## Introduction

If you have never heard of it, Dancing Links is one of the techniques that Knuth describes to implement a recursive backtracking algorithm. If you truly want a good treatment as to what the technique is, I can not recommend going straight to the source heavily enough.

I originally thought of the "Dancing Links" technique as one that is solely suited to solving "exact cover" problems. This actually caused me to not attempt implementing it for a time, as I confess it intimidated me. Also, I'm not entirely sure I have ever had to work such a problem in my day jobs.

Regardless, I have been intrigued by the technique, if only for the name, for quite a while. This, then, is my attempt at implementing the algorithm that was used to introduce the technique. I'm hoping to use the technique in another algorithm later.

## N-Queens

I'll be exploring dancing links in solving a straight forward "exact cover" problem. This will be, directly, the DLX algorithm that was used in introducing the technique.

For those that do not know, the "$$N$$-Queens" problem is one of how to place $$N$$ queens that are not attacking each other on an $$N \times N$$ chessboard. Simply stated, we want to pick a solution where we have a queen in every row, column, diagonal, and reverse diagonal.

The basic algorithm for solving this is to place one piece, then see if you can place another piece on the spaces remaining. Continue until you place $$N$$ pieces or can not place another. In either event, you simply take off pieces that you have placed and try a different placement.

It can be seen that this is a depth first search across all possible piece placements. Simply start with all possible positions, pick a starting point, remove all positions it eliminates, and repeat.

### Basic Recursion

A ridiculously naive solution follows. My strategy is simply to go as stated above. Start by enumerating all positions on the board. Pick one, add it to our list of "moves taken," remove all positions it eliminates, and call ourselves starting over. When we return from the recursive call, try the next position in our list.

This method is short enough that I will just put the whole thing here. I will also add that that is a benefit for this approach. The DLX version turns out to be a lot more code.

function solveRecursively(n, showSteps) {
var r = n, c = n, allPositions,
steps = [];

function initializePositions() {
var i, j;
allPositions = [];
for (i = 0; i < r; i++) {
for (j = 0; j < c; j++) {
allPositions.push({
r: i,
f: j,
a: i + j,
b: n - 1 - j + i,
//To get our placements the same as the DLX one...
toString: function() {
return "R"+this.r+" F"+this.f+" A"+this.a+" B"+this.b;
}
});
}
}
}

function placePiece(availablePositions, position) {
return availablePositions.
filter(function(v) {
return v.r !== position.r &&
v.f !== position.f &&
v.a !== position.a &&
v.b !== position.b;
});
}

function testSolution(availablePositions, candidate) {
if (showSteps)
steps.push(candidate);

if (availablePositions.length) {
var childSolutions = [];
for (var p = 0; p < Math.min(n, availablePositions.length); p++) {
var position = availablePositions,
remainingPositions = placePiece(availablePositions,
position),
recurseResults = testSolution(remainingPositions,
candidate.concat(position));

availablePositions = availablePositions.slice(1);
childSolutions = childSolutions.concat(recurseResults);
}
return childSolutions;
} else if (candidate.length === n) {
return [ candidate ];
}
return [];
}

initializePositions();
var solutions = testSolution(allPositions, []);
if (showSteps)
return steps;
return solutions;
}


Quickly running this, we see that it has the expected results for various small values. I did not try for much larger values as I don't exactly have high hopes for speed there. Just running for 10 nodes already takes a bit of time. (Well, a second or so. Feels like forever. Bumping up to 11 starts taking tens of seconds.)

 $$N$$ Solutions 1 1 2 0 3 0 4 2 5 10 6 4 7 40 8 92

Now that we've looked at solving this with a naive recursive solution, how would this look with dancing links? Not going to lie, this is more involved. Luckily, it isn't that terribly scary.

In fact, the actual dancing links code is rather short and not too terribly involved. Because I did not bother to clean up the code that generates the data for the $$N$$-queens problem, this section is large.

We'll look at this in pieces, then. If you want to see it all tangled into a single file, peek over at dancingLinks.js.

#### Outline

The basic outline is to create a method that will take in the number of queens we want to solve, and then create the necessary structures to make it happen.

function solveWithDancingLinks(n, showSteps) {
var headers, solutions = [], O = [];

<<dlx_search>>

<<dlx_cover_uncover>>

<<dlx_utilities>>

search(0);
return solutions;
}


#### Search

We'll begin by looking at the search code. This is fairly straight forward. First, check to see if there are any columns left uncovered and return the current solution if not. Otherwise, pick a column, cover it, then for each row in the current column, cover all connected columns and continue the search.

