“Bad” puzzle #6: solution
Here is the solution for the humanly unsolvable(?) variant of #6 I posted for its curiosity value. [EDIT: it has now been solved logically, much to my surprise/admiration -- see the comments to this post...]
#6 ALT:
5 3 9 8 6 7 1 2 4
6 8 1 4 2 3 7 9 5
2 7 4 9 5 1 6 8 3
9 5 3 7 8 2 4 1 6
1 4 7 3 9 6 8 5 2
8 6 2 1 4 5 9 3 7
3 9 6 2 7 8 5 4 1
4 2 5 6 1 9 3 7 8
7 1 8 5 3 4 2 6 9
January 19th, 2006 at 11:39 am
I had a go on this puzzle but didn’t finish it… a bit hard to get motivated when you know the creator isn’t sure if there is a logical solving route… but I vow to try later when I have some more leisure time!
The route I tried started from nonet-2 like the “good” version, and then locked the {789} on c9 with 4 cells, then I did a 45-rule on r5-9 and determined that r567c4+r5c9=8, so either {123}+2 or {124}+1… then I went up to nonet-1 and with the 45-rule determined that r3c1+r23c3 is less than r1c4 by 1, so the sum of those 3 cells must be 6, 7 or 8… Then I got stumped… hopefully will have a better crack next time!
January 19th, 2006 at 3:39 pm
Hm, I suspect you’ll have to resort to trial & error with this one: but I’m wondering if there’s a “minimal t&e” path to use…
January 20th, 2006 at 1:08 pm
hey, can’t wait for your next one. LOVE your puzzles. keep it up and you are awesome for caring about us addicts out here!
January 20th, 2006 at 1:47 pm
Udosuk, I guess you forgot to say that, after the {789} in the three corners in nonet 2, r2c3=1 is prttey obvious. After that, apart from what you said, I found straightforward to place 1 in r3c5 or r3c6, {7,8,9} in nonet 3, (36)cage, {3,4,5,6} in r3c9, 1 and 2 in nonet 6, (12) cage, and 1 and 2 in nonet 3, (19) cage. Since I have not progressed any further in half an hour, I think I am quitting…
January 21st, 2006 at 12:41 pm
Enrique, please include more details on your solving… such as how you fixed the {789} in nonet-3… I’ve only been able to limit them in the 2×2 region R23C78. I’ve probably missed some obvious moves, again…
January 21st, 2006 at 1:43 pm
Udoku, what you did is what I meant in my post. Sorry for using the brackets wrongly, I am new and I am getting used to your language. I have not tried further, I want to see you guys making progress. I have the feeling that the puzzle beats me this time. I am waiting number 8…
January 22nd, 2006 at 12:42 pm
Turns out this puzzle is solvable by “logic”, but with amazingly complicated steps! Man, what a workout! Here it is:
———
Complete walkthrough for ND’s Killer Sudoku 6 alternate version:
1. r1c5+r2c456+r3c56 is at least 21 (6-cell cage) and it must be less than 22 (13-cage+9-cage).
Therefore the sum of these 6 cells is exactly 21. Hence r2c3=1, r1c46+r3c4=45-21=24={789}.
r1c5+r2c45=13-1=12={4(26|35)}, the 9(3)-cage r2c6+r3c56={1(26|35)} with the 1 within r3c56.
2. Now 1 in r1 is locked in r1c789, so the 19(5)-cage in n2/3 must contain a 1.
2 must also be in that cage, otherwise we’ll get at least 1+3+4+5+7=20, exceeding 19. So eliminate 2 from the rest of n3. Also, r1c789+r2c9 cannot be more than 19-7=12, so 7,8,9 are eliminated from these 4 cells.
3. Looking at the 12(4)-cage in n3/6, we can directly eliminate 7,8,9 (since 7+1+2+3=13 exceeds 12).
SC shows that 3,6 or 4,5 must be eliminated, so r4c8+r345c9={12(36|45)} -> 1,2 must be in r4c89+r5c9, and not rest of n6.
4. From the 16(4)-cage in n6/9, r7c9 must be 1 or 2, otherwise the cage will be at least 3+4+5+6=18 exceeding 16.
Now we get a hidden triple of {789} in r689c9. But it’s obvious r6c9 cannot be 9 (1+3+4+9=17 exceeds 16).
So r6c9={7|8} and r89c9={(78)9}, thus 9 is eliminated from the rest of n9 and r9c6 (29(5)-cage).
Also, r56c8 is at most 16-7-1=8 now, so can only be {3(45)}, and 3 is eliminated from there rest of n6/c8.
5. 45-rule on r5..9 -> r567c4+r5c9=8={12(34)}+{2|1}, so 1,2 are eliminated from r123489c4, r3c3 & r4c5 (34(7)-cage).
Also, r57c9 is a naked pair of {12} -> r4c8={1|2}, r1234c9={3456} with r34c9=12-1-2=9={36|45}, so r12c9=9={36|45} too.
Then r1c78={12}, so r1c6=19-1-2-9=7 -> r13c4={89}, a naked pair on r4. Also, r14c8 is a naked pair {12} on c8.
