Amazing New Rules = Amazing New Puzzle
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- čas přidán 26. 06. 2024
- ** TODAY'S PUZZLE **
A new constructor, Jolly Rogers, has created a really remarkable ruleset based around introducing more squares to the sudoku grid (there clearly aren't enough already!) The upshot is a sudoku that has several distinct parts to the solve - all of which are really interesting!
Play the puzzle at the link below:
sudokupad.app/7FngrT9ftq
Rules:
Normal sudoku rules apply. Divide the grid into non-overlapping square-shaped regions. The sum of a square's four corner cells must be divisible by the number of cells that square contains. E.g the sum of the corners for a 3x3 square must be divisible by 9. Every cell must belong to a square and a 1x1 square region is not valid. The cell containing a red circle cannot be a corner of any square region. Orthogonally adjacent cells separated by a square region border must have the same parity (odd or even). Square region borders divide each blue line into segments with the same sum. Each line must be divided at least once.
** NEW SUDOKU HUNT ON PATREON **
We're delighted to share with you a brand new sudoku hunt themed around snake egg puzzles by Glum Hippo. If you've never tried snake egg logic puzzles before, prepare for serious fun - these sudokus are outstanding! Finish by May 20 to enter the competition!
Other treats on Patreon include:
- the Sumgeons & Diagrams sudoku by sunnyjum;
- Simon's latest forays into the world of Islands Of Insight;
- Mark's video looking at the new OneUp puzzle from Rodolfo Kurchan;
- his solve of Region Geometry by Emre Kolotoğlu (3hr 36min long...!);
- and Mark's latest solve of The Times Club Monthly cryptic crossword
/ crackingthecryptic
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▶ Contents Of This Video ◀
0:00 Theme music & puzzle intro & competition
1:12 Snake Egg Sudoku Competition
2:05 Happy Birthdays
3:14 Rules
7:10 Start of Solve: Let's Get Cracking
▶ Contact Us ◀
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We also post the Wordle In A Minute videos on TikTok. - Zábava
33 minutes in and "I've not even thought of numbers yet" these are really enjoyable puzzles to watch you solve 😊
These types of puzzles are the best ones!
Sudoku by itself is very boring.
Sometimes I read a rule set and I think "this puzzle is probably genius, and I refuse to even approach it." I'm glad we have Simon and Mark to brilliantly walk us through them.
Agree!!
This puzzle had it all; construction, coloring, parity, line-rules, arithmatic, normal sudoku
It was a joy to watch
Can't believe Simon didn't notice that, with all the boxes, there were *6* 3's deserving of a song!
I love how he started with the red circle and ended with the red circle
There's also the beautiful pattern of the parity colors to notice, they are symmetrical along the downgoing diagonal and there's a 4x4 of odds, which is fitting for the theme.
Ironic how in a puzzle all about squares, that little red circle did so much. Never let 'em tell you you need to conform to be worthy. ❤⭕
😸💚
I used that circle a lot less. It is needed, but not nearly as much as Simon used it. I had my focus on the R9C1 region as the logic of it spread across the bottom. Nice to see that the title wasn't the only key way to break in.
I like how this puzzle shifts gears from one part of the rules to the next
"I don't know if this is how it's meant to be done, I'm having fun" - well, I guess than it IS meant to bi done that way :D having fun is one of the main reasons we do puzzles like this, isn't it?
_"The upshot is a sudoku that has several distinct parts to the solve - all of which are really interesting!"_ (Video description)
True, and fun to figure out as well. 😏👍👍👍👍👍👍
Maverick always knows.
Of course! Maverick may be Phistomefel...
I believe Samuel turned 37! Congratulations!
Before he said the multiple of 7 last year part I was thinking damn, this guy is either 17, 37, or 101 😂
No, clearly he's 2 years old. What an impressive toddler, to know so much about numbers already!
I spent at least a minute trying to figure out how this could be possible with the "multiple of 7" before realizing he was talking about the 35 and not the 36 🤦♀️
Thank you for spelling it out! I was thinking "well, a square multiple of seven has to be 49 but 50 isn't prime... Unless they mean _14 squared_ since 196 fits and 197 is prime, but then they're nearly two centuries old... *_What?!"_* makes much more sense that "the year before" means two years ago not last year lol
401, Methusela?
The Red circle was the first clue to get you going. It was also the last digit he filled in. That's pretty
At one point early in the video I thought "oh so that must be a 4x4" and skipped ahead to see if I was right, saw all the orange and blue with 1x1 regions and not-square regions coloured in and thought Simon had completely lost his mind 🤪
41:05 for me, albeit with a logical error, overlooking a possibility that turned out not to be in the correct solution. Really enjoyed it regardless!
An easier(?) way to see what's going on around the red circle:
The square that contains r9c9 cannot be 3x3 because of the circle. It's either 2x2, or it's 4x4 or larger. Can it be 2x2? Well, consider the square above it, which would have its lower right corner at r7c9. That square:
- can't be 4x4 or larger or the blue line's sum in that square would be way too big for r8c9.