One thing to note here is if we pick a column that has zeros rows on it, we immediately backtrack up and try a different path.

#### Covering and Uncovering Columns

This is actually the heart of the "dancing" in the dancing links technique. Not much to offer on them other than that they have to run in reverse order from each other for our purposes. Both methods are included here.

function cover(c) {
var r = c.down;
c.right.left = c.left;
c.left.right = c.right;
while (r !== c) {
r = r.right;
while (r.col !== c) {
r.up.down = r.down;
r.down.up = r.up;
r.col.size--;
r = r.right;
}
r = r.down;
}
}

function uncover(c) {
var r = c.up;
c.right.left = c;
c.left.right = c;
while (r !== c) {
r = r.left;
while (r.col !== c) {
r.up.down = r;
r.down.up = r;
r.col.size++;
r = r.left;
}
r = r.up;
}
}


#### Utility Methods

The rest of the methods we need are fairly self explanatory.

function smallestColumn() {
var h, c, s = Number.MAX_VALUE;
if (h.size < s) {
c = h;
s = c.size;
}
h = h.right;
}
return c;
}
function printRow(r) {
var s = r.col.name + ' ', e = r;
r = r.right;
while (r !== e) {
s += r.col.name + ' ';
r = r.right;
}
return s;
}
function copySolution() {
var solution = [].concat(O);
return solution;
}


And now, the only part of this code that is actually specific to the queens problem, generating the rows and columns of our data structure.

The basic idea is to generate a primary column for each rank and file, then secondary columns for each diagonal. Primary columns are doubly linked to the 'root' column, whereas secondary ones are not. Knuth points out that you can get further speed increases by creating the headers in "pipe organ" order. I confess I had to just use his method for generating said order, as I did not know it.

This does make a very interesting example where it is not just the data structure that matters, but how you initialize it. Using the "pipe organ" order can cut the running time by half, compared to the straight forward ordering.

The only trick this required is that I do keep an array of all columns while making them so that I can find the ones that are not hooked into the main header list. (After I did this, I took a look directly at Knuth's code to see how he does it. Kind of sad to see how much easier this is to do in C. I can't say I would have done it the easy way, though. I do not think of pointer tricks quickly.)

function initializeHeaders() {
var i, j, k, rows=[];
cols = [];

name: 'root',
right: null,
left: null,
up: null,
down: null
};

for (i = 0; i < n; i++) {
var t = ((i & 1) ? n - 1 - i : n + i) >> 1;
var cur = {
name: 'R' + t,
size: 0,
down: null,
up: null,
};
cols.push(cur);
cur.up = cur;
cur.down = cur;

cur = {
name: 'F' + t,
size: 0,
down: null,
up: null,
};
cols.push(cur);
cur.up = cur;
cur.down = cur;
}
for (i = 0; i < 2 * n; i++) {
var cur = {
name: 'A' + i,
right: null,
left: null,
size: 0,
up: null,
down: null
};
cols.push(cur);
cur.left = cur;
cur.right = cur;
cur.up = cur;
cur.down = cur;
}
for (i = 0; i < 2 * n; i++) {
var cur = {
name: 'B' + i,
right: null,
left: null,
size: 0,
up: null,
down: null
};
cols.push(cur);
cur.left = cur;
cur.right = cur;
cur.up = cur;
cur.down = cur;
}

for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
var a, b, c, d;
a = {
up: null,
down: null,
left: null,
right: null,
col: null
};
b = {
up: null,
down: null,
left: null,
right: null,
col: null
};
c = {
up: null,
down: null,
left: null,
right: null,
col: null
};
d = {
up: null,
down: null,
left: null,
right: null,
col: null
};
a.left = d;
a.right = b;
b.left = a;
b.right = c;
c.left = b;
c.right = d;
d.left = c;
d.right = a;

var colIdx = 0;
var aCol = cols[colIdx++];
while (aCol.name !== 'R' + i)
aCol = cols[colIdx++];
aCol.size++;
a.col = aCol;
a.down = aCol;
a.up = aCol.up;
a.down.up = a;
a.up.down = a;

