There were quite a few comments from our visitors, especially on the Daily Killer Sudoku page, which makes that and other pages load slow. Therefore, I have moved some of the old comments to this new category.
If you’d like to read what other people had to say about our puzzles you may find the old comment by clicking on the “comments” link below or here.
UPDATE (some 5 years later): I reopened the comments on all pages. Feel free to add your comments there, it means a lot to me!
Udosuk – I just uploaded the 3 missing solution files that you requested. Thanks for keeping the log – if you think there are other solutions that are still missing, just let me know.
Frank – you can try talking to the publisher directly: viaoptima@mail.telepac.pt
I’m not sure what the conditions are for ordering a copy of the book.
Cheryl, please let me know which solutions you are missing and I will upload them in to the archive.
The solutions to Daily Killer and “Daily” Sudoku (which don’t actually appear “daily”) can be found in the archive.
The solutions to Samurai, Killer Samurai, mini-Killer Samurai and Harakiri are always attached to the same post as the puzzle itself. These can be found under “samurai” and “overlapping puzzles” categories, found in the right-hand sidebar.
Jean-Christophe – Great puzzle!
I convinced myself the 4 corner posts were {1,2,2,3} and carefully positioned them accordingly, only to be confronted by an 8(2) cage from which the {1,2,3} were excluded. Oops.
After a restart it took me about 90 minutes.
Many thanx.
Hi,
I’d like to show my solution to Jean-Christophe’s great killer, “The-Frame”, because it involves solving techniques that I never used before, and it might be of interest to some people….
It is also the first time I’m using the killer lingo, so prepare yourselves…..
***SPOILER***
How to solve Jean-Christophe’s “The frame”.
1.
Some name-calling…
r9c1=X, r9c2=A, r8c1=B, r1c1=C, r1c8=D, r1c9=E.
X+A+B=6={1,2,3}.
2. “45” on R1,R9,C1,C9 (aka “the frame”) -> r1c1+r1c9+r9c1+r9c9=8.
This can be in only 3 options: 1124, 1133, 1223. Or, more graphically:
1 4
2 1
1 3
3 1
1 2
2 3
All options can be rotated and/or mirrored.
The 5(2)-cage in C9 rejects the 1133 option.
3. “45” on C1 -> r1/2/3c1 = 12+A -> r1c2/3 = 18-A.
4. “45” on R1 -> r1c1/8/9 = C+D+E = 5+A = 6,7 or 8. In all cases D is less than 6.
5. “45” on C9 -> r7/8/9c9 = 12+D -> r9c7/8 = 19-D. This means that D is at least 2.
6. If D is 2 then r9c7/8 is 17={9,8}, which will reject all options for the 15(2)-cage in R9.
7. If D is 3 then r9c7/8 is 16={9,7}, which will also reject all options for the 15(2)-cage. So D is 4 or 5.
8. Since D is at least 4, then C+D+E is at least 7. And since C+D+E = 5+A, then A is at least 2. So A is 2 or 3, and B or X is 1.
9. If D is 4 then r9c7/8 is 15, and together with the other 15(2)-cage in R9, the only possibility for the 8(2)-cage is {5,3} (both 6 & 7 are used by the 15(2)-cages..) and we’ll get B=3, X=1.
If D is 5 then r9c7/8 is 14, which is {8,6} or {9,5}. The {8,6} option will reject all options for the 15(2)-cage. The {9,5} option will force the 15(2)-cage to be {8,7}, and the only possibily left for the 8(2)-cage will be {2,6} and we’ll get B=2 and X=1.
So…., whether D is 4 or 5, X is 1 !!! Got our first number !
10. According to step 2, since X is 1, E can be only 1 or 3. If E is 3 then Since C+D+E is at most 8, and D is at least 4, we’ll get C=1, which cannot be, becasue both C & X are in C1. Therefore E=1.
11. According to step 2, since X=E=1, C is 2 or 4. If C is 4 then C+D+E is bigger than 8, and it cannot be. So C=2, and we get A=2, B=3, D=4, r2c9=8, r9c9=4 (step 2), 8(2)-cage in R9 is {3,5}, 5(2)-cage in C9 is {2,3}.
The rest is more simple:
12. Because r2c9=8, then the 15(2)-cage in C9 is {9,6}, and therefore r7/8c9 is {5,7}, and therefore r9c7/8 is {9,6} and the 15(2)-cage in R9 is {8,7}.
