Honestly being able to understand where you went wrong is WAY more impressive than not going wrong in the first place :P I know for 100% sure I can't spot where I go wrong!
I am never sorry to see a longer-than-average-Mark-video length from you, Mark, because it promises to deliver something interesting. Tonight was no exception. I used to be entirely intimidated, yet fascinated, with any of the doubler/halver/etc. kinds of rules, but I am now getting more accustomed to thinking about them. I bet I will try this puzzle someday before long! Thanks for the daily sudoku puzzles even through the holidays!
@45:03 Mark was soooo close to saving himself 15 minutes and being able to simply sudoko it all out. He looked at the 9s in box 7 but missed that he had a single naked 9 elsewhere after which it all falls in place. ;-) I only know because I never could have done it except by following along with his solve - I got stuck so many times! Thank you Mark for puzzling it out for me!
I finished in 99:50 minutes. This was a tough one, but the math was so fun to work through. I got the coloring done pretty early, but figuring out the trick to narrow down the numbers was much trickier. Thinking about it for a while finally solidified how limited the prey cells were and that gave me major progress. I think my favorite part was near the end, figuring out that r4c4 had to also appear in the cage that has r7c3. This forced where that digit goes in box 1, forcing a 6 into r1c3, which disambiguates a lot of cells. This was a really fun puzzle, but it did require some thinking on my part. The title is very cute by the way. Great Puzzle!
23:30 there's a "simpler" way of doing this step. assume the 34/3 cage has only two non-zero cells. then you need v-cells of 8+9 AND p-cells of 8+9. this breaks the p-v requirement in box 3. 24:00 there's a "simpler" way of doing this step. the 9-cell cage needs 3 non-cancelling p-cells. this restricts the v-cells in boxes 1,2, and 4, and PLACES the v-cell in box 5. (into the 9-cell cage) now we must remove at least 22/3 from that cage, which must include a 9. --> box 3. this now forces a p-cell in that cage to be 8, with 6 or 5 left to find.
42:49 finish. I had a major snafu in the middle. I knew that there couldn't be a vampire in the 9-cell cage in boxes 1-2-4, but I never considered that there could be one in r4c4. This got me stuck for a while, before thinking about the possibility. Always good to see math in the puzzles!
At 28:05, I didn't get why "it can't add more than a value of 1"? Why couldn't it add 2? (Since 45 - 9 - 8 - 6 + 2 = 24) (So it follows quickly by deductive reasoning that it can't be 2 because you need a 6 elsewhere to make up the 10 or 11, but it didn't seem clear why you had so immediately discarded 2 as an option.)
He knew he only had one degree of freedom from the earlier calculation of 6 prey cells combined, had to be at least 38 and cant be more than 39. If you add 2 you would need 40.
I did actually have the idea for a Complex Sudoku using Gaussian Integers (of the form m + jn). You could have a bit of fun with constraints applying differently to real and imaginary parts (like, outside clues that act as Little Killers for real numbers and X-Sums for J-coefficients; lines that behave as Renbans for the Reals and German Whispers for the J's; or a rule that in a size 4 or smaller cage, a digit cannot appear both in real and imaginary forms). But ultimately, it would still be just a "two for the price of one" Sudoku. I suppose you could also have a Yin-Yang sudoku where shaded digits were real, unshaded were imaginary and cages provided clues like "7+j3" or "?+j?" (which might still tell you something, if it was within the perimeter .....)
I hadn't noticed. But I suppose it would be a consequence of computers using * to represent multiplication in spreadsheets and programming languages. Anyone who uses these tools extensively (me included) learns to subconsciously treat them as synonymous. I could easily see myself making a deliberate effort to use x to avoid confusing those unfamiliar with this computing quirk; but occasionally forgetting and reverting to the habit of using *. Once the mistake is made, I'll have very little chance of noticing the inconsistency.
65:15 for my time. That was a really intimidating problem, but actually rather approachable as I never really felt stuck. The logic at the beginning was actually beautiful. That said, it was challenging until the end as specially the sudoku part was not easy. Very nice puzzle!
If R6C6 was yellow, the red-yellow pairs in boxes 6 and 8 would be 8-9. Wouldn't this clash with the red-yellow pair in row 1 and rule out R6C6 from being a prey cell?
Tremendously clever puzzle even down to the Nosferatu word play which I doubt I would have spotted without seeing a comment. Messed with my head a bit, but very enjoyable indeed. I found the ending surprisingly tough.
