24:40 you missed my favorite bit of logic with this puzzle. the 13 and 21 clue share 2 cells and each have 1 additional cell. 21-13 = 8, the only way to make two cells differ by 8 is 1 and 9. You can actually use this logic to also grey out the R1C9 corner
First we had the Phistomefel ring, now we have the Phiistomefel wall and then Simon looks at a cell and says, "That must be something, I'll put a 6 in it." I love this channel, now where is that chocolate teapot?
It's really interesting from a psychological perspective watching how Simon's mind works. Simon spends the first bit explaining the 8 clue while missing the 9 clue completely. It's fun watching how he'll get stuck on one bit of logic and not see the other one mixing with it but then figure it out through a larger means. I love watching these videos.
Once you have the first grey cell in box 1, the 7 gets all its sum horizontally, so the additional vertical spur of the 11 must sum to 4. So the 7 can't be a 3 cell 124, as it would eliminate both 4 and 13. Then once the grey bounds the 7, the greens must get out, making the 11 spur 13
In the break-in with the 11 and the 7 clues, another way to analyze it is the following. Since a grey cell is right underneath the 7 clue, and since the 11 and the 7 clue see exactly same cells in the x direction, they both see 7 in that direction. That means that the 11 clue sees 4 in the y direction, not counting the cell with the clue. So, underneath it is either 4 or 1 3. 4 breaks the rules, since the 7 clue would have to see exactly 3 cells for the reasons Simon mentioned, and can't be 1 2 4, because Sudoku. Not that much of an improvement as I find this logic with two clues seeing the same cells in some direction cute, and Simon used a similar logic with the 9 and the 19 clues.
As soon As I see Phistomefel I get very excited to see what work of sublime genies he has come up with this time. This man deserves a award for the amount of stunning puzzles he’s created
Simon spends a lot of time on the box 1 logic, but he missed that it's actually very simple: The 11 and 7 clues are connected, and must take at least 1 more cell, and the 7 can't see down. That means the 11 sees 7 going right starting at R1C1, and 4 going down from R2C1. The 4 cannot be a single 4, as that would mean a 3 cell 7 clue that doesn't include a 4 and is impossible, so it must be a 13 pair and the 7 gets cut off on the right. I loved this puzzle when I solved it, but Simon just overlooked some simple logic in favour of complicated logic here and made it look harder than it is. I'd argue its rating should be about 3.5/5.
Making the 11 at the top left so much more complicated than it has to be. The 11 shares a row with the 7 clue next to it, so you need to add 4 more in the column. If you just add a 4 in R2C1, there's no way to make 7 with three numbers (would need a 4), so it has to be 1 and 3 in R2C1 and R3C1. If you then want to make 7 in 3 numbers, you'd need another 1, so it has to be 2 numbers, and it can only be 2 and 5.
Another Phistomefel masterpiece. Poirot's always going on about the little grey cells, and there was some enjoyable sleuthing in here. There were some lovely interactions between the clues. I particularly liked the roles played by the 36 clue. I also liked the way the 15 & 16 clues interacted. Whatever hung down from the 15 had to add up to one less than R3C7, so it had to be a single-cell 1 or 2, and couldn't be a 2 because that would force R4C7 to also be a 2, giving us three digits. Simon's thinking on the 11 clue in box 1 was somewhat laboured. The grey in R2C2 meant the 7 was restricted to horizontal cells only, so the 11 clue saw all 7, and the balance going down must be 4. Regardless of whether it's a single 4, or a 31 pair, it ruled out a 3-cell 7, so the corner break-out must be via the 11, forcing it to be a 31 pair, and the 7 a 25 pair.
As early as 16:00 you could used the 13 next to the 21 to deduce that the difference in the cells they could see must be 8 which means that r2c9 must be 9, r1c7 must be 1, and r1c9 must be gray
It seems Simon didn't really get into the thinking of "what's the difference between what these two clues can see" for the whole solve. He used it only at the end of long deliberations a few times.
It's more do-able than many of his puzzles. No single too-complicated-for-words stumbling block, but requires the use of many different logical techniques any one of which can trip a person up for a while.
@@aceldamia9114 How is this dictation? It's a help to those who want it, and can be ignored if you want to watch the whole video. Who are you to dictate how I comment?
23:30 he's driving me crazy! With the 13 and 21 adjacent, we know that R2C9 MUST be 8 greater than R1C7. AKA, they must be 9 and 1. The same logic could have been applied LONG ago and would have dealt with the gray in R1C9 as well.
