"Aztec Diamond" analogue for Square-Octagon graph. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:59:39Z http://mathoverflow.net/feeds/question/110169 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110169/aztec-diamond-analogue-for-square-octagon-graph "Aztec Diamond" analogue for Square-Octagon graph. John Mangual 2012-10-20T16:56:46Z 2012-10-22T02:57:01Z <p>I have been reading David Speyer's <a href="http://arxiv.org/abs/math/0402452" rel="nofollow">Perfect Matchings and the Octahedron Recurrence</a>, trying to carry out his "cross-wrenches" generalization of the Aztec diamond. In what follows, I'm asking for a construction of $G_{n_0,i_0,j_0}$ in the case where the infinite graph $\mathcal{G}$ gives <a href="http://en.wikipedia.org/wiki/Truncated_square_tiling" rel="nofollow">square-octagon tiling</a>. </p> <p>I found Speyer's notation very difficult. Maybe section 4 of <a href="http://arxiv.org/abs/math.CO/0501521" rel="nofollow">Perfect Matchings and Perfect Powers</a> by Mihai Ciucu will be easier to use for this special case. <hr> In section 1.2 the "Aztec Diamond" theorem is stated $f(n_0, i_0, j_0) = \sum m(M)$ </p> <ul> <li>$f(n_0,i_0,j_0)$ is the solution to the octahedron recurrence. $$f(n,i,j)f(n-2,i,j)= f(n-1,i-1,j)f(n-1,i+1,j)-f(n-1,i,j-1)f(n-1,i,j+1)$$</li> <li>The sum over matchings of generalized Aztec diamond graphs $G(n_0, i_0, j_0)$ is called $\sum m(M)$. These graphs are embeeded in an infinite grid $\mathcal{G}$.</li> <li>For now, every matching is counted with unit weight: $m(M)=1$. </li> <li>$n_0+i_0+j_0 \equiv 0 \mod 2$</li> </ul> <p>What is the sequence of shapes corresponding to Speyer's "crosses+wrenches" construction for the square-octagon lattice? Section 3.7 some relevant info.</p> <p>The faces of his lattices are indexed by pairs of integers $(i,j) \in \mathbb{Z}^2$. He defines a "level" ( I really want to avoid using his word "height", which has another meaning in terms of dominos.) $$h(i,j) = \left\{ \begin{array}{rc} 0 &amp; \text{if }(i,j) \equiv (0,0) \mod 2\mathbb{Z}\times 2 \mathbb{Z} \\ 0 &amp; \text{if }(i,j) \equiv (1,1) \mod 2\mathbb{Z}\times 2 \mathbb{Z} \\ 1 &amp; \text{if }(i,j) \equiv (0,1) \mod 2\mathbb{Z}\times 2 \mathbb{Z} \\ -1 &amp; \text{if }(i,j) \equiv (1,0) \mod 2\mathbb{Z}\times 2 \mathbb{Z}<br> \end{array} \right.$$ The Octahedron recurrence has initial conditions $f(h(i,j),i,j)=1$ and specializes here to powers of <strong>5</strong>: \begin{array}{rlc} f(2n,i,j) &amp; = 5^{n^2} &amp; \\ f(2n+1,i,j) &amp; = 5^{n^2+n} &amp; \text{if } i \equiv n \mod 2\\ f(2n+1,i,j) &amp; = 2 \cdot 5^{n^2+n} &amp; \text{if } j \equiv n \mod 2 \end{array} I'm not even sure this list of possibilities is exhaustive. For $(2n+1,i,j)$ only one of $i,j$ can be odd.