📕Get my FREE Solving Guide that will help you solve over 80% of all Sudoku puzzles🧩to include NYT Hard👉👉www.buymeacoffee.com/timberlakeB/e/125822 Timestamps 0:00 Intro 00:23 It’s Solving Time 00:27 Mistake #1 03:30 Mistake #2 06:39 Mistake #3 07:45 Cool Combination 09:09 Mistake #4 11:01 BONUS Tip 12:13 Mistake #5
"and because of the 4,8 it can't go here (55)" The question I have is: Why? I know that a 4 in 55 busts the puzzle leaving two 9's in 37 and 77. But how did that 4,8 in 53 and the 4,5,8 in 57 tell you this?
Okay I just figured out a coding path to logically get rid of the 4 in 55. The key is column 7. Since 37 and 77 have that 5,9 that gets rid of them in 57 and 97 which is what you show in screen. So once those are gone then that leaves the 4,8 that you point out thus eliminating that 4 in 55. That then leaves 95 the only place the 4 can go.
Got it! My code now methodically completes this puzzle. So naturally I revisited Intermediate Tutorial 5 #1 but it still gets stuck at block 4 where the 5,8 and 4,6 pairs are the only choices. I'll post a comment on that video once I re-watch to see how you overcame that.
Pen, paper, pencil. 15 min. 48 , 59 pairs, pointing 4s bx5 made it easy. Now the watching time. Hope it will nice. Good puzzle Thanks. Saw vedio and mostly the same.
Started off crosshatching the same as you - and immediately spotted a naked single 2 in R9C3, and then 2 has to be in R8C5. Same pointing pair of 3's that you spotted. Single 7 in column 3, then the single 1 in the column. 59 pairs in column 7 just by centermarking the two rows. Those two claim the column, placing 48's in the remaining cells. Then the same dance with the 4's that you did. Attacking box 8, found a 13 pair in columns 4 and 6, leaving a single 5 in R7C5 and disambiguating the 59s in column 7. Crosshatching finds the 7 in box 2, and now it's the endgame with singles, singles everywhere. 17 minutes, not a great time, but I was eating dinner at the same time that I was solving.
Nice and relatively easy puzzle. Solved it in 12:54. The 59 pair was very easy to notice by filling in candidates. I tend to do that whenever I have at least 4 digits in a row/column/box. For me that is the only way to notice such pairs/triples or singles. The rest didn't need higher level techniques to finish.
In column 3, I got the 2 in row 9 among my first digits at the start of the puzzle, so completing the column with the 48 pair and the 1 was rather obvious. The 3s at the start were also quickly placed. From there, dividing column 7 between the 48 and 59 pairs was quickly spotted. I lost quite a bit of time due to neglect at considering column 5. When I did, the 4 being isolated in row 9 was the last straw. Very well constructed puzzle, lot's of fun doing it, thanks.
I probably made several of the mistakes identified here. If one of the mistakes is not to notice something, well... In my case, I didn't notice a pointing pair of 4s. I spent some time searching for restricted cells in vain -- I'd already found a 59 pair and a 48 chain of pairs. I finally resorted to my standard practice. As I was centermarking blocks, I noticed the pointing pair of 4s in block 5, and that cracked the puzzle. I thought I'd broken the puzzle when I discovered that neither 9 nor 8 could go in a bivalue 89 cell in block 2. I blanked that cell out, and continued, hoping against hope... Eventually, I placed 5 in that cell and managed to finish the puzzle. Mistake #1: Bull's eye, for me with the pointing 4s. But how does one avoid *not* noticing something? I don't notice something until I notice it. 1:40 Just as you mentioned "gotcha", you fell for one. Continue with 2s, blocks 1 and 4, 2s are in columns 1 and 2. So place 2s in blocks 7 and 8. 4:10 Right after placing the 2s, I placed the 1 in the green cell because of the pointing 1s in block 4. Okay, I just backtracked and realized that my pointing 1s was a well-placed blunder. Instead, I should have (and would have) placed 1 in the green cell upon attacking the five-digit row 7. 5:50 Frustration? Not at this point in the screen's grid -- even without the naked 1 and 7. I attack rows and columns 3 and 7, and get those digits and 48 and 59 pairs. 9:20 Block 5 there is where I made my big mistake, especially since I had 4 and 8 cornermarked. I removed 4 and 8 from the central cell, but didn't notice the resulting pointing pairs -- specifically the pointing 4s. I would have solved this a lot sooner otherwise. My final mistake: once I noticed the pointing 4s, it was mostly unraveling centermarks, although the grid wasn't completely centermarked. I guess that I thought I had 5s pointing at R2C4, and removed the 5 leaving 89.