Simon's break-in can be made slightly simpler by taking it in two steps, the first being to ask whether R1C3 can ever be green. (It can't because there's no way to put 124 across R1 and still complete the 11 clue.) After that Simon's logic is more concrete, since one of the 11 clue is already anchored. And ooh, ooh! There's a more elegant way to resolve the 21 clue in box 3. If R1C8 and R2C8 are any less than 12, then the 21 clue breaks, because the third cell in it has to equal at least 10. And if they're any greater than 12, the 13 clue breaks since it still needs one more cell. So R1C8 and R2C8 do add to 12, and thanks mostly to plain old Sudoku (also a tiny bit of killer math), it's quickly evident that 4/8 is the only pair that works. I can't believe I saw that and Simon didn't, that sure doesn't happen often. Sure wish he did though because he would've about fallen out of his chair with delight I think.
Took me a little over 1h30m. The entry really had me stumped for a while. I was just staring at it and going "What am I supposed to do???". I used quite a bit of logic that was different than Simon. So I'm quite happy that I saw different ways to solve this.
That was so much fun to solve. No time, too many interruptions, but the 4/5 Simon mentions sounds about right. Probably took about 90 minutes? Not sure. My prediction: Simon is going to fall all over himself at how it's put together and rightly so. It was almost magical how it unfolded.
Made a mistake yesterday that I couldn't figure out what went wrong, so I was disheartened and gave up, but I came back to take another crack at it and solved it smoothly in just under an hour. So very satisfying to finish something after a roadblock like that.
24:00 it's quite a bit easier at this point if you just compare the sums... the difference between 13 and 21 is 8, so the digit that's only in the 13 sum has to be 8 less than the digit that's only in the 21 sum. The only two digits that fulfill this are 1 and 9, so the 13 cage has 1 and the 21 cage has 9 and the two digits that they share have to add up to 12 (12 + 1 = 13, 12 + 9 = 21).
It's is rare that I feel the same level of exhilliration that Simon seems to show a bit too often for my taste. But this puzzle was just so feel-good to go through.
This is somewhat similar to Phistomefel's Cave Sudoku, only we have pillars instead of walls, and sum cells instead of counting. I found what I learned from Cave Sudoku very useful here as well.
Well, that was crazy cool, and definitely one for me to just watch your solve! Feedback on the alpha app: The corner marks are way too far in, to the point where they can be confused easily with center marks. I understand trying to avoid the upper left corner, but just look at how much wasted space there is at the right and bottom of the box. Also, since the dark grey is available, maybe take away the black and give us another usable colour choice?
Can you imagine making a puzzle so complex that you made a slightly smaller puzzle with a solution to ensure people could complete your puzzle? It’s like learning to drive a car by watching someone drive a go-cart. Absolutely nuts.
Wow, that was something all right! I managed it with a couple times to back up, and a couple times I thought a line of thinking would rule something out but it went a bit far and was more bifurcation-y. But I'll take what I can get, and here that's 1:21:26.
Regarding the 11/7 clue and how far the 11 goes down. Since the 7 is blocked vertically we know the horizontal line sums to 7 that leaves the 11 clue the sum 4 for the vertical line and that is impossible with 3 digits. That forces r2c1 and r3c1 into a 13 pair which forces r1c1 and r1c2 into a 52 pair since 1 or 3 is already fixed.
The 21 being joint to the 13 while the 13 included an extra cell to the left implied they had to add up to 12 vertically but the 21 including a maximum of an extra cell to the right implied the had to add at least 12 vertically, so they has to add 12 and the digits left and right were actually 1 and 9. That could be deducted even before r3c8 was greyed out.
Nice puzzle, only saying instead of grey if you use blue i.e. it would be better for seeing it in the phone. Tks 4 the time you spend doing that really nice videos :-)
I'm proper chuffed I solved this in about 35 minutes; could have been even better if I hadn't missed an easy placement at the start of the home stretch. I think my Corral/Bag experience came in handy.
4 out of 5 Difficulty seems right for this one -- lower than most Phistomefel's on here. There was no blinking hard break-in, but each step took some thought, and continued to all the way to the end.
There is a Kuroma (I think?) technique where if two of the clues are connected, you know that the difference between the clues must be the difference of all the cells that one clue sees and the other doesn't. I.E. all the cells orthogonal to the connection. For example, the 13 and 21 that are connected mean that the difference is 8, and as soon as you see that 21 has one horizontal cell, that must be a 9 and the 13 must have a 1 horizontally. Same with the 15 and 16 in box 6, where the difference between R3C6 and R5C7(plus down) is one.