Moreover, 3 must be locked inside r123c9, so is eliminated from the rest of n3.
6. In n3, 7,8,9 are locked inside r23c78, so r4c67 cannot have 7,8,9 anymore (36(6)-cage).
So r4c79={(45)6} and 6 is eliminated from r4c1..6. Also r56c7+r6c9={(78)9}+{8|7} forms a naked triple.
7. This is the first critical step! 45-rule on n3 -> r4c6789+r5c9=22-7=15. Now combine these 5 cells with the 16(4)-cage!
We have this 9-cell region that sum to 31. Of these, r4c789+r5c89+r6c8={123456} sums to 21. So the rest sums to 10.
These are 3 cells: r4c6+r67c9. Since r6c9={7|8} and r7c9={1|2}, r4c6 must be less than 3, i.e. {1|2}.
We now have r4c68 a naked pair on r4. Combining with the naked pair r57c9, We have 2 pairs of {12} that sum to 6 together.
Of these, r4c8 and r5c9 is a pair of {12} in n6 which sums to 3, so r4c6+r7c9=6-3=3, and r6c9=10-3=7! Brilliant!
Now we have a naked pair of {89} in r56c7 -> r23c8={89} (hidden pair) -> r789c8={(45)67}, so eliminate 6,7 from r789c7. Also, r89r9 now is the naked pair {89} and 8,9 are eliminated from r9c6 (29(5)-cage).
8. Since r4c6={1|2}, r567c4={12(34)} can only have 1 or 2 in n5, i.e. r56c4={(12)(34)} and r7c4={1|2}.
And then we have a naked pair of {12} in r7c49. Also, r56c4 and r4c6 forms a “complex naked pair” of {12} in n5.
9. Now take a look at n1. The 5 cells outside the 26(4)-cage sums to 45-26=19. But r1c23=20-8|9=11|12 and r2c3=1.
So r3c13=19-1-11|12=6|7, and 6,7,8,9 are all eliminated from there, leaving {(2345)}.
10. Here comes the second critial step! Consider r3c34+r4c45. These 4 cells together with r567c4 forms the 34(7)-cage.
Since r567c4={12(34)}, the sum of these 4 cells is 34-6|7 = 27|28, coming from the candidates {(345789)}.
Subtraction combo shows a pair totalling 8 or 9 must be eliminated. This is either {35} or {45}. So 5 must be eliminated!
Now r3c3+r567c4 form a naked quad of {1234} within the 34(7)-cage! So eliminate 3,4 from r3c4+r4c45, and r4c4=7. Great!
11. Now the 19(4)-cage r3c1+r4c123 must be from {234589}. SC eliminates a pair totalling 12, which is 3,9 or 4,8.
So 2 & 5 stays in the cage, and r3c1=2, r4c123={5(39|48)}. Now r4c79 becomes the naked pair {46}, and this is big progress!
First, we now have r4c5=8 and r4c123={359}, a naked triple in n4. Then r56c8={35}, and r789c8={467}, a naked triple in n9.
More importantly, we now have cracked the series of {12} pairs: r7c9=16-3-5-7=1, so r5c9=r7c4=2, r4c68=[21], r1c78=[12]!
Also, r56c4=8-2-2=4={13}, a naked pair on c4/n5/34(7)-cage! So r3c3=4, r1c23=19-1-2-4=12={39}, r1c4=[89], r23c8=[98]!
Eliminate 3,9 from the rest of r1 & n1. Also, we can eliminate 5 from r23c7 (r789c7 is the naked triple {235}).
12. Focus on n2, the 9(3)-cage must be {135} now. So from this naked triple r2c5=2, r1c5+r2c4={46}.
Also, we have r3c569 a naked triple {135} on r3, making r3c27 a naked pair {67}. The top 4 rows are very clear now!
13. Now move on to the 31(5)-cage in n5/8. Since r56c56 is the naked quad {4569}, r5c7=31-4-5-6-9=7.
Also, notice the 5 in c4 is locked in r89c4, so we can eliminate 5 from the rest of n8.
14. Now is time for the third and last critical step! Notice r7c78=26-8-9=9=[36|54] (26(4)-cage).
Now apply 45-rule on n9, and we have the 4-cell region r8c5+r789c6=9+1+28+29-45=22 being from {134689}!
SC eliminates a pair totalling 9, which is 1,8 or 3,6. So 4 & 9 stay in these 4 cells, and is eliminated from rest of n8!
Now we have the naked pair {56} in r89c4. Eliminate them from rest of n8, c4 and the 23(5)-cage, and we’re almost done!
First, now the 4 cells region becomes {1489}, and r9c5=3. Also, r2c4=4, r1c15=[56], r1234c9=[4536], r4c7=4, r2c6=3.
15. Go to the 29(5)-cage in n8/9 now, r9c8 cannot be 7 because we cannot make r9c67=[(14)(25)] sum to 29-7-8-9=5!
Thus r8c8=7 (hidden single) and r9c678=29-8-9=12 coming from {(12456)}, and SC eliminates 1,5 or 2,4, giving r9c8=6.
