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Re: Step by step solution of a 7x7 subtractions only
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Posted on: Wed Sep 21, 2011 11:36 pm

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
Re: Step by step solution of a 7x7 subtractions only
[quote="clm"]Step by step solution of a 7x7 subtractions only.
(This is a retransmission of the topic with larger graphics and enhanced resolution).
For this example I will use the wednesday Aug 10, 2011, 7x7 difficult puzzle (9 points) (Puzzle id: 335050). This puzzle was solved by only 63 players (less than a 20% of the active players in that moment). However it does not require excessive analysis.

I will use pairs of numbers (in black and small size in the lower part of the cell) when those are the final candidates for that cage, in big size (in the center of the cell) for the final solution for that cell, and I will use colors in general (and in one side of the cages to show the different options discussed). The small numbered “ellipses” refer to the steps along the procedure.

Different ways or methods may be followed. The method of “test and error” will be applied during the process of solution. A good inicial strategy is determining the “isolated” cells like E2 and/or discussing the cages “n-“ where n has a high value (less number of combinations). First we see that the cage “6-“ contains the combination 1-7 (only one possibility, obviously a cage “n-“ has p - n possibilities, being p the size of the puzzle, i.e., “3-“ in a 9x9 has 9 - 3 = 6 possibilities, that is, 1-4, 2-5, 3-6, 4-7, 5-8 and 6-9; in our case 7 - 6 = 1, only one possibility the obvious 1-7). So we write 17 (1 or 7) in cells F5 and F6.

Step 1. In general it is easy to determine the parity of “isolated” cells like E2 that must be odd to complete the even parity of the first two rows (see my topic “The parity”); in this particular puzzle it is obvious anyway that the parity of the cell is odd since 2-6 is not a possible combination for the cage E2-E3, so the other two possibilities, 1-5 and 3-7, are made with odd numbers. Since 3-7 is not possible (a 6 could not be located in column E because another 3 would appear), the solution for the cage is 1-5, so E2 = 1 and E3 = 5.

Step 2 (shown in light blue). 2-6 is not valid in C3-C4 because the 7 of row 3 should go to A3, with a 6 in A4, and consequently no 5 could be located in row 4. So the 3-7 is the only possibility for this cage “4-“ and we write 37 in cells C3-C4.

Step 3 (shown in red). 1-5 is not valid in C6-C7 because the 7 of column D should only go to D5, with a 6 in C5, and no 5 could be located in column D (now the 3-7 in C3-C4 inhibit any 5 to be in D1 or D2). Then, since 3-7 and 1-5 are not possible, C6-C7 must be 2-6 and we write 26 in these cells.

Step 4. A 1 is not possible in C5 (it would require a 2 in D5 but there is a 2 already in column D, in D3) so C1= 1 and consequently D1 = 3.

Step 5 (shown in orange). 7 is not valid in G1 because it produces a 5 in G2 and no 7 could be located in row 2. So the 7 of row 1 must be placed (with a 6) in A1-B1 and we write 67 in these two cells.

This is the actual situation:

Now let’s continue with the following graphic:

Step 6 (shown with numbers in yellow).

1) We solve now the cage “4-“ in C3-C4. A 7 in C3 is not possible (with C4 = 3) since the 6 of row 3 should necessarily be in A3 (could not be inside F3-G3 because of the contiguous numbers, 5 and 7, already present in row 3), and since 6-7 would not be possible now in A3-A4 (otherwise there is a conflict with A1-B1), A4 = 5, the cage “3-“ in F4-G4 would be 4-7 so G4 = 7, F4 = 4, and the 4 of row 3 would go to G3 (F3 = 3) making impossible to have any 4 in row 1. Then C3 = 3 and C4 = 7.

2) Now the 4 of row 3 is not inside the cage “1-“ of F3-G3 (the contiguous numbers necessary for “1-“, 3 or 5, are already present in row 3). So A3 = 4. Also G3 = 7 and F3 = 6.

