A 6x6 Sudoku Puzzle today - But we Know they aren't Always Easy!
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- Опубликовано: 13 май 2024
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Place a subset of the digits 1-9 in each row, column, and box (the SAME set must be placed in each). Adjacent digits on a green line differ by at least 5. Digits on a purple form a set of consecutive digits, in any order. Box borders divide the blue line into segments with the same sum. Digits in a cage sum to the indicated total.
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▶ Contents ◀
0:00 Theme Music & News around the channel
2:28 Rules of today’s puzzle
3:50 Start Of Solve - Let's Get Cracking! 
Развлечения
really really thrilled to see you playing another one of my puzzles! the commenters seem to think it's quite easy but when i went to resolve it before submitting i got stuck right where you did for ~20 minutes :P. sorry about not including a solution, i changed the colors of the lines before submitting and must have messed it up, but i can confirm you got it right!
The part that made it super easy (for me anyway) was seeing that the blue line sum and the vertical renban in the middle kept pushing the minimum digit on the renban up for each digit removed from the single cell joining them. In other words, you can quickly rule out 6, 7, and 8 as the sum leaving you with 9 and from there it just plays out very quickly as you have all the digits.
More detailed explanation:
If the sum is 6, then the other three are 1, 2, and 3, but you can only hide two of the digits from the renban line, so the minimum is 3, 4, 5, and 6. The 5 can't go on the German Whispers line, so it is either 3 or 4 with the outer digits being from 8 or 9, but 8 and 9 aren't available in this scenario.
If the sum is 7, then the other three are 1, 2, and 4, but you can again only hide two of the digits from the renban line, so the minimum is 4, 5, 6, and 7. The 5 can't go on the German Whispers line, so either 4 or 6 with the outer digits being from 1 or 9. The 9 isn't available, and you can't duplicate the 1 on the line, so this scenario doesn't work, either.
If the sum is 8, then the other three are from 5 and below. Again, you can only hide two of the digits from the renban line, so the minimum is 4, but you can't reach the 8 from the 4, so it is 5, 6, 7, and 8 on the renban with the set of digits being 1, 2, 5, 6, 7, 8 in this scenario. The 5 can't go on the German Whispers line, so either 6 or 7 with the outer digits being from 1 or 2. Here that is possible with the 7, but column one becomes problematic due to the 10 cage. The 10 cage can be 19, 28, 37, or 46, but the 9, 3, and 4 are unavailable leaving only 28. Likewise, the 8 cage can be 17, 26, or 35, but the 1, 2, and 7 are taken on the German Whispers line and the 3 isn't available for the 35 combo, leaving only 6 in r4C1 and 2 in R3C1. Now the consecutive of the 6 on that renban can't be 5 because it is needed on the blue sums line making R5C1 a 7. The consecutive for the renban sticking out of the 10 cage is either 1 or 7 (3 and 9 are not available) and we already put the 7 in the column making it a 1 leaving only the 5 for the blue sums line cell in R6C1. Now where does the 8 go in column 1? The only cell left is R1C1, but the 10 cage in that box is already using a 2 and an 8, so it is broken, meaning that the blue sums line sum must be 9. From here the solve is pretty straight forward.
Since we know the blue sums line sum is 9, we know the renban goes 9, 8, 7, 6 in some fashion. That tells us the sums line has to have a 6 on it paired with a 1 and a 2 to make 9. (5, 3, 1 and 4, 3, 2 both introduce 3 different numbers, but we only have two slots not filled by the digits on the tall vertical renban.) The 6 can't go on the German Whispers because it would require a duplicate 1 in the row, so R4C4 is from 78 while the other two cells beside it are from 12. We get a 789 triple completing the bottom left box, so R5C4 is the 6. The renban in the 8 cage makes R4C1 either 6 or 7 with the other cell in the 8 cage from 12, removing the 9 from R5C1. The 10 cage is now from 19 or 28 as 37 and 46 are not available. This makes the renban cell outside the 10 cage from 1, 2, 7, 8, or 9. Now we have a problem if R5C1 is 7 as that forces a 6 and 2 above forcing a 1 below leaving only the 8 and 9 as options for the renban cell outside the 10 cage as well as an 8 or 9 for the cell above it. This 89 pair leaves only a 1 and 2 in the 10 cage which most certainly does not add up to 10.