So r7c78=[54], r89c7=[32], r89c4=[65], r9c6=29-8-9-6-2=4, r9c23=23-3-5-6=9=[18], r89c9=[89], r9c1=7.
16. From here on in the steps are almost trivial! The naked triple r7c123={369} will lead to r7c6=8 and r8c1=4, r8c23={25}.
Then the 22(4)-cages will give r67c1=[83] and r67c2=[69], and the rest are all naked singles! Amazing puzzle!
———
The 3 critical steps in (7), (10-11), (14-15) are the most complicated and quite innovative… I’m very proud of working them out… hope ND & the readers could understand and follow the logic! Any comments are welcomed!
January 22nd, 2006 at 5:59 pm
udosuk — wow, so it is solvable! Thanks for putting in the work — I’ve just edited together your original post + subsequent corrections: could you take a look to make sure I did this right? In any case I’ll be printing out the puzzle & your solution & giving them a walkthrough…. Hm, I think that the collective efforts of people on this site & on the DJape site have really raised the bar for killer “difficulty”: puzzles which would have been hopeless a few months back now seem relatively tamable, & the list of useful techniques & shortcuts continues to grow apace. I recently went back to the Times Killer book & worked through one of the puzzles in the last (”deadly”) section which had resisted my efforts last fall & now it seems if anything of only moderate difficulty.
January 22nd, 2006 at 6:10 pm
ND, glad you are working on another one.
I found this site http://www.killersudokuonline.com/
for lovers of killers. Hope I can post it here - I can’t on DJs, fair enough.
There is so few killer sites about I wanted to share it.
They do a daily one, easy getting to extreme and then a weekly one which appears really hard.
Anyways I hope you can share this with killer lovers.
January 22nd, 2006 at 6:40 pm
Yes, I’ve corresponded a little with the guy who runs that site — I didn’t initially post a link to it because the first puzzles he posted were pretty simple, but the latest ones he’s posted are pretty hard & well worth checking out. I can’t really get the online version of the puzzle to work well in IE but I prefer doing things on paper anyway…
I have no problem with people posting links here as long as it’s not, you know, “BUSTY GIRLS LOOKING FOR HORNY GUYS” or other spam-crud. DJApe’s got a specific product he’s trying to promote & preserve the value of, so I can understand why he might not want to permit links to competing sites; but this is just a blog to post up whatever I feel like. The more “professional” areas of my activities are on other areas of the blog (the press/mag & the music review section).
January 23rd, 2006 at 2:01 am
Udosuk,
You’re the king.
January 23rd, 2006 at 12:18 pm
OK: before I went to bed last night I worked through the solution: that’s great stuff indeed. If you’d been born a few generations back I’m sure you’d have been in Blechley Park cracking the Enigma Code….! You can see why I gave up & redrafted the puzzle originally: the “correct” solving path I had in mind would have required locating the 1 in r7c9 immediately after step 4, whereas in udosuk’s solution this issue is not resolved until step 11, amazingly enough. That’s 7 intervening steps! Basically, udosuk’s working against everything I’d set up against his solving path (since I’d anticipated/planned for a different path, but had left an unintended loophole). In other words, this is a pretty unusual puzzle. With my “good” puzzles, & I feel also with the Japanese ones I’ve seen, there’s a sense of “friendly camber” to the path, guiding you along — it may be hard to find the right “key” but once you do, the puzzle does lead you along the path at each step, if you think about it. With computer-generated puzzles like DJApe’s, the feeling is basically “neutral”–it’s more like sifting/panning for gold. In the case of #6-alt, we have a puzzle that is solvable without trial & error (udosuk’s solution never uses any hypotheticals, remarkably enough), but which is about as “unfriendly” as it can be!
The steps at 7 & 10 are very interesting indeed, as they combine familar techniques in novel ways which I certainly haven’t used myself in solving or creating puzzles & I don’t think I’ve seen any walkthroughs using them either. I think this demonstrates why no-one has yet created a computer solver that can solve these especially hard puzzles logically, rather than simply using brute-force numbercrunching: killer sudoku solving techniques are “modular” in a way that is unlike regular sudoku techniques — in other words, you can “nest” two or more techniques (as in a computer program) to produce results you couldn’t get from each technique on its own. (An instance being the modified “45 technique” which is instead a “21 technique” in step 7, & the combination of subtraction combo + range restriction in step 10.) This could conceivably be programmed in a computer, but it’d be unbelievably complex & cumbersome, & it would take vastly more computing power/time than just to have the program blast through all the combinations by trial & error.
There’s one typo in the walkthrough (step 11: “r1c4=[89]” should be “r1c4=8″), a few steps I feel could use some clarification, & one bit where there’s a small shortcut — I’ll type these in on my lunchbreak, & if udosuk OKs them I can incorporate them into the text above. Anyway, thanks for typing all that in: I do recommend any denizens of the site interested in these puzzles sit down & work through it: it’s as gripping a read as any mystery novel…!
