Solution to Tough Sudoku of 12/21/07 Page One


The following is an illustrated solution for the sudoku.com.au tough puzzle of 12/21/07. This particular tough puzzle is somewhat diabolical. However, compared to the most diabolical Easter Monster, it is trivial!. Hopefully, some of the Sudoku tips & tricks used in this solution will be helpful. I believe some of the steps that I found are at least interesting.

If this is your first visit to this blog, WELCOME!!

Previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries.

The illustrations of steps shown on this page will share this key:

  • black line = strong inference performed upon a set (strong link)
  • red line = weak inference performed upon a set (weak link)
  • black containers define a partioning of a strong set(s)
  • candidates crossed out in red = candidates proven false
Please be aware that, for me, strong and weak need not be mutually exclusive properties.


Given Puzzle - 23 givens


Puzzle as given








One Unique Possibility exists:

  1. (7)r9c4 %box&row
Thus UP24.


Steps 2a,2b,2c


First Three Easy Steps


Above, find three easy steps:

  • 2a) Locked 8's r23c1 => r56c1≠8
  • 2b) (7): r6c3=r6c8-r2c8=r2c2 => r1c3,r4c2≠7
  • 2c) (2=9)r8c8-(9)r8c2=(9-3)r7c1=(3)r7c8 => r7c8≠2
Note that step 2c could have been ignored, as that 2 will be eliminated by a host of other steps, including possibly the next one. However, I decided to let it stay. Finding this easy step certainly helped to find the next one. It is often the case that similar pieces of the puzzle lead to multiple elimination steps.


Step 2d - Short Chain using an Almost Hidden Triple


AHT step 2d


One way to write the step shown above:

  • (2=9)r8c8-(9)r8c2=(Hidden Triple 359-1)r7c138=(1)r8c3 => r8c3≠2
If one wishes, one can alternatively use the Almost Naked Quad:
  • (2=9)r8c8-(9=Quad1246)r7c4569-(1)r7c13=(1)r8c3 => r8c3≠2
In either case, the 2 from step 2c could be eliminated by this step. Just consider the chain without the strong inference set on 1's in box 7. The standard puzzle mark-up that I use heavily indicates parts of this chain, as cell r8c3 has candidate 1 circled, while 35 are circled in row 7, columns 138. The circled 9's in box 7 plus cell r8c8 complete the indicaters. Note that the mark-up that I use is prejudiced towards Hidden Sets, thus in most cases I will find the Almost Hidden Set deductions first. After that, the existence of a complimentary Almost Naked Set deduction is certain.


Step 2e A valuable net using Kraken box 7s and an Almost Hidden Triple.


After marking this puzzle lightly, the circled candidates made it hard to ignore the fact that candidates 7 & 8 are practically tripping over each other. That theme will present itself more than once. In the case below, one needs to consider all the possible locations for candidate 7 in box 3. The four locations are divided into three groups.

Kraken box 7's net


The eliminations depicted above are justified with the following net-like Alternating Inference Chain:

  • (7)r3c9=*
  • [(8)r6c5=(8-7)r6c3=(7)r6c8-(7)r12c8=*(7)r1c9-(7)r1c6=(Hidden Triple 357)r345c6]
  • -(8)r5c6=(8)r8c6-(8)r8c9=(8)r3c9
  • => (7=8)r3c9 => r3c9≠249
The partitioning of the AIC shown above indicates that this wrap-around chain only proves the following 2 weak links are strong:
  1. (7-8)r3c9 => the conclusion written above
  2. (8)r8c6-(8)r8c9 => (8)r8c6=(8)r8c9, but we already have that as a native strong inference set. (sis)
One could alternatively use cell r7c6 containing (1=6) and cells r1c6,r8c6 as two Almost Naked Pairs instead of the Almost Hidden Triple illustrated above. It seemed more straightforward to me to use the Almost Hidden Triple in this case. However, there is really no increase in complexity using two Almost Naked Pairs. Also, one could just think in terms of one Almost Naked Triple (1678) using the same cells: r178c6.

No matter how one approaches it, the almost sets in column 6 are key in realizing that one should consider the 7s in box 7 as a sis group. Moreover, they help to determine the grouping. One could of course place (7)r1c8 into a column group or a row group within that box (or even both!). The choice made above was completely arbitrary, and quite irrelevant towards the final deduction.

After making the depth 8 eliminations shown above, an easy deduction follows. I will not illustrate it to save some space:

  • 2f) Locked 4's r3c12 => r1c23≠4


Step 2g Another Almost Hidden Triple chained


It seems like an early theme here is the consideration of Almost Hidden Triples, or their Almost Naked conjugates.

AHT 148


The newly found sis(7=8)r3c9 and the newly stronger 4's in box 1 are easy keys to locating this one:

  • (7=8)r3c9-(8)r8c9=(8-1)r8c6=(1)r8c3-(1)r7c1=(Hidden Triple 148)r23c1r3c2 => r3c2≠7
The conjugate almost naked set counterpart (ALS):
  • (7=8)r3c9-(8)r8c9=(8-1)r8c6=(1)r8c3-(1)r1c3=(Naked Triple 267)r1c3r12c2 => r3c2≠7


Step 2h Reusing the Almost Hidden Triple from step 2e as an Almost Hidden Pair


The newly stronger 7's in in row 3 make this one a breeze, given step 2e. One could use the same Almost Hidden Triple, but it seems more clear in this case to consider the Almost Hidden Pair 35.

AHT 357 again


This one is a wrap around forbidding chain, also called a Continuous AIC, and it practically writes itself:

  • (Hidden Pair 35)r45c6=(5-7)r3c6=(7-8)r3c9=(8)r8c9-(8)r8c6=(8)r5c6 =>
  • All the weak links above form sis =>
    1. (HP35)r45c6=(8)r5c6 => r5c6≠6
    2. (5=7)r3c6 => r3c6≠1
    3. (7=8)r3c9 , but we already had that
    4. (8)r8c9=(8)r8c6, but we already had that


Step 2i Another Wrap Chain primarily focused on candidates 7,8


I am still beating up on candidates 7 & 8 tripping over each other. I just needed candidate 6 a bit.

Continuous AIC using primarily 78 again


One could start anywhere, but I saw it looking first at the newly stronger 6's in box 5:

  • (6)r6c5=(6)r56c4-(6)r2c4=(6-7)r2c2=(7)r2c8-(7)r6c8=(7-8)r6c3=(8)r6c5
  • => Continuous AIC, thus all weak links are proven strong:
    1. (6=8)r6c5 => r6c5≠49
    2. (6)r56c4=(6)r2c4 => r7c4≠6
    3. (6=7)r2c2 => r2c2≠2
    4. (7)r2c8=(7)r6c8 => r1c8≠7
    5. (7=8)r6c3 => r6c3≠45

I am almost done beating up on 78. However, several wrap around chains using 78 exist that could forbid 7 from r1c9. I will not execute that step now, as that 7 will fall in a more powerful wrap around chain on a later page.


Notes

It is very common that once one identifies a particular group suitable for attack in a puzzle that one can continue to focus on that group for quite some time. In my opinion, when that group tends to produce continuous AIC, or wrap around chains, then the deductions are easier to find. The added bonus of elegance and multiple eliminations makes such a focus often particularly rewarding, and more importantly, fun!

This concludes the first page of this solution. The next page should be soon forthcoming.




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