July 10
I know you folks tend to prefer pure mathematics here, but I have a little "applied" question. I am looking for a mathematical description and/or analysis of a roundabout. For someone familiar with the parameters it should be a well defined linear algebra problem. I know that traffic patterns are often modeled mathematically, but I can’t seem to find anything specific to roundabouts. 199.172.246.196 12:14, 10 July 2007 (UTC)[reply]
- I think this can be modelled as a sort of Continuous-time Markov process, but without additional details about your assumptions and goals, I can't take it much farther than that. -- Meni Rosenfeld (talk) 12:37, 10 July 2007 (UTC)[reply]
- I don't know what sort of assumptions to make either. I am quite sure that it has been modeled before and I was wondering whether you knew of any existing examples. If all else fails I'll take a crack at modeling it via the Continuous-time Markov process as you have suggested. 199.172.246.196 14:09, 10 July 2007 (UTC)[reply]
- Here's one (rather simplified) way to go about it. Suppose there are n exits placed equidistantly, and that each segment of the roundabout between consecutive exits can contain only one vehicle. Suppose further that crossing a segment takes time which is distributed exponentially (far from reality, but simplifies the model) with mean and that it takes no time for a waiting car to enter the roundabout. We also assume that cars reach the line at the entrypoints by a Poisson process of rate . Our state space will be , where each copy of represent the segment before an exit and what car is in it (no car, or a car with destination k), and each copy of represent the number of cars waiting at an entrypoint.
- The different transitions are somewhat complicated to present exactly, but they all follow naturally from our assumptions: For example, if some segment is occupied, there is a transition of rate for the entry before it to increase its car count, but if the segment is vacant, this transition will be to occupy this segment with a car of random destination.
- Does this help in any way? -- Meni Rosenfeld (talk) 17:23, 10 July 2007 (UTC)[reply]
- What is the purpose you are pursuing that is aided by having a mathematical model of a roundabout? Designing better roundabouts? Showing that roundabouts are a roundabout way of getting somewhere? What aspects and details need to be modelled and which may be omitted depends on the use to which the model will be put. --LambiamTalk 07:59, 11 July 2007 (UTC)[reply]
I am working on a text layout application, and find myself dealing with mathematics I had not bargained for. I have asked a very similar question above, but some experimentation has proven that my original objective was not quite what I needed. Here is a rephrasing, this time with a new objective:
Let A be a sequence. Given an integer n (n<Length(A)), divide A into n consecutive subsequences, A1 through An, the union of which is A (a division, D, can be thought of as a sequence of n indices, the last of which is Length(A)). Calculate the sum of the elements of each sequence, Si. Find the maximal such sum, Max(S1,...,Sn), which we'll call Smax.
My question is this: how may we find which division yields the smallest Smax? (Without going over each and every possibility.) — Itai (talk) 15:57, 10 July 2007 (UTC)[reply]
- Can you explain directly what text layout problem you are working on, rather than trying to "mathematize" it for us? In particular, if you're trying to break a paragraph into lines, you really need to read Donald Knuth and Michael Plass's paper "Breaking Paragraphs into Lines" (originally published in Software — Practice and Experience 11 (1981), 1119–1184; reprinted in Knuth's Digital Typography, CSLI, 1999). See our word wrap article too.
- The problem you give can be solved in O(n|A|) by a straightforward dynamic programming approach (compute Mij — the maximal sum that you can get by splitting the first j elements of the sequence into i subsequences — for all i up to n and j up to |A|). Gdr 17:06, 10 July 2007 (UTC)[reply]
- I've read a synopsis of the paper you mentioned, and it seems very interesting. I'll read it sometime should I obtain a copy. My own problem is that of dividing p indivisible paragraphs (well, truth be told, each consists of one question and four possible answers, but that hardly matters) between c columns. I admit I was uncertain as to what is the proper division criterion. At first I thought the best thing is to minimize the standard deviation, to make the page look prettier. Then I settled for a lesser, and seemingly much easier to accomplish, objective of minimizing the difference between the longest and shortest columns. At last I came to the conclusion that what I'm really after is "conserving paper" - making the longest column as short as possible, making for the least space-consuming page. — Itai (talk) 22:30, 14 July 2007 (UTC)[reply]
- (after edit conflict) I'll avoid the term subsequence, which often does not imply contiguity (like the sequence of primes 2, 3, 5, ... is a subsequence of 1, 2, 3, 4, 5, ...). I'll use segment instead for a contiguous subsequence. Also, let a k-split of a sequence be a split into k consecutive segments, so you want an n-split.
- Let m = Length(A), and let A|i denote the prefix (initial segment) of A of length i, so A|0 = (), the empty sequence, and A|m = A. Define mM(i, k) to be the minimum, over all k-splits of A|i, of the maximum of the segment sums. You can compute, incrementally, an optimal k-split of A|i, 0 ≤ k ≤ n, with the length i of the prefix going from 0 up to and including n. (This is essentially the usual algorithm for a version of dynamic programming, using a vector instead of a single value.) You only need to include 0 in the range for k when i = 0, and n only when i reaches m.
- To start, mM(0, k) = 0 for all k. Assume we already have computed mM(i–1, k) for i > 0 and k in the appropriate range. We can then determine mM(i, k+1) for each relevant value of k (i.e., with k+1 in range) as follows. Each ( k+1)-split of A|i consists of a k-split of a shorter prefix A|j followed by a final segment of A|i of length i–j, so it is an extension by one segment of an initial k-split. For 0 ≤ j < i, let Tj be the sum of the final segment of A|i of length i–j. Among all ( k+1)-splits of A|i with that final segment, the best score is then max(mM(j, k), Tj). So then mM(i, k+1) is the minimum of these scores for 0 ≤ j < i.
- The above only gives you the value mM(m, n) of the optimal m-split of A, not the split itself. But as you store mM in an array, you can store with each new entry mM(i, k+1) a backpointer, being the value of j that minimized the value of max(mM(j, k), Tj). Then you can trace a path back from the entry mM(m, n) to mM(j, n–1) (using the "j" stored at mM(m, n)) to ... to finally mM(0, 0).
- The same procedure can be followed for many different optimality criteria, including your earlier minimization of the spread. The only requirement is that the principal of optimality is met: the score of a (k+1)-split can be expressed as a function of the score of its initial k-split and the final segment, where that function is monotonically increasing in its first argument. Then an optimal (k+1)-split can be obtained by extending a shorter optimal k-split.
- Specifically for your min-max criterion, if you let j run backwards, once Tj reaches the best score you already have for mM(i, k+1), you can break from the j-loop, since it's not going to get any better. --LambiamTalk 17:26, 10 July 2007 (UTC)[reply]
- Thanks a million. This seems like a very good algorithm, and, amazingly, one I can understand - at least as far as this recent criterion is concerned. If by "minimization of the spread", above, you mean minimizing what I've referred to before as "standard deviation", I do not see how this can be achieved using a variation on the algorithm specified above. I suppose this will give me something to think about. I should really read more about dynamic programming. I doubt whether my own code will be worthwhile posting, but should I come up with something even barely legible, I'll put it up in a subpage of my user page, and place a reference here. Anyway, I am, of course, much obliged. — Itai (talk) 22:30, 14 July 2007 (UTC)[reply]