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!
Hi killer maniacs,
I got some idea for a technique, so I created a killer sudoku based on it. I created it by hand using logic (my controted logic !). I checked it with Perfect Sudoku which rate it as “Insane”. It has a unique solution. I will let you search for a few days before telling the technique.
Here is the grid :
http://jcbonsai.free.fr/sudoku/KillerJC20060104.png
And the data file for Perfect Sudoku :
http://jcbonsai.free.fr/sudoku/KillerJC20060104.txt
It’s fun solving killer sudoku. It’s even more fun to create them !!!
Kind regards,
Jean-Christophe
Wow, Jean Christophe, a truly great killer.
Really Insane…
Took me almost 2 hours !!
I really loved your technique, it’s quite unique. Alot of alternative thinking involved.
I guess that I solved it the way you intended, but i’d still like to hear about it.
Hope to see more killers from you,
Keep up the good work.
Shai, Israel
Jean-Christophe says:
Great puzzle, Jean-Christope!
49 minutes, with frequent use of a table of killer combo values
45 rule on top row reveals the combinations of 17 and 27 cages, then worked downwards.
Thanks! Bob USA
Cannot solve Times today (5th). Anyone else having problems?
Thanks
Paul UK
I messed up yesterday’s Times at the first attempt but today’s was straightforward enough – DJApe’s puzzles are always more challenging and satisfying!
Think I am going mad. Tried today’s Times killer a second time and did it but when I tried to put it into ‘perfect sudoku’ it came up with no solutions available. Txt detail below 3×3::k:1536:1025:3330:2052:2052:4358:4358:
2567:2567:1536:1:3330:2828:3853:3853:2319:5392:5392:3346:3604:
3604:2828:6422:3853:2319:1049:5392:3346:5660:5660:5660:6422:
3360:2319:1049:2595:4644:4644:5660:807:6422:3360:6962:2595:2595:4644:
2350:3887:807:6422:6962:6962:6962:3125:5430:2350:3887:3641:6422:
1339:828:828:3125:5430:5430:3887:3641:3641:1339:3397:4422:2119:
840:840:3146:3146:2636:2636:3397:4422:2119:
Any thoughts DJ?
Kind rgds
Paul UK
Greetings DJ & everyone,
Just back from a long holiday trip. Hope everyone had a great CNY holiday!
After a bit of effort, I’ve organised all the overlapping puzzles (excluding all samurai shaped ones) into the following list:
24/11/2005: Double Trouble X #1
26/11/2005: Wing X
28/11/2005: Triple Trouble X #1
30/11/2005: Triple Trouble X #2
03/12/2005: X-Ring
06/12/2005: Double Trouble X #2
08/12/2005: Double Trouble Killer X #1
14/12/2005: Double Trouble X #3
20/12/2005: Butterfly
21/12/2005: Butterfly X #1
26/12/2005: Double Trouble X #4
31/12/2005: Butterfly X #2
Also, there are 3 Killer-Samurai puzzles on Dec 16,23,31 and 3 Mini-Samurai puzzles on Dec 14,23,31, plus 11 normal Samurai puzzles from Oct 2 to Dec 31.
Of these, I find I miss 3 solution grids:
14/12/2005: Double Trouble X #3
26/12/2005: Double Trouble X #4 (the link in the archive only gives the partial solution, not the full one)
31/12/2005: Butterfly X #2 (should be available on next update)
So if possible, could DJ please post them up next time? Especially the Dec 14 one. A million thanks!
Also, I think a “Killer Butterfly” (not “Killer Bee” ^_^) could be quite interesting albeit the amount of work involved! Just a thought.
Hi,
Glad you liked my puzzle !
The real difficulty for me was to design a puzzle without any obvious innies/outies/combination so that you cannot start somewhere else and have to follow the way I created it, at least at the begining. 😉
Bob is correct, the start is somewhere at the top. The technique I used for the very start has something to do with the “45 rule”. Not innies/outies as such, but something else. Quite simple indeed. Think about the geometry of the cages at the top…
Then, one must play with combinations. A table with all possible combinations is probably a necessity to discover numbers that must or cannot go into some cage. You’ll find a combination tool on this site :
http://www.djapedjape.com/index.php?cat=8&paged=2
There is also a table with all combinations on Miyuki Misawa’s web site :
http://www7a.biglobe.ne.jp/~sumnumberplace/31642733/
BTW, Perfect Sudoku will take a loooooong time to open the data file. I first thought it was bugged, but it will finally open it after about 10 seconds !
Jean-Christophe
I’d like to be able to eliminate pencil marks manually – will that be possible in future versions? Also for some kind of solver log to explain why a number can be placed if you’ve “cheated” and asked for a hint!
PS When’s that PDF of your book going to be available?
Sadly, after a lot of attempts, I have to admit that Jean-Christophe’s killer has defeated me. After using the tip of starting with the possible combinations of 17 and 27, I’m still stuck. I’ve never resorted to tables before, but I must be missing a possible combination as all the ones I’ve tried have meant that the other cages don’t work. I’ve even been moving pieces of paper with numbers on them around, but to no avail. Too difficult for me (at the moment). I’m glad DJApe’s aren’t as difficult, though have been doing them too quickly recently. A happy medium would be nice.
Me too! I got as far as working out that r1c5 – r2c4 – r2c6 = 1, which means r1c5 has to >=4 and a few candidates in other cells, e.g. the 17(5) cage has to contain 1,2,3 and the 28(4) cage has to contain 8,9. But I’m stumped too – DJApe’s Brain today was a breeze by comparison! Look forward to some more information about the technique soon.
Pat is stumped with Jean-Christophes which is good since Djz Brain took 15 minutes.
Are you still in holiday mode DJ, making easy puzzles, lol
Adele
Australia
Nice puzzle Jean-Chistophe. Look forward to the next one.
Mike Japan
Hi and have a great mindbending 2006!
This site is fantastic, I have oodles of fun with your daily killers but am behind a bit now!
But I love the puzzles and think your site is the best. Thanks for all your hard work.
Could you tell me what the order of the ratings is? Is it Easy, Cool, Brain, Thinker, IQ?
Cheers,
Myriam
Hey Jean-Christophe: nice puzzle. Is that your only one so far or have you got more in your bag?
If you’re having trouble solving it: here’s my opening path–
45 rule shows r1c4/5/6 = 23, i.e. {6,8,9}, then r1c4 can only be 6, then the rest of that cage must add to 11 = {1,2,3,5}; this means the 11(2) cage underneath must be {4,7} (to avoid crashing into the 1,2,3,5); since r1c7/8/9 must contain {4,7}, then r3c4/5/6 must similarly also contain {4,7}, which leaves only two possibilities for that cage (97421, 87431)……
Anyway, the new Killer is oout on my site:
http://www.ndorward.com/blog/?p=86
Another very hard one: good luck!
Yes it’s the first one I created. Hummm, too bad I didn’t notice one could solve it the easy way using that famous 47 pair… 😉
Anyhow, as promised, I wrote a Q&D page explaining the technique I used and a step step solve of puzzle :
http://jcbonsai.free.fr/sudoku/
I think that Jean-Christophe’s killer is a cross between a Kakuro and a Killer. Personally I prefer the normal killers. I did finish it with the help of the table (thanks for the link, though there are some mistakes in the possible combinations for 17 over 4 cells), but prefer to rely on my own logic not on juggling possible combinations. Personal preference I know, I’m sure others disagree. Will read other comments with interest.
Pat uses his own logic, too Gaynor.
Lots of pen and paper gets used but he doesn’t see the point of using programs and the like.
More personal satisfaction that way.
Thanks ND – looks like Pat will be kept amused with your new one.
I must say this sudoku rage has been a blessing. He has always done cryptic crosswords and we subscribed to numerous worldwide papers to get them. Now we don’t have to.
Keep up the good work
Adele
Australia
8.20 pm Friday
Hi Myriam – you nearly got it right.
The order is Easy, Cool, Thinker, Brain, IQ and Insane, although this shouldn’t be so important, because difficulty is a personal thing and it’s very arbitrary.
It does give you a general idea how difficult a puzzle is, though.
Jean-Christophe,
In your killer, there’s also another way to find the first numbers, which does not include your (COOL !) double counting technique.
One can simply sum all the cages which do not appear (even partly) in nonet 2.
The difference from that sum and 8*45 will give you the sum of the top rows of nonets 1 & 3, from which you will get the sum of the top row of nonet 2 (23=6,8,9….).
Hope to see another killer from you soon.
