As we know, many tough combinatorial problems are based on a very basic step or strategy which can simplify the problem upto a tremendous extent. One of our important strategies in proving combinatorial problems is using colourings; i.e., to use colours to paint the problem and then proving it using a semi-contradictory argument.
Before proving colouring problems we must first come to some important terms that we need to know.
1. Colourings Of Graphs
First of all, I will introduce you briefly to the colourings of graphs. Though it is not very related to problem-solving, still we need to have some basic knowledge on this.
Before that, we need to know what a graph is. I will introduce us to the idea of a graph in brief. Actually; a graph is defined as a figure, which can be 2-Dimensional (planar) or 3-Dimensional, which must consist of a set of vertices, a set of edges. We denote as an edge with as endpoints; and in order for the figure to be a graph we must have a mapping from each sets of vertices to the endpoints.
An edge is called a loop whenever we have identical with . A graph is called simple when it has no loops, and no two edges having the same set of endpoints. Not that every polygon can be called a simple graph.We will come to graphs later again.
We say that a graph is properly coloured if we have no edge having two endpoints of the same colour. We denote the set of colurs with and hence for a polygon with an odd number of sides we have where denotes the cardinality of or the number of elements in
We denote i.e. the minimum number of colours required for a proper colouring. If we denote as the polygon with number of sides , then we have This is said to be the chromatic number of a graph We also have the Four Colour Theorem, which can be stated equivalently as; if is planar we must have We will discuss about this later too, but now let us come to our main focus.
In case of combinatorial problems, the simpler, the better. Therefore we will show why colurings are simple.
We can cover up a chessboard with numbers of dominoes in finitely many ways. A physicist M.E. Fischer calculated this number to be But that is not our point of concern. We try to modify the problem a bit. If diagonally opposite squares are cut off from this chessboard, then in how many ways can we do the covering with dominoes?
Though it looks tough, we can explain it with a simple combinatorial argument. Let us consider the number of black and white squares. First of all, we had squares; black and white. If we take out two diagonally opposite squares, then let both the squares be white in colour (WLOG). In that case, we see that in our newly arranged chessboard, we have white squares and black squares. If we were to cover it up with dominoes, it would have covered white and black squares. but we don’t have white squares, we have only of them left. Therefore the colouring is not possible.
Like this, we can prove many problems by just colouring and then contradicting.
A chessboard cannot be covered by straight tetrominoes.
We colour our board with (Orange, green, blue, ash/grey) as follows:
Here each horizontal straight tetromino covers four squares of different colours, while the vertical ones cover four squares of the same colour each. After all horizontal tetominoes have been placed, we are left with squares of colours respