Albcell.D Guide

See below for instructions on how to play Albcell.D.

In the following, we’ll solve an example Albcell puzzle together. To keep things simple, we’ll work out the odd and even numbers separately. Once youre more familiar with the process, you can always mix and match odd and even to your taste, which should make the puzzles more fun to do. Now lets try one …

Look at the first diagram. It contains Roman numerals that identify the 10 cell groups in this particular puzzle. Each group contains 10 cells. If youre wondering where these cells are, look at diagram 2, which shows the starting grid for the puzzle. Circles contain even numbers; diamonds contain odd numbers. For ease of reference, Ive also given each column a letter and each row a number. (This is just to make the solution easier to follow – the actual puzzles don’t have this notation).

In the third diagram (the Notation Chart), you can pinpoint the location of any particular cell according to its corresponding column and row. Keep in mind that although each puzzle contains a different picture and a different ordering of numbers, the underlying cell structure is always the same from one puzzle to the next.

Now that youve got the essentials, grab a pencil and lets look at the puzzle itself. At this point its worth repeating the basic rule:

Place the numbers between 0 and 9 once in each row, column and cell group, putting the odd numbers (1, 3, 5, 7, 9) into the diamonds, and the even numbers (0, 2, 4, 6, 8) into the circles.

(Remember, a number can be used only once in each row, column and cell group.)

At first the puzzle may seem daunting – but there are a few techniques you can use to solve it. Lets start from the beginning…

Use the diagram on the next page as a worksheet on which to keep track of your progress, by adding numbers until the puzzle is completed.


Odd numbers First, look at the odd numbers in the last column, j. 1, 3 and 9 are given, but we still need 5 and 7. Luckily for us, there is already a 5 in group V, cell f3, which means there can’t be a 5 in j31.This leaves 7 as the only possible 1 candidate for that cell. This also enables us to fill in the last diamond in column j, j7, with the number 5. So grab your pencil and fill in cells j3(7) and j7(5).

Next, add the remaining odd numbers in cell group III, since by the same logic 5 can only be in i2 and the remaining 1 in h3. This then allows us to complete the second row by filling in the remaining odd number (1) in c2.

With me so far? Now you can practice what you’ve learned by completing column g. Use the number 7 in e8 as a guide. This will allow you to fill in the cells at g4 and g8 easily. Then, to reinforce what we’ve just learned, look at what’s missing in column e. Thanks to the 5 in g10, you should be able to see that 1 belongs in e10 and 5 in e6.

Once youve done that, try to identify the missing odd numbers in group VIII, in cells b9 and f9. Since there is no way to be sure just yet which is which, mark both candidates lightly in the cells, as shown in the diagram below. You dont have to do this, of course – if you prefer you could always #write” them in your memory. Whichever method you choose, our ultimate goal here is to “complete” the row for the single purpose of finding h9.

We’re making progress. Now we can consider a new approach that becomes clear if we look at cell i10. At first glance, it seems that this cell could contain any odd number except 5 and 1. But if you look closely enough, there is a catch! You should notice that the 7 in g4 covers off i4

and i6, and there’s no way it can be in i8. So we can say with certainty that in column i there is no other home for the 7 except i10.

Next, focus on the two number 3s in cells d5 and e4. Draw two imaginary arrows toward cell group VI, so that they line up with the two diamonds in that group, h5 and i4. Can you see now why the number 3 in group VI can only go in i6?

This last discovery also enables us to fill in h7 and i8 in group IX.

Look upwards at the first number we found, the 7 in j3. Where do you think the 7 in column d can go, since d3 is now unavailable? Next, by combining imaginary lines from j3 (horizontal) and a2 (vertical), can you deduce where the 7 is located in group IV? If you said b5 you were correct! This also helps us to complete column d (d3 being the last odd number remaining in that column), and then the third row.

From here we can resolve the uncertainty around cells b9 and f9. Thanks to the number 3 that we’ve just placed in the third row, in b3, we can now deduce both numbers as 9(b9) and 3(f9).

We can then finish column b by adding the remaining 5 in b1. While we’re here, let’s remain at the top a little longer, and fill in f1 and h1. (To do this, refer to the 7 in h9). Once you’ve completed that, add the last number in column h. Now we’re certainly making some serious progress! This is how our Albcell puzzle should now look:

(Hopefully your version looks the same! If it doesn’t, retrace your steps and try to figure out where you went wrong.)

