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!
Paul, in the current version of Perfect Sudoku, you can print classic Sudoku puzzles but you cannot print Killer Sudokus.
This probably won’t sound very intelligent, but I’ll try to explain why I had trouble with yesterday’s and not todays. Yesterday’s started out fairly well and I got to the point where I had a good many numbers placed. But then I got to a point where my powers of deduction would not work and I needed to do some “math” to figure things out. (For example, ruling out numbers from cages of 3 by pairing one possibility with another possibilty and then seeing if the leftover number is in the remaining possibilities). Easier for me if I can go down and across the rows and rule in/out numbers that way. This was basically the case with today’s after getting the first few numbers and filling in the rest of the possibilities. As I said, not very intelligent, but that’s all I can come up with! Thanks again for the daily killers!
OK – having done the Insane in less than an hour (and cooked dinner at the same time!), I’ve been trying to do yesterday’s IQ and I would agree that it is more difficult. I’ve been looking at it for nearly 2 hours now and not completed. I’ve got the obvious ones in r2c2 and r8c8 and eventually spotted the innies from boxes 2&3 and 7&8 which has led to a few other entries and candidates. I also recognised that r4c3 and r6c4 have to be the same number although I don’t know what that is yet, also r4c6 and r6c7 have to be the same (thankfully I have worked out what they are!). Nonetheless there is still a lot of work to do and I don’t think I’m going to finish it tonight unless inspiration strikes and I spot something I’ve missed so far. It seems harder because there are no givens in terms of candidates unlike the 2-cell cage of 4 in today’s Insane, and fewer new cages that I can draw on my print-out to work out other candidates or entries. I’ve got 20 entries so far in the IQ but they don’t help a great deal yet in working out the others. I had a similar number at the point I posted earlier about the Insane, but the rest of that was relatively straightforward because the first few entries enabled you to work out lots of other candidates and thus the puzzle.
Does that help? And I’d still like a Sunday Bonus please!!!!
Thanks for all your effort Zauspon
Hi, DJ. I started exactly the same as Chris and Cathy and from there everything fell neatly into place. I can’t be sure how long it took me as I had two phone calls during the puzzle (don’t these people THINK before they ring me????!!!)but I’ll estimate under half an hour.
I think perhaps the puzzle had a few too many cages of two. They do tend to simplify things.
I did the 11/11 INSANE in ~40 minutes last night. i don’t recall how, but I don’t remember it being difficult at all. For all but the ones you label EASY, I tend to take 35 – 55 minutes, so I’m with the majority who thought this one wasn’t even as difficult as the average IQ/BRAIN/THINKER.
P.S. Your puzzles are great, so I’m not really complaining!
Didn’t solve the IQ last night and in fact went wrong somewhere so I’ve just reprinted it to have another go on my train journey into London today. Now not so sure about r4c3 and r6c4 having to be the same! Will let you know how I get on.
Started with the innies much the same as above. Then stuck.
Read Cathy’s idea about r4c3 innie being the same asr6c4 – its an attractive idea and things seemed to flow from there. Until this flow created a duplicat in nonet 6. Guess this step is unsound. Will have to reprint and start again. Would appreciate DJ’s comment on soundness of this step.
btw isn’t nice (above) to see dj saying, “I just don’t get it”. Nice to know he struggles too (if over different things).
That’s a nice comment Brian. 🙂
I’m not sure how Cathy concluded that r4c3=r6c4 but I can reveal that it isn’t correct.
As the others say, Friday was relatively easy. I always start your puzzles by listing all innies and outies, and in this one we had:
TR nonet 2-I: 5
L1 2-I: 17
BL nonet 2-I: 7
L9 2-I: 11
These two pairs worked well together.
note in cathy’s solution the 7 cage in box 3 has to be 1:6 as on immediate inspection, the MR and BR block cells in col 7 cannot be a 1.