13. “45” on N9 -> r7c6=r8c7 -> the 14(4)-cage is {1,2,3,8}.
14. “45” on R8 -> r8c2/8/9 = 20 = {9,8,3} / {9,7,4} / {9,6,5} / {8,7,5}. {9,8,3} is rejected because r8c1=3, and according to step 13, only {8,7,5} is possible, and we get r8c8=8. r8c7 cannot be 3 (r8c1) or 2 (will cause r8c6 to be 8), and has to be 1, and we get r7c6=1, r8c6=9, 12(3)-cage={2,4,6}.
15. “45” on R7-9 -> r6c1/3/4=17. “45” on N4 -> r4c2+r6c1/3=9, so r6c1/3=8, r4c2=1, r6c4=9.
16. 12(2)-cage in C1 is {4,8} or {5,7}. It cannot be {4,8} because r6c1/3=8. r6c1 cannot be 7 (it will cause two 1’s in N4), so 12-cage is [5,7], and we get r6c3=3, r8c2=5, r7/8c9=[5,7], 15-cage in R9 is [8,7], 15-cage in C1 is {9,6}, r2/3c1=[4,8]…
And you carry on….
The times has produced a Killer on a Sat 14 Jan. This is a change for them.
See http://www.timesonline.co.uk/article/0,,18209-1984510,00.html
Hi,
Different persons, different solving approaches. The way Shai started to solve this killer is indeed very interseting.
I’ve written a page with the solving steps, well my solving steps 😉
http://jcbonsai.free.fr/sudoku/TheFrame.html
1. Thanks for djape’s kind response to repost the 3 solutions I acquired. Successfully downloaded & completed the “collection”.
2. Regarding Shai’s killer lingo (January 13th, 2006 at 7:21 pm), I agree the most of them but could I suggest the omission of “/” in cell combinations & “,” in sets/ordered sets?
That’s because we generally are dealing with single digits here so I think separators are not that necessary. For example, r1c1/2/3 could be r1c123 could be {1347}. That would look more concise I reckon…
Also, for consecutive sequences I suggest to use “..”, e.g. {12345} could be {1..5}. This is important when we’re doing range manipulations.
Finally I must praise Nate for the innovative idea of using “*” to denote necessity in probable sets e.g. {67*8*9} meaning {689}/{789}.
3. For the sequence question, the answer (if anyone’s still interested) is 41110 because the whole sequence is in fact the powers of 2 in “permutational base”. The next 2 entries are 122220, 250110. To recap, the sequence asked is:
1,10,20,110,220,1110,2220,10110,20220,…
So we can go like this:
2^0 = 1 = 1*1!
2^1 = 2 = 0*1! + 1*2!
2^2 = 4 = 0*1! + 2*2!
2^3 = 8 = 0*1! + 1*2! + 1*3!
2^4 = 16 = 0*1! + 2*2! + 2*3!
2^5 = 32 = 0*1! + 1*2! + 1*3! + 1*4!
2^6 = 64 = 0*1! + 2*2! + 2*3! + 2*4!
2^7 = 128 = 0*1! + 1*2! + 1*3! + 0*4! + 1*5!
2^8 = 256 = 0*1! + 2*2! + 2*3! + 0*4! + 2*5!
2^9 = 512 = 0*1! + 1*2! + 1*3! + 1*4! + 4*5!
2^10 = 1024 = 0*1! + 2*2! + 2*3! + 2*4! + 2*5! + 1*6!
2^11 = 2048 = 0*1! + 1*2! + 1*3! + 0*4! + 5*5! + 2*6!
Hope it wouldn’t appear too messy…
Correction: In (2) para 2, I meant to say {1,3,4,7} could be {1347}… somehow it didn’t come out right.
Also, in (3) the “permutational base” it could also be called “factorial base”. Basically we use 1!, 2!, 3!,… to replace 10^0, 10^1, 10^2 as in decimal base system.
Now I could download Jean-Christophe’s new puzzle and try it out!
I wonder why nobody is discussing about the IQ on Fri Jan 13 (156781). I managed to solve it but only after very finesse analysis in the 13-cage in r6..9c9, which I don’t like. Wonder anybody has a clear & elegant route?
Udosuk – IQ on Fri Jan 13 (156781).
The trick is to place the number 6 in nonet 9. It cannot go in the 13 cage. It cannot go in the 12 cage. It cannot go in the 10 cage. So it goes in the 21 cage. That then becomes {6,7,8}. The 12 cage then becomes {3,9}. The 2 remaining options for the 13 cage both contain a 3, so that goes in r6c9. That then makes r9c6 = 8 since these 2 sum to 11.
Hi,
About the killer lingo, I agree w/ Udosuk’s suggestions. They are OK, and we can use both syntaxes, it shouldn’t be a problem.
“r1c1/2” = “r1c12”
“{1,3,4}” = “{134}”
“[2,5]” = “[25]”
I’ll update my file,
About ND’s suggestion I found it a bit problematic. A good lingo should be quite readible even if you don’t know the “rules”, and besides, Udosuk’s suggestion interferes with Nate’s:
“r1/2/3c2 = {6,7,*8,*9} would mean either {6,8,9} or {7,8,9}” is OK, but once you write {67*8*9}, it gets complicated (need to remember if the * comes before or after the digit), and not quite readible for the new players.
I’m still waiting to hear DJ’s opinion on the whole lingo issue.
About Udosuk’s riddle:
This is a truly debolic riddle from hell. I’d like to say you are crazy, but then Mike from Japan would probably suggest we fight it out….:)
Do you know of anyone who solved it ???
And finally,
Hints on Friday’s IQ (156781)
1. “45” on C1 -> r9c1/2 = [6,5].
2. “45” on C9 -> r1c8/9 = [7,9].
3. “45” on N1 -> r1c3=r4c1+3. The 11-cage in nonet 1 is {1,2,3,5}. r4c1 must be 3, beacuse r4c1=1 will cause r1c3=1+3=4 (can’t have a 4 in a 8-cage of 2 cells), r4c1=2 will cause r1c3=5 and then we have 2 5’s in nonet 1 (the other one is in the 11-cage), r4c1=5 will cause r1c3=8, which is also impossible in an 8 cage of 2 cells.
4. Now we get r1c3/4 = [6,2], 15-cage in nonet 1 = {8,7}, 16-cage in nonet 1 = {3,4,9}.
5. Now in nonet 4: 23-cage is {9,8,6}, so 11-cage is {4,7}, and r5c3,r6c2,r6c3 = {1,2,5}, and therfore looking at the 10-cage will show that r7c2 is 3,4 or 7.
6. “45” on N7 shows r8c4=r7c2+3. so r7c2 cannot be 7 (7+3=10).
7. 17-cage in nonet 7 = {8,9}. If r7c2 is 4 then r8c4=7, and then we get two 7’s in the 20-cage (the 7 in nonet 7 must go in the 20-cage!), so r7c2=3 and r8c4=6.
And you carry on.
Hm, well, you could use underline or boldface to indicate necessity (i.e. {6789})but then that gets into HTML stuff rather than “pure” notation….. yeah, I think if necessity is to be indicated via asterisks then commas in sets are needed. On the other hand they are definitely not needed in columns & rows. — I do think that the “necessity” notation is useful as it’s a concept that comes up again & again in solving puzzles.
udosuk’s use of double dots for ranges is a good choice–I often use a hyphen (1-5 = 1,2,3,4,5) but that could easily get confused with a minus sign whereas the double-dot is unambiguous.
I’m not very good at keeping track of my solutions–I did both Jean-Christophe’s puzzle & the Friday IQ (156781) but forget my exact steps!
Hm, one idea on removing the commas from unordered sets: use internal brackets:
{67[89]} would work.
But the problem is to my eye it looks like it’s suggesting that 678 or 679 are the possibilities….. hm.
Nate- I took your suggestion & did ND 3,4, & 5 with only minor difficulty. I just downloaded ND 6 & will then try ND 7 again.
Well todays Brain took Pat just 10 minutes to complete. Are there any tough ones out there?
Adele
Australia
8.45pm
How about something like:
{67}+8+9
Even if you don’t know the lingo, it’s clear that there must be 8 and 9.
With ordinary sudoku (not killer),does anyone know what the least amount of numbers given at the start is? Obviously the puzzle must be able to be solved legitimately without T & E. The least I can find is 24.
Mike Japan
The minimum nb of given for a unique regular sudoku is 17. There is a site entirely dedicated to this question. See :
http://www.csse.uwa.edu.au/~gordon/sudokumin.php
Thanks Jean-Christophe.
Re your puzzle, I knew that r1c1=1, or r1c9=1. Using the former it made it impossible to make the 8 cage on r9 so the latter had to be 1. The next two numbers I got were r2c3=1 and r4c2=1.
Great puzzle, better than #1.
Mike Japan
Will wait till you all decide on how to write details of how you did them before wading through all the suggestions. Only look at them when I’m stuck, and recently that hasn’t happened (expect ND’s 7th!!). Today’s BRAIN (57190) was more like an ‘easy’ (agree with Pat, Adele). I strung it out as long as possible but I didn’t have to stop to think once. PLEASE can we have a more difficult one tomorrow? It’s raining out and I’M BORED! Can’t find any other sites on the Internet other than those mentioned here. I like those with >,
I agree with Gaynor. While I still enjoy the other puzzles, the only ones I find truly challenging are the ‘IQ’ and ‘insane’ ones.
I agree.
Todays Brain was very easy. Didnt think about any of it. It just slipped into place.
Dave
1. Thanks for Frank & Shai’s rundown on the “Black Friday IQ”. I think Frank’s idea of elimination of all choices of cages the 6 could go in nonet 9 is very ingenius, and solve my bottleneck where I needed minor T&E to break down the 13-cage… And of course Shai’s walkthrough is classy as ever…
2. Regarding the “notation for necessity”, I think we could do it this way: {(67)89}, and avoid square brackets against confusion with ordered sets. Also, we could put all “possible but uncertain candidates” inside a pair of round brackets at all times. For example, if we know a 15/3-cage being 3,4,8 then we can use {348}, but if it must come from 1,2,4,5,9 then we can show it as {(1245)9} since 9 is the only locked candidate in this cage. And if they are 2,3,4,6,7,8 then make it {(234678)} since nothing is certain here. Just a thought…
3. For Mike’s question about minimum clues, I suggest you go to the “General topics” forum at http://www.sudoku.com/forums where there are very in-depth discussions about this topic and many other mathematics issues. As far as I know right now 17 is the minimum for non-symmetric sudoku and 18 for symmertic ones. Also there is a 16-clue “pseudo sudoku” (with exactly 2 solutions) discovered which was still unique from what I looked up last time.
4. To Shai about the “crazy” riddle, I must say I didn’t invent it nor did I solved it myself when I first stumpled upon it. But I have quite a few of these up on my sleeves and if you’re interested I could send you a link to a forum thread dedicated to these kinds of riddle where I made my share of contributions (under another alias)…
Hi
I am just getting into these Killer Sudoku puzzles!! Can someone explain how the nonet 45 rule is used? I understand 1-9 adds up to 45 but so what?
In particular I am struggling to get started on Brain 97589.
Thanks in advance.
Gary
Udosuk,
Give me the link (and tell me what your other alias is….). tnx.
Gary, I suggest you to go through the “solving tips” category found in the right-hand sidebar. Look for explanation of “innies” and “outies”. You might need to go through a couple of pages to find it because it’s one of the earliest solving tips posted.
Also, you need to start with some easier puzzles (find a “cool” or “easy” in the archive). Ignore these lads here who claim “brains are easy”, a few months ago they wouldn’t know where to start either! 🙂
Good luck!
Shai,
The link for riddles/brain teasers is
http://www.techenclave.com/forums/brain-teasers-187.html
The rule is simple: you answer the current question right, you get to post the next one.
There are about 30 pages now, but the thread has gone relatively inactive for quite some time…
I used the alias “yakshaman” and didn’t start posting until about 20 pages…
There were quite a variety, some were quite silly and some were awfully hard, which were mostly posted by me… ^_^
Hi,
after reading udosuk’s list of overlapping puzzles I checked my own saved solutions and found out that I miss the one from 06/12/2005: Double Trouble X #2.
I don’t know if it is anywhere where I haven´t looked, in that case, could you please tell me where?
If it isn’t, I would be very pleased if you could put it in the archive again or e-mail it to me.
I would also like to see the solution for Double Trouble X Tuesday, January 10, (which is the current daily sudoku puzzle).
Kind regards,
Karin
Several solutions on Brain 57190?
Is it more than one solution to the Brain 57190?
(I found quite a few…)
Jallaman, I bet you didn’t pay attention to the diagonals, right? Well, both diagonals must have all numbers from 1 to 9 on them.
This is indicated in the puzzle by thin grey lines across the puzzle (and there is an explanation up the top of this page).
If you follow that rule, then there is only one solution to the puzzle.
After yesterdays relatively easy brain, I thought I would have a go at the killer called “The Frame”
After well over an hour I still havent managed to get a single number!
It seems that all the logic skills I have acquired over the last few months for killer, do not apply to this puzzle.
However I will persevere. I am determined to get at least one number!!