I'm watching since I make mistakes, a common one being to work our solution numbers then misplacing them which is infuriating. Another one when working mentally is to juggle an incorrect combination of numbers in my head. Btw, I noticed you could fill in all the twos at the outset here.
-Square 57 can also be a 5 not just a 4 or 8. And 4 can also be in 55, 86 and 96 so I am not following how you determined 4 to be in 95.- Okay strike all that. I see now what you pointed out that since there are only two choices for the 4 in column 5 it has to be either of those. Sow how does the 4,8 in 53 and 57 foretell that a 4 in 55 busts the puzzle leaving two single 9's in 37 and 77?
Great question. Since 4,8 are only 2 possibilities for 5,3 and 5,7, that makes them a naked pair. If you put a 4 in 5,5 then the 8 would be only possibility for 5,3 and 5,7. So you know you can eliminate 4 from all other cells in row 5. Does that help?
I call them "helper bits" since that is the name of the variable I am using in the code I am writing. No clue who "snyder" is. Schneider was Pat Harrington. Learning Sudoku techniques _One Day At A Time._ btw, the code I have so far solves every example I have found in my local paper (every step eventually creates at least one "only" possibility) so I came to YT for examples where a more complex algorithm is needed. So far I added code for pointing pairs to remove bit possibilities. But most pointing pairs are in rows or columns where there are no more bits in other blocks to remove. So the result is minimal help if any. I then coded X-wing but it had unintended consequences that busts the puzzle so I guess I don't get the concept of them yet or am missing a requirement to make them valid.
This is quite fascinating Bill Keith. Snyder is Thomas Snyder, 3 time world Sudoku champion who started the marking to help find restrictions in the puzzle better. Tell me more about your coding. Are you trying to make a solver or puzzle generator? What is the end goal?
@@SmartHobbies The original intent was to write code to turn on all the helper bits at the start (I guess I came up with the same idea as Snyder) to make it easier instead of doing it all manually. But I can see my code became a solver if you just keep hitting the enter key. The boxes in the newspaper are printed so small so i had to write really tiny using my fine point marker. Since I was using a marker, making a trial and error mistake was not an option otherwise the puzzle becomes a mess so I wanted to be 100% right before filling in a number. At some point I got annoyed once I decided a program would be more efficient (had no clue there are a ton of apps out there) and decided to do something else while sitting on the toilet than Sudoku. 6 months later I now have free time and decided to get back into coding after 15 years and felt this would be a nice project to get it all in my head again. The logic has been all straightforward so far but now I am getting into the tough stuff where complex decisions have to be made to whittle down the helper bit choices for my code to decide an "only" choice. I search the puzzle then highlight the squares that have only one choice. Hit enter and it auto fills in all the correct numbers and updates the bits for what is left. The steps are repeated until solved or it can't find any more obvious "only" choices. The endgoal is the complete the program then re-write it from C/C++ to Java to help me learn Java/HTML more. I originally learned C by coding on a MUD in 1998.
I tried this puzzle through 2 separate sittings, and butchered it both times. My first time was 40-something minutes, the second was 20-something minutes, when I got error messages that I had made a mistake. I was trying to get fancy early on with the 5,6,9 combinations in rows 3&7, but I didn't think it all the way through. I correctly deduced that at least one end cell had to be a 6, because if neither end cell contained a 6, that would put two 6's in column 5. But then I moved on before considering the possibility that both end cells could be 6's, and that messed up the rest of the puzzle. And as for the suggestion at 11:15, I can't do that. I have no memory whatsoever, and I have to write everything down.
