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 Step by step solution of a difficult 8x8 mod function puzzle 
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Posted on: Mon Jun 29, 2020 8:45 pm




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Post Step by step solution of a difficult 8x8 mod function puzzle
(Please, Patrick, it would be helpful if you can add the graphic of the puzzle for future discussions)(pnm edit: here it is, thanks for posting the guide [smile])
(pnm edit II: for reference, here is the original thread about having to use trial and error for this puzzle: viewtopic.php?f=16&t=1248)

Image

In the thread “8x8 difficult Thursday 25/06/2020”, Section “Specific puzzles / your own puzzles”, viewtopic.php?f=16&t=1248 , some puzzlers (tusky, ineedaname, …) expressed the opinion that this puzzle was particularly difficult and they had some problems with logic and had to follow Trial and Error procedure to solve it. In my opinion TAE is not advisable, I think that “all puzzles can be solved analytically”. Here is my “step by step” full analytical solution (not enough time to provide intermediate graphics, so the digits must be directly entered in the grid, if puzzle is still available on the site or, otherwise, written in a separate printout). My concept of the analysis is based in the idea that, once you arrive to any point, if you have two branches (rarely three), you must eliminate one of them by pure logic, etc., then you bold the selected digits and continue to the next point.

1. We fill all “given” numbers (including pairs of candidates, etc.): d1 = 7, e1 = 1, h1 = 4, d4 = 2, b5 = 1, d6 = 8, a7 = 2, d7 = 1, h7 = 8 and a8 = 1. Also e2-f2 = [3,7] and a4-a5 = [7,8]. Now we know that a 3 cannot be inside a “4mod” cage and, since a3 cannot be a 3, a1 = 3. As we know, cage “3mod” in f1-g1 must be [5,8] since it cannot be [4,7] or contain a 3 inside. Then, b1-c1 = [2,6] and b2 = 8. Consequently (cage “12+”), a3 = 5, b3 = 7.

2. “192x” is not a multiple of 5, so 5 in column d must be in d5 or d8. Let’s eliminate one of these options: If d5 = 5, c5-c6 = [2,4] with c5 = 4, c6 = 2, since “4mod” contains a 4. Now we observe that in these conditions cage “2-” = [6,8] (unique) (with a5 = 7) but then there is no place for a 1 in row 6 (since a 1 cannot be inside a “2mod” cage, that is, in g6). So we have concluded that d8 = 5.

3. Cage “3mod” in row 8 has only two possibilities: c8 = 3 or c8 = 8. Let’s eliminate c8 = 8 >>> g7 + g8 + h8 = 14 (addition rule to rows 7 and 8, 72 in total); then g7 should be odd (parity rule, since “1-” is odd); now 3 is not possible since “1-” = [5,6]; 5 is not possible since “1-” = [4,5] so g7 = 7 >>> g8-h8 = [3,4] with g8 = 4, h8 = 3, and now the 7 of column h goes to h4 >>> a5 = 7, and then there is no place for a 7 in column c since c6 = 7 is not possible due to c5-d5 = [1,3] but there is a 1 in b5. So we have concluded that d8 = 3.

4. The 8 of column c must go to c3, since it is not possible in c5 (>>> d5 = 1 or 2) or c4 (b4 should be 1 or 7 both impossible). So c3 = 8.

5. Now we go with the position of the 3 in column d: It must be in d3 or d5. If d5 = 3 >>> d2-d3 = [4,6] >>> c2 = 1 but there is no way of obtaining a sum of 8 in c5-c6, since [1,7], [2,6] and [3,5] are all impossible. So we have concluded that d3 = 3. Now c2 x d2 = 8 = [2,4]. That is, c2 = 2, d2 = 4, b1 = 2, c1 = 6, a6 = 4, a2 = 6. And g2-h2 = [1,5]. Additionally, d5 = 6, c5 = 4, c6 = 1.

6. The 5 in column c must be in c4 or c7. If c7 = 5 >>> b7 = 3, b8 = 6, then b4 = 4, c4 = 7 which is impossible for cage “1mod”. We have concluded that c4 = 5, c7 = 7, and then b7 = 3, b8 = 4, b6 = 5 and b4 = 6.

7. Considering a4-a5 we conclude that if a5 = 7 there is no place for an 8 in row 5 because if e5 = 8 it’s not possible to obtain a sum of 7 in e6-f6, that is, either [1,6] or [2,5] or [3,4] are impossible. So a5 = 8 and a4 = 7.

