – is it when the lower-right cell is taken up by a cell in the goal piece, or is it when the goal piece is as far to the lower-right as it can be? If we take the latter to be the definition, then additional ambiguities pop up when faced with certain problems, such as the following one often encountered by SBPSearcher:

Because of problems like these, SBPSearcher uses the former definition, which means that puzzles where the goal piece takes the shape of an R aren’t processed. (In actuality, it’s SBPFinder that does this checking, when it checks if a puzzle is in the ‘justsolved’ state). If we say that the first definition is stricter than the second, then it could be said that SBPSearcher searched only through the “Strict simple simple simple 4×4 Sliding Block Puzzles”. While I don’t think that any of the results would change other than the p=7 case, it would probably be worth it to modify a version of SBPSearcher so that it works with non-strict simple simple simple sliding block puzzles.

A few last interesting things: The 13-piece case appears to be almost exactly the same as the Time puzzle, listed in Hordern’s book as D50-59! (See also its entry in Rob Stegmann’s collection) Additionally, the same split-a-piece-and-rearrange-the-pieces-gives-you-extra-moves effect is still present, if only because of the chosen metric.

The Algorithm, In Slightly More Detail Than Last Time

There are only two ‘neat’ algorithms contained in the entire SBP collection of programs, these two in SBPFinder and SBPSearcher respectively. The first of them is used to find all possible sliding block puzzle ‘justsolved’ positions that fit in an NxM grid, and runs in at most O(2*3^(N+M-2)*4^((N-1)*(M-1))) time. (Empirical guess; Due to the nature of the algorithm, the calculation of the running time is probably very hairy).

where the numbers increase going from top-right to lower-left, and moving to the next column over every time the path hits the left or bottom edges. (Technical note: For square grids, the formula x+y<N?TriangleNumber(x + y + 1) – x- 1:N*M – TriangleNumber(N + M- x – y – 1) + N – x – 1 should generate this array)

Then, the algorithm starts at cell #0, and does the following:

Given a partially filled board: (at cell 0 this board is all holes)

Take all the values of the cells from (cell_number-1) to the number of the cell to the left of me, and put that in list A.

Remove all duplicates from A.

If the following values do not exist in A, add them: 0 (aka a hole) and the smallest value between 1 and 4 inclusive that does not exist in A (if there is no smallest value between 1 and 4, don’t add anything except the 0)

Run fillboard (that is, the current function), giving it cell_number+1 and the board given to this level of the function with the value of cell_number changed to n, for each value n in A.

However, if the board is all filled (i.e, cell_number=25) , check to see that the board has holes and that it is justsolved, and if so, standardize the piece numbering and make sure that you haven’t encountered it before, and if so, sort the board into the appropriate piece number “box”.

Once the fillboard algorithm finishes, you should have N*M-1 boxes with all the possible justsolved positions that can be made on an N*M grid! There are a number of other possible methods that do the same thing- all it basically needs to do is generate all possible ways that pieces fitting in an NxM grid can be placed in a grid of the same size.

For example, you could potentially go through all possible 5-colorings of the grid (4-colors and holes), and remove duplicates, but that would take O(5^(N*M)) time, which isn’t the most feasible option for even a 4×4 grid. You could also proceed in a way that would generate the results for the next number of pieces based on the results of the current number of pieces and all possible rotations and positions of a single piece in an NxM grid by seeing which pieces can be added to the results of the current stage, but that would take O(single_piece_results*all_justsolved_results). While that number may seem small, taking into account the actual numbers for a 4×4 grid (single_piece_results=11505 and all_justsolved_results=12295564) reveals the expected running time to be about the same as the slow 5-coloring method. However, it may be possible to speed up this method using various tricks of reducing which pieces need to be checked as to if they can be added. Lastly, one could go through all possibilities of edges separating pieces, and then figuring out which shapes are holes. The time for this would be somewhere between O(2^(3NM-N-M)) and O(2^(2NM-N-M)), the first being clearly infeasible and the second being much too plausible for a 4×4 grid.

In practice, the fillboard algorithm needs to check about 1.5 times the estimated number of boards to make sure it hasn’t found them before, resulting in about half a billion hash table lookups for a 4×4 grid.

The second algorithm, which SBPSearcher is almost completely composed of, is much simpler! Starting from a list of all justsolved puzzles (which can be generated by the fillboard algorithm), the following is run for each board in the list:

Run a diameter search from the current puzzle to find which other positions in the current puzzle’s graph have the goal piece in the same position;

Remove the results from step 1 from the list;

Run another diameter search from all the positions from step 1 (i.e consider all positions from step 1 to be 0 moves away from the start and work from there), and return the last position found where the goal piece is in the upper-left.

Step 2 is really where the speedup happens- Because each puzzle has a graph of positions that can be reached from it, and some of these positions are also in the big list of puzzles to be processed, you can find the puzzle furthest away from any of the goal positions by just searching away from them. Then, because the entire group has been solved, you don’t need to solve the group again for each of the other goal positions in it and those can be removed from the list. For a 4×4 board, the search can be done in 3 hours, 27 minutes on one thread on a computer with a Core I7-2600 @3.4 Ghz and a reasonable amount of memory. In total, the entire thing, finding puzzles and searching through all of the results, can be done in about 4 hours.

Of course, where there are algorithms, there are also problems that mess up the algorithms- for example, how would it be possible to modify SBPSearcher’s algorithm to do, say, simple simple puzzles? Or, is it possible to have the fillboard algorithm work with boards with internal walls or boards in weird shapes, or does it need to choose where the walls are? An interesting thing that would seem to point that the answer to the first question might be yes is that to find the pair of points furthest apart in a graph (which would be equivalent to finding the hardest compound SBP in a graph) requires only 2 diameter searches! Basically, you start with any point in the graph, then find the point furthest away from that, and let it be your new point. Then, find the point furthest away from your new point, and the two points, the one you’ve just found and the one before that, are the two points farthest away from each other. (See “Wennmacker’s Gadget”, page 98-99 and 7 in Ivan Moscovich’s “The Monty Hall Problem and Other Puzzles”)

Metrics

a.k.a. redefining the problem

Through the last 3 posts on sliding block puzzles, I have usually used the “Moves” metric for defining how hard a puzzle is. Just to be clear, an action in the Moves metric is defined as sliding a single piece to another location by a sequence of steps to the left, right, up and down, making sure not to pass through any other pieces or invoke any other quantum phenomena along the way. (The jury is out as to if sliding at the speed of light is allowed). While the majority of solvers use the Moves metric (my SBPSolver, Jimslide, KlotskiSolver, etc.), there are many other metrics for giving an approximation to the difficulty of a sliding block puzzle, such as the Steps metric, and the not-widely used sliding-line metric. The Steps metric is defined as just that- an action (or ‘step’) is sliding a single piece a single unit up, down, left, or right. The sliding-line metric is similar: an action is sliding a single piece any distance in a straight line up, down, left or right. So far as I know, only Analogbit’s online solver and the earliest version of my SBPSolver used the steps metric, and only Donald Knuth’s “SLIDING” program has support for the sliding-line metric. (It also has support for all the kinds of metric presented in this post except for the BB metrics!)

Additionally, each of the 3 metrics described above has another version which has the same constraints but can move multiple pieces at a time in the same direction(s). For example, a ‘supermove’ version of the Steps metric would allow you to slide any set of pieces one square in any one direction. (As far as I know, only Donald Knuth’s SLIDING program and Soft Qui Peut’s SBPSolver have support for any of the supermove metrics) In total, combining the supermove metrics with the normal metrics, there are 6 different metrics and thus 6 different ways to express the difficulty of a puzzle as a number. Note however, that a difficulty in one metric can’t be converted into another, which means that for completeness when presenting results you need to solve each sliding block puzzle 6 different ways! Even worse, the solving paths in different metrics need not be the same!

For example, in the left part of the image above, where the goal is to get the red piece to the lower-right corner, the Moves metric would report 1, and the red piece would go around the left side of the large block in the center. However, the Steps metric would report 5, by moving the yellow block left and then the red block down 4 times. Also, in the right picture both the Moves and Steps metrics would report ∞, because the green block can’t be moved without intersecting with the blue, and the blue block can’t be moved without intersecting with the green, but any of the Supermove metrics would report a finite number by moving both blocks at once!

Various other metrics can be proposed, some with other restrictions (You may not move a 1×1 next to a triomino, etc.), some which, like the Supermove metrics and the second puzzle above, actually change the way the pieces move, and you can eventually get to the point where it’s hard to call the puzzle a sliding block puzzle anymore. (For example, Dries de Clerq’s “Flying Block Puzzles” can be written as a metric: “An action is defined as a move or rotation of one piece to another location. Pieces may pass through each other while moving, but only one piece may be moved at a time”.

Suppose, however, that for now we’re purist and only allow metrics which generate numbers based on the step, sliding-line, moves, super-step, super-sliding-line, and super-moves metrics. It can be seen , quite obviously in fact, that these metrics don’t in all cases show the actual difficulty of the puzzle being considered. For example, take a very large (say 128×128) grid, and add a 126×126 wall in the center. Fill the moat that forms with 507 1×1 cells, all different pieces, and make the problem be to bring the block in the upper-left down to the lower-right. If my calculations are correct, the resulting problem should take 254*507+254=129,032 steps, sliding-line actions, and moves to solve, which would seem to indicate that this is a very hard puzzle indeed! However, any person who knows the first thing about sliding block puzzles should be able to solve it -assuming they can stay awake the full 17 hours it would take!

Because of this discouraging fact- that is, 6 competing standards, none of which are quite right, I would like to introduce a 7th, this one based on a theoretical simulation of a robot that knows the first thing about sliding block puzzles, but nothing else.

Nick Baxter and I have been working on a metric which should better approximate the difficulty of getting from one point to another in a position graph. The basic idea is that the difficulty of getting from node A to node B in a graph is about the same as the average difficulty of getting from node A to node B in all spanning trees of the graph. However, finding the difficulty of getting to node A to node B in a tree is nontrivial, or at least so it would seem at first glance.

Suppose you’re at the entrance of a maze, and the maze goes on for far enough and is tall enough such that you can only see the paths immediately leading off from where you are. If you know that the maze is a tree (i.e, it has no loops), then a reasonable method might be to choose a random pathway, and traverse that part of the maze. If you return from that part to the original node, then that part doesn’t contain the goal node and you can choose another random pathway to traverse, making sure of course not to go down the same paths you’ve gone down before. (Note that to be sure that the goal node isn’t in a part of the maze, you need to go down all the paths twice, to walk down a path and then back up the path). For now, we take the difficulty of the maze to be the average number of paths you’ll need to walk on to get to the goal node or decide that the maze has no goal(counting paths you walk down and up on as +2 to the difficulty). Because of the fact that if the node you’re in has no descendant nodes which are the goal node you’ll need to go down all of the paths leading from that node twice, the difficulty of the part of the maze tree branching off from a node A can be calculated as

sum(a_i+2,i=1 to n) (eq. 1)

where n is the number of subnodes, and a_i is the difficulty of the ith subnode. Also, if the node A is on the solution path between the start and end nodes, then the difficulty of A can be calculated as

a_n+1+1/2 sum(a_i+2,i=1 to n-1) (eq. 2)

where a_n is assumed to be the subnode which leads to the goal. This basically states that on average that you’ll have to go down half of the subpaths and the path leading to the goal to get to the goal. Because root nodes are assumed to have 0 difficulty, you can work up from the bottom of the maze, filling in difficulties of nodes as you go up the tree. After the difficulty of the root node has been calculated, the length of the path between start and end nodes should be subtracted to give labyrinths (mazes with only a single path) a BB difficulty of 0.

Perhaps surprisingly, it turns out that using this scheme, the difficulty of a tree with one goal node is always measured as V-1-m, where V is the number of nodes in the tree (and V-1 is the number of edges, but this is not true for graphs) and m is the number of steps needed to get from the start node to the end node in the tree! Because of this, the difficulty of getting from one point to another in a graph under the BB metric is just the number of vertices, minus one, minus the average path length between the start and end nodes in the graph.

A few things to note (and a disclaimer): First of all, the actual graph of which positions can be reached in 1 action from each of the other positions actually depends on the type of move chosen, so the BB metric doesn’t really remedy the problem of the 6 competing standards! Secondly, the problem of computing the average path length between two points in a graph is really really hard to do quickly, especially because an algorithm which would also give you the maximum path length between two points in polynomial time would allow you to test if a graph has a Hamiltonian Path in polynomial time, and since the Hamiltonian Path problem is NP-Complete, you could also do the Traveling Salesman Problem, Knapsack Problem, Graph Coloring Problem, etc. in polynomial time! Lastly, I haven’t tested this metric on any actual puzzles yet, and I’m also not certain that nobody else has come up with the same difficulty metric. If anybody knows, please tell me!

One last note: Humans don’t actually walk down mazes by choosing random paths- usually it’s possible to see if a path dead-ends, and often people choose the path leading closest to the finish first, as well as a whole host of other techniques that people use when trying to escape from a maze. Walter D. Pullen, author of the excellent maze-making program Daedalus, has a big long list of things that make a maze difficult here. (Many of these factors could be implemented by just adding weights to eqns. 1 and 2 above)

Open Problems

a.k.a. things for an idling programmer to do

What are the hardest simple simple simple 3×4 sliding block puzzles in different metrics? 2×8? 4×5? (Many, many popular sliding block puzzles fit in a 4×5 grid)

How much of a difference do the hardest strict simple^3 puzzles have with the hardest simple^3 SBPs?

How hard is it to search through all simple simple 4×4 SBPs? What about simple SBPs?

(Robert Smith) Is there any importance to the dual of the graph induced by an SBP?