January 23rd, 2006 at 1:42 pm
OK: corrections/suggestions:
step 5: minor point: you say “r1234c9={3456}” but it’s maybe worth noting that r4c9 is actually just {456}.
step 6: it required a moment’s thought to see why the 6 was “locked” in r4c79. Perhaps add a comment that r4c79+r56c8 is a naked quad {3456}.
step 9: again, small point: “leaving {(2345)}”: more exactly, in r3c3, only {(345)}.
step 10: typo in “critical”. Now, here’s a point where there’s a slight omission of a shortcut: the last sentence would work better as just “So r3c4+r4c45=34-4-3-2-1=24, i.e. {789}. So r4c4=7, r4c5={89}.” [we already know r3c4={89}].
step 11: having done this, we can revise a couple things here. The sentence after you locate {359} in r4c123 could read: “Then r4c5=8, r56c8={35}, and r789c8={467}, a naked triple in n9.” Delete “r1c4=[89]” later on in this step.
I also feel that there’s a minor clarification here that is useful in order to make step 12 obvious: insert somewhere after you place the naked pair {46} in r4c9: “So r3c9 = {35}.” This makes the comment on the naked triple in step 12 obvious.
step 14: there is an important step missing here after you whittle down the possible omitted pairs to {18} or {36}, since you then immediately proceed under the assumption that it’s {36}. What is needed is a clarifying sentence like: “But 8 is already blocked from r789c45, so r89c4={56}, r9c5=3.” (& rewrite the next sentence a little to reflect this).
And that’s it — do the above comments seem OK? If so I’ll paste them in to produce a definitive version of the solution.
January 23rd, 2006 at 6:39 pm
Nate,
you wrote:
> There’s one typo in the walkthrough (step 11: “r1c4=[89]” should be “r1c4=8″).
I think he meant r13c4=[89]….
January 24th, 2006 at 3:42 am
Thanks for the proof reading! I’ve actually did a lot of corrections since my last session on-line and I’ll post a new version up later… However I don’t have time now to read ND’s fixing and the proposed shortcut, so I’ll try and see if I’d improve after I read them. Will post it later tonight…
One thing is sure… I gained a lot in the process of solving this puzzle too… all credits to ND… and thanks for the complements Shai…
January 24th, 2006 at 7:59 am
New corrected version:
———
Complete walkthrough for ND’s Killer Sudoku 6 alternate version:
1. r1c5+r2c456+r3c56 is at least 21 (6-cell cage) and it must be less than 22 (13-cage+9-cage).
Therefore the sum of these 6 cells is exactly 21. Hence r2c3=1, r1c46+r3c4=45-21=24={789}.
r1c5+r2c45=13-1=12={4(26|35)}, the 9(3)-cage r2c6+r3c56={1(26|35)} with the 1 within r3c56.
2. Now 1 in r1 is locked in r1c789, so the 19(5)-cage in n2/3 must contain a 1.
2 must also be in that cage, otherwise we’ll get at least 1+3+4+5+7=20, exceeding 19. So eliminate 2 from the rest of n3.
Also, r1c789+r2c9 cannot be more than 19-7=12, so 7,8,9 are eliminated from these 4 cells (7+1+2+3=13 exceeds 12).
3. Looking at the 12(4)-cage in n3/6, we can directly eliminate 7,8,9 (7+1+2+3=13 exceeds 12).
SC shows that 3,6 or 4,5 must be eliminated, so r4c8+r345c9={12(36|45)} -> 1,2 must be in r4c89+r5c9, and not rest of n6.
4. From the 16(4)-cage in n6/9, r7c9 must be {1|2}, otherwise the cage will be at least 3+4+5+6=18 exceeding 16.
Now we get a hidden triple of {789} in r689c9, so r67c9 is at least 1+7=8. Now r56c8 must not exceed 16-8=8.
Therefore r56c8 could only be {3(45)}, and 3 is eliminated from the rest of n6/c8.
5. 45-rule on r5..9 -> r567c4+r5c9=8={12(34)}+{2|1}, so 1,2 are eliminated from r123489c4, r3c3 & r4c5 (34(7)-cage).
Also, r57c9 is a naked pair of {12} -> r4c8={1|2}, r1234c9={3456} with r34c9=12-1-2=9=[36] or {45}, so r12c9=9={36|45}.
Then r1c78={12}, so r1c6=19-1-2-9=7 -> r13c4={89}, a naked pair on c4. Also, r14c8 is a naked pair {12} on c8.
6. In n3, 7,8,9 are locked inside r23c78, so r4c67 cannot have 7,8,9 anymore (36(6)-cage).
Now r4c79={(45)6}, so eliminate 6 from the rest of r4. Also r56c7+r6c9 is now a hidden triple of {789}.
7. This is the first critical step! 45-rule on n3 -> r4c6789+r5c9=22-7=15. Now combine these 5 cells with the 16(4)-cage!
We have this 9-cell region that sum to 31. But the 6 cells r4c789+r5c89+r6c8={123456} total 21, so the rest must total 10!
These are 3 cells: r4c6+r67c9. Since r67c9 is at least 7+1=8, r4c6 cannot exceed 10-8=2, i.e. r4c6={1|2}.
