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Re: 9 x 9 Tuesday May 3rd
http://www.calcudoku.org/forum/viewtopic.php?f=16&t=788
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Author:  firefly  [ Sun May 08, 2016 7:57 pm ]
Post subject:  Re: 9 x 9 Tuesday May 3rd

pnm wrote:
Nice one :)

Was this one easier/harder/the same?


Depends what you mean. Easier to solve, but harder (so far) to find a simple and elegant way to move through this last section.

On the whole, though, I'd say easier.


clm wrote:
I followed practically the same way up this point (step 15): maximums and minimums, etc., except that, having determined via the parity rule, in the three bottom rows, that the 3240x cage is odd (so its sum is 27 or 29), g6 must be odd so f6 = 5, g6 = 1, and now the rest of the subtraction to the three central rows is 17 (instead of 22) what permits to more quicly see that f4 = 2 since f4 = 8 forces 4-5 in 1- at f7 which is impossible ...

See, you say "more quickly", but I think you're ignoring the fact that you had to first factor out the 3240x, then check parity (sum) the bottom three rows to find the parity of the 3240x, and then check parity (sum) the right three columns to find the parity of 6g, all just to find the 5 in 6f, in order to make it "more quick" to find the 2 in 4f after summing columns 4-6.

I mean, you could more easily have just used parity on columns 4-6 directly at this point to find that 6f had to be odd and thus had to be 5 (which I actually considered), but it's ultimately an unnecessary step to solving the [4f,6f,7f,8f] block, because you can basically do a mini-parity by eye while solving for 22 there anyway.

I think you might fail to see this because you do a lot of your parity checking early on, or in smaller increments as you go, but when you lay it all out there when you need to use parity for the first time in your description of the solution, it kind of showcases how many steps it takes.

When you consider all that work, I still think my way was more efficient, though it may just be a subjective thing.


clm wrote:
Very curiously, the green area that I first obtained is exactly the same as yours, with c8 undefined yet to complete a perfect simmetry. The rest is a little more difficult I think.

Yeah. So far, I've found a couple ways to solve from this point, but they either require an extremely long set of logic proofs to get enough individual cells to provide more leverage, or a many-step elimination of one of the 3240x options that might as well be TAE by its length. Neither are what I consider to be anywhere close to an elegant solution, so I'll try to find something more concise.

But once again, it's time for work, so it'll have to be much later tonight.

Author:  clm  [ Sun May 08, 2016 10:47 pm ]
Post subject:  Re: 9 x 9 Tuesday May 3rd

firefly wrote:

clm wrote:
I followed practically the same way up this point (step 15): maximums and minimums, etc., except that, having determined via the parity rule, in the three bottom rows, that the 3240x cage is odd (so its sum is 27 or 29), g6 must be odd so f6 = 5, g6 = 1, and now the rest of the subtraction to the three central rows is 17 (instead of 22) what permits to more quicly see that f4 = 2 since f4 = 8 forces 4-5 in 1- at f7 which is impossible ...

See, you say "more quickly", but I think you're ignoring the fact that you had to first factor out the 3240x, then check parity (sum) the bottom three rows to find the parity of the 3240x, and then check parity (sum) the right three columns to find the parity of 6g, all just to find the 5 in 6f, in order to make it "more quick" to find the 2 in 4f after summing columns 4-6.

I mean, you could more easily have just used parity on columns 4-6 directly at this point to find that 6f had to be odd and thus had to be 5 (which I actually considered), but it's ultimately an unnecessary step to solving the [4f,6f,7f,8f] block, because you can basically do a mini-parity by eye while solving for 22 there anyway.

I think you might fail to see this because you do a lot of your parity checking early on, or in smaller increments as you go, but when you lay it all out there when you need to use parity for the first time in your description of the solution, it kind of showcases how many steps it takes.

When you consider all that work, I still think my way was more efficient, though it may just be a subjective thing.



It's not necessary at all to factorize 3240x to simply find it's odd parity since you get it directly by observing the three bottom rows.
And anyway sooner or later you will need the factorization, then concluding that the sum is 27 or 29 as I mentioned, and the numbers involved, though perhaps I should not mention that sum in that point of my comment (because certainly "detailing" the solution in a post is quite different than "obtaining" the solution), you need to be more precise since you are writing the solution, and in fact you do the factorization a little later, in step 18, when is it required in the development of the solution.

