should not throw spokes.

As I was rather critical of spokes puzzles in my last entry, I thought it was only fair to write one of my own that I felt satisfied my desire to have all the necessary steps fall from just logical analysis and have eliminations be close and simple. I think as a result of trying to write some that I've gotten a little better at seeing some chained eliminations through 6's and went back and redid some of the puzzles that I did not enjoy before with just logic. While solvable, they'd still make my list of least fun and fit more into the Hitori (I just don't like doing them) class. Maybe its the way each of those two types seems to have you "test" eliminations for the consequences, but neither does much for me when these tests are hard to find. Anyway, here is my first ever blog-posted puzzle that I wrote and my first Spokes; I've used my "0" hub variant which I haven't seen before but seems obvious so I will not take any credit for it as I'm sure its been done somewhere.

As with all spokes puzzles, the circles in the grid represent "hubs" from which you draw lines, or "spokes", to adjacent circles. The numbers inside of the hubs indicate the number of spokes that connect to that hub. Spokes cannot cross. The puzzle has one solution and can be determined using logic alone.

(SPOILER warning: skip the following section if you have not yet done spokes puzzles 13/15 in the Tuller/Rios orange book as I give away some solution methods).

On the other end of the spokes spectrum, here is an example of one I did not like.

I'm solving now basically to test my speed and so you do things like "assume uniqueness" even though this feels unsatisfying and use light pencil-shading when close to finished to "guess" the solution as opposed to always doing logic, just because the time is more important in competition than feeling satisfied by your solution. I much prefer puzzles that do not reward inspired guesses, sort of why I did not like the final at the WSC being a classic vanilla sudoku that turned into which of the finalists could be luckiest at solving by bifurcation because the logical step, on large placards, was harder to see than the T&E.

Anyway, consider Spokes 13 in the orange-covered Tuller/Rios book. If you are pretty good, it won't take you long to see the whole left column of 2 4 4 4 2 must have vertical spokes. You can then use a forced chain on the 6's in row two (basically because the 1 3 3 pattern in C5 at the top prevents both an upper right and lower right spoke from the 6 in R2C4) to help you guide all of the spokes on the 2's in the top row to the correct place. Once you see the pattern, you can actually progress sort of fast as the same reasoning is used each time, but it took me awhile and I don't think the forcing is immediately obvious. Using just logic at the time I first did this puzzle in that way took about 6 minutes. Now, imagine an alternative version of myself who uses the following reasonable guess that is quickly rewarded on this puzzle. The top row has lots of small numbers and the second row has lots of big numbers, so probably the 2 in R1C1 connects to the 4 and 6 in R2. This lets you quickly solve all of column 1's hubs. Repeat this hypothesis on column 2, then on column 3, etc. You finish the puzzle quickly. My clocked time was actually 1 minute 12 seconds on my first run-through of this puzzle using just this method. I guess I did not like this puzzle because the logical route, at least from my spokes experience, was slow versus the "fairly reasonable guess" route. I'm sure many others used just logic and were very satisfied by finding the pattern, but I personally get dissatisfied when I feel spotting a gimmicky pattern and going with it is a reward over thinking things out. I give this same reason for why Paint By Numbers is not really enjoyable.

A better puzzle then, although very hard, would be number 15 in the orange book. I did the whole thing by just logic and it is solvable, but somehow I'm still not that excited by my method to solve it which I give below. If someone has a cleaner way to solve it, I'd be glad to hear it.

