I think there's a mistake in Simon's logic concerning box 5 (around the half hour mark). You can have both "sticks" sum to 9 but placing a 93-pair in r4c3 and r4c7, with one arrow being a 234-sum and the other being 135.
@@renedekker9806 I was able to get everything to work this way in boxes 2 and 5. I put a 3 in r4c3 and a 9 in r4c7. Then I have the following down the middle 3 columns: C4: 386149257 C5: 125678349 C6: 497235168 This satisfies all the lines in the middle 3 columns, except for the one in box 8-9. I have 68 as the two numbers in box 8 on that line, so the number in box 9 has to be 2, but it can't because of the 12 pair in that row (row 7). This is definitely bifurcating. I'm sure it's not the intended path.
Yes, solving the puzzle by saying "it must be that way" when the logic so far doesn't prove that was very disappointing. This is the first time I feel like Simon failed the solve.
Jay Dyer is one of the very special constructors who can make lovely puzzles with simple rules. Thanks for solving this, Simon. (And just since your video appeared the solves have gone over 500 as I am looking at it now - very well deserved!)
If I may comment just one thing: to alert that there is an error in logic is OK, but to 'shout' it is normally considered at least unnecessary, and often considered rude. (Perhaps you did not realize that in typing the convention has become that all caps are used to indicate shouting or yelling?)
is there not a mistake around 31:00, couldn't the 2 lines going through box 5 be 2349 and 1359 with something like 9 in R4C3, a 34 pair in R5C4,R6C4 and a 2 in R7C4, with a 3 in R4C7, a 59 pair in R5C6,R6C6 and a 1 in R7C6
That would put two 3s in box 5. But if you meant to put the 3 in box 6, that could potentially work (at first glance) - and you could also swap the locations along the two lines. I can't see why that wouldn't work (at the moment), but I'm sure that it will eventually break.
It does eventually. But not yet where Simon misses that option. It gets sorted out once you get the line in box 8. And the logic to get there is nearly the same as what Simon does. There are a few extra pencil marks on a few cells, but the path is basically the same.
Not usually one to bay for Mea Culpas, but I would really appreciate a relooking over the logic in box 5 where Simon neglects the possibility of a 9 in the box, and a 9 in box 4/6 on the wings. I couldn't find a nice way of doing it, and given that this is a Jay Dyer puzzle, I'm sure one exists. Sadly I'm not smart enough to find it, yet I'm sure Simon is! I'm fairly sure it is something to do with the constraints it puts on box 8 and box 2 via box 4 and 6, but I couldn't find it without just case testing.
I needed to prove a lot of things in boxes 2 and 8 before I could resolve that. We get an X wing on 1s and 2s putting them in the central column of box 2, and they have to go on the 4 cell lines making them a pair. 3, 4 and 9 cannot go on the 3 cell line in box 2 and have to go on the wings in box 5, therefore they form a triple in the central column of box 8. Then the 3 cell line in boxes 8 and 9 has to be a 58 pair in box 8 and a 3 in box 9. The 3 gets placed in box 4 since it has to be repeated on the 4 cell lines in box 5. R3c5 can only be a 5 or 8, therefore r4c6 has to be a 2 resolving the 12 pairs. Then we can eliminate the possibility of the lines in box 5 being 2349 and 1359 because the 1 and 5 can't be on the same line.