colIdx = 0;
var bCol = cols[colIdx++];
while (bCol.name !== 'F' + j) {
bCol = cols[colIdx++];
}
bCol.size++;
b.col = bCol;
b.down = bCol;
b.up = bCol.up;
b.down.up = b;
b.up.down = b;

colIdx = 0;
var cCol = cols[colIdx++];
while (cCol.name !== 'A' + (j + i))
cCol = cols[colIdx++];
cCol.size++;
c.col = cCol;
c.down = cCol;
c.up = cCol.up;
c.down.up = c;
c.up.down = c;

colIdx = 0;
var dCol = cols[colIdx++];
while (dCol.name !== 'B' + (n - 1 - j + i))
dCol = cols[colIdx++];
dCol.size++;
d.col = dCol;
d.down = dCol;
d.up = dCol.up;
d.up.down = d;
d.down.up = d;
}
}
}
}
}


#### Running for small values

And finally, we run for some quick examples to see what we can see. I'm willing to show quite a few more results for this case, as they go much faster than the recursive solution I created.

 $$N$$ Solutions 1 1 2 0 3 0 4 2 5 10 6 4 7 40 8 92 9 352 10 724 11 2680 12 14200 13 73712

### Visualized

I had originally thought to run some fairly sophisticated visualizations of these two algorithms. Truth to tell, I just can't think of any amazing visualization that is that enlightening.

However, running the visualization of my DLX algorithm where I just see what board positions it tries did help me spot a bug in my code. So, I'll include what I did. If you want to just play with some larger visualizations, simply run:

document.body.appendChild(makeVisualization(N, solveWithDancingLinks));


To see the boards that the recursive solution inspects, use "solveRecursively."

#### Recursive solution

First thing I was curious on was if I could get a good grasp on the speed difference at a lower value of $$N$$. Sure enough, there is enough to look at. We'll begin by looking at the recursive solution.

The basic problem is this algorithm hits a lot of tree states where it is not possible to fully cover. In particular, notice that from position 2 to position 3, it is possible to place another queen, however, it is not possible to place a queen on the last file. So, the algorithm should be able to backtrack early. Instead, it places the queen.

Otherwise, this algorithm inspects the pieces in a very straight forward left to right, top to bottom method of placing pieces.

Contrast the previous behavior with the dancing links technique. Here, the ordering heuristic of Knuth's accounts for starting on the third row, but the natural DLX behavior is as soon as a rank or a file is impossible to place, the algorithm will backtrack.

This shows in the marked decrease in number of board configurations tried. For DLX, this is a nice 12 positions instead of the 31 previously.

To see the specific scenario that helps, notice from configuration 6 to 7, there is a place a queen could be placed, but the system does not bother trying, as the third rank is already impossible to place.

## Conclusion

After all of that, what is the reason to use the "dancing links" technique? Mainly for speed. It seems to be a classic case study in using a fair bit more memory for the main data in order to more easily backtrack on modifications to it.

It should be possible to use something akin to the same ordering heuristic in the naive recursive solution that the dancing links version uses. However, the linked nature of the nodes in the DLX algorithm makes it straight forward to find what position of the board to try next. There is plenty of following links, but there is relatively little "searching" to find whichc piece to modify.

Also, please note that the point of the technique is to show how modifying a datastructure can sometimes easily be undone. All in all it is more difficult to build up the main data structure, but manipulating it is very fast.

Finally, I am considering using this technique elsewhere. There are a few oddities to it that will likely stonewall this effort. Mainly, I am definitely more comfortable with "immutable" data structures. At least the simple ones.

I should also put a big plug in to the literate programming ideas, again. To note, this document is not just excerpts of the code, but rather the full source code for everything I did. It can be tangled into the different files.

I'm torn on really recommending this style for full projects. I think it sadly lends itself better to pieces where one person does it all. However, even working in a team it is not uncommon for a few people to develop sections on their own. Perhaps it could work better in a team than I am giving it credit.

I will say that reading Knuth's code directly is both easier and more of a learning experience than was originally anticipated. There is definitely a mythos that surrounds his work that convinces many to think they can't read it. I will not claim the heavier math is easy. Nor will I claim all of the exercises are. I will say that it is easier than it is typically portrayed. Especially just the programming sections.

