From 3666deaed5b0baf0a74f14db5872105c9e7865f9 Mon Sep 17 00:00:00 2001 From: JSDurand Date: Wed, 13 Jan 2021 13:01:34 +0800 Subject: A temporary intermeidate step Now I got almost every functionality that we need, including pdf, mu4e, magit, et cetera. --- comb/suffiex-tree.txt | 217 -------------------------------------------------- 1 file changed, 217 deletions(-) delete mode 100644 comb/suffiex-tree.txt (limited to 'comb/suffiex-tree.txt') diff --git a/comb/suffiex-tree.txt b/comb/suffiex-tree.txt deleted file mode 100644 index 51908a3..0000000 --- a/comb/suffiex-tree.txt +++ /dev/null @@ -1,217 +0,0 @@ -Title: Suffix Trees -Author: JSDurand -Created: 2020-01-03 -------------------- - -====================================================================== - Motivation to implement this algorithm in Emacs -====================================================================== - -The reason I want to implement this algorithm is a problem I -encountered in using the package "orderless", which provides a -completion-style for the built-in completion system in Emacs. The -problem is that the function "orderless-try-completion" does not -handle completion aggressively in a way that conforms to the -documentation of "try-completion". To be more precise, if there is -only one match for the current text input, then this function returns -that match (rather than comforming to the requirement in the -documentation of "try-completion" by returning t, since it wants to -highlight the matches by itself). This isn't a problem from the -perspective of the user, and if there are no matches, then this -correctly returns nil. But in any other cases, this function returns -the original string, instead of the longest common substring, as a -user might desire. - -Initially, this does not seem like a serious concern: the user could -still select a completion from the "*Completions*" buffer once the -list of candidates becomes small enough to easily manage. But as time -goes by, this starts to frustrate me. For example, when there are only -two candidates left, but one is a substring of another, then I cannot -use the completion feature to quickly select (or "complete to") the -shorter candidate, unless I type out the full candidate string, or to -choose from the "*Completions*" buffer. For me this kind of defeats -the main purpose of the built-in completion system. - -So I start wondering, is it possible to fix the problem by finding the -longest common substring in all the matches? From a naïve first -impression, this seems to be what the user might expect in most cases. - - - -====================================================================== - Choice of the algorithm -====================================================================== - -After some thinking and searching through the internet, I found that -perhaps the most flexible and performant solution to the problem of -finding the longest common substring(s) of multiple strings is to -build a "generalized suffix tree" of them, and then use a tree -traversal to find the longest common substring(s). Well, this is all -fun and great. The only problem is that it is difficult to build a -suffix tree (or a generalized one). - -So I decide to implement the seemingly fastest algorithm to construct -a suffix tree of strings and hope that this can not only solve my -problem but also help others out in some other problems in the future. - - - -====================================================================== - Definitions -====================================================================== - -I describe the basic definition of a suffix tree briefly below. - -In this document a string is a sequence of alphabets. In particular, -for the case of Emacs, these alphabets are just numbers. We begin by -considering a string S of length n. A suffix of S is a substring of S -of the form S[i..n] for some i from 1 to n. And we also say the empty -string is a suffix of S. A suffix tree of S is defined as a rooted -tree T (so it has a node called "root" that is the ancestor of every -other node) whose every edge has a label which is a substring of S -that satisfies the following conditions: - -- Starting from the root of T, walking down any path to a leaf, and - concatenating the labels along the way, then we will get a suffix of - S. And every suffix of S is obtained in this way as well. -- Every node has at least two out-going edges. -- For every node, every two out-going edges cannot have labels that - start with the same letter. (So two edges with labels both starting - with 'a' cannot emanate from the same node.) - -Intuitively speaking, this is to list all suffixes of S as an edge -from the root to a leaf, and then "merge" these suffixes