Calcudoku puzzle forumhttps://www.calcudoku.org/forum/ Can anyone show an analytical solution for the 10 March 6x6?https://www.calcudoku.org/forum/viewtopic.php?f=3&t=167 Page 1 of 1

 Author: [ Sun Mar 11, 2012 1:51 am ] Post subject: Can anyone show an analytical solution for the 10 March 6x6? I would be very interested to see an analytical approach to solving yesterday's 6x6 (10 Mar 2012). I've been reading and learning from the many excellant posts in this section. But nothing that I tried seemed to work for me and I eventually had to use brute force to solve it.

 Author: clm  [ Sun Mar 11, 2012 9:04 pm ] Post subject: Re: Can anyone show an analytical solution for the 10 March fb_1811 wrote:I would be very interested to see an analytical approach to solving yesterday's 6x6 (10 Mar 2012). I've been reading and learning from the many excellant posts in this section. But nothing that I tried seemed to work for me and I eventually had to use brute force to solve it. I will try this "6x6 difficult" (Puzzle id: 430420) (it has a solver rating of only 50 and 218 solvers) (by the way, which country are you from?) (your doubts, if any, welcome, though in about a week I will be travelling for another eight days though I hope anyone would answer). There are different ways to arrive to the solution. In the process I have followed I have used some times the “international terminology” (below the graphics, for instance, but I believe it’s easy to follow) and the conventional language in other places to make the things easier. First we write the candidates for the cages “6+” and “7+”; observe that since a 5 can not be either in f2 (it is not a divisor of 72) or in f4 (there is already a 5 in row 4, in a4) it must be “7+” = [2,5]. Row 2: b2 = 5 (the 5 is not a divisor of 24 or 72) so that’s the only valid position for a 5 in row 2 and it follows that c1 = 5 (unique position for a 5 in row 1). Here we arrive to an interesting point. First observe that the blue and yellow areas must have a sum of 13 (since columns 1 and 2 must sum 42 = 2 x 21, that is, two lines). Once we have obtained the two possible combinations for “288x”, 2-4-6-6 and 3-4-4-6, we can follow three lines of thinking: First. Deciding that a 1 is not possible in b1 because we would have left to 22 a sum of 11 (5 + 6, unique) for a3-b3 which is impossible (due to a4 = 5 and b2= 5) so b4 = 1 (unique). Consequently the yellow area (which sum would be 12) is 2-4-6 (>>> a5 = 6, a3 = 3) (with 3 numbers from 3-4-4-6 you can not obtain 12) and c6 = 6 (the number left) (with b5-b6 = [2,4]) (perhaps someone could say that this process is of the type “guessing”). Second. Deciding “directly” that a 6 cann’t go to a3 because then b1-b3 = [2,4] (unique, to have the sum 22) but since now a5 =3 that situation is in conflict with the rest of the combination, 4-4-6, because we would need the two 4’s in b5 and c6. Consequently a3 = 3, a5 =6 once more (also “guessing”?). Third. This time I will apply the property described in the topic “The use of maximums and minimums in the sum of cages": b4 + Y = 13 (being Y the sum of numbers in the yellow area). The minimum for b4 is 1 (anyway 1 is the minimum value for any cell always) so the maximum value for Y is 12; it cann’t be 11 (like in the case of 3-4-4 coming from the combination 3-4-4-6 due to the 4 already present in column 1, in cage “6+”) so Y = 12 driving to 2-4-6 for those cells and consequently defining the 2-4-6-6 as the only possibility for “288x”. In summary: b4 = 2 + 3-4-4 (for Y) is not possible, leaving only b4 = 1 + 2-4-6 (for Y, from the combination 2-4-6-6) as the solution for that part of the puzzle, and then c6 = 6, a5 = 6, a3 = 3, b5-b6 = [2,4]. (Inversely, the minimum for b4 is 1 but, would it be possible b4 = 2? … clearly no because then Y = 11 but 3-4-4 [the only group within those two combinations to produce a sum of 11] is not allowed in the yellow area). Once we know the 3 is outside the cage “288x”, that is, a3 = 3 >>> b1 + b3 = 9 >>> b1 = 3, b3 = 6 (unique). Now (this part shown in blue colour) that we have a 3 in b1, we decide that “72x” (that initially had four possible combinations: 1-2-6-6, 1-3-4-6, 2-2-3-6 and 2-3-3-4) must be [1,3,4,6] because a 1 is required in row 1 but 1-2-6-6 repeats a 6 (or, alternately, if you prefer this way of thinking, because two 2’s or two 3’s or two 6’s are not allowed now inside the cage “72x” permitting only the 1-3-4-6). So e2 = 6, f2 = 3 (f4 = 4), d1-e1 = [1,4] what forces a1 = 2, a2 = 4 and consequently c1-d1 = [1,2]. In the cage “24x”, the product of c3 by c4 must be 12, and being c6 = 6, it follows that c4 = 3 and c3 = 4 thus d3-e3 = [2,5]. The rest is straight forward (in brown colour): In row 4 the 6 goes to d4 >>> e4 = 2 this forces d3 = 2 (and e3 = 5) forcing d2 = 1, c2 =2 and c5 = 1 (additionally d1 = 4, e1 = 1). Inside “17+” we need a 3 in d5 (green) to complete the sum >>> d6 = 5. And the final steps in violet colour. Final comment: As mentioned above, the analysis could have been started in other place, following a different process, i.e., (with the 5’s in b2 and c1) by studying the two possibilities 1-3-4-6 and 2-3-3-4 for the cage “72x” (the other two, 1-2-6-6 and 2-2-3-6, inmediately suppressed because of the repetition of the 2 or the 6), etc.. And the official solution:

 Author: beaker  [ Sun Mar 11, 2012 10:09 pm ] Post subject: Re: Can anyone show an analytical solution for the 10 March clm: this is a great explanation.......as I said, I too had trouble with this puzzle but the starting explanation with the 5's should have been obvious to me and wasn't and then I didn't think of using the total of 21 for each column (something sneaklyfox told me about months ago when I saw her video)....I am sure with this type of logic I will have fewer problems in the future.....as for the guessing discussion, I think you have taken the guessing out of the "equation" by using some established mechanisms (ie: the placement of the 5's and the total of 21 for each row)......so while thinking of the possibilities in your mind, you eliminate the possibilities that don't fit and "go" with those that, logically, do fit (the "catch 22" is the word possibilities as that word in itself suggests guessing......a real quandry!!!!!)

 Author: [ Mon Mar 12, 2012 4:39 am ] Post subject: Re: Can anyone show an analytical solution for the 10 March Thank you clm. Much appreciated. I got the 5's in c1 and b2 sorted out and even managed to discern the 1-3-4-6 for 72x but then couldn't quite see my way forward to solve the rest.FYI, IMHO guessing is throwing in a number in and seeing what happened. Stating a systematic hypothesis and then seeking to disprove it still falls in the analytical camp. So it counts for me!