The double-count technique is nice: I don’t think there are any cases where it would get results that the usual innies/outies approach wouldn’t, but it has the merit of simplicity & elegance. The simpler the calculations, the better, I think: much less room for making mistakes in adding & subtracting (I do everything by hand, not calculator, since I don’t want to drag one out on the busride home where I usually do the puzzlemaking & -solving).
One point re: J-C’s step 2: there’s a much simpler way to put 6 in r1c4: it can’t be 8 or 9 because that would make the remaining 4 cells sum to less than the minimum (10), so it’s got to be 6. — Though J-C’s solution makes use of combination-crunching, I think there’s no point in the puzzle where you need to deal with more than 2 or 3 possible combinations, if you solve it in the right order (the placement of 4/7 in the left top 11(2) cage is a useful initial step in this regard).
For those of you with an applied maths background, a christmas cracker joke:
What do you get when you cross a mountain-climber with a mosquito?
Nothing: you can’t cross a scaler with a vector.
Happy New Year
Hatman
P.S. I am currently wrapping up a New Year’s present for you all – sould be ready sometime next week.
A new year puzzle for all you maths geeks.
Although I am no mathematician, my Dad was a maths teacher and he told me about this;
The number 142857 is the most amazing number in the world, why?
Mike Japan
1/7 = .142857142857142857142857142857…
Hey Babs,
I was reading your comment from Jan 1st. I didn’t see the killer in question. How can 4 possibly be 2+2 ?
fdkr
Not the correct answer
Mike Japan
142857 is an interesting number, because when multiplied by either 1,2,3,4,5 or 6, the product will contain the same digits, and in the same (cyclic) order.
142857, 285714, 428571, 571428, 714285, 857142.
I don’t think this will qualify it as the most amazing number in the world, though. Most mathematicians are pretty much in love with Pai or e.
Here’s the best math riddle I heard in the last year. It may seem nonsense at first, but it’s not:
What’s the next number in the series:
0, 1, 2, 720!, …
(720! is the factorial of 720, i.e. 720 * 719 * 718 * … * 3 * 2 * 1)
Well done Shai, but you can carry on the progression. If you multiply by mutiples of 7 you will get all 9s. If you multiply by bigger numbers then the outer two digits will add up to the missing digit of 142857 eg; 142857 x 11 = 1571427 etc.
Mike Japan
The next number in Shai’s sequence is 42. If you multiply any of these numbers by 0 the answer is 0 (not funny!)
1/7 = 0.142857 142857 142857 142857 14285714…
2/7 = 0.285714 285714 285714 285714 28571429…
3/7 = 0.428571 428571 428571 428571 42857143…
4/7 = 0.571428 571428 571428 571428 57142857…
5/7 = 0.714285 714285 714285 714285 71428571…
6/7 = 0.857142 857142 857142 857142 85714286…
DJ – the natives are getting restless. I think we need more puzzles to keep us busy 🙂
> How can 4 possibly be 2+2 ?
By not being in the same row/column/nonet.
Mike,
You can carry on the progression, but you have to be more accurate.
First, multiplying by multiples of 7 doesn’t give all 9s (except for 7 itself and other weird mulitples of 7 – like 7000007), you have to add the outer digits in order to get all 9s with almost all multiples of 7. Example: 142857*14=1999998.
Second, while multiplying by bigger numbers, you might need to add more than 2 outer digits. Example: 142857*69=9857133. Here you must add the first digit (9) to the last 2 digits (33), 9+33=42, in order to get the original digits.
>The next number in Shai’s sequence is 42. If you multiply any of these numbers by 0 the answer is 0 (not funny!)
I know it’s very difficult to take this riddle seriously, but I promise there’s a logical solution, don’t give up.
Shai, Israel.
I got the answer!
0 = 0!!!
1 = 1!!!
2 = 2!!!
720 = 3!!!
So the next number (no matter how big it is) is
4!!!
Or wait, since your sequence is 0, 1, 2, 720!, then the next number is 4!!!!
Hmmm 4 = 2+2? I thought you could not repeat a number within a cage?
Martin – that is true but the puzzle in question (IQ 679028, 26-12-05, still in the archive today) required two cells in different rows/nonets to add to 4. 1 and 3 led to a contradiction so the only possible solution for these particular cells was 2 and 2!
