Well done Bremster! I'm really impressed with this solve; you definitely understood the puzzle better as you went on and found a really nice path through. There are a few ways to do a couple of steps, in particular I really liked how you eliminated 5 from R9C3 as that wasn't one of the steps on my own path. Overall, I thought you explained your thought process really well, particularly in C7 (everything you said there is true, whether the total is 30 or 40!). You found most of my Sudoku steps quickly too, particularly in R1 and R5. I see a couple of comments about C9, but I wouldn't worry. You worked out the key logic that it needed 3&4 and even if you were hasty placing them, you didn't jump ahead as the 5 in R6 is the next step and that then resolves those digits. Hard to make three cells sum to 16 with 4 when they can't include 3 or 5. Thanks so much for giving it a go! I knew it would be a longer one, but you got through very smoothly and explained the logic very well as you worked through.
I appreciate you taking on more challenging puzzles from time to time! I nearly completed this on my own but had to get help once from watching you. In the central column I looked at the 16 sum, but I didn't think about how that implied a 9 sum above it.
Just watched again today (eye surgery went well), and you did an awesome job solving, BremSter! I suspect I would have struggled a bit more. 🙂 Thanks for doing this, and props to fjam for a very cool puzzle!
I haven't watched ahead to see if you catch this but at 14:10 when you figure out the 16s in column 9, you correctly determine that the 4 digits go on top and 3 on bottom, however you place the 4 and 3 as if they were normal Xsums when in this puzzle they could be swapped. It does turn out that you cannot make the sums work doing it that way but you missed that step of logic.
@@林老師-i5d To expand on that, 1 can't go in the leftovers, since 1+X will never sum to 13. It also can't go in a 3-cell 16 total, since 1+3+X or 1+4+Y won't reach 16. So the only way for a 4-sum 16 to include a 1, 2, and either 3 or 4 is 1+2+4+9. I did it a slightly more complicated way, by figuring out that the cell closest to the 22 had to be a 5 or 6, then the only 4-sum 16 that has a 2 and 3 or 4, but not both, and not both of 5 and 6 is 1+2+4+9.
In this variation of X sums, if the column has identical X sum clues outside the grid as in column 9 with 16 clues, the sums in the column must be different sequences of digits. That is to say they can't overlap, like in column 5 of this puzzle. If you add an additional cell to a three digit sequence that already adds to 16, you will break a four cell sequence since 0 is not a valid sudoku digit in this puzzle. Therefore as Bremstar correctly pointed out that 7 cells in that column add to 32 which means the two unused cells in column nine add to 13. Given the 2 in row 4 column 9, the two unused cells in column have to be row5 and row 6 since 2 and 11 sum to 13 , 11 not being a valid sudoku digit. Since Bremstar made his column 9 deductions after row 4 deduction, I don't believe he made a logical jump.
I got the first step (row 4), and then ran into a wall that would not budge. Then I watched you walk through it... just wow. Whenever you doubt your skills, look back at this puzzle's stats when you solved it.
I too have struggled with the pack- I managed the first 4 but couldn’t do the remaining puzzles I am hoping Bremster will show us some tips and tricks when the entry date has passed as I too struggled with the Dutch whispers
@14:14 I think you either made an unjustified logical leap or else I have mis-understood the rule. You stated that r1c9 had to be 4 and r9c9 had to be 3, based on the fact that the run of cells attached to them in the column is for 3 or 4 cells - this is ordinary x-sums logic. However, with this variation of the rule, why could not r1c9 be a 3 referring to the last 3 cells in the column and r9c9 be a 4 referring to the first 4 cells in the column? Have I mis-understood the rule? As it happens, you were right because you could not put a 4 in r9c9 as you couldn't make the the remaining sums in the column add up, but I don't think this was deduced.
I dont see how you can assume in the ninth column that the 4 goes in row 1 and the 3 goes in row 9. The rules call for either the first or last X digits from the clue.
You did assume the positions of the 3 & 4 in column 9, but of course in theory they could have been the other way round, because although of course the 4 X sum was at the top. The X number didn’t need to be.
Very very hard one. I tried but could not make it withou your help. But your laugh at the end make it everything better! 🤣🤣🤣 Did you say "nuts burger?" What does this mean? I did not understand, is a slang?
This one was way out of my ability and I still dont understand it lol. Interesting puzzle tho. Thanks for the solve. As for the pack , dont know if they are just super tough or I just cant find the way through.
C7 took me longer because 16 + 24 = 40 not 30. 4 5 wasn't possible because of no gap, 3 5 wasn't possible because of the 5 in box 6. 3 4 was apparently possible but proved to have no fill in box 9. Took me an age to consider the full overlap.
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Well done Bremster! I'm really impressed with this solve; you definitely understood the puzzle better as you went on and found a really nice path through. There are a few ways to do a couple of steps, in particular I really liked how you eliminated 5 from R9C3 as that wasn't one of the steps on my own path. Overall, I thought you explained your thought process really well, particularly in C7 (everything you said there is true, whether the total is 30 or 40!). You found most of my Sudoku steps quickly too, particularly in R1 and R5.