Except for the forced line between the centers of 4 hubs that touch on the edges (R34C1, R23C5), there are no other simple spokes to place. What you can see though are two necessary non-spokes, the 1 in R5C1 cannot touch the 6 and the 1 in R1C5 cannot touch the 5. The latter does not provide much yet but the lower left corner gives you a start. The 6 hub in the lower left now has just one unidentified non-spoke. Assume that it is neither of the spokes that point to the upper left and upper right. Then you can propagate upper-left and upper-right spokes all the way through the column of 6's and see that at the top, with 1 2 1 each being hit by the spokes from the 6 at R2C2, you cannot satisfy the 2 hub at R1C2. Therefore, the missing non-spoke in the lower-left 6 is one of the ones pointed up and to the left/right. Fill in the other spokes from that 6. The remaining 6's in this column can all be described as missing one upwards diagonal and one downwards diagonal spoke as a result of repeating this analysis and so you can actually draw in every horizontal and vertical spoke that could occur for these 6's in C2. Now R4C1 and R5C2 must be connected as if the 3 did not use this connection, you could not satisfy the 4. Consider if the 1 in R5C1 touched the 3 in R5C2. Now, you fill-in the 4's along the left column which are forced until you get to the 3 near the top of the column. Consider any placement of spokes around the 3 in R2C1 - you will basically have to have the 6 in R2C2 have a spoke pointing down and to the right. As there is also a spoke between R3C1 and R2C2 at this point, this forces all the 6's below it to have both lower left and lower right pointing spokes, but then that cannot work as I've already mentioned. So, you can say that the 1 in R5C1 touches the 4 and the last spoke from the 3 connects to the 2, completing that hub as well. The 4 in R4C1 has all four spokes drawn, so it does not touch the 6 in R3C2 and you can say that the diagonal spokes for the 6's there are from R3C1 to R4C2 and R3C2 to R4C3. Look at the 3 in R5C4. It cannot connect a spoke to the 1 to the upper right as then you cannot fill the 2 hub in the corner. Draw in the only other three potential spokes. The 2 in the lower right corner must connect to the 6 now as otherwise you cannot complete the 6 hub. Notice that the 6 in R3C4 cannot connect to the 4 in R4C3 or the 1 in R4C5. Draw in the other 6 spokes. The 4 in R3C5 now has just 4 possibilities, so fill them in. This then forces the 6 in R4C4, which forces the 5 in R3C3 which you can continue to force the 6's in C2. As the 5 in R2C4 cannot connect to the 1 in the upper right corner, it is also fully defined now and just follow from here to finish the solution.

As I was rather critical of spokes puzzles in my last entry, I thought it was only fair to write one of my own that I felt satisfied my desire to have all the necessary steps fall from just logical analysis and have eliminations be close and simple. I think as a result of trying to write some that I've gotten a little better at seeing some chained eliminations through 6's and went back and redid some of the puzzles that I did not enjoy before with just logic. While solvable, they'd still make my list of least fun and fit more into the Hitori (I just don't like doing them) class. Maybe its the way each of those two types seems to have you "test" eliminations for the consequences, but neither does much for me when these tests are hard to find. Anyway, here is my first ever blog-posted puzzle that I wrote and my first Spokes; I've used my "0" hub variant which I haven't seen before but seems obvious so I will not take any credit for it as I'm sure its been done somewhere.

As with all spokes puzzles, the circles in the grid represent "hubs" from which you draw lines, or "spokes", to adjacent circles. The numbers inside of the hubs indicate the number of spokes that connect to that hub. Spokes cannot cross. The puzzle has one solution and can be determined using logic alone.

(SPOILER warning: skip the following section if you have not yet done spokes puzzles 13/15 in the Tuller/Rios orange book as I give away some solution methods).

On the other end of the spokes spectrum, here is an example of one I did not like.

I'm solving now basically to test my speed and so you do things like "assume uniqueness" even though this feels unsatisfying and use light pencil-shading when close to finished to "guess" the solution as opposed to always doing logic, just because the time is more important in competition than feeling satisfied by your solution. I much prefer puzzles that do not reward inspired guesses, sort of why I did not like the final at the WSC being a classic vanilla sudoku that turned into which of the finalists could be luckiest at solving by bifurcation because the logical step, on large placards, was harder to see than the T&E.

