Wednesday, January 26, 2011

Entanglement Basic Strategy

Basic Entanglement Strategy

I discovered a simple to learn puzzle game, gopherwood studio's "Entanglement"
http://entanglement.gopherwoodstudios.com/
while checking on Google's Chromium's "Web Store".

As a 1300+ player, with a record in the best 25 ever, I think I might have something to say about the basic strategy of the game.

First, the scoring is very simple: When you place a tile, you get an arithmetic progression of points per tile crossed (1 + 2 + 3 + ...). Since the total number of points progresses quadratically on the length of the paths, it makes all the sense in the world to prefer to play single-steps if that way you prepare very long paths that you hope to eventually traverse. The principle is this: let's say you have the option to place two tiles to make a path of 20 squares, either make two steps of 10 squares each, or a sequence of length 1 followed by length 19: the first case will give you 5*11*2 = 110 points, the second will give you 1 + 19*10 = 191. So, there you have it, it is most efficient to prepare one long path even while making 1-length extensions to your line. This is the cardinal rule. For the sake of example, consider a 4-tile sequence of total length 20 and individual lengths 1, 1, 1, 17: You would get 1 + 1 + 1 + 17*9 = 156 points; versus a sequence of 5, 5, 5, 5 = 4 * 5 * 3 = 60.

Whenever you have freedom to place any of your two tiles and many rotation options, chose the one that connects the longest two "free" paths [[ I call a path "free" when neither extreme is a wall ]] inbound to the square you are placing. This strategy also minimizes the total number of paths.

In the early stages, try to keep the tiles that when placed at the six corner positions would leave no standing non-free path (a corner position is one of those that have three walls). Since in the corners your options are reduced to the minimum it is worth keeping those tiles around.

Try to connect non-free paths together as soon as possible.

In the later stages, make sure you have an exit path when you get into an isolated hole. Obviously try to keep the option to go over all the hexagons, but if going to an hexagon and back would give you two short paths at the price of not joining two long paths you could traverse later, then it's preferable to forgo that hexagon altogether.

Lastly, this game is stochastic, if you aim for an absolute record, focus on joining the longest free paths and joining non-free paths and assume you will get the hexagon you need; I don't think you can reliably get 2000 points or more playing it safe, playing safe, with my current understanding of the game, I get around 500 points reliably but it is extremely hard to get more than 1000.

Fellow blogger Nathaniel Johnston blogs about the theoretical maximum score:
http://www.nathanieljohnston.com/2011/01/the-maximum-score-in-the-game-entanglement-is-9080/

One non-related recommendation: For Christ's (or Alah's, Buddha's, etc) sake, please use Google's Chromium rather Google's Chrome, the fellow blogger "Manuel Jose" explains: http://www.techdrivein.com/2010/05/why-cant-we-all-use-chromium-instead-of.html

12 comments:

Anonymous said...

Very good the article.

Eddie said...

You are welcome

Anonymous said...

Sorry, could you explain this with a picture? I'm a bit confused...

"Whenever you have freedom to place any of your two tiles and many rotation options, chose the one that connects the longest two "free" paths [[ I call a path "free" when neither extreme is a wall ]] inbound to the square you are placing. This strategy also minimizes the total number of paths."

Eddie said...

Pictures are a real pain in blogspot. Let me explain the idea:

In the early game you have freedom to use any of your two hexagons, there is not much of a difference between which one you use. But if you can, join the longest "free paths".

A free path is a path that can be extended both sides.

The main objective is to connect free paths so that you have the longest path possible to traverse in your last hexagon.

If you do this, to connect free paths together and non-free paths as soon as possible, you minimize the number of free paths, ideally, that number is just 1.

og said...

i wish i knew how to do this... worst part is my gf is scoring higher than me. I really suck at this. pls help

og said...

any help will be very much appreciated

Noige said...

Ok my tips: first you gotta make sure you get the tiles with a u-shape circling the centre tile. So there are ZERO dead-ends into the centre tile other than the one on your initial tile.

Then, I like to cover a certain region first, ie NOT go all around the centre and leave the perimeter hexagons until the end. If you do the latter, you're stuck with no options at the end. You need to take on a whole quadrant and fill it up, choosing whether to go to the perimeter or stay inside, depending on what tiles you get. When you get a fantastic tile that is perfect for the corners, save it until you get to the corner - it's worth it. You'll know what should go where after about twenty games and studying each tile pattern. Bloody great game!

-MG said...

I agree with all the OP's advice and can get 500 without much work but the only way I seem to get 1000 is with luck plus extra patience...my record is around 1250. But the top scores are well over 2000 plus, I see Shiny Wet Llama on the leaderboard getting over 1000 regularly so I suspect there is something more than our strategy, we are missing something I think.

Kevin Quigley said...

I am hoping to make my own set of physical tiles, but would love to see if anyone's determined the total set of unique tiles. If it's more than 200, it's probably not worth the trouble of designing them, printing and cutting them out... Has anybody out there calculated the total tiles or even come up with a set?

Eddie said...

A while ago I generated a program to generate all possible tiles, the number was not big; however, I did not figure out the *distribution* of the tiles, I think it is not uniform

Jeff Severson said...
This comment has been removed by the author.
Anonymous said...

+1 to first by Anon :)