That means that R5C1 is an 8 making the 8 cage have the 7 and 1 above it. This leaves R1C1 to be from 2, 6, and 9. Now R6C1 can't be 6 because that would leave a 29 pair in the column in the same box as the 10 cage leaving only 1 and 8, which again doesn't add up to 10. That makes R6C1 the 2, R2C1 the 9, and R1C1 the 6. We now have a 28 pair in the 10 cage, a 16 pair in the bottom left box in row 6, and a 79 pair in that same box in row 5. This leaves a 78 pair in row 6 in the bottom right box and another 12 pair in row 5 in that box as well. More importantly, the row 4 German Whispers line is sorted as 8 in the middle, 2 on the left and 1 on the right. We also get 7 in R3C4 to complete that renban line. The 12 pair in the bottom right box is also sorted.
The 9 in R2C1 makes R2C2 the 8 and R1C2 the 2 sorting the 12 pair in column 4 and placing a 7 in R2C5 to go with the 2 on that German Whispers line. This sorts the 78 pair in the bottom right box. The 6, then the 8, and then the 9 get placed in the top right box going counter clockwise. The 6, then the 2, and then the 9 go in clockwise for the box below it. The 7 and 1 finish the top left box, sorting the bottom left box pairs. Finally, the 6, then the 9, and then the 8 go into the remaining box (center left) in a clockwise order to complete the solve.
Once you see this key piece of logic, the solve can easily be done in under 20 minutes. It took me only 12 minutes to run through all this logic and type it up at the same time. The only slow down points are the two times you have to use column 1 to get a contradiction in box 1 with the 10 cage sum.
@@jdyerjdyer thanks for your explanation! the part i didn't notice while setting was being "immediately" able to eliminate 8 as an option for the highest digit (aka eliminating 5 as the lowest digit). my intended solve path was much closer to how mark did it - coloring each digit in order to eventually place them in the cages and disambiguate, but i hadn't considered just running through the cage candidates like that lol, great work :D
It is really refreshing that world class setters such as yourself make the same silly mistakes that I do
@@nathanmays7926 (assuming you're referring to me and not j dyer, who really is a world class setter): wow i would not refer to myself as a world class setter, in fact i'm quite new to this whole world! but thank you :D silly mistakes are just a part of the process lol
I must say Mark's meta-logic about *X* was stratospherically clever, as I pointed out in my separate comment, but my solve was based on testing the values of the *blue line sum.* Rejecting *6* and *7* was easy, as explained by *Jay Dyer* above. Here is my way to reject *8:*
🔹Since the long renban must contain *X* (a digit from the left segment of the blue line) then it must contain *5, 6, 7, 8* and...
🔹...its intersection with the German whisper line must be *7,* with *1* and *2* on its left and right sides.
🔹Hence, the *10-cage* must contain *2* and *8*
🔹By sudoku,
*r3c3 = r5c1 = 8.*
🔹By the short renban in column 1, the *8-cage* must contain *7* and *1,* with *r4c1 = 7* , which is incompatible with the *7* at the intersection between whisper and renban.
QED 😏👍
By the way, @13:01, Mark missed the logic step that would have immediately cracked the puzzle, making his whole solve a logic masterpiece (see reply to my separate comment for details), and this proves he is human as well
QED 😏👍
First time I have ever got a better time than Mark! More pencil markings on the 2 digit renban lines would have made it much easier rather than all the colouring. Once you realise that the renban in c1r45 contains a 7 then it all solves quite quickly. Dont worry Mark you are still beating me about 400 to 1 but I am now on a roll!
Simon calling him "Pencil Mark" so often must have gotten to him 😁
I feel Mark could have easily dismissed the 7 from R6C4 on the identical grounds he dismissed 6 (if a 7, then the arrow is 124, so the renban must be 7654, and none of 654 can be on the intersecting whisper; impossible, so no 7) and would be ahead.
Once you place all the 1-2 pairs in the grid, you can deduce that the renban in column 4 must be 6789; if it were 5678, the cage in Box 1 would be a 2-8 pair, and the 8 would need to have a 7 next to it on the renban, but placing a 2 in the 10 cage forces the other cage to have a 1 in it, forcing a 7 beneath it in the same column.
You sort of stumbled into this, mentioning that in both possibilities for the box 1 renban it has to have a 9 on it. Once you get the 6789 line in column 4, the sum line is forced to be 1+2+6=9, meaning light blue is 6. Dark blue is therefore 7, which disambiguates the 1s and 2s, which then gives you an 8 in the 10 cage, and from there it's easy to fill in the rest.