As others already pointed out, Simon jumped to a correct conclusion using incorrect logic at about 31:00. Here is what I did to reach the same conclusion. In short, after pencilmarking the *1-2* pair in *box 5* (as Simon did @21:53): 1) I ruled *2* out of *r4c4* (easy) 2) I showed that, *in box 8,* (easy) *r7c4 = 2* *r7c6 = 1* 3) I restricted *r4c3* to either *3* or *4* (hard) 👉 See below for details
*STEP 1* As shown by Simon, *box 2* contains three circles. This implies that the long lines contain *2* circles and *5* _"arrow digits"_ that can be either: 1 + 2 + 3 + 4 + 5 = *15* (option 1) 1 + 2 + 3 + 4 + 6 = *16* (option 2) If *r4c4 = 2* then option *2* would be ruled out (it would force the corresponding two circles to be 9 + 9 = 18) and the corresponding two circles would add up to *r4c4 + 15 = 2 + 15 = 17* Hence, they would contain *8 + 9 = 17.* and the short line segment in *box 2* would contain the remaining two digits: *6* and *7.* However, *r3c5 = either 6 or 7* is impossible as it would force either *6* or *7* on one of the line segments in *box 5,* which would push the correspondind arrow total to at least *6 + 3 + 1 = 10*
*STEP 2* It can be easily shown that, in *box 8,* *r7c4 = 2* *r7c6 = 1* By sudoku, this is their minimum value, and any value higher that that would force the corresponding arrow total to be at least *3 + 3 + 4 =10*
*STEP 3* Let's now focus on the *4-cell* line with endpoints in *boxes 4* and *8.* As shown above, its endpoint in *box 8* must be *r7c4 = 2* By sudoku, its other two _"arrow digits"_ must add up to at least *3 + 3 = 6.* Hence, the total value of its "arrow digits" is either: *r7c4 + 3 + 3 = 8* or *r7c4 + 3 + 4 = 9* and the possible candidates for its endpoint in *box 2* are: *r4c3 = {3, 4, 8, 9}.* However, *r4c3 = 8* can be easily ruled out because, by sudoku, the other cells would add up to at least: *3 + 4 + 2 = 9* Moreover, if *r4c3 = 9* then, by sudoku: 🔹the digits *3* and *4* would be locked in the upper two cells of *column 6* 🔹the minimum possible values for the other arrow in *box 5* (with endpoints in boxes 6 and 8) would be *3 + 5 + 1 = 9* (with r4c7 = 3) 🔹the lower two cells of *column 6* would contain digits higher than *5:* *r8c6 = r9c6 = {6, 7, 8}* As a consequence, by sudoku, the *3-cell* arrow with endpoints in boxes *8* and *9* would add up to at least: *4 + 6 = 10,* which is impossible. Hence, *r4c3 = {3, 4}.* QED 😏👍
@@Paolo_De_Leva I actually did this a bit different. I noticed that 67 is in C5 in box 5 and with some simple deductions I noticed that R3C6 has to be either a 6 or a 7 If it was a 7, then R3C5 has to be either 5 or 6, 6 doesn't work so R4C6 is a 2 If it was a 6, then R3C5 has to be 7 or 8, 8 doesn't work so R4C6 is a 2 So R4C6 is always a 2 This resolved the 12 pairs in boxes 5 and 8 Also R3C5 being a 5 or an 8 makes a 5678 quad with the digits in box 5 C5 -> so the digits in C5 box 8 are a 349 triple. The triple in box 8 helps. The min value of R7C7 is 3, the min value in R8C6/R9C6 is a 5. So the sum is 8. You get a 58 pair in R8C6/R9C6 and a 3 in R7C7 and R4C3 The 58 pair stops the sum to be in R4C7, so you get a 4 in R4C7
I loved the way the initial head-scratching and break-in finally gave me... just a 12 pair. And yet that 12 pair becomes extraordinarily powerful and does so much work. I did struggle a lot in the middle and probably didn't spot some things I was supposed to, so I'll be interested to see what Simon picks up.
21:40 Once you conclude that the circles for the two long lines in box 2 must be in box 2, then the highest these circles can be is 8 and 9, which would mean that the sum of the two long lines is only 34. If the circle for the short line is outside the box and is 9, then the sum of the two cells on the short line in box 2 would be 9, and the sum of the whole box would, therefore, be less than 34+9, which is only 43 (less than 43 because there's still one cell outside the box on one of the long lines).
I'm so proud of figuring this one out! It was an immense struggle over several hours, grappling with what seemed like such ambiguous information. I still don't think I fully "got" it, but managed to work it out in the end.
The incredible run of astonishing sudokus continues... 👉 *Jay Dyer* = Cosmic class constructor 👉 Only one mistake by *Simon* (at about 31:00) = Cosmic class solver This puzzle was so hard that by mistake I hit dead ends several times before figuring out the correct logic path for the break-in (which I described in a separate comment).
I started this puzzle by concluding that every line that was 4 cells long needed at least either a 1 or 2, or both. I colored in those cells and that led me to get through the first few steps of the break in.
Sorry I don't have time to watch this right away as normal. I am waiting for Zetamath to start one of his unmissable reactions to you solving his puzzle, the one you featured yesterday. Five minutes to go. Popcorn ready.
The toughest CTC solve for me for some time. I had to look at it for several days, until this morning I immediately spotted something not that difficult but which had previously eluded me (in box 7). What a clever puzzle, as one would expect from Jay Dyer.
I finished in 124:49 minutes. This was an excellent puzzle as usual from Jay Dyer. She has this uncanny ability to create these spectacular geometrical showcases. It took me too long to even notice that box 2 was the obvious place to start. That limited r4c4&6, which finally gave me the in that I was looking for. Calculating box 5 was so fun to do. I was surprised when I saw that no matter the orientation, column 5 of box 8 always had to contain the same digits. This created a feedback loop that in turn affected boxes 5 and 2. That was so awesome. I think if it weren't for Simon telling a secret long ago that three celled arrows with different digits have to contain at least two digits from 123, then this puzzle would have taken me significantly longer, if that is possible. Lovely job as usual from Jay. Great Puzzle!
Definitely a mistake but I'm following it through and eventually the 3-long line between boxes 8 and 9 disproves it - 1/2/3/4 can't be on the box 8 segment (regardless of the way you set up those box 5 lines), and the box 9 part of the line is forced to 3/4, so the pair 6/8 can't be on the box 8 part, which it would have to be with the 951 line above.
@@jacklinton4885don't see why it can't be 3 or 4 on the line in box 8. It can't be both, but if R4C3 is a 9 then only 2, 4 and 9 are excluded from the line in box 8.
@@calebu2 So first, from where the mistake is made, add in the 1359 option as well as the 3489 pattern that Simon gets. Then look at 1s and 2s in box 2 - given where the 1/2 pairs are, they are forced to be in column 5 in box 2. Now the 3-length line in box 2 we've already determined has a 7/8/9 on it, so we can't have the 1/2 in the r3c5 cell (as 1 or 2 + 1 or 2 = maximum 4). By the same logic, we can't have the 3/4 in that cell (max 6). Also, in box 5, we either use both the 3/4 on the lefthand line, or we use one on each line, so we can't have 3/4 in column 5 in box 5. Therefore, the 3/4 in column 5 must be in box 8 - so the part of the 3-length line in box 8 can't have 3/4. Quite a long logic chain, would have loved to see Simon work through it!
Wow! Great puzzle; loved it. Did not feel confident in my solve while I was going through it, but the time was quite good in the end ---- 44:15. Happy with that.
Simon may miss silly things occasionally but he seems to have a magical ability to divine the starting logic and then to push through to the end no matter how tough the logic along the way!
When Simon started to think about 1*179 as an option for it, I was like no don't waste time on that, it would be a 1*56 for best, leaving 8 with 234 being 1 too many.
The way I figured out box 2 plus its sticky-oiy bits is to consider that two lines would have 9 and one would have 8. That would nake the three lines sum to 52, leaving only 7 for the sticky-out bits, making 6 the largest digit that could go on them (in all cases). With that proven, the largest numbers that can be on the kines are 9, 8 and and therefore they sum to 48, leaving 3 outside the box to be 1 and 2.
Great puzzle! The 349 logic around box 5 resolves later, as many have pointed out. Pretty hard cruxes, mixed with nice-flowing sequences. I liked the 123 triples. :-)
Simon missed something. You forgot about the 9-9 combo for the two lines in box 5. This allows a line to have a 5. It breaks for other reasons later on, but you saved yourself a lot of headache. However, I know that you don't want to save yourself headaches :D
Haven't watched the solve yet, but this one took over 2 hours for me to complete. Still don't know whether I was dealing with an arrow puzzle, a parity puzzle or a simply a maths problem...
