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July 10

24-cell

How can I visualize the 24-cell? Unlike all other 4-polytopes, it isn't related to any regular polyhedra. --134.10.114.238 (talk) 15:08, 10 July 2011 (UTC)[reply]

Our article 24-cell has some nice pictures; I've inserted one here. Looie496 (talk) 17:03, 10 July 2011 (UTC)[reply]

That doesn't help me visualize it. What does it actually look like? --134.10.113.106 (talk) 17:15, 10 July 2011 (UTC)[reply]

It has six octahedrons meeting at each vertex, a sphere round a vertex will be cut to look like a blown up cube. I must admit I think pictures like that are a bit misleading as they show a three dimensional perspective rather than a four dimensional one, the closest in 4D would be somewhere near the centre and should have a larger yellow ball and thicker and longer lines around it. Dmcq (talk) 20:14, 10 July 2011 (UTC)[reply]

It doesn't make sense to ask what something actually looks like when it can't exist in three dimensional space. All you can ask is what projections look like, and a picture is basically a projection onto a 2D space. Looie496 (talk) 22:18, 10 July 2011 (UTC)[reply]

You can use your imagination. Dmcq (talk) 23:03, 10 July 2011 (UTC)[reply]

I find that doing the face first view helps the most. At the W=1 and W=-1 level you have an Octahedron and at the W=0 you have a Cuboctahedron. The 8 Octahedra that border it are the two at W=1/-1 and the 6 formed of one point each in the two Octahedra that you first considered and a square in the W=0.Naraht (talk) 04:52, 11 July 2011 (UTC)[reply]

You can look at it as a tiling of positively curved 3-space. [1] — If you manipulate a 2D projection of a 3D object, it's not hard to get a good idea of the 3D shape. It may be possible to train one's 4D imagination by manipulating 3D projections in stereoscopic view. —Tamfang (talk) 23:06, 15 July 2011 (UTC)[reply]

Nash equilibrium in a two-player zero-sum game

Is it possible to have multiple Nash equilibria in a two-player zero-sum game? By multiple I'm assuming non-trivial solutions with different payoffs in the equilibria, i.e. a payoff matrix of for example ((5,-5), (5,-5), ..., (5, -5)) isn't what I'm looking for. Rymatz (talk) 15:24, 10 July 2011 (UTC)[reply]

No. If you have a Nash equilibrium at strategy pair (x,y) and at (x',y'), consider what the relationships must be between the payoffs at those equilibria and the payoffs at (x,y') and (x',y). Rckrone (talk) 17:31, 10 July 2011 (UTC)[reply]

A two player zero sum game with multiple Nash equilibria is somewhat easy to construct:

	a cooperates	a defects	a alter-defects
b cooperates	a wins 0, b wins 0	a wins 2, b loses 2	a wins 1, b loses 1
b defects	a loses 2, b wins 2	a wins 0, b wins 0	a wins 0, b wins 0
b alter-defects	a loses 1, b wins 1	a wins 0, b wins 0	a wins 0, b wins 0

The four (0,0) outcomes in the lower right are all in Nash equilibrium. I'm not sure if this is non-trivial, but it could be extended to any number of choices with arbitrary labels and you can add (x,-x) to the diagonal and add (y,-y) and (-y,y) to corresponding pairs of cells for arbitrary values of x and y as long as some equalities and inequalities are preserved. 99.24.223.58 (talk) 21:59, 11 July 2011 (UTC)[reply]

The OP excluded the case where the payoffs are equal. With that condition it's not possible to have multiple Nash equilibria. Rckrone (talk) 00:19, 12 July 2011 (UTC)[reply]

Of course I missed that. I wonder if

	a1	a2	a3
b1	(0, 0)	(2, -2)	(1, -1)
b2	(-2, 2)	(0, 0)	(-0.5, 0.5)
b3	(-1, 1)	(0.5, -0.5)	(0, 0)

...would qualify. 99.24.223.58 (talk) 04:59, 12 July 2011 (UTC)[reply]

What is supposed to be the equilibrium other than (a2,b2)? From (a2,b3) b will defect to b2, etc.

Try proving that this is impossible using Rckrone's hint. -- Meni Rosenfeld (talk) 08:58, 12 July 2011 (UTC)[reply]

Tetrahedral pyramid

What is a tetrahedral pyramid: a) a 5-cell; or b) a 5-simplex? --84.61.182.251 (talk) 16:35, 10 July 2011 (UTC)[reply]

The article says that it is a 5-cell and a 4-simplex.

"It is a four-dimensional object bounded by 5 tetrahedral cells"
"It is a 4-simplex, the simplest possible convex regular 4-polytope"

— Fly by Night (talk) 16:51, 10 July 2011 (UTC)[reply]