Why hasn’t anybody found the hardest 15 puzzle position yet? (According to Karlemo and Östergård, only 1.3 TB would be required, which many large external hard drives today can hold! Unfortunately, there would be a lot of reading and writing to the hard disk, which would slow down the computation a bit.) (Or have they?)

Why 132?

What’s the hardest 2-piece simple sliding block puzzle in a square grid? Ziegler-Hunts & Ziegler-Hunts have shown a lower bound of 4n-16 for an nxn grid, n>6. How to do so isn’t too difficult, and is left as a puzzle for the reader.

Is there a better metric for difficulty than the BB metric?

Is there a better way to find different sliding block puzzle positions? (i.e, improve the fillboard algorithm?)

Is it possible to tell if a SBP is solvable without searching through all possible positions? (This question was proposed in Martin Gardner’s article on Sliding Block Puzzles in the February 1964 issue of Scientific American)

(Robert Smith) How do solutions of SBPs vary when we make an atomic change to the puzzle?

Are 3-dimensional sliding block puzzles interesting?

Would it be worthwhile to create an online database of sliding block puzzles based on the OEIS and as a sort of spiritual successor to Edward Hordern’s Sliding Piece Puzzles?

Readers of this blog may notice that I haven’t been updating for the last 4 months. The purpose of this “filler” post is for me to say why.

First of all, I could easily blame the season. Summer, as is well known, is a time to “kick back and relax” as well as forgetting about important things you should be doing and reading webcomics instead.

I could also blame the wonderful 3D printer I’ve bought, which has quickly filled up my desk and emptied my wallet with tens of tiny plastic models and puzzles.

However plausible that may seem, I would more truthfully blame the projects I’ve been working on, especially my “work” in the field of sliding block puzzles.

As you may know from a post I made about a year ago, I’m quite interested in really hard sliding block puzzles such as the Panex Puzzle (30,000? moves) or Sunshine (729? moves). James Stephens, of puzzlebeast.com, is certainly a pioneer in making really hard sliding block puzzles by computer. Using his “Puzzlebeast” software, he’s made at least 23 progressively harder sliding block puzzles, ranging from 18 to 148 moves! What’s more, he restricts many of these puzzles to only having a few pieces, and in at least 11 of them the goals are “Simple Simple”, that is, move a single piece to a single corner! Even the puzzle which fits on a 4×4 grid and uses only 4 pieces and is only 18 moves to a solution, “Simplicity”, is really quite hard for a human to solve! Oskar van Deventer, maker of manymechanicalmonstrosities, has designed a 3d-printable version of the Simplicity puzzle, calling it the “Hardest sliding piece puzzle on a 4×4 grid”.

Gauging from Stephens’ description of Puzzlebeast, Puzzlebeast uses a genetic algorithm to create its puzzles. So far as I can tell, it starts with a random set of puzzles. It picks the hardest out of that set, then for the next generation it creates random “mutations” of the best puzzles of the last generation. For example, it may add a piece, move a piece, modify a piece, or remove a piece. Repeat this process over and over again, and eventually you’ll have a hard sliding block puzzle. I think because it seemed to work very well for him, I eventually tried to make a sliding block puzzle evolver for myself.

And so the SBP project began: The earliest version of my SBP evolver were coded in Mathematica, and was very manual. I would start with 5 randomly generated 4×5 rectangular puzzles, without any restrictions on the number of pieces, and then for each puzzle I interpreted it as a “Simple simple simple sliding block puzzle”(see footnote 1) and then fed it into Analogbit’s online sliding block puzzle solver, which counts moves in steps, meaning that each “step” is sliding one piece one grid space in the puzzle. Once that was finished, I took the best two puzzles, converted them to 1-d arrays, “interleaved” them, and then influenced random mutations in their cells (That is, each cell had about a 1/5 chance of being added 1 to or being subtracted 1 from). Since that probably isn’t too clear, here’s an example:

Suppose the best two puzzles from the last generation were

and the “mutants” would be generated from there, and the process would repeat.

Notes: 1. “Simple simple simple sliding block puzzle” is my own term, based off of Nick Baxter’s definitions at the Sliding Block Puzzle Page . For me, it means that each piece is different from another only by its shape, and that the goal is to get the top-left-most piece down to the lower-right. The original definition was somewhat ambiguous, it actually means that if there is no piece with a square occupying the upper-left corner, it is not a valid simple simple simple sliding block puzzle. Additionally, many of the puzzles in this first section are copied straight from the Mathematica notebook, and so there is a distinct lack of proper number parsing. Parsing manually is simple: If two numbered pieces have the same number and are orthogonally adjacent, they both belong to one piece. Otherwise, they belong to different pieces. (Eventually I got a parser working to make these puzzles conform to standards, so the reading should get easier). Lastly, steps and moves are different, as one involves moving one piece one space and the other involves moving one piece any number of spaces. Edward Hordern’s book, Sliding Piece Puzzles, does a good job to clear this up.

Anyways, the very first incarnation of the SBP Evolver produced quite fantastic results! In about 10 generations, with populations varying from 4 to 10, the puzzles had progressed from 4 impossible puzzles and 1 8-stepper to 3 impossible puzzles and 7 possible, one a 58-stepper! Now of course this is no big result, as such puzzles as Bob Henderson and Gil Dogon’s “Quzzle-Killer” take 121 steps to solve, and so this early experiment was really just a proof of concept.

Eventually, I got tired of feeding all these puzzles through Analogbit, so I decided to write a C# program to do it for me, except instead of using Analogbit, it would use Jim Leonard’s very speedy Jimslide software. Furthermore, it would have much larger populations (perhaps 100 or so) and do many more generations. Some time later, after writing many, many lines of code to process the puzzles for Jimslide’s use, and then processing the output, I finally got it working! Almost immediately, I set it going on a huge number of 4×5 puzzles (10 million) and let it run overnight. It actually took around 4 days, but the best it found was a 176-move puzzle!

The SBP Evolver also displayed some behavior which I conjecture is common to most poorly-written genetic evolution programs: Once the program found a really good optimization, if there weren’t any tiny optimizations which would improve the puzzle, it would tend to stay with that puzzle for a while unless, by pure chance, the next generation happened not to contain that puzzle. As such, the 4 days might actually not have been long enough to find the best puzzle. Additionally, Bob Henderson’s Gauntlet #2, another simple simple sliding block puzzle, trumps the 176-mover with 235 moves!

As far as I can tell from his description on Andrea Gilbert’s clickmazes site, Bob Henderson uses a very different method of generating sliding block puzzles. He inputs a collection of shapes, and then his packing program generates all possible ways that those blocks could be placed in a sliding puzzle. He then feeds those through his sliding block puzzle solving program, which returns the best results. I expect it produces results much faster, albeit only with the blocks the designer chooses.

Having been thus discouraged, I set out to go through the much easier 4×4 puzzles instead. This time, after about a day, it generated a 125-move simple simple puzzle! However, it turned out that the puzzle could be backtracked to make a 132-move puzzle, presented below. Interestingly, when I ran the same computation again, it generated another descendant of the 132! Now, of course, the question was: Is this the best 4×4 simple simple simple SBP?

The problem could be easily solved by just generating all possible 4×4 positions, interpreting a simple simple simple SBP out of them, and then solving each and every single one. But how would you do the former? After all, there can be 16 possible block types in a 4×4 (including holes) and each space could be one of the 16, so you’d have 16^16=18,446,744,073,709,551,616 possible combinations! My answer to this was to invoke the 4-color theorem, using 0 for the holes, and 1,2,and 3 for colors for the other pieces. After the 4-coloring of the board, a parser would assign each block a unique number, and then remove any duplicates (This was in fact WRONG; The holes could be separate, and so the 4-color theorem applied only to the blocks. I in fact had to eventually use 5-colors for the whole thing). After around a day or so, the puzzle-finder would process 4294967296 boards, and it returned only 52,825,604 different boards!

Two other methods which might have worked as well or better might be to either generate all 1-piece boards, then generate all 2-piece boards from all nonoverlapping combinations of the 1-piece boards, and so forth. There’s also a method based on dynamic programming which I currently use, which (while it is too long to write out here) is included in the SBPFinder source code, available at the bottom of the post.

After the 52,825,604 different boards were separated into files depending on the number of pieces, the host software would load up each file into a hash table, then solve the first puzzle. After the puzzle was solved, the software would process the output and remove the puzzles mentioned in the solution from the hash table. It would then solve the new first puzzle, and so on. While it isn’t very fast at all to use this algorithm (it took 48 days), it at least had an advantage: By separating the puzzles into files labeled with the number of pieces, it was possible to return the “hardest” puzzle for an arbitrary number of pieces in a 4×4 grid! Below is a table of the hardest simple simple simple SBPs in a 4×4 grid, from 2 to 15 pieces. Keep note, though, that since the computation missed out on a few million puzzles, all the below are unverified and the ones which are very dubious are removed. Also, because SBP determined which piece was the goal piece by going line by line, some of the puzzles below are simply simple simple SBPs and the ones which are just simple SBPs are removed. (Confused? Check http://www.johnrausch.com/SlidingBlockPuzzles/4×5.htm)

Note that these results are not to be trusted; for one, bugs have been found in both the problem finder and the puzzle-to-Jimslide converter itself, and so the results certainly aren’t rigorous. Also, the SBP Evolver (which also runs the puzzle-to-Jimslide converter and searcher) was originally designed to treat any puzzle as a simple simple simple, by choosing the goal piece as the first piece encountered going line by line through the board. Lastly, there’s a curious phenomenon around 12 pieces: Breaking a 1×2 into 2 1×1 pieces and shuffling the remainder about creates 9 more moves!

Since then, I’ve been trying to do a verification of the 4×4 computation, specifically by creating another program to rerun the entire search process, return the best results, and then check to see if the results are the same. The catch: I need it to be a few times faster. The optimizations I’m working on including are:

•Using one of the alternative methods to find potential SBPs faster,

•Only storing the “justsolved” puzzles, that is, puzzles in which the goal piece is already in the lower-right and can move out of it’s current position. (Based on an idea from “Rush Hour Complexity” , by John Tromp and Rudi Cilibrasi. It turns out that about 1/4 of 4×4 SBP positions are justsolved, a rather large value, mostly because in most positions there is a piece in the lower-right corner, and the only real limitation is that it must be able to move), and

•Using a custom-written sliding block puzzle solver to find the hardest puzzle in the entire group of positions the justsolved puzzle is linked to. (This can be done in 2 diameter searches, one to find the other justsolved puzzles in the group, and the other to find the starting position the furthest away from all the justsolved positions. My custom solver is about 3x as slow as Jimslide, but it makes up for it by that it’s solving the entire group of puzzles and removing the other justsolved puzzles from the list. There can be a stunning amount of other justsolved positions reachable from a puzzle- some have been found with over 14,000 !)

However, I’ve hit upon a gypsy moth in my solver causing it not to search the puzzle completely! Preliminary tests of the program though, have revealed the expected running time to be about a day, so stay posted!

Anyways, that’s my explanation as to what’s been going on, and I apologize for my relative silence.

Our tale starts with the PDP-1, a $120,000 room-sized system which, when released in 1960, was state-of-the art computer technology: 4096 18-bit words of memory, a 200 Khz clock speed, and came with only a Macro assembler and a DDT debugger, but what it did have was a paper tape reader, and a Type 30 precision CRT display. As such, many of the first programs of the PDP-1 were “Display hacks”: Programs using only a few lines of code, but when run, create intricate patterns on the screen. A prime example of this is the Munching Squares program by Jackson Wright, described in the MIT A.I. Lab’s HAKMEM:

foo, lat
adm bar
rcl 9s
xor bar
dpy
jmp foo
bar, .

This creates a sequence of images corresponding to where the bitwise XOR function of the x and y coordinates of every point on the screen is less than the frame number. If this happens to be a bit complicated, it’s a bit easier to understand when you see the animation:

A goal of a member of the Research Lab for Electronics (and later the head of the MIT AI lab), Marvin Minsky, was to develop a program to make curves in a square grid, in as few lines as possible. When he was attempting to get the system to draw a spiral, he stumbled across a most curious thing: A two-statement program which would draw stable circles! Decoded from machine language, it reads:

loop: y = y – 1/16 * x

x = x + 1/16 * y

Note the lack of temporary variables; In fact, with a temporary y variable, the points spiral out of the circle! However, the program does not draw a perfect circle, but rather a very round ellipse, which becomes rounder as 1/16 gets closer and closer to 0.

You can generalize the Minsky circle algorithm by replacing the first 1/16 by δ and the latter by ε, to get circles that take longer or shorter to generate:

x = x – δ * y

y = y + ε * x

It turns out that this can even be solved recursively! Using a substitution and a “guess and-by-gosh” method, the authors of the book “Minsky’s and Trinsky’s” (which most of the content for this article was lifted from, as of now privately published by the authors: Corey Ziegler Hunts, Julian Ziegler Hunts, R.W. Gosper and Jack Holloway) were able to prove that, for the Nth step:

Xn=X0 cos(n ω)+(X0/2-Y0/ε) sin(n ω)/d

Yn=Y0 cos(n ω)+(X0/δ-Y0/2) sin(n ω)/d

where d=sqrt(1/(δ ε)-1/4) and ω=2 arcsin(sqrt(δ ε)/2) , which happens to actually be an ellipse! Note how if δ*ε>4 or δ*ε<0, ω becomes imaginary and the ellipse becomes a hyperbola. However, with anything so seemingly simple, there is something that makes the subject much more complex. In this case, it was the nearly imperceptible (but strictly periodic) wobble of the ellipse.

