HOVERDIA EIGHTEEN is a First-Of-Its-Kind Two-In-One logic-number puzzle, which consists of 64 boxes (8x8 boxes).

This unique puzzle is also very suitable for exercising not only your left-brain but the right-brain as well simultaneously.

This puzzle is usually presented with some pre-set numbers and leaving some empty boxes. The main objective is to fill up the remaining empty boxes with one of the number ranging from 1 to 8  inclusively in each box without repeating any thereof for all the horizontal and vertical blocks in accordance with the guidelines of the HOVERDIA EIGHTEEN.

There are two major steps in decoding this puzzle:-

Identify the pre-set or filled-up numbers; and determine a list of the remaining missing numbers by applying Rule One on the long block (8 boxes); and


Fill up the empty boxes with the missing numbers from the list of the remaining missing numbers by applying Rule Two on the short block (4 boxes).


In decoding this puzzle, Rule One and Rule Two may be applied interchangeably.

Hint One and Hint Two act as early error detection only. You may not have to memorize these two Hints but it would be an added advantage for you.

This tutorial will walk through the entire Hoverdia Eighteen puzzle till completion.

Happy puzzle decoding.

In actual fact, you can directly and easily fill up both the upper-short blocks on the long vertical blocks (V4) and (V5) as these two blocks are having only one empty box by adding up the three digits (1+6+3) and (5+4+7) respectively; and subtracting each of the resulting sub-total from 18 (Rule Two).

But you move on to the next long horizontal block (H5) which is quite similar to long horizontal block (H4) for the sake of practice; and for an alternative approach.

With the pre-set numbers of 7, 8, 5 & 6, the remaining missing numbers are 1, 2, 3 & 4.


For the left short block, 7 & 8 give a sub-total of 15 and with a variance of another sub-total of 3 making up from two empty boxes. From the missing numbers listing of 1, 2, 3 & 4, the only combination of two numbers that gives a sub-total of 3 is 1 & 2; and not from any of such combinations : 1 & 3, 1 & 4, 2 & 3 and 3 & 4.

Having decided on the appropriate set of two numbers (1 & 2), 1 can only be placed into an empty box at the intersection of H5:V3 and not at the intersection of H5:V4 as another 1 has already been filled in one of the box on the long vertical block (V4). The remaining 2 will be safely placed at the intersection of H5:V4.

Likewise, with the pre-set numbers of 5 & 6 on the right short block, which give a sub-total of 11, the remaining needed sub-total is 7. From the list of remaining missing numbers, 1 & 2 have already taken up, the remaining missing numbers would be 3 & 4 and it is perfectly fit into the remaining two empty boxes.

This is just another approach in decoding this puzzle where both long horizontal blocks ( H4) and ( H5) are having something in common. 

This is the most straight forward one. It just works with your mental calculation by applying Rule Two.

You may skip the Step One and start with  Step Two for placing the remaining missing numbers on the long vertical blocks (V4) and (V5).

On the upper-left short block within the long vertical block (V4), the filled up numbers are 1, 6 & 3.

You just add up the three digits (1+6+3) and would get a sub-total of 10. The difference between 10 and 18 (Rule Two) is 8. So the empty box at the intersection of H3:V4 should be placed with 8.

Continue onto the lower-left short block within the same long vertical block (V4), 2, 4 & 7 have been filled up and give a sub-total of 13. By subtracting 13 from 18, you would get 5. So 5 should be perfectly placed at H6:V4 in order to  comply with Rule Two.

You can now move on to the next long vertical block (V5). The upper-right short block has already been filled up with 5, 4 & 7. You would get 2 by adding up the three digits (filled-up numbers) and subtracting it from 18. The needed missing number to be placed at the intersection of H3:V5 is 2.

The same applies to the lower-short block. The needed missing number for the only empty box can be easily found and it should be 6 before placing it at the intersection of H6:V5.

Well, you have just completed that four short blocks on each of the four quarters under Rule Two.

Just before you more on to the next challenge, you are reminded to counter check with both the long vertical blocks (V4) and (V5) for the compliance of Rule One as well. 

As this is a diagonal block which consists of 4 boxes, you can only use Step Two approach.

When proceeding to Step Two with the application of Rule Two, you must also take into consideration of the Rule One for the long horizontal blocks (H2) and (H3) as well as the long vertical blocks (V2) and V3).

With the pre-set numbers of 6 & 1, which give a sub-total of 7; and the remaining two empty boxes have to be made up of another sub-total of 11.

The possible combination of two numbers from the numbers listing ranging from 1 to 8, which could be added up to a sub-total of 11, would be (3 & 8), (5 & 6) and (4 & 7).

Let's go through it one by one.

You may start with the first pair by placing 8 at H2:V3; and 3 at H3:V2 as shown on your left.

It might work fine for the long horizontal blocks (H2) and (H3) as long as there is no repeating number in any of the boxes in that blocks but it is impossible for both the long vertical blocks as another 8 has been placed on each of the two long vertical blocks (V2) and (V3).

The Rule One for both the long vertical blocks (V2) and (V3) is being violated.

For the sake of practicing, you may go through this again. But in actual decoding, you would easily fill up the single empty box at H3:V1.

With the pre-set numbers of 5 & 2, what would be the likely two numbers that give a sub-total of 11 for that diagonal block.

The possible combination of two numbers for the remaining two empty boxes would be (3 & 8), (5 & 6) and (4 & 7); and also give a sub-total of 11.

Of all the three pairs of possible numbers, only the first pair (3 & 8) is applicable.

As  both the long horizontal blocks (H2) and (H3) together with both the long vertical blocks (V6) and (V7) have been filled with 5 and 6; and 4 has already been filled up on both the long horizontal blocks (H2) and (H3). So these two pairs are out of question.

You go back to the first pair (3 & 8). 8 can be easily filled at the intersection of H2:V6 as another 8 has been filled up at H4:V7. With that, 3 to be placed at H3:V7.

You can easily tackle these two without much problem.

By adding up all the three digits on the left-side of the long horizontal block (H3), you would get a sub-total of 17 (4+5+8). The number for the empty box at H3:V1 should be 1, if you subtract the resulting sub-total of 17 from 18.

Similarly, for the right-side of the long horizontal block (H3), you should be able to get 7 for the empty box at H3:V8 by adding up all the three digits (2+6+3) before subtracting it from 18.

This would be in compliance with Rule One for the long horizontal block (H3).
 

You may start on the lower-left quadrant where part of the diagonal block has already been filled up with 7 & 7. It is permissible under Hint Two. Repeating number can only be happened in diagonal block and not in any other blocks may it be horizontal or vertical.

In that diagonal block, 7 & 7 give a sub-total of 14 and with the variance of 4 by subtracting 14 from 18. You have to get two numbers that would give another sub-total of 4 for the other two empty boxes.

Two possible pairs which would give the needed sub-total of 4, are (1 & 3) and (2 & 2).

The lower-left quadrant would be having Triple 2 if the second pair (2 & 2) is chosen. It is not in compliance with Hint One.

As 1 has already been filled therein at the intersection of H7:V5, 1 from the first pair of numbers can only be placed at the intersection of H6:V2 and whereas 3 should be placed at H7:V3.

Now, you can move on to the lower-right quadrant whereby the diagonal block is having not only  the same format in term of the empty boxes; and the sub-total of 14, but also the same possible two pairs of numbers.

So, with the first pair of numbers being taken up, leave you with the only second pair of numbers (2 & 2).

These two same numbers can be simply placed into two empty boxes without having to crack your head anymore.

With the least empty box on these blocks, it is quite straight forward.

You may just apply your mental calculation in decoding it, in this case.

You should be able to fill in 2 at the intersection of H1:V3 by adding up all the three digits (7+5+4); and subtracting the derived sub-total from 18.

With the same method as above, you should be able to find these:-
- 3 for the empty box at H1:V6, if not, this would get that [ 18-(8+6+1)];
- 4 for the empty box at H6:V1, 18-(1+8+5);
- 3 for the empty box at H6:V8, 18-(6+7+2);
- 6 for the empty box at H8:V3, 18-(1+8+3); and
- 5 for the empty box at H8:V6, 18-(4+7+2).

