Author 
Message 
pharosian
Posted on: Mon Jun 11, 2012 4:26 pm
Posts: 116 Joined: Sat May 14, 2011 3:18 am

Re: More fuel to the fire
clm wrote: Hi, jomapil, first have a look to this “analysis”, which in fact does not essentially differ from the one detailed by arjen. It is clear that c3 = 2 has been determined using the parity rule (analysis). This is where I get frustrated. You've lost me on your very first statement! I read your post several months ago about parity, and I am sometimes able to use it now, especially in the 7x7 puzzles. But in a situation like this where there are still large swaths of blank cells in both the row and column containing c3, some of which belong to large "sums" (such as the 29+), how in the world can it be "clear that c3 = 2 has been determined using the parity rule"???? Thanks for any insight you can provide here.




clm
Posted on: Mon Jun 11, 2012 6:23 pm
Posts: 713 Joined: Fri May 13, 2011 6:51 pm

Re: More fuel to the fire
pharosian wrote: clm wrote: Hi, jomapil, first have a look to this “analysis”, which in fact does not essentially differ from the one detailed by arjen. ... It is clear that c3 = 2 has been determined using the parity rule (analysis). This is where I get frustrated. You've lost me on your very first statement! I read your post several months ago about parity, and I am sometimes able to use it now, especially in the 7x7 puzzles. But in a situation like this where there are still large swaths of blank cells in both the row and column containing c3, some of which belong to large "sums" (such as the 29+), how in the world can it be "clear that c3 = 2 has been determined using the parity rule"???? Thanks for any insight you can provide here. A cage of the type "n" is even (that is, the sum of the numbers inside is even) if n is even and it's odd if n is odd, regardless of the number of cells. Once this is understood (if you need a further clarification on this pse tell me), see that the four cages in the top three rows, of the type "", are all even and then its sum even, now applying the parity rule to the three upmost rows (with a total sum of 135) we have: 17 + 16 + 18 + 2 + 5 + 1 + 29 + b3 + an even number (the result of the four cages "") = 88 + b3 + even number = odd number, because b3 is odd, no matter if it's a 3 or a 7; consequently, to maintain the odd parity for the three upmost rows, the only left number in this case, the alone cell c3 must be even and the only even number in the combination [2,3,5] for "30x" (obviously unique, being invalid, due to the 1's in a4 and e3, the other possibility [1,5,6]) is the 2, that's why we place the 2 there.




jomapil
Posted on: Mon Jun 11, 2012 6:37 pm
Posts: 246 Location: Lisbon, Portugal Joined: Sun Sep 18, 2011 5:40 pm

Re: More fuel to the fire
Pharosian: I tried to follow the reasoning of Clm , but I applied the parity rule to the 3 first columns + d1 + d4. But this way I only conclude that the 2 must be in c3 or c4. But with the 3 first rows you must be elucidated ( and me ).
_________________Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.




pharosian
Posted on: Tue Jun 12, 2012 2:49 am
Posts: 116 Joined: Sat May 14, 2011 3:18 am

Re: More fuel to the fire
clm wrote: A cage of the type "n" is even (that is, the sum of the numbers inside is even) if n is even and it's odd if n is odd, regardless of the number of cells. Once this is understood (if you need a further clarification on this pse tell me), see that the four cages in the top three rows, of the type "", are all even and then its sum even, now applying the parity rule to the three upmost rows (with a total sum of 135) we have: 17 + 16 + 18 + 2 + 5 + 1 + 29 + b3 + an even number (the result of the four cages "") = 88 + b3 + even number = odd number, because b3 is odd, no matter if it's a 3 or a 7; consequently, to maintain the odd parity for the three upmost rows, the only left number in this case, the alone cell c3 must be even and the only even number in the combination [2,3,5] for "30x" (obviously unique, being invalid, due to the 1's in a4 and e3, the other possibility [1,5,6]) is the 2, that's why we place the 2 there. Ohhhhhhhhhh, now I get it! I've only used parity in a single row/column before; I didn't know how to use it in a case like this. The parity rules might be a big help in my future struggles with the larger puzzles! Thanks so much for your detailed explanation.




clm
Posted on: Tue Jun 12, 2012 12:28 pm
Posts: 713 Joined: Fri May 13, 2011 6:51 pm

Re: More fuel to the fire
pharosian wrote: ... The parity rules might be a big help in my future struggles with the larger puzzles!
Thanks so much for your detailed explanation. Welcome, it’s a pleasure. An additional comment: Obtaining all sums (cages, lines, etc.) is tedious and tired (as the puzzle size increases, well, easier if the numbers are big , and if the numbers are small subjected to mistakes). Instead, I use this technique (shown in red colour): I assign a 0 to the even cages and a 1 to the odd cages. Now you only need to count the 1’s to see if we are in an even case or in an odd case (obviously every two 1’s count as a 0). Let’s put an example (the calculation made by jomapil using the three leftmost columns): d1 + d4 + 1 (this one represents the value 135, odd, for the full three leftmost columns) = 1 (the result of ones and zeros for all cages affected since we have a total of 5 ones), then: d1 + d4 = 0 (this represents the even), but d1 is odd (all even numbers are being used in row 1) so d4 must be odd = [3,5] and the 2 of cage “30x” must go to c3 or c4 (as jomapil says). But, jomapil, you still have an uncertainty, it’s very important to “cut” the puzzle by the appropriate lines to have success and, sure easier, if we leave outside (or inside), like an external (or internal) appendix, only one cell (however in some other cases we would need to deal with more cells), in our case we need to “cut” the first three rows. I’m sure you probably found the c3 = 2 (that you showed in your initial graphic) by other means.




jomapil
Posted on: Tue Jun 12, 2012 12:53 pm
Posts: 246 Location: Lisbon, Portugal Joined: Sun Sep 18, 2011 5:40 pm

Re: More fuel to the fire
Hi Pharosian. The puzzle 9x9 of today is a very good opportunity to train the parity rule in several of its aspects. After you fill in the diagram with the usual, you can apply the rule to the first four rows and to the last four rows. Clm also likes very much to discoverer, at first, the cells c3, c7, g3, g7. It's the first time I solved this weekly patterned diagram exclusively with analysis ! On the contrary, I think the 10x10 of today must be the first time I solve one 10x10 by TAE ! P.S.  After all I succeeded to solve the 10x10 analytically
_________________Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.




marblevolcano
Posted on: Sun Jul 24, 2016 4:33 pm
Posts: 244 Joined: Sun May 22, 2016 2:17 pm

Re: More fuel to the fire
jomapil wrote: My resolution, ( although with a " little " TAE ), can be considered Analytical or TAE?
In other words, what is TAE method and Analytical method? What is the border between the two methods? I would label them Extrapolation vs Experimentation. If you're experimenting with numbers and you have no clue what will result from it, then that's TAE. However, if you try to extrapolate with one cell or cage as a pivot point to see if any of the options ran into a roadblock, but without actually plugging in numbers, then that is analysis. Really, in my opinion, it's all about whether you actually plug in the numbers or not.






You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum

