More efficient columnar transposition decryption

Question

Below, I have created a rudimentary script to decipher a columnar transposition. I am wondering if there is a more efficient way to determine the number of the rows in a column? Also, is there a way to build the resulting string without creating a matrix in the interim?

Overview

Quick refresher for columnar transpositions.

Method 1

Given a key of 3,

the text 'Attack at dawn' is transposed into three columns...

1 2 3
A t t
a c k
a t D
a w n

...joining each column; left->right, top->bottom; will result in the following text:

Aaaa tctw tkDn

Method 2

Given a key 'table',

the text 'Attack at dawn' is transposed into five columns (length of the key)...

t a b l e
A t t a c
k a t D a
w n

...joining the columns, as described above, result in the following text:

tAkw atan btt laD eca

Code

Is there a better way to determine how many rows will be present for each column?

function decrypt(ciphertext, key) {
 var chars = normalizeText(ciphertext);
 var len = chars.length;
 var isStr = typeof key === 'string';
 var cols = isStr ? key.length : key;
 var rows = Math.ceil(len / cols);
 var mod = divMod(len, cols)[1];
 var matrix = Matrix(rows, cols, '');
 var counter = 0;
 for (var col = 0; col < cols; col++) {
 var remainder = mod !== 0 && col >= mod ? 1 : 0;
 for (var row = 0; row < rows - remainder; row++) {
 matrix[row][col] = chars[counter++];
 }
 }
 return (function(plaintext, startIndex) {
 return plaintext.substr(startIndex);
 }(matrix.reduce(function(result, row) {
 return result + row.join('');
 }, ''), isStr ? cols : 0));
}

Demo

[ [ 'Aaaa tctw tkDn' , 3 ],
 [ 'tAkw atan btt laD eca', 'table' ]
].forEach(function(cipher) {
 println(decrypt(cipher[0], cipher[1]));
});
function decrypt(ciphertext, key) {
 var chars = normalizeText(ciphertext);
 var len = chars.length;
 var isStr = typeof key === 'string';
 var cols = isStr ? key.length : key;
 var rows = Math.ceil(len / cols);
 var mod = divMod(len, cols)[1];
 var matrix = Matrix(rows, cols, '');
 var counter = 0;
 for (var col = 0; col < cols; col++) {
 var remainder = mod !== 0 && col >= mod ? 1 : 0;
 for (var row = 0; row < rows - remainder; row++) {
 matrix[row][col] = chars[counter++];
 }
 }
 return (function(plaintext, startIndex) {
 return plaintext.substr(startIndex);
 }(matrix.reduce(function(result, row) {
 return result + row.join('');
 }, ''), isStr ? cols : 0));
}
function divMod(y, x) {
 return [~~(y / x), y % x];
}
function normalizeText(ciphertext) {
 return ciphertext.replace(/ +/g, "").split('');
}
function Matrix(rows, cols, defaultVal) {
 defaultVal = defaultVal !== undefined ? defaultVal : 0;
 var arr = [];
 for (var i = 0; i < rows; i++) {
 arr.push([]);
 arr[i].push(new Array(cols));
 for (var j = 0; j < cols; j++) {
 arr[i][j] = defaultVal;
 }
 }
 return arr;
}
function print(str, targetEl) {
 return (function (el) {
 return el.html(el.html() + (str || ''));
 }(targetEl || $('#out')));
}
function println(str, targetEl) {
 print((str || '') + '<br />', targetEl)
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="out"></div>

Answer 1

Allow me to provide a bit of a different approach. I know it does not exactly answer your questions, but I believe it makes them somewhat obsolete.

This is how I would do the decrypt:

function decrypt(sentence) {
 var result = '';
 var pos = 0;
 var words = sentence.split(' ');
 words = words.map(function(word) {
 return word.split(''); 
 });
 while(true) {
 char = words[pos].shift();
 if (! char) {
 break;
 }
 result += char;
 pos = pos > words.length -2 ? 0 : pos + 1;
 }
 return result;
}

For a working example: http://jsfiddle.net/35jyyv8g/1/

Note that I left the key out, since it is not really necessary with the spaces in the input.

Answer 2

For me, all it takes are two for loops, the first one with i going from 0 to cols (stricly inferior), with increments of 1, and the second loop (inside the first one) with j going from i to len (stricly inferior) with increments of rows . And the characters you want will be at the position j of the initial message. Of course, a better naming for i and j would be a good thing.

Answer 3

Interesting question,

first off, your encryption approach has some serious problems: To know the encryption key length, you just need ciphertext.split(' ').length

Worse, the cipher-text even has the key in clear text:

tAkw atan btt laD eca

This means @Pevara's approach is correct, and far more elegant. (It does also output the key for strings encrypted with the second method.)

My personal approach uses a bit of ES6 and is a bit more functional:

function simpleDecrypt2(cipher) {
 let out = '',
 parts = cipher.split(' ').map(function(part) { return part.split(''); });
 while( parts[0].length ){
 out += parts.reduce( (out,part)=>out+(part.shift()||''), '' ); 
 }
 return out;
}

As for your first question, you could have calculated the row count by

ciphertext.split(' ')[0].length