Thanks so much for featuring my puzzle! I look forward to watching later when i get the chance Edit: sorry the ending gave you so much trouble! Spotting the naked 9 in box 4 earlier would have saved you so much grief :(
For me, the problems started one step earlier. I had fully pencilmarked the whole puzzle and didn‘t see the 123456/789 split in column 3. Ended up using a whole lot of complicated logic to prove that 9 in row 5 and 8 in row 2 had to always be in the same column and used that to put the 8 in r8c1, which turned out to be enough.
@@screwaccountnames Yeah, this is where I got stuck, too. I had solved quite a bit, lots of pencil marks, and that was where I stalled. Had to come to the video for the column 3 bit, which unlocked everything else. Loved the puzzle, though!
Did it twice lol. 2.5 hours and then... 2 hours. Forced myself to logic every step, repeated to make sure I could still do it based on logic. P/V colouring was quick, but took time to calculate permutations that worked between 24 & 34 cages.
i got a digit, the single cell cage in box 3 that obviously needs to be the vampire cell means the other one cell cage in box 3 must be it's true value... that's it... that's all i found i have no idea how to hunt these vampires. i found the vampires in box 6 and 8 (and 3) and that's all i found
Hello sir hope you you had a wonderful holiday question for you. I’m currently stuck on a puzzle by Penndel puzzles. The puzzles called Word math word, arithmetic I was wondering if you could crack that and do some hints in solving strategies for that puzzle.
![Primer | Sudoku Series - RAT RUN Phase 1 [1]](http://i.ytimg.com/vi/sba1T9vCRig/mqdefault.jpg)
![Primer | Sudoku Series - RAT RUN Phase 1 [1]](/img/tr.png)
Honestly being able to understand where you went wrong is WAY more impressive than not going wrong in the first place :P
I know for 100% sure I can't spot where I go wrong!
Exactly! I'm in awe
Noz four eight two. Nosferatu 😂
Ah! I didn't spot that! Mark
A great puzzle and great solve though. I highly enjoyed watching!
I am never sorry to see a longer-than-average-Mark-video length from you, Mark, because it promises to deliver something interesting. Tonight was no exception. I used to be entirely intimidated, yet fascinated, with any of the doubler/halver/etc. kinds of rules, but I am now getting more accustomed to thinking about them. I bet I will try this puzzle someday before long! Thanks for the daily sudoku puzzles even through the holidays!
@45:03 Mark was soooo close to saving himself 15 minutes and being able to simply sudoko it all out. He looked at the 9s in box 7 but missed that he had a single naked 9 elsewhere after which it all falls in place. ;-)
I only know because I never could have done it except by following along with his solve - I got stuck so many times! Thank you Mark for puzzling it out for me!
I finished in 99:50 minutes. This was a tough one, but the math was so fun to work through. I got the coloring done pretty early, but figuring out the trick to narrow down the numbers was much trickier. Thinking about it for a while finally solidified how limited the prey cells were and that gave me major progress. I think my favorite part was near the end, figuring out that r4c4 had to also appear in the cage that has r7c3. This forced where that digit goes in box 1, forcing a 6 into r1c3, which disambiguates a lot of cells. This was a really fun puzzle, but it did require some thinking on my part. The title is very cute by the way. Great Puzzle!
I can't believe how quickly you figured out the error in such complicated logic. Seriously impressive.
I like Happy Mark 😁
23:30
there's a "simpler" way of doing this step.
assume the 34/3 cage has only two non-zero cells. then you need v-cells of 8+9 AND p-cells of 8+9.
this breaks the p-v requirement in box 3.
24:00
there's a "simpler" way of doing this step.
the 9-cell cage needs 3 non-cancelling p-cells.
this restricts the v-cells in boxes 1,2, and 4, and PLACES the v-cell in box 5. (into the 9-cell cage)
now we must remove at least 22/3 from that cage, which must include a 9.
--> box 3.
this now forces a p-cell in that cage to be 8, with 6 or 5 left to find.
Stunningly clever.
42:49 finish. I had a major snafu in the middle. I knew that there couldn't be a vampire in the 9-cell cage in boxes 1-2-4, but I never considered that there could be one in r4c4. This got me stuck for a while, before thinking about the possibility. Always good to see math in the puzzles!
49:09 ... very thematic; I prefer this over the paying for the new movie
Nice puzzle!