We now have r4c68 a naked pair on r4. Combining with the naked pair r57c9, We have 2 pairs of {12} that total 6 altogether.
Of these, r4c8 and r5c9 is a naked pair of {12} in n6 which sum to 3, so r4c6+r7c9=6-3=3, and r6c9=10-3=7! Brilliant!
Now we have a naked pair of {89} in r56c7 -> r23c8={89} (hidden pair) -> r789c8={(45)67}, so eliminate 6,7 from r789c7.
Also, r89r9 now is the naked pair {89} and 8,9 are eliminated from r9c6 (29(5)-cage).
8. Since r4c6={1|2}, r567c4={12(34)} can only have either 1 or 2 within n5, i.e. r56c4={(12)(34)} and r7c4={1|2}.
And then we have a naked pair of {12} in r7c49. Also, r56c4 and r4c6 now form a “complex naked pair” of {12} in n5.
9. Now take a look at n1. The 5 cells outside the 26(4)-cage sums to 45-26=19. But r1c23=20-8|9=11|12 and r2c3=1.
So r3c13=19-1-11|12=6|7, and 6,7,8,9 are all eliminated from there, leaving {(2345)}.
10. Here comes the second critical step! Consider r3c34+r4c45. These 4 cells together with r567c4 form the 34(7)-cage.
Since r567c4={12(34)}, the sum of these 4 cells is 34-6|7 = 27|28, coming from the candidates {(345789)}.
Subtraction combo shows a pair totalling 8 or 9 must be eliminated. This is either {35} or {45}. So 5 must be eliminated!
Now r3c3+r567c4 form a naked quad of {1234} within the 34(7)-cage! So eliminate 3,4 from r3c4+r4c45, and r4c4=7. Great!
11. Now the 19(4)-cage r3c1+r4c123 must be from {234589}. SC eliminates a pair totalling 12, which is 3,9 or 4,8.
So 2,5 stay in the cage, thus r3c1=2, r4c123={5(39|48)}. Now r4c79 becomes the naked pair {46}, and this is big progress!
First, we now have r4c123={359}, a naked triple in n4 and r4c5=8. Then r56c8={35}, and r789c8={467}, a naked triple in n9.
More importantly, we now have cracked the series of {12} pairs: r7c9=16-3-5-7=1, so r5c9=r7c4=2, r4c68=[21], r1c78=[12]!
Also, r56c4=8-2-2=4={13}, a naked pair on c4/n5/34(7)-cage! So r3c3=4, r1c23=19-1-2-4=12={39}, r13c4=[89], r23c8=[98]!
Eliminate 3,9 from the rest of r1 & n1. Also, we can eliminate 3,5 from r23c7 (r789c7 is the naked triple {235}).
12. Focus on n2, the 9(3)-cage must be {135} now. So from this naked triple, r1c5+r2c4 is the naked pair {46} & r2c5=2.
Also, we have r3c569 a naked triple {135} on r3, making r3c27 a naked pair {67}. The top 4 rows are very clear now!
13. Now move on to the 31(5)-cage in n5/8. Since r56c56 is the naked quad {4569}, r5c7=31-4-5-6-9=7.
Also, notice the 5 in c4 is locked in r89c4, so we can eliminate 5 from the rest of n8.
14. Now is time for the third and last critical step! Notice r7c78=26-8-9=9=[36|54] (26(4)-cage).
Now apply 45-rule on n9, and we have the 4-cell region r8c5+r789c6=9+1+28+29-45=22 being from {134689}!
SC eliminates a pair totalling 9, which is 1,8 or 3,6. So 4,9 stay in these 4 cells, and is eliminated from rest of n8!
Now we have the naked pair {56} in r89c4. Eliminate them from rest of n8, c4 and the 23(5)-cage, and we’re almost done!
First, now the 4 cells region becomes {1489}, and r9c5=3. Also, r2c4=4, r1c15=[56], r1234c9=[4536], r4c7=4, r2c6=3.
15. Go to the 29(5)-cage in n8/9 now, r9c8 cannot be 7 because we cannot make r9c67=[(14)(25)] sum to 29-7-8-9=5!
Thus r8c8=7 (hidden single) and r9c678=29-8-9=12 coming from {(12456)}, and SC eliminates 1,5 or 2,4, giving r9c8=6.
So r7c78=[54], r89c7=[32], r89c4=[65], r9c6=29-8-9-6-2=4, r9c23=23-3-5-6=9=[18], r89c9=[89], r9c1=7.
16. From here on in the steps are almost trivial! The naked pair r8c56={19} will lead to r7c6=8 and r8c1=4, r8c23={25}.
Then the 22(4)-cages will give r67c1=[83] and r67c2=[69], and the rest are all naked singles! Amazing puzzle!
———
I haven’t added ND’s modifications, but I read them all and think they are all okay and would help the reader to understand it more easily. So if you want you can append them to my version above and then produce a final version to publish…
BTW I’m interested (and I think others are too) about the original “correct” solving path in your mind which requires T&E… Also look forward to #8!