Now I wonder what's an intended solution for this puzzle. Cause I didn't find it, I looked at same logic as Simon, just saw that there can be two 9s there and didn't know what to do next. Of course, if you try two 9s it fails eventualy, but you need to do a lot of work for it to fail, so I don't think this is intended path. I guess it has something to do with column 6, but I'm not sure
I like the short ruleset and looking forward to trying the puzzle for myself (before watching Simon's entire video). I think I'll need to think of each line as a single arrow with an unknown position for the circle, if that makes sense.
I'm not sure how, but managed to do it. Maybe I'm starting to learn something. :D I did had to use notepad to keep track of the possibilities though. My brain just doesn't relate to numbers. Now to watch the video. :) And see where my logic was sound and where it was "just luck"... LOL
Solved in 51:08. How I worked out the 12 pair in box 5 was that if one of them was the arrow circle then we have 7 digits on the arrows in box 2, and therefore need 3 digits summing to at least 28+1 on the arrow circles.
How do people even start to construct any of these Sudoku's that you do? Do they start with a solved puzzle then add clues till it makes logical sense?
My opening logic was that on box 2 you have at least 5 consecutive numbers. You have to include both 1/2 or you get 3+4+5+6+x>18 and that is impossible for 2 numbers.
This is such a weird way to write the rules? Isn't it just that 1 digit is the sum of every other digit on its line, because its own presence will always repeat. With the "digits may repeat if allowed by Sudoku".
Or, to put it in the way the rule is worded, the sum of those digits (3+4+5+6) is 18. The line must contain a digit equal to half the sum, so would need to contain a 9.
Usually watch these religiously, only Simon though as I find mark doesn’t explain as well, that’s fine I wouldn’t have a hope of solving these so need the extra explanation but tbh this one left me cold, seemed too hard for Simon too which is unusual and I couldn’t follow his logic which from comments seems to have been a little flawed anyway
not addressing the fact that every line has to sum to an EVEN number (as you cant put something like 5.5 in a cell if the sum is odd) and the cells it is made from to a maximum of 18, being half of 9, the maximum number you can put in a cell. 🧀🐁
@@TemujinReads You'd have to write at least a 13 in one of the cells for that to function as one large line, I'm not sure that clarification is necessary.
I think there's a mistake in Simon's logic concerning box 5 (around the half hour mark). You can have both "sticks" sum to 9 but placing a 93-pair in r4c3 and r4c7, with one arrow being a 234-sum and the other being 135.
That would put an 68 in the middle column of box 5, making box 2 impossible. But you are correct that only comes much later in Simons solve.
@@renedekker9806 I was able to get everything to work this way in boxes 2 and 5. I put a 3 in r4c3 and a 9 in r4c7. Then I have the following down the middle 3 columns:
C4: 386149257
C5: 125678349
C6: 497235168
This satisfies all the lines in the middle 3 columns, except for the one in box 8-9. I have 68 as the two numbers in box 8 on that line, so the number in box 9 has to be 2, but it can't because of the 12 pair in that row (row 7).
This is definitely bifurcating. I'm sure it's not the intended path.
And I just paused the video at that point to see if those could be two lines summing to 9. Glad to see someone else did that, too!
I kept that option in my solve path. After getting the box 8 line sorted, the double 9 possibility became impossible.
Yes, solving the puzzle by saying "it must be that way" when the logic so far doesn't prove that was very disappointing. This is the first time I feel like Simon failed the solve.
The set of rules should be described as "The highest digit on a line equals the sum of the other digits on that line".
Jay Dyer is one of the very special constructors who can make lovely puzzles with simple rules. Thanks for solving this, Simon. (And just since your video appeared the solves have gone over 500 as I am looking at it now - very well deserved!)
"It's either something new or it's sudoku"
Yes, Simon. Those are always the 2 options.
LOGIC MISTAKE:!
Cant we put 234 on one stick with a 9 in R4(Box 4 or 6). And 135 with a 9 on the other stick in Box 5???