I ended up with pair candidates in all the outer 8 boxes and was able to solve the centre one without using the centre box cage sums. I am reasonably sure I did this rigourously.
Post another insane on Sunday please.
I agree. Yesterday’s insane was easy compared to thursday’s IQ. I don’t really remember HOW I solved them, but I do remember that the IQ puzzle involved a lot of maths.
Cheers,
I agree with Babs that the Insane did have too many cages of two (and three). It was quite hard to find an entry, but from there on it was very easy because there were a lot of cages for which only one combination of numbers in the sum was allowed. After that it was “search and replace”.
I don’t think that the difficulty of the entry has much to do with the overall difficulty of the puzzle. E.g. Thursdays IQ was easy to start but I stuck two or three times after having half of the cages filled. Then it’s best for me to take a long break 😉
I think the insane on Fri was definitely easier than the IQ on Thurs and even the brain on Wed. I remember djape once said his rating system is based on how many different types of techniques you need to apply to solve the puzzle, and while the insane might require more different types of tricks they were all quite simple and easy to spot. In contrast the IQ and brain needed more advance tricks, especially maths ones to get some of the harder entries (it certainly helped that I applied the subtraction combo vigorously and completed both of them within the hour). I guess using the calculator comprehensive will achieve the same effect. Judging from the response here I’d say most users here rely more on their “sudoku skills” than their “combination skills”. And don’t remember for normal sudoku puzzles the difficulty is most likey rated on the most advanced technique requires but not the quantity or variety, e.g. one needing a single swordfish and a whole bunch of singles is rated more difficult than one with a whole heaps of trios and locked candidates.
Sorry for my long comments about my view on djape’s rating system. Although I wouldn’t mind whether he posted an extra insane tomorrow (it wouldn’t hurt ^_^), I hoped in the future he could try make some puzzles which you need more advanced sudoku techniques like x-wing, xy-wing, swordfish, turbot fish, coloring, nishio etc. to get past a critical point. That would surely be “insane”…
And some tips on the IQ on Thurs:
Cathy’s claim that r4c3=r6c4 & r4c6=r6c7 were both incorrect unfortunately, and the only identical cells I think useful is R3C1=R4C2, which will help solving the 10-cage. For starters Cathy’s tips are very useful to narrow down C1 and C9 a lot. And the 5-cell 15-cage in the BL region also helps. But I wonder why she didn’t mention the obvious innie pairs in nonet 1 & 9?
So, I guess many people should be stalled after all these. The next step would be quite tricky:
You should get R9C4=1, R9C[56]={3,5}, R8C6=7, so the rest of nonet 6 are {2,4,6,8,9} totalling 29. And R7C6 should be one of 2,6,9. Using the 32-cage, it’s obvious R6C4=R7C6+3, i.e. 5,9,12. You’ll see 5 is the only possible option, so R6C4=5 and R7C6=2. And then applying subtraction combo on the R4 13-cage will have the middle 3 rows opened up.
mea culpa – I realised after I went wrong again on the IQ that my assumption about r4c3=r6c4 & r4c6=r6c7 was incorrect. Haven’t had too much time to look at it today but I haven’t given up yet!!
And thanks for the extra Insane DJ – looks interesting.
Thank you for the extra puzzle 🙂
Definitely not INSANE though. Easier than Friday’s INSANE.
Thanks for the extra killer and I agree it was easier. Call me ignorant, but I have absolutely no idea what udosuk is on about. What are these aquatic/ornithological related techniques? (And Nishio translates as “west king”)
Mike Japan
You can read more about these techniques several places including http://www.simes.clara.co.uk/programs/sudokutechniques.htm I am disappointed that Perfect Sudoku does not employ more of these techniques in order to create more difficult sudokus. Nisio is actually named after a sudoku master, so it doesn’t really mean anything on its own.
