**Step by step: very difficult 8x8 mod function (2012-03-22).** Answering to the jomapil’s question (see the thread “Hard patterns” in the Section Calcudoku General) I will try, in this topic, to describe a possible “analytical” solution for that 8x8 mod function (thursday Mar 22, 2012, Puzzle id: 432882) (though it was officially rated only 72.2 and it was solved by 63 puzzlers, some other calcudokers considered it as a really difficult puzzle whitin the 8x8’s):

Though we are dealing with a very concrete puzzle I have included this topic in this Section “Solving strategies and tips” (better than in the Section “Specific puzzles / your own puzzles”) in order to have in this Section an “step by step” solution of an 8x8 mod function. I will be using conventional language in many parts of the solution process for an easy understading.

Step 1. Initially we write the canditates and solve some determined cells: “12+” = [4,8] (since [5,7] is not posible) then (in blue) c1 = 8 and d1 = 4 >>> c8 = 6; now, since “72x” = [3,3,8], [3,4,6], [2,6,6], we arrive to the only possibility (in green) “72x” = [3,3,8] >>> d6 = 8, d7 = 7; also d8 = [1,2] and a6 = [2,6]. This last pair of candidates 2-6 for the cell a6 inhibit e6 of being 4 (or 2) (this is important and we will use it later because we would have 2-4-6 in cells a6-e6-f6 making impossible to solve the cage “128x” = [1,2,8,8], [1,4,4,8], [2,2,4,8] due to, i.e., g6 forced to be 3, 5 or 7).

Step 2. Different strategies or ways may be followed to solve any puzzle but, usually, if we have only two possibilities for a cell (like in the case of d8), a simple and good “strategy” is to analyze if one of them is not valid (using any “tactical” means), then as we eliminate uncertainties we quickly advance to the final solution. This is what we are going to do in the steps 2 and 3 and that will practically solve the puzzle. We will start demonstrating that a 2 is not possible in d8. Next graphic:

Let’s name a, b and c the content of the cells d8, e3 and e6, respectively (for simplicity) so: a + b + c = 10, since two lines, in this case columns d and e, must have a total sum of 72 (= 2 x 36). So if a = 2 >>> b + c = 8 >>> e3-e6 = [1,7], [2,6], [3,5]; the pair [2,6] is inhibited by the 6 in e7, the pair [3,5] is not valid (the 3 and the 5 should go “together” to d2 in the cage “15x”) then [1,7] would be the only valid combination with e3 = 1 and e6 = 7 >>> f6 = 5 >>> g8 = 5 (the 5 of row 8 must be inside the cage “2mod” since h8 <> 5 due to b7 = 8 and c7 = 2). Finally the 5 of row 7 to h7 and then there is not a solution for h8 since the 2 and the 8 are already set in row 8. Consequently,

**d8 = 1**.

Step 3. Now we will continue from this situation:

Since now b + c = 9, we have the combinations [2,7], [4,5] (obviously the [1,8] and the [3,6] are not possible due to e7 = 6 and e8 = 8). Let’s see that the [2,7] is not allowed. As commented in Step 1, in this case, the 2 could only go to e3 (with e6 = 7). Next graphic:

Now (f6 = 5) the only position for a 3 in column f should be f3 (it cann’t be inside the cage “6mod”, see my topic “Full tables: “mod function” and “bitwise OR” (in the 8x8’s)” in this same Section “Solving strategies and tips”). And this (see the arrows) permits to complete columns d and e with the final numbers and candidates for this hypothesis. Only by observing the pair [4,5] in e4-e5 we could conclude that this is an invalid setting (assuming the unicity of the solution and because of the other pair 4-5 in c4-c5).

But, if we prefer so, we can quickly see other conflicts, for instance: The combinations [1,4,4,8] (two additional 4’s for rows 4, 5, 6) and [2,2,4,8] (>>> g6 = 2 and then f4-f5-g5 = [2,4,8] which is not possible because there is no place for a new 4 in rows 4 or 5) would not be allowed. So the only left combination would be the [1,2,8,8] (as shown in red in the graphic) and, in these conditions, we had: a) There is no place for a 2 in column f; b) the cage “6mod” = [6,7], [6,8] has not a solution.

So we are left with the only

**pair [4,5] for the cells e3-e6** and, as commented in Step 1, necessarily, e3 = 4 and e6 = 5. We have arrived to this situation:

Step 4. An advanced puzzler would say that the calcudoku is solved, but we will continue to the end, as a practice for beginners or some medium players, considering the special characteristics of the “mod function”. Initially the candidates for f3 (to produce “1mod” with a 4) are 1-3-5. Now (next graphic, numbers in blue) the only place for a 2 (rows 4 and 5 inhibited and a 2 is not possible inside “6mod”, same topic “Full tables: …”) in column f is f8 >>> g8 = 5 (a 5 cann’t be in h8 as commented in Step 2) >>> h8 = 7 >>> h7 = 4; also f7 = 5 and g7 = 1.

Now f3 cann’t be a 1 because no combination would be possible for the cage “128x”: the two 1’s in f3 and g7 would inhibit the combinations [1,2,8,8] or [1,4,4,8] and, obviously, the [2,2,4,8] is forbidden due to the two 2’s already present in rows 4 and 5. So f3 = 3 >>> c3 = 7; c2 = 3; e2 = 1; e1 = 3. Also, f3 = 3 >>> f6 = 7 >>> “6mod” = [6,8] >>> f1 = 6; f2 = 8; f4-f5 = [1,4] >>> “128x” = [1,4,4,8] >>> g5 = 8; g6 = 4. Also a2 = 7 besides h6 = 3.

And what other things to be considered from this situation?:

Step 5. The only place for a 4 in column b is b2 (in orange colour) (being b4 or b5 inhibited by the pairs 4-5 in c4-c5 and 1-4 in f4-f5). Now we can see that a 5 is not possible in b1 (it would require a1 = 10 to obtain ”1-“) so the 5 of row 1 must go to h1 (a1 inhibited by the 5 of cage “280x” and g1 inhibited by g8 = 5). In these conditions, since g7 = 1, the 1 of row 1 must be in a1 or b1, that is, a1-b1 = [1,2] to obtain “1-“ with b2 = 4 (since [1,6] is inhibited by the 6 in f1). Then g1 = 7.

The 3 in column g (in violet colour) has the only place in g4 >>> g3 = 2. Then g2 = 6 and h2 = 2. Since a 1 cann’t go inside a cage “2mod” >>> h5 = 1 and we complete the numbers shown in violet in rows 4 and 5.

Now (in green) h4 = 6 and h3 = 8 >>> a3 = 5; b3 = 1. And a4 = 8.

Finally (in blue) b5 = 7; a5 =6 >>> b6 = 6; a6 = 2. Additionally a1 = 1 and b1 = 2. The last cell (gray): b4 = 5.

In summary, only steps 2 and 3 are difficult and require a more deep analysis. For the rest of the puzzle we have used the properties of the mod function (in some parts) so we must be familiarized with these; the property of the addition has been used but the parity property is not generally useful for this type of puzzles.

And the official solution: