Zeven is even Dutch for Seven!

Here's where you can play Zeven by James Kopp: app.crackingthecryptic.com/v8...
Normal sudoku rules apply. Adjacent digits along an orange line have a difference of at least four. The small numbers in each cage show the sum of the digits in the cage and digits cannot repeat in a cage.
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Hi! We're Simon Anthony and Mark Goodliffe, two of the UK's most enthusiastic puzzle solvers. We have both represented the UK at the World Sudoku Championships and the World Puzzle Championships. We're also "cryptic crossword" aficionados. Mark is the twelve-time winner of The Times crossword championship and twice Senior World Champion in Sudoku, and Simon is the former record holder for most consecutive correct solutions to The Listener crossword. We hope we can help your puzzle solving while also introducing you to some of the world's best puzzles.
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▶ Contents ◀
0:00 Theme Music & News around the channel
2:38 Rules of today’s puzzle
3:30 Start Of Solve - Let's Get Cracking!

[@Oscarefraim]

Groetjes uit Nederland we houden van jullie!❤

[@chocolateboy300]

I finished in 42 minutes. I'm so used to German Whispers that this puzzle kept messing with me about 5s and 46s. Once I stopped myself from doing that, this puzzle was quite enjoyable. Great Puzzle!

[@jdyerjdyer]

About the same for me, too, in time and the brain twisting. I definitely prefer German Whispers to Dutch, but that said, Dutch gives a chance to learn more about these difference lines. Luckily I only made one slip, having to back out of my solution about 12 steps. I instantly knew where I'd made a false assumption (and yes it was on the line in column 9) dealing with the placement of the 5 due to ruling out a few possibilities erroneously because I had slipped into thinking about German Whispers differences regarding the 6. (I had said, this can't be a 1, so 6 can't go here or here, but there was a perfectly good 2 in those cells, which in this case does allow a 6 where I had ruled it out.) C'est la vie.

[@smartass4758]

An easier way to exclude only 2 5s on the long orange line is: 5 in r1c1 would need a 1 either in column 1 or column 2. And suddenly there are three 1s in columns 1 and 2.

[@bobh6728]

Excellent deduction. Almost as easy was to check the blue digit in column 1, r4c1, and go to the blue digit in row 1, r1c8. If you alternate they end up in the same group, so there can’t be just 1 change. So the 5 in box 1 is not in the corner.

[@RecreationalCynic]

Solved it in 38:22. Looking forward to Acht next year!

[@tomgalli1188]

I had to let Mark give me a push, just to the point of realizing there had to be four 5s in the outline. After that, what fun!

[@MattYDdraig]

14:21 Really nicely themed and perfectly constructed. Not too taxing but enjoyable.

[@davidh.4944]

1:13:48@#1758. Even though my solve time is long, this is the first one in a while where I didn't feel like I was tricked by the video time. I recognized that it would take me a good amount of time going in. It has been ages since I've tackled a Dutch whispers puzzle, and I had to re-work the logic of them. There was also not too much going in to distinguish them, either, and I had to piece out the polarity by trial and error, and of course coloring.

[@Yttria]

Very relaxing and straightforward puzzle. 18:27 solve time for me.

[@cjbarthvdm]

Congratulations / gefeliciteerd. Thanks for the puzzle and video. I started immediately with 6159 in the bottom row, because there is no other way to place the 159 sequence in row 9.

[@RichSmith77]

Did you consider the possibility of no 1-5-9 in row 9? Possible, in theory, with 9-5(-1) around the corner in box 9, no? Mark had to rule this option out first.

[@TurquoizeGoldscraper]

17:56 for me - that was a fun Dutch whispers puzzle.

[@benjaminrealy5661]

17:59. Quickly learned the 159 had to be in order on each outer line. By placing candidates in the cages and realizing the 2 cell had to be 16 but 1 couldn't be corner was the break in I needed

[@geezerama]

Great puzzle. I'm well chuffed with 32:42.