- can't be 3x3 because it would have the red circle in its corner.
- can't be 2x2 because whatever square is directly above *it* would then have four different digits on the same blue line that has a single digit in r8c9.
Therefore, the bottom-right square can't be 2x2 because there could be no possible valid square directly above it, so it must go past the red circle.
oh wow, I didn't even consider that as a possibility. I guess I got unlucky. I just centered a 3x3 square on the red circle and expanded it into r9c9 to prevent 1x1 squares and didn't think that the circle could be on the edge of a square.
That's neat logic
I don't think a 5x5 square would ever be possible in a 9x9 grid, since it keeps you from making anything other than 2x2 or 4x4 squares in the rest of the grid, and you could never produce a full tiling. And I'm pretty sure having only 3x3 squares along at least two edges is similarly forced.
you can get away with only 1 edge of 3x3s (same structure as puzzle, but shift 3 2x2s to the edge and replace them with the 3x3s). It's part of why that straight line near the top is important
Region sum lines actually being region sum lines for 1h3m27s.
There it is! I've been refreshing for ever - worth it! :D Take care Simon ❤
I loved the nested nature of this puzzle. It was such a joy watching you solve it.
I love how simon's first digit was totally obscure, while I just plopped a 9 in the seemingly most obvious spot after getting the coloring
Jolly Rogers - well done! Finding the squares was fun and straight forward. I fluffed up when I tried to make all the digits in the 4x4 divisible by 16 instead of just the corners. Great Puzzle!
Well, fluff you! .....hmm, doesn't quite work, does it?
Two greens in the bottom row!!
thanks, now I can't unsee it and therefore won't sleep tonight x(
@@MaierFlorian Sorry!
After discovering this channel and having an absolute blast following along as the puzzles get solved, I have tried myself to do a few puzzles from logic-masters on the sudokupad. It has been great fun, although I stick to the 1-star difficulty, beacuse while I mostly have no issues following Simon and Mark's reasonings (thank you for the constant great explanations), I don't have nearly that kind of capacity to figure out which question to ask next. And also, I don't often have the luxury of sitting down for very long at a time, so I need to be able to either finish relatively quickly, or be able to pick a puzzle back up after not looking at it for a while.
With that in mind, I personally thought it might be cool to see them breeze through 3-4 easy puzzles in a video, to see how fast they work on puzzles where I can actually compare myself to them, relatively fairly. Just a suggestion for a one-off thing or something.
This is a wonderful puzzle, and a wonderful solve!
What a puzzle. It seems we have 3 puzzles in one! 😆
Absolutely brilliant setting! Thanks for a wonderful gem of a puzzle!
Also it is nice to see that you don't have to "cheat" and look at the title for the break in. Simon used the circle for the main break, but I used the region in box 7's corner. (The red circle is used for some logic, but isn't my main focus.)
Spoiler warning: Any way you try to do it as you work your way across the bottom it rules out one possibility after another until you are left with it being a 3x3 with a 2x2 next to it and a 4x4 in the far right (which turns out to be the case) or you are left with a 4x4 next to it and a 2x2 in the far right. Either way around you are left with a blocked off rectangular region and from there you can work up the left side filling in 3x3s and the top filling in 3x3s with the 2x2s surrounding the 4x4 region forced.
I found it interesting that you can get all the regions colored before even looking at parity and you can deduce all the parity except for six cells in columns 1 and 2 between boxes 4 and 7 (4 in box 4 and 2 in box 7) along with the eight cells in rows 1 and 2 on and above the blue line there, but you also can know that of those 8 cells, you either end up with two evens in the center of the blue line, or two odds with the opposite parity on the ends forcing either two evens between the borders in row 1 or two odds with the other parity again on the outside two cells, and due to the other parities in the column, you know that they can't both have the same parities across the border, i.e. if row 1 has the evens in the middle, then row 2 has to have the odds in the middle and vice versa.
Even more surprising is that it still doesn't just fall apart into simple sudoku. That is real expertise in setting as a lot of setting can accidentally leave logic that just opens up the whole puzzle instantly, especially with somewhat more complex rulesets such as these. Also, I loved how these rules didn't feel contrived or just set for the simplicity of preventing the need for bifurcation. They do that, but they flow nicely in that you don't really struggle to see where to go next (at least not too badly). You almost always immediately see where you need to look next, whether it be a corner sum (including which corner sum to look at), a pair/triple, or a single digit forcing limitations on other cells.
