Blog > The Maximum Score in the Game “Entanglement” is 9080

The Maximum Score in the Game “Entanglement” is 9080

January 21st, 2011

Entanglement is a browser-based game that has gained a fair bit of popularity lately due to its recent inclusion in Google’s Chrome Web Store and Chrome 9. The way the game works is probably best understood by actually playing it, but here is my brief attempt:

  • You are given a hexagonal tile with six paths printed on it, with two path ends touching each side of the hexagon. One such tile is as follows:

  • You may rotate, but not move the hexagon that has been provided to you.
  • Once you have selected an orientation of the hexagon, a path is traced along that hexagon, and you are provided a new hexagon that you may rotate at the end of your current path.
  • The goal of the game is to create the longest path possible without running into either the centre hexagon or the outer edge of the game board.

To make things a bit more interesting, the game was updated in November 2010 to include a new scoring system that gives you 1 + 2 + 3 + … + n (the nth triangular number) points on a turn if you extend the length of your path by n on that turn. This encourages clever moves that significantly extend the length of the path all at once. The question that I am going to answer today is what the maximum score in Entanglement is under this scoring system (inspired by this reddit thread).

On a Standard-Size Game Board

The standard Entanglement game board is made up of a hexagonal ring of 6 hexagons, surrounded by a hexagonal ring of 12 hexagons, surrounded by a hexagonal ring of 18 hexagons, for a total of 36 hexagons. In order to maximize our score, we want to maximize how much we increase the length of our path on our final move. Thus, we want to just extend our path by a length of one on each of our first 35 moves, and then score big on the 36th move.

Well, each hexagon that we lay has six paths on it, for a total of 6*36 = 216 paths on the board. 35 of those paths will be used up by our first 35 moves. It is not possible to use all of the remaining 181 paths, however, because many of them lead into the edge of the game board or the central hexagon, and connecting to such a path immediately ends the game. Because there are 12 path ends that touch the central hexagon and 84 path ends that touch the outer border, there must be at least (12+84)/2 – 1 = 47 unused paths on the game board (we divided by 2 because each unused path takes up two path ends and we subtracted 1 because one of the paths will be used by us).

Thus we can add a length of at most 181 – 47 = 134 to our path on the 36th and final move of the game, giving a total score of at most 35 (from the first 35 moves of the game) + 1 + 2 + 3 + … + 134 = 35 + 9045 = 9080. Not only is this an upper bound of the possible scores, but it is actually attainable, as demonstrated by the following optimal game board:

Paths in red are unused, the green line depicts the portion of the path laid by the first 35 moves of the game, and the blue line depicts the portion of the path (of length 134) gained on the 36th move. One fun property of the above game board is that it is actually completely “unentangled” – no paths cross over any other paths.

On a Larger or Smaller Game Board

Other than being a good size for playability purposes, there is no reason why we couldn’t play Entanglement on a game board of larger or smaller radius (by radius I mean the number of rings of hexagons around the central hexagon – the standard game board has a radius of 3). We will compute the maximum score simply by mimicking our previous analysis for the standard game board. If the board has radius n, then there are 6 + 12 + 18 + … + 6n = 3n(n+1) hexagons, each of which contains 6 paths. Thus there are 18n(n+1) lengths of path, 3n(n+1)-1 of which are used in the first 3n(n+1)-1 moves of the game, and we want to add as many as possible of the remaining 15n(n+1)+1 lengths of path in the final move of the game. There are 12 path ends that touch the central hexagon and 12 + 24n path ends that touch the outer edge of the game board. Thus there are at least (12 + 12 + 24n)/2 – 1 = 11 + 12n unused paths on the game board.

Tallying the numbers up, we see that on the final move, we can add at most 15n(n+1)+1 – (11 + 12n) = 15n2 + 3n – 10 lengths of path. If T(n) = n(n+1)/2 is the nth triangular number, then we see that it’s not possible to obtain more than 3n(n+1)-1 + T(15n2 + 3n – 10) = (225/2)n4 + 45n3 – 135n2 – (51/2)n + 44 points. In fact, this score is obtainable via the exact same construction as the optimal board in the n = 3 case – just extend the (counter)clockwise rotation of the path in the obvious way. Thus, the maximum score for a game of Entanglement on a board of radius n for n = 1, 2, 3, … is given by the sequence 41, 1613, 9080, 29462, 72479, … (A180667 in the OEIS).

