I'm not sure if anyone else has pointed this out, but I've found a way to consistently reach the end with at most 21 Eyes of Ender. I think that this is the most efficient way to consistently find a stronghold. Here are the steps:
Go to x=0, z=0 and throw an Eye of Ender.
Travel 896 blocks in that direction. If you can, find the precise angle of the Eye and use ratios to determine the coordinates to head for.
Throw another Eye of Ender and travel 128 blocks in that direction.
Repeat step 3, dividing the number of blocks in half each time. So move 128 blocks, then 64, 32, 16, 8, 4, 2 and 1.
If you were precise enough, you should be on the exact block of the Stronghold (or close enough for comfort). Use 12 Eyes of Ender to open the portal. Steps 1-4 should have taken at most 9 Eyes of Ender.
How it works: Strongholds appear 640-1152 blocks from x=0,z=0. This is a ring of radius 512. Since log2512=9, an optimal method will always find the stronghold in 9 guesses. This should actually take only 4 or 5 Eyes of Ender, since some of them can be reused. In an absolute worst-case scenario, a precise method requires 21 Eyes to make it all the way to the End. That's 9 for guessing, 12 for opening the portal.
What do you think? Has this been suggested before?
Can't you also use 2 ender eyes and some good old trigonometry to find the stronghold? :] Ive never seen this method though but I think there are better ways
How? I can see how trig would help find the angle of a stronghold numerically, but just finding how far away it is would be very challenging. A ring of choices and a known angle really boils down to game theory, and guessing a number in a range X always takes log2X tries in a case like this. I may be wrong though.
Can't you also use 2 ender eyes and some good old trigonometry to find the stronghold? :] Ive never seen this method though but I think there are better ways
I've seen someone try that and he did his math wrong so it didnt work anyway but maybe ill try it sometime...
And to answer the question it sounds like a pretty good method, ill try that sometime and tell you if it works
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1. Trig is not needed if you triangulate, because you're not working with angles. Just find the the slope of a line and you've got y=mx+b. Then go a few hundred perpendicular and do it again. Then solve two equations in two unknowns. You only need basic algebra.
2. Log2(512) is overkill, because you don't need the one exact block. Strongholds are big. I'll bet once you narrow it down to about 32, you're there. But triangulation really only takes two.
I think your ring of choices is true for data sets, not surfaces.
The question here is how many data you need to completely determine a triangle.
Two angels(given by the Eyes) and one line (the one you travel between the throws) are sufficient.
So, if you travel far enough between both trows (say, 512 meters but that is just guessed, not optimized), note both coordinates and the angels the Eyes flew and which you can look up after pressing F3, its a simple mater of plugging these values into the correct formulas.
I don't know that much about game theory, but I think the log2X-borderline only applys if you are regarding a data set, not if you are looking at a 2D-geometrical problem. (Yeah, the surface of MineWorld is nominally 3D, but because one is able to reach the Stronghold by walking about a surface one can just assume that the problem is 2D.)
Edit afterthought: Nice work, nonetheless, Isometrus! I'm usually to lazy for trigonometrie, so I will look for my Stronghold in your proprosed way
Wow, I never thought of it that way...thanks! I'm totally using that from now on.
As for whether the data set is 1D or 2D, remember that you know the direction of the stronghold. The data set is a 2D ring, but you know the planar cross-section to examine, so it becomes a 512-block line. The 2D method is better, though.
What do you think? Has this been suggested before?
I've seen someone try that and he did his math wrong so it didnt work anyway but maybe ill try it sometime...
And to answer the question it sounds like a pretty good method, ill try that sometime and tell you if it works
2. Log2(512) is overkill, because you don't need the one exact block. Strongholds are big. I'll bet once you narrow it down to about 32, you're there. But triangulation really only takes two.
3. This is not game theory. It's just math.
Wow, I never thought of it that way...thanks! I'm totally using that from now on.
As for whether the data set is 1D or 2D, remember that you know the direction of the stronghold. The data set is a 2D ring, but you know the planar cross-section to examine, so it becomes a 512-block line. The 2D method is better, though.