Step 7 (shown in dark blue). We analyze now the cage “3-“ in F4-G4. The combination 3-6 is not possible because it requires A4 = 5, E4-E5 would be 4-7 with E4 = 4 and E5 = 7. Observe that now no 7 could be placed in row 7 (the 5’s in A4 and B6 would inhibit the 7 to be in A7, B7 or D7, the other cells or row 7 “erased” by 7’s). So the cage “3-“ in F4-G4 must be 2-5 and we write 25 in those cells. And as a consequence A4 = 3 (to complete the cage A3-A4).

Step 8 (shown in violet). The combination 1-5 is not possible for the cage “4-“ in A5-A6. Observe that a 5 can not be located in F2 or G2 (since it must be a 5 in F1 or G1 and F4-G4 contain a 5 so “erasing” columns F and G simultaneously); a 5 is not possible in D2 (3 and 7 in C3-C4), so the 5 of row 2 would go to C2 with a 7 in D2 and then no 5 could be positioned in column D (a 5, i.e., in D7 would require a 3 or a 7 in D6 both already present in column D). So A5-A6 is 2-6 and we write 26 in those cells.

Additionally A7 =1, B7 = 3. The 7 of row 7 goes to E7 (E6 =4).

Step 9 (numbers, although shown in green this time, are final numbers). We now complete the cage “3-“ in E4-E5. E5 = 3, E4 = 6. And also B4 = 4 with B5 being 2 or 6 (we write 26 in cell B5).

The 5 of row 7 must go to the only free seat D7 (not inside “2-“ in F7-G7 due to the contiguous numbers 3 and 7 already present in row 7). Now D6 = 7 (3 in D1). Then F6 = 1 and F5 = 7. The 1 of row 5 goes to G5 (1’s already in columns C and D) and G6 = 3.

This is the situation after step 9:

Now let’s finish the puzzle with two simple steps (next graphic):

Step 10 (shown in orange).

A1 = 7 (2-6 in A5-A6), B1 = 6.
A2 = 5, B2 = 7 (3 in B7).
Row 5: C5 = 5, D5 = 4.
Row 2: C2 = 4, D2 = 6, F2 = 3 (3 in G6), G2 = 2.
Row 4: F4 = 2, G4 = 5.

Step 11 (shown in light blue).

The 6 in B1 forces B5 = 2, A5 = 6, A6 = 2, C6 = 6, C7 = 2.
G7 = 6 (6 in F3), F7 = 4, G1 = 4 and F1 = 5.

And this is the “official” solution:

Posted on: Wed Sep 28, 2011 3:38 pm

Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: Step by step solution of a 7x7 subtractions only
Thanks for your good work, clm.
It's a little complicated but enough elucidatory. I have learned so much with you.
I'm waiting for your next help.
Thank you very much, again.

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Posted on: Thu Sep 29, 2011 10:13 am

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
Re: Step by step solution of a 7x7 subtractions only
jomapil wrote:
Thanks for your good work, clm.
It's a little complicated but enough elucidatory. I have learned so much with you.
I'm waiting for your next help.
Thank you very much, again.

You are very welcome, thank you for appreciating it. It is a little complicated because it is difficult to show the variety of steps (many ways or alternatives...) that a solver is following with the pencil and the eraser. I will see if I can go to higher puzzles in the future, like 8x8's ... The main idea, though, is transmitting confidence in the sense that every puzzle can be completed by everyone (regardless of its difficulty), and that there is a lot of information inside the puzzle itself to arrive to the unique solution. You are in danger of becoming soon (as me) a "calcuholic" (this game is very addictive for people who love numbers, and there must be a few old members in a detoxification centre).

Posted on: Fri Sep 30, 2011 10:09 am

Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: Step by step solution of a 7x7 subtractions only
I have just two weeks I joined to calcudoku and I'm already a calkuholic. I only had some experience with the simpler kenken. But calcudoku has much more interest and much more variety. From the 12x12 and 10x10 puzzles to the exponentiation or bitwise OR... But the Modulo Puzzle is the more difficult on account of the many possibilities of each cage that become it very compact, confusing and heavy. It's the only special puzzle I never succeed to solve.