“The Worst Circles I Ever Drew”

In machine arithmetic, integers, no matter what happens to them, are always integers. While this may seem a triviality, machines with only integer arithmetic effectively use the floor function on each operation they do. For example, if you divide 1 by 2, in many programming languages the answer will be 0. The reason for this is because the bit routines for dividing integers always return integers: They work to no more than the precision given. In order for us to understand what’s going on, we can just pretend that the computer works out the number, then finds the largest integer less than or equal to the answer. No problem, right? Inserting floors into the Minsky recursion, like this:

x = x – floor(δ * y)

y = y + floor(ε * x)

shouldn’t hurt the overall mechanics, right? Wrong. When x or y is small enough, the floor will actually make the result of the multiplication be 0, losing precision. While this may not seem as big a problem, when we chart the plots for strange values of delta or epsilon we get something which is definitely not an ellipse. Corey Ziegler Hunts and Julian Ziegler Hunts discovered this behavior (discovered earlier by Steve Witham), and began to dig deeper. It turns out that if you calculate the Periods (how long it takes before the points reach X0 and Y0) of the Minsky recurrence starting at different X0,Y0, δ and ε, the periods can vary wildly. For example, when δ and ε both equal 1, all X and Y (other than 0,0) loop with period 6. On the other side, the X0,Y0,δ and ε corresponding to {-19/24,-1015/121,7381/5040,5040/7381} has a period of no less than 2,759,393! (Even with the floor function, the algorithm is exactly reversible, so it must return to (X0,Y0) or never repeat any point, unless the machine integers overflow.)

Another one, x=1,y=0,δ=9/17 and ε=15/2, has been followed for no less than 104 trillion steps in both directions (the Minsky recurrence can be modified to run backward) without looping! A logical question might be to ask: What do these periods look like when plotted? Well, since there are 4 variables: x,y,δ, and ε, there are 6 different ways to plot the periods: x and y, x and δ, x and ε, y and δ, y and ε, and lastly δ and ε. The Zieglers started with the x-y plots. Now, choose a constant δ and ε, such as 1 and 0.467911, and plot the periods mapped to colors for integer x and y, and you get what may as well be a Persian rug! Okay, but maybe that’s just a special case of the Minsky recurrence, and if we try something like δ=1 and ε=0.91939, we’ll just see a blob? Wrong.

Coloring methods also have effects on Minsky plots. For example, the Minsky plot δ=100, ε=1/99, which when rendered with a simple linear coloring from Dr.Fibdork’s version (programmers, see comments) looks like this: However, the authors of the book use a different method to bring out some of the finer details, which shows this: At some point, though, we have to stop messing with numbers and get down to math. If we return to the original Minsky solutions:

Xn=X0 cos(n ω)+(X0/2-Y0/ε) sin(n ω)/d

Yn=Y0 cos(n ω)+(X0/δ-Y0/2) sin(n ω)/d

where δ=sqrt(1/(δ ε)-1/4) and ω=2 arcsin(sqrt(δ ε)/2) ,we notice that this ellipse returns to its original point when n*ω mod 2*Pi =0, because when n=0, n*ω=0, and also because the periods of the cos and sin functions are 2*Pi . Now, let us define the theoretical period (denoted as P) of a Minsky recurrence as the first n greater than 0 for which n*ω mod 2*Pi =0, which is the same as saying “The n for which n*ω=2 Pi” . (Note that n can be a non-integer) N can be trivially solved for, and expanding ω we get that P=Pi/arcsin(sqrt(δ ε)/2) . We can write this in two ways: The already mentioned one, or by solving for δ ε we get δ ε=4 (Sin(Pi/P))^2, which we can use to get a product of δ and ε from the period. Earlier on, readers may have looked at the figures for δ and ε and seen something suspicious about them. As it turns out, the first “Rugplot” had a period of 9, and the second had a period of 2 Pi. In general, when P>4, we can generate very symmetrical plots by setting δ to 1 and ε to the value given by 4 (Sin(Pi/P))^2 . For P<5, the method generates very monotonous images (both delta and epsilon are integers), but when P=5 we get what Holloway terms “Rastermen”: Koch Snowflake-like fractals but with only five arms repeated infinitely with sizes according to the Fibonacci numbers!

It’s even possible to make animations of Minskyplots, such as along the line δ=3/2+t/200, P 40/139. It turns out that Minsky space is rather epileptic:

Different configurations arise with different functions of time for δ and ε, such as when δ=t and ε=1/(t-1), and δε approaches 1:

The horizontalness of the boundaries at the end are due to the fact that the slope of major axes of the ellipses in a Minsky x-y plot is approximately ε/δ , because in the original Minsky solutions the amplitude of Xn (roughly the “run” in rise/run) is larger when ε is larger, and the amplitude of Yn (the “rise”) reacts the same way to δ.

Minskyspace

At this point you may be asking what the other 5 arrangements of Minsky plots look like. I don’t have the space in this post to talk about them all, but I can describe one of the most interesting Minsky plot methods, sort of the opposite of the x-y plot: The δ-ε plot. Recall from earlier that the simple Minsky solutions become imaginary if δε>4. It actually turns out that this is generally not the case. Suppose we use a simple Minsky period plotting method to draw the periods with x=1 and y=0 , where -4<δ<4 and -4<ε<4:

The gray areas are where there was no Minsky loop with a period less than 5000, and the white areas are period 1. As you can see, although the outline of the colored region resembles the function y=4/x, in many places there are regions of periodicity extending out into the “dangerous” area, such as on the upper right corner, really small. (I should note here that it has been proven that the shapes of periods are always rectangles) Furthermore, the authors of Minsky’s and Trinsky’s have discovered that some areas inside the two “Islands” are nonperiodic, such as x=1,y=1/2,δ=9 and ε=1/3. (Plot) Even more, any Minsky point of the form x=1, y=1/2, δ=3^(n+1) , ε=3^(-n) , where n is an integer greater than 0, is unbounded. Not much is known about these δ-ε plots: We can prove that the shapes are always rectangles, and we can find a few general periods, but in general it’s a bit like trying to find the value of an arbitrary point on the Mandelbrot Set. Which brings me to my next point: Unlike the Minsky x-y plots, you can take a δ and ε and zoom into it! The video below shows a sequence of period 5 x-y plots with δ ranging from 99/100 to close to 1:

Some of the areas can seem fractal, and it’s certainly possible to find patterns that seem to converge to a series of infinitely thin rectangles, such as the top-right edge of the x=1,y=1 δ-ε space (δ*ε≤4): Other areas, such as this one near x=0, y=8, δ=173/80 , ε=137/96 , display more localized behavior: However, in many of the places where the differences in size between rectangles go to 0, the periods appear to approach ∞, such as when you approach x=1, y=0, δ=1 , ε=1 from the upper-left-hand side. These places, called “Accumulation points”, seem to be ubiquitous throughout any δ-ε plot . As a great example of this, take the neighborhood of x0=0, y0=8, ε=10/7, δ=37/17 (zoomed in and recolored from the image above) , which Bill Gosper has termed the “Neighborhood of a Monster” because of the “monsters” (points with gigantic periods) that occur. In this case, even though the center appears at first to be devoid of accumulation points, there are some particularly nasty periods- right in between two large blocks!