Dave
Regarding J-C’s “The Frame”, I’ve finally done a thorough analysis and think that I’ve got a very simple & clear 24-step route the solve the major part of the puzzle (i.e. until the trivial steps at the end). One of the most important move is to very easily determine r9c9=4 at the very beginning of the solving process. My reasoning goes like this: we know r9c1 & r9c2 are both from {123}. Also, the 8(2)-cage on r9c56 must contain one of 1,2,3 (1+7/2+6/3+5), so all the {123} are accounted on r9. Therefore r9c9 must be 4 (the corners cannot be more than 4). Simple isn’t it?
So I’ll include my 24-step walkthrough in the next post to avoid confusion… Wish J-C, Shai, ND, Mike and all interested killer players would take a look… Notice I’ve applied SC (subtraction combo, my signature move) 3 times in it, so players who aren’t familiar with this move are suggested to refer to ND’s tips page in his blog (http://www.ndorward.com/blog)…
I’ll try to post in tiny fonts so only players in real need can copy & paste it to notepad to read it, and hopefully it won’t spoil the funs for others who haven’t solved it. However if failed please be warned that they’re spoilers!
1. 45-rule on r1,r9,c1,c9 -> r19c19=8 -> each of these 4 cells must be naked triple in nonet-7
3. r9c56 must contain one of 1,2,3 (8(2)-cage) -> 1,2,3 eliminated from r9c34789
4. r9c9=4 (naked single) -> r1c19+r9c1=4 -> r1c1=2, r1c9=r9c1=1, r8c1=3, r9c2=2
5. r9c56={35} (8(2)-cage) -> naked pair in nonet-8/r9
6. r56c9={23} (5(2)-cage) -> naked pair in nonet-6/c9
7. 45-rule on r9 -> r9c78=15 -> r78c9=31-15-4=12={57} -> naked pair in nonet-9/c9
8. r9c78={69} (15(2)-cage) -> naked pair in nonet-9/r9 -> r9c34={78}
9. r34c9={69} (15(2)-cage) -> naked pair on c9 -> r2c9=8 -> r1c8=13-8-1=4
10. 7 & 9 could not be in r1c4567 (11(2)-cages) -> r1c4567={3568} -> naked quad on r1 -> r1c23={79} -> naked pair in nonet-1
11. r23c1=30-9-7-2=12=[48] -> r45c1={69} (15(2)-cage) -> naked pair in nonet-4/c1 -> r67c1={57}
12. 45-rule in nonet-4 -> r6c1+r4c2+r6c3=9, so r6c1 cannot be 7 -> r67c1=[57], r9c34=[87], r78c9=[57]
13. r4c2+r6c3=9-5=4={13} -> naked pair in nonet-4
14. r34c6={(6789)} (15(2)-cage), while r3c6 cannot be 8 (r3c1=8)
15. naked pair r4c19={69} -> r4c6={(78)} -> r34c6=[78]
16. 45-rule in nonet-9 -> r78c78=14 -> r7c6=r8c7=1 or 2 (comparing r78c78 with the L-shaped 14(4)-cage)
17. r8c67=[91] (10(2)-cage) -> r7c6=1
18. SC (subtraction combo) on r8c345=12={(24568)} -> 5,8 eliminated -> r8c345={246} -> naked triple on r8 -> r8c28=[58]
19. SC on r6c3+r7c23=23-5=18={(13469)} -> 1,4 eliminated -> r6c3=3, r4c2=1, r56c9=[32]
20. naked pair r23c2={36} -> r7c23=[96], r8c3=4, r1c23=[79]
21. naked pair r7c78={23} -> r7c45={48}, r6c4=21-4-8=9
22. r1c67={56} (11(2)-cage) -> naked pair on r1 -> r1c45={38} -> naked pair in nonet-2
23. SC on r2c234=10={(12356)} -> 1,6 or 2,5 eliminated -> r23c2=[36], r34c9=[96], r45c1=[96]
24. r3c34=16-6-1=9=[54] (16(4)-cage) -> r2c34=[16] (10(3)-cage)
From now on everything is trivial!
Oopsy… I wanted to make it into tiny fonts using the “font size” html codes and obviously they aren’t working… perhaps djape would be able to fix it?
Correction on steps 1 and 2: (the “less than” symbol is being a nuisance)
1. 45-rule on r1,r9,c1,c9 -> r19c19=8 -> each of these 4 cells must be less than 4
2. r8c1+r9c12={123} (6(3)-cage) -> naked triple in nonet-7
Of course… in step 1 it should be “less than 5” or “less than or equal to 4″…
5 posts in a row… that must be some record!