BTW I’m pleased to say I finally cracked Jean-Christophe’s Killer at the 3rd attempt, thanks to the tip on double counting which led to the 23(3) in nonet 2! Now catching up on the Killer Samurai’s that DJ produced on 23rd Dec and 1st Jan. And I’ve just twigged that the 23rd Dec one is an X which should help enormously to fill in lots more cells!
DJ 0! = 1
So how about
0
1!
2!!
3!!!
4!!!!
fdkr, i think you’re quite right and i think your solution is correct! oh well, at least i was on the right track 🙂
fdkr,
You are correct. 0,1!,2!!,3!!!,4!!!!.
DJ, you were on the right track (and you found the next number anyway…:) )
Cool
Bonus question – how many digits are there in 4!!!!
J’aime bien me passer le temps avec les sudoku
any hints on today’s (brain 97589) ‘INSANE’ in disguise??
Hints on the today’s Brain (truly INSANE..):
(**Spoiler…**)
1.45 rule on columns 1 & 2 reveals r4c3 to be 1.
2. 45 rule on columns 8 & 9 reveals r6c7 to be 6.
3. 45 rule on 4 bottoms rows revels r5c1+r5c3+r5c8 to be 8, that is {1,2,5} or {1,3,4}. both options contain a 1.
4. Since nonet 4 already contains a 1 (from step 1) then r5c8 is 1 (and r6c8 is 4).
5. Now we see that r5c1+r5c3 is 7, and r4c2+r5c2 is also 7. Since we already have a 1 in nonet 4, these cells must conatin {2,3,4,5}, and then the only cell in nonet 4 which can be 6 is r4c1.
6. Now comes the tricky part. 45 rule on 4 left rows shows r1c3+r1c4+r5c4=21. This will show you that r5c4 is 8 or 9 (cannot be 7 or less because of the 14 cage at the top), and r1c5 is 1 or 2, accordingly.
7. The 22 cage in nonet 5 is either {9,8,5} or {9,7,6}. The 6’s we found in steps 2 & 5, force it to be {9,7,6}, which means r5c4 is 9 (cannot be 8) and r1c5 is 2.
And you carry on…
Shai, Israel
HTML…. what to do.
I meant:
7. The 22 cage in nonet 5 is either {9,8,5} or {9,7,6}. The 6’s we found in steps 2 & 5, force it to be {9,7,6}, which means r5c4 is 9 (cannot be 8 ) and r1c5 is 2.
I did it slightly differently. r1c7 and r6c9 must add to 5. Working through the combinations for the 13 cage that r6c9 is in leaves only one that allows two combinations that add to 12 for the 9th nonet.
Today’s Brain (97589) did seem harder than usual and definitely took me longer – probably about an hour and a half on and off. I didn’t keep track of every step I took but I had seven split cages which helped eventually (2 in row 2, 2 in row 5, 2 in row 8 and 1 in column 5) especially r5c1+r5c3+r5c8 = 8.
fdkr
Great question. How many digits in 4!!!!
I cant believe anybody can do that!!!
Would be very interested to know the answer though
Regards
Dave
Why the sudden interest in maths pop quizzes?
Anyway for fdkr’s bonus question, using Stirling’s approximation ln(n!)~=n*ln(n)-n
#digit(4!!!!)
= ln(4!!!!)/ln(10)
~= 4!!!*(ln(4!!!)-1)/ln(10)
~= 4!!!*(4!!*ln(4!!)-4!!-1)/ln(10)
~= 4!!!*1.44927e+25
and 4!!! has 1.44927e+25 digits…
My question in return: 1,10,20,110,220,1110,2220,10110,20220, what is next?
Did anyone else struggle with today’s killer (Brain 97589)? Took me ages to get going then it all fell into place. After the recent easy ones it was quite a change.
Gave up with ND’s latest, but have downloaded his solution and started to work my way through it – got the first steps right, so I was on the right track.
I’m still struggling with today’s Gaynor !
Took Pat a full day to get NDs 7th.
This Brain one is being a challenge though.
Jean Christophes was difficult to say the least. Pat found his steps really informative – he had to double check he was on the right track.
Adele
Australia
Udosuk
I reckon it’s 100110 or 110110.
Mike Japan
PS. As a mere mortal with the mathematical knowledge of jelly, I didn’t understand Shai’s 0,1,2,720! puzzle answer.