Below is my break-in including coloring all the regions for anyone interested (note that I use different colors than Simon):
Each corner cell must be in a square region, and because the four corners of a square region must sum to a number divisible
by the number of cells in that region, and a 1 cell region isn't allowed, we get 2x2 regions summing to a number divisible
by 4, 3x3 summing to a number divisible by 9, 4x4 cells summing to a number divisible by 16, and 5x5 summing to a number
divisible by 25, but there can't be a 6x6 region as that would have to sum to a number divisible by 36, but even with the
allowable repeats on the two opposite sets of corners we get only 9+9+8+8=34. Each blue line must be divided at least once
by one or more square regions so the R1C1 square region can only be of size 2x2 or 3x3 as 4x4 would consume the entire line
going from R4C1 to R4C4 to R1C4. The R9C1 region can likewise only be size 2x2 or 3x3 due to the blue line poking out of that
box. The R1C9 region can be 2x2, 3x3, or 4x4, but not 5x5 as that would consume the blue line poking out of box 3. Finally,
the R9C9 region can be any of the sizes, 2x2, 3x3, 4x4, or 5x5, but the rules state that the red circle can't be the corner
of a region, so 3x3 is eliminated. So is 5x5 because that would divide the box 5 zig-zag blue line such that R6C4 and R4C5
are single cells, but both being in the same box can't be the same digit. That means that the R9C9 cell region is either 2x2
or 4x4. Now the R1C9 can't be 4x4 because that would either leave a 1 cell area between the lower 4x4 R9C9 region, or it
would leave a 1 cell region between the 2x2 R5C9 and the 2x2 R9C9 regions left over as R5C9 can't be 3x3 due to the red
circle. That leaves R1C9 either a 2x2 or 3xe3 region.
If the R9C1 region is 2x2, then R9C3 is in another region. It can't be 2x2 because That would leave either another 2x2 beside
it and a 1 cell region between that and the R9C9 region, or it would have a 3x3 putting the red circle in the corner, which
is forbidden. It can't be a 3x3 because that splits the blue line in box 7 into three single cells with two of them in the
same box having the same digit, which is forbidden. It can't be a 4x4 because that would leave a single cell between it and
the R9C9 2x2 region. It can't be 5x5 because that would require a 2x2 region in R9C9, another 2x2 above it for R7C9, and
another 2x2 above it for R5C9. That creates a problem for box 2's R1C9 region as if it is 2x2, then there is a 1 cell region
between those, and if it is 3x3, then there is a single cell isolated in R4C7. That means that R9C1 can't be 2x2 and must be
a 3x3 region. Now the R9C4 region is either 2x2 or 4x4. It can't be 3x3 because that would leave a 1 cell region between it
and the R9C9 region. Now, both the R9C4 and R9C9 regions can't be 2x2 as that would make R9C6 a 2x2 region, and now R7C9
needs a region, but a 2x2 would put the red circle in the corner of a region, 3x3 would do the same, but 4x4 for that region
means that R8C9 as a single cell is the same sum as the four digits above it which would already sum to 10, but also the other
two digits on that line, which is totally impossible. That means that either the R9C4 is 2xw2 and R9C9 is 4x4, or R9C4 is 4x4
and R9C9 is 2x2. This makes either R7C4 a 2x2 region or R7C9 a 2x2 region completing a rectangle between the three regions
in R6C4 to R9C9. Either way around, R7C1 is a region, but it can't be 2x2 as that leaves a 1 cell region in R7C3, so it is
a 3x3 region. This sorts the R1C1 region to be 3x3 for a similar reason.
Now R1C4 can't be a 2x2 region as that would make R1C6 a 2x2 to avoid a one cell region between the R1C4 region and the R1C9
3x3 region, and this makes the R1C9 a 2x2 region as well, but that would make R2C5 the same sum as R2C8 putting two of the
same digits in the row. The R1C4 region can't be 4x4 either as that would leave a once cell gap in row 5. That makes R1C4
a 3x3 region along with R1C9 also a 3x3 region to avoid the 1 cell gap in column 7. This forces 2x2 regions for the three
remaining areas in rows 4 and 5. Now that there is a region containing four cells in the same box on the blue line in column
9, the R9C9 region can't be 2x2 as that would imply R8C9 is the sum of four cells which at minimum add to 10, impossible.
That makes R9C9 a 4x4 region meaning that R9C4 is 2x2 and R7C4 is another 2x2 region.
To summarize, box 1 is a region, box 2 is a region, box 3 is a region, box 4 is a region, box 7 is a region. R9C9 is a 4x4
region, and the rest form 2x2 regions from R4C4 to R5C9 and R4C4 down to R9C5. I have made box 1 red, box 2 gray, box 3 green,
box 4 green, box 7 purple, the 4x4 R9C9 region gray, R4C4 to R5C5 purple, R4C6 to R5C7 blue, R4C8 to R5C9 red, R6C4 to R7C5
red, and R8C4 to R9C5 blue. This gives 11 regions total with five 3x3, five 2x2, and one 4x4.