[@piarittersporn]

Very beautiful and relaxing puzzle. Still tired this morning I could solve it in 24 minutes while enjoying my coffee. I loved it. 🙂

[@Paolo_De_Leva]

Yes, the perimeter is fascinating. Quite easily cracked if you know the *1-5-9* trick, but nice. All the cages are very restricted. I would say this is a great introduction to *Dutch whisper lines* for unexperienced solvers. Almost genuinely approachable. Great and insightful meta-logic by Mark to disambiguate the cage containing *1-6.* 👏👏👏 I used a less elegant approach, but I did it more quickly. I immediately ruled out *r9c1 = 1* because it would require *5* and *9* above it, and a second *1-5-9* sequence in *row 1,* which would be incompatible with the high digit in the opposite corner of the grid (r1c9). In short, I concluded there could be no *5* in the upper left corner of the grid (r1c1). I did not need to generalize to the opposite corner, as Mark brilliantly did.

[@poslfit]

At 03:38, you can then write in 6 in the top right, because the Dutch whisper lines would force the top and right sides to start with the same 72..., 84..., or 94... if the top right were 7, 8 or 9.

[@EelcoWind]

Zayven would be the pronunciation, I think.

[@never7848]

more like Zayve with a silent n i think

[@Dido01]

@@never7848 And now I wonder who is going to try and explain the also extant pronunciation of Dutch "zeuven" for that number 😂

[@Rach881101]

36:17 for me. Nice puzzle!

[@titusadduxas]

53:49 - I really struggled with the break in even though I knew the 159 trick!

[@anaayoung9142]

Nice! Your idea about 5's (2 or 4 of them), helped to understand the puzzle better and I was able to do it! 😁

[@Kirbyfan87827]

Finished in 40:24 on my own. Not too proud of my time, but I'm not really ashamed, either. Fine puzzle.

[@David_K_Booth]

My time was very close to yours, and I agree. That sums it up nicely. It's not one for a complete novice, but in my opinion a good confidence-builder for anyone who has a few successful solves under their belt.

[@Gonzalo_Garcia_]

10:35 for me. Great puzzle!

[@emilywilliams3237]

I did this without watching first in 21 minutes- entirely credited to your excellent teaching about Dutch whispers in many videos, Mark. Thanks for this excellent video and happy birthday, Isaac!

[@johnbell9074]

24 mins for me. Thanks.

[@darthrainbows]

Oooh nice, I beat Mark's time! Solved in 17:20.

[@jdyerjdyer]

Thanks for the quick and easy one. I should have followed suit and colored high/low like you did. I would have shaved a good twenty minutes off my time, probably. Still hovering around twice your and Simon's times on most puzzles, so I haven't gotten any worse. I, too almost made a ribbon, but like you, I quickly realized that it could and indeed would have to swap out in places. To get my break, I used the 16 cage and asked if there was a problem with one of the orientations. Sure enough, one quickly lead to an impossibility along row 9 with the 6789s and 2345s (1 being in the box meant that 5 either lived there or in the opposite corner) making 4 cells from 234 in the row. From there, because I didn't have the aid of high/low, I did more pencil marking the possibilities. I also didn't think about the 5 being the only place for a polarity switch for this reason, but it falls out from the logic anyway. I sort of did do high/low, just without the colors. Silly me. I hope you and Simon get a laugh out of my follies, sometimes. As frustrating as they can be for me, I do learn from them (albeit, slowly). Cheers!

[@jinkela7295]

@12:10, 5 is actually placed in the highlighted cell

[@srwapo]

25:00 even. Super fun solve! I broke it at one point, but managed to figure it out (on an orange line, 3 can be next to 7 in addition to 8 and 9!).

[@mudscuffer]

30:43 for me. Quite nice.

[@sunflower499]

I realized that the only way to place a 5 on any of the outside lines is to have 1 and 9 on both sides: 159 or 951. This is surprisingly restrictive in column 1 and got me my first digits.

[@ayakoa2096]

8:53 for me, nice approachable puzzle

[@SarahLivne]

Nice one!

[@raysouth1952]

Very enjoyable. Took me an age to determine the number of 5s around the perimeter because I kept trying to fit three of them. Duh!! Very smooth from there.

[@tonybrooms]

00:46:12

[@robertcousins2274]

22:51 for me

[@awilliams1701]

I think the 2 digit cage is 1/6 by means of sums AND dutch whispers.

[@awilliams1701]

lol yup mark got it. LMAO

[@dwdei8815]

But two is their cutest number by far. It is very twee.

[@_-_-Sipita-_-_]

15:00 for me.

[@bait6652]

So close to a dualsolve on the outer ring...only breaks at.one point in B2.... And if u change the top left 7sum to 8sum(4.to 5 in B4)...8 of the 9 boxes create a valid solution...unfortunately B2 is invalid and alot of chaos... It's why there are setters...nice rules.