  1. bmvaughn
    January 22nd, 2011 at 10:46 | #1

    Seems there is a solution that could yield even larger numbers should you be able to go from the 3rd ring to the 1st ring and back on every step on your final move. Wouldn’t that generate an even higher score?

  2. January 22nd, 2011 at 11:13 | #2

    @bmvaughn – There are 169 lengths of path on the board that you can use, 35 of which you must use in the first 35 moves, leaving at most 134 for the final move. If you count the length of the blue path on the posted image, you’ll see it has length 134 and so it is indeed optimal — there is no way to do better.

    Of course, there may be many, many other optimal boards out there. It doesn’t really matter *how* the final blue line crosses over the rest of the image — we could have crossed from the 3rd ring to the first and back repeatedly. But as you can see the posted image already uses all 12 entrances/exits of each tile, so there’s no way another strategy could do better.

  3. Michael
    January 22nd, 2011 at 16:34 | #3

    here’s a different optimal path i found yesterday! nice mathematical analysis!

  4. January 22nd, 2011 at 17:45 | #4

    @Michael – Awesome, thanks for sharing 🙂 It would be neat to somehow derive the number of optimal paths that exist, but it seems like a rather messy problem so I haven’t had the nerve to try my hand at it yet.

  5. Ross Presser
    January 24th, 2011 at 00:58 | #5

    The odds are extremly high against an actual game resulting in an actual optimal path. The hexagons dealt to you are random, not all the same type as in your optimal path. There is no look-ahead either, so you can’t really plan your path out to the end.

    However, the actual all-time high score for path length is 134, which is amazing, only 35 short of the optimal 169.

  6. Nic
    January 24th, 2011 at 12:19 | #6

    impossible — my highest score was 227 with 85 segments

  7. January 25th, 2011 at 03:37 | #7

    Hey, very nice analysis!
    I think the question about the number of optimal paths is interesting, and as said, very difficult. However, once an optimal path is found (like the one here) one could ask what the chance of getting the right pieces in the right sequence if they are selected at random. Given the regularity of the optimal path found here (you need to get many long sequences of the same path!), the probability is pretty low.

    Is anybody able to find the most non-regular optimal path? By regular I mean that it requires the shortest sequences of the same path for its construction.

    Keep going!

  8. Jason
    January 27th, 2011 at 16:02 | #8

    Wouldn’t that be a path length of 85, far less than 169?

  9. Spinoza
    January 27th, 2011 at 19:09 | #9


    I am pretty sure your path is THE best if the scoring system was points= 1*2*3*….*n (where the best thing to do would be to score big on the last one, obviously). Im also pretty sure your path is optimal for each scoring system where having a long path is favoured more than having a ton of smaller paths. In short, there is some kind of “sur-optimality” in that path. Have to work on that.

  10. SereneSelene
    January 29th, 2011 at 02:35 | #10

    Hey, for a game that’s simple on the surface and a heck of a lot of fun, it sure is complex! They should include this Entanglement in mathematics courses.

  11. Mihail Sergeev
    February 2nd, 2011 at 17:10 | #11

    You should publish the sequence of maximal scores per game radius on OEIS, they don’t seem to have it. 🙂

    Great work!

  12. yagdash
    February 2nd, 2011 at 18:29 | #12

    Gotta be some hell’s combinatorics problem using Bernoulli and the fitness function T you derived.

  13. yagdash
    February 2nd, 2011 at 19:05 | #13

    Wrong address, “Nathaniel derived” the T function.
    Probability of one given path is pretty straight forward (1/6)^(3n(n+1)), n being field radius.
    6 possible positions to every tile, probability of getting one specific position for each is 1/6. For k tiles it’s 1/6 at power of k.

  14. Niki
    February 7th, 2011 at 19:20 | #14

    It’s over 9000!

  15. joshshat
    February 8th, 2011 at 16:57 | #15

    Get Smart 😀

  16. CT
    February 9th, 2011 at 19:47 | #16

    Finely people who proses the world the world the way i do

  17. Dsplayname
    February 10th, 2011 at 19:26 | #17

    I’m fairly certain that it is written into the game that the same piece will never appear 4 times consecutively, thereby rendering this “Optimal path” nonexistent. I have not done the math, but based upon actually observing the game and how the pathes of the tiles are chosen, the highest score possible is in the 4000 range. Of course, we could always just ask the developers for a definitive solution.

