How often should you clean your room?

We introduce and study a combinatorial optimization problem motivated by the question in the title. In the simple case where you use all objects in your room equally often, we investigate asymptotics of the optimal time to clean up in terms of the number of objects in your room. In particular, we prove a logarithmic upper bound, solve an approximate version of this problem, and conjecture a precise logarithmic asymptotic. INTRODUCTION Suppose you have n objects in your room which are totally ordered. For simplicity, let us say they are books on shelves alphabetized by author and title. If you are looking for a book (assume you remember the author and title, but not its location on the shelves), the most efficient algorithm is a binary search. Namely, look at the book in the middle of the shelf, and because of the ordering, now you can narrow your search by half. Repeat this process of halving your search list, and you can find your book in about log2 n steps. (Here is perhaps a better model for humans naturally search: go to where you think the book should be, scan that area, and if need be jump to a different area based on the ordering. However a logarithmic cost still seems like a good model for this process.) The theory of searching (and sorting) algorithms is of course well studied in computer science—what is not, however, is what happens after that for humans. Namely, after you are done with your book, you can do one of two things: either put it back on the shelf, which we will also say takes about log2 n time, or leave it on your desk, which takes no time. The latter is of course more efficient now, but if you keep doing this, eventually all of your books will wind up as an unsorted pile on your desk. Then when you search for a book, you essentially have to go through your pile book by book (a sequential, or linear, search), which takes about n2 time, and thus is not very efficient for n large. The question we are interested in here is: when is the optimal time to clean up? That is, over the long run, what is the optimal value mopt(n) of m (1 6 m 6 n) at which you should put all the books in the pile back, in order, on the shelf, in the sense that the the average search plus cleanup cost (per search) is minimized. Here we assume the cleanup algorithm is to simply go through the pile, book by book, and find the right location for each book on the shelves via a binary search (see Remark 1.2 for a discussion of other cleanup algorithms). The paper is organized as follows. (See Section 1.3 for a more detailed overview.) In Section 1, after first formulating this problem precisely, we will discuss four different models Date: May 11, 2015. ∗ Supported in part by a Simons grant. † Supported in part by an NSF grant. 2 K. MARTIN AND K. SHANKAR and focus on the (generally unreasonable) case of the uniform distribution, i.e., where you use all objects in your room equally often. It might be more realistic to consider a power law distribution, but even the simple case of the uniform distribution is not so easy. The different models correspond to having either complete or no memory of what is in the pile, and having numbered shelves (each object has a designated location on the shelves) or unnumbered shelves (only the relative order of books is important). In Section 2, we analyze the search and cleanup cost functions in some detail for each of these models. Our first result is that, in each of these models, one should not clean up immediately (see Proposition 1 below). In fact, if n is small enough, one should never cleanup (see Remarks 4.6 and 4.7). In Section 3, we restrict ourselves to complete memory with numbered shelves for simplicity, and prove that one should clean up before about 4 log2(n) objects are in the pile (see Proposition 2). A good lower bound for the mopt(n) is not so easy, and so we instead consider an approximate problem in Section 4. Based on the analysis from Section 2, we expect the optimal value m̃opt(n) of m for the approximate problem to be a lower bound for mopt(n). We essentially determine exactly the optimal value of m for the approximate problem (Theorem 3), which is about 3 log2(n), and then based on numerics conjecture that mopt(n) ∼ 3 log2(n) (Conjecture 4). In fact we expect that for all four models with arbitrary distributions, mopt(n) grows no faster than 4 log2(n). Therefore, we humbly suggest you clean your room before 4 log2(n) objects are out. Since we use a fair amount of (often similar looking) notation, we provide a notation guide at the end for convenience (Appendix A). Acknowledgements. It is a pleasure to thank our colleague Alex Grigo for helpful comments and suggestions. We thank the referee whose careful reading has helped improve the exposition. We also thank our parents for never making us clean up our rooms. 1. GENERAL SETUP 1.1. The Statement of the Problem. We now make a general formulation of our problem, which we call a search with cleanup optimization problem. Let X = {X1, . . . , Xn} be a finite set of distinct well-ordered objects, which we view as a probability space with probability measure μ. We consider the following discrete-time Markov chain, depending on a parameter 1 6 m 6 n. (1) At time t = 0 each Xi is in a sorted list1 L, and there is an unsorted pile P which is empty. (2) At any time t ∈ {0, 1, 2, . . .}, each X ∈ X is in exactly one of L and P , i.e., X is a disjoint union X = L t P . (3) At any time t > 1 with |P| < m, exactly one object X = Xi is selected, and X is selected with probability μ(X). If the selected X ∈ P , nothing changes. Otherwise, then X is removed from L and added to P . 1By list, we mean an indexed list, rather than a linked list. HOW OFTEN SHOULD YOU CLEAN YOUR ROOM? 3 (4) At any time t, if |P| = m, we stop the process. This process has finite stopping time with probability 1 provided at least m elements of X have nonzero probabilities, which we will assume. Note the state of the process at time t is described simply by a subset P of X, together with a marked element Xit (the object selected at time t > 1). The set of possible states is then simply all subsets P of X, together with a marked point Xit , of cardinality at most m. Associated to this process are two (nonnegative) cost functions, S(X;P) and C(P), which do not depend upon t. Here X ∈ X and P ⊂ X. The functions S(X;P) and C(P) are called the search and cleanup costs. Let Xm = Xm,n denote the set of finite sequences χ = (Xi1 , . . . , Xi`) in X such that (i) the underlying set {Xi1 , . . . , Xi`} has cardinality m, and (ii) Xij 6= Xi` for j < `. We extend the measure μ to a probability measure on Xm by