This is more or less the same way I cracked it, although I pencil-marked 89 as the options for the bottom of the 10 cage, saw that if it was 8, then 1+7 would end up in the 8 cage because of the 1-2 colouring, forcing 9 to partner 8 on the renban in r2. So there was always a 9 on that renban, either way. Therefore long renban in c4 had to be 6789 to get a 9 in the puzzle.
@@RichSmith77 This was also my path, and I find it way more satisfying than all Mark's grunt coloring work
I loved this - a great demonstration of the use of color! Thanks, Mark, as always!
Nice! After I "try" the blue line with the purple line together with all the combinations for 6789, I get that we need 12 pair on the blue, and either 5 or 6, making a 8 or a 9. Then I pencilmark a lot and used colours to get the solution right! 🎉
Ps: I see, we did almost the same way 😊
Because that's the way to solve the puzzle...
Wow what a thrill. I believe that this is the first timethat I have beaten Mark. 14:04 no with guesses. Lovely puzzle.
What I did is realize that 6 had to be on the renban, but 5 and 6 couldn’t both be on the renban, or it would break the region sum. That gives you a 78 pair and makes the region sum a forced 9
How does 1258 on the blue region sum line and 6758 down the renban break the region sum line?
6:53 for me. Blimey, I've never "beaten" Mark by this much. It's crazy how much a puzzle can just fall into place for some but not others. Usually I'm one of the others.
A thrilling 4:34 for me! The four-long renban to blue line logic might as well have been made for me with how quickly that came to me. And the rest of the puzzle came together quickly once I had the intersection sorted out.
At 13:00, you can ask where the R2C2 digit goes in column 1. It can only go in R5C1, allowing only a 7-8 pair on the column 1 renban. The rest of the puzzle then resolves quickly.
Yes. A simpler way to say it is: if the 10 cage has a 9 in it, you can't put 9 in C1.
I finished in 14:35 minutes. This puzzle agreed with me very much and I flew quickly through the logic. I very much enjoyed this one, short and sweet. Also, it always feels good to beat Mark's time, even though he has to stop and explain the logic to the audience. Great Puzzle!
4:54 for me. Very cool puzzle!!
Another fun, nice, and quick one! Thanks!
That purple vertical line pulls a lot of weight in this puzzle. :)
8:28
Breaking in from the intersection of the renban and region sum lines was relatively straightforward but still a very neat puzzle.
8:25 finish. A very enjoyable little puzzle, with some nice logic. Excellent!
7:44. Nice, short puzzle that took some thinking.
Finished in 5:55
With all possible moves in the blue line (123/6, 124/7, 125/8, 134/8, 126/9, 135/9, 234/9), and the relationship with the renban in col.4, it's pretty easy to find that there is only one remaining way to do it.
After finding it, it's quite easy to finish it.
11:03 for me! I'm kinda shocked how smoothly it went
I don't normally come anywhere close to Marks times but I did this in 3:39. Putting 6 or 7 in r6c4 breaks it quickly, so that must be 8 or 9 and 8 doesn't work. That may be bifurcation but in this case it's much quicker than colouring!
This puzzle I like, yes I do.
And now that the puzzle is through
All the work that I did
Made a beautiful grid,
Each digit a differing hue.
10:28 for me. Usually I don't bother with the subset 6x6's, but the setup was compelling enough to give it a go. After some key deductions, the puzzle unraveled quite nicely!
Okay, I didn't do this logically. I just ran with the first set of digits to come to mind: 126 adding to 9 (blue line), 9876 on long purple line, and 12 getting 5 away from 6789 mostly. It helped that 12 + 89 gave 10 and 12 + 67 gave 8. I kept going, and going, and going, until the sign lit up saying my solution was correct. I had to color 12s.
The end of the video: your solution is the same as mine, which the window said was correct.
Finished in 6:06. Not sure if it's just me, but this seemed like a fairly straightforward sudoku solve. Maybe not good as an introductory puzzle, but if you understand the limitations, it's flows pretty quickly.
Fun puzzle!
I did manage to do this puzzle entirely on my own, without watching the video (which I usually can't do) but I did so in a very inelegant way. I started by realizing I needed at least one high (6-9) digit to be the single cell of the blue line. I then looked at what would happen if the center cell of the 3-length Whisper was high or if it was low, and concluded that in all cases I could up my requirements to at least 3 high digits and at least 2 low digits.