When solutioning the puzzle, all these things are very subjective, but particularly for me the use of the parity is so easy (I am very accustomed to) that many times I go thru different columns or "areas" (when there is segmentation like in the 6x6's with quadrants) to arrive at the same point (it's true that in this case it's shorter using the three central columns), and I go very fast, simply assigning a 0 to the even cages and a 1 to the odd cages (as I mentioned in a post long time ago IIRC), every two 1's I make again a 0, with this method one goes "as fast as the light" to isolate the parity, you do not need keeping in mind the parity. Additionally, at the same time, I annotate, specially in the big puzzles, the 0 or the 1 assigned to the multiplication, division, exponentiation, mod, parity, etc., cages, when the parity can be determined, since this info can be useful later.

In this particular puzzle the rule of "the use of maximums and minimums in the sum of cages" becomes specially useful, along with the "addition rule", ... , and that's very interesting, I use a lot "the maximums and minimums" and some times looks like the "only" way, I consider this rule very powerful as well, and it can be used even inside the same cage, as mentioned in other places in the Forum. In summary, what I try to say is that all possible tools must be considered, depending on the particular puzzle.

(BTW, Patrick, I cann't find in the Forum, Section "Solving strategies and tips" the original post on this subject, some date in 2010, though there is another post viewtopic.php?f=3&t=39 referring to it)

Author:  firefly  [ Mon May 09, 2016 11:47 am ]
Post subject:  Re: 9 x 9 Tuesday May 3rd

clm wrote:
It's not necessary at all to factorize 3240x to simply find it's odd parity since you get it directly by observing the three bottom rows.
And anyway sooner or later you will need the factorization, then concluding that the sum is 27 or 29 as I mentioned, and the numbers involved, though perhaps I should not mention that sum in that point of my comment (because certainly "detailing" the solution in a post is quite different than "obtaining" the solution), you need to be more precise since you are writing the solution, and in fact you do the factorization a little later, in step 18, when is it required in the development of the solution.

And that's precisely the crux of what i meant. I was saying that when drafting (writing) a solution to post, the goal is to make it as simple and efficient as possible. So I'm always looking for a way to eliminate unnecessary steps, even if those steps are valid and helpful.

And that's often what parity is. It's valid, helpful, but very rarely necessary. It's a very useful tool to have and use when you're initially trying to solve the puzzle, because it's fairly easy to see when it can be applied, but the best it usually does is eliminate a few stray bits and suggest the next logic tool to be applied. But looking back, you often could have done without that step, if you'd been able to see the following step without its help.

But it can sometimes be extremely difficult to see that next step without trimming the clutter, so to speak, so I do use parity somewhat frequently, but it's usually a last resort after I've spent a while looking for other options.

But like I said earlier, I'm a logic snob. My goal isn't simply to solve the puzzle, it's to challenge myself to see the trickier logic steps with as little help as possible.


But back to the puzzle at hand. I've found what I think is just about the most concise way to move past the next part, though it's still fairly tedious...

19. 3a and 3i must both be [6,9].
20. With the 9s, 8s, 6s, 5s, and 3s blocked in columns b-c, the only choices for 9b-c are [7,4,2,1].
21. 9b-c must contain a 7 because no other combination allows for usable values in 8-9a. ([1,2] leads to sum 18 which is impossible, [1,4] leads to sum 16, which is blocked by 7a, and [2,4] leads to sum 15, which is blocked by 7a and 3a.)
22. 8a cannot be a 6 because it allows for no usable values for 9a. (With two 7s and a 6 in the 28+ cage, 9a and the non-7 in 9b-c would sum 8: [1,7] is blocked by both 7s, [2,6] is blocked by the 6, and [4,4] is impossible.)
23. The only place for a 6 in row 8 is 8i.
24. The only option for the 3240x cage is now [5,4,9,6,3] which makes 8h 5 to balance of the sum of rows 7-9.
25. This forces a resolution of the cells in the 3240x at 7i, the 15+ at 2g, as well as 3a and 3i.

26. 8a cannot be a 3 because it allows for no usable values for 9a. (As in step 23, the remainders in the 28+ cage would sum 11: [1,10] is impossible, [2,9] is blocked by the 9 in row 9, and [4,7] is blocked by both 7s.)
27. 8a must therefore be 9 and 8e must be 3.
28. 9a-c must be [1,4,7] due to the blocked 3 in row 9.
29. The remainder of [2,6,8] are easily resolved in 9d-f.

