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!
Loved the butterfly killer X!
Mike Japan
Cindy,
A cage 18/4 can’t have both {97} : 18-9-7=2. Two cells can’t add up to 2.
HATMAN,
Did your KIMO 2A:
I like it because there are some new thoughts to be thought. But it was not very difficult. On many occasions I like this too, so I solved it while I prepared the Sunday meal.
Thanks! Waiting for an other KIMO
zauspon, Germany
Udosuk
Would you please send me a copy of the notation rules that you and the rest of the gang have worked out (to hatman@diceboard.co.uk). I have not had time to follow the debate but would like to look them through in terms of KiMo. I will then put them up on the site for everyone’s reference.
Ever Onwards
Hatman
DJ,
We all love your killers very much, and we do them on a regular basis (I don’t think I missed even one killer in the last 2 months).
However, some of us are getting better and better as time goes by, and look for harder challenges.
Will it be too much to ask for an INSANE killer once a week ?
Shai, Israel
After Friday’s IQ (396437) the Killer Butterfly (4_12_12_20060126 etc) was ‘a piece of cake’. ND’s 8th, and J_C’s latest are either very difficult, or I’m missing something in both. I’ll have to read all the comments to see what others think.
Hatman,
Sorry is currently very busy right now. You could check out the old comments by clicking the “comments” link on the right hand side column of this page. Also, JC’s site contains a very good “killer lingo” page:
http://www.freesoft.be/jc/KillerLingo.html
Hope these could help!
PS: On the spreadsheet issue, I found that grouping 9×9 tiny squares as a single cell in the grid would produce very nice pics for both puzzles and walkthroughs… The resulting pic would be about 490×490 pixels. When I have time I’ll try to write a very simple “walkthrough” for producing a killer sudoku pic using Excel…
Where there two solutions to the Sunday 22 January 2006 Samurai Sudoku in the Washington Post? I had a different solution to the lower left hand 9X9.
Thanks,
Dan
Hi
I keep trying all sorts of variations of sudoku and killer (like the ‘equality/inequality’ ones, they are just too much for me!) but think these daily killers are still the best, especially the iq and insane ones and X’s. I do other killer’s too but they are always either too easy or somehow not as good as Djape’s, which I think has something to do with its symmetry. They are so beautifully perfect!
Cheerio,
Myriam
Thanks for the clarification, JC. I should have seen that. In step 20, you use a moderately complicated method (remote naked pair) to determine that R6C5 is
Dan,
All puzzles that I create have only one solution.
If you think there were two solutions, please copy both solutions (use dots for empty places in the Samurai puzzle) and I will tell you where you went wrong.
… to determine that R6C5 can’t have {89}.
I just wanted to speak of the techniques. it’s not necessary to solve the puzzle.
Only just solved Friday’s IQ! Definitely a challenge. I need more spare time for these puzzles – still got JC’s B2 and ND’s 8th to do.
Hmmm … looks like a less-than-or-equal-to symbol caused my last message to be truncated. Here is the full text:
Thanks for the clarification, JC. I should have seen that. In step 20, you use a moderately complicated method (remote naked pair) to determine that R6C5 is less than 8. I thought you might like to know that there’s a simpler way to do that. In step 16, you determined that R7C9 = 1, so 5 of the 27/6 cells cannot contain 1, so the remaining cell (R6C5) must be less than 8.
DJ – I agree with Shai. I would love to see an INSANE killer once a week.
-Cindy-
Hatman,
It is possible to have a killer puzzle that has more than one KiMo solution. Take this puzzle fragment:
1 6
2 7
=== (nonet boundary)
6 1
7 2
Put a 3 cage around each {12} pair and a 13 cage around each {67} pair. Voila!
