paulv66 wrote:

I found yesterday's difficult killer sudoku very hard to get to grips with. I managed to solve it eventually, but I'd say I spent around 3 hours in total working on it. I see that it's only been solved just over 100 times, which is much lower than usual for a killer sudoku.

There were two 5s that could be input straight away, but after that I played around for ages before making any further progress. I finally managed to get my hooks into it by sorting out the options for column A and then solving row 9 when I determined that the cage in the bottom right corner had to be 126.

It's certainly the most difficult killer sudoku I can remember seeing on this site. Did anyone manage to find a quick key to solving it?

Hi, paulv66, yes, also in my opinion the puzzle is quite difficult (of course there are KS’s much more harder).

I have not enough time to provide the graphics (it would be useful if Patrick could simply add the proposed puzzle to this thread for a future use) but I’ll try to explain the way I have followed.

First I placed (as you did) the 5’s in a9 and e6 (this last as an “outie” of the sum ( 180) of the 4 leftmost columns). Cage “17” in b7 is [8,9] so the 8 of column 1 must be in cage “13” (a1-a2-a3) considering that an 8 in a6 is not possible due to a7-a8 = [2,4] what would make impossible the cage “10” in a9.

To determine that c1 = 9 we depart from the fact that the sum of c1, b3 and c3 is 20, and having an 8 in cage “13” consequently the combinations contain a 9 so: [4,7,9] or [5,6,9]. A 9 is not possible in c3 because the rest of cage “20” would be 11 = [1,2,3,5] regardless of the sequence. A 4 (>>> c1 = 7) could not be in b3, and in case of 5 and 6, the 5 would go to b3 and the 6 to c1 >>> d1 = 9 and this situation is impossible because then the cage “22” in c4 could not be formed due to two simultaneous 9’s in c3 and d1. Then c1 = 9 >>> d1 = 6.

The first consequence of this is that the 9 of the referred cage “22” is in d4 or d5; now looking to the nonet number 8 (that is, the 3x3 box in the middle of the three bottom rows), since the 9 could not be in f8 because in that case there is no way of placing a 9 in row 9 (g9 is unable due to repetition of 9’s in cage ”30”), then the cage “15” must contain the 9, that is: “15” = [6,9].

And now cage “9” in i8 is not [1,3,5] or [1,2,6] because in both cases the cage “10” in a9 could not be formed so cage “9” in i8 = [2,3,4] and then “10” in a9 is [1,4,5], but not [2,3,5], that is, i8 = 4 and h9-i9 = [2,3].

Now we look at cage “22” in c4. The case [5,8,9] must be eliminated because an 8 in d4 or d5 would require the 8 of the nonet number 8 to go to f8 with similar consequences as in the case of the 9 previously studied, that is, an 8 could not be placed in row 9. So, cage “22” in c4 is [6,7,9] with c4 = 6 and d4-d5 = [7,9]. Three consequences: 1) Cage “13” in a4 = [4,9] (>>> cage “13” in a1 = [2,3,8]); 2) cage “14” in column a is [1,6,7] with a6 = 1 and a7-a8 = [6,7]; 3) a 7 could not go to f8 using the previous logic (a 7 could not be placed in row 9) so the L-shape cage “10” is [1,2,7].

From this point, clearly, g9 = 7 and d9 = 8. Now looking to row 7, d7 = 4 and obviously since d8 cann’t be a 5 >>> f8 = 5 and d8 = 3 >>> c8 = 2 >>> c7 = 3. Since we have left a total of 10 in the 5-cell cage “22” necessarily c6 = 8 and d6 = 2 and d2-d3 = [1,5]. Also b4 = 3 and b3-c3 = [4,7] (we are left with 11 in cage “20”, but [5,6] would repeat a 5). Additionally, since “14” in b5 is [2,5,7] >>> b6 = 7, b5 =2, c5 = 5. Now b3 = 4, c3 = 7 (and b9 = 1, c9 =4). And since cage “12” is [1,5,6] >>> c2 = 1 (>>> d2 = 5, d3 = 1), b2 = 6, b1 = 5.

I will continue a little bit: g7 + i7 = 11 = [5,6] then g7 = 6 (due to c6 = 8), i7 = 5 >>> i6 = 9. Also a7 = 7 and a8 = 6. And e8 = 7 with e7-f7 = [1,2].

Due to the non-repetition of 6’s in cage “19” h6 = 6 >>> f6-g6 = [3,4] >>> g4 + g5 = 6 = [1,5], that is, g4 = 5, g5 = 1. Now we observe that h5 + i5 = 12 so = [4,8], that is: h5 = 4, i5 = 8 >>> a4 = 4, a5 = 9, d4 = 9 and d5 = 7. Also g6 = 3 and f6 = 4. Cage “9” in e5 = [3,6]. Additionally, h4-i4 = [2,7], the rest of cage “32” being 23 (BTW with a 6 in i3 observing row 3) then: g1 + g2 = 6 = [2,4].

The 1 of nonet 9 is in h8, … .

Last comment for Patrick:

From time to time it's sane to have a difficult KS like this

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