This is a brilliant puzzle. I made a stupid mistake near the end and had to restart, it's annoying that it's hard to know where one went wrong. I didn't feel this was a particularly difficult puzzle, the logic flowed beautifully. Perhaps I'll blame the new software or fat fingers.
23:37 There is one key not in the 13 that is in the 21, and vice versa. So the two common have to be at the most 12 (13+1) and at least 12 (12+9) So the gray cell follows, as well a 1 and 9 in the box. Also, the 14 in box 2: three cells, 34 in one. 15 in the other. The options are 14-3-1=10 (no) 14-3-5=6 (works) 14-4-1=9 (works) 14-4-5=5 (no) So that is a 69.
I'd recommend using the black in the new software for the wall next time. Gives more visible color difference between green and also the text inside is visible clearly, unlike darker grey or light grey.
r1c6 and r2c9 were a given earlier if 2 cells + another cell add to 21 and the same 2 cells + yet another cell add to 13, the difference between the last 2 cells is 8, therefore one is 9, one is 1
I'd love to give this one a shot and tried watching for the break-in, but I just don't understand the logic of 'this cage sees 11 this way' when it just sees the 11 clue... the values in one axis aren't necessarily the total of the whole thing, though... I tried following along and coloring but it just seems silly when I don't really get how the logical leaps are made to place some of the greens at the beginning. Another dud for me - it's like I'm collecting them. Looking forward to the day when I can solve some of the more interesting puzzles I see on here.
There was an easier way to figure out the 13- and 21-clues in Box 3 ... . . . ... once you know that r1c7, r1c8, r2c8, and r2c9 are green (label these A, B, C, D, respectively), since both clues see B and C, it follows that D is 8 more than A; thus, D=9 and A=1.
Well chuffed that I got as far as Simon did in 40 mins (even though it did take me 1 hr 40 mins!) For some reason I took out the option of 5-7 for the 12 clue and creamed in at the end...
As usual Phistomefel's puzzles are beyond my ability. Is this logic correct? At 23:11 On the 13 and 21 clues, the difference between the clues is 8 therefor the number in r2c9 must be 9 and the number in r1c7 must be 1 to make the 8 difference?
Took me over two hours and two restarts. First, because I forgot about multiple digits for 19; second because I neglected to see a second option for the 22. That second one did a number on me.
Before the box 5 interaction, I started with the box 1 logic (corresponding to around 12:00 in the video) - after concluding that 7 and 8 do not connect, we can see that 11 sees everything 7 sees. So if the 7 is a 3 cell sum (124), there is no way for 11 to be satisfied. So 7 is a 2 cell sum, and the walled-off corner means that 11 is a 4 cell sum, with the two cells not seen by 7 being 13 pair
You can get rid of the highlighted cell by clicking it again. You don't have to click off the board. Just don't click it again too soon, the computer will see it as a double click.
It took me 1 hour 15 mins to determine the grey/non-grey cells. It took me another hour to reach a point right near the end where I had two 4s in the same row :(. I had to check the video to see that I'd somehow placed the 2,4 and 5 in box four in the wrong order and after fixing the related cells everything else was correct. So disappointed. This was a great puzzle though.
Couldn't the 36 clue be only 5 cells? If you place 9, 8, 7 and 6 on column 9 (6 being in box 6) and another 6 left to clue, that would add up to 36 in only 5 cells. I know it doesn't make a difference to the solve, but I'm just asking.
Seems to me that he forgot the possibility of repeated digits when looking at the options for the 36 clue. Not sure if there was some easy way to rule that out.
I think he just gets away with it. it could in theory be more than 8 cells with repeats - but if there are 8 cells all in one colum, then they have to add up to at least 36 (if there is no 9). since the 9 is not at the edge of column 9, the 36 cant have 8 cells in that column - so I think him ignoring the possibility of more than8 cell works.
The column sum already exceeds 36, meaning R8C9 must be grey, and from there, the digits in the 36 sum are restricted to 1 row and cannot repeat. So he didn't really need to. But you are right that he didn't consider it earlier on
You can't have repeated digits in a row or column. When he made R8 C9 gray it was green all the way to Row 2. which had a 9 in it. If all those cells are green, they add up to 38.
By the time he has enough information to use the 36 clue, he has enough information to put a grey cell directly above it, which therefore eliminates any repeats since all its digits are in the same row.