I was fighting with the puzzle before watching the video and came to this conclusion. I gave up and hit play hoping this reality would be ruled out
If I may comment just one thing: to alert that there is an error in logic is OK, but to 'shout' it is normally considered at least unnecessary, and often considered rude. (Perhaps you did not realize that in typing the convention has become that all caps are used to indicate shouting or yelling?)
is there not a mistake around 31:00, couldn't the 2 lines going through box 5 be 2349 and 1359 with something like 9 in R4C3, a 34 pair in R5C4,R6C4 and a 2 in R7C4, with a 3 in R4C7, a 59 pair in R5C6,R6C6 and a 1 in R7C6
That would put two 3s in box 5. But if you meant to put the 3 in box 6, that could potentially work (at first glance) - and you could also swap the locations along the two lines. I can't see why that wouldn't work (at the moment), but I'm sure that it will eventually break.
It does eventually. But not yet where Simon misses that option. It gets sorted out once you get the line in box 8. And the logic to get there is nearly the same as what Simon does. There are a few extra pencil marks on a few cells, but the path is basically the same.
It's never too late to do the right thing! Glad to see this puzzle surface after so long, worthy of a broader audience.
Not usually one to bay for Mea Culpas, but I would really appreciate a relooking over the logic in box 5 where Simon neglects the possibility of a 9 in the box, and a 9 in box 4/6 on the wings. I couldn't find a nice way of doing it, and given that this is a Jay Dyer puzzle, I'm sure one exists. Sadly I'm not smart enough to find it, yet I'm sure Simon is! I'm fairly sure it is something to do with the constraints it puts on box 8 and box 2 via box 4 and 6, but I couldn't find it without just case testing.
I needed to prove a lot of things in boxes 2 and 8 before I could resolve that. We get an X wing on 1s and 2s putting them in the central column of box 2, and they have to go on the 4 cell lines making them a pair. 3, 4 and 9 cannot go on the 3 cell line in box 2 and have to go on the wings in box 5, therefore they form a triple in the central column of box 8. Then the 3 cell line in boxes 8 and 9 has to be a 58 pair in box 8 and a 3 in box 9. The 3 gets placed in box 4 since it has to be repeated on the 4 cell lines in box 5. R3c5 can only be a 5 or 8, therefore r4c6 has to be a 2 resolving the 12 pairs. Then we can eliminate the possibility of the lines in box 5 being 2349 and 1359 because the 1 and 5 can't be on the same line.
1:04:30 Simon completely ignores the 8 in the column, despite me calmly screaming at him to fill in the 5.
As others already pointed out, Simon jumped to a correct conclusion using incorrect logic at about 31:00. Here is what I did to reach the same conclusion.
In short, after pencilmarking the *1-2* pair in *box 5* (as Simon did @21:53):
1) I ruled *2* out of *r4c4* (easy)
2) I showed that, *in box 8,* (easy)
*r7c4 = 2*
*r7c6 = 1*
3) I restricted *r4c3* to either *3* or *4* (hard)
👉 See below for details
*STEP 1*
As shown by Simon, *box 2* contains three circles. This implies that the long lines contain *2* circles and *5* _"arrow digits"_ that can be either:
1 + 2 + 3 + 4 + 5 = *15* (option 1)
1 + 2 + 3 + 4 + 6 = *16* (option 2)
If *r4c4 = 2* then option *2* would be ruled out (it would force the corresponding two circles to be 9 + 9 = 18) and
the corresponding two circles would add up to
*r4c4 + 15 = 2 + 15 = 17*
Hence, they would contain *8 + 9 = 17.*
and the short line segment in *box 2* would contain the remaining two digits: *6* and *7.*
However, *r3c5 = either 6 or 7* is impossible as it would force either *6* or *7* on one of the line segments in *box 5,* which would push the correspondind arrow total to at least
*6 + 3 + 1 = 10*
*STEP 2*
It can be easily shown that, in *box 8,*
*r7c4 = 2*
*r7c6 = 1*
By sudoku, this is their minimum value, and any value higher that that would force the corresponding arrow total to be at least
*3 + 3 + 4 =10*
*STEP 3*
Let's now focus on the *4-cell* line with endpoints in *boxes 4* and *8.*
As shown above, its endpoint in *box 8* must be
*r7c4 = 2*
By sudoku, its other two _"arrow digits"_ must add up to at least
*3 + 3 = 6.*
Hence, the total value of its "arrow digits" is either:
*r7c4 + 3 + 3 = 8*
or
*r7c4 + 3 + 4 = 9*
and the possible candidates for its endpoint in *box 2* are:
*r4c3 = {3, 4, 8, 9}.*
However, *r4c3 = 8* can be easily ruled out because, by sudoku, the other cells would add up to at least:
*3 + 4 + 2 = 9*
Moreover, if *r4c3 = 9* then, by sudoku:
🔹the digits *3* and *4* would be locked in the upper two cells of *column 6*
🔹the minimum possible values for the other arrow in *box 5* (with endpoints in boxes 6 and 8) would be
*3 + 5 + 1 = 9*
(with r4c7 = 3)
🔹the lower two cells of *column 6* would contain digits higher than *5:*
*r8c6 = r9c6 = {6, 7, 8}*
As a consequence, by sudoku, the *3-cell* arrow with endpoints in boxes *8* and *9* would add up to at least:
*4 + 6 = 10,*
which is impossible.