Continuing from the region section, here is the logic I used for the parity bits (I don't know if it matches Simon's logic completely as I've yet to finish the video.):
Now let's look at the parity (even/odd) rule. It says that if two cells are adjacent orthogonally across a region boundary
then those two cells share the same parity. Looking at R5C5 we get a 2x2 region of four cells that share the same parity
from R5C5 to R6C6. Now if that parity is even, then we get a problem as all the other cells in box 5 would be odd, but that
would put 6 odd digits into rows 3 and 4 or columns 3 and 4. (The odds in box 5 make their adjacent cells in row 3 and column
3 odd, but the corner cells in box 2 and box 4 make the adjacent cells in boxes 1, 3, and 7 odd, which in turn make the
adjacent cells in R7C4 and R4C7 odd, giving odds from R3C3 to R4C7 and R3C3 to R7C4.) That is too many odds, therefore
R5C5, R5C6, R6C5 and R6C6 are all odd. That makes the shared sum on that line odd, so to get a two cell sum that is odd
using an odd digit (which we have in R5C5 and R6C5) the other cell has to be even making R4C5 and R6C4 both even.
Next, the four cells from R7C5 to R8C6 have to have the same parity, but they can't be even because that would make the other
cells in box 8 all odd, which would make the adjacent cells in box 7 odd, but R7C3 being odd makes R6C3 odd, but we know it
is even because of the even in R6C4. That makes R7C5, R7C6, R8C5, and R8C6 all odd. Two odds make an even total, so to get an
odd total for the remaining sum on that blue line we need R6C7 to also be odd. This makes R5C7 odd as well. As we already
stated, R6C4 being even makes R6C3 even, which makes R7C3 even, which makes R7C4 even. This continues into R8C4 being even
and R8C3 becoming even, too.
R4C5 being even makes R3C5 even. Also, R3C3, R3C4, R4C3, and R4C4 form a 2x2 same parity region, but due to the three evens
below them in the column already, they can't also be even, so they become odd. This completes the odds in box 5 making the
other two cells and their adjacent cells both even. This completes the evens in columns 3 and 4 making the other cells there
all odd. This completes the odds in box 8 making the other two cells even. The even cell in R6C3 and the even cell in R7C3
make R8C2 even to maintain the even sum on that side of the line. Next, the even in R3C6 makes R3C7 even and that makes
R4C7 even. This continues into R4C8 and R3C8, both becoming even. That completes the evens in rows 3 and 4 making the other
cells all odd. That completes the odds in box 1 making the other cells all even. The odd in R5C7 makes R5C8 odd which makes
R6C8 odd as well. That is five odds in box 6 making the last two cells in column 9 there both even. The remaining cells in
column 9 on the blue line either have to both be odd or both be even and the two blue line cells in row 2 box 2 have to be
one of each, even and odd because two evens give the four evens in the row, but the border would force a fifth even, and two
odds would give a fifth odd in the row, but two odds make an even sum and an odd plus an even is odd, so there is no
possibility for R2C8. That does mean that the evens are used up in the row on the line in some order making R2C9 an odd.
The 4x4 region's four corners must be divisible by 16. It has an odd cell and two even cells making an odd sum so far, but
you can't evenly divide an odd sum by an even number evenly, so R9C9 becomes odd. Looking at the blue line cells left
unfilled, we know the row 2 line has two evens and two odds where the ones crossing the border are the same and the outside
two are the other type. We know that the two cells above in box 2 must be in the opposite arrangement due to column 6. This
also means that there is exactly one even and one odd in the two cells in column 7 in box 3. This puts two odds and an even
into column 7 box 9. That gives three odds already into box 9. We also know that to maintain the even sum for the three cells
of the column 9 blue line with the even already in R6C9 that it can't be both odd as that would finish the odds in the box
making the three cells of column 8 in the box all even putting 5 evens into column 8. That makes R7C9 and R8C9 both even.
Now column 9 has four evens making R1C9 odd. That now forces the remaining cells in box 3 to be one of each in the two
columns, and the border forces them to be one of each in the two rows in an X type pattern as putting two of the same
type in box 3 row 1 would force a third across the border, but there are exactly two of each left to place in the row.
That means that for both columns 7 and 8 now we have one even and one odd giving three evens and three odds in both
columns. That leaves two evens to place, one in each column in box 9 with the other four cells being odd.
Row 8 already has the four evens, so all the rest are odd (R8C1, R8C7, and R8C8). Now, if we put one of the evens into
row 7 that puts the other even into box 7 in that row where it finishes the evens in that box making R9C1 and R9C2 both
odd, but that forces two evens into box 9 in row 9 to finish the evens for the row where we now have five evens in the
box. We also can't put both evens into row 7 box 9 as that would put five evens in the row, so that means there those
two cells (R7C7 and R7C8) are both odd and the remaining two cells in box 9 (R9C7 and R9C8) are even. This finishes the
evens in row 9 making R9C1 and R9C2 odd. That is as far as the parity can go at this point as the remaining cell are all
in 3x3 regions giving no help with the corner cells and the only remaining blue line has two cells of each parity and
while we know that the two middle cells are the same and the two outer cells are of the other type, that is all we know.
For the rest, from here it just uses the lines, the corner sums, and such:
Now we can look at R4C4 which sees three cells on the line and is a single cell sum making it at least 6, but being odd
makes it 7 or 9. Likewise, R5C6 is a single cell sum in a box with 5 cells on the line in three regions making those
five cells add up to at least 15 and 15 divided by 3 gives a minimum value of 5, but it is odd as well, so R5C6 is from
579. Now R4C4 can't be 7 as it would require the other cells on the line to be from 124, but they are all odd, so it is
9 with the other cells on the line from 135. This also takes the 9 out of R5C6 and places a 7 in both R4C9 and R9C4.
Next, we know that the red 2x2 region containing the blue line in column 9 must sum to a total divisible by 4, which
means that the three even cells in the gray region column 9 on the blue line must also be divisible by 4. Now 246 works
as that sums to 12, while 248 does not being 14, 268 does being 16, and 468 does not being 18. That makes those three
cells either from 246 or 268, but more importantly, that makes the remaining digit in R5C9 from 4 or 8. Now it can't be 8
as that would make 8+7+1 already 16 with nothing to fill into the other cell to sum to 12 (let alone 16). That means
that R5C9 is 4. This gives a 268 triple in the even cells below it. It also means that the even and odd digits in the
red 2x2 region in box 6 must sum to 5 (7+4=11 leaving 5 to reach 16) making them either 41 or 23, but we can't duplicate
the 4 in the box, so it is 2 in R4C8 and 3 in R5C8 leaving a 68 pair in the remaining two even cells in box 6 and a 159
triple in the odd cells there. Now R6C7 can't be 9 as that is already larger than the single cell sum on that blue line
not even adding the other two cells in box 8. It can't be 5 either as it would require two 1s on the line in both in box 8
to make 7. That means R6C7 is 1 surrounded by the 59 pair in R5C7 and R6C8.
Continuing with this line, if R5C6 was 5, then it couldn't be 23 in the purple region as 3 is eliminated from both cells
so it would have to be 14 with the 1 going into the odd cell R5C5 and the 4 going into the even cell R4C5. Now if R5C6
was 7, then it could be 16 with the 6 in the even cell and 1 again in the odd cell, it can't be 25 prevented by the 2 in
row 4, and it can't be 34 again because of the 3s being eliminated in both cells. That makes R5C5 a 1 with R4C5 from 46.
The red region in that box R6C4 and R6C5 can't be 14 if R5C6 is the 5, but it can be 23 with the 2 in the even cell R6C4
and the 3 in the odd cell R6C5. For the case where R5C6 is 7, the 16 option is eliminated, but the 25 possibility is
possible with the 2 again in R6C4 and the 5 going into R6C5. Here, the 34 possibility also works with the 3 in R6C5 and
the 4 in R6C4. That makes R6C4 from 24 and R6C5 from 35. To finish this line, with the 1 in R6C7, we need R7C6 and R8C6
to be either 13 to make 5 or 15 to make 7 in some order, so those two cells are from 135. Also R6C6 is from 357.
The blue line in box 7 at minimum adds to 6 in the two cells there, so R6C3 is either 6 or 8 forming a 68 pair in the
row with R6C9. Now the only region that is not 2x2 where we know most of the corner cells is the green region in box 4.
The remaining cell R6C1 is either from 24 if even, or 79 if odd. Now if we have 1+3+6 in the other cells, then we need
8 for the sum to be divisible by 9, but we can't have 8, so that isn't possible. If we have 1+3+8, then we need 6, which
again we can't have in R6C1. If it is 1+5+6, we would need 6 again, and if it is 1+5+8, then we need 4, which is finally
a possibility we can do. If it is 3+5+6, then we would need 4 again, and if it is 3+5+8, then we would need 2. That makes
R6C1 the last even in the row from 24 putting the last odd in R6C2 from 79. In the three possibilities that work, we have
the corners as 1+5+8+4, 3+5+6+4, or 3+5+8+2. In all three there is a 5 in one of the corners, meaning that R4C2 can't be
a 5, a small deduction, but still a deduction.
Now if R5C6 is 5, we end up with 2 and 3 in the red region in box 5 and we end up with a 6 in R5C4 making the evens below
form a 48 pair. The 5 sum also makes R7C6 and R8C6 a 13 pair making the odds beside those from 59 as there is already a
7 in the box. Now 2+3+4+5=14, 2+3+4+9=18, 2+3+8+5=18, and 2+3+8+9=22, none of which are divisible by 4 which the red
region 2x2 must be. That means that R5C6 can't be 5 and in fact must be 7. This makes the two odds on the line in box 8
from 15 as 1+5+1=7, making R6C6 a 3. This sorts the line in the red region in box 5 to be 5 in R6C5 and 2 in R6C4. A 6 goes
into R4C5 making R4C7 an 8. This puts a 6 in R6C9, the 5 we placed a second ago makes sorts the 59 pair in box 6. We have a
28 pair left on the line in column 9, a 4 goes into R4C6 and an 8 into R5C4. Row 6 is now finished off with a 4, 7, and 8.