  18. def
    February 12th, 2011 at 20:23 | #18

    I think you are wrong. the pieces are chosen randomly, there isn’t much sense in preferring one piece over another, though the probability of the same piece to appear in 4 consecutive times is (really) small.
    Plus, I don’t think the developers care much about what is the highest score possible, they probably got more important things to do.

  19. ColoradoHermit
    February 13th, 2011 at 16:12 | #19

    My best is only 235 and 96. And this so called highest score is flat impossible. Since there is no way of controlling which tile comes up next for you to use it cant be done. Besides that the funky patterns on the tiles used to show your “highest score possible” don’t exist in the first place. At least I’ve never seen any with any of those patterns to choose from. I would just like to see a screen shot of the final pattern for those scores shown that are over 1000. I’d really like to see the final pattern for today’s high of 1856 that’s listed, cuz I don’t belive that’s possible without some way of changing the tile selections, so that you have more than just the 2 to choose from, and being able to back up a move and change it.

  20. February 13th, 2011 at 18:58 | #20

    @ColoradoHermit (and everyone else who mistook this post as meaning that 9080 is easily-attainable or some such thing) – I got all of the tiles from the game itself. Each and every tile in the image above is from a screenshot of the game; they all exist. I’m not so good at image manipulation that I was able to recreate them myself.

    The reason that you perhaps don’t recognize some of them is because there are hundreds of different tiles in the game (there are lots of ways to arrange 6 paths between 12 nodes). This is also the reason why in most games you don’t see the same piece appear multiple times in a row. As far as anyone can tell though, they are indeed randomly-generated on the fly as the game is played (though feel free to e-mail the developer and ask if you believe otherwise).

    Also, this post is about the mathematically best-possible score. Not the best score that you expect to attain on a given playthrough, or the best score that a “good” player will get, or the best score that you will see on the high-score boards. It is about the best *possible* score that you could ever, ever see, assuming someone got extraordinarily lucky. It means that if someone said “I bet you $1 trillion that I got a score of 9081 on Entanglement”, you should take that bet, whereas if they said “I bet you $1 trillion that I got a score of 9080” then you should maybe think twice.

    It’s just like how the best possible score in Jeopardy! is $566,000. Of course no one has actually ever won that much, but that’s the most that anyone could ever hope to win in a single episode.

    As for how likely it is to get a string of tiles that could potentially lead to an optimal game with a score of 9080, I simply don’t know. The image included in this post is not the only optimal board – it is just one that is particularly “nice”, has relatively easy-to-see optimality, and is easily generalized to boards of different radius. There are plenty of optimal boards out there that use very different tiles and have no apparent symmetry, but they’re not particularly fun to draw or come up with, and they’re not particularly enlightening to look at either.

    Anyway, I’m glad that so many people found this post interesting. Cheers 🙂

  21. February 19th, 2011 at 23:56 | #21

    dig the math, brother. based on the first few games i have played….i wondered about the highest possible score.

    ftl, heavyier than a planet, and way past the point of know return….

    rock on!

  22. sandeep
    February 21st, 2011 at 05:00 | #22

    @Nic i scored 319 on my 15th attempt with a path length of 89.i think much higher score can be attained.I can share my scrrenshot if you want.

  23. lickster
    February 27th, 2011 at 03:00 | #23

    So now we know that for a given set of random hexes(for one full game), that the computer dishes out one by one, there are many optimal paths with maximum score. What do we try to achieve in each step to get as close to the maximum score as possible?

  24. vovikb50
    March 11th, 2011 at 04:21 | #24

    possible, mine was 304 with 95

  25. March 11th, 2011 at 05:31 | #25
  26. Jon
    March 24th, 2011 at 16:44 | #26

    I did some quick work, and there are 12 connections on each tile. for the first connection there are 11 possible positions, for the second there are 9, third 7, fourth 5, fifth 3, and sixth 1. 11*9*7*5*3=10395 possible combinations of paths through a tile. unfortunately I’m not mathematician enough to know if this number includes non-unique rotated paths or not. if it does, then I’m not sure how to find the number of unique paths, because 10395/6 doesn’t give an integer quotient

  27. March 30th, 2011 at 21:06 | #27

    I started playing Entanglement a few weeks ago, and while I can’t seem to break my score of 673, I don’t think I should be ashamed about “only” getting 16th on the daily top 100. It did take a lot of luck to get that far.