You will be getting used to mental calculation very often as from now on in analyzing and decoding the rest of this puzzle.

Starting with the pre-set numbers 2, 1, 5 & 3 on the long horizontal block (H1), the missing numbers are 4, 6, 7 & 8.


With 2 & 1 on the left short block, you need two more missing numbers which can be added up to another sub-total of 15 for that two empty boxes.

With that four listed missing numbers, only 7 & 8 can meet the requirement of 15. 

Another 7 and 8 have already been filled-in along the long vertical block (V1) and (V2) respectively, so just switch the vertical block.  7 for the long vertical block (V2) should be placed at H1:V2; and 8 for the long vertical block (V1) to be placed at H1:V1.

With these two numbers being determined, you can easily get the other numbers for the empty boxes at H2:V1 and H2:V2 by doing addition and subtraction. You should be able to get 3 for the empty box at H2:V1 and 2 for the empty box at H2:V2. 

Besides these, not only the diagonal block containing 8, 2, 5 & 3 is in compliance with Rule Two but the left short block of the long horizontal block (H2) containing 3, 2, 7 & 6 is also in compliance with Rule Two.

Now, you move on to the right side of the long horizontal block (H1). With 7 & 8 being taken up from that four listed missing numbers, the remaining missing numbers are 4 & 6.

6 should be placed at H1:V7 so that it will not be repeated in block (V8) as 6 has already appeared along the long vertical block (V8); and 4 to be placed at H1:V8.

By applying addition and subtraction on the upper short blocks (V7) and (V8), you would  easily get 1 for H2:V7 and 5 for H2:V8.

The remaining puzzle can be decoded differently. You may start on two short blocks either horizontally or vertically.

Let's start on vertical blocks (V1) and (V2). With pre-set numbers 8, 3, 1, 6, 7 & 4 on the long vertical block (V1), you should be able to get the needed two missing numbers and there are 5 & 2 for the compliance with Rule One.

Similarly, apply the same on the long vertical block (V2), the needed two missing numbers are 6 & 3.

Before placing two missing numbers into the long vertical block (V1), you would notice that another 2 has already been filled up at H7:V6; and 5 at H8:V6. You can easily place the missing numbers 5 & 2 by switching places along the long vertical block so that both missing numbers would not be repeated on the same horizontal blocks (H7) and (H8) accordingly.

For the next vertical block (V2), it is even more obvious as reflected with 3 & 6 on the vertical block (V3). So the remaining missing numbers 6 should be placed on  horizontal blocks (H7); and 3 on horizontal block (H8).

You are now coming to the last part of this puzzle. You may choose to work on either the long horizontal blocks or the long vertical blocks. Perhaps, you should start with  horizontal blocks as vertical blocks have been tried previously as about.

As usual, you have to find its respective missing numbers on these two long horizontal blocks (H7) and (H8). There are 7 & 8 for block (H7) and 4 & 1 for block (H8) by applying Rule One.

As easy as before, for the empty boxes in block (H7), 7 for vertical block (V7); and 8 for vertical block (V8) as 7 & 8 have appeared in block (V8) and (V7) respectively.

For the last horizontal block (H8), a pre-set 4 appeared along the long vertical block (V8) as such the missing number 4 should be placed on the other block (V7). The last missing number 1 is perfectly fitted into the last empty box at H8:V8 for the diagonal block to be in compliance of Rule Two as well.

It would be advisable to counter check with Rule One and Rule Two together with Hint One and Hint Two for a complete compliance of the entire HOVERDIA EIGHTEEN.

Hoping that these would give you the needed information about HOVERDIA EIGHTEEN so as to get you started on this First-of-its-kind Two-in-one 8x8 Logic-number  puzzle. 

Happy Puzzle decoding for sharpening your mind

"There's always an opportunity, if there is an alternative"

Always begin with RULE TWO for this style of puzzle

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