Next, follow these steps:

  1. Complete group VI by adding 1 at i4.
  2. With the help of the 9 in j5, and the 1 in b7, you should be able to complete column f
  3. In which cells do you think the 1 and 7 in row 6 can go? They must be a6 and c6, of course – and the positions relate directly to the numbers in a2 and c2.

Continue with the missing odd number in group IV. Finding this should help us finish row 4. From here the only odd numbers still missing are the four digits in the bottom-left corner. H o p e f u l l y y o u shouldn’t need any help to find them by yourself …

Even numbers Now let’s focus on the even numbers. Look at d6 in the diagram on the right, in which I have included all the odd numbers we’ve found so far.

Cell d6 is empty, but the surrounding cells provide lots of hints as to what it might contain! T h e numbers 8(d4), 2(b6) and 6(d10) all “cross” at cell d6, which means that d6 cannot contain any of them. How about the remaining even numbers, 0 and 4? But wait!

Doesn’t group V already have a 4? This tells us that d6 can only contain a 0.

Now look at the 0 in i9. By applying the elimination technique we used before, we can deduce what’s missing in f8. (The question mark can only be 0, because group VIII still needs a 0, and the ninth row can’t have a second one!)

Now look at 2(c1), 0(c7), 6(i3) and 8(d4). Can you find the number in c3 in group IV based on this information? Great! Lets remain with group IV a little longer. Although we only know two of the five even numbers required to complete the group, we can clarify the picture a little further by penciling in the missing numbers in all their probable cells. Look at the vertical diagram. Youll notice that in this particular case, the positions of 0, 2 and 6 in group IV are all in column a. Because of this, we can safely assume that cells a1 and a9 must contain either 4

or 8. Although its not clear immediately which is which, as you focus more on the puzzle, you might see that the number 8 in column h covers off f2. This means that the 8 in cell group II can only be located somewhere in the first row. So we can safely assume that a1 must be 4! And because we now know that a9 is 8, this gives us another 8 in c5! We can also guess what j8 is, with the help of h2, since these two numbers make it clear where the 8 in group IX must be. This, in turn, gives us the number 8 in group V (f6). Why? Because both j6 and h6 are covered off. If you mark 4 and 6 in b4 and b8 respectively, the other two missing numbers in column b can be found straight away. So far, your puzzle should look like this:

Now complete the following steps:

  1. Find f2 using a1. (Where else can 4 find a home in group II?)
  2. Next deduce f10. It should be easy, since we can cross-reference the even numbers we have in column f (4, 8, 0) and row 10 (8, 6). In other words, since none of those numbers can be in f10, it must contain a 2. This step enables us to complete group V and column b (see figure below).

Write 4 and 6 in b4 and b8 respectively, then complete the eighth row.

Next, consider all the circles we filled with multiple candidates in group IV. Now that weve solved group V, the number 2 in g3 should help us complete the third row. Then we can fill in the missing numbers (2, 6) in group I by first completing column d. Then, with the help of 6 in e5, we can figure out a5 and a7. This is how the puzzle should look now:

Take a deep breath. Were almost there!

Now, having worked out most of the numbers in column c, we can add c9 and then complete group VIII by adding a 2 to e9. This comes in handy for completing both the ninth row and column e! Next, add the last number in group IX. (I’m talking about the 6 in h6, of course.) After that, let’s keep things interesting and take a look at the letter N of the word ‘FUN’. This remains the main unexplored area, which we can complete using our newly acquired skills. Of course, we can approach this problem from many angles – that’s the advantage of having accumulated so much data!

For example, you could draw an imaginary line along the fifth row, and use the 2, 6 and 8 there, together with the 4 and 0 in group IX, to find g5 and i5.

Then, by making use of the position of the 2 in g3, you could deduce g7 (8) and i7 (2). From here, finishing columns g and i is a mere formality. Now add a 6 to the end of the second row (j2). Then refer to the 2 in h8 to figure out why h4 can only contain a 0.

What you do from here is up to you! The remaining four empty circles could be deduced in several ways. You could choose to continue with the 2 or the 4 in group III (where we started our adventure with the lucky 7 in j3), or in the tenth row, by adding 4 and 0.

Either way, well done!