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August 5

3D noughts and crosses

What is the chance of the player who takes the first turn, winning? For clarity, the board consists of a 27-cell cube, wherein alignments can be made vertically, horizontally, diagonally across a plane, and diagonally with respect to both planes. Plasmic Physics (talk) 00:26, 5 August 2013 (UTC)[reply]

I think I worked this out once (actually, I started with 4D, but I think it applies to 3D as well), and if I recall correctly, the first player has a guaranteed win by taking the center cube. In fact, I think if you disallow the first player from taking the center, then the second player can always win by taking the center. It's a little more interesting if you make the rule that you can take the center cube only to win on that move. --Trovatore (talk) 02:20, 5 August 2013 (UTC)[reply]

So, that would be a 100% probability, if the first player always makes the optimal move. Different probabilities would apply for sub-optimal or even random moves. StuRat (talk) 13:34, 5 August 2013 (UTC)[reply]

All finite-length perfect-information games are determined, so the probability can only be 100% or 0%, assuming perfect play by both sides. Note though that chess (with an inessential modification) is such a game, so the theorem doesn't always apply in practice, because "perfect play" is too hard to figure out. --Trovatore (talk) 17:47, 5 August 2013 (UTC)[reply]

Your rule modification intrigues me, how does it affect the fairness of the game, assuming both players have an equivalent handicap? Plasmic Physics (talk) 04:57, 6 August 2013 (UTC)[reply]

Well, what do you mean by "fair"? It's still a determined game, so if neither player makes a mistake, the outcome will never vary. By a simple strategy-stealing argument, it can't be a win for the second player, so either the first player can always win, or perfect players will always draw — I suppose you could argue that the second possibility is "fair".

In practice, a determined game can still be more or less "fair" if the optimal strategy is too difficult to calculate, but I doubt that's going to be the case here. I think my modification makes it a little harder to calculate, so maybe that's a little fairer. --Trovatore (talk) 05:41, 6 August 2013 (UTC)[reply]

By 'fair', I mean that if both players make a predetermined number of random strategic errors, that either player has a ~50% chance of winning - so assume an perfectly imperfect game. Plasmic Physics (talk) 07:21, 6 August 2013 (UTC)[reply]

Oh. Well, I'd suggest you set up some simulations for that. Throwing in random errors by the players (are they just completely random moves, or are they somehow supposed to be plausible errors?) isn't something I normally think about when analyzing a game, and I don't have much intuition for what would happen. --Trovatore (talk) 18:41, 6 August 2013 (UTC)[reply]

Well, I did only ask for an empirical answer. (Or meant to.) Plasmic Physics (talk) 23:54, 6 August 2013 (UTC)[reply]

Oh, gotcha — you want to know what happened when I played it with my friends, right? Never did that. It didn't seem interesting enough to bother teaching people and trying it out; I just spent a little idle time thinking about it on my own. --Trovatore (talk) 07:30, 7 August 2013 (UTC)[reply]

My understanding is 3x3x3 is rather trivial, but 4x4x4 is good and it was quite popular at my school. See Qubic. RDBury (talk) 04:07, 5 August 2013 (UTC)[reply]

I used to pass the time in boring classes playing the 4-d version (4×4×4×4). This required careful thought. Over time we developed ambush strategies and counter-strategies (really just play patterns) that we used, and had to look out for the other using. My memory on this is not great, but I think that the 3-d version (4×4×4) almost invariably ended in a draw if both players were reasonably careful. The smaller versions (3×3 and 3×3×3) were pointless (respectively effectively always draw and always first win). There seems to be a relation between the row length and number of dimensions that leads to balance – no marked advantage for either player, but still enough room for either side to win (assuming imperfect play). — Quondum 18:59, 6 August 2013 (UTC)[reply]

A 4D game sound too complex, I can't even imagine half of the possible alignments. For instance, how would you use the extrapose and intrapose faces of the tesseract to make alignments which do not involve the centrapose face? Plasmic Physics (talk) 07:53, 7 August 2013 (UTC)[reply]

Just get a big sheet of paper and write down 16 4x4 squares, in a 4x4 arrangement. Then figure out what "four in a row" means — it's not difficult to figure out. --Trovatore (talk) 07:58, 7 August 2013 (UTC)[reply]

It's the 'four in a row' part that is difficult for me to figure out, which cubes connect with which to form an alignment. If you arrange the cubes in a 4 x 4 pattern, you are forced to remember a lot more spatial information. I would prefer to arrange mine in an isometric net. Plasmic Physics (talk) 08:22, 7 August 2013 (UTC)[reply]

You get used to that pretty quickly: putting it all laid out flat on a sheet of paper as Trovatore suggests works pretty well. You get used to all the alignments: they all occur within a 4×4 section, or on a horizontal, vertical or diagonal row of 4×4 sections, each entry stepping the same amount across in a constant direction. The alignments actually end up equally spaced on straight lines on the paper. An isometric net sounds challenging to construct if not impossible in 3 d, whereas to draw this arrangement is quick and robust. As to complexity, this is easy enough to get used to.

An aside: as an "old hand" at it (at the age of about 16) I introduced the game to a friend's 8-year-old sister, who from the second game when she'd understood my hand-wavy explanation invariably trounced me. I'll blame it on her exceptional intelligence... (she was clearly super-bright). — Quondum 11:32, 7 August 2013 (UTC)[reply]

10 9 20 8

What's the next number? 163.202.48.125 (talk) 14:42, 5 August 2013 (UTC)[reply]

30, 7, 40, 6... Now stop misusing the reference desk. Plasmic Physics (talk) 14:46, 5 August 2013 (UTC)[reply]

According to A205450 the next numbers are 30, 32. ;-) Dmcq (talk) 15:46, 5 August 2013 (UTC)[reply]

That's for an obscure sequence that begins 2, 2, 4, 4 ... All such questions have more than one answer, and it is usually expected that the answer with the simplest explanation (as provided by Plasmic Physics) is "best". Dbfirs 16:19, 5 August 2013 (UTC)[reply]

It's absolutely impossible to tell. You tell me what you want the next two terms to be and I'll give you a quintic sequence with the first four terms 10, 9, 20, 8 and the fifth and sixth terms of your choosing. Suppose you want the fifth and sixth terms to both be zero then

a_{n}:=-{\tfrac {17}{24}}n^{5}+{\tfrac {317}{24}}n^{4}-{\tfrac {735}{8}}n^{3}+{\tfrac {6979}{24}}n^{2}-{\tfrac {4877}{12}}n+205

is a sequence for which

a_{1}=10

,

a_{2}=9

,

a_{3}=20

,

a_{4}=8

,

a_{5}=0

and

a_{6}=0

. In fact, you can specify as many terms as you like after the

10,9,20,8

and find a sequence that matches that data. You'd just have to solve a lot of simultaneous equations. — Fly by Night (talk) 17:35, 6 August 2013 (UTC)[reply]

But only a few choices will yield a polynomial with small coefficients. To find the "correct solution" this way, you can use integer relation algorithms, such as the Lenstra–Lenstra–Lovász lattice basis reduction algorithm. Count Iblis (talk) 00:06, 7 August 2013 (UTC)[reply]

Brass crosses, the sequel

Referring back to this question, I'm going to try another approach.

The actual base for each cross has a rectangular top (sides s1 and s2) and a rectangular base (sides s3 and s5), an isosceles trapezoid in the front and back, and a trapezoid on each side which is not isosceles.

So let's make the assumption the two sides are isosceles, making the length of the base s4 rather than s5. The center has volume hs1s2. The front and back added together have volume hs1(s4-s2). The two sides added together have volume hs4(s3-s1). Finally, the four corners added together form a pyramid with a rectangular base. Its volume is h/3(s3-s1)(s4-s2). I think that works.

The original question was about the difference in volume with the actual base. The front would be (hs1/2)(s4-s2), while the back would be (hs1/2)(s5-s2). The pyramid would look different now. Okay, I'm stuck.— Vchimpanzee · talk · contributions · 19:59, 5 August 2013 (UTC)[reply]

I forgot about this until I saw them again on Sunday. Let's say the volume of the back for an isosceles trapezoid is (hs1/2)(s4-s2).

But if not isosceles, it would be (hs1/2){s5-s2), or (hs1/2)((s5-s4)+s4)-s2). The difference would be ((hs1/2)((s5-s4)+s4)-s2))-( (hs1/2)(s4-s2)).— Vchimpanzee · talk · contributions · 21:00, 13 August 2013 (UTC)[reply]