sneaklyfox wrote: picklepep wrote: This is very interesting, but I am not sure that it is generally useful. I think that most solutions can be worked out with quicker more straightforward methods. I would agree with you.
The corners and the border in the 5x5’s and 6x6’s (answer to Sneaklyfox and Picklepep and a practical example of some utility of this rule). The Sneaklyfox demo for a 4x4 is abridged and concise. However all that paraphernalia I used was because I was really interested in doing some systematic analysis (mainly oriented to beginners). The idea is: First deriving the rule, second obtaining the corollary and practical use, if any. Let’s see the case of a 5x5. If we consider the equation of my first article: I = C + (p - 4)s, when applied to a 5x5, we get: I = C + (5 - 4) x 15 = C + 15. But in the core of a 5x5 we have 9 cells, why not deducing then intuitively that the corners and each of the five numbers 1, 2, 3, 4 and 5 (that have a sum of 15) could be inside?. Certainly that is what really happens: In every 5x5, the inner area (the core 3x3) is made by the four corners and once the numbers 1 thru 5. Lets see an example (the 5x5 difficult, July 05, 2011, Puzzle id: 314715):
Consider one of the numbers, i. e., the 3. Now we can affirm that, since it is impossible to have five 3’s in the border (only four lines) then one 3 must be necessarily inside (and the same happens with the other numbers 1, 2, 4 and 5). Now we have four idle cells in the “heart”. Let’s see the corners (following the Sneaklyfox outline): If a number “n” is in a corner, an additional “n” must go to the core (three in total to the moment) (because in the L-shaped “z” we can only locate the other two “n’s”, that makes the total of five n’s). And this is valid for any of the other three corners (see for instance the location of the two’s in the example, etc. (I use green for the corners, and blue for the numbers 1 thru 5 in the “heart”).
nxxxx x???z x???z x???z xzzzz
Now let’s apply the property to a 6x6. The equation gives: I = C + (p - 4)s = C + (6 - 4) x 21 = C + 42. Observing the equation we could initially estimate that, being 21 the sum of all number in a line, perhaps we would have in the inner area (16 cells), the 4 corners plus twice the six numbers 1, 2, 3, 4, 5 and 6, and this is what really happens in every 6x6. The demonstration is exactly the same: Any number can not be more than 4 times in the border so it must be twice in the inner area (12 numbers occupied) and for a corner “n” since we have 3 n’s already located (the corner and two more inside) we must locate an additonal “n” in the inner area since no more than two (among the pending three) can be situated in the L-shaped “z’s”. See an example (the 6x6 difficult, June 14, 2011, Puzzle id: 305714):
(I use the green again for the corners and blue and purple to show the two series of 1 thru 6). The rule of course can be extended to any size. What really happens in a 4x4 is that there is no “additional space” in the core (like a mathematical point of dimension zero) and then the corners go just once to the inner area, so the case of the 4x4 looks trivial, but it is curious to observe in general how the core grows when the puzzle grows. I agree that, in a 4x4, in very few cases, the rule is practical (the example I used in my first topic was an exception, in fact the number of all possible 4x4’s, excluding symmetries, is not too big, it is mainly a calculation with "sudokus", neither the number of different operations applied to them) and one in general will use other means. But now I am going to apply the property to a practical case in a 6x6 (the 6x6 difficult, July 07, 2011, Puzzle id: 318137).
Solution: First, of course, we place the “144x”, that is, 4-6-6, 4 in B2, 6 in C2 and 6 in B3. Now C = 1 + 6 + 4 + 3 = 14, so I = 14 + 42 = 56 (even number) (we may say that the “addition” of the border can be calculated directly and then obtain I subtracting from 126, the full puzzle, but this way is faster). Applying the parity rule the cage “2: “ is an odd cage, so the operands are 1-2-4 (sum of 7) (since 1-1-2 and 1-3-6 are even). The 4 must go to E3. Now the cage “0-“ has a sum of 6 ( to the total of 56 in the inner area) so the combination is 1-2-3. The cage “14+” has three possibilities 6-6-2 (not valid since only three 6’s are in the core as we have seen; here we may object : ”Well, we can see that because otherwise no 6’s could be placed in row 6"), the 5-5-4 is not possible due to the 4 in E3, so the only valid combinaron is 6-5-3. Only two 2’s (2 is not in a corner) can be inside so, since we already have them, “5+” is made with 1-4 and “8+” is made with 3-5 (not 2-6; also because another 5 must be in the core and the “8+” is the only possible place; once again we can argue: "Well, four 6’s are not possible in core due to row 6") (vice versa, if “8+” is 3-5, “5+” can not contain a fourth 3 “due to row 1” and due to the presence in core of two 2’s an three 3’s according to our exposition). Now we place the 6 of the “14+” in D5.
Now we go with the numbers in red: 6 in E6 and A4, 4 in D1, 1 in E2, 2 in E1, 2 in D2, 3 in A2 and 5 in FE2, 3 in D3 (see the 3’s in A2 and F6), 5 in D4, 3 in E4, 5 in E5 and 1 in D6. Now we situate the purple numbers of rows 3, 4 and 5 consecutively and quickly. The rest is trivial.
Conclusion: Of course there are many other ways of arriving to the solutions, but all possible tools may be used or can aid.
This is the “official” solution:
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