math / array / matrix-question

[-hans-]

Hallo everybody,

first of all: it's not a indesignscripting-  but general math-javascriptquestion. please be patient

I've got a first (matrixlike-)array (won't change)

	var containers = [
		'container11', 'container12', 'container13', 'container14', 'container15',
		'container21', 'container22', 'container23', 'container24', 'container25',
		'container31', 'container32', 'container33', 'container34', 'container35',
		'container41', 'container42', 'container43', 'container44', 'container45',
		'container51', 'container52', 'container53', 'container54', 'container55'
		];

and I've got a second array:

["container14", "container25", "container34", "container44", "container54"] //this array may contain 3 up to 8 items

My aim is to check if a part of 5 to 3 items of the second array is part of or equal to a row or column of the matrix-like-array.

For example: "container34", "container44", "container54" or "container11", "container12", "container13", "container14" (as part of second array) would be a result I#m looking for. Note: I only want to find the 'biggest charge'!

Hope it's getting clear and one of the math-cracks will have a idea.

Addittional: there's no MUST to work with arrays. I can also store the data to a object or mixture ... and may fill it with numbers instead of strings ...

To get it visible:

https://dl.dropboxusercontent.com/spa/3ftsuc9opmid3j4/Exports/fourWins/fourWins.html

Items can be dragged and dropped. After every dropp the arrays have to be compared ... and I#m searching for a nice and elegant solution

May be someone's interested

Hans

[-hans-]

Good morning,

splitted the sourcearray to its 'cols' and 'rows' to compare each with the resultarrays.

Solved

Hans

[Marc Autret · Accepted solution]

Hi Hans,

Just a quick note although your question is solved.

Provided that your matrix is 5×5 you could easily map any element to a single character in the set { A, B..., Z } (for example).

Then your problem can be reduced to some pattern matching algorithm, that is, finding the longest part of the input string within a 'flat string' that just concatenates the rows and the columns of the search matrix in the form ROW1|ROW2...|COL1|COL2...|COL5

And you can create RegExp on the fly to compute the solution(s) with almost no effort:

const MX_ORDER = 5;
const MIN_MATCH = 3; // We need at least 3 contiguous items
var bestMatrixMatch = function F(/*str[]*/ROWS, /*str*/ND)
//--------------------------------------
// NB: No check is made on ROWS, so make sure you supply
//     MX_ORDER strings, each being MX_ORDER-sized
{
    // Put in cache some subroutines
    // ---
    F.RES_TO_STR ||(F.RES_TO_STR = function()
        {
            return localize("'%1' found in %2", this.result, this.location);
        });
    F.ROWS_TO_HS ||(F.ROWS_TO_HS = function(R, C,i,j)
        {
            for( i=0,C=[] ; i < MX_ORDER ; ++i )
            for( C='',j=0 ; j < MX_ORDER ; C+=R[j++] );
            return R.concat(C).join('|');
        });
    // Vars
    // ---
    var haystack = F.ROWS_TO_HS(ROWS),
        candidates = ND &&
            haystack.match( new RegExp('['+ND+']{'+MIN_MATCH+',}','g') ),
        t, p;
    if( !candidates ) return null;
    // Sort the candidates by increasing size
    // ---
    candidates.sort( function(x,y){return x.length-y.length} );
    // Grab the matches and keep the best
    // ---
    while( t=candidates.pop() )
        {
        if( 0 > ND.indexOf(t) ) continue;
        p = 1+~~(haystack.indexOf(t)/(1+MX_ORDER));
        return {
            result:   t,
            location: (p<=MX_ORDER)?('Row #'+p):('Col #'+(p-MX_ORDER)),
            toString: F.RES_TO_STR,
            }
        }
    return null;
}
// =================
// Sample code
// =================
var rows = [
    "ABCDE",
    "FGHIJ",
    "KLMNO",
    "PQRST",
    "UVWXY"
    ];
var needle = "EKLMINSX";
// get the result
// ---
var result = bestMatrixMatch(rows, needle);
alert(
    "Searching the longest part of '" + needle + "' in:\r\r" +
    ' '+rows.join('\r').split('').join(' ') +
    '\r\r===============\r\r' +
    (result || "No result.")
    );

@+

Marc

[-hans-]

Hi,

sorry for the late reply. works like a charm

Perhaps, some day I'll even understand how ... 😉

Hans