Hardest puzzle yet?? | Gene - Three Legged Spider
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- Puzzle Link - sudokupad.app/...
Let me know in the comments how you do!
Thank you for watching please let me know if you have puzzle recommendations/requests or can teach me something!
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I think a tip would be to slow down when you get to the pencil marking the whole grid stage. If you're putting in pencil marks due to column logic, just take a moment to consider if that does anything with the row/box/other clues etc. before moving on to the next lot of pencil marks. It sometimes seems like if you're thinking about rows you ignore the columns and vice versa.
To give a specific example, though you found it eventually, when you first pencil marked the options for the R2C5 arrow (arrow logic), noting that all of those cells see the R4C5 arrow (box logic) could have let you eliminate a 3 in that position much earlier. Likewise when you were examining Column 5 later in the puzzle, you noted that R3C5 and R4C5 couldn't be a 1-3 pair (Column logic), but didn't note that a 3 in R3C5 forced a 1 into R4C5 due to Arrow Logic
You're making really good progress in terms of picking up new rulesets and new types of logic. I just think a bit of focus on learning to shift between different types of logic with a bit more fluidity and combining them together will help as you move on to harder puzzles
This puzzle almost looks like it is designed to be an intro to Set Equivalence Theory (SET). (In this case it's helpful for the break-in but not necessary). Many great examples on CTC, also a complex example in a video Simon did for Numberphile. Personally I find achieving break-ins using SET as among the most gratifying solving experiences.
In this case, briefly, color rows 1 and 6 blue. Color columns 3 and 4 green. Knock out the cells shared by both, then knock out the cells where the arrow is in one color and the circle in another. You are left with 5 green cells and 2 blue cells which must add up to the same value, which can only be achieved by minimizing the open green cells (12s) and maximizing the blue cells (56s).
at 8:04 you see that 4 and 6 are still possible for r2c4 with 13 and 15, so 2 can come off that arrow, placing 2 in Box 2, and placing it on the other arrow in r1, which has further consequences. And later, you found the 3 on the 5-arrow breaks. Good job sticking with it!
29:55 really struggled to get it going, and to find the new place to look at
The 46 pair in b2c4 forces arrow to have a 1,so 13/15 meaning arrow in b1 must be 23/24 to add up to 56. A 4 in r1c3 will force arrow in b2 to be 4, and arrow in b1 to be 5 so it breaks...now we have a 56 pair in b1c3 and a 46 pair in b2c4..these 2 pairs will solve about half of the puzzle in the lower part, and when working on the way up, the arrow in r2c5 is forced to be 5 that will clear up all the upper part also. The trickiest spot for me was that r3c5 being a 3 will force a 1 in r4c5 and breaks r1c5..Awesome puzzle
I think you ended up finding pretty much all the main things I used. From experience, you can do a bit more with the combined arrow at the top. Starting out, the sum being 10 or 11 means there's a 6 in one of the circles, which pairs up with the 6 in the two squares above them. And then from there, finding the possibilities for 10 (24 | 13) vs. 11 (23 | 15) reveals that 1 must be on the right arrow and 2 on the left, along with crossing out a couple more digits.
I think your long "what if" was actually already solved by you afterward, with the 35 circle not being able to be 3. I think that's what really cracks into the rest of the puzzle without the "what if" deduction being necessary for it.
I looked at the ending for a while and ended up doing what you did, noticing that it couldn't be 1&3 and then realizing that 3 forced 1&3. I knew it couldn't be something with the left side because the only way to resolve those pairs is to use the 34 from the right side. Something equivalent would be saying either r3c4 is a 1 or else r1c5 is forced to be a 1, making r4c5 not a 1 because it sees both of those.
17:54 for a 6 * 6 that was a tough one.
Fun puzzle tho.
7:41. I got lucky with the arrows in the top row. I neglected the possibility of a 56 in the circles and went directly to a 46. Everything else was straightforward until I got to the 1-1 arrow in B46. It wasn't until I figured out its value affected the arrow across B24 that I realized the arrow was restricted.
Sometimes you just miss things. It happens. For instance: talking about the arrow in the top right region, you even said "if its 6 then it would have to be 1 and 5. If it's 4 it would have to be 1 and 3". Missing that the 2 is never used and the 1 is used both times. You could have done a lot with that.