With this knowledge, I turned to brute force and simply made a list of every possible combination of 6 digits. Then I crossed off any sets which didn't have the right number of high&low digits, crossed off the ones without a run of 4, crossed off the ones where the lowest 3 digits summed to more than 9. Luckily this worked out and left me with just two possibilities, and a bit of clever thinking about what could go in R5C1 narrowed it to a naked single and cracked the puzzle wide open.
I guess I can fall back on the old adage that "it ain't stupid if it works."
Nice puzzle. Pretty straight forward with colouring. Had fun yelling at the screen when Mark missed, for quite some time, the obvious sudoku placing r5c1 in r2c2.
Done in 12:16. For me a good time. Great fun.
solved in 4:25 - cool setup
Amazing puzzle! ❤ I love it!
Easy puzzle, but Mark's stratospherically smart meta-logic deserves a standing ovation 👏👏👏👏👏 👏👏👏👏👏👏👏
Impressively smart use of *X* as a placeholder for the shared digit, which was brilliantly shown to be at the same time:
🔹the smallest on the long renban
🔹the largest on the left segment of the blue line.
🔹3 apart from the blue line sum
I used a much dumber method to test and reject the following hypotheses:
🔹blue line sum = *6*
🔹blue line sum = *7*
🔹blue line sum = *8*
@13:01, Mark missed the logic step that would have immediately cracked the puzzle, making his whole solve a logic masterpiece: by plain sudoku, *r2c2* must be mirrored into *r3c3* and *r5c1.*
In short,
*r2c2 = r3c3 = r5c1 = 8* or *9*
8:51 for me, loved the idea! Can there be something similar for 9X9. I am going to try making one
I did that in 4:59 but had a wild guess based on the sum of the blue line
Short and sweet. 12:36.
"The constructor is called *Teal,* so maybe we should use shades of blue and green"_ (Mark @13:55)
🙃👍
My first discovery was that I needed 1 and 2 on the blue line because the digit above the place where purple and blue meet is at most 3 different from the digit where they do meet.
10:30 for me. Felt pretty straightforward.
6:34 for me, but I didn’t use pure logic. Somehow very early I started speculating that the equal sums line was 9 and played with what that would force. Once it started clicking I just kept going. I’ll chalk my quick solve up to a lucky early guess. Still fun!
Nice puzzle. got it in 11 minutes by using letters for the 4 highest digits for a while
I got there by filling in the renban on row 2 as 789, 89, then noticing that 7 on that renban forces 2 in both cages, leaving no place for 2 in box 5. That means the renban has to be an 89 pair, and therefore 9 is the "real" digit.
8:38. The intersection in box 6 was crucial for the break-in
15:46 for me
nice puzzle
the run either goes 5678 or 6789. but once you get to the stage you were at @17:36 with the two boxes. no matter which way round the 1/2s are, there can never be a 7 in R2C1 on the purple line because either theres already a 7 looking at it from the 8 cage, or its neighbor is 9. so the line is an 89 pair. meaning 9 is on the grid and the full number set discovered as 6789. after that they all fall into place
20 mins for me. Thanks.
5 OR 6 was the big breakthrough here. Fun solve!
18:28 for me. Nice puzzle!
15:06 for me, got stuck for a bit figuring out 125/8 vs 126/9.
I noticed the 6-9 combo on the two left renbans failed pretty quickly because of lack of space to fill in the 9s. But Mark otherwise schooled me in the initial logic. Very nice puzzle.
At around minute 15 when Mark identified the cafe with sum 8 have digit 67 at the lower cell. Immediately you can deduce that cage with sum 10 cannot have 19 pair. If 9 was part of sudoku digit then where would 9 go in column 1. Only place it could go in column 1 was at box 1. So cage with sum 10 cannot have 9 in it. That deduction makes the solve much quicker.
20 minutes. That cost me a few brain cells. Seeing the middle of the German Whisper couldn't be the 7 because it would break the 8box to it's left opened it up for me.
Blue line in box 5 can’t have 125, because that makes the dark blue cell a 6, and then both the 8 and 10 cages need a 2, which doesn’t work. Also makes the renbans in boxes 1 and 3 need a 7, which also doesn’t work.
11.09 for me. Interesting puzzle.
This was painful, but seing Mark struggle for once is actually enjoyable in some sense.
i want to see mark do birthdays and other things like Simon does.
Let's Get Cracking: 02:21
And how about this video's Simarkisms?!
Sorry: 7x (08:45, 09:05, 13:38, 16:36, 16:43, 19:08, 20:49)
Hang On: 4x (04:45, 04:48, 04:48)
By Sudoku: 2x (19:20, 22:43)
Surely: 2x (03:45, 21:20)
In Fact: 2x (20:14, 20:18)
What Does This Mean?: 2x (03:25, 17:01)
Clever: 1x (24:32)
Lovely: 1x (00:16)
Beautiful: 1x (04:41)
Obviously: 1x (22:19)
Weird: 1x (05:24)
Most popular number(>9), digit and colour this video:
Seventy Eight (4 mentions)
One (52 mentions)
Blue (38 mentions)
Antithesis Battles:
Low (4) - High (1)
Even (2) - Odd (0)
Lower (2) - Higher (0)
Lowest (4) - Highest (3)
Column (6) - Row (3)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
This 6x6 is super-easy, though :)
And very smart!
7:38 for me
Spoiler: Another way to reveal the identity of the digits is to realize that Mark's analysis of the digits in the 8 and 10 cages leads there to be a unique pair of digits occupying the renban line in Box 1.
Damn... 7:20
Lucky guess with set 1 2 6 7 8 9
Cool, done in 5:55
Also, Don't forget that 6x6 sudokus all have roping.
Horizontal roping, but not vertical roping.
What’s roping?
@@Zommbaquincy For a 3x3, it's when 3 digits in a row/column appear in an adjacent box's row/column, causing a pattern where you can map a consistent pattern for all the digits for all 3 rows/columns
Simon can explain it way better than I can: ruclips.net/video/Nv2B5juhC4Q/видео.htmlsi=9vWEeg4gdD3NyMan&t=263
Roping is used with 3 strings to make the rope 🪢 stronger. In 9x9 sudoku grid you may have roping when you have 123 on the first row in the first box and 123 on the second row in the second box... It makes you have 123 in the third row in the third box.
In 6x6 grid you have only 2 boxes in row 1 and they occupy only one more row.
@@joshuahitchins1897I see, thanks!
I found a much simpler way I found compared to coloring- once you figure out that r2c2 is 89, then r1c2 is 789, and r4c1 is 67 BUT-
if it's 6, then r5c1 is 7, and the 4 digits are 126789 and no digit will fit in r1c1 in that situation.
00:11:24
Marc is going for a way too hard break-in. With the max difference of 3 in the long purple line, it was quite obvious that R6C4 is the biggest number on the line, and R5C4 the smallest because it will also be on the blue line with a 1 and 2 on that line creating the maximum 3 difference allowed.
11:32 for me.
5:07 for me.
Whew, done in a blistering 5:18. Almost never beat Mark on shorter puzzles. Hurrah! 🎉
Not difficult at all but an interesting diversion during my lunch break. Fun puzzle.
15:14, I restarted once and stared at the puzzle for like 5 minutes the second time before realizing that I was treating the purple line in column 1 as a green line. 🤦♂️
what a complicated solve
It's all about discovering the break-in which, to me, looked very obvious. 5 mintues for this one.
A lot of people posting fairly fast times in the comments. But how many read the rules carefully enough, and took the time to prove that repeated digits arent used? Mark overlooked this in the video too. There's a ton of potential for the German whisper line in R4 to contain a pair of 1's.
By the mathematical definition of subset, it cannot contain repeats, because the mathematical definition of sets cannot contain repeats either. Given a set S of 9 elements, a subset of S of 6 elements is comprised of 6 distinct elements picked from S.
@@hisham_hm as the setter of the puzzle i'd like to second this: my intention was that the rules excluded repeats in the digits used, though perhaps i should have made that explicitly clear.
Yarg. Yet another "easy" (short video) puzzle that gave me a nightmare of difficulty. I have had such a string of them recently.
I managed to get things down to the final 5/6 question, but then no matter what I tried I kept getting conflicts. I restarted and got the same problem. Only after another age of man did I finally notice I had a wrong pencil-mark on my second run, and got to the end.
I don't really deserve to post a time today, but my second run came out at 39:27@#2420, with about another equal amount for the first run. Frustrating.
Haha, 've beaten Mark's time, was about of twice faster!
5:35 for me. The easiest trick was to note that in order for whatever digit was on both the renban and sum to work, it had to have three adjacent digits, but one of which maximally had to appear on its sum, with a 1 and 2. So that forced two of the digits to be 1 and 2, and then 6789 fell into place.
Took me 10:13. I should have solved it faster. It was extremely simple.