-Cindy-
Udosuk
Thank you for the pointer on notation. Shai also sent me his notes. (working on the spreadsheet)
I have now loaded up some further KiMos. 3 and 4 are based on TiKis. With 5 I got confident enough to try one of Djape’s Easies, No 7607,it went quite well, not exactly straightforward but it made me confident. Then came KiMo 6 -Djape’ Easy 14493. I solved it eventually but this required a new adaptation of the innies and outies method followed by an unfortunately long four branch elimination.
I simplified the puzzle in 6A, found this overdone so created 6B as a halfway house.
Hope you enjoy them.
Ever Onwards
HATMAN
Is Monday’s Cool (27438) legit?
(1) r1c3=9, r1c7=3, r9c3=8, r9c7=1, straightforward
(2) 16 cage in r4c9 must be 79
(3) 14 cage r1c9 must be 68
(4) 17 cage r2c7 must be 1457 (since it can’t contain any of 368)
(5) 11 cage r1c8 must be 29
(6) So rest of column 8 can’t contain 29
(7) So rest of column 9 can’t contain 6789
(8) 18 cage r6c9 must be 378/468/567 since it can’t contain a 9. Each possibility has two cells > 5.
But, given (7), only one cell of this cage (the one in column 8 ) can have candidates > 5.
We have a contradiction.
Beldar,
Point 4 in your analysis: 17 cage can also be 1+2+5+9, so your point 5 isn’t correct causing problems in your further analysis.
Thank you. My cheat sheet of cage combinations is in error. My apologies.
Cindy
Unfortunately this is not possible as the 8 cell placement shown is unsolvable.(see spreadsheet on the dicebard site http://www.diceboard.co.uk/excel/unique1.xls)
This is because you would then have to then fit all the numbers 1267 in the 3 cells of the third column in the Nonet below these two.
I am begining to believe that it is unique, however a proof is likely to be messy, like those for the colour placement problems. (Note, I am only thinking in terms of the 6-colour proof not the 4-colour one.)
Ever Onwards
Hatman
Has anyone completed today’s killer? I’m convinced it’s impossible!
I end up with r1c7=4 c7 in bottom right nonet as 8,2,9 leaving 5 to get in (r5c7, r6c7) but having used up a 2 and 4 already in r7.
Just can’t see where I’ve gone wrong!!!
DJape,
Is todays puzzle (433384) Insane or IQ? On top of the page it says INSANE, but on the puzzle itself – IQ.
Olga, London, UK
Forbes, your first conclusion (r1c7) is wrong. The puzzle does have a (unique) solution.
Olda, today’s puzzle is INSANE according to one of my difficulty rating algorithms and it’s IQ according to another :). Does it really matter, anyway? It’s very difficult, that’s what’s important :).
It feels like an INSANE to me.
djape,
Oh, OK, I was just going to find out why I can’t work it out :-))) If it’s Insane, I can live with it :-)) So far I have managed to work out only 3 cells and I am well and truly stuck.
Olga,
Probably the same 3 cells as me !
Solution on today’s INSANE:
*** SPOILER ***
1. “45” on N4 -> r3c34+r7c1=26 -> r7c1=9, r3c34={98}, r56c1=10, r45c3=8.
2. “45” on N7 -> r9c34=[42], r9c12={15}, 12(2) in R9 = {39}.
3. “45” on N1 -> r123c3=19=952/961/871/862/853.
4. “45” on N5 -> r7c4+r3c6=10.
5. “45” on N12 -> r13c6=9 -> r3c6=r1c7.
6. “45” on N3 -> r1c7+r3c9=6.
7. “45” on C12 -> r678c3=17=926/827/836.
8. “45” on R89 -> r7c23589=17=12347/12356, r7c467=19=856/847.
9. “45” on N6 -> r3c9+r7c67=15. r3c9 can’t be 5 (will contradcit the conclusion from step 8). According to step 6, r3c9={124}.
10. According to steps 4,5,6,8 and 9, the series is one of 3 options: A) , B) , B) .
11. “45” on R1 -> r2c1=r1c5+4 -> r1c5={1234}, r2c1={5678}. Let’s check this options.