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Posted on: Mon Oct 03, 2011 9:05 am

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
Re: Step by step solution of a 7x7 subtractions only
jomapil wrote:
... But the Modulo Puzzle is the more difficult on account of the many possibilities of each cage that become it very compact, confusing and heavy. It's the only special puzzle I never succeed to solve.

Here is the full table:
7mod: 78 (1 combination)
6mod: 67 68 (2)
5mod: 56 57 58 (3)
4mod: 45 46 47 48 (4)
3mod: 34 35 36 37 38 and additionally 85 74 (7)
2mod: 23 24 25 26 27 28 and 86 83 75 64 53 (11)
1mod: 12 13 14 15 16 17 18 and 87 76 73 72 65 54 52 43 32 (16)
0mod: 12 13 14 15 16 17 18 and 84 82 63 62 42 (12).
Total of possible combinations: 56.

The main tips are:
1) A number x can never be inside a cage ymod if x < y, for instance, 2 can never go inside a cage 3mod, 4mod, 5mod, 6mod or 7mod.
2) In a cage 0mod or 1mod "any" number (from 1 to 8) can initially be present.
3) In cages 4mod, 5mod, 6mod and 7mod, the 4, 5, 6 and 7, respectively, must be present; the cages 1mod, 2mod and 3mod are possible without the respective 1, 2 or 3. The cage 0mod has pairs that are multiples or a 1 is inside, what in fact is a particular case of the multiples.
There are a few combinations (if you forget any when solving, the full puzzle may collapse) but less than in the "bitwise OR". I am sure next thursday you will solve it.

Posted on: Tue Oct 04, 2011 3:40 pm

Posts: 116
Joined: Sat May 14, 2011 3:18 am
Re: Step by step solution of a 7x7 subtractions only
Thank you very much for posting those tips, clm!

The term "mod" was taught to me in school as a "remainder" function when any number is divided by a smaller one. So I was really in the dark for a long time about how a valid entry in a "4mod" cage could be 5,4 -- I was convinced that had to be a 1mod.

The stuff I read online was just confusing. This really makes it much clearer.

Posted on: Tue Oct 04, 2011 10:39 pm

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
Re: Step by step solution of a 7x7 subtractions only
pharosian wrote:
Thank you very much for posting those tips, clm!

The term "mod" was taught to me in school as a "remainder" function when any number is divided by a smaller one. So I was really in the dark for a long time about how a valid entry in a "4mod" cage could be 5,4 -- I was convinced that had to be a 1mod.

The stuff I read online was just confusing. This really makes it much clearer.

Very welcome, pharosian. Yes, it is always the "remainder", and this is what produces the confussion because in the "calcudoku" "any" of the two numbers in the cage can be the dividend and the other the divisor (changing the role, following the same philosophy of the "-" cages, etc., ...) so 4 / 7 gives a quotient of 0 and a remainder of 4 then 47 is a valid combination for 4mod, but the "natural" division (7 / 4) would produce the 3mod, in fact, obviously, 47 is a valid combination (among others) for 3mod. In the table I have shown first, for each "xmod", the pairs of numbers in the sequence that produce a quotient 0, followed by the natural divisions.

In the case of the parity (your post under "Trial and error"): In today's 10x10 (Puzzle id: 368234), you will probably see quickly that cell b2 is even (2 upper rows = 110; we have 78 + 2 cages "2-" with an even total, so 110 - even = even); a 2 in b2 would produce a 4 in b3 and there is already a 4 inside "7+" in d3-e3; a 4 is not possible since there is already a 4 inside "5+" in e2-f2; a 6 is not possible because (for the "2-") the roommates, 4 or 8, are already present in row 3, so ... b2 = 8 (and b3 = 6 since b1 = 10) (I hope Patrick does not object for this little help). So, using the "parity rule", you have determined the value of a cell, in this case b2.

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