Hi Udosuk,
Do x-wings, xy-wings, swordfish, turbot fish, coloring, nishio etc have any relevance in the Samunamupure (‘Killer’) puzzles, or are you talking about the Sudoku puzzles on this site? I just can’t see how an x-wing or above could be applied to find a solution in this style of puzzle given the ‘maths’ restrictions that each cell/cage must consider.
Anyway, the weekend ‘Insane’ killer… I mucked up somehow and have to start again. I hang my head in shame.
Later: I have redeemed myself. Puzzle solved!
Regarding the so called “advanced” Sudoku techniques:
1. I consider everything beyond Swordfish (X-wing) to be limited trial and error. Those techniques are geared towards explaining how a puzzle can be solved. I really find them useless for generating puzzles – I could limit my generator to create puzzles with one level of trial and error only and then you would almost certainly find some “coloring” that would solve the puzzle.
2. It is questionable if a Killer Sudoku could be created in such a way that it MUST be solved by using some of those techniques. I think we should try and find more new Killer techniques – I’m sure there are some that we haven’t found yet.
3. I will implement Swordfish soon and try to generate a puzzle that requires solving it.
All this has given me an idea which I’m not going to reveal now. I have to give it some more thought but I think you’ll be happy with it – eventually :).
Cheers!
thanks for this insane gift ¡¡ I resolved it in 1 hour and I enjoyed it a lot
Cheers Shade.
Mike Japan
At last! I solved Thursday’s IQ on the 4th attempt. Now for the extra Insane.
Thanks for everyone that replied to my last posts. I hope most people understand all those terminlogy now thanks to Shade’s link… and I totally support whatever djape’s decide on, and am looking forward to the puzzle which requires swordfish… (Yes Babs, all those techniques could apply to KILLER sudoku as well, and I believe there can be a puzzle where you’re required to do one of them before reverting to trial and error of certain cages… after all there are killer puzzles where the solver must apply locked candidates/naked triples as some stage before returning to normal arithmetical maneuvers.)
Now back to the tips for the Nov 10 IQ. Sorry last time I totally mixed up with the rows and columns. Was in half-sleep state… In light of that I redo the last paragraph and then add another to complete the “walkthrough”:
You should get R4C9=1, R[56]C9={3,5}, R6C8=7, so the rest of nonet 6 are {2,4,6,8,9} totalling 29. And R6C7 should be one of 2,6,9. Using the 32-cage, it’s obvious R4C6=R6C7+3, i.e. 5,9,12. You’ll see 5 is the only possible option, so R4C6=5 and R6C7=2. And then applying subtraction combo on the R4 13-cage will have the middle 3 rows opened up.
Because both the 1 and 9 of nonet 3 must be locked inside the 22-cage, R1C6 cannot be 1. Now focus on R[1-3]C6, an innie trio summing to 10. This trio has to contain a 1 because {235} is not available (R4C6=5), which must be in R2C6 now (R3C6 cannot be 1 because R3C2=1). Now R1C6+R3C6=9, and 4,8,9 are quickly eliminated, leaving R3C6 being 1 of {236}. Now do a subtraction combo on R3C[6-8], which can only be {2356} and sum to 13. Then we’ll get R3C7=5 and R3C[6,8]={26}. Next, perform subtraction combo on the 22-cage, which must be from {123679} now and 6 is eliminated, leaving R3C8=6 and R1C6=7. From now on it’s all trivial, especially if you can spot the naked pair {48} on C4.
Assume you have a virginal grid apart from R1C123 which is a cage of 16 made up of a combination of the numbers 268. And R1C456 which is a cage of 14. After erasing all the 2s,6s and 8s you can also erase a further 12 numbers. What is this technique called?
Mike Japan
Mike, I would love to answer your question (and even more to learn if that’s a new technique you’re proposing) but I don’t understand what you’re saying. Not sure what 12 numbers you’re referring to and from where you’re erasing what and why.
Please send a more detailed explanation to sudoku@djapedjape.com
Thanks!
