How to Solve Hoshi Puzzles

Hoshi, or Star Sudoku, first appeared in the Russian puzzle magazine Krot. Your goal is to fill the triangles in the grid with the numbers 1-9 so that there are no duplicates in any large triangular region or along any row of cells , even those that skip over the middle. Because the lengths of these rows may differ, not all numbers may be used in any given one.

Here are some example lines showing what are considered “rows” in a Hoshi puzzle. Some variants include the outer tips of the triangle as part of a row, as shown by the light gray line at the bottom right.

To solve a Hoshi puzzle, you will be using all the same techniques outlined in the Sudoku tutorial. The only thing that really changes is the geometry.

When scanning for any given cell, you will be looking in three directions, rather than just horizontally or vertically. Let’s start with basic process of elimination scanning. This is what you would use to simply place pencil marks of what’s possible in each cell.

Looking at this cell, out of the numbers 1-9, we already know that it can’t be 1, 3, 4, 6, or 8, because those already exist in its region. This leaves a set of 2, 5, 7, or 9.

If we follow the horizontal line to the left, we can eliminate a 5. Then, we can follow another row down and to the left, where we see a 2 is also impossible for this cell. That only leaves 7 or 9 as possible numbers here.

Now that we’ve established how that works, we can start with the basic single-direction scanning. Here, we have an 8 that can’t be duplicated along this row of cells. Because the purple triangle has its bottom row completely filled out, that leaves only the very top cell able to contain an 8 in this region.

Here’s another example of one-direction scanning. This 2 can’t be repeated along this row, and since all the other cells in the region are filled out, it leaves only the tip of the red triangle as an option to contain a 2.

Most often, you will need to scan in more than one direction to discover what cells you can fill.

Here, we check along these two rows to show where a 2 cannot be placed in the orange region. Once again, this leaves only the tip as a placement option.

And here’s an example when you scan in three directions. Together, these 3s eliminate all possible cells except the highlighted one in the blue triangle, so it must contain a 3.

If you take the time to add pencil marks, assuming you don’t accidentally miss some eliminations, they can reveal a lot of information. Here, the two white cells can only contain a 5 or 9.

That means that neither 5 nor 9 can be anywhere else in their region. So in the purple cell, that leaves only a 6 as an option, and in the dark blue cell, if you remove 5, 6, and 9, it has to be a 4.

Then, if you look along the row that the two white cells share, you find that cell we discussed at the beginning as only being a 7 or 9. Because we know the 5 and 9 have to be in the two white cells, we now know that this red cell must be a 7.

Now that we know the basic techniques, we’re going to use pencil marks for the remainder of the tutorial to make it easier to see what’s going on we we move a little more quickly, doing multiple steps at once.

The pink triangle has only three cells left, which must contain the numbers 5, 4, and 9. That means we can eliminate these as options along the rest of the row.

So, in the yellow triangle, starting at the far right cell, removing 5 and 9 leaves only a 2. One cell to the left, and we’re left with only a 7. Then go all the way to the beginning of the row, and if you take away 5, 9, and 2, the only option left is a 3. After that, just to the right of the 1, we can remove 3 and 9, which means a 6 must be in that cell.

Here in the red triangle, the two faded cells can only be a 4 or 9, which means that neither number can appear elsewhere in the region.

If we eliminate the 9 from the white cell on the right, it leaves only a 3. Then, taking away 3 and 9 from the white cell on the left, we see that only a 1 can still be placed there.

Next, in this upward-slanting row, we know the two cells which must contain a 4 or 9, shown as faded cells in the red and blue triangles. So that means we can eliminate 4 and 9 in the rest of the cells in the row.

Starting at the bottom, taking away 4 and 9 leaves a 5, which is going to give us a nice chain reaction in a moment, because of all those cells in other regions that have to be either a 5 or 9.

Moving upward, when we take away the 9 and now the 5, we’re left with a 6. And finally, at the top, taking away 5 and 9 leaves us with a 2.

Next, let’s see what placing that 5 next to the 8 will cause to happen.

First, eliminating the 5 leaves only a 9 in these three cells, whether because they share a region or a row with the 5 we placed.
Next, the 9s leave only a single digit in these cells. That means we can no place those numbers.
Again, the newly-placed numbers leave cells in the same rows or regions with only one possible digit.
One more time, and these 9s force the last possible number in these cells.
The completed puzzle.

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