If r1c5=1, r2c1=5 then 15(3) in N1 = {168} -> 14(3) in N1 = {257} -> r123c3={349}, contradicts the 4 in r9c3.
If r1c5=2, r2c1=6 then 14(3)={167} -> 15(3)={924} -> r123c3={835}.
If r1c5=3, r2c1=7 then 14(3)={167} -> 15(3)={834} -> r3c123={925}.
If r1c5=4, r2c1=8 then 14(3)={158} -> 15(3)= {762} -> r3c123={349}, contradiction.
So 15(3) in N1 = {834|924} & r123c4 = {835|925}.
So r45c3={17|26} (no 5) -> r78c3={26|36|17}. 17 is rejected (1 in 6/2), so there’s no 6 in r45c3 -> r45c3={17} -> 10(3) in N4 = {235}, r56c1={46}.
15(3) in N1 = {924|834}, in both cases 4 must go in r3c2 (r2c2=8|9). And now we see that only option A from step 10 is possible !
12. r8c3=6.
13. “45” on N9 -> r789c7=19=937/928/946. Since r7c67={68},only {928|946} are possible -> r9c67=[39] & there’s no 5 in the 31-cage.
14. “45” on R9 -> r8c9+5=r9c5 -> r8c9={123}, r9c5={678} -> 16/3 in N9 = {178|268|367} -> 5 in N9 must go in r8c8 -> r7c89 = [14]|{23}. {23} is rejected (will cause no place for 1 & 4 in R7) -> r8c7=2, r7c67=[68], r7c23={23}, r7c5=7, r8c23={87} -> r9c5=8, r8c9=3, r9c89={67}.
15. r56c7={13}.
16. 5 in N1 must fo in r2c3. 5 in N5 must be in 20-cage -> 5 in N4 must be in r4c12 -> 5 in N6 must be in r6c9 -> r56c8=[84] -> 13/2 in N6 = {67} -> r45c3=[17] – > 20(3) in N5 = [659] -> r56c1=[46], r45c9=[92], r6c456={127} -> r56c7=[13], r5c2=3, r7c23=[23] -> r1c34=[27] -> r1c5=3, r3c1=3, 14(3) in N1 = [167], r2c2=8, r3c34=[98], r6c23=[98].
And you carry on…
I had some HTML problems with the previous post in steps 9 &10. Here They are:
*** SPOILER ***
9. “45” on N6 -> r3c9+r7c67=15. r3c9 can’t be 5 (will contradcit the conclusion from step 8 ). According to step 6, r3c9={124}.
10. According to steps 4,5,6,8 and 9, the series (r7c4, r3c6, r1c6, r1c7, r3c9, r7c67) is one of 3 options: A) (5,5,4,5,1,{68}), B) (6,4,5,4,2,{58}), C) (8,2,7,2,4,{56|47}).
DJ, if you’re willing, you can edit this 2 posts into 1….
Do we really need to have someone posting the solutions so quickly to avoid temptation, or can we leave it to DJ to post solutions a couple of days later.
Sometimes, depending on where you are in the world, the solution is there to see [hard not to take a peek] at the same time you print the puzzle out.
Also are the ones solving these so quick doing it by mental logic or a programme?
Adele
Australia
9am Wed.
Today’s insane.
Started like Shai with 45s
Came to conclusion C2C1 was {78} (don’t remember how…)
-> 14/3 in N1 = {167|248|347}
Since 9/2 in N23 = {27|45} & 9/2 in N12 = {27|36|45} -> 14/3 in N1 = {167}
which unlocks the puzzle.
Hi Adele
I did Tuesday’s insane in under an hour with only a pen and paper. I never read anyone’s walkthroughs and I also never check the solutions – when I finish a puzzle it must be right because I’ve finished it. If it was wrong, I couldn’t have finished it!