8. At this point we have the left side of the grid, rows a, b, c and d, completely filled, and g7 + g8 + h8 = 19 (addition rule to rows 7 and 8, 72 in total). Here we have five “a priori” possibilities: [3,8,8], [5,7,7], [6,6,7], [4,7,8] and [5,6,8]. We see immediately that [3,8,8], [5,7,7] and [5,6,8] are impossible (this last due to “1-” with a 6 and an 8). Then, we eliminate one of these: [6,6,7] or [4,7,8]. Let’s start with [4,7,8]: If g8 = 8, h8 = 7 and g7 = 4 >>> h5 = 2, h6 = 6, h4 = 3, h3 = 1, h2 = 5 and g2 = 1 but then cage “2:” cannot be accomplished, that is, [1,2], [2,4], [3,6] and [4,8] are all invalid. Consequently, the only possibility is [6,6,7], in L-shape, that is, g7 = 6, g8 = 7 and h8 = 6.

9. The rest seems very easy. In row 6, among 2, 3, 6 or 7, only g6 = 2 is possible. Also “4-” = [3,7], h3 = 2, h4 = 1, h2 = 5, g2 = 1. So, cage “2:” = [4,8] >>> g3 = 4, g4 = 8 >>> f1 = 8, g1 = 5 >>> g5 = 3, f5 = 5 >>> h5 = 7, h6 = 3, then e5 = 2 and e6-f6 = [6,7]. We finish the puzzle completing row 3: e3 = 6, f3 = 1 (1 cannot be inside the cage “2mod” in e3-e4) and then e4 = 4 (but not e4 = 3 to complete cage “2mod”), f4 = 3 >>> e2 = 3, f2 = 7 >>> e6 = 7, f6 = 6. And, finally, e7 = 5, f7 = 4, e8 = 8 and f8 = 2.


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Posted on: Tue Jun 30, 2020 2:43 am




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Location: France
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Post Re: Step by step solution of a difficult 8x8 mod function pu
Hello clm and thank you for taking the time to make this very detailled step by step solution!

I used the same global strategy, although far less efficiently. In step 2, I didn't see this quick cul-de-sac and had to dig deeper to find the correct position for the '5'. But in step 3, I had to dig a LOT deeper. Completely missed the '7's, so no quick and easy clearing of the '8' hypothesis. Argghhh... I'll try to better observe the puzzle next time! Thanks again!


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Posted on: Thu Jul 02, 2020 5:02 am




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Joined: Fri May 13, 2011 2:21 am
Post Re: Step by step solution of a difficult 8x8 mod function pu
Thankyou CLM much appeciated as always.


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Posted on: Thu Jul 02, 2020 12:35 pm




Posts: 106
Joined: Sun Jan 31, 2016 7:52 am
Post Re: Step by step solution of a difficult 8x8 mod function pu
I think clm as made it more difficult than it needs to be. Below is my method.

It should be straight forward to get to this stage:

Image

From there follow these steps.

1. c1 is not 2, refer to cages in columns 3&4

2. d5 is not 5, refer to row 6 would have no cell that could be 1

3. add rows 7 & 8, this eliminates 3 & 5 from g7, and 2 from g8 & h8

4. g8 is not 4, refer to row 6, no cell has a 7

5. b8 is not 6, as above, no cell in row 6 has a 7

6. e8 or f8 must be 2, therefore e8 & f8 can not be 4, cage b7(14+) has no possible values

7. e5 is not 5, this would give us 3&7 in cells e6 and e7, this creates a parallel cell value conflict with e2 & f2

From this point it is all basic methods to complete the puzzle.


Last edited by michaele on Sat Jul 11, 2020 8:41 am, edited 1 time in total.



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Posted on: Fri Jul 03, 2020 4:59 am




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Location: France
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Post Re: Step by step solution of a difficult 8x8 mod function pu
michaele wrote:
1. c1 is not 2, refer to cages in columns 3&4

(I think you meant columns c&d) Could you expand, please? I fail to see how you deduced that.


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Posted on: Fri Jul 03, 2020 7:23 am




Posts: 106
Joined: Sun Jan 31, 2016 7:52 am
Post Re: Step by step solution of a difficult 8x8 mod function pu
Quote:
(I think you meant columns c&d) Could you expand, please? I fail to see how you deduced that.


Yes I did mean columns c & d,

Quote:
1. c1 is not 2, refer to cages in columns c&d


If we try 2 in c1, we can test the cages in columns c & d as follows:

cage c2 (192x) would become c2=1, c3=8, d2=4,6, d3=4.6
cage c5 (14+) would become c5=5, c6=3, d5=3
cage c8 (3mod) then has no possible values, therefore c1 can not be 2

This image shows what I mean.

Image


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Posted on: Fri Jul 03, 2020 11:39 pm




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Joined: Thu Jan 02, 2020 1:31 am
Post Re: Step by step solution of a difficult 8x8 mod function pu
Thank you.


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