There’s tons more to study of course, but in reality all of the pictures we see of Minsky plots, whether x-y,δ-ε, or some combination of the two, are all slices of a 4-dimensional, jagged, infinite object, called Minskyspace. The 4 dimensions come from the four parameters, and while slices from this object have not even been rendered in a dimension higher than 2, we can tell a few things about it:

It goes forever in the x,y,δ, and ε directions. However, in the δ and ε directions, it takes on a rather hyperbolic shape, due to the original, non-rounded Minsky circle algorithm.

Nearly half of it seems to be missing, due to δ*ε being less than 0.

Certain “congruence regions”, that is, areas in Minskyspace where the periods are the same which produce identical orbits modulo translation, are shaped like polyhedra instead of infinitely thin slices when sliced through with a 3-dimensional plane! (some of the faces are curved, though)

At irrational δ*ε , there can be infinitely fine structure around that area, but a conjecture is that there is no irrational δ*ε which has infinite period.

All accumulation points, infinitely near their centers, have unlimited period.

Conjecture: All congruence regions are Cartesian products of two polygons with at most 6 hyperbolically curved sides.

That’s a bit of what we know for sure. There are tons of conjectures, and I’ve hosted a version of the Minsky “To do” list, “Problems posed and solved”, at neilbickford.com/Minsky/problems.txt.

In summary: Although some algorithms may seem simple, there can be great areas of mathematics behind them. We still don’t know all about the shapes that the Minsky circle algorithm creates, and there are still many problems waiting to be uncovered and solved. Even if you don’t want to tackle recursive formulae, just discovering areas in Minskyspace is a fun and entertaining pastime.

All major bugs have been fixed, the integrated help system is now functional, and it even comes with a tutorial! Described in more detail in my pi post…

Currently only for x64 Windows: download here (0.416 MB)

Pi is one of the greatest numbers of all time, having been known for thousands of years and over that time gaining quite a bit of popularity in the form of celebrations such as Pi Day and others, all from a number that came from the simplest smooth object: A circle. Suppose you have a circle with a width of 1 inch, and then take a measuring tape and measure around the edge. You’ll find that it comes out to 3 inches and a bit, and if you increase the inch to a foot, you might get 3.14 if you look carefully enough. On the more extreme scale, you could go out to a crop circle, measure it, and perhaps get 3.1415926 . Now, suppose you have a perfect circle, and an infinitely precise ruler (for lengths shorter than an atom) , and do the same thing once again. You would get the number 3.141592653589793238462643383… which is expressed as the Greek symbol

One of the first mentions of pi is in the Bible, where in Kings 7:23-26 it states:

“And he [Hiram] made a molten sea, ten cubits from the one rim to the other it was round all about, and…a line of thirty cubits did compass it round about….And it was an hand breadth thick….” This states that pi=3, a definite approximation, but a terrible one nonetheless. A slightly better approximation was made by Archimedes, when he developed a formula for computing pi by using polygons with large numbers of sides, and getting two approximations for the area of a circle ( pi*r^2) , like this:

Using this method, he drew 2 96-sided polygons and got 3 10/71<pi<3 1/7 , an approximation accurate to 2 decimal places: 3.14… Ptolemy later updated this with 3.141… and this was updated by Tsu Ch’ung Chi to 355/113 , correct to 6 places. Later on, in the 1600s, Gottfried Leibniz/James Gregory found an infinite sum for pi: pi=4*(1-1/3+1/5-1/7…) The proof of this requires calculus, but takes up less than a page. Leibniz’s/Gregory’s formula is rarely used because it takes exponentially many terms to create more digits, which would slow down even the fastest computers. A slightly better formula, but much more amazing, was found by Francois Viete in 1593, using only the number 2!

A quite beautiful formula for pi was found by John Wallis, in the form of

Notice how the numerators and the denominators appear to “carry over” to the next fraction!

Shortly thereafter, a much better formula was found by John Machin in 1706:

This formula, when expressed in radians, can be computed rapidly using Arccot(x)=1/x-1/(3x^3)+1/(5x^5)-1/(7x^7)… Formulas of this type, arctans of fractions, are now called “Machin-like formulae”. The simplest of these is Pi/4=Arctan(1), followed by

and

The arctans with bigger denominators produce more digits per series term, so the efficiency of a Machin-like formula is limited by the arctan with the smallest denominator. For example, the 2002 Pi decimal places record was set by Yasumasa Kanada on a supercomputer using Kikuko Takano’s

Even more complicated Machin-like formulae exist, such as Hwang Chien-Lih’s 2002

However, in the computer age, the length or the elegance of the formula don’t count: it’s the rate at which the formula converges. Snirvasa Ramanujan, Indian matematician and nemesis of Bill Gosper (“Every time I find an identity, he’s found it before me!”), created a number of formulae for pi, including the following:

where

denotes f(a)+f(a+1)+f(a+2)…+f(b). Note not only the factorials (n!=1*2*3*4*5…*n) but also the large terms both on the outside and on the inside, especially the factorial to the 4th power and the 396^(4k), which can be shown to mean that the sum converges exponentially rapidly (digits/term), as opposed to exponentially slowly as in the Gregory-Leibniz formula, which makes it one of the fastest algorithms known for computing pi. An even faster algorithm, which has been used to break the pi record many times, is the formula found by the Chudnovsky brothers in 1987:

This rather monstrous formula gives about 14 digits per term, and was used most recently by Shigeru Kondo and Alexander Yee to calculate 5 trillion digits of pi, billions of times more than enough to estimate the area of your wading pool to the atom. There are even formulae that give an exponential number of digits per iteration, with the drawback that each calculation is exponentially hard. One of these, the Brent-Salamin algorithm, only uses simple arithmetic and would take about 35 iterations to break the record:

First, start with a_0=1,b_0=1/sqrt(2),t_0=1/4,and p_0=1. Then iterate: a_(n+1)=(a_n+b_n)/2, b_(n+1)= sqrt(a_n*b_n), t_(n+1)=t_n-p_n(a_n+a_(n+1))^2, and p_(n+1)=2 p_n. Then when you’ve iterated enough, the estimate for pi is given by (a_n+b_n)^2/(4 t_n).The best of these iterative formulas that I know of is Borwein and Borwein’s, which converges like 9^n (Effectively, it requires about 11 iterations to beat the current record):

Start with

and then iterate

Then the estimate for pi is given by 1/a_n .

A fairly significant formula, found in 1995 by Simon Plouffe, is the Bailey-Borwein-Plouffe formula, which can be used to compute any bit in the hexadecimal representation of pi-without needing to know the previous digits, which can then be used to compute binary bits. In decimal-finding form, it is:

This formula was used by PiHex, an ended distributed computing program, to determine that the 1 quadrillionth bit of pi was 0. Yahoo later used the same to find that the 2 quadrillionth bit of pi was also 0.

Of course, the reasons of finding decimal digits of pi are not only to show how great your new supercomputer is, but also to attempt to find a pattern. In base 10, this is probably unlikely, as there are an infinite number of other bases to test, including the non-integer bases(i.e. 7/5ths, sqrt(2),6*e/19…) This makes it practically impossible, and even if base 10 or base 11 or base 16 had a pattern, we might have to look any number of places to find it, as in Carl Sagan’s novel Contact, where (spoiler) after a few trillion digits in base 11, one of the main characters finds a field of 0s and 1s the size of two prime numbers multiplied together. Plotting the 0s and 1s as black and white dots, she plots it on her computer screen to find- a picture of a circle! This is actually possible (though very unlikely) as one of Hardy and Wright’s theorems state that any sequence of digits you can think of can be found in pi. In fact, there’s a website (http://www.dr-mikes-maths.com/pisearch.html) which will search for names in pi expressed in base 26! (end spoiler)

However, there’s a way to express pi in such a way that it doesn’t depend on the base: Continued fractions! Continued fractions are “infinite fractions” which are in the form of

and are usually expressed as [a0,a1,a2,a3,a4,a5,…] or as [a0;a1,a2,a3,a4,a5,…] with all an positive integers. Many numbers, such as integers and fractions, have rational continued fractions: For example, 1=[1], and 355/113=[3,7,15,1]. Of course, if 355/113 were expressed in decimal, you’d have to use an infinite number of digits to get the actual fraction. A significant advantage that continued fractions have over decimal notation is that often irrational numbers can be expressed as repeating continued fractions. For example,

sqrt(2)=1.4142135623730950488016887242097… but in continued fraction notation

Much simpler. In fact, you can go to your friends, claim you know more digits of the square root of 2 than them, and you can simply expand the continued fraction out to beat them no matter how many decimal digits they know! Possibly the most elegant of these repeating continued fractions is the one for the Golden Ratio, (1+sqrt(5))/2:

Also, sometimes transcendental numbers can be expressed as simple continued fractions. For example, Euler’s Number, e, is equal to lim(n->infinity) (1+1/n)^n and is often used in exponentiation and calculus. In continued fraction form, it is equal to [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1…]! Decimal is not as elegant, e being about 2.71828182845904523536…

However, despite any hope, Pi is not as pretty in continued fraction form, though it is invariant of base: [3,7,15,1,292 (ack!),1,1,12,1,3,1,14,2,1,1,2…] There have been only a few attempts for the continued fraction of pi; Tsu Ch’ung Chi’s 355/113=[3,7,15,1] was the first nontrivial one, and Euclid’s algorithm can be used for computing the continued fraction of pi, though his GCD algorithm just throws the terms away. The first major record that I know of was made by Bill Gosper on August 19,1977 when he computed 204103 terms using his own algorithm in Macsyma, an early computer algebra system. Later, he beat his own record in 1985 with a whopping 17001303 terms, again using his algorithm. Later, in 1999 Hans Havermann beat Gosper’s record by using Mathematica to compute 20,000,000 terms. He later beat this in March 2002 to make 180,000,000 terms, the previous record.

Now might be a good time to tell why I haven’t been blogging recently.

Over the past few months, I have been working on a C# program, PiCF (not released yet, current alpha source code here) which can calculate the continued fraction of any number, not just pi, using Gosper’s algorithm. On October 17th, I calculated approximately 458,000,000 terms of pi in about 3 hours on a 64-bit machine running Windows on a Core 2 Duo @ 3.00 Ghz. This was later verified using Mathematica (taking up much more memory than the calculation did!). The program was coded in C#, has a command-line interface (with menus!), and uses Emil Stefanov’s wrapper of GNU MP for the BigInteger multiplications. The maximum term is still 878783625, originally found by Bill Gosper during the 17M calculation. Other stats: The minimum term is one (naturally), the terms take a 1.4 GB file (download here if you dare) and I was very nervous.

Pi has, over the years, gained a huge following: March 14th is Pi Day (in decimal) on which MIT ships their student acceptments/declines and on which people make pie, 1:59:26 of that day is Pi Second; May 22nd is Pi Approximation Day, and Kate Bush even wrote a song about pi. Many jokes about pi have surfaced on the Internet, including that:

This may be because over the thousands of years, pi has become so far removed from its original purpose: measuring circles. To start out, almost every round object has a volume and surface area that involves pi. The volume of a cone is one-third r^2 *h*pi, where r is the radius of the base and h is the height. The volume of a torus is (pi^2)/4*(b+a)(b-a)^2 where a is the inner radius and b is the total radius.

What about the volume of a crescent? Well, that’s quite a different story…

This is the 6th and final post in a series of posts about Gathering For Gardner: 12345

I started the last day of Gathering For Gardner 9 by waking up late.

As a result, I completely missed the first talk and entered the conference room in the middle of the second talk, by Karl Schaffer, about “Dancing Tessellations”. It consisted mainly of a few videos in which a normal dance would be reflected along certain axes so that it effectively makes a video tessellation. The next one was a short talk on extending the Side-Angle-Side (SAS) similarity theorem to three-dimensional shapes using the least possible number of measurements. One of the especially interesting talks in the first session was by Linda Zayas-Palmer, on why the infinite number, 0.999999… , is actually not just equivalent to, but greater than, 1. The talk directly after that was about some of Salvador Dali’s puzzles in some of his paintings, in which you have to find a certain phrase. Turns out his puzzles are rather easy. My favorite talk in the session, though, was by Burkard Polster on how to balance your laptop on a bedside table such that it occupies the minimum amount of space on the table and doesn’t fall off. It starts out by a simple dissection of a square by Martin Gardner, and then exteds it to show how to balance the laptop just right so that it’ll balance, thus leaving room for puzzles or whatever you would put on a bedside table.

During the break, Mom and I went up to the exhibit room, where the exhibitors were packing up their exhibits. Most of the exhibits were already gone, but a few were there that weren’t the first time. For one of these, involving curious polyhedral sculptures made with egg cartons, Mom actually got to talk with Jeannie Mosely, the person who made them. In fact, she actually got a model of the octahedron which had fairly novel methods of holding together as an example to make more.

After the break, the talks continued with a talk on 3-dimensional packing puzzles made out of various spheres glued together to make polyomino-like shapes. Due to the spherical nature of the pieces, some of the puzzles need the polyspheres to “snap” together, something which doesn’t normally happen. A rather good talk in this session was by Ed Pegg, about the Wolfram Demonstrations Project, and some of the best demonstrations that had been posted, such as a program to find the smallest box that squares of size 1/n can fit in:

The talk after Ed Pegg’s talk was one of the most anticipated talks of the entire conference: Finding a single shape that covers the entire plane aperiodically, which was an unsolved problem. Joshua Socolar managed to find a single hexagonal tile (with matching rules, though) which creates a Sierpinski Gasket-like shape when made. He also made a 3-dimensional tile which acts the same.

Afterwards, Vi Hart did a great talk on making music with music boxes with the music scores in Mobius strips, or the music boxes arranged in such a way so that the music played by one music box is played by another music box seconds later.