No, I think he followed all the logic correctly. The first time the clue is dealt with is in making R8C9 gray, and that's completely valid as 8 green cells along C9, with the 7 at the top already in place, would have added to 38 and broken the clue with or without additional cells from R9. After that the clue can only ever involve R9, so no repeats.
You can make 10 in 2 cells but in this puzzle you can see that you can only count a cell once.. he has already used r2c4 to that's why he got 10 from the clue in r2c4 - r3c4. He can't now use another in the same column as the 2 clues and he can not use r2c2 because if that cell is green then you can see there will be 3 green cells connected vertically in column 2 with 2 clues which is 7 and 8. This is not possible for 3 cells to have 2 different sums.
@@Sakamori14... To put a person behind the puzzle... It wouldn't spoil the puzzle. Sometimes Simon or Mark tell some story of how the puzzle came to be...
@@Tferdz This is a nice idea! It's a lot more work for Simon and Mark though. They can even do the interview afterwards after all the struggle to solve them
bring Phistomefel for an educative video about how he builds his puzzles
This is excellent idea
I would also like to learn how to
That would easily be the most epic crossover man have seen
^^^^ THIS ^^^^
Seconded, thirded, and fourthed!
24:40 you missed my favorite bit of logic with this puzzle. the 13 and 21 clue share 2 cells and each have 1 additional cell. 21-13 = 8, the only way to make two cells differ by 8 is 1 and 9. You can actually use this logic to also grey out the R1C9 corner
That was my first digit in the grid :-)
Similar logic applied back at 10:25 to see that the 11 in r1c1 has to see exactly 4 in c1 outside of r1, and so can be at most 2 cells.
Commander Keen!
@@davidjh1978 And it stops the 7 in r1c2 being 3 cells as it would break that extra 4.
First we had the Phistomefel ring, now we have the Phiistomefel wall and then Simon looks at a cell and says, "That must be something, I'll put a 6 in it." I love this channel, now where is that chocolate teapot?
Simon's crushing hard on Phistomefel
@@moarkrabspointlesschanneld8467 what in the bloody hell
@@KusaneHexaku She recently subscribed for the promise of naked singles
not only him. i mean im surprised hes not putting his name in video title. when i see his puzzle i drop everything to do it. that man is genius
43:26
"There's a 4,9 pair here, which is probably resolved, but I'm just not spotting it..."
*9 in r2c9*: "Am I a joke to you?"
"I'll put 6 in" this just tickled me. Thanks for the giggle
It's really interesting from a psychological perspective watching how Simon's mind works. Simon spends the first bit explaining the 8 clue while missing the 9 clue completely. It's fun watching how he'll get stuck on one bit of logic and not see the other one mixing with it but then figure it out through a larger means. I love watching these videos.
Life would be empty without knowledge bombs from Cracking the Cryptic.
Once you have the first grey cell in box 1, the 7 gets all its sum horizontally, so the additional vertical spur of the 11 must sum to 4. So the 7 can't be a 3 cell 124, as it would eliminate both 4 and 13. Then once the grey bounds the 7, the greens must get out, making the 11 spur 13
In the break-in with the 11 and the 7 clues, another way to analyze it is the following. Since a grey cell is right underneath the 7 clue, and since the 11 and the 7 clue see exactly same cells in the x direction, they both see 7 in that direction. That means that the 11 clue sees 4 in the y direction, not counting the cell with the clue. So, underneath it is either 4 or 1 3. 4 breaks the rules, since the 7 clue would have to see exactly 3 cells for the reasons Simon mentioned, and can't be 1 2 4, because Sudoku.
Not that much of an improvement as I find this logic with two clues seeing the same cells in some direction cute, and Simon used a similar logic with the 9 and the 19 clues.
Woot! ... under 40 minutes (39:56 to be exact) to solve a Phisto puzzle. My day is now 'made'.
Nice puzzle!
Well done. I made a careless mistake but I totally feel I had the ability to solve it...
As soon As I see Phistomefel I get very excited to see what work of sublime genies he has come up with this time. This man deserves a award for the amount of stunning puzzles he’s created
sublime genies hehe
Simon spends a lot of time on the box 1 logic, but he missed that it's actually very simple: The 11 and 7 clues are connected, and must take at least 1 more cell, and the 7 can't see down. That means the 11 sees 7 going right starting at R1C1, and 4 going down from R2C1. The 4 cannot be a single 4, as that would mean a 3 cell 7 clue that doesn't include a 4 and is impossible, so it must be a 13 pair and the 7 gets cut off on the right.
