Culica is a puzzle and a multi-player game played on the surface of a 3×3×3
cube. It consists of a cube base an pegs of four different colours that click into the
cube. There are many different games and puzzles that can be played on this board. If
you have two or more Culicas, then you can even connect them together with the pegs to
make a larger playing board.
Each face of the cube consists of a 3×3 square, so there are 6·9 = 54
playing locations. Every location has exactly 4 neighbouring locations, some of which
may lie on a different face of the cube. I will not describe all the different multi-player
games that are possible, and restrict myself to some of the more interesting solo puzzle tasks.
CuColours: Fill the whole cube with pegs such that pegs of the same colour are never
neighbours, nor diagonally adjacent.
Jump to solution
CuSnakes: Find a snake's tour on the cube. Place the first peg somewhere on the cube.
Each subsequent peg must be adjacent to the previously placed peg. Try to place as
many pegs as you can before no more moves are possible. It is possible to fill the whole
cube this way. If you have done that, try to make it so that the final peg is adjacent
to the starting peg (for which you can use a different colour peg so as to keep track of
it more easily).
Jump to solution
CuFrog: Find a frog's tour on the cube. This is like the snake's tour, but using 'frog'
moves. A frog moves exactly two squares straight in any direction (but not diagonally). So between
two consecutive pegs you place must lie exactly one other square, which may be empty or full.
Again try to fill as many squares as possible, preferably such that the final location
is a frog's move away from the first.
Jump to solution
CuKnights: Find a knight's tour on the cube. Place the first peg somewhere on the cube.
Each subsequent peg must be a chess-knight's move away from the previously placed peg, that
is to say that it goes two squares forward in one direction and then one square sideways
(at right angles to the forwards direction). You are only allowed to go from one face of the cube
to another at most once per move. An easier way to understand this is to imagine a 2x3
rectangle wrapped around the surface of the cube, and a knight move goes from one corner
of this rectangle to the diagonally opposite corner. Try to place as
many pegs as you can before no more moves are possible. It is possible to fill the whole
cube this way. If you have done that, try to make it so that the final peg is a knight's
move away from the starting peg, so that you have a 'closed' knight's tour.
Jump to solution
CuMadness: Place as many pegs as you can using the following rules: Every red peg must be next
to exactly one yellow peg, no more no less. The colours of its other neighbours do not matter, and
can be any number of red, green or blue. In the same way every yellow must be next to exactly one green,
every green next to exactly one blue, and every blue next to exactly one red. This is remarkably difficult.
Jump to solution
CuRing: Place as many pegs as possible such that for every peg there is no peg of the same colour
within the next three squares in any of the four directions. Two pegs of the same colour may be close together
diagonally, but if they lie in a straight line, then there must be at least three squares between them.
Jump to solution
CuLoops Consider the loops of squares around the cube - a loop of 12 squares in a straight line
going around four faces of the cube. There are 9 of such loops. Fill as much of the cube as possible with
the four colours of pegs such that no loop has more than 3 pegs of the same colour. Note that this means
that if a loop is to be completely filled it will have exactly three of each colour.
Jump to solution
CuLook: Similar to CuLoops, but using a single colour. Place as many pegs as possible such that
every loop has at most one peg. Then try again such that every loop has at most two pegs. Then try with
three, with four, etc.
Jump to solution
CuWizard: Consider four squares in a row, adjacent to each other, and crossing over an edge of
the cube such that two squares are on one face and two squares on another face of the cube. There are
36 of these quadruplets on the cube. Put pegs on the cube such that as many of these quads as possible
contain four pegs, one of each colour.
Jump to solution
CuNitrogen: Find a molecule consisting only of Nitrogen atoms, that is to say place pegs such
that every peg has exactly three neighbouring pegs leaving its fourth adjacent location empty.
Apart from the trivial empty cube, there are only two solutions.