After Vi Hart’s talk the lunch break started, but instead of going to lunch I went immediately up to the Gift Exchange area, where I would wait in line to get a bag full of exchange gifts from nearly everybody, repeated for everyone, who would get one also.

Apparently everybody else had the same idea of waiting in line early. I ended up behind Bram Cohen, inventor of BitTorrent as well as a whole bunch of super-duper-hard twisty puzzles which have some rather ingenious ideas behind them. He happened to have a few non-twisty, but still hard, puzzles while waiting in line, and I managed to almost solve one of them (the Cast Rattle), which involved getting the right pieces in the right place (That isn’t a spoiler, is it?). The other one, Cast Marble, I couldn’t get immediately, but I might have been getting close. Soon, I got to the front of the line to get my bag of exchange gifts and got the huge bag (I could barely carry it) as well as a few other miscellaneous items from people at a few nearby tables, such as a perplexing wooden object which looked like a gear, or a few pictures from Caspar Schwabe (I presume) of large inflatable solids, including the huge 59th stellation of the icosahedron seen on day 3.

I brought the huge bag of gifts up to the hotel room to open, and even though the bag’s heaviness was an indicator of the number of things inside, it still felt like Christmas when I opened it. There were mathematical dice with the sides only using the number nine and a set of plastic rings with interchangeable art based on the Traveling Salesman problem; there was a CD containing rather high-quality pictures as well as a digital copy of the exchange book from G4G8; A set of pieces for an unsolved puzzle and a key to open doors with using a hammer; A book about formulae that changed the world and a second copy of A New Kind of Science(Ah, so that’s where the heaviness came from);Even a mysterious back-scratcher and tapper. These are just a few of the things in the bag, and to list them all out would certainly lengthen this blog post by quite a lot.

By the time I had gotten to the bottom of the bag, the third session was about to start so I had to rush down to see the talk by Gordon Hamilton (of the Magical Mathematics Museum) about having problems in K-12 classrooms involving unsolved problems such as the circle-packing problem (In how small a square can you pack n circles?), the 3n+1 conjecture (Do all Collatz sequences end in 4,2,1?), and others. Also in the same session was Solomon Golomb’s talk about the Pentomino Game on nxn boards. The “Pentomino Game”, is a rather interesting two-player game in which players alternate turns placing pentominoes onto an 8×8 board. The first player who can not place another piece loses. One of the most interesting talks of the session was “Fun with Egg Cartons” by Jeannie Mosely. In the talk, she described how she made most of the Platonic Solids – out of egg cartons! The process of making these is pretty easy- just interleave strips of egg cartons at the vertices to make the edges- but the results are still interesting.

Immediately following was a talk not relating to mathematics at all (but still cool), about restoring various ancient text adventures. The talk was by Adam Atkinson, and it was about cross-compiling old text adventures (running on mainframes) so that you could play them on newer computers. Many of them are stored at ifarchive.org (the Archive for Interactive Fiction), including Acheton, probably the third text adventure ever made. Just don’t get eaten by a grue.

After the last short break, the last session of G4G9 began. Kate Jones started of with a philosophical talk involving pentomino puzzles,followed by Bill Mullins, who talked about Martin Gardner’s search for the person who wrote “The Expert at the Card Table”, probably the most important book ever on sleight of hand with cards, who wrote his name as “S.W.Erdnase”. It can be reversed to make E.S.Andrews , but from there it’s much harder. So far, they’ve found 5 suspects as to who S.W.Erdnase might be, 2 by Gardner and 3 by others, but the writer remains hidden. Hirokazu Iwasawa (also known as Iwahiro) then followed that with a talk about the subclass of “Hat-Team Puzzles” , how to solve them, as well as other variations on the problem. A bit later, Colin Wright did a double talk about “How far is the Moon?” and about notations for juggling. The former was left out (the pdf is available here) , but the talk about juggling was amazing. Not only did he juggle normally with up to 5 balls, but he also showed how to use a notation for juggling to make up your own tricks, some insanely complex and others trivial. Following that, George Hart talked about his new sculpture, “Comet!” which involves multiple smaller versions of the main model (a puzzle-like polyhedral-ish form) with different colors. It’s so big that it has to be hung on the ceiling of an atrium. After that, Mike Keith did the second-to-last talk about his book, Not A Wake, in which every word of the text- including the subtitle and the title- has the same number of letters as the nth digit of pi does. The book goes on for 10,000 digits, with 10 stories followed by the digits of pi in that story, each story being a different style than the others.

After the last talk and closing notes, G4G9 was over. However, the fun(at least for me) continued. I was invited to dinner (along with Bill Gosper, Mom, Julian and Corey Hunts) by Dick Esterle to the Varsity Jr., an old-style fast food diner that had been operating for 45 years. Bill and Julian declined, but the rest of us went.

The diner had good food (especially the burgers) , and talking with Esterle was quite interesting. I brought a box puzzle (the same from the prelude) , and he managed to solve it in record time just by shaking it hard. He also, using the materials that were available, gave me 2 versions of the same puzzle. First, arrange 3 cups in an equilateral triangle such that a knife can reach from any cup to any other cup. Then, use the knives to balance a salt shaker in the middle of the triangle above the table. (This can be done using 2 knives) Then, set the cups so that they are just a bit too far for the knives to reach, and once again balance the salt shaker using 3 knives. Corey and I eventually solved it and put a few straw decorations on (from my solution to the problem using straws to extend the knives and only using 2 knives). To prevent spoilers, it’s at http://daftmusings.stattenfield.org/wp-content/uploads/2010/04/Neil-and-Corey-at-the-Varsity-copy.jpg .

After dinner, Dick Esterle drove us back to the Ritz-Carlton, where we went back up to our hotel rooms and played with puzzles until bedtime.

The day after that, I was waken up very early to get packed up for the airplane trip back to San Jose. We met Bill Gosper in the hotel lobby, and took a cab to the airport, at which point we waited until dawn-ish. On the airplane trip back (with an exchange in Chicago), I looked at all of the exchange gifts in the bag and Bill Gosper programmed on his laptop. Eventually, after having 2 breakfasts due to time zones, we landed in San Jose, drove over to my house, whereupon Bill drove back to his house. And because of time zones, I still had the rest of the day to play with puzzles.

Thus the epic of Gathering for Gardner 9 ended.

It was an absolutely great experience from before it even started to it’s end, and I met a lot of new people and saw a lot of puzzles and magic tricks and optical illusions. I would certainly go the next time it happens, and the time after that. Certainly, it was one of the best events that I attended ever, if not the best.

As an afterword, on May 22, Martin Gardner died. Hearing the news of this was incredibly saddening to me, as well as sudden. He was one of the most important people that ever lived, for mathematics and as well as for many other subjects. He helped popularize M.C. Escher and Godel,Escher,Bach , and introduced mathematics in a fun way to at least everybody at any of the G4Gs. He was truly amazing.