    Anyway, I was actually thinking about this subject while I was playing earlier today, and the equation that I came up with is a lot simpler than anything I’ve seen on this page. I got the same total as you, so here’s my theory for any move in the game–


    If you take that and apply it to the “maximum possible score” question, it looks like this–


    As you know, that 35 is the initial 35 that’s used to fill up the board, and the 134 is that epic final move.

    I’ve tried it with a lot of path lengths, and it seems to work in all cases. So if anyone’s brain starts to hurt when they’re thinking about points, it’s (n+1)(n/2)

  28. March 30th, 2011 at 21:12 | #28

    Actually sorry- I found that equation up there right before the beginning of the comments@Derek

  29. Nick
    April 4th, 2011 at 13:41 | #29

    There are pieces in your solution that never appear in the game, is there an optimal solution with only possible pieces ?

  30. Mark
    April 13th, 2011 at 04:22 | #30

    You should calculate the probability of getting all of those tiles (assuming the piece generator is truly random)

  31. Stereotyped Blue Strawberry
    May 9th, 2011 at 01:09 | #31

    A good solution. A somewhat more difficult problem will be, what is the highest attainable score in the Sakura Grove mode. When I have some more time next week I will give that a shot.

  32. Kevin
    May 23rd, 2011 at 11:41 | #32

    None of the pieces used in this path are made up. The board is made entirely of possible game pieces. Of course, it is very unlikely that a person is given these pieces in the correct order, but it is still possible.

  33. Jack3d
    May 23rd, 2011 at 21:19 | #33

    The highest score to date is 3551. I had a personal best score of 951 with 41 path segments on the last move, on a different game with a lower score I got 92 as the longest path. To get a high score you must construct a very long path separate to the one you are currently on which you then use in your last move.

  34. May 24th, 2011 at 18:00 | #34

    I have been dabbling in this game a bit, and it is certainly addictive. My all time best is only 359 with longest path of 98. Nowhere near a record.

    I tend to build huge paths, but then get screwed and lose out when I can’t get the final piece needed to connect to it! Oh well. Would be interesting to play on a larger board too bad you cant just adjust it in options or something.

  35. Jof
    June 5th, 2011 at 05:14 | #35

    I was intrigued by the comment:

    “The reason that you perhaps don’t recognize some of them is because there are hundreds of different tiles in the game (there are lots of ways to arrange 6 paths between 12 nodes).”

    So I thought I would try and work out how many different tiles there were. After several hours and much hair-pulling later, I eventually came up with an answer of 1,799! Can this really be correct?

  36. June 6th, 2011 at 00:45 | #36

    @Jof – Yep, 1799 is correct 🙂 That number can be computed using Burnside’s lemma or via brute-force (I’ve done it both ways just to be extra sure). It’s also worth pointing out that not all of the tiles are equally probable — tiles with lots of rotational symmetry occur less frequently. For example, the tile in the top-left corner with six half-loops on it only has probability 1/10395 of occurring (and no, it’s not a coincidence that 10395 showed up in Jon’s comment as well).

  37. Jof
    June 7th, 2011 at 07:23 | #37

    I went the brute-force route. Wrote a program to create all the different sets of paths (yes, I got 10395 of these). Then came up with a notation that would allow me to ‘rotate’ these tiles to see if they matched any of the others and used this to eliminate the duplicates. Good to know my logic was correct. Thanks for the very interesting blog.

  38. dainichi
    August 3rd, 2011 at 23:21 | #38


    Maybe I’m just stating the obvious here (or using fancy names for something obvious), but it seems to me that if we consider the board (excluding path ends facing walls/the central hexagon) an undirected graph, regarding hexagons as nodes and connections as edges, every Eulerian cycle (a cycle using all edges exactly once) whose first 35 edges form a Hamiltonian path (a path that visits all nodes exactly once) would be an “optimal path”.

    I might not really constructively be getting us closer to the number of optimal paths, but since counting Eulerian cycles is known to be #P-complete (, at least I hope to substantiate that it’s hard! Not sure how the regularity of the graph and the restriction that the first 35 edges have to be Hamiltonian adds to/subtracts from the complexity, though.

  39. Anpadh
    August 20th, 2011 at 17:08 | #39

    Not sure if anyone has noticed and/or agrees, but it seems to me that Entanglement is at least partly based on chess. You can move only one “square” (well, hexagon) the first move. Later, you can create a Knight-move (2 and a half steps). Obviously, you cannot create a Bishop or Queen move (diagonally) because you would hit the centre, but you can definitely make a Rook move. The walls are angular, no doubt, but it IS possible to move in a kind of a “straight line” and “connect the dots” if you see what I mean. Aside from the moves, the logic is also chess-like. The centre (King) can never be compromised. The longer you control the centre (the line of hexagons between the line closest to the centre and the line closest to the wall) you increase your chances of maximizing your score. I just thought it was interesting that there were so many similarities between chess and Entanglement.

  40. ajb123
    September 16th, 2011 at 18:50 | #40

    Great analysis Nathaniel! Being like-minded, I had been thinking of figuring this out myself, but couldn’t find the time due to time spent playing the game instead. 🙂 Just scored my best score to date, 754, path length 131. I’m not sure that I’ll beat that too often…

  41. ajb123
    October 11th, 2011 at 00:15 | #41

    1821. 128 length. Gotta quit… I’d love to see a breakdown of the probabilities of getting different kinds of tiles, e.g. the oh-so-valuable non-intersecting types good for the outer ring, or of horribly intersecting (i.e. six straight lines).

  42. Pythius Jang
    November 15th, 2011 at 20:59 | #42

    Okay, I will accept (because I’m lazy) that there are 1799 distinct organizations of 6 non-intersecting paths connecting 12 nodes. Now, like most players, we sometimes have to gamble that in the next tile we will get at least one path configuration that we desire. What is the probability of that? Release the Mathletes!

  43. Pythius Jang
    December 19th, 2011 at 19:12 | #43


    Okay, I will accept (because I’m lazy) that there are 1799 distinct organizations of 6 non-intersecting paths connecting 12 nodes. Now, like most players, we sometimes choose to gamble that in the next tile we will get at least one path configuration that we desire. What is the probability of that? Release the Mathletes!

    Would someone mind computing, then posting the probabilities of each path configuration? Nathaniel, that sounds right up your alley! 😉 Thanks!

  44. December 20th, 2011 at 00:14 | #44

    @Pythius Jang – Each tile has either a 1/10395, 2/10395, 3/10395, or 6/10395 probability of appearing at any given stage, depending on the rotational symmetry of that tile. Tiles that are completely rotationally symmetric (e.g., the piece at the top-left of the image in this post that has the six U shapes on it) are the least likely to occur (i.e., probability 1/10395). Shapes that are 120-degree rotationally symmetric have probability 2/10395. Shapes that are 180-degree rotationally symmetric have probability 3/10395. Finally, shapes that have no rotational symmetry are most common, with probability 6/10395 each.

  45. Skeptic
    May 21st, 2012 at 01:18 | #45

    The high scores I’ve seen published could only be achieved by someone who had hacked into the program to make manipulation possible.

  46. H Pereira
    September 17th, 2012 at 19:11 | #46

    Givin’ a try to at least guess the number of routes leading to the optimum score (a bit over midnight, so I’m not too sharp, but I think I’m right; please comment if I’m making any mistakes.)

    Starting the first path passing all tiles (the green one above), the first step is irrelevant (due to symmetry), the next gives you 5 possible moves, and after that, following the inner ring and then the second gives you 16 moves where each moves gives 4 possible directions (going directly outwards will lower that, but I’ll ignore that for the guess). The outer ring leads only to two possible ways to pass all consecutive tiles: clockwise and counter-clockwise. That gives 5 x 2^16 x 2, comes down roughly to 10^10.

    Then ‘entangling’ the way back passing all 134 possible steps that are left (the blue line). Let’s first look at the number of possible paths to follow per tile: for the tiles in the second circle, that’s 9 (12-2-1=9: -2 are the points where the green line runs through, -1 for where you enter the tile). Similarly, the number of paths per tile are 3 for the corner tiles, 5 for the outer ring, and 7 for the center ring. I hope you get where I’m going so cutting my story short would yield a final number of (9!)^12 x (7!)^6 x (3!)^6, and that multiplied by the 10^10 I mention above, which comes down to 4 x 10^103.

    Even if this guess is over-estimated, it will still give an increadibly high number. It must be possible to get one of these possible solutions with the random tiles given 🙂

    But, to disencourage you all from trying, there must be roughly 11^(12*36) = 10^450 routes possible that give any lower score.

  47. Jim
    April 13th, 2013 at 23:31 | #47

    Just today, 2013-04-13, “Smart Familiar Mandolin” has posted two impossible scores: 1234567 (which is clearly hacked) and 10,119 (which must be a cheat, thanks to Nathaniel’s theorem). Also, “Free New Flag” has posted 6972 today. This is also suspicious, since it occurred today, and since the 4th all-time high score is 5353.

    Have such fake scores appeared before, and been erased?

  48. Jim
    April 14th, 2013 at 00:08 | #48

    I have checked and “Free New Flag”‘s score was not posted today. It was posted this week.

  49. Rhys
    July 24th, 2013 at 22:38 | #49

    Great maths there! I miss the days when I understood what you said.

    on a side note: Lofty Prancing Llama currently holds 39 of the top 100 all time scores. Sadly s/he peaked at 8th a mere 1111 points from the current highest of 5195.

    My record is 191 on 83. Long way to go yet.

  50. frustranglement
    July 1st, 2014 at 18:41 | #50

    My best is 263, my wife’s is 243, my 9 yr old son got 288 on his 3rd game. Scary thing is the wife and I both have 130+ Iq, scares me to think what his might be. He’s even smart enough to sit back and watch his frustrated parents try to best his score rather than try to show off and go higher.

  51. loqk
    January 15th, 2016 at 12:51 | #51

    I was going to leave a comment congratulating you on your analysis, empathising over those who had to complain about what they had misinterpreted, and laughing at all the people who feel the need to show off how much maths they don’t quite understand, but have heard of.

    Then I spotted the “Assistant Professor” and saw that at the time of this post you were working on your Ph.D. in mathematics.

    My kudos on your analysis seem somewhat trite now, so, may I say, congratulations on your appointment 😀

  52. JDarc
    June 7th, 2016 at 01:32 | #52

    @Jon Not sure if anyone answered this, but you have to divide by n-1, if n is the number of unique node to node paths with a turnable tile, or 5 and not 6, making 2079 possible turnable tiles. But I seem to know so many of the tiles that show up, so maybe the programmers picked somewhat less than 2079 to put in the game.

  53. JDarc
    June 7th, 2016 at 01:40 | #53

    @JDarc Oh well, I wasn’t taking symmetry into account. Darn. Back to my cut out tiles to test my math. 🙂

  54. JDarc
    June 7th, 2016 at 03:13 | #54

    Why am I staying up so late for this? I find patterns irresistible – that has to be it.

    In my preliminary research on symmetry, I’m finding the math will be much more complex when limiting tiles that look completely different to a human.

    For example, let’s say 4 paths will use this algorithm (node-x + 4). The 4 paths are identified as nodes 1-5, nodes 2-6, nodes 3-7, nodes 4-8. Then the two remaining paths could have their own repeatable pattern of (node-x + 1) or they will not follow any repeatable algorithm. Those remaining two paths can only be situated in two ways. For the example, I’ll say the paths are nodes 9-12 and 10-11. If you turn the tile and write down the new resulting paths, mathematically there will be 4 unique path sets. But the tile “looks” the same to a human, so there is only 1 tile that will be used in the game. Even if you look at the mirror image of this tile, it’s the same, again leaving only one tile.

    Now for an example where the mirror image is different. The tile is not symmetrical, yet some paths will use the same algorithm. For 4 of the paths, I used two different repeatable algorithms (node-x + 1 and node-x + 5) and the 2 remaining paths had non-repeating algorithms. The first set of these paths are:
    Nodes 1-2
    Nodes 3-4
    Nodes 5-10
    Nodes 6-11
    Nodes 7-9
    Nodes 8-12

    Mathematically there are 6 sets of unique paths in that example. But the tile looks the same any way it’s turned. Because the tile is not symmetrical, the mirror image looks slightly different – this results in 6 more unique path sets. So there will only be two tiles used in the game, not 12 as math would suggest.

    Math calculations for the true number of tiles in the game can be found based on these 3 dynamics:
    -The number of paths that use the same algorithm
    -The number of paths that use different algorithms than any other path
    -The decision about symmetry (the mirror image thing)

    However, I have not figured out the math. I have only played with little hexagonal cutouts and noted my human-view findings. 🙂

  1. January 22nd, 2011 at 06:26 | #1
  2. January 23rd, 2011 at 02:01 | #2
  3. February 14th, 2011 at 10:28 | #3