## Appendix

### Board Highlighting

The code I'm using for the board highlighting is here. I really just needed a few things. Probably could have just pulled in a library, I'll use the excuse of doing most of this while on a train. (Which, sadly is not true. I certainly started this while on the train.)

function element(name, children) {
var el = document.createElement(name), i;
if (children) {
for(i = 0; i < children.length; i++) {
var child = children[i];
if (typeof child === "string") {
el.appendChild(document.createTextNode(child));
} else {
el.appendChild(child);
}
};
}
return el;
}
function div()   { return element("div", arguments);   }
function table() { return element("table", arguments); }
function tbody() { return element("tbody", arguments); }
function tr()    { return element("tr", arguments);    }
function td()    { return element("td", arguments);    }
function span()  { return element("span", arguments);  }
function withAttribute(element, attr, value) {
element.setAttribute(attr, value);
return element;
}
function withClassname(element, cls) {
return withAttribute(element, "class", cls);
}
function withMouseMoveListener(element, listener) {
element.onmouseover = listener;
return element;
}
function slider() {
return withAttribute(
withAttribute(element("input"), "type", "range"),
"min", "0");
}

function makeVisualization(n, method) {
var board = makeBoard(n),
input = slider(),
curState = span(),
states = method(n, true);

input.setAttribute("max", states.length - 1);
curState.innerHTML = "0 / " + (states.length - 1);
input.value = 0;
input.oninput = function () {
curState.innerHTML = input.value + " / " + (states.length - 1);
clearQueens(board);
states[input.value].forEach(function (s) {
placeQueen(board, "."+s.toString().trim().replace(/ /g, '.'));
});
}

return withClassname(div(curState, board, input), "visualization");
}

function makeBoard(n) {
var r = n, c = n;
var rows = tbody();
for (var i = 0; i < r; i++) {
var row = tr();
rows.appendChild(row);
for (var j = 0; j < c; j++) {
var cell = td();
var cls = "";
cls += " R"+i;
cls += " F"+j;
cls += " A"+ (i + j);
cls += " B"+ (n - 1 - j + i);
cell.setAttribute("class", cls.trim());
row.appendChild(cell);
}
}
return withMouseMoveListener(withClassname(table(rows),
"chessboard"),
hoverListener);
}

function clearHighlight(board, cls) {
var i, cells = board.querySelectorAll('.'+cls);
for (i = 0; i < cells.length; i++) {
var cell = cells[i];
cell.setAttribute("class",
cell.getAttribute("class")
.replace(new RegExp(cls, 'g'), "").trim());

};

}

function getAttackingClasses(cls) {
if (! cls)
return null;

cls = cls.replace(/.*(R.*B\d+).*/, "$1"); cls = cls.replace(/((R|F|A|B)\d+)/g, ".$1");
cls = cls.replace(/ /g, ",");
return cls
}

function highlight(parent, cls, highlightCls) {
if (! highlightCls)
highlightCls = "highlight";
var i, cells = parent.querySelectorAll(cls);
for (i = 0; i < cells.length; i++) {
var cell = cells[i],
curCls = cell.getAttribute("class");
if (curCls.indexOf(highlightCls) === -1) {
cell.setAttribute("class",
curCls + " " + highlightCls);
}
}
}

function hoverListener(e) {
if (e.target.tagName === 'TD' &&
e.target
.parentElement
.parentElement
.parentElement.getAttribute("class") === "chessboard") {
var rows = e.target.parentElement.parentElement;
clearHighlight(rows, 'highlight');
var toggleCls = getAttackingClasses(e.target.getAttribute("class"));
if (toggleCls) {
highlight(rows, toggleCls);
}
}
}

function placeQueen(board, posSelector) {
var position = board.querySelector(posSelector);
//Yes, this is the unicode for the queen symbol...
position.appendChild(document.createTextNode("\u2655"));
var highlightCls = getAttackingClasses(position.getAttribute("class"));
highlight(board, highlightCls, 'attacked');
}

function clearQueens(board) {
var i, cells = board.querySelectorAll("td");
for (i = 0; i < cells.length; i++) {
var td = cells[i];
if (td.firstChild)
td.removeChild(td.firstChild);
};
clearHighlight(board, 'attacked');
}


Additionally, I used the following css.

div.visualization {
text-align: center;
}

table.chessboard {
margin: auto;
margin-bottom: 1em;
}
table.chessboard td {
width: 1em;
height: 1em;
font-size: 1em;
line-height: 1em;
border: solid thin black;
}
.highlight {
background-color: #FAA;
}
.attacked {
background-color: #F66;
}
.solution {
background-color: grey;
}
input[type='range'] {
-webkit-appearance: none;