Pavel's Puzzles

This is the example puzzle exactly as presented on the ænigma paper. We are given that we should draw a single loop, horizontally and vertically connecting the dots of the grid, that neither crosses nor touches itself. Further, each grid square that contains a number must contribute exactly that many of its sides to the loop's path.

We begin by examining the square labelled “3” in the upper right corner. By the rules, we must draw segments of our looping path along three of the four sides of this square. Because this square is in the corner of the grid, the path must actually go around that corner: if it did not, we'd have to leave out both of the outer edges of the square, and that would leave us shy of our quota of three sides.

Naturally, the very same logic applies to the “3” in the lower right corner.

Now let us consider how the path segment we drew on the top edge of the grid might continue onward. From its current endpoint, it must go either leftward or downward; at present, we cannot know which of these is correct, but one of them must be right. Can we make productive use of this modicum of knowledge?

Indeed we can. Although we don't know precisely which way our path will eventually continue, it must be either along the top or right side of the square labeled “1&rdquo. By the rules, that square may only contribute a single side to the path, so we will definitely not be drawing any loop segments along the bottom or left edges of that square.

Again, we may also apply this reasoning to the path segment we drew along the bottom of the grid.

Remaining in that lower-right corner of the grid, now let us give our attention to the “3” square that's up and left of the two X's we just drew. Those recent marks have effectively placed this “3” square into a corner-like situation. Just as before, when we considered “3” squares that were actually in corners, we can see that the path must go around this induced corner.

We now turn our attention to the two “3” squares stacked vertically along the right edge of the grid. As we shall shortly see, this pattern of adjacent “3”s yields quite a number of useful deductions, even in isolation.

First, suppose that we did not draw a loop segment between the two “3” squares. Take a moment to think about what consequences would follow from such a choice before reading onward.

If no loop segment is drawn between the “3” squares, then each of them is left having to contribute all of its remaining edges to the path, and that leads to us drawing a tight little loop around these two squares. The rules say that we must draw just a single loop over the whole grid, so this exclusive little looplet cannot be allowed.

We can therefore conclude that our original hypothesis was incorrect, and deduce that there must indeed be a loop segment between the “3” squares.

Let us next consider the top edge of the upper “3” square: suppose that there was no loop segment there. What would the consequences be?

In that case, we would be forced to draw loop segments on both remaining sides of that square, and that would compel us to leave undrawn both the left and right sides of the bottom “3” square. That, however, would violate the rules, leaving the bottom “3” without its full quota of three path segments.

Once again, we have seen that our original hypothesis must be false, and that there must, indeed, be a loop segment along the top edge of the upper “3” square.

Of course, the same logic applies equally well to the bottom edge of the lower “3” square.

Another common pattern is illustrated towards the bottom centre of the grid. We have determined two of the three drawn sides of the “3” square we worked on earlier. Could the third drawn side be on the right of that square?

If we did draw that path segment, we would then be required to X out the two nearby sides of the lower “3” square, violating the three-segment quota for that square. The pattern here is that a path approaching the corner of a “3” square mayn't turn away from that square.

We can thus place an X on the left side of the upper “3”, thereby forcing us to draw its third path segment on the one remaining edge.

That then leads to X's on the two edges just outside the path corner just introduced.

We've seen that the path leading up to the corner of the bottom centre “3” square could not turn up from there. That leaves just two other possibilities, exactly one of which must be taken. Whichever of these two edges is not taken must, therefore, be the one and only X on edges of this “3” square.

We can then deduce that the other two edges of that “3” square are not X's.

That has drawn in the one path segment allowed around the adjacent “1” square, so we may confidently add X's to its remaining edges.

That leads once more to a pattern we've seen multiple times before: a “3” square in a corner-like situation. Again, that compels us to draw in segments on the two edges leading into that corner.

Another familiar pattern has again arisen: we have the path forced to traverse one of two sides of a “1” square. One of those two sides must be the one and only path segment drawn around this square.

