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HOVERDIA EIGHTEEN
is a FirstOfItsKind
TwoInOne logicnumber puzzle,
which consists
of 64
boxes (8x8
boxes). This unique puzzle is also
very suitable for exercising
not
only
your
leftbrain but
the
rightbrain as well
simultaneously.
This
puzzle is
usually
presented
with
some
preset
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 preset or filledup
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.
You
may take a look at this Hoverdia Eighteen
puzzle
on
your
right starting
with the
long horizontal
block
(H4)
as
indicated
with 4
small
squares.
6,
5,
8
&
2
(Preset numbers)
have
already
been filled in.
For
compliance
with
Rule One on
this long horizontal
block, a list
of missing
numbers is 1,
3,
4
&
7.
You always start with short block, which
consists of 4
boxes, for
Rule
Two. Two
missing
numbers
are
needed
on
the
short block,
which
is
on the right
side of
the long
horizontal
block
(H4). With
the
preset
numbers of
8
&
2,
which give a
subtotal of 10,
the
remaining
two empty
boxes must give
another
subtotal
of 8
(1810)
in order to
comply
with
Rule
Two,
where
the
short
block
with
4
boxes
must
be
added up
to an exact
sum of
18.
From
the list
of
missing
numbers,
which
set of
two
numbers
would give
you the needed
subtotal of 8. These sets of
(1
&
3),
(1
&
4)
and (3
& 4)
do not qualify it
but the only
available
set is (1
&
7).
Having narrowed
down to
these
two
numbers 1 &
7,
1 can be placed
into the
empty box at
the intersection of
H4:V6 as
another 1 has
already been filledin
on the long
vertical block (V5).
And
the remaining
7
can be easily
placed
into the empty
box at
the intersection
of
H4:V5.
Alternatively,
you
can
counter check
this by
referring
to the
long
vertical
block
(V6),
whereby
another
7
has
already
been filled
in.
You can
now move on to
the left
side of the
long
horizontal
block
(H4).
With the
remaining
missing
numbers of
3
&
4,
you
may
apply
the
same method
by
referring
to
the
long
vertical
block
(V4) whereby
another
4
has
been filled
therein. So
4
can only
be placed at the
intersection of H4:V3
and 3 go straight into
the empty box at
H4:V4.
The set of
numbers
(6, 5,
4, 3, 7, 1, 8, 2)
on
the
long
horizontal
block
(H4)
is
in
compliance
with Rule
One
whereby there is no
repeating number from a
set
of
numbers 1
to
8
inclusively.
Both
the
left
short
block
(6+5+4+3=18)
together
with
the
right
short
block
(7+1+8+2=18) are
in
compliance with
Rule
Two.
It
is
just
the
beginning
of an
excitement
in
decoding
the
Hoverdia
Eighteen
puzzle.
In actual fact, you can directly and easily fill up both the
uppershort
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 subtotal 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 preset
numbers of
7,
8,
5 & 6, the remaining missing numbers are
1,
2,
3 & 4.
For
the left
short
block,
7
&
8
give a subtotal
of 15
and with
a
variance of another
subtotal 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
subtotal
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 preset
numbers of
5 & 6 on the right short block, which
give a subtotal of 11, the remaining
needed
subtotal 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
upperleft 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
subtotal 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 lowerleft
short block within the
same long vertical block
(V4), 2, 4 & 7 have been filled up and give a
subtotal 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
upperright short
block has already been
filled up with
5,
4
& 7.
You
would
get 2
by adding
up
the
three
digits (filledup
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 lowershort 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
preset numbers
of
6
&
1,
which
give
a
subtotal
of
7;
and the
remaining
two
empty
boxes have to
be
made up of
another
subtotal
of
11.
The possible
combination
of
two
numbers
from the
numbers listing
ranging
from 1 to
8, which
could be
added up
to a
subtotal
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.
As for the
second pair of numbers (5 & 6), 5 has already been placed in both the
long vertical blocks (V2) and (V3); and
6
has already been
placed
in both the
long
horizontal blocks (H2)
and (H3).
So
these two numbers
are
not
in compliance
with
the
Rule
One for
two
long
vertical
blocks as
well
as
two long
horizontal
blocks.
Besides
these, you
will
be
introduced with
something new which is
called Hint
One where
each
quadrant (4x4 boxes)
shall
not
contain
more than 2
sets
of numbers
1 to 8
inclusively. In
other
word, with
these
two
missing
numbers to
be
placed
therein,
this
quadrant
would
be
having
Triple
5
and
6.
So this
pair is out
again.
It has
to be
this last
pair of
numbers
(4
&
7).
As both
the long horizontal
block
(H2) and the long
vertical
block
(V3)
have been
placed with
another
4,
so
4
cannot be
placed at
the
intersection of
H2:V3 but
it can be
placed at
H3:V2.
With
this,
the remaining
missing
number
7
should be
placed
at
H2:V3.
This
set
of
numbers
for this
diagonal block
will be
perfectly
added up to a
sum of
18
which is in
compliance
with
Rule
Two.
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
preset numbers
of
5
&
2,
what would be
the
likely two
numbers
that
give
a
subtotal
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 subtotal 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
leftside
of
the
long
horizontal
block
(H3),
you would get a
subtotal of 17
(4+5+8). The number for the empty box at
H3:V1 should be 1, if
you
subtract
the resulting
subtotal
of
17
from
18.
Similarly,
for
the
rightside
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 lowerleft 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
subtotal of 14
and with
the
variance
of
4
by
subtracting
14
from 18.
You have
to
get two
numbers
that would
give
another
subtotal
of
4
for the
other two
empty
boxes.
Two
possible pairs
which would
give
the
needed
subtotal
of
4,
are
(1
&
3)
and
(2
&
2).
The
lowerleft 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
lowerright
quadrant
whereby
the
diagonal
block
is
having not only
the same
format
in
term of
the
empty
boxes; and
the
subtotal 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
subtotal 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 preset
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
subtotal
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 filledin 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
preset
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
preset
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 Firstofitskind Twoinone 8x8 Logicnumber
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
Please
email us
at
comments@hoverdia.com, if
you have any abuse report or
complaint. If
you
like our puzzle,
please tell your
friends
to
visit
our web
site at www.hoverdia.com