Hence,
*r4c3 = {3, 4}.*
QED 😏👍
STEP 3 was quite hard❗
I would love to know whether you were able to find some elegant shortcut to reach the same conclusion.
@@Paolo_De_Leva
I actually did this a bit different. I noticed that 67 is in C5 in box 5 and with some simple deductions I noticed that R3C6 has to be either a 6 or a 7
If it was a 7, then R3C5 has to be either 5 or 6, 6 doesn't work so R4C6 is a 2
If it was a 6, then R3C5 has to be 7 or 8, 8 doesn't work so R4C6 is a 2
So R4C6 is always a 2
This resolved the 12 pairs in boxes 5 and 8
Also R3C5 being a 5 or an 8 makes a 5678 quad with the digits in box 5 C5 -> so the digits in C5 box 8 are a 349 triple.
The triple in box 8 helps. The min value of R7C7 is 3, the min value in R8C6/R9C6 is a 5. So the sum is 8.
You get a 58 pair in R8C6/R9C6 and a 3 in R7C7 and R4C3
The 58 pair stops the sum to be in R4C7, so you get a 4 in R4C7
I loved the way the initial head-scratching and break-in finally gave me... just a 12 pair. And yet that 12 pair becomes extraordinarily powerful and does so much work. I did struggle a lot in the middle and probably didn't spot some things I was supposed to, so I'll be interested to see what Simon picks up.
Always love watching you solve puzzles.
21:40 Once you conclude that the circles for the two long lines in box 2 must be in box 2, then the highest these circles can be is 8 and 9, which would mean that the sum of the two long lines is only 34. If the circle for the short line is outside the box and is 9, then the sum of the two cells on the short line in box 2 would be 9, and the sum of the whole box would, therefore, be less than 34+9, which is only 43 (less than 43 because there's still one cell outside the box on one of the long lines).
I'm so proud of figuring this one out! It was an immense struggle over several hours, grappling with what seemed like such ambiguous information. I still don't think I fully "got" it, but managed to work it out in the end.
"I nearly had a good thought" has to be T-shirt fodder
I might just get that tattooed
Done in 31:58, turned out easier than I thought but it didn't affect the impression - very nice ruleset.
47:16 ... I greatly enjoyed solving this exquisite sudoku!
Nice puzzle!
The incredible run of astonishing sudokus continues...
👉 *Jay Dyer* = Cosmic class constructor
👉 Only one mistake by *Simon* (at about 31:00) = Cosmic class solver
This puzzle was so hard that by mistake I hit dead ends several times before figuring out the correct logic path for the break-in (which I described in a separate comment).
I started this puzzle by concluding that every line that was 4 cells long needed at least either a 1 or 2, or both. I colored in those cells and that led me to get through the first few steps of the break in.
Sorry I don't have time to watch this right away as normal. I am waiting for Zetamath to start one of his unmissable reactions to you solving his puzzle, the one you featured yesterday. Five minutes to go. Popcorn ready.
The toughest CTC solve for me for some time. I had to look at it for several days, until this morning I immediately spotted something not that difficult but which had previously eluded me (in box 7). What a clever puzzle, as one would expect from Jay Dyer.