Thanks for yesterday’s “insane” – not too difficult after 1 false start! I think some have been harder.
As´far as I Mike understood he described the following situation:
If r1c123 has cage sum 16 (cs(r1c123)=16) with the only valid triplet {268}
and if cs(r1c456)=15
then cs(r1c789)=14
because the cs(r1)= 45.
The only possible triplet for cs(r1c789) is {159}
and therefore the only possible triplet for cs(r1c456) is {268}.
Now you can remove {159} in r2c789 and r3c789 (Naked Triplets) and {268} in r2c456 and r3c456.
Simplified, a cage of 3 numbers covering R1C123 totals 14, and each box contains a possible combination of 134579. The only way to make 14 with these numbers is using combinations of 347 or 149. Therefore 5 can be deleted from this cage. Also (this is the main point)because the number 4 appears in every conceivable permutation, it must be in the cage. Therefore the other 6 boxes of the nonet can’t contain 4 and the rest of row 1 can’t contain a 4 either.
Zauspon saw much more than I realised was there!
INSANE (497818) is unsolvable. Have you guys tried it?
Mayundo:
It’s definitely solvable, and many regard it as quite easy. Perhaps you didn’t try hard enough. I’ll show you some hints at the end of this post.
mike & Zauspon,
Perhaps it’s language barrier or something, but I couldn’t understand the 2 first posts of you guys at all. But mike’s latest post was very clear and I use that technique described by you quite a lot. I don’t know how to name this trick though. Perhaps djape could formalize it more?
Regarding INSANE (497818):
Okay, after redoing the Sunday insane I have to admit it’s not as difficult as it first looked. Surely we don’t need x-wing or even the outies of any whole bands/stacks to solve it. But to spot the easiest route in the first try is certainly a luck thing, or you need to be extremely patient to refrain from using any fancy techniques and keep searching and searching for simpler moves. So I stick to my first opinion that it deserves the insane rating, or at least an IQ. Surely it’s more difficult than the original insane posted on Friday.
In a grand scheme, first you work on the 3 nonets on the right, then focus on column 1, followed by nonet 1 & 7 and everything should fall into place soon after. The outies of nonet 1 & 9 are extremely useful, as well as the innies in C1.
And regarding djape’s idea about developing techniques for killer puzzles, I’m proposing another one called “candidate synchronization”. That happens when we deduce 2 cells to be of equal value or differ by a certain constant. Then we can keep “synchronizing” those cells to have the identical set of candidates. Eliminating on one cell has an immediate effect on the other. For example, in my walkthrough on the IQ on Nov 10, when a 5-cell 32-cage and 5 cells of nonet 6 (totalling 29) have 4 cells overlapping, then we conclude that 2 of the cells must differ by 3, and because one of them is down to 3 candidates, we could even determine the other one immediately. This technique could be applied again in the Nov 13 insane in nonets 1 & 9 too.
Can’t wait to hear other people’s views on this or any other technique proposed.
> Then we can keep “synchronizing†those cells to have the identical set of candidates.
I meant same number of candidates… the 2 cells have identical set of candidates only when they’re meant to be equal, not when they’re meant to differ by a positive value…
Perfect Sudoku is the only program that works with killer sudokus. I use it for solving puzzles. I have a couple suggestions for the program. First, I think it needs a pause. I see comments here where people say they are not sure how long it took because they had to leave it. If they could at least pause the timer, they could tell how long it took.
About printing, if you were to combine the grid drawing and print drawing methods into one, then you could print anything the program can display.
Actually, one more thing. Some times when I create a new killer puzzle, it doesn’t work right. I have noticed two different problems. If you change a cage after setting the total, the cage ends up bad. Occasionally, I get a cage without a total. It seems to be connected to another cage somewhere, but I don’t know why.