Anyway, consider Spokes 13 in the orange-covered Tuller/Rios book. If you are pretty good, it won't take you long to see the whole left column of 2 4 4 4 2 must have vertical spokes. You can then use a forced chain on the 6's in row two (basically because the 1 3 3 pattern in C5 at the top prevents both an upper right and lower right spoke from the 6 in R2C4) to help you guide all of the spokes on the 2's in the top row to the correct place. Once you see the pattern, you can actually progress sort of fast as the same reasoning is used each time, but it took me awhile and I don't think the forcing is immediately obvious. Using just logic at the time I first did this puzzle in that way took about 6 minutes. Now, imagine an alternative version of myself who uses the following reasonable guess that is quickly rewarded on this puzzle. The top row has lots of small numbers and the second row has lots of big numbers, so probably the 2 in R1C1 connects to the 4 and 6 in R2. This lets you quickly solve all of column 1's hubs. Repeat this hypothesis on column 2, then on column 3, etc. You finish the puzzle quickly. My clocked time was actually 1 minute 12 seconds on my first run-through of this puzzle using just this method. I guess I did not like this puzzle because the logical route, at least from my spokes experience, was slow versus the "fairly reasonable guess" route. I'm sure many others used just logic and were very satisfied by finding the pattern, but I personally get dissatisfied when I feel spotting a gimmicky pattern and going with it is a reward over thinking things out. I give this same reason for why Paint By Numbers is not really enjoyable.

A better puzzle then, although very hard, would be number 15 in the orange book. I did the whole thing by just logic and it is solvable, but somehow I'm still not that excited by my method to solve it which I give below. If someone has a cleaner way to solve it, I'd be glad to hear it.

Except for the forced line between the centers of 4 hubs that touch on the edges (R34C1, R23C5), there are no other simple spokes to place. What you can see though are two necessary non-spokes, the 1 in R5C1 cannot touch the 6 and the 1 in R1C5 cannot touch the 5. The latter does not provide much yet but the lower left corner gives you a start. The 6 hub in the lower left now has just one unidentified non-spoke. Assume that it is neither of the spokes that point to the upper left and upper right. Then you can propagate upper-left and upper-right spokes all the way through the column of 6's and see that at the top, with 1 2 1 each being hit by the spokes from the 6 at R2C2, you cannot satisfy the 2 hub at R1C2. Therefore, the missing non-spoke in the lower-left 6 is one of the ones pointed up and to the left/right. Fill in the other spokes from that 6. The remaining 6's in this column can all be described as missing one upwards diagonal and one downwards diagonal spoke as a result of repeating this analysis and so you can actually draw in every horizontal and vertical spoke that could occur for these 6's in C2. Now R4C1 and R5C2 must be connected as if the 3 did not use this connection, you could not satisfy the 4. Consider if the 1 in R5C1 touched the 3 in R5C2. Now, you fill-in the 4's along the left column which are forced until you get to the 3 near the top of the column. Consider any placement of spokes around the 3 in R2C1 - you will basically have to have the 6 in R2C2 have a spoke pointing down and to the right. As there is also a spoke between R3C1 and R2C2 at this point, this forces all the 6's below it to have both lower left and lower right pointing spokes, but then that cannot work as I've already mentioned. So, you can say that the 1 in R5C1 touches the 4 and the last spoke from the 3 connects to the 2, completing that hub as well. The 4 in R4C1 has all four spokes drawn, so it does not touch the 6 in R3C2 and you can say that the diagonal spokes for the 6's there are from R3C1 to R4C2 and R3C2 to R4C3. Look at the 3 in R5C4. It cannot connect a spoke to the 1 to the upper right as then you cannot fill the 2 hub in the corner. Draw in the only other three potential spokes. The 2 in the lower right corner must connect to the 6 now as otherwise you cannot complete the 6 hub. Notice that the 6 in R3C4 cannot connect to the 4 in R4C3 or the 1 in R4C5. Draw in the other 6 spokes. The 4 in R3C5 now has just 4 possibilities, so fill them in. This then forces the 6 in R4C4, which forces the 5 in R3C3 which you can continue to force the 6's in C2. As the 5 in R2C4 cannot connect to the 1 in the upper right corner, it is also fully defined now and just follow from here to finish the solution.

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