Mike Japan
Thats excellent Mike, under an hour. Good stuff. Pat made a boo-boo and had to start again [usually does if hungover].
I would be so tempted to peek to get a start. Thankfully its all double dutch to me so I can’t even offer him a clue or drop a hint.
Adele, Australia
Today’s puzzle was excellent! It took me a bit under two hours to do it with paper and pencil. (Today, I double checked my work all through the puzzle, after flubbing yesterday’s puzzle twice.) Here are some minor hints that might help those still working on the puzzle:
1) Remember that a split cage might allow cominations that a normal cage would not. For example, if you figure out that R1C1 + R4C4 = 8, it is possible that they both equal 4, since the two cells are in different rows, columns and nonets.
2) Figure out the relationships of *many* groups of cells. I used innies, outies and the 45 rule over and over again, to figure out the relationships of groups of 2 and 3 cells. (e.g. A + B + C = 12) It’s a pain doing that much addition (especially with paper and pencil), but it allowed me to keep chipping away at the puzzle. Any time I started to feel stuck in the puzzle, I was able to look for (and find) more places to determine relationships.
3) I found it helpful in this puzzle to note and use relationships that were not exact, such as these two cells add up to 7 or 8, allowing me to eliminate some possibilities.
4) I found it useful to refer to a sum table for some of the larger cages. Gradually, I could reduce the number of possibilities and take advantage of the new information as it became available (e.g. this cage cannot contain both a 2 and 3, cross out the ones that have both – hmmm, now I see that cage must contain an 8)
Thanks again for such an enjoyable, and challenging puzzle, DJ!
-Cindy-
I usually post a solution if people ask for clues. That’s usually a sign that’s it out for enough time for people to get stuck.
I also think that using the killer lingo helps people learn new ways of solving puzzles 😉
I enjoy the helpful suggestions of contributors.
Only read them if in trouble and usually one or two missing steps are enough to get me going again – don’t think I have ever followed through a long analysis.
Many thanks to those who post hepful hints.
I also feel no ‘guilt’ at reading hints. It is all part of the learning process.
on on
Hasher
Has anyone got the solution to THINKER (47938) from 12th January. I get 2 solutions and it isn’t an X.
I haven’t caught up yet and am determined to do them all, without cheating. (Still haven’t done the one from 13th December yet).
Really enjoy your daily killers although they can demand a lot of time.
Have looked at a number of your and other variants but prefer the basic killer or Xkiller.
Not having the time or inclination myself, I wonder if any of your correspondents might care to explore a concept for a variation on your puzzle which I would call a Mystery Killer.
This uses a standard Killer grid but has a zero substituted for one of the numbers. The cage total would include the zero so totalling up all the cages would show which number has been substituted – but not its whereabouts.
The natural progression from this would be double or triple mystery where 2 or 3 zeros are included. ” I feel could present quite a challenge even on a relatively easy full grid puzzle.
Thanks again for a great site.
Stuartdm, I made a topic for this puzzle in the forum: http://www.djapedjape.com/sudoku/forum/viewtopic.php?t=3
leggedit, the answer to your question is here: http://www.djapedjape.com/sudoku/forum/viewtopic.php?t=4
Those of us who have this killer page in their favourites, rather than the home page may have missed this:
“A new feature on this site – FORUM
Filed under: General — djape @ 9:01 am
Well, after nearly 1000 comments have been written to various posts on this site, I think it’s time to provide you with a better way to communicate.
Therefore, I just opened up a forum.
I’m pretty sure you are all familiar with phpBB forums so you won’t have problems using it. One thing, though, you must register to post messages.
In the next few days I will close comments on the Daily Killer and Daily Sudoku pages so please start moving your discussions into the forum.”