January 24th, 2006 at 5:50 pm
Hi everyone,
There’s a new site for killer sudokus:
http://www.killersudokuonline.com/
They post a daily killer, a weekly killer which is quite extreme, and also a weekly “Greater Than” sudoku, which is also challenging.
Their weekly #3, is impossible.
So Udosuk,
Here’s you next challenge:
http://www.killersudokuonline.com/#weekly
January 24th, 2006 at 5:53 pm
Yeah, I’ve been working on that one (#3) & though I’ve made some progress (it’s easy to locate 9 in n2 for instance via 45 rule) it’s tough sledding. Not “impossible”, though, I hope!
January 25th, 2006 at 9:47 am
That #3 from hell is exactly what I’m working on & off for the last couple days. So it’s not only me who has trouble with it, eh? It actually gives me pains and makes me think that I’m missing something obvious again, like working on the whole “table of combinations” (I usually only remember the pairs and use subtraction combo to sieve out large groups of candidates)…
But I know it’s valid and has a unique solution, because I recently acquired a simple program (from dukuso of sudoku.com forums) which solve killers by T&E and checks if solutions are unique, etc. That one only costs a few seconds on that program, while Nate’s #7 takes 5 mins or so. And very interestingly, that program says #6 alt is unsolvable! I think it might be because of the 7-cell cage there, but I’ll notify dukuso about it…
Will let you guys know when I finally crack it…
January 25th, 2006 at 10:15 am
I think the thing with #3 is that (to my eye) it looks like, if I weren’t so stubborn about not using T&E it would be quickly solvable — once you use the 45 rule to slice it up, there are plenty of areas of the puzzle with only two or three combinatorial options. But I’m holding out for a satisfying solution not a guessing-game… I hope that the folks who run that site will actually post a walkthrough rather than just the bare solution.
That’s rather amusing about dukuso’s program & 6-alt! Incidentally, there’s an excellent, if ad hoc, way of “grading” difficulty for the hardest puzzles: the Perfect Sudoku program indicates how many levels of trial & error it needed to solve the ones it can’t do by pure logic. This is what you get:
#4 - 4 levels
#6 - 5 levels
#7 - 7 levels
#6-alt - 9 levels
which is about the right ranking for them in terms of difficulty.
January 25th, 2006 at 10:20 am
Here is a little pic I created on that weekly #3 puzzle:
http://i1.tinypic.com/mk7hfo.png
Where innies/outies are highlighted but no candidates yet. Maybe next time…
January 25th, 2006 at 10:35 am
ND, when you said “it’s easy to locate 9 in n2 for instance via 45 rule”, could you elaborate more? I know the 9 would be inside the 22(3)-cage but I can’t see where exactly it should be…
That one is totally opposite to your puzzles because it has a lot of small cages (2..3 cells) while you aim to make them at least 4..5 cells. So it’s surprising for being so difficult… It shouldn’t be call “mind-bending”, but “mind-wrecking”… It’s hard but it’s not enlightening or inspiring (to me)… I truly hope they hadn’t just picked that puzzle out with a program that makes sure it has a unique solution but hadn’t checked if there’s indeed a perfectly logical solving route…
January 25th, 2006 at 11:31 am
One thing I see in #3,
is that r4c1256=26, which, together with the 11(3) in N4 (r4c12+r5c1), show that r5c1={12} and there’s a 9 in r1c56.
Maybe this will help you guys.
January 25th, 2006 at 11:39 am
Yeah, just checking it out now I think I goofed up in the logic for that placement of the 9 in n2. Back to square one. I think I’ll just skip KSO#3, it doesn’t to my eye look like there are any more solution pathways to try beyond trial & error. Time better spent on finishing #8. I’ve been busy with trying to write a feature article on the pianist Michiel Braam, as well as going to the Cinematheque last night for the last instalment of The Apu Trilogy, but have been assiduously working on #8. What I’ve been trying to do is work out a more efficient method for creating puzzles with less grief (i.e. screwups) — I think I have it now. It involves working forwards & backwards (from cages -> solution, from solution -> cages) via facing pages with duplicate puzzle grids. So I can doublecheck my work from both directions instantly.
Yeah, I’ve been trying to avoid 2-cell cages entirely & keep 3-cells ones to a minimum. It’s something of a mannerism, & there are certainly many puzzles which have 2-cell cages which are very hard (like KSO#3!, or, more positively, some of JC Godart & DJApe’s puzzles) so it’s not like they necessarily make a puzzle easy, but they kind of limit the kinds of techniques that you can use, I find. & also, computer programs tend to generate lots of 2 & 3 cell cages, so I try to create puzzles that have a more “planned” (human-made, quirky) feel to them.
January 25th, 2006 at 1:22 pm
I just solved #3.
I’ll post the detailed solution later if asked for, don’t have much time now.
I’ll just say that the 9 in N2 was located only in the later phases of the solution….
basically what I did is: “45″ on R1 -> r2c46 - r1c3 = 14.
This means that r1c3 is {123} and r2c46 is {(96|87)(97)(98)} respectively.
Checking these combinations TOGETHER with the possible combinations for the 12 & 7 cages in R1 gives only two possible results, in both of them 12-cage={48}, 7-cage={25}, r2c6=8, 22-cage={976}, and from there the road is bumpy but but steady…
January 26th, 2006 at 10:27 am
Yep, I’ve also solved it… The first cell I determined was r6c4=1, and then worked thru n5 then after a short while the “avalance” came because of the amount of 2- & 3-cell cages/hidden cages there… The beginning is definitely evily difficult but the rest couldn’t compare to ND’s ones. Though range manipulation, synchronization & overlapping sums are still required so it’s still of “demonic” rating (i.e. like Misawa’s & ND’s ones). So I guess they still deserve a lot of credits…
But I still want to stress that ND’s puzzles are definitely much more attractive, as the difficulties are “even out” during the whole process, unlike the KSO ones which are only extremely hard at a certain stage (for
this one the first stage before any cell is filled in), but will be much easier at the other parts. ND’s ones are like a great blockbuster movie that’s exciting from start to end while the KSO ones only have certain spectacular scenes at times. And that’s why ND said my walkthrough here was like a mystery novel - the puzzle provided all the thrillers…
This time surely Shai deserves the full credit as his first tip yesterday set off my solving, making me realised that range manipulation was required there (before I’ve underestimated the difficulty a bit). So like before let’s both post our walkthroughs later and compare… Later…
For an excuse, I could only dedicate a very limited amount of leisure time for this one (around 1 hr for each of the last 3 days) on this puzzle, as I’m watching a lot of Aussie Open tennis these couple weeks. (Some visitors here know that I’m both an Aussie and a tennis fan). Pity my favourite Martina Hingis lost in the quarters but I found another exciting player in Baghdatis from Cyprus who just won another entertaining match tonight…
Oh, and lastly, any chance ND has the #1 to #4 puzzle from the KSO? I couldn’t find them in their archives.
January 26th, 2006 at 7:51 pm
OK,
Here’s my solution to Killer-sudoku-online-weekly #3.
First of all I want to thank ND for letting us doing this on his site. Our nice killerholic community cannot do it on DJ Ape’s site for obvious reasons, and since ND’s is also trying to solve their puzzles, everybody can benefit from moving part of our killer-discussions here…
Secondly, I want to join Usdosuk in his admiration for ND’s killers, and express my undying anticipations for his No. 8…..
Thirdly, I do like Killer-sudoku-online, because they post a killer everyday, which is harder than Times’, and are a bit different from all other sources available. I think this is one of the greatest thing about killers, every manufacturer has something different to offer. The thing with KSO-Weekly-killers, is that I still can’t make up my mind if they’re just hard in the Trial & Error sense, or in the logical sense, I wonder if there’s something we keep missing.
About Usdosuk’s searches for KSO’s-weekly #1 & #2, They’re available in their archive with the daily killers of those weeks, at the bottom of the page:
http://www.killersudokuonline.com/apage270870.html (First weekly, called H0)
http://www.killersudokuonline.com/apage270887.html (second weekly, called H1).
Enough said, let’s look at the solution:
I assume that all the hidden cages (from Udosuk’s link) are known (if not, check a few posts before).
1. “45″ on R4 -> r4c1256=26={9872|9863|9854|9764|8765}.
In all options, considering the 11(3)-hidden-cage in N4, we see a 9 in r4c56, and that r5c1={12}. ({8765} is rejected because no two of its digits could be a part of a 11(3)-cage), {9872} is rejected, will cause two 2’s in the 11-cage). The 9 in r4c56 forces the 9 of the 22-cage in R5 to be in r5c78. Also remember that r4c12 = 9 or 10.
2. “45″ on R1 -> r2c46 - r1c3 = 14. This means that r1c3 is {123} and r2c46 is {(96|87)(97)(98)} respectively. Let’s look into these options, considering both the 12 & 7 & 22 cages of R1.
A. First option, r1c3=3, r2c46={98}. Check both options for r2c46:
r2c46=[98] -> r1c45={76} -> 12-cage={48} -> 7-cage={25}, r1c67=[19]. Looks fine.
r2c46=[89] -> r1c45={95}, and we get two 9’s in N2.
B. Second option, r1c3=2, r2c46={97}.
r2c46=[79] -> r1c45={69}, and we get two 9’s.
r2c46=[97] -> r1c45={85} -> 12-cage={39} -> 7-cage={16} -> r1c67={47}, and we get two 7’s in the 18-cage.
C. Third option, r1c3=1, r2c46={96}.
r2c46=[69] -> r1c45={97} and we get two 9’s in N2.
r2c46=[96] -> r1c45={85} -> 12-cage={39} -> no options for 7-cage.
D. Fourth option, r1c3=1, r2c46={87}.
r2c46=[87] -> r1c45={95} -> 12-cage={48}, and we get no options for the 7-cage.
r2c46=[78] -> r1c45={69} -> 12-cage={(48)|(57)}.
12-cage={57} -> 7-cage={34} -> r1c67={28}, and we get two 8’s in the 18-cage.
12-cage={48} -> 7-cage={25} -> r1c67=[37], looks fine.