Working out the blue line: what is the possible difference between r6c4 and whichever digit appears on both its purple and blue lines? Only 3, as you need a difference that is a sum of two digits. Later between the two two-cell renbans, can you possibly have four different digits in those renbans, and given the placing of all the 1s and 2s, the answer is "no", and the r2 renban must use the top two digits, while the c1 renban must use the two middle digits of the four high digits.
Fun puzzle. Took me just over five minutes. The key observation that Mark seemed to miss is that r4c1 or r5c1 must be 7.
What a fun one! short and sweet but not too simple.
wouldnt it be funny if it was in the software and it was wrong?
Note to self, if the video is longer than 20 minutes, don't even bother trying.
_ed_ Stop telling me how easy it is. The break in requires some kind of second degree logic that goes right over my head, and the '6 out of 9' rule is completely foreign. I could probably beat 90% of you at picross.
_ed2_ There's no way to fit the 123 a 6 requires on the green/purple cross without breaking one line or the other. 7 Fails for the same reason once you narrow down the candidates for column 4 via other means. These in turn remove three and four as candidates from the purple line. After that pigeonhole principle limits r4c3 and r4c5 to 123 via the blue line, and by killing 4 as a candidate, it limits the blue line candidates, taking out 3 in the process.
This one is ridiculously easy and most of us are beating Mark by several minutes.
@@TheMeanderingduck6 I'm not. :( It took me blinkin' ages to narrow down the 4567-5678-6789 possibilities though. In the end I did it with the ten cage at the top left. Working out that there HAD to be a nine - since if it was a 1-9 pairing then obviously there's a nine, but if there's a 2-8 pairing then the renban on the left must be a 1-7 and therefore the only thing that the 8 can be consecutive to is a 9 (since the 7 is now taken) - turned it into a fairly straightforward wrap-up for me.
I feel like Mark's solution was maybe overcomplicated, but a lot prettier than mine.
@@TheMeanderingduck6have you considered that some of us are stupid (it's me, I'm stupid)
Only way to get good is to try! I like to attempt them all myself, and then skip through the video when I get stuck to see the next step. I've recently started getting sub-video-length times, all from just solving alongside Simon and Mark!
@@H0lyMoley So what I did was started by assuming 6789 were the candidates for R6C4, and realizing that the only way to have a digit fit on both the sum and renban (since at least one had to overlap) was having the digits on the sum be 1, 2, X-3, and the remaining renban digits be X-1, X-2, X-3, giving 6 distinct digits of X, X-1, X-2, X-3, 1, and 2. Now since one of the X values was on the renban, the german whisper intersection had to be flanked on both sides by 1 and 2, meaning that intersection had to be 7, 8, or 9. Now knowing that 7 had to be on that renban, 3 couldn't possibly be anymore, which meant X-3 had a minimum of 4, ruling X out of being 6. Now, X can also be ruled out of being 7, since the whisper intersection would then have to be greater than 7, and we already proved that X is the highest value on our renban (since one of the other digits on the renban is 3 less than X). So now we have X being 8 or 9. From here, we can see that if X was 8, it would get forced into box 1 in column 1, and we wouldn't be able to make 10 with any other combination. The rest sort of falls into place.
i am not ashamed to bifurcate ... not by any means. one big advantage to bifurcation is that it often reveals the essential PATTERNS.
0) "trivially" obvious is that r6c4 is HIGH.
a) the lower part of box 5 shows up as the upper part of box 6.
b) r5c4 must be WITHIN 3 of r6c4. now SUBTRACT these 2 digits (yes, it only goes one way). the difference must be 3 OR LESS. but the 2 OTHER digits on the long part of the blue line must add to AT LEAST 3. therefore they must be EXACTLY {12}.
c) the remaining 2 digits on the long renban must FILL IN the gap between the other 2. therefore the {12} must be ABOVE the long renban.
to continue, simplest is to just do some easy casing. consider the long renban.
d1) {36} and {45} breaks c4 r4
d2) {47} and {56} breaks c4 r4
d3) {58} and {67}
d4) {69} and {78}
therefore, the digits in the puzzle must be {12678} and ONE of {59}.
the entire grid can now be colour-coded as (definite) LOW and "HIGH" (which might include a 5).
there are several ways of proceeding now. one is to track the "maybe-5" cells around the grid - there are only 6 possible such cells. (*) NOTE that these are only "possible 5s", and when they get converted to HIGH at the next step, they still have multiple candidates.
the cages and remaining renbans can now be examined. the col1 renban must have a 7 on it. therefore the box 1 renban must now be (89). this eliminates 5 from the 6 available puzzle digits.
etc.