Thus, all other sides of that “1” square must gain X's.

That, in turn, limits us to just one possible continuation of the path around the “3” square to the right.

Here we are presented once more with a path approaching the corner of a “3” square. We recognize this pattern from earlier, and know that the path mayn't turn away from the “3”.

Further, recall that this pattern also leads inevitably to the other two sides of that “3” square having path segments.

That in turn leads to another “3” in a corner.

Turning our attentions back to the right side of the grid, note that our earlier work has led to a “1” square in a corner. This new pattern has quite a different outcome: if either edge leading into that corner had a path segment, then the other such edge would also require one. That would lead to two path segments around a “1” square, an illegality, so both such edges must instead gain X's.

That leads to an interesting new deduction. Suppose that the remaining loop segment around the upper “3” square were long its left side, as shown. Then the two edges outside the corner thus drawn would need to acquire X's, and that would leave the nearby “1” square with no segments at all.

The upper “3” must therefore have its third segment along its right edge.

That provides us with X's on the edges outside the newly drawn corners.

We are then forced to complete the “3” square at the top of the grid as well as the lower “3” on the right edge of the grid.

The loop's path is now forced, in two separate areas of the grid.

That leaves us only one possible placement for the path segment around this “1” square.

That path can then only continue one way, which in turn decides the remaining sides of the centre “3” square.

We have long been neglecting the lower right corner of the grid; let us now return to it. Suppose that the third segment of that corner “3” square were along its left edge, as shown. Take a moment to visualize the consequences before continuing.

First, of course, that would force us to add an X on the top edge of that square. What then?

Then the path along the right side of that “3” square would have no choice in its continuation, and that would lead to no continuation being possible for the path along the left side!

The third path segment around that corner “3” square cannot, therefore, be along the left edge.

That leaves only the top edge for that third path segment, and its forced continuation.

We now have more forced continuations along the bottom edge of the grid.

This leads to completion of the “3” square in the bottom centre of the grid.

That gives us three X's around the upper-right grid point of the nearby “E” square, so we can confidently fill in the fourth.

From the bottom, let us now move our attention to the top of the grid. Here, we have a path that must continue along one of two edges of a “1” square. As we've seen previously, that implies that the other two edges must need X's.

That provides the one and only X around the “3” square just below that, thereby forcing us to draw path segments around the other three edges.

The path along the right edge of the nearby “T” square now has but one continuation.

The same now applies to the path along the top of the grid.

If there were to be a path segment along the left side of the “C” square, it would close off the loop, but it's not yet time to do that, since there are multiple path segments still left to connect together.

That then leads to more forced path continuation.

Let us now suppose that the path along the bottom edge of the “C” square were to turn downward. It should not be difficult at this stage to see all of the consequences pictured in this diagram. The most important aspect to notice here, however, is that there is now no way by which the main body of the loop could ever join up with the segment stranded in the lower left quadrant of the grid.

That hypothesis is thus disproved.

This leads to more forced path continuation and another occasion to block premature closing of the loop.

This process repeats: more forced continuation and blocked closure.

One more repetition of this process.

Now, the path along the left edge of the grid has a forced continuation and that draws in the one allowed segment around the “1” square. In turn, that leads to a forced completion of the adjacent “3” square.

We then return to forced path continuation and blocked loop closure.

Again, the path along the left edge of the grid has a forced continuation, leading to a familiar pattern: a path that will continue along one of two sides of a “1” square, thus forcing X's on the other two sides.

That leads to completion of the adjacent “3” square.

The rest is a familar tale of forced path continuations, finishing off our grid!

Here we pause for just a moment to savor our completed grid. (I have here erased all of the crosses marking known-undrawn path segments, in the interests of aesthetic clarity.)

We are told to walk our completed loop clockwise, beginning in the lower right corner, making note of each lettered square that contributes three edges to the path. The first such square that we come to is the “P” in the second row of the grid, after which we continue, encountering in order “H”, “O”, “B”, “I”, and finally “A”.