Early on, when you have the arrowhead in box 2 down to 46 and the values on he arrow down to 1235, it is impossible to use a 2 to make either 4 or 6 on that arrow. Removing 2 from those cells leaves only one cell for the 2 in box 2.
I don't know if this opens up the puzzle.
I think row 2, columns 1 & 2 could have been a 1-2 pair, as long as the 1 or 2 in box 2 didn't go on the arrow.
Completed in 9:55 with a lucky bifurcation at the end as I wasn't seeing a logical way to resolve the pairs. It took me a while to resolve the top as well, I suspect I was missing a few obvious things. Still very cool puzzle from Gene, and good solve from Kathy!
At 13:10 the only thing that matters is if there can be a 3 in the 35 pair. You dont have to look at box 1 or anyting. Just the 35 pair in box 2 and compare it to box 4. When you see that 3 won't work you could just place the 5 and take it from there.
This was a tricky one for sure. The way I saw that last deduction was by noting that because b4 had 1-2-3-4 left, and two of them (r4c4/5) added to 5 thanks to the arrow, the other two (r5c5/6) must also add to 5. Then, looking up at the 1-3 pair in r1c5/6, I worked out that a 1 in c5 forced two 3s into c6 by its own pair and the hidden 5 sum. Not easy! Good solve and great video as always 🙂
Oh yeah, and my time was 23:45, love a sequential time
07:50 for me, the final deduction took me a while though it should have been obvious.
This one was pretty brutal. Didn't really find what I would consider a "clean" solve path myself. Found the solution in about 7:30, but didn't fully prove that it was unique until 12:34
I got to the final pencil markings at the end pretty fast but also got stuck on what to do to finish… if anyone knows a better method than running through the different options lmk!
I had trouble with this one. But then again, I always have trouble with arrow sudokus.
Solved this in 5:08, neat puzzle. I used some SET theory to pencilmark the bottom of c34 at the start, and then started solving from the bottom of the puzzle to figure stuff out.
The way to see the thing was to think about 14 from box 6 in box 4
4:46
A clever little puzzle.
A useful thing to note at the start is that the 4 squares R1/2C3/4 must sum to 21 and be 4, 5, and a pair of 6s because the two circles are analogous sunwise to the rest of row 1 which sums to 21, of course.
Removing the 5 from r2c3 with your what-if resolves that whole set. A slightly quicker view of that is that placing a 5 I'm that circle forces a 3 into r2c5 which needs a 12 pair on its arrow removing all your candidates from the circle at r4c5. That's basically what you did anyway with a few added steps mixed in.
You can also note that the one cell arrow on the left can't be a 5 and must be in c3 in box 1.
The final big step was resolving the other one cell arrow. Placing a 1 there immediately forced 4s into C6 in both boxes 4 and 6 so was ruled out and the rest unfolds exactly how you did it.
Still a good solve from you though, but saving time on those what-ifs will help you down the line I hope.
I really like the way you thought about the break-in! That's very elegant :)
I really feel like that top left could have been solved sooner, using that r2c5 arrow, but my brain can’t figure out why 😅
There's a simpler way to ascertain the circles in r2c3 and r2c4... They MUST sum to at least 10... and must use 1 and 2.
As arrow sudoku go this is fairly easy. Completed in 2m57s.
Cool shirt! Also good job on this one, it was a tricky one for sure, but you managed to solve it nonetheless! 👏👏👏
@20:35 Exactly. "This is not 3, 1." So IF (5,3) was 3, then (5,4) needed to be 2 and (4,3) needed to be 1, which would break the arrow logic. Hence (5,3) cannot be 3 and must be 4.
solved in 19:58. this one definitely gave me a test for my reading skills, it took me a while to understand what was going on with the top two arrows.
also, I misread this title and thought it was "gene the three legged spider", lol
Yes, I thought that was pretty tough - I couldn't find a way to break in other than bifurcating on the 3 or 5 possibilities on r2c5, and finding that 3 doesn't work.
That's not bifurcation, that's the logical step. If r2c5 is a 3, so is r4c5.
00:06:04