Todays text file has huge numbers in it. I can’t figure out how to fill a cage with a sum of 4864. 🙂
Well done udosuk … as usual 😉
Your step 3 is probably the shortest way to unlock the 4 corners !
BTW you could have recognized early that r7c678 forms a hidden set on {123} wich gives r8c8=14-(1+2+3)=8.
Kind regards,
Jean-Christophe
JC’s very nice puzzle has been the recipient of a number of very involved and elegant solutions from Udosuk, Shai and others. Am I missing something, or isn’t it easier to start with Nonet 4, after which everything follows reasonably easily (i.e. roughly an IQ standard).
r4c2 + r6c1 + r6c3 = 9 (45 rule Nonet 4)
r6c1 + r6c3 + r6c4 = 17 (45 rule Nonets 7-9)
r6c1 + r6c3 cannot be less than 8 (17-9)
Therefore r4c2 must be 1 and r6c4 must be 9.
From here all the 1s and most of the 9s are fairly easily located and it all follows from there.
Much more boring as a solution, but probably quicker!
Love the classic killers, djape.
Mathew, you’re not missing anything.
We all missed your solution, which is by far the quickest and most simple. I tried it out, and solved the killer in a few minutes. Everybody else tried to find the way to solution looking at the “frame”, and its corners. Only you were not distracted by the name of the puzzle in your observations of it….
I also think that Usoduk’s solution is interesting, and quicker than mine, and over all, I think this puzzle enabled us to observe a few new solving techniques, and a new idea of puzzle building.
DJ, I overpaid optima on the killer book so they kindly sent me the other book.
Since I am beginning to learn these puzzles that was good BUT as it is all in Spanish can you give me a very simple rundown on anything special I should be doing with it please.
I noticed some greyed in squares in some puzzles.
Many thanks
Adele
Australia
Adele, the “other book” contains “sequential” puzzles. I posted one such set of puzzles on this site. You can find it here: http://www.djapedjape.com/?p=94
Do that one first and then try the ones from the book.
Good luck!
Just getting ready for an overhaul sometime in the next few days of the tips page — any suggestions very much appreciated. One question: I was trying to think of a felicitous phrase for the kinds of “hidden pairs” that arise in killer puzzles–see my example at the very bottom of the tips page. (my example is the fact that a row with {12} and a 5(2) cage side by side can have {12} eliminated from the rest of the row, which could not be done in “regular” sudoku). Suggestions for names for this?
(just to clarify: by the first “{12}” I meant “one {12} cell”. My point is that this configuration–
{12} {1234} {1234}
in normal sudoku doesn’t get you anywhere, but if the last two cells are “bound together” as a 5(2) cage then you can eliminate {12} in the rest of the row.)
ND – since you are inferring the hidden pair, how about “hidden inferential pair”?
ND & Frank,
I wouldn’t use the term “hidden pair” for this. Hidden pair (or set) is used in regular sudoku to designate a pair of value that can go only in two cells in some r/c/n. Thus we can remove all other possibilities from the two cells.
This case is more like a naked pair (or set).
How about calling it a “killer naked set” ?
BTW, that’s the way udosuk spotted very fast that R9C9=4 in my puzzle “the frame”. In this case, it is a naked triple on {123}.
Back to the lingo for “necessity” values. How about:
{8(4|5)}
wich means {} = an unordered set
with 8, a necessity
then the parenthesis for grouping
with 4 or 5, an alternative
The vertical break bar is the usual symbol for “OR” on several computer languages : C, Java…
{98(27|36|45)}
would mean 9 & 8 are necessities, then 3 alternative pairs
could work also for ordered sets :
R1C12 = [(1|2)5]
is equivalent to:
R1C1={12}, R1C2=5
DJ,
The Killer Butterfly was great.
G’day Adele(Shiela)
Re the other book, the grey squares aren’t Vegemite stains. Complete puzzle 1, then in puzzle 2 fill in the grey squares with the numbers from puzzle 1 that correspond to those squares. At the end of the book, before the solutions, there is a list of numbers from the previous puzzle to be added.
Hasta la vista
Mike Japan
PS I think it’s Portuguese not Spanish
Oops,
thanks Mike – that makes it a bit easier for me.
Adele
Australia
ne1 have more old puzzles? teh arcive only has 5.