I loved this puzzle when I solved it, but Simon just overlooked some simple logic in favour of complicated logic here and made it look harder than it is. I'd argue its rating should be about 3.5/5.
Making the 11 at the top left so much more complicated than it has to be.
The 11 shares a row with the 7 clue next to it, so you need to add 4 more in the column. If you just add a 4 in R2C1, there's no way to make 7 with three numbers (would need a 4), so it has to be 1 and 3 in R2C1 and R3C1. If you then want to make 7 in 3 numbers, you'd need another 1, so it has to be 2 numbers, and it can only be 2 and 5.
Another Phistomefel masterpiece. Poirot's always going on about the little grey cells, and there was some enjoyable sleuthing in here. There were some lovely interactions between the clues. I particularly liked the roles played by the 36 clue. I also liked the way the 15 & 16 clues interacted. Whatever hung down from the 15 had to add up to one less than R3C7, so it had to be a single-cell 1 or 2, and couldn't be a 2 because that would force R4C7 to also be a 2, giving us three digits.
Simon's thinking on the 11 clue in box 1 was somewhat laboured. The grey in R2C2 meant the 7 was restricted to horizontal cells only, so the 11 clue saw all 7, and the balance going down must be 4. Regardless of whether it's a single 4, or a 31 pair, it ruled out a 3-cell 7, so the corner break-out must be via the 11, forcing it to be a 31 pair, and the 7 a 25 pair.
As early as 16:00 you could used the 13 next to the 21 to deduce that the difference in the cells they could see must be 8 which means that r2c9 must be 9, r1c7 must be 1, and r1c9 must be gray
It seems Simon didn't really get into the thinking of "what's the difference between what these two clues can see" for the whole solve. He used it only at the end of long deliberations a few times.
Simon: “Do have a go, there’s link in the description”
Me seeing’s video length and Phistomefel’s name: “I’ll pass”
It's more do-able than many of his puzzles. No single too-complicated-for-words stumbling block, but requires the use of many different logical techniques any one of which can trip a person up for a while.
2:20 Rules
4:32 Let's get cracking!
New T-Shirt idea: "I always trust my pencil marks"
Quit hijacking others videos, dicatting when people start watching. Who the hell do you think you are?
@@aceldamia9114 no need to be rude, many people will skip the intro regardless. Its not like we're skipping an ad....
@@aceldamia9114 actually if you want to be mad at someone there's a person (possibly a bot) advertising his own channel.
@@aceldamia9114 How is this dictation? It's a help to those who want it, and can be ignored if you want to watch the whole video. Who are you to dictate how I comment?
Brilliant loved the solve especially "That's a something... I'll put 6 in!"
23:30 he's driving me crazy! With the 13 and 21 adjacent, we know that R2C9 MUST be 8 greater than R1C7. AKA, they must be 9 and 1. The same logic could have been applied LONG ago and would have dealt with the gray in R1C9 as well.
The Werefrog love the "That's a something".
Yep. It's definitely something.
Whoa, this is seriously intricate. Not that I expected anything else...
"You know what I'm going to say now, don't you? Well, I'm still going to say it."
Bobbins, he said it.
He said it, he said the line.
Simon's break-in can be made slightly simpler by taking it in two steps, the first being to ask whether R1C3 can ever be green. (It can't because there's no way to put 124 across R1 and still complete the 11 clue.) After that Simon's logic is more concrete, since one of the 11 clue is already anchored.
And ooh, ooh! There's a more elegant way to resolve the 21 clue in box 3. If R1C8 and R2C8 are any less than 12, then the 21 clue breaks, because the third cell in it has to equal at least 10. And if they're any greater than 12, the 13 clue breaks since it still needs one more cell. So R1C8 and R2C8 do add to 12, and thanks mostly to plain old Sudoku (also a tiny bit of killer math), it's quickly evident that 4/8 is the only pair that works.
I can't believe I saw that and Simon didn't, that sure doesn't happen often. Sure wish he did though because he would've about fallen out of his chair with delight I think.
Took me a little over 1h30m. The entry really had me stumped for a while. I was just staring at it and going "What am I supposed to do???". I used quite a bit of logic that was different than Simon. So I'm quite happy that I saw different ways to solve this.
A brilliant one. Solving the example problem first helps
That was so much fun to solve. No time, too many interruptions, but the 4/5 Simon mentions sounds about right. Probably took about 90 minutes? Not sure. My prediction: Simon is going to fall all over himself at how it's put together and rightly so. It was almost magical how it unfolded.