What I love about this site and killers generally is that I find I have to continually re-invent my approach to sudoku. Even over the last couple of weeks I have discovered new techniques. One that I would like to share is where you have common squares within two separate sums eg. you might have three squares adding to 24 and three squares adding to 27. If two of the squares are common (for instance) then you know that the remaining squares have a difference of three. This technique can also be used where you have more than one sqare not common to the group. I have used this technique a few times recently to solve the hardest puzzles.
Shoot was I beaten to the draw! I assume that is the same technoque udosuk is referring to.
Haha… sorry for just beating you out Phyllis… but it’s a nice trick isn’t it?
Again sorry for being pricky, can’t help but noticing that in your example when “three squares adding to 27” won’t that immediately implies that they must all be 9’s? So the other set of 3 squares must be {6,9,9} in this case, and of course both sets here cannot be cages or along the same line or inside the same nonet…
Please excuse my ramblings of yesterday, I was trying to be too clever for my own good. Anyway, how about this for a suggestion. Once a month on a Saturday at 2pm UK time (11pm Japan) post an insane killer. Everyone tries to solve it as quickly as possible. As proof of completion, choose one row or column and people email only these numbers to you at a newly created email address. Post the top 5 winners names and email times. No prizes, only kudos. I want to see just how good Udosuk is.
Mike Japan
Mike, this is a great idea! Would love to see the times, too. However your proposed time is not too appropriate for me (Sunday would be better) – but anyways, I think I would not be on the list 😉
I must confess I dislike this obsession with times. I like to solve killers in a relaxed fashion and don’t even look at the clock. Each to their own of course.
Phyllis, the beauty of my suggestion is that we would get an extra killer at weekends. You might not want to join in the race but you might want the extra puzzle.
Mike Japan
had a busy weekend. only got to do the insane yesterday. LOVED IT!! thanks!
Reposting message on this page as no-one’s yet responded from the comments on the Bonus Insane. 🙁
It’s no good – I’m obviously missing something here. Please can someone give me a little extra clue, I’ve only got 5 numbers so far: r3c7=2, r4c9=4, r5c9=2, r7c7=1, c7c8=2. Lots of candidates in the right hand stack (box 6 with naked pairs and a triplet) and a number of split cages but they don’t seem to help.
Thanks.
Cathy,
I don’t know if this is Trial and Error or not, but:
In Nonet 6 (r4-r6, c7-c9)…you know that the rest of the 11 cage is 135, which makes the 14 cage 68, leaving 7 and 9. Also, you know that based on the 3s in r2C9 or r3C9 and the 3 in C8…that a 3 must be in C7 of r8 or r9, which means that one of the 12 cages is 39. Now in Nonet 9 (the lowest right), a 4 cannot be in C9 because of r4c9. A 4 cannot be r8c8 beacuse of the 8 already in C7 from the 14 cage. Nor can a 4 be in r9c8 because that would require a 49 in the 13 cage, but we already know a 9 in one of the 12 cages, so that’s out.
Therefore, 4 is in r8-r9c7, making it 48 in one of the 12 cages.
Therefore, the only possibility for the 13 cage is 67.
The combinations for 21 if r6c9 = 9 are r7-r8c9 = 48 or 57, which are both impossible.
Therefore, r6c9 must be 7 and r6c7 must be 9.
From there, it was much easier.
There’s probably an easier way…
Thanks Miguel – done it now 🙂
I’d be interested to know if there was anything more obvious that helped to crack this one. I had managed to place 8 in r6c1 through the split cage in box 4 but without getting the 9 in r6c7 and 7 in r6c9 by “if, then, so” (or T&E if you prefer) I think I’d still be stuck!
Hi Mike,
I’d be in the competition, but 11pm in Japan is 1am here in Australia!! Yikes!