30. 1g must be a 2.
31. The only place for a 2 in column c is 8c, resolving the 4: at 8b, the 15+ at 2b, and 9c.
32. The only place for a 2 in column b is now 5b.
33. 1c must be a 4, which means 1b must be 1, further resolving 9a-b.
34. 4-6a must be [3,4,2] and 1-2a must be [8,5].

35. The rest of the puzzle is fairly simple.

I will edit this into the original solution, with pictures.

If anyone comes up with a better path than this, please let me know.

Author:  clm  [ Mon May 09, 2016 10:17 pm ]
Post subject:  Re: 9 x 9 Tuesday May 3rd

firefly wrote:
...
If anyone comes up with a better path than this, please let me know.


Different way: First I decide who is the classmate of the 4 in “1-” (g8-h8) using only the two bottom rows.

1. If h8 = 3 >>>” 3240x” = [2,4,5,9,9] >>> i8 = 9 (one of the 9’s) >>> a8 =6, e8 = 5 >>> g9-h9-i9 = (2,5,9), now a 3 is not possible in e9 since d9 + f9 is different that 11 due to the impossibility of forming [2,9], [3,8], [4,7] or [5,6] ([4,7] because all 4’s are in use in columns d and f), so a9 = 3, but now b9 + c9 = 12 while [3,9], [5,7] and [4,8] are not available (this last due to all 8’s filled in columns b and c). Thus: h8 = 5 >>> i9 = 5.

2. h8 = 5 >>> “3240x” =[3,4,5,6,9] >>> g1 = 2. Initially i8 = (3,6,9) but a 3 is not possible here since, apart of forcing g9-h9 to be (6,9), breaking the unicity of the solution {due to “15+” (g2-h2) = (6,9)}, it would force a8 = 6 and e8 = 9 and now an 8 is only possible in a9 >>> b9 + c9 = 7 but [1,6], [2,5] or [3,4] are not available (this last due to all 3’s filled in columns b and c). And the 3 is not possible in a8 >>> e8 = 9 (i8 = 6) then again a9 = 8 >>> b9 + c9 = 10 but [1,9] {there is a pair (3,9) in g9-h9}, [2,8], [3,7] and [4,6] are all blocked (this last due to all 6’s filled in columns b and c).

3. Consequently e8 = 3 and i8 = (6,9). Looking at column i, a 3 is not possible in i4 since to add to the cage an amount of 20 we need an L-shape 7-7-6 (unique combination possible) with i5 = 6 >>> i8 = 9 which is impossible due to the pair (6,9) in a3-i3. Then i1 = 3 (only place in row 4) >>> a4 = 3 >>> a5 = 4, a6 = 2, b5 = 2 >>> b8 = 8 >>> b2 = 7 >>> c2 = 8 >>> c8 = 2. Also i5 = 8, i2 = 2 (only place in column i) and still in column i: i4 = 1 (only place in column i), i6 = 7 >>> h5 = 7. Also e4 = 7 and e6 = 4 and since d5-e5-f5 = ((1,6,9) >>> d5 = 6.

4. Since a8 <> 8 because either 6 or 9 in a8 makes impossible to have a sum of 7 or 4 for b9 + c9 >>> a1 = 8 (unique place for the 8 in column a), and being a3-a8 = (6,9) >>> a2 = 5 >>> a9 = 1 >>> e2 = 1 >>> e5 = 9 >>> f5 = 1. Also a1 = 8 >>> d9 = 8, d1 = 7 >>> c9 = 7>>> c1 = 4 >>> b9 = 4 and b1 = 1. Also g1 = 2 >>> e9 = 2 and e1 = 5.

5. f1 y f9 are filled to comply with sums “22+” and “19+”, that is, f1 = 9, f9 = 6 >>> h9 = 3 >>> g9 = 9 >>> g2 = 6 >>> h2 = 9 >>> h1 = 6 solving all pending 6’s and 9’s distributed along the rest of the puzzle except for i8 whose uncertainty is solved by the “3240x” product since now: i8 = 6 >>> i3 = 9 >>> a3 = 6 and finally a8 = 9.

Ok, it’s not my pretension that this path is better, more elegant, more logic, or even less TAE, etc., simply it’s a different way, as there are probably a couple (I must study the segmentation in this type of puzzle, what I consider an interesting method for the future).

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