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.

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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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It should be straight forward to get to this stage:

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.

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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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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,

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

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

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

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

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

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

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,

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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The blank spaces in the main diagonal because obviously we cann't repeat the same number twice in a 2-cell cage.

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Because I'm afraid of making mistakes...

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fzpowerman47 wrote:

cool, thank you clm, it's clear in my head, now !

your table is perfect for me, I'll succeed to finish my 4 book puzzles, I know it now, it's nice, thank again

You are welcome, I am happy if the table can help. It's not difficult to extend it to - 7 to 7, for instance, so covering the - 7 to 1 or the -1 to 7 book puzzles.

However, do not forget that you could find situations (depending on the software or, in our site, until Patrick clarifies about the sign of the remainder, ... ), when dealing with mod operation and negative numbers, where two different results can be obtained, both correct, and both complying with |r| < |d| (as exposed in the Wikipedia), so there is, let's say, a "duplicity" or "duality" (in the remainder and in the quotient).

Those two solutions are part of the same class of "equivalence". For instance, you can see in the table that -4 mod -5 = - 4 (choosing the "natural" "integer" 0 as the quotient) but if you choose a quotient of 1 then 1 x (- 5) = - 5 and the remainder is: - 4 - (- 5) = - 4 + 5 = 1. In the class of equivalence created by the divisor - 5, the

Consequently in the table provided each result has another "colleague", "in the shadow", as a possible solution of the mod operation (both absolute values below the absolute value of the divisor). The modulo operation is easily understood with positive numbers while with negative numbers the "field of possibilities" expand to all numbers "to the left" of zero .

Best, clm.

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skeeter84

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your table is perfect for me, I'll succeed to finish my 4 book puzzles, I know it now, it's nice, thank again

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fzpowerman47 wrote:

I tried to do many times puzzles with mod and negative numbers but I make mistakes everytime.

For example, a cage of 2 cells with "0mod", what are the possible solutions with negative numbers ? I don't know...

-1 and -2 ? -1 and -3 ? -1 and 0 ?

Or a cage with "1mod", we can have -1 and -2 ? -1 and -3 ?

I don't know if I'm speaking clearly...

I have no problem with positive numbers but with negative numbers, I'm lost...

If you divide -1 by -2 the quotient is 0 then as 0 x (-2) = 0 the remainder is -1, but if you divide -2 by -1 the quotient is 2, then 2 x (-1) = -2 and the remainder is 0 (-2 - [-2] = -2 + 2 = 0).

Case -1 and -3, quotient is 0, remainder is -1, while in case -3 and -1 quotient is 3, 3 x (-1) = -3 and the remainder is 0 (-3 - [-3] = -3 + 3 = 0).

Case -1 and 0: Since the division by 0 is not allowed, -1 and 0 can only be considered as 0 divided by -1, quotient 0, remainder 0.

For the other question I would answer that, consequently, in no case, with -1 and -2 or with -1 and -3 you will have “1 mod”, only “-1 mod” or “0 mod” (in both cases).

I will propose you a table and later I will make some comments:

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To get a result you use the coordinates, that is, i.e., 3 mod -4 = 3. To use the table inversely we look for the result inside, i.e., “-4 mod” could be [-5, -4] or [-4, 5]. For “3 mod” you find in the table four possible combinations, eight combinations for “-2 mod”, etc..

To obtain this table I have followed a “natural” thinking, making the division, …, instead of successive subtractions … .

For instance, in the Code table provided by Patrick, for the first 7 languages (from C to Javascript) the operation “-13 mod -3” gives a result of -1, this is because you have a quotient of 4 (with positive sign since you divide two negative signs), so 4 x (-3) = -12 and the remainder is -13 – (-12) = -13 + 12 = -1. Similarly, “-13 mod 3”, quotient is -4, then (-4) x 3 = -12, remainder: -13 – (-12) = -1.

With languages Ruby or Python it seems that, for instance, in “13 mod -3”, a quotient of -5 (lower than -4) is taken, then (-5) x (-3) = 15, remainder 13 – 15 = -2, etc..

Since there are dozens of languages, Patrick should specify more precisely the way the modulo operation is evaluated in his software, I mean

Generally speaking, the modulo operation between two numbers D and d, was conceived for positive integers, such that “D mod d” = r, with a dividend (D), a divisor (d) (different from zero), a quotient (q) and a remainder (r). The result of the modulo operation is the remainder r, with r < d, and being both positive, obviously, |r| < |d| (the bars indicate the absolute value). This creates the “classes of equivalence”.

For instance, if you go to the Wikipedia you find this:

The concept of the modulo operation may be extended, as exposed by jpoos in a previous thread (the link provided by paulv66), to negative numbers (dividend, divisor or both) or even fractional numbers. And this brings new uncertainties and complexity, i.e., considering only integers you can find two results for the operation, one positive and one negative, though both accomplish the condition |r| < |d|.

See this comment in the Wikipedia:

q is a member of

D = dq + r

|r|<|d|

However, this still leaves a sign ambiguity if the remainder is nonzero: two possible choices for the remainder occur, one negative and the other positive, and two possible choices for the quotient occur. Usually, in number theory, the positive remainder is always chosen, but programming languages choose depending on the language and the signs of a(D) or n (d)”

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