There should be something with "if I trust my pencil marks, and I do"
Sounds like a t-shirt waiting to happen
@@AntonyCurrington If I trust my pencil marks... on the front and ...and I do on the back. :)
They should make CTC pencils with that stamped on the side.
I did it!! Great approachable puzzle!!
Made a mistake yesterday that I couldn't figure out what went wrong, so I was disheartened and gave up, but I came back to take another crack at it and solved it smoothly in just under an hour. So very satisfying to finish something after a roadblock like that.
I really enjoyed solving this one! Took me a bit to figure out how to start, but then it flowed nicely
24:00 it's quite a bit easier at this point if you just compare the sums... the difference between 13 and 21 is 8, so the digit that's only in the 13 sum has to be 8 less than the digit that's only in the 21 sum. The only two digits that fulfill this are 1 and 9, so the 13 cage has 1 and the 21 cage has 9 and the two digits that they share have to add up to 12 (12 + 1 = 13, 12 + 9 = 21).
It's is rare that I feel the same level of exhilliration that Simon seems to show a bit too often for my taste. But this puzzle was just so feel-good to go through.
08:37 I think the 36 can be a 5 cell clue also:
9
3
7
9 8
This is somewhat similar to Phistomefel's Cave Sudoku, only we have pillars instead of walls, and sum cells instead of counting. I found what I learned from Cave Sudoku very useful here as well.
"Hang on i'm being slow here"
Me, having lost him for 10 minutes by now: ahh yes, this is obviously a 1 there
Well, that was crazy cool, and definitely one for me to just watch your solve!
Feedback on the alpha app: The corner marks are way too far in, to the point where they can be confused easily with center marks. I understand trying to avoid the upper left corner, but just look at how much wasted space there is at the right and bottom of the box.
Also, since the dark grey is available, maybe take away the black and give us another usable colour choice?
I would love to see more available colors! Why only 9 colors?
also colored numbers would be cool.
Can you imagine making a puzzle so complex that you made a slightly smaller puzzle with a solution to ensure people could complete your puzzle? It’s like learning to drive a car by watching someone drive a go-cart. Absolutely nuts.
many of Phistomefel's puzzles with unique rulesets actually come miniature examples so that there is no confusion about how the ruleset works
@@Nighthound pure genius. Just amazing.
My first phistomefel solve I spent an hour solving the example, before taking 3 tries and 6-7 hours to solve the actual puzzle
Wow, that was something all right! I managed it with a couple times to back up, and a couple times I thought a line of thinking would rule something out but it went a bit far and was more bifurcation-y. But I'll take what I can get, and here that's 1:21:26.
Regarding the 11/7 clue and how far the 11 goes down.
Since the 7 is blocked vertically we know the horizontal line sums to 7 that leaves the 11 clue the sum 4 for the vertical line and that is impossible with 3 digits.
That forces r2c1 and r3c1 into a 13 pair which forces r1c1 and r1c2 into a 52 pair since 1 or 3 is already fixed.
I put my finger on what makes this channel so good. You have a great way of explaining what you're thinking, that so many people can't replicate.
The 21 being joint to the 13 while the 13 included an extra cell to the left implied they had to add up to 12 vertically but the 21 including a maximum of an extra cell to the right implied the had to add at least 12 vertically, so they has to add 12 and the digits left and right were actually 1 and 9.
That could be deducted even before r3c8 was greyed out.
Great new variant! I hope to see more!
Nice puzzle, only saying instead of grey if you use blue i.e. it would be better for seeing it in the phone. Tks 4 the time you spend doing that really nice videos :-)
Mmm, this was a slow burn...and I absolutely loved it!
Phistomefel puzzles are always an interesting watch.
thank you Simon, I share your joy of solving.
I'm proper chuffed I solved this in about 35 minutes; could have been even better if I hadn't missed an easy placement at the start of the home stretch. I think my Corral/Bag experience came in handy.
Great puzzle! 🤩
4 out of 5 Difficulty seems right for this one -- lower than most Phistomefel's on here. There was no blinking hard break-in, but each step took some thought, and continued to all the way to the end.
There is a Kuroma (I think?) technique where if two of the clues are connected, you know that the difference between the clues must be the difference of all the cells that one clue sees and the other doesn't. I.E. all the cells orthogonal to the connection.
