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Killer difficult (Sep 16, 2012): Step by step solution
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Posted on: Fri Sep 21, 2012 12:54 am

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Killer difficult (Sep 16, 2012): Step by step solution
Killer Sudoku difficult (Sep 16, 2012): Step by step.

Graphic 1 shows the Killer problem where we have represented some of the candidates and determined the values of cells a4 = 3 (outie of nonet 1) and i7 = 6 (innie of nonet 9) (as we know any of the nonets, the 3x3 boxes, has a sum of 45, the sum of numbers 1 thru 9). Due to i7 = 6, the candidates of cage “15” in i3-i4 are obviously [7,8].

Step 1: Let’s observe that h6 + h8 = 3 + 15 + 15 + 21 – 45 = 9 (where 45 is the sum of numbers in column i). Now we conclude that h6 can only be 1 or 2, then h8 = [7,8]; for instance, h6 = 3 >>> i5 + i6 = 6 but either [1,5] or [2,4] are impossible due to [1,2] already present in cage “3” (i1-i2).

Step 2: Now we are going to define h6 and h8. Graphic 2, in red colour: if h6 = 2 (h8 = 7) >>> i5-i6 = [3,4] but this forces “13” (h4-h5) = [5,8] and then cage “10” (g9-h9) becomes impossible (it should be [2,8] once a 7 is in g8).

So (Graphic 3, in green colour) h6 = 1, h8 = 8, consequently i8-i9 = [4,9], i5-i6 = [3,5], “10” (g9-h9) = [3,7] and “8” (g7-g8-h7) = [1,2,5].

Step 3: (Graphic 3, in red colour). The combination [6,7] is not possible in h4-h5 because then [6,7,9] would be invalid for cage “22” (g2-h1-h2) (both 6 and 7 cann’t go “together” to the same cell g2) forcing “22” = [5,8,9] (“22+” in three cells, with different numbers, only admits these two combinations); the 5 inside “22” forces “9” (g3-h3) to be [3,6] which is impossible because now we have a 6 and a 3 in column h ([6,7] in h4-h5 would force a 3 in h9). As a consequence “13” = [4,9] >>> “22” = [6,7,9] (g2 = 9, h1-h2 = [6,7]) >>> “9” = [4,5] with g3 = 4, h3 = 5 >>> h7 = 2, g7-g8 = [1,5]. We have arrived to the following situation (Graphic 4, numbers in black colour):

In this Graphic 4 we have also placed an 8 in i3 (due to a 7 in h1-h2) >>> i4 = 7, also g1 = 3 and then d3 + f3 = 5 (sum of innies of nonet 2) = [2,3] since g3 = 4; and since f1 + f2 = 6 >>> f1-f2 = [1,5].

Step 4: (Graphic 4, in blue colour). Let’s observe (considering the two bottom rows) that c8 + g8 = 90 – 18 – 31 – 21 – 10 = 10. In these conditions if g8 = 5 >>> c8 = 5 which is impossible then g8 = 1 (g7 = 5) and c8 = 9. Now we need a 9 in row 7 so f7 = 9 (the 9 in c8 does not permit a 9 in the three cells a7-b7-c7 and a second 9 is not possible inside the same cage “27” (d7 or e7, making use of the “Killer”’s property, the non-repetition of numbers inside the same cage ***). Now the 9 of column “d” cann’t be in the segment d7-d8-d9, or d6 (for the same previous reason), or d5 (>>> [1,3] in c5-c6 and there is a 3 in a4) so d1 = 9. Finally the 9 of column “e” must go inside cage “10” (e4-e5) (obviously e6 cann’t be a 9); d1 = 9 >>> “15” (e2-e3) = [7,8] with e2 = 8, e3 = 7. And, since a3-b3-c3 = [1,6,9] >>> a2 = 2.

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*** Note: The non-repetition of numbers inside a cage is a convention widely accepted for this type of puzzle (consequently the maximum possible size of a cage would be 9 cells, containing numbers 1 thru 9); actually the Patrick Min’s software follows this rule and assures the non-repetition; however, outside this site, one can find Killer’s with repetition, there were discussions in the past about this subject and not all authors agree with this restriction, so the reasoning we have followed here could fail in other places.

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Also, let’s see why the cage “14” must be [6,8]. Cage “20” (a5-a6-a7) has only three possibilities: [4,7,9], [5,6,9], [5,7,8], but “14” = [5,9] would inhibit any combination with a 9 or any combination with a 5 due to c8 = 9 and g7 = 5.