Having a bit of difficulty following the shorthand in some of the discussions above – can someone please point me to somewhere where the basics of such lines as:
(from Shiia above] “45″ on N1 -> r123c3=19=952/961/871/862/853 can be interpreted
My reading is that Nonet 1 rows 1,2 and 3 col 3=19 – if that’s right, then I’m baffled as in puzzle 433383 Nonet 1 rows 1,2 and 3 col 3=16.
Perhaps I’m missing the thing completely – if so then perhaps the following can be explained in ‘longhand’ please:
“45″ on N4 -> r3c34+r7c1=26 -> r7c1=9, r3c34={98}, r56c1=10, r45c3=8.
Thanks very much. Sorry to appear thick, but yesterday’s 433384 has left me puddled 🙁
Graham
Graham,
You can see the “Killer Sudoku lingo” discussion here: http://www.djapedjape.com/sudoku/forum/viewtopic.php?t=6
Graham,
That “19” in step 3 from Shai was a typing mistake, should be 16 – look at possible combinations she listed (952/961/871/862/853). they all add up to 16. So your understanding is correct.
Olga
Thanks Jean-Christophe for your hint on yesterday’s Insane. Only after you pointed out 14/3 in N1 = 167, I understood that there couldn’t be a 4 in r1c7 and r3c6! And that unlocked this devilish puzzle indeed, I’d been strugling for hours on it!
Graham wrote:
>from Shiia above] “45″ on N1 -> r123c3=19=952/961/871/862/853 can be interpreted
>My reading is that Nonet 1 rows 1,2 and 3 col 3=19 – if that’s right, then I’m baffled as in puzzle 433383 Nonet 1 rows 1,2 and 3 col 3=16.
You are correct, it should be 16, other than that you understood. Note that 952/961/871/862/853 all sum to 16..
>“45″ on N4 -> r3c34+r7c1=26 -> r7c1=9, r3c34={98}, r56c1=10, r45c3=8.
this means that applying the 45 rule on Nonet 4 shows that r3c3+r3c4+r7c1=26. if r7c1 would be less than 9 (8 or less), then r3c3+r3c4 would be at least 18, which is impossible for 2 neighbouring cells, so r7c=9 and the rest sum to 26-9-17 which is {9,8). We also get that r5c1+r6c1=10 and that r4c3+r5c3=8.
I hope that the lingo is usually clear.
Olga, I’m a man. Shai is a very common Israeli name, It appear 3 times in the bible, and means ‘a gift’ 😉
Shai
Shai,
Ooops, sorry about that! I’ll remember now 🙂
Olga
Thanks for the help – I’ve now found the forum where it’s all explained in detail if anyone else has the same probs.
Many thnaks agan Shai.
DJ,
The Jan. 31 IQ/Insane puzzle was frustratingly fun. However, I came up with two solutions. I’ve looked all over the alternative solution frontwards and backwards, and can’t find the error. Could you find it for me? This is the answer I got:
+—+—+—+
|726|314|598|
|581|792|436|
|349|865|721|
+—+—+—+
|215|438|679|
|473|659|182|
|698|127|345|
+—+—+—+
|932|576|814|
|867|941|253|
|154|283|967|
+—+—+—+
Stuartdm – Solution to THINKER (47938) from 12th January.
312456789
659187324
874392561
791834256
483625197
526719843
167248935
235961478
948573612
Check that your other ‘solution’ doesn’t have a repeated digit – especially in a cage.
Mel-o-rama,
Your solution is incorrect because in the cage with the sum 26 you have number 1 twice.
This isn’t according to this rule which is clearly indicated on the Daily Killer page:
Numbers CANNOT be repeated within a cage, even if it is possible by other Sudoku rules.
This is a general rule and I know you may object to it, but that’s how it it.
No objection – I’m just being dense. I didn’t realize the non-repeating rule within a cage, and I’ve been doing these for over a month now! I guess I’ve been making these too hard for myself. Keep up the good work!
BTW, has there ever been any debate on the non-repeating in cage rule? Or is it just understood universally to be that way?
Thanks,
Mel-o