For example, the 13 and 21 that are connected mean that the difference is 8, and as soon as you see that 21 has one horizontal cell, that must be a 9 and the 13 must have a 1 horizontally.
Same with the 15 and 16 in box 6, where the difference between R3C6 and R5C7(plus down) is one.
Phistomefel + rules that need to be read with a magnifying glass = I am not bothering
9:10 if r7c1 is green then r8c3 is green and r8c4 is grey - but if r7c1 is grey then the 22 clue could be satisfied with as few as 3 cells
This is a brilliant puzzle. I made a stupid mistake near the end and had to restart, it's annoying that it's hard to know where one went wrong. I didn't feel this was a particularly difficult puzzle, the logic flowed beautifully. Perhaps I'll blame the new software or fat fingers.
These Phisto puzzles are getting out of hand. I have no clue the rules on this one. I think I may be regressing in brain power.
23:37 There is one key not in the 13 that is in the 21, and vice versa.
So the two common have to be at the most 12 (13+1) and at least 12 (12+9) So the gray cell follows, as well a 1 and 9 in the box.
Also, the 14 in box 2: three cells, 34 in one. 15 in the other. The options are 14-3-1=10 (no) 14-3-5=6 (works) 14-4-1=9 (works) 14-4-5=5 (no) So that is a 69.
DO NOT LET YOUR GREY CELLS TOUCH THEMSELVES. More fun from ctc!!!
I'd recommend using the black in the new software for the wall next time. Gives more visible color difference between green and also the text inside is visible clearly, unlike darker grey or light grey.
Yeah, that light grey is only distinguishable from white on my monitor/eyeballs combo if they're in adjacent cells...
16:16
*me screaming that you can fill in actual numbers*
17:16
*yes basically 1 minute after me, im good... or something* xD
Green and gray were bad choices for colors. I'm not color blind, but it was still quite hard to differentiate.
r1c6 and r2c9 were a given earlier
if 2 cells + another cell add to 21 and the same 2 cells + yet another cell add to 13, the difference between the last 2 cells is 8, therefore one is 9, one is 1
I'd love to give this one a shot and tried watching for the break-in, but I just don't understand the logic of 'this cage sees 11 this way' when it just sees the 11 clue... the values in one axis aren't necessarily the total of the whole thing, though... I tried following along and coloring but it just seems silly when I don't really get how the logical leaps are made to place some of the greens at the beginning. Another dud for me - it's like I'm collecting them. Looking forward to the day when I can solve some of the more interesting puzzles I see on here.
There was an easier way to figure out the 13- and 21-clues in Box 3 ...
.
.
.
... once you know that r1c7, r1c8, r2c8, and r2c9 are green (label these A, B, C, D, respectively), since both clues see B and C, it follows that D is 8 more than A; thus, D=9 and A=1.
Well chuffed that I got as far as Simon did in 40 mins (even though it did take me 1 hr 40 mins!) For some reason I took out the option of 5-7 for the 12 clue and creamed in at the end...
Grey is a very difficult color to see
in the new software version, black, dark grey and light gray are lighter, so the lighter dark gray is the normal gray in the old version
I can't see it at all
As usual Phistomefel's puzzles are beyond my ability.
Is this logic correct?
At 23:11
On the 13 and 21 clues, the difference between the clues is 8 therefor the number in r2c9 must be 9 and the number in r1c7 must be 1 to make the 8 difference?
yup, that's a nice way to see it
I was too slow to stop him from saying it
Took me over two hours and two restarts. First, because I forgot about multiple digits for 19; second because I neglected to see a second option for the 22. That second one did a number on me.
BEST
18:20 The 4 in box 1 is saying you have to put a 4 in the 9 clue which forces it to be 234.
2:02:36
I think this was the first Phistomefel puzzle I completed without checking the video once :)
1h19m, the start was brutal, but after that it flowed quite smoothly, albeit slowly still.
Nice puzzle! Loved the logic!
Before the box 5 interaction, I started with the box 1 logic (corresponding to around 12:00 in the video) - after concluding that 7 and 8 do not connect, we can see that 11 sees everything 7 sees. So if the 7 is a 3 cell sum (124), there is no way for 11 to be satisfied. So 7 is a 2 cell sum, and the walled-off corner means that 11 is a 4 cell sum, with the two cells not seen by 7 being 13 pair
Holy crap ... That was incredibly difficult ... Damn ....
Wow I just spent 3 hours and then noticed "non grey cells must be orthogonally connected."