Now to get to a sum divisible by 9 in the green region for box 4 we have 12 and need 6 more meaning R4C1 and R4C3 are from 15
making R4C2 the 3. We get a 26 pair on the line in box 7 to sum to 8. We get a 46 pair to finish the evens in column 4 and
we get a 39 pair to finish the odds in box 8. The evens left in box 8 form a 28 pair making the evens remaining in box 9
a 46 pair. R8C3 is now a 4 sorting the 46 pair sorting the 26 pair sorting the 28 pair in rows 7 and 8. The 2+5+4=11 in the
2x2 red region in boxes 5 and 8 requires the 9 to be divisible by 4, placing the 3 below it in R8C5. Now 3+6+7=16 requiring
the 8 to be divisible by 4 in the 2x2 blue region in box 8 making R9C6 the 2. We now have 3+6+2=11 for three of the four
corner cells in the 4x4 gray region. That means that R9C9 must be 5 to be divisible by 16. The 3 in box 9 is now forced into
the red circle R7C7, the 9 forced into R8C7 leaving a 17 pair in R7C8 and R8C8. A 139 triple finishes column 9 in box 3.
A 247 triple finishes column 5 in box 2 with the 7 not able to be in R3C5. Likewise, a 689 triple finishes column 6 in box 2
with the 9 not able to be in R3C6.
Now the remaining odds in box 3 are either 5 or 7 and the line needs one odd, so it is either 7 in R2C7 or 5 in R2C8 due to
the 5 in column 7 box 6 and the 17 pair in box 9. It also needs an even with R2C7 not able to be 8 and R2C8 not able to be 2.
This gives R2C7 from 2467, and R2C8 from 4568. Now if the center digits are odd we get 9+2=11 or 9+4=13 on the left and
7+4=11 or 7+6=13 on the right as 7+8=15 can't match the left side sums. However, if the odds are on the outside cells on that
blue line, we get 7+6=13 or 7+8=15 on the left and 5+6=11 at most on the right, which means that the odds can't be on the
outside cells on the line. That makes R2C6 and R2C7 odd being 9 and 7 respectively. It also makes R2C5 and R2C8 even with
R2C5 from 24 and R2C8 from 46 to make the sums work on that line. The 9 also comes out of R2C9 and R1C6. Now the 3x3 gray
region in box 2 has 6+8=14 in two corner cells meaning the other two must add to 4 to be divisible by 9 making R1C4 and R3C4
from 13 putting a 5 into R2C4.
Back to simpler logic, R5C3 is a 2 leaving a 69 pair to finish the row. Next, R2C3 is from 13 and R9C3 is from 139. In fact
the last three cells in row 9 box 7 form a 139 triple with 3 not able to be in R9C2. That makes R8C1 from 157, but it can't
be 1, so it is from 57. The last two cells in box 7 are from 578 with R7C2 not able to be 7. Looking at the 3x3 purple
region in box 7 now, we have a 6 and we can't make 3 in three cells to get 9 (being divisible by 9 in the corners), and the
most we can get is 9+3+8=20, so we can't get the 21 we would need to reach 27 (being divisible by 9 in the corners), so the
corners sum must be 18 needing 12 more besides the 6 we already have. If we use 9 at all, we get 9+3 which is already 12
leaving nothing for the last cell, and 9+1=10, but there is no 2 available in the last cell, so 9 is not in the corner cells
R9C1 or R9C3 putting the 9 in R9C2. Now we have 6+1+3=10 leaving 8 to reach 18 (the divisible by 9 sum) putting the 8 into
R7C1 and making R7C2 a 5. We also get the 69 pair in box 4 sorted by the 9 we just placed and we get a 7 into R8C1 which sorts
rows 7 and 8 pairs in boxes 8 and 9.
Looking at row 1, we now know that R1C6 is even forcing R1C7 to be even and the other two cells left unfilled to be odd. That
makes R1C5 a 7 (besides the 24 pair in the column already doing the same). Now the evens in box 3 column 7 are from 246 since
there is already an 8 in the column giving a 246 triple in the box putting 8 into R3C8 sorting the 68 pair in box 2. R1C8 is
a 5, the only odd left for that column and the 6 comes out of R3C7. The 8 in box 1 is forced into R2C2 making R1C2 a 4,
taking the 4 out of R1C7. That just leaves a 1 to go into R3C2 making R2C3 a 3 and R3C4 a 3 making R1C4 a 1. This sorts the
139 pair in box 3. To get a sum divisible by 9 in the green 3x3 region in box 3 we now have 3+9=12 requiring 6 more being
2 and 4 with 2 in R1C7 and 4 in R3C7. This puts a 6 in R2C8 giving 6+7=13 making R2C5 the 4 which can also be done by using
the 4 in R3C7 we just placed to sort the 24 pair in box 2. A 2 finishes row 2, a 6 goes into R1C1 with a 9 in R1C3. The 5
and 7 to finish the box, the 15 pair now sorted in box 4, the 13 pair now sorted in box 7, and the 46 pair in box 9 finishes
the whole puzzle!
Btw, sorry for the slightly odd formatting in my solution. I typed it up in notepad with manual line breaks and just copy/pasted it here where YT uses word-wrapping. (And, yes, I know notepad has that option, but I tend not to set it because sometimes I use it for a quick programming scratch pad when doing some macro work for data processing tasks and the word-wrap feature causes some issues with the recorded key sequences.)
Oh wow, from 40:00 onward, Simon and I solved this in completely different ways. That's fascinating.
Very impressive solve and puzzle!
It's fascinating. I had like 10 minutes for all regions and parity shading but then had about 40 minutes for the rest. Let's just say I'm not that good ad Sudoku^^
3 in the corner twice? I should play the lottery
Aha! He admits it! 20:35 "Oh, that doesn't work, but it's complicated. Yeah, ok, well I'm going to do this because it is complicated."
The letter tool isn't limited to just the letters from A to I. You do have to be careful that you are in letter mode as some of the other letters are used as shortcuts in the other modes, but instead of removing all the coloring like Simon did, I switched to the letter tool and the corner mode putting O for odd and E for even. :)
Beautiful puzzle but now I'm cross-eyed from trying to differentiate the regions and boxs 🤪
I enjoyed Mark's subtle acknowledgement around 29:00 that his island coloring wasn't particularly well-received 😂
So did I, but this is Simon, not Mark.
Lesson never learned. When having great trouble finishing off REVIST the rules, even the ones I used earlier in the puzzle and forgot about. I loved creating the squares and started fine with the digits, then forgot all about the corner of the squares rule until I eventually reread them.
this was super great, was done in less than 1 hr
Happy 37th (or 2nd?!) birthday, Samuel!
He's 401! 😲
51:57
Very clever arrangement unravelling into the "new" squares, and then still so much to be done. The corners logic was a really nice touch.
I finished in 179 minutes. At first I was completely lost, so I started to watch the video before realizing as pointed out that corner must extend, that the entire grid had to be entirely populated with squares. After that realization, I was able to slowly make me way through and completely colored the grid. I switched to parity colors next. I got through most of those and proceeded to lose my mind as I made false logic after another. I made r5c6 a 5, broke it. I made r6c8 an 8, broke it. I made r4c2 a 5 due to miscounting, broke it. I made r3c2 a 9 because I missed my pencil marks, broke it. I enjoyed the puzzle, but I don't know what's wrong with me that I keep doing this. Great Puzzle!
42:30 so weird to see Simon having fun doing sudoku!
I've had a few ideas for puzzles that have tiling it with squares, but this general idea was more elegant than mine
I started by noticing that every row and column would have to have an odd squared box in them. Then I deduced that there was no way for an 5x5 to exist and that lead to the left and top edge to be all 3x3 boxes. Felt easier then how Simon deduced the 4x4 in the bottom right.
This was brilliant! I did grind to a halt around the point I wanted to put numbers in (after finding the squares and the odd/evens), so that wasn't as fun, but overall very interesting. 90 minutes.
Wow! Samuel is 401?! He's ANCIENT!!!
Surprised nobody seems to have commented that Simon got "lucky" again at 39:58... why wasn't r6c4 considered to be a 4?
Because if so, box 4 has no place to put 4 in it
And also because the 4 in C9 puts a 4 in either C8 or C9 of box 9, leaving 4 to go in C4 in box 8.
The symmetry about the main diagonal was quite pretty.
41:13 finish. Never turn down math in my puzzles! I wound up keeping my colored boxes and using different colored circles with the pen tool for my odd/even parity. Excellent!
I wish I had seen the parity logic on the border in boxes 4 and 7. That would have given me 6 more parity cells, but glad to see all my parity logic otherwise flowed similarly to Simon's except the beginning box 5 logic where I used the 2x2 cell borders putting 6 odd digits into rows and columns 3 and 4 to rule out the 4x4 region in box 5 from being all even. Having that logic also would have helped me place the 4 in box 4 making a bit of my other logic easier, too. Oh, bother. lol, Simon schooling me again. :)
What an absolutely wonderful puzzle.
I am glad to see Simon used colours tge same way as me for zoning first, then replace them by frames and use colours for parity. But gosh did it make scanning the Sudoku boxes a challenge.
Finished in 130:11. That was tough, but I liked it!
Thanks again.
Fun puzzle!
62:26. I tried to just use pencilmarking for the odd/evens instead of recolouring the grid, however I made a few typos doing that which came back to bite me later and forced me to backtrack a few times.
The way to finish off the odd/even pairs earlier is to notice that the corners of the 3x3 squares need to sum to 18 or 27, and once you have the 135 triple in box 2 then it's not possible to reach 27 in that square.
51:30 Simon needs to practice saying "Which is very important for a reason that will become apparent" and/or "..which is very important for reasons that are too obvious to mention"
40:00 R4 needs a 4 but since Simon only puts that in afterwards I don't know if this is why he ignored R6C4
72:23, spun my wheels for about 20 minutes in the end until I looked at the video and saw how simple the logic on the line in column 9 was. Like, one of the first things Simon did and I ignored the entire time! 😂
It's just squares all the way down.
so frustrating, for some reason i kept trying to make every 4 corners of a 9 cell box to be divisible by 4, and the4 corners of the 16 cells box to be divisible by 9, gave up at some point, slept on it, and finished it when i woke up
Took me 82 minutes this time… mostly because I put off colouring the 6s and 8s for so long.
