A temporary intermeidate step

Now I got almost every functionality that we need, including pdf,
mu4e, magit, et cetera.

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-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.