I see a couple of comments about C9, but I wouldn't worry. You worked out the key logic that it needed 3&4 and even if you were hasty placing them, you didn't jump ahead as the 5 in R6 is the next step and that then resolves those digits. Hard to make three cells sum to 16 with 4 when they can't include 3 or 5. Thanks so much for giving it a go! I knew it would be a longer one, but you got through very smoothly and explained the logic very well as you worked through.
That was a STUNNING solve and puzzle!!
I appreciate you taking on more challenging puzzles from time to time! I nearly completed this on my own but had to get help once from watching you. In the central column I looked at the 16 sum, but I didn't think about how that implied a 9 sum above it.
Just watched again today (eye surgery went well), and you did an awesome job solving, BremSter! I suspect I would have struggled a bit more. 🙂 Thanks for doing this, and props to fjam for a very cool puzzle!
I haven't watched ahead to see if you catch this but at 14:10 when you figure out the 16s in column 9, you correctly determine that the 4 digits go on top and 3 on bottom, however you place the 4 and 3 as if they were normal Xsums when in this puzzle they could be swapped. It does turn out that you cannot make the sums work doing it that way but you missed that step of logic.
I was wondering if that assumption was based on normal X sums
Sum of R1C9~R4C9=16, I should be included and causes 1+2+4+9=16, so R1C9=4
@@林老師-i5d To expand on that, 1 can't go in the leftovers, since 1+X will never sum to 13. It also can't go in a 3-cell 16 total, since 1+3+X or 1+4+Y won't reach 16. So the only way for a 4-sum 16 to include a 1, 2, and either 3 or 4 is 1+2+4+9.
I did it a slightly more complicated way, by figuring out that the cell closest to the 22 had to be a 5 or 6, then the only 4-sum 16 that has a 2 and 3 or 4, but not both, and not both of 5 and 6 is 1+2+4+9.
@@blargo thanks for your sharing
In this variation of X sums, if the column has identical X sum clues outside the grid as in column 9 with 16 clues, the sums in the column must be different sequences of digits. That is to say they can't overlap, like in column 5 of this puzzle. If you add an additional cell to a three digit sequence that already adds to 16, you will break a four cell sequence since 0 is not a valid sudoku digit in this puzzle. Therefore as Bremstar correctly pointed out that 7 cells in that column add to 32 which means the two unused cells in column nine add to 13. Given the 2 in row 4 column 9, the two unused cells in column have to be row5 and row 6 since 2 and 11 sum to 13 , 11 not being a valid sudoku digit. Since Bremstar made his column 9 deductions after row 4 deduction, I don't believe he made a logical jump.
Excellent puzzle and great solve. Solved it in 47:41.
I got the first step (row 4), and then ran into a wall that would not budge. Then I watched you walk through it... just wow.
Whenever you doubt your skills, look back at this puzzle's stats when you solved it.
nice puzzle and solving
Can't believe you just handed out the secret like that... :-)
In regards to the pack, I've tried them all, but only broke into two, only to then break them. Dutch Whispers are way tricky for me...
I too have struggled with the pack- I managed the first 4 but couldn’t do the remaining puzzles
I am hoping Bremster will show us some tips and tricks when the entry date has passed as I too struggled with the Dutch whispers
It feels hilarious that sudoku community would think that the sum of the digits from one to nine should be considered some kind of secret
Well done, BremSter!
@14:14 I think you either made an unjustified logical leap or else I have mis-understood the rule. You stated that r1c9 had to be 4 and r9c9 had to be 3, based on the fact that the run of cells attached to them in the column is for 3 or 4 cells - this is ordinary x-sums logic. However, with this variation of the rule, why could not r1c9 be a 3 referring to the last 3 cells in the column and r9c9 be a 4 referring to the first 4 cells in the column? Have I mis-understood the rule? As it happens, you were right because you could not put a 4 in r9c9 as you couldn't make the the remaining sums in the column add up, but I don't think this was deduced.
He just got lucky that 3 wont work on top
Sum of R1C9~R4C9=16, I should be included and causes 1+2+4+9=16, so R1C9=4
I dont see how you can assume in the ninth column that the 4 goes in row 1 and the 3 goes in row 9. The rules call for either the first or last X digits from the clue.
You did assume the positions of the 3 & 4 in column 9, but of course in theory they could have been the other way round, because although of course the 4 X sum was at the top. The X number didn’t need to be.
Very very hard one. I tried but could not make it withou your help. But your laugh at the end make it everything better! 🤣🤣🤣
Did you say "nuts burger?" What does this mean? I did not understand, is a slang?
This one was way out of my ability and I still dont understand it lol.
Interesting puzzle tho.
Thanks for the solve.
As for the pack , dont know if they are just super tough or I just cant find the way through.
Super difficult…. Did in 48 mins… consider it more as did it with all ur help…. Column 8 was tricky,…. Took lot of time to realise…
C7 took me longer because 16 + 24 = 40 not 30. 4 5 wasn't possible because of no gap, 3 5 wasn't possible because of the 5 in box 6. 3 4 was apparently possible but proved to have no fill in box 9. Took me an age to consider the full overlap.