Painted Cube and Cuboids - Basics

I was wondering how to keep things interesting and thought of changing up the domain of our recent discussions and elucidate something which is useful yet refreshing. Hence today, I hope to present what should be a very colorful illustration of one of the confusing concepts that appears in MBA examinations.

The name is Cuboid..Painted Cuboid. Well Cube was neglected as it is a special extension of cuboid only..offspring if you will ;)

Expect this one to be a slightly longer article since I feel the need to lay down some foundation before I start generalizing the concepts and formulas. Let's kick off then..


First Observation : If the cuboid is painted on all faces.
The red marked cubes(corners of the cuboid) will be painted 3 sides.
The green marked cubes(along the edges) will be painted 2 sides.
The blue marked cubes(centre of each face) will be painted only 1 side.

While there would be some no sides painted cubes at the centre of the cuboid which are as yet unseen.Your job primarily in these types of questions is to find how many of each of these categories above. Option 1 is to visualize and count. Tough job for most. Option 2 is the one to go for and is the one we discuss here as we move onto second observation.

Second Observation :
Visualize this cuboid layer by layer moving from the face looking at you and going backwards away from you cross section by cross section along the breadth(Like a Zoom in and you pass through the front face reaching the opposite one).

You should be able to understand by now that we would have 5 layers..right? Moving on..


The cubes that are part of the faces of the cuboid will be exposed more to the paint than the ones more internal to the structure.


What you need to do now is number each cube of the layer you consider for the number of sides it will have exposed/painted when outer surface of the cuboid is painted and jot that down.
So for the first layer(L1, nearest to your viewpoint), you will have :                                                                                            

If you observe carefully this configuration will be same for the farthest layer as well, L5 as mentioned in the image itself.
Now when you look to move onto L2 and zoom in along the breadth into the cuboid. Note the changes that will occur. Basically, you are taking out the external surface of cuboid out of picture while rest of the configuration remains same. Hence what you will get for the next three layers(the internal layers), L2,L3 and L4(say) is a configuration similar to L1 and L5 with each cube having 1 less side exposed. Hence you get the layer as :
                                 
So we basically have 2 layers of one kind and 3 layers of another kind.
If we just manually count up the n-side painted cubes now, we should be getting the following results :
3 face painted : 2*(4 corners) = 8
2 face painted : 2*(12) + 3*(4) = 36
1 face painted : 2*(8) + 3*(12) = 52
No side painted : 3*8 = 24
Total : 120 = 6*5*4 as it should have been the case.

Now let's try and generalize this for sides : length l, breadth b, height h
3 face painted :

Always 2 layers with 4 corners = 2*4 = 8
No side painted :

Say we took layers along breadth b,then we will have 2 layers : first and farthest ones ; (b - 2) layers internal. So number of such cubes will be : (b - 2)[(l - 2)(h - 2)] as the no face painted cubes form a grid at the centre of each internal layer with each dimension reduced by 2 units(1 from each end).

So, we get, no face painted cubes = (l - 2)(b - 2)(h - 2)

1 face painted :

Two outer layers will have (l - 2)(h - 2) grid at centre if we stick to our along breadth b cross-section. Internal layers will have :l*h - (4 + (l - 2)(h - 2)) cubes
= lh - (4 + lh - 2l - 2h + 4) cubes
= 2l + 2h - 8 cubes = 2(l - 2) + 2(h - 2) cubes

Adding them all up,
2(l - 2)(h - 2) + (b - 2)[2(l - 2) + 2(h - 2)] giving us :

1 face painted cubes = 2[(l - 2)(b - 2) + (b - 2)(h - 2) + (l - 2)(h - 2)]
2 face painted :

Two outer layers : l*h - (4 + (l - 2)(h - 2)) = (2l + 2h - 8) cubes
Internal layers : 4 corners. So, 4(b - 2) = 4b - 8
Adding them up, we get,
2[2l + 2h - 8] + 4b - 8
= 4l + 4h + 4b - 24
= 4(l + b + h - 6) cubes for 2 face painted

Summary :

Cuboid : l x b x h
3 face painted : 8
2 face painted : 4(l + b + h - 6)
1 face painted : 2[(l - 2)(b - 2) + (b - 2)(h - 2) + (l - 2)(h - 2)]
No face painted : (l - 2)(b - 2)(h - 2)

Cube : Nothing but a cuboid with l = b = h = n(say). So,
3 face painted : 8
2 face painted : 12(n - 2)
1 face painted : 6(n - 2)^2
No face painted : (n - 2)^3


That will be all for this article on Painted Cubes and Cuboids. Next and the final article on this topic, I will discuss some variety of problem that exist on this topic and also the famous Polya Enumerization Theorem.


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Cheers!
AS