Covering a $d$-dimensional integer lattice by repeating a minimal set of deterministic moves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:19:51Z http://mathoverflow.net/feeds/question/108085 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108085/covering-a-d-dimensional-integer-lattice-by-repeating-a-minimal-set-of-determin Covering a $d$-dimensional integer lattice by repeating a minimal set of deterministic moves T.R. 2012-09-25T18:11:56Z 2012-09-25T23:16:55Z <p>Imagine I place a turtle on some desired vertex, $v_i$, of a bounded $d$-dimensional integer lattice, $Z^d$, with dimensions $(l_1, ..., l_d)$. The turtle is able to travel from vertex to vertex along the edges of the lattice, but is forbidden from ever returning to a previously visited vertex (i.e. it "burns" the vertices it visits with probability $p = 1$).</p> <p>We want to teach the turtle to cover the $d$-dimensional integer lattice by repeating a series of $P$ deterministic moves along the set of $2*d$ possible direction vectors. For example, in two-dimensions ($d = 2$), we might label the four possible directions in which the turtle can move as $(N, W, E, S)$ and train the turtle with an instruction set like the following: {{Step 1, GO NORTH}, {Step 2, GO SOUTH}, ..., {Step P, GO SOUTH}, {GOTO Step 1}}. With the right instruction set, after some number of GOTO loops, the turtle will visit all possible vertices and quit.</p> <p>Provided that the turtle can be initialized anywhere one desires on the lattice, and provided dimensions of the $d$-dimensional integer lattice, $(l_1, ..., l_d)$, what is the smallest value of $P$ that permits the turtle to cover the lattice?</p> <p>Note - The turtle is forbidden from leaving the grid, and any instruction that leads to this will be counted as illegal (not simply ignored).</p> <hr> <p>Let me provide a few trivial examples:</p> <p>For $d = 1$, the optimal solution is to place the turtle at the far left or right-side of the lattice, and then (assuming the turtle is placed on the far left), program the turtle with the two-line $(P = 1)$ instruction set: {{Step 1, GO RIGHT}, {GOTO Step 1}}.</p> <p>For $d = 2$, with a bounded lattice that has dimensions $N$ by $M$, if $M &lt; N$, we can program the turtle to: (1) first move from $(i, 1)$ to $(i, M)$, (2) move from $(i, M)$ to $(i+1, M)$, (3) move from $(i+1, M)$ to $(i+1, 1)$, and finally (4) move from $(i+1, 1)$ to $(i+2, 1)$, then GOTO (1) until we sweep through the 2D lattice. This constitutes a value of $P = (M - 1) + 1 + (M - 1) + 1 = 2M$. I wonder if it's possible to do better?</p> http://mathoverflow.net/questions/108085/covering-a-d-dimensional-integer-lattice-by-repeating-a-minimal-set-of-determin/108116#108116 Answer by Sergei Ivanov for Covering a $d$-dimensional integer lattice by repeating a minimal set of deterministic moves Sergei Ivanov 2012-09-25T23:16:55Z 2012-09-25T23:16:55Z <p>Yes your example is optimal provided that the longest side is longer than 2. Let me stick to dimension 2, the higher dimensions are similar. I assume the vertices have integer coordinates $(i,j)$, $1\le i\le M$, $1\le j\le N$.</p> <p>The turtle changes its position by a fixed vector $v$ after each cycle of execution. Let $X_1$ be the set of first $P$ vertices visited, $X_2$ the set of vertices from $(P+1)$-th to $(2P)$-th, and so on. The last set $X_k$, $k=[MN/P]$, may contain fewer than $P$ vertices if the last cycle is not executed completely. Note that $X_{i+1}=X_i+v$ (i.e., the translation of $X_i$ by $v$) for every <code>$i&lt;k-1$</code> and $X_k\subset X_{k-1}+v$.</p> <p>First observe that $v$ cannot have two nonzero coordinates if $P\le MN/2$. Indeed, assume w.l.o.g. that both coordinates are positive. Then consider the corner $c=(1,N)$. It must belong to $X_1$ because $c-v$ is outside the grid. Since $c+v$ is also outside the grid, $X_2$ is incomplete, so $P>MN/2$.</p> <p>The case $v=(0,1)$ or $v=(1,0)$ is impossible (if $M,N>1$). Indeed, suppose that $v=(0,1)$. Then every vertex $x$ of the form $(i,1)$, $1\le i\le M$, belongs to $X_1$ because $x-v$ is outside the grid. Then $x+v$ belongs to $X_2$, $x+2v$ belongs to $X_3$ and so on. Therefore no points other than $(i,1)$ belong to $X_1$. Let $x_0$ be the original vertex, then $x_0+v$ is the first visited vertex of $X_2$. So the $P$-th move is from some point of $X_1$ to $x_0+v$. But in $X_1$ only $x_0$ lies at distance 1 from $x_0+v$, a contradiction.</p> <p>So $v$ has a form $(0,m)$ or $(m,0)$ where $m\ge 2$. Then a similar analysis show that $X_1$ consists of first $m$ rows or columns of the grid, so $P=mM$ or $P=mN$. Therefore $m=2$ and $P=2M$ is optimal.</p>