Also resolved r5c6 by a somewhat fiddly method rather than sudoku; realising that making it a 5 would force 7 and 8 into the top corners of the 4×4 square, making r9c6&9 add up to either 17 or 1.
Yay. I was zeroth like today!
Green 2 makes me itch
"I'm going to do this because.... it is complicated" (Simon Anthony, 2024, 20:43)
42:14 for me. a bit hard for new rules, but it might be just unfamiliarity.
Do you see "purple" as purple or pink?
Man, the region/box overlaps messed with my head. I kept making mistaken deductions and having to back up to find the error. I'm exhausted.
39:06 for me. A little bit confusing, but I managed to finish it.
Why does it look like there's a chalk outline in the middle of this grid?
I just found an article called 'cracking the courtyard crypto' about the James Sanborn 'Kyrptos' sculpture, made me think of this channel and wonder what the inspiration was for the title of the channel. Also, I had a good time solving the moth-eaten sudoku while watching the video, it was interesting that I had started on some of the last spots you got to at the top of the board which helped me figure out how many numbers each cell had to contain, but I missed some of the parts you got to earlier on which revealed the first 5! Had a blast because I could just about do it and follow along with your logic.
I'm not 100% certain, but I think one of them did an interview once and talked about how the channel started with solving puzzles called cryptic crosswords, before they started including sudokus as well. Over time, the sudokus took over as the primary focus of the channel, but they still do crosswords sometimes too.
@@AllenMS828 Thanks!
@@NoaWinds You're welcome!
38:08 for me
It's no good, how can orange be odd when it's so much warmer than steely blue?
1:02:00 That's three in the NOT corner! :D
29:18 for me.
The center line looks like a person taking a dump, with the circle being the poo….
71:20 for me
66:38 for me.
Poor last three in the spotlight.
You had your first digit at 30:20
Correction first 3 digits
Looks like sums of squares. e.g. 81 is (5x9) + (9x4), 14 squares. Perhaps elementary number theory.
The geometry of the grid is telling you more than simple arithmetic. E.g. it is perfectly valid to write
81= 2 x 16 + 5 x 9 + 1 x 4,
but there is no way to fit 2 squares of length 4 into a 9x9 grid and to fill the rest with 2x2s and 3x3s.
@@pR0stYp3 Exactly. There is a component which makes number theory visible. Squares can only be of size 2, 3 or 4. Start with the 3s until a smaller square shows up. Keep subtracting 9. 81-9-9-9-9-9 = 36. 36 can be further sub-divided. ( 5s don't work as 81-25 doesn't leave a square and 2-5s overlap.)The solution appears to have only two possibilities: one with 9-4s or 5-4s with a 16.
(Should that be 5-9s, 5-4s and a 16?)
I'm not sure your reason for saying 5's don't work is a comprehensive enough reason. It doesn't really matter that 81-25 isn't a square. What matters is whether 81-25 can be further split into multiple squares. E.g. 81 - 25 = 56 = (5x4) + 36 = (10x4) + 16 = (5x5) + (3x9) + 4.
So number theory alone would suggest your 81 cells could be divided into,
one 5x5, five 2x2s and one 6x6; or
one 5x5, ten 2x2s and one 4x4; or
two 5x5s, three 3x3s and one 2x2.
Only the geometry of trying to fit these into 9x9 grid rules these options out.
@@RichSmith77 Yeah that's the visual component.
"Saying 5's don't work is [not] a comprehensive enough reason." Visually 2 5x5s can't be used due to overlap. Using 1 5x5 square leaves sides of odd length 9. So a 3x3 needs to be used. But that leaves an odd number of cells so a 2nd 3x3 is needed. Now try and place the 5x5 and 2 3x3s into the 9x9 grid without leaving a gap of 1 cell anywhere. Using number theory may be a slight misnomer. 81 is a square so Pythagoras might be a better way to go. e.g. 5(3 squared) plus 6 squared equals 9 squared. (Prime factorization will subdivide 6 squared down to squares involving 3 and 2.)
chess battle advanced
Around 1:00:30 why do you add the green and purple flairs to the six and eights? Haven’t they fulfilled their purpose? I find all the added flairs so cluttered.
Just comment for comment
First happily so
This solution is wrong.
There is a green 2 in line 9 😝
(Ha! Made you scroll!)
Hi. Can you fix the search engines in youtube? When I write something like modular, youtube offers me all other sudoku puzzles that do not contain modular lines, which is frustrating.
how is he supposed to fix the search engine of youtube?
There is a fan-made catalogue which may meet your needs. docs.google.com/spreadsheets/d/1rVqAjm-l_Urjd3TNmIc3SmTmz_OlgSoBuhY7RPgiuRg/edit?usp=sharing