[@leickrobinson5186]

Fun puzzle! And quite approachable! :-D

[@Orenotter]

Egregious mistakes I was I was making. My poor otter heart they were breaking. My sixes did clash And my sevens did crash Too bad I can't get by with faking.

[@erikbakker6974]

At 17:46 why can’t the 2 cells contain a low number and a 5?

[@erikbakker6974]

Because 5 was already pencil marked in box 3

[@craigroberts7013]

11m 05s

[@rhoadeshouse]

I might just be making fun of myself with this confusion, but i did not understand at all, why there couldn't be 3 5s, instead of eighter 2 or 4. And that i thought would be by having 1 of them in a corner and also 2 more not in a corner.

[@stangerrits6712]

Good question. There is an even number of digits on the perimeter. If there weren’t any 5’s, the alternation between high and low would work. Say you start in R1C1 with a low digit and follow clockwise going low-high-low-high-etc, you’ll end up in R2C1 with a high digit, which you need in order to close the loop with R1C1. When introducing a single five, the alternation would change, skip a step, and if you again follow the perimeter, R2C1 would end up being low, which breaks the puzzle because it can’t connect to R1C1 being low as well. If you would add a second 5, the alternation changes twice, so R2C1 would again be high and your error is fixed. So, because there is an even number of digits on the perimeter, and the perimeter is a closed loop, you’ll need an even number (or zero) 5’s to make it work.

[@RichSmith77]

It's also relevant that there's nowhere you could have 1-5-1 or 9-5-9 around the perimeter, so every 5 forces a change of polarity alternation. And there has to be an even number of them for the reasons above.

[@rhoadeshouse]

​@@RichSmith77 Thanks to Both of you, i think i now understand the work of thinking, Mark must've gone through too but it was so elementary for him (and probably most people on the channel), he didn't think someone would need it to be explained. Thanks a lot.

[@rhoadeshouse]

​@@stangerrits6712Thank you for explaining this so detailed to me, i assume i finally get it.

[@rhoadeshouse]

Is it normal for the puzzle in the android app to not include the solution, when the puzzle does so in the desktop-app? Didn't notice till today.😂

[@highpath4776]

No 5s on the 7s but three 7s makes me think of fruit machine

[@poslfit]

At 07:00, why does it have to be 16 and not say 26 or 15?

[@erikbakker6974]

Because it is in a 7 cage. Needs to sum to 7.

[@poslfit]

@@erikbakker6974 Lol, thanks!

[@Leo-ct9pm]

Why it can't be 3 5s around the perimeter?

[@felipeguimaraesvasconcelos3772]

There are 32 cells around the perimeter, which can be low, high, or 5. Except for where there's a 5, they'll alternate in high-low polarity. A five can only be adjacent to 1 or a 9, and because of the geometry of the loop it must be adjacent to both. Note that this breaks the polarity alternation: you have a low digit, 5, and a high digit. Note also that if there are 3 fives on the loop, then the line will switch polarity exactly 3 times. That can't happen, because the loop has an even number of cells, so it must have an even number of polarity-switching 5's.

[@Leo-ct9pm]

@@felipeguimaraesvasconcelos3772 Thanks for the explanation, I get it now.

[@laurv8370]

Nitpicking: "almost always when 5 appears on the line it changes the alternation, the only way it doesn't is when it appears between the same digit", well, as 5 can only appear close to 1 or 9, it does NOT change the alternation in 67% of the cases :D (i.e. 1-5-1 and 9-5-9) and it only changes it in 33% (i.e. 1-5-9) 😝

[@RichSmith77]

I think you would need to factor in the proportion of times the two ends digits on a three cell line see each other. In those instances, you couldn't have 1-5-1 or 9-5-9, but could have 1-5-9. Also, surely 1-5-9 and 9-5-1 each occur as frequently as either 1-5-1 or 9-5-9. That immediately doubles the probability, no? "Almost always" no doubt over sells it, but I could see polarity switching more times than not, when a 5 appears on a Dutch whisper. 😉

[@bramjanssen2056]

Zeven is Dutch for seven, so quite a fun double pun

[@TriforceOfCourage97]

its pronounced ZAY-VIN

[@Pritchie45]

53 minutes

[@LednacekZ]

made a complete mess of the solve. i made the mistake of german/dutch wispers.