I finished in 124:49 minutes. This was an excellent puzzle as usual from Jay Dyer. She has this uncanny ability to create these spectacular geometrical showcases. It took me too long to even notice that box 2 was the obvious place to start. That limited r4c4&6, which finally gave me the in that I was looking for. Calculating box 5 was so fun to do. I was surprised when I saw that no matter the orientation, column 5 of box 8 always had to contain the same digits. This created a feedback loop that in turn affected boxes 5 and 2. That was so awesome. I think if it weren't for Simon telling a secret long ago that three celled arrows with different digits have to contain at least two digits from 123, then this puzzle would have taken me significantly longer, if that is possible. Lovely job as usual from Jay. Great Puzzle!
31:00 couldn't we put 9 in box 2 with 432 on the rest of the line and then 3 in box 6 with 951 on the line?
Yes! That's where I'm stuck at right now...
Definitely a mistake but I'm following it through and eventually the 3-long line between boxes 8 and 9 disproves it - 1/2/3/4 can't be on the box 8 segment (regardless of the way you set up those box 5 lines), and the box 9 part of the line is forced to 3/4, so the pair 6/8 can't be on the box 8 part, which it would have to be with the 951 line above.
@@jacklinton48859 on the right in box 6 fixes that
@@jacklinton4885don't see why it can't be 3 or 4 on the line in box 8. It can't be both, but if R4C3 is a 9 then only 2, 4 and 9 are excluded from the line in box 8.
@@calebu2 So first, from where the mistake is made, add in the 1359 option as well as the 3489 pattern that Simon gets. Then look at 1s and 2s in box 2 - given where the 1/2 pairs are, they are forced to be in column 5 in box 2. Now the 3-length line in box 2 we've already determined has a 7/8/9 on it, so we can't have the 1/2 in the r3c5 cell (as 1 or 2 + 1 or 2 = maximum 4). By the same logic, we can't have the 3/4 in that cell (max 6). Also, in box 5, we either use both the 3/4 on the lefthand line, or we use one on each line, so we can't have 3/4 in column 5 in box 5. Therefore, the 3/4 in column 5 must be in box 8 - so the part of the 3-length line in box 8 can't have 3/4. Quite a long logic chain, would have loved to see Simon work through it!
@15:44: "I nearly had a really good thought, then."
Wow! Great puzzle; loved it. Did not feel confident in my solve while I was going through it, but the time was quite good in the end ---- 44:15. Happy with that.
I started Nurikabe World - it’s great. Even though my 6-year old decided it was for “us.”
R7C2 was really the bane of Simon in this puzzle. He basically missed it everytime he looked at it. Never change :)
Simon may miss silly things occasionally but he seems to have a magical ability to divine the starting logic and then to push through to the end no matter how tough the logic along the way!
A brilliant looking puzzle, can't wait to watch Simon battle Jay Dyer.
When Simon started to think about 1*179 as an option for it, I was like no don't waste time on that, it would be a 1*56 for best, leaving 8 with 234 being 1 too many.
Ambiguous Arrows, Unambiguous Rules, Superb Puzzle
I love short rulesets
38:33 finish. A fun puzzle, loved the logic!
The way I figured out box 2 plus its sticky-oiy bits is to consider that two lines would have 9 and one would have 8. That would nake the three lines sum to 52, leaving only 7 for the sticky-out bits, making 6 the largest digit that could go on them (in all cases). With that proven, the largest numbers that can be on the kines are 9, 8 and and therefore they sum to 48, leaving 3 outside the box to be 1 and 2.
Thank you Simon, that was too tricky for me but Wanted to know what I missed and your pathway was beautiful.
Great puzzle! The 349 logic around box 5 resolves later, as many have pointed out. Pretty hard cruxes, mixed with nice-flowing sequences. I liked the 123 triples. :-)
Simon missed something. You forgot about the 9-9 combo for the two lines in box 5. This allows a line to have a 5. It breaks for other reasons later on, but you saved yourself a lot of headache. However, I know that you don't want to save yourself headaches :D
Haven't watched the solve yet, but this one took over 2 hours for me to complete.
Still don't know whether I was dealing with an arrow puzzle, a parity puzzle or a simply a maths problem...
Only 18 solutions in 600 days. 1259 solutions in 602 days. What a push in solutions you made by posting your solving! WOW.