Jump to solution
CuOxybon: Find molecules consisting only of carbon and oxygen atoms. In other words, place
red pegs ('Oxygen') such that they have exactly two neighbouring pegs and two neighbouring empty locations,
and pegs of any other colour ('Carbon') such that they have four neighbouring pegs. A completely filled
cube would be a valid molecule containing only carbon, so let's use only molecules that have at leasy one
Oxygen atom. Fill up as much of the cube as possible using one or more of these molecules that contain
at least one red peg.
Jump to solution
CuOxygen: Find a molecule consisting only of Oxygen atoms, that is to say place pegs such
that every peg has exactly two neighbouring pegs leaving its two other adjacent locations empty. Use
as many pegs as possible to create one or more of these molecules.
Jump to solution
CuSol: Solitaire. Fill the whole cube with pegs, except for any single square. Then try to
do as many moves as possible, where a move consists of jumping one peg over an adjacent peg to an
empty square immediately behind that, and removing the peg that was jumped over. So each move involves
three adjacent squares in a row, which start full/full/empty before the move and are empty/empty/full
afterwards. Each move removes one peg, so try to do 52 moves so that only one peg remains.
Jump to solution
You can find these and other puzzle tasks on the Culica site.
Solutions:
CuColours:
If you put three colours in a row on any face, then the rest of the colours
are forced, and that leads to a dead end. Therefore every face must have its corner
squares the same colour, as well as each pair of opposite edge squares. This
leaves only one solution.
CuSnakes:
The easiest way is to fill each face in turn, which can lead to a closed
tour such as this:
CuFrog:
I decided to split it up into six congruent parts, which simplifies the problem
to finding only a 9 move tour. Like CuSnakes, this construction leads to a closed
tour.
CuKnights:
If a knight jumps around on a single face, it can fill everything but the centre. To get a full
knights tour, I therefore decided to fill the centres first, by necessity also alternating with corners.
After that, I fill the remaining squares on each face in turn.
If you want to have a closed tour, this method does not quite work, because you will
have used up 7 corners when filling up the face centres (you have to get to the first face
centre from a corner). This means that one face has its centre and two corners filled,
and this face cannot be later filled up in one visit. That face can be filled up with
two visits however, which leads to a solution like the following:
CuMadness:
This is a very difficult puzzle. I think that what makes it so hard is that it is so tempting to make
long RYGBRYGB chains. When I set my computer program to the task it spit out hundreds of solutions per
second, so theoretically it should be relatively easy once you set aside that chain-building idea.
Here is one of the solutions my computer found:
CuRing:
The solution is unique (except for its mirror image).
Once one column is filled, there is almost no choice left if you want to fill all 48 non-centre spots.
CuLoops:
My solution was to take this pattern:
and repeat it with each colour a quarter turn of the cube apart, to get this:
There are many other less symmetric solutions. Note however that you can also take a solution to CuRing
and just fill four of the centres with one peg of each colour. It is not possible to place more
than 52 pegs, because it is impossible to place more than 13 of any colour. See CuLook below for a
proof.
CuLook:
It is not possible to fill all the loops with an odd number of pegs. If there are k pegs in each loop,
then there are 9k/2 pegs on the cube because each peg lies in exactly two loops. This is not a whole
number when k is odd, so it is impossible to fill all the loops with an odd number of pegs. You can
however fill it with (9k-1)/2 pegs in those cases.
Here is another way to look at the problem.
Each loop contains at most k pegs. Three parallel loops therefore contain at most 3k pegs.
Let's use A to represent the total number of pegs in the front and back faces, and B the number
of yellow pegs in the left and right faces. Since the front, right, back, and left faces together
consist of the three horizontal loops, then there can be at most 3k pegs in those faces. In other words A+B<=3k.
