(RK - 23-Oct-2017)

fuzzy embedding

So, I am thinking through the issue of how to organize different embeddings of a substring in a larger string. First step are the attached variants using the Levenshtein algorithm.

But I am now thinking of a simpler, broader case.  The "larger string" is imagined as a loop and the issue is different ways to embed the substring in that loop, but over any number of iterations.  Then, those embeddings can be ordered by the number of loops they require to complete.

About Margins

(HHR - 23-Oct-2017)

So it was funny that I had used I think the same metric, the Levenshtein distance for an 'edit transcript' when experimenting with the morphing between words and text phrases.

 

⬀Here is my implementation of the algorithm.

// levenshtein distance
~ld = { | s,t|
    var m,n, d, cost;
    m = s.size;
    n = t.size;
    d = m.collect {| i |  0.dup(n).addFirst(i + 1) };
    d = d.addFirst( (0..n) );
    n.do { | j |
        j = j + 1;
        m.do { | i |
            i = i + 1;
            if (s[i-1] == t[j-1]) {
                cost = 0
            } {
                cost = 1;
            };
            d[i][j] = (d[i-1][j] + 1)                 // deletion
            min: (d[i][j-1] + 1)                     // insertion
            min: (d[i-1][j-1] + cost);                // substitution

        }
    };
    d;
};

~study = { |s, t| var sa, ta, d;
    d= ~ld.(s, t);
    d = d.collect({| l, i | l.addFirst((" " ++ s)[i]) });
    d = d.addFirst(("  " ++ t).collectAs({|x| x}, Array));
    d
};


// take the matrix from the ld calculation and work out what chars from the source are in the target
// display this by making them caps in source and target
~parse = { | table, s, t |
    var choices;  // 0 substitute, 1 insert, 2 delete
    var held = [];
    var spos = table.size -1;
    var tpos = table[0].size - 1;
    while{ (spos >0) && (tpos >0) } {
        // hack in a preference for contiguous characters
        // so txtest -> test takes txTEST rather than TxtEST
         if (s[spos -1] == t[tpos - 1]) {
            choices = 0
        } {
            choices = [table[spos -1][tpos-1], table[spos -1][tpos], table[spos][tpos-1]].minIndex;
        };
        case
        {choices ==0 } {
            if(table[spos -1][tpos-1] == table[spos][tpos]) { held = held.add([spos, tpos]) };
            spos = spos -1; tpos = tpos-1 }
        {choices == 1 } { spos = spos -1}
        {choices == 2 } { tpos = tpos-1 };
    };
    s = s.copy;
    t = t.copy;
    (held - 1).do { | p|
        s[p[0]] = s[p[0]].toUpper;
        t[p[1]] = t[p[1]].toUpper;
    };
    [s,t]
};

// do the whole thing
~replace = { |s,  t|
    ~parse.(~ld.(s,t), s,t);
};

// a few tests, not methodical
// these match towards the end of the source string
~replace.("abtsttest", "test")
~replace.("abtesttst", "test")
~replace.("test", "abtsttest" )
~replace.("testab", "test")
~replace.("txtxtyyy", "txyyy")
~replace.("ttxtyyy", "txyyy")


// reversing the string matches towards the beginning
// might  be preferable for Steno
~rReplace = { |s,  t|
    s = s.reverse; t = t.reverse;
    ~parse.(~ld.(s,t), s,t).collect(_.reverse)
};


~rReplace.("abtsttest", "test")
~rReplace.("abtesttst", "test")
~rReplace.("test", "abtsttest" )
~rReplace.("testab", "test")
~rReplace.("txtxtyyy", "txyyy")
~rReplace.("ttxtyyy", "txyyy")