so that any -common prefix among some of them is in only one edge. - - - -====================================================================== - Description of the algorithm -====================================================================== - -Below is a breif description of the algorithm. For a description in -"plain English", see the accepted answer to this Stack Overflow post. - -https://stackoverflow.com/questions/9452701/ - -For a more detailed survey on the principle behind the algorithm and -on many other related topics, see the book "Algorithms on Strings, -Trees, and Sequences: Computer Science and Computational Biology" by -Dan Gusfield, or if you prefer reading the original paper, then the -original paper by Ukkonen is as follows. - -https://link.springer.com/article/10.1007/BF01206331 - -(I am fortunate enough to be able to access the article. If you want -to read that PDF and don't want to pay Springer, let me know and I can -send you the file.) - -Given a string S of length n, we will first append a symbol that is -not present anywhere in S, in order to ensure that no suffix of S is -also the prefix of another suffix of S; otherwise S cannot have a -suffix tree. I refer to this terminating symbol as $. - -Also an edge of the tree is not labelled explicitly by strings. To do -so would violate already the linear time constraint. Instead, we -represent each edge with a pair of integers, interpreted as the -indices of the starting and the ending points of the associated -substring in T. - -The algorithm has n + 1 iterations. We start with the following -variables: - -- s = root -- k = 1 -- i = 0 -# - A root node. -# - An "active point" with value (root, nil, 0). -# - "remainder" with value 1. - -Then we want to add n + 1 symbols to the tree iteratively. - -In the i-th iteration we look to add the i-th letter S (i) of S. In -Emacs Lisp this is expressed as (aref S i). - -# substring -# S[(i-remainder+1)..i] - -And in each iteration we do the following things: - -- (setq i (+ i 1)) -- (let ((result (update s k i))) - (setq s (car result)) - (setq k (cadr result))) - We update by adding the i-th letter S(i) to the current active point - indicated by (s, k, i - 1), see below. -- (let ((result (canonize s k i))) - (setq s (car result)) - (setq k (cadr result))) - We canonize the new active point returned by the update function. - The process of canonization is to find the closest node to the - point. - -Then we repeat until S(i) equals the terminating symbol (this way we -avoid calculating the length of the string beforehand, since in Emacs -Lisp the string does not have the length pre-calculated. Though this -does not affect the overall time complexity, it might affect the -practical performance. - -The function "update is described below. - -It taks three arguments: s, k, and i. First let oldr = root. Then let -(end-p, r) be the result of (test-and-split s k (- i 1) S(i)). - - - - - - - - - - - - - -# begin a "remainder loop". Whenever the remainder -# loop ends, we go to the next iteration. - -We compare the letter S(i) with the labels of edges going out from the -active point. That is, if the active point is (node, edge_label, m), -then this is the length m point in the edge going out from node with -the first letter of label given as edge_label. - -If the letter is the prefix of some edge label, then we set the -active point to (node, edge_label, m+1) and increment remainder by 1. -If m+1 is greater than or equal to the length of the current edge, -then set the active point to follow that edge to the new point. - -Then we end the remainder loop and skip to the next iteration. - ---- - -If the letter is not the prefix of any edge label emanating from the -active point, and m > 0, then we split the point (node, edge_label, m) -into a node and add a new leaf to that node, with label S(i). And then -we decrement remainder by 1. - -If this is not the first time we split a node in the current -iteration, then we add a "suffix link" from the previously created -node to the newly created node. - -If the "node" in the specification of the active point is the root, -then we set the active point to (root, S(i-remainder+1), m-1). - -If the "node" is not the root, then if the node has a suffix link to -other_node, then we set the active point to the following: -(other_node, S(i-remainder+1), m-1) - -If the node has no suffix link, then we set the active point to the -following: (root, S(i-remainder+1), m-1) - ---- - -If the letter is not the prefix of any edge label emanating from the -active point, and if m = 0, then we add a new leaf with label S(i) to -node, and decrement remainder by 1. Then we follow the same rules to -reset the active point. -- cgit v1.2.3-18-g5258