I’m feeling a little bit regretful about the bragging… Anyway about mike’s suggestion, I wouldn’t oppose to a weekend insane once every month but regarding the competition – it might be fun but it won’t prove anything. And I’m sure you won’t be able to see “how good Udosuk is”, but instead might be “how fast somebody’s computer is”… Although I personally can’t make djape’s perfect sudoku software working on my computer, and I don’t know how fast does it solve the insane puzzles, I’m fairly sure it’s quite impossible for any human to beat it… I also know a web page with online java applet that solves killers using brute force and I’m sure with some effort I could write a program to do it effeciently too… So, to really prove anything we must at least give a more detailed walkthrough/outline of the routines to solve the puzzle, like the following…
A walkthrough for the Sun Nov 11 insane (without trial and error):
Firstly you have to work on nonets 3,6,9 to eliminate a lot of candidates as Cathy suggested. Not much to say, except that in the 28-cage at the top we can eliminate 1,2,3 and notice that 8,9 must stay inside there.
In column 1, r1c1+r5c1+r9c1=7, so they must be {124}. That leaves the 21-cage to be {56789} with a fixed 7 and the 17-cage to be {35689} with a fixed 3.
Now since R6C9+R9C6=16, they must be {79} (so R6C7=R9C6). Similarly R1C4+R4C1=7, so R4C1 must be 5 or 6. That makes one of R[23]C1 being a 7, and we can eliminate 7 from other cells in nonet 1. That leads to the elimination of 1 in the 8-cage and 4 in the 11-cage. Doing a subtraction combo on these 2 cages combined ({23568} summing to 19 => 5 eliminated) makes the 8-cage being {26} and the 11-cage being {38}. Also, the 5-cage now must be {14} (only place for 1). So R1C4=2, R4C1=R1C3=5.
Also, you should have worked out that R[23]C6=12 and so R1C4+R3C4=45-27-12=6, so now R3C4=4, forcing R3C[678] to be 5,2,6 and R2C[67] to be 7,5. From here R9C6=9 and R6C9=7 and the rest should be all trivial…
I assume that the majority of killerholics like me, have never even contemplated using a computer to solve any puzzle. What’s the point? (It just isn’t cricket old boy!). And anyone who would stoop to using a computer to help them win an insignificant competition with no prize, is in need of serious help. Plus, if you have to give a detailed walkthrough of a puzzle, who is going to check it?
Mike Japan
Mike you seem so keen to bring the competition on… ^_^ Even if it’s “insignificant with no prize”… But that’s exactly my point: “What’s the point?” There are lots of factors… time difference, some prefer solving on paper and some on computer (by no means it means letting the computer to solve it), the efficiency of the email server… It just won’t mean anything except for an extra puzzle for most, and isn’t it exactly what we’re having right now?
An online race for any form of sudoku just isn’t feasible IMHO, even if it’s on demonic ones like the ones decribed here:
http://www.sudoku.com/forums/viewtopic.php?t=2165
And somehow I think your assumption that “the majority of killerholics like me, have never even contemplated using a computer to solve any puzzle” is a bit far-fetched, depends on how you define “killerholics”. I’m probably not one with your definition. ^_^ But I still enjoy the klllers a lot solving them with the help of my normal sudoku solver (without using the “hint” features at all). ^_^ For easier ones I even turn off the candidate features and solve it simply by pressing the digits in step by step, just like you’d do it on paper without making any pencilmarks. I solved the Monday & Tuesday puzzles in this manner (which I have to ask djape: is today’s puzzle a easy or a cool?).
The Hidden Puzzle?
Has anyone spotted DJ’s hidden puzzle in all this?
Today’s ‘Thinker’ is number 38272. The last Thinker was 42013. Yesterdays Cool (or easy) was 24909.
In the archive section there is an additional number (e.g. 20051115) but that is clearly the date.
So non-sequential, not date related, and no apparant ‘toughness’ code. Knowing DJ as we do it will all be entirely logical – absolutely requiring no guess work at all.
Has anybody cracked his Killer numbering code?
btw not personally interested in competion but woud appreciate an extra puzzle.