At 28:05, I didn't get why "it can't add more than a value of 1"? Why couldn't it add 2? (Since 45 - 9 - 8 - 6 + 2 = 24)
(So it follows quickly by deductive reasoning that it can't be 2 because you need a 6 elsewhere to make up the 10 or 11, but it didn't seem clear why you had so immediately discarded 2 as an option.)
He knew he only had one degree of freedom from the earlier calculation of 6 prey cells combined, had to be at least 38 and cant be more than 39. If you add 2 you would need 40.
I love the use of the joker in the thumbnail in place of nosferatu!
@15:03, "this was done without real numbers."
Uh-oh, now there's going to be a sudoku made up with imaginary numbers. 😮
I did actually have the idea for a Complex Sudoku using Gaussian Integers (of the form m + jn). You could have a bit of fun with constraints applying differently to real and imaginary parts (like, outside clues that act as Little Killers for real numbers and X-Sums for J-coefficients; lines that behave as Renbans for the Reals and German Whispers for the J's; or a rule that in a size 4 or smaller cage, a digit cannot appear both in real and imaginary forms). But ultimately, it would still be just a "two for the price of one" Sudoku.
I suppose you could also have a Yin-Yang sudoku where shaded digits were real, unshaded were imaginary and cages provided clues like "7+j3" or "?+j?" (which might still tell you something, if it was within the perimeter .....)
Why is multiplication in this puzzle represented by both × and *?
I hadn't noticed.
But I suppose it would be a consequence of computers using * to represent multiplication in spreadsheets and programming languages.
Anyone who uses these tools extensively (me included) learns to subconsciously treat them as synonymous.
I could easily see myself making a deliberate effort to use x to avoid confusing those unfamiliar with this computing quirk; but occasionally forgetting and reverting to the habit of using *. Once the mistake is made, I'll have very little chance of noticing the inconsistency.
65:15 for my time. That was a really intimidating problem, but actually rather approachable as I never really felt stuck. The logic at the beginning was actually beautiful. That said, it was challenging until the end as specially the sudoku part was not easy. Very nice puzzle!
7/6th of a mark, 70 minutes solve here ;) very very good puzzle
If R6C6 was yellow, the red-yellow pairs in boxes 6 and 8 would be 8-9. Wouldn't this clash with the red-yellow pair in row 1 and rule out R6C6 from being a prey cell?
Tremendously clever puzzle even down to the Nosferatu word play which I doubt I would have spotted without seeing a comment. Messed with my head a bit, but very enjoyable indeed. I found the ending surprisingly tough.
Thank you Mark, always a pleasure!
Got stuck on column 3 -well done spotting that one
Thanks so much for featuring my puzzle! I look forward to watching later when i get the chance
Edit: sorry the ending gave you so much trouble! Spotting the naked 9 in box 4 earlier would have saved you so much grief :(
About 15-18 minutes of it
For me, the problems started one step earlier. I had fully pencilmarked the whole puzzle and didn‘t see the 123456/789 split in column 3. Ended up using a whole lot of complicated logic to prove that 9 in row 5 and 8 in row 2 had to always be in the same column and used that to put the 8 in r8c1, which turned out to be enough.
@@screwaccountnames Yeah, this is where I got stuck, too. I had solved quite a bit, lots of pencil marks, and that was where I stalled. Had to come to the video for the column 3 bit, which unlocked everything else. Loved the puzzle, though!
Did it twice lol. 2.5 hours and then... 2 hours. Forced myself to logic every step, repeated to make sure I could still do it based on logic. P/V colouring was quick, but took time to calculate permutations that worked between 24 & 34 cages.
28:59 for me. Great puzzle, I really enjoyed the solvepath on this one!
How the cage clues were written gave me an itch.
i got a digit, the single cell cage in box 3 that obviously needs to be the vampire cell means the other one cell cage in box 3 must be it's true value... that's it... that's all i found i have no idea how to hunt these vampires. i found the vampires in box 6 and 8 (and 3) and that's all i found
Could have placed 9 in box 7 before all the coloring of low digits, might have helped because 9, 8,7 get done
Hello sir hope you you had a wonderful holiday question for you. I’m currently stuck on a puzzle by Penndel puzzles. The puzzles called Word math word, arithmetic I was wondering if you could crack that and do some hints in solving strategies for that puzzle.
That's a brilliant puzzle that I really enjoyed solving. Thanks very much for sharing it, a top rate Christmas gift 😅
Not even going to attempt this one...
I was afraid to attempt it, and almost didn't even try, but tried it and found that it was surprisingly doable for me.
30:35 for me and solver #383.
64:47 for me.