Step 5: Now it’s very easy to see that cage “12” (b6-b7) cann’t be [5,7] (Graphic 4, in red colour). The 5 in b6 and the 5 in g7 force the combination [4,7,9] to be the only possible for “20”, and b7 = 7, c8 = 9 >>> a7= 4 but in this case c7 should be another 7 because a7 + b7 + c7 = 18 (45 – 18 – 9, in nonet 7). So “12” = [3,9] and we place (in violet colour) the 3 and the 9 in c7 and c6, respectively. And, inmediately, a3 = 9, b3 = 1 and c3 = 6.

Now (in brown colour), as no 9’s are allowed in cage “20”, the only available combination is [5,7,8] with the 8 outside the nonet 4, that is, a7 = 8 and a5-a6 = [5,7]. Then c7 = 7 (18 – 8 – 3) and d7 = 1, e7 = 4 (a 1 and a 4 are required in row 7). Also d6 = 6 (the difference, 27 minus the numbers already defined in cage “27”). Additionally cage “10” (b2-c2) must be [3,7].

Step 6. Suddenly several numbers can be defined now (Graphic 5, in black colour): c8 = 9 >>> i8 = 4, i9 =9; a2 = 2 >>> i1 = 2, i2 =1; i2 = 1 >>> f1 = 1, f2 = 5; b6-b7 = [3,9] >>> b2 = 7, c2 = 3; a7 = 8 and b4-b5 = [6,8] >>> c1 = 8; a5-a6 = [5,7] >>> b1 = 5 and nonet 1 is completed with a1 = 4. Also, e1 = 6, d2 = 4 (considering a1 and h2 or, alternately, d6 and e7).

Comment: We have arrived to the present situation following a process of “analysis” where some times we had to decide between two possibilities reasoning in the sense that one of them drives to an absurd thus taking the other way, this can be considered TAE (trial and error) by some puzzlers or simply FAS (full analytical solution) by another group of puzzlers, anyway it’s useful (let’s think that this thread is oriented to beginners and medium players so if those decisions could be based in more complex equations falls beyond this particular process). Something to underline is that it’s important to think in terms of “analysis” as much as we can, and that we must be very sure when taking decisions, otherwise any error can have “a catastrophic propagation” in the solution process. But a small correct step advances the solution a lot though we must be patient until we place the 81 numbers.

Step 7. Now (Graphic 6 below, in green colour) we place a few more numbers:

Let's continue: g8 = 1 >>> a9 = 1, a8 = 6; i8 = 4 >>> b9 = 4, b8 = 2; and to complete the nonet 7, c9 = 5 (also to complete the sum in “18”).

Step 8. Now (Graphic 6, new numbers in blue colour) we observe (very frequent trick in the Killer’s) that having 5’s in a5-a6 and i5-i6 inhibits from a 5 the cells a4-b4-c4 and g4-h4-i4 so the 5 of row 4 must go to d4 since f4 is not available due to f2 = 5. But this 5 (encircled) means d3 = 2 and c4 = 1 (f3 = 3). As a consequence c5-c6 = [2,4] >>> d5 = 7. And the rest of column d can be completed: h8 = 8 >>> d9 = 8, d8 = 3. Also we complete row 8: f2 = 5 >>> e8 = 5, f8 = 7.

Step 9 (Graphic 6, in violet colour). Next “trick” (also frequent): Let’s demonstrate that g4 must be equal to f5: g4 + g5 + g6 = 16 (2 + 6 + 8, whichever is the order), so g4 + (16 – f5) = 16 >>> g4 = f5 (16 – f5, being now 16 the value of cage "16" represents the sum of the two cells g5 and g6). Now if we place an 8 in g4 another 8 would go to f5 and this would break the cage “14”, the same would happen with two 6’s so necessarily both must be a 2 (encircled). And we finish the row 9: f5 = 2 >>> e9 = 2, f9 = 6.

We resume these results in the above Graphic 7 (in black colour).

Step 10. As we can see in Graphic 7 (in blue colour): d5 = 7 >>> a5 = 5, a6 = 7; a5 = 5 >>> i5 = 3, i6 = 5.

In green colour: The required number in f4 is an 8 (encircled) and then b4 = 6 >>> b5 = 8 >>> g5 = 6 >>> g6 = 8 (indicated with green arrows).

In brown colour: f5 = 2 >>> c5 = 4, c6 = 2.

In purple colour: c4 = 1 >>> e4 = 9 >>> h4 = 4 >>> h5 = 9 and e5 = 1.

Finally we finish the puzzle with f6 = 4, e6 = 3 (in dark green colour).

And the official solution:

Reedited Sep 22, 2012 to complete the process.

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