And mine said 1338 solved in 603 days!
I miss rat Fridays, but this ruleset looks interesting 😋
I could see a variation on this where you got to place the arrow circles but they also have the "counting circles" rule.
Awesome puzzle, always fun to wrap yr head around new concepts. Finished in just over 1 hour.
Now I wonder what's an intended solution for this puzzle. Cause I didn't find it, I looked at same logic as Simon, just saw that there can be two 9s there and didn't know what to do next.
Of course, if you try two 9s it fails eventualy, but you need to do a lot of work for it to fail, so I don't think this is intended path.
I guess it has something to do with column 6, but I'm not sure
I like the short ruleset and looking forward to trying the puzzle for myself (before watching Simon's entire video).
I think I'll need to think of each line as a single arrow with an unknown position for the circle, if that makes sense.
I'm not sure how, but managed to do it. Maybe I'm starting to learn something. :D I did had to use notepad to keep track of the possibilities though. My brain just doesn't relate to numbers. Now to watch the video. :) And see where my logic was sound and where it was "just luck"... LOL
Solved in 51:08.
How I worked out the 12 pair in box 5 was that if one of them was the arrow circle then we have 7 digits on the arrows in box 2, and therefore need 3 digits summing to at least 28+1 on the arrow circles.
46:23 just look down 😂😂😂
23:32 for me. Very nice puzzle!
What an extraordinary and amazing puzzle.
49:12 for me. I needed a small bit of reassurance and correction from the video.
Wow. What a pizzle. Got there eventually. 91 minutes.
How do people even start to construct any of these Sudoku's that you do?
Do they start with a solved puzzle then add clues till it makes logical sense?
The channel has a list of brillant videos from setters explaining how they work, including a master class by Clover.
My opening logic was that on box 2 you have at least 5 consecutive numbers. You have to include both 1/2 or you get 3+4+5+6+x>18 and that is impossible for 2 numbers.
If, like me, you were confused, in this video, the low digits for Simons are 1, 2, and 3. There is no 4.
4 would be a middly(TM) digit!
1:16:11 - I screwed up some pencil mark somewhere and had to back out a bunch.
This one looked diabolical. I gave up real quick.
I just got the notification even tho this video is from 4 hours ago
I'd rather that you did her nabner doubler. Anyone solved that?
This is such a weird way to write the rules? Isn't it just that 1 digit is the sum of every other digit on its line, because its own presence will always repeat. With the "digits may repeat if allowed by Sudoku".
Don’t even know how to start this one 😱
I don't get it. Why does a four-cell line with no repeat have to contain two low digits ???? Couldn't it be (for instance) 3456 ?
One of the digits must be the sum of the other digits.
Or, to put it in the way the rule is worded, the sum of those digits (3+4+5+6) is 18. The line must contain a digit equal to half the sum, so would need to contain a 9.
Usually watch these religiously, only Simon though as I find mark doesn’t explain as well, that’s fine I wouldn’t have a hope of solving these so need the extra explanation but tbh this one left me cold, seemed too hard for Simon too which is unusual and I couldn’t follow his logic which from comments seems to have been a little flawed anyway
Getting cranky about column 6.
She? I always imagined Jay Dyer was Danny's lesser known uncle.
Guess she's his aunt, then
@@PersonWhoExists50306 Could be :)
not addressing the fact that every line has to sum to an EVEN number (as you cant put something like 5.5 in a cell if the sum is odd) and the cells it is made from to a maximum of 18, being half of 9, the maximum number you can put in a cell. 🧀🐁
Read the rules: not going to attempt this one
125 minutes
Twitter? Leave x 😮
Nope X is the last good social platform, unless you're pro-censorship or like handing out on tiny obscure sites.
A thing of utter beauty
Something that should be clarified: there are two lines at the bottom left that cross over each other.
They wouldn't have been drawn this way if, eg, the line starting in r9c1 had either stayed in box 7 or gone straight to r6c1 :)
True, though my comment was meant more as a clarification that it's not one big branching line.
@@TemujinReads You'd have to write at least a 13 in one of the cells for that to function as one large line, I'm not sure that clarification is necessary.