</p> <p><img src="http://oi46.tinypic.com/35i88zk.jpg" width="200"></p> <p>The shapes $G(n_0, i_0, j_0)\subset \mathcal{G}$ are planar graphs with "open" and "closed" faces. He defines the "lattice", "edges", and "faces": \begin{eqnarray} \mathcal{L} &amp;=&amp; \{ (n,i,j) \in \mathbb{Z}^2 : n = i + j \mod 2\} \\ \mathcal{E} &amp;=&amp; \{ (i,j) \in \mathbb{Z}^2: i + j \equiv 1 \mod 2\} \times \{ a,b,c,d\} \\ \mathcal{F} &amp;=&amp; \mathbb{Z}^2 \end{eqnarray} The faces are indexed by pairs of integers. The edges are labelled <strong>a,b,c,d</strong>. </p> <p>In section 2.1 Speyer defined some cones in $\mathbb{Z}^3$: \begin{eqnarray} p_{(n_0,i_0,j_0)} &amp;=&amp; n_0 - |i - i_0| - |j - j_0| \\ C_{(n_0,i_0,j_0)} &amp;=&amp; \{ (n,i,j) \in \mathcal{L}: n \leq n_0 - |i - i_0| - |j - j_0| \} \\ \mathring{C}_{(n_0,i_0,j_0)} &amp;=&amp; \{ (n,i,j) \in \mathcal{L}: n &lt; n_0 - |i - i_0| - |j - j_0| \} \\ \partial C_{(n_0,i_0,j_0)} &amp;=&amp; \{ (n,i,j) \in \mathcal{L}: n = n_0 - |i - i_0| - |j - j_0| \} \\ \mathcal{I} &amp;=&amp; \{ (i,j, n) \in \mathcal{L}: n = h(i,j) \} \\ \mathcal{U} &amp;=&amp; \{ (i,j, n) \in \mathcal{L}: n > h(i,j) \} \end{eqnarray}</p> <p>I have been unable to sort out definition of $G_{n_0,i_0,j_0}$ in section 3.3</p> <p>It would be really amazing if one could show how the dominos actually "shuffle".</p> http://mathoverflow.net/questions/110169/aztec-diamond-analogue-for-square-octagon-graph/110265#110265 Answer by John Mangual for "Aztec Diamond" analogue for Square-Octagon graph. John Mangual 2012-10-21T20:09:16Z 2012-10-22T02:57:01Z <p>The faces are indexed by $\mathbb{Z}^2$, but $\mathring{C}_{(n_0,i_0,j_0)} \cap \mathcal{I} \in \mathbb{Z}^3$. The closed faces of $G = G_{(n_0,i_0,j_0)}$ centered at $(n_0,i_0,j_0)$ satisfy an inequality:</p> <p>$$\mathring{C}_{(n_0,i_0,j_0)} \cap \mathcal{I} = \{ (i,j,n): n = h(i,j) &lt; n_0 - |i - i_0| - |j - j_0| \}$$</p> <p>How do we get a planar graph? Speyer defines a projection map that drops the last coordinate:</p> <p>$$\alpha(\mathring{C}_{(n_0,i_0,j_0)} \cap \mathcal{I}) = \{ (i,j): h(i,j) + |i - i_0| + |j - j_0|&lt; n_0 \}$$</p> <p>This number $h(i,j) + |i - i_0| + |j - j_0|$ seems to be important for building the graph. <hr> One simple height function" is an alternating pattern of 0's and 1's. We then overlay <a href="http://en.wikipedia.org/wiki/Taxicab_geometry" rel="nofollow">taxicab distances</a> from various points to get the height function. The Aztec Diamond patterns emerge</p> <pre> 4 010101 3 444 101010 333 44244 010101 33133 4422244 101010 3311133 442202244 010101 33133 4422244 101010 333 44244 3 444 4</pre> <p>Let's try this procedure for square octagon lattice.</p> <p><img src="http://s16.postimage.org/acg6tmkmb/matrices.gif" /></p> <p>We get a clear general sense of the shape. The Mathematica code is:</p> <pre><code>h[x_, y_] := Switch[{Mod[x, 2], Mod[y, 2]}, {0, 0}, 0, {1, 1}, 0, {1, 0}, 1, {0, 1}, -1] cut[x_, y_] = If[ x &lt; y, x, "."] b = Table[ cut[Abs[k] + Abs[l] + h[k + 1, l], 6], {k, -6, 6}, {l, -6, 6} ]; MatrixForm[b] </code></pre> <p>It seems to be possible to take off the corners and still be an acceptable "diamond".</p> <hr> <p>The analogue of "domino" shuffling in these case would be an "inductive" way of moving from one Aztec Diamond to the next. <del>Not sure how to do this at the moment</del>. There seems to be more than one way</p> <p><img src="http://oi50.tinypic.com/2rhr9ud.jpg" alt="alt text"></p>