Killer for the new year

For fans of Killer Sudoku, Samurai Sudoku and other variants

Killer for the new year

Postby nd » Fri Jan 06, 2006 2:07 am

Just finished work on Killer #7:

It's a pretty hard one, I think. Happy solving!
Posts: 28
Joined: 11 September 2005

Postby udosuk » Mon Jan 09, 2006 8:00 pm

The following is my complete walkthrough to this puzzle. You could copy & paste the "ants" text to notepad and read if you're hopelessly stuck... But I suggest everyone to try at least once to crack this marvelous puzzle before referring to the spoilers!

Here is my complete walkthrough to ND's Killer Sudoku 7.
I think it is quite simple and clear enough and hopefully it's understandable to everybody!

Step 1:
Since R[156]C4=8 and R6C[4589]=28, R6C4={45}, R[15]C4={12} or {13}, R6C[589]={689} or {789}.
Since R6C[12367]=45-28=17, R[45]C3+R[45]C7=26+23-17=32=15+17 or 16+16 or 17+15. So R[45]C3 & R[45]C7 are both from {6789}.
Since R6C[89] and R[45]C7 are both from {6789}, they form a "naked quad" in nonet 6, so the remaining 5 cells are {12345}.
Since R[345]C[89]=14+15=29 and R3C[89] is at most 17, R[45]C[89]=(1+2+3+4+5)-R6C7 is at least 12, so R6C7=1, 2 or 3.

Step 2:
Since R6C[89]+R[45]C7=6+7+8+9=30 and R[45]C7 ranges from 15 to 17, R6C[89] ranges from 13 to 15.
On the other hand, R6C[45] cannot exceed 5+9=14, so R6C[89] must be at least 14 to make up the total of 28.
So R6C[89]=14..15, R[45]C7=15..16, R6C[67]=7..8 to make up the 23/4 cage.
Now that R6C7={123}, R6C6 must be from {4567} (1,2,3 are eliminated to make sure the pair sums to at least 7).
Further, 6,7 are also eliminated because R[45]C7 & R6C[89] are in the same cage/row of R6C6 and they already cover {6789}.
So R6C6={45}, forming a naked pair with R6C4!

Step 3:
Last step we had R[45]C7=15..16, so R[45]C3=16..17 to make up 32, i.e. {79} or {89}.
Also, R6C[123]=9..10 to make up the 26/5 cage which now must be from {1236789}.
"Subtraction combo" shows that a pair totalling 10 must be eliminated.
These can be {19},{28} or {37}. But 6 must stay in the cage, which must be within R6C[123]. So R6C[123]={126} or {136}.

Step 4:
Now R6C[89]=14..15 coming from {789}.
This can only be {78}, forcing R[45]C7={69}, R6C5=9, R6C4=28-7-8-9=4, R6C6=5, R6C7=23-6-9-5=3.
Also, Now that R6C[123] must be {126}, R[45]C3 must be 26-1-2-6=17, i.e. {89}.

Step 5:
Now we look at nonet 3. Since R[45]C[89]={1245}=12, R3C[89]=14+15-12=17={89}.
Also, R[123]C7=45-14-17=14 and must be from {12457} now. Subtraction combo eliminates {14}, leaving {257}.
Comparing the 17/3 cage and these 3 cells, R1C6=R3C7+3, and the only remaining possibility is 8=5+3. So R1C6=8, R3C7=5.

Step 6:
Focus on C4 now. First we have R[15]C4={13} to make up the 8 with R6C4.
Next, we must have R[789]C4=19. However, R[78]C4 belongs to the 14/4 cage, and cannot exceed 11.
So R9C4=8..9, R[78]C4=10..11, R7C[23]=3..4 i.e. {12} or {13}, so the 1 in R7 must remain in these 2 cells.

Step 7:
Next we go to nonet 9. We already have the naked triple {148} in R[789]C7.
Comparing with the 13/3 cage, we have R7C6=R9C7 and the only common candidate is 4. So R[789]C7=[814] & R9C7=4.
Now we have the part of the 25/5 cage in this nonet summing to 10, which is easily determined as {235}.
The 22/3 cage is of course what's left, {679}.

Step 8:
Since R[345]C6=21-5=16 comes from {12367}, subtraction combo quickly eliminates 1 & 2, leaving the naked triple {367}.
Going down to the 16/4 cage at the bottom, we should see that the 1 in R9 must be in R9C5 or R9C6.
This together with R8C6 being {29} now fixes the cage as {1249}. So 1,2,9 could be eliminated from the rest of nonet 8.
Therefore R9C4=8, R[78]C4=11={56}, R7C[23]=3={12}.
Now we have the naked pair {35} on R7C[89] which basically cracks down the bottom 3 rows!
Still a few twists and turns before we get to the finish line...

Step 9:
Look at the 16/4 cage in the centre. Now it must be from {123468}.
Subtraction combo eliminates a pair summing to 8, i.e. 2 & 6, leaving {1348}.
So, R[45]C5={18} and this naked pair almost cleans up the central 3 columns!

Step 10:
Go to nonet 1 now. R3C[123] is the naked pair {126}.
More importantly, since R[45]C[12]=45-26=19, comparing to the 17/4 cage gives R4C1=R3C2+2.
With R4C1={3457}, we must have R4C1={34}, R3C2={12}, giving us a naked pair in C2 and forcing R6C2=6.
This in turn gives us the naked pair R[67]C3={12} which forces R3C3=6.
To complete the 36/6 cage, we must have R2C[23]=12, and 8+4 is the only option left!

The puzzle is as good as finished now!
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Joined: 17 July 2005

Postby nd » Mon Jan 09, 2006 10:45 pm

Glad you liked this one! Definitely my hardest so far, but I think the solving path flows logically & clearly once you get the hang of it.
Posts: 28
Joined: 11 September 2005

Postby PaulIQ164 » Mon Jan 09, 2006 10:56 pm

What's the notation used in the solution there? I'm afraid I'm not following it.
Posts: 533
Joined: 16 July 2005

Postby nd » Tue Jan 10, 2006 12:19 am

In udosuk's solution I take it you mean? He uses a similar notation to mine, with a few variants:

26(5) = a 5-cell cage that adds to 26

R = row

C = column

so R[156]C4 means the three cells R1C4, R5C4, R6C4.

Curly brackets mean "unordered set" (of numbers): so {689} could be 6, 8, 9, or 6, 9, 8, or 8, 9, 6, or....

There is some shorthand that refers to common strategies: "45 rule" and "subtraction combo" for instance. The former is explained in the tips page of my site ( I'll add "subtraction combo" to that later this evening...
Posts: 28
Joined: 11 September 2005

Postby PaulIQ164 » Tue Jan 10, 2006 1:43 am

Ah yes, I see. It turns the notation meant what I thought it did, I was just being dumb and not seeing the reasons for the steps written. All better now.
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Joined: 16 July 2005

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