Let C be the total number of yellow pieces in the top and bottom faces. Looking at the other sets of parallel
loops we find in the same way that A+C<=3k, and B+C<=3k. The total number of yellow pieces on the whole cube
is A+B+C. The largest value of A+B+C, given that A+B<=3k, B+C<=3k, and A+C<=3k, is when the pegs are as
evenly divided as possible amongst A, B and C. For even k this means that A = B = C = 3k/2. For odd k, the
best we can do is A = B = (3k-1)/2 and C = (3k+1)/2.
For at most 1 peg per loop, this means you can only place 4 pegs, and that one pair of opposite faces
must contain 2 pegs and the other two pairs 1 peg each. This is easy to do, for example:
To achieve 2 pegs per loop, we must place 9 pegs, with each pair of opposite faces containing 3 of them.
We can restrict ourselves to three adjacent faces by putting a diagonal line on each:
For at most 3 pegs per loop you can place no more than 13 pegs, and one pair of opposite faces
must contain 5 pegs and the other two pairs 4 pegs each. We already saw one such pattern in my
solution to CuLoops:
You can also combine the 3-face solutions above for 1 and for 2 pegs to give this:
To achieve 4 pegs per loop, we must place 18 pegs, with each pair of opposite faces containing 6 of them.
We can simply duplicate the 2 peg solution that used only three adjacent faces, leading to a solution where
all six face have one filled diagonal. The nicest pattern is to choose the diagonals so that they form
a tetrahedron:
You can instead also invert the three faces used in that 2-peg solution, so that each of those
three faces have two pegs in each row and column instead of just one.
For at most 5 pegs per loop you can place no more than 22 pegs, and one pair of opposite faces
must contain 8 pegs and the other two pairs 7 pegs each. Adding some pegs to the 4-peg solution gives:
To achieve 6 pegs per loop, we must place 27 pegs, with each pair of opposite faces containing 9 of them.
The simplest is to just fill three adjacent faces:
For a best possible solution to having at most k pegs per loop, where k is 7, 8, 9, 10, 11 or 12, just take a
solution for k-6 pegs that uses only three faces and then completely fill the three empty faces.
CuWizard:
There are only 2 full solutions (up to rotation/reflection/colour-permutation). Looking only
at one 2x2x2 corner, there are 6 overlapping CuWizard quadruplets. Once you choose the 3 colours
to use on the three face centres, there is essentially only one way to arrange the 4th colour
(as well as its mirror image). This then fully determines the rest of that 2x2x2 corner.
From this it follows that there are essentially only two ways to cover the whole cube. We could have a
fully symmetric pattern, reflecting that 2x2x2 block to the rest of the cube. Each pair of opposite face
centres has a single colour:
If we choose one pair of opposite centres to have different colours, then the colours of the other face centres are forced,
which in turn forces the rest of the cube.
CuNitrogen:
The two solutions are:
and
CuOxybon:
The fewest possible empty spots is 9, and this can be achieved in two ways.
and
CuOxygen:
The maximal number of pegs is 32, and this can be achieved in 3 distinct ways.
Firstly you can do it using 1 long closed loop in two ways:
and the following nicely symmetrical pattern
You can also use 3 closed loops:
CuSol:
I constructed this solution in two parts, each part clears three
adjacent faces. Number the squares as follows:
Start with square 11 empty.Then do the first part of the solution shown below, which clears
the first three faces apart from one peg remaining at 14.
Then do the two connecting moves, so that you now have a space in the last three faces
at square 29.
Finally do the second half of the solution, clearing those last three faces, leaving a
single peg at 32 (or 29).
| Part 1 | connection | Part 2 |
23-11
8-17
2-8
25-23-11-5
19-17
9-18
3-9
27-25
4-6-12-24-26
10-4
22-10
1-7-16-18
15-27-25
14-12-24-26-14
|
29-15
14-28
|
31-29
33-31
45-33
51-45
44-32-30
43-31
46-48
52-46
30-32
33-31
41-43-49
54-52-46
28-30-42
45-51-49-43-41-29
40-28-30-32
|
