In the **games section** of the site, I’ve just posted six examples of a puzzle I invented, called Jum Noli. It’s inspired by Japanese-style number puzzles like Kakuro and Sudoku, but in keeping with my tastes, I’ve tried to make it “denser.” The small grid and tight restrictions mean that most Jum Noli puzzles require just one or two leaps of logic to get the first few boxes filled in, after which the solution usually comes quite quickly. Making that breakthrough, however, can be quite challenging.

As explained in the **puzzle PDF**, the objective is to fill in a 4×4 grid of boxes, placing one, two or three dots in each box.

There are two restrictions in placing the dots. The first is indicated by the numbers next to the rows and columns; these indicate the total number of dots that there must be in that row or column. The second is indicate by the dots inside the diamonds. These indicate that the surrounding four boxes must have a *clear majority* of that number of dots – thus, two or more of the boxes must contain that number of dots, and no more than one may contain any other particular number. That is, if the diamond shows three dots, the surrounding boxes could contain [3, 3, 3, 1] or [3, 3, 2, 1] but not [3, 3, 2, 2].

Each puzzle has only one solution. Finding it typically involves applying logic to determine with certainty how many dots must go in specific boxes, then using that information to make similar conclusions about other nearby boxes, and so on. Usually, once 6-10 boxes are filled, the rest will be much easier, but how does one go about filling in those first few boxes? It’s easiest to explain by means of an example. Let’s start with the grid to the left.

One trick to apply is to look for the highest and/or lowest row and column numbers in the grid, because there are fewer possibilities for how to arrive at those totals. As an extreme example, if a row showed a 4, you would know immediately that all the boxes in that row contained a single dot. Likewise, if it showed a 12, all the boxes would contain three dots.

This puzzle doesn’t have any 4s or 12s, but it does have an 11, which can only be made with three sets of three dots and one set of two. The obvious question to ask, then, is which box contains the two dots. Looking at the two diamonds at the top of that column, we can see that the group of two dots must be in one of the top two boxes – if both of those boxes contained three dots, it would be impossible to obey the requirements imposed by those diamonds.

The left diagram shows one possibility. The right diagram shows the other, but we can eliminate this as an option. If the top box of the second column contains 3 dots, then the only way to make the top row add up to six is for the remaining boxes all to contain a single dot. However, the top right diamond requires that it be surrounded by a two-dot majority, which creates a contradiction. Thus, we can firmly place two dots in the top box of the second row, and fill the rest of the row with three dots in each box.

The next step is easy. The top left diamond requires that the surrounding four boxes contain a clear majority of one-dot groups. Since two boxes have now been filled in with a two-dot and a three-dot group, the remaining two boxes must clearly contain a single dot.

After this, however, the puzzle gets a little trickier.

Looking at the third column, we see that it must contain a total of six dots. There are only two ways this can be accomplished; either with one group of three dots and three singles, or else two groups of two and two singles. However, the key thing to notice is that none of the middle three diamonds contains a single dot. This means that nowhere in this column can there be two consecutive single dots! This rules out the first possibility, and also combinations like [1,1,2,2]. The only possibilities are [1,2,1,2] and [2,1,2,1].

All we can do is try out these two combinations and look for a contradiction. It turns out that it’s easy to find one. If we try [2,1,2,1], we find that the top right corner must also contain a 1, in order to satisfy the top row total (6). This conflicts with the top right diamond, which states that it must be surrounded by a clear majority of two-dot groups.

Eliminating this possibility, we arrive at the following result, at which point the rest of the puzzle should be very easy to solve, since most of the rows and columns are almost complete. I’ll leave it to you to finish it:

Hopefully this example helps illustrate the sort of techniques needed to solve the puzzles. Of course, they vary widely in difficulty; some will only require reasoning at the level of the first two steps of the example – others will require even greater leaps of logic than in the third step. I’ve even managed to come up with puzzles that I myself can’t solve (though I’ve written a simple computer program to verify the existence of a unique solution).