So we got only two possiblilties, in both of them:
12-cage={48}, 7-cage={25}, 22-cage={976}, r2c6=8, 6 in r1c45. And we also see that r1c3={13}, r3c23={59} (hidden 14-cage) -> r2c5=5.
3. Since r3c23={59}, there’s no 5 in the 11(3)-hidden-cage of N2, so the 26-hidden-cage from step 1 is {9863|9764}, so there’s a 6 in r4c12 -> r7c2=6 (23(3)={986}), r4c1=6, r2c3=6, r6c23={89}.
4. 8 in N6 cannot go in top row (10(3)-hidden-cage) or bottom row, must be in the middle. If it’s a part of the 22-cage, then r5c6=5 will force r6c6=8 (13(2)-hidden-cage), which is rejected because r2c6=8. So r5c9=8, And we get 22-cage={976}, 13(2)-hidden-cage in N5 ={76}, r3c8=8, r3c9=6.
5. “45″ on R3 -> r3c1=7.
6.
A little trick:
10-cage in R5 = {145|235} (no 6 or 7) -> r5c1234={(125)34} ->
r5c5={34} -> 7(2)-hidden-cage in N5 = {25} (no 6, no{34}) -> r4c56=[89] (only one place for the 8 in N5) -> r4c2=3 (26-hidden-cage from step1), r5c1=2, r45c4=[25], r4c3=7, r5c23={14}, r6c1=5, r6c45={14}.
7. r4c789={145}, SC on the 17-cage shows that r4c78={45}, r4c9=1 -> r2c9=4.
8. r9c89=[69] -> r78c1={49} (must be a 9 in r789c1) -> r2c1=1 (1 in r9c9 will cause two 9’s on N7) -> r1c3=3, r1c12=[84], r2c4=9, r1c45={76}, r9c12=[37], r5c23=[14], r2c78={37} -> r3c4567=[3241], r1c67=[19], r8c23=[81], r79c3=[25] (can’t have a 5 in 10(2)-cage),….
And you carry on…
January 26th, 2006 at 8:04 pm
Udosuk, I thought you meant KSO’s first 2 weeklys, now I understand you meant their first 4 dailies…NM
January 26th, 2006 at 10:11 pm
Yeah, I’ve no problem with people posting here on whatever strikes their fancy, as long as it’s not spam or otherwise objectionable content. — #8’s in a funny state: I filled out the grid today at last but I made a logical goof at one point & so I’m not sure it’s solvable, at least in the way I wanted. If I can’t patch it up I will simply leave it as an “alternate” version again & then post the revised version.
Re: udosuk’s query about #6 — the solving-path I’d had in mind would have had the solver, once the troublesome {12} cell in nonet 9 was placed, go work on nonets 1-2, which is more or less the path that one takes for the corrected version (or at least, which I take).
I think the main reason I can’t quite figure out the Killer Sudoku Online site is just because it’s so bare there’s little information to go on: there is no comments function or forum discussion, the info on the site proprietors is slim, there is no indication of how the puzzles were created (pure human effort? computer generation? a mix of both?), & no walkthrough solutions are given. I have had a little email correspondence with Seth, the guy who runs it, & my impression from that is that they’re human-made. Anyway, it’s a good source of puzzles, but it’s not really a place to hang out because of the lack of interactivity.
Well, thanks for the kind words on my own creations! Yeah, a lot of the effort goes into ensuring that they don’t crumble easily: I hate it when there’s an extended fill-in-the-blank stage. In part this is just covering my ass: I figure, even if I goof up somewhere & there’s an easier entryway to the puzzle than I’d intended (it’s impossible to anticipate all angles), then eventually the solver WILL come up with something hard.
January 28th, 2006 at 11:50 pm
Thanks, Shai for the walkthrough and ND’s explainations…
Now that I solved some more KSO’s puzzles, I also think I begin to admire their (or his if “Seth” is the only guy responsible) efforts to produce so many Killer puzzles with reasonable to great difficulties… Credits should be given when they’re due… I’ve solved H0 with some hard work and am currently working on H1… I think they both warrant a complicated walkthrough…
As for weekly #3, last time I made a blunder so ignore my claim that “the first cell was r6c4=1″… Actually the first critical step is to eliminate 1 from r6c7, which I’ll explain in detail when I get to write the walkthrough, hopefullly sometime next week…
So I take it ND also doesn’t have the daily #1-4… it’s alright so my collection starts from daily #5 and the 3 weeklys + the “greater than” one…
ND’s #8 has been released and I’ve posted a link in case anyone couldn’t save the pic with the browser (which happened to me, but the link works fine)…
January 31st, 2006 at 8:52 pm
Have you all noticed the killersudokuonline ones get much harder towards the end of the week?
I can now keep Pat totally UNbored in his spare time.
I also appreciate being able to discuss this ‘other’ site here Nate. I tried a couple of times, accidently, at DJs and they were pulled.
My way of thinking is that if you enjoy these killers why not share the info available amongst the legion of loyal fans. You are not going to lose a fan as obviously there is not enough sites posting them for lovers of killers to be kept happy.