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I have been working on my research as a student at Wilbur Wright College on Topological Complexity. We solved the problem of two robots moving on a circle and letter $T$ using Farber's theorem but having difficulty finding $TC$ for two robots moving on a number $8$. Our group founded $2 \leq TC \leq 3$ using Farber's theorem.

What will be the exact $TC$ for one/two robots moving on a number $8$ using Farber's theorem?

(Note: We founded configuration space for two robots moving on $8$ is a bouquet of seven circles(seven petals)

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    $\begingroup$ It might help if you would explain or cite Farber's theorem. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 11, 2017 at 18:38

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The topological complexity of $X$ is the minimum number of open sets needed to cover $X\times X$, on each of which the path fibration $X^I\to X\times X$ admits a local section. If I understood your question, then you are asking about the cases $X=S^1\vee S^1$, and $Y=F(S^1\vee S^1,2)$, the $2$-point ordered configuration space.

The wedge of circles $X$ has LS-category $2$ (by Jeff Strom's argument), and hence $TC(X)\le 2\,cat(X)-1 = 3$. The space $Y$ is the configuration space of a graph with one essential vertex, therefore $TC(Y)\le 3$ also: this follows from the theorem of Michael Farber you mention.

To see that both of these space have topological complexity $3$, you can apply a result of Greg Lupton, John Oprea and myself:

Grant, Mark; Lupton, Gregory; Oprea, John, Spaces of topological complexity one, Homology Homotopy Appl. 15, No. 2, 73-81 (2013). ZBL1277.55001.

The main result of the cited paper is that if $TC(Z)=2$ then $Z$ has the homotopy type of an odd sphere (the "one" in the title of the paper is because we are using the reduced version of topological complexity which is one less than your definition). Thus if your space has $2\le TC(Z)\le 3$ and is not an odd sphere, then in fact $TC(Z)=3$. The arguments use little more than the standard zero-divisors cup-length lower bound.

The space $X$ is clearly not an odd sphere (it has nonabelian $\pi_1$, for example). You seem to have convinced yourselves that $Y$ is not an odd sphere, either.

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  • $\begingroup$ This is interesting! My argument is wrong, but I don’t see why. $\endgroup$
    – Jeff Strom
    Commented Dec 3, 2017 at 21:21
  • $\begingroup$ @JeffStrom: it looks like you are using a non-standard definition of topological complexity. The standard definition uses covers of $X\times X$, whereas in your answer you find a cover of $X$. $\endgroup$
    – Mark Grant
    Commented Dec 4, 2017 at 7:09
  • $\begingroup$ Thank you for all your answer. This proof also works. You can also check how our research group came up with the solution. $\endgroup$
    – contemplator
    Commented Dec 9, 2017 at 21:27
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If I understand your question properly, you're asking for the minimum number of open subsets of a wedge of seven circles such that you have unambiguous planning to get from point to point in each subset. I think you can do this for any finite wedge of circles with two sets: divide each circle into two arcs that contain both the wedge point and its antipode--I'll refer to these as the $A$-arcs and a $B$-arcs. Then let $$ A = \fbox{small open set around the wedge point} \cup \fbox{all $A$-arcs} $$ and $$ B = \fbox{small open set around the wedge point} \cup \fbox{all $B$-arcs}. $$ Each of these is an open `asterisk' and with a well-defined and continuous motion-planning rule (shortest path).

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