I should have read the instructions twice too. >
This one hurt my head!😋
You can get rid of the highlighted cell by clicking it again. You don't have to click off the board. Just don't click it again too soon, the computer will see it as a double click.
Love the videos!!!
Let him who hath understanding reckon the number of the beast, for it is a human number... Something something something, or other.
Nice reference there 😉
It took me 1 hour 15 mins to determine the grey/non-grey cells. It took me another hour to reach a point right near the end where I had two 4s in the same row :(. I had to check the video to see that I'd somehow placed the 2,4 and 5 in box four in the wrong order and after fixing the related cells everything else was correct. So disappointed. This was a great puzzle though.
Couldn't the 36 clue be only 5 cells? If you place 9, 8, 7 and 6 on column 9 (6 being in box 6) and another 6 left to clue, that would add up to 36 in only 5 cells. I know it doesn't make a difference to the solve, but I'm just asking.
You are correct. This way allows us to repeat the 6. Only after placing the gray square in R8C9 can we dismiss this possibility.
Thank you Once Again 👍🏻👍🏻
Phistomefel's devil graphic was missing in the video thumbnail... I almost missed it! 😆
Tried it twice, always end up in a dead loop, can't solve this xD
Next step - make the clues able to see through the border
Damn, that looks hard.
8:24 couldn't the 36 clue hypothetically be up to 11 cells? 123456 in c9 and 543216 in r9 would do it, overlapping with the 6 in r9c9
What if the cells with the clues have imposter vision(sry for the stupid joke).
Anxiously awaiting Among Us, the Sudoku.
Amazing how Simon can take a Phistomefel so far so smoothly.
Seems to me that he forgot the possibility of repeated digits when looking at the options for the 36 clue. Not sure if there was some easy way to rule that out.
I think he just gets away with it. it could in theory be more than 8 cells with repeats - but if there are 8 cells all in one colum, then they have to add up to at least 36 (if there is no 9). since the 9 is not at the edge of column 9, the 36 cant have 8 cells in that column - so I think him ignoring the possibility of more than8 cell works.
The column sum already exceeds 36, meaning R8C9 must be grey, and from there, the digits in the 36 sum are restricted to 1 row and cannot repeat. So he didn't really need to. But you are right that he didn't consider it earlier on
You can't have repeated digits in a row or column.
When he made R8 C9 gray it was green all the way to Row 2. which had a 9 in it.
If all those cells are green, they add up to 38.
By the time he has enough information to use the 36 clue, he has enough information to put a grey cell directly above it, which therefore eliminates any repeats since all its digits are in the same row.
No, I think he followed all the logic correctly. The first time the clue is dealt with is in making R8C9 gray, and that's completely valid as 8 green cells along C9, with the 7 at the top already in place, would have added to 38 and broken the clue with or without additional cells from R9. After that the clue can only ever involve R9, so no repeats.
CTC Pseudo-Cockney Slang: Batman
Batman & Robin - Bobbins.
Why is R2C5 green? You could make 10 in the row in just two cells cant you?
R2C4 is already part of the 9. So you need 10 more for 19 and that needs R2C5 aswell.
Yes, but you can't include r2c4 in that, because it's already being counted for the vertical 9 total.
You can make 10 in 2 cells but in this puzzle you can see that you can only count a cell once.. he has already used r2c4 to that's why he got 10 from the clue in r2c4 - r3c4.
He can't now use another in the same column as the 2 clues and he can not use r2c2 because if that cell is green then you can see there will be 3 green cells connected vertically in column 2 with 2 clues which is 7 and 8. This is not possible for 3 cells to have 2 different sums.
Ahaa, Thank you all
Fun enough to watch but (for me) not to solve
Why not have some quick interview with some constructors before solving some puzzles? To have some insight on how they've done it
What's the point beforehand?
@@Sakamori14... To put a person behind the puzzle... It wouldn't spoil the puzzle. Sometimes Simon or Mark tell some story of how the puzzle came to be...
@@Tferdz This is a nice idea! It's a lot more work for Simon and Mark though. They can even do the interview afterwards after all the struggle to solve them
If they'd do this, it would have to happen *after* the puzzle, so they can actually say anything about it without spoiling the solve.
@@SimpleAmadeus or, just talk about some key points in the begging as an appetizer (entrée) and at the end they can go to the main course.
OH! I gave up after about 45 minutes, but I forgot about the green cells having to be connected.
Cave & wall rule
not a fun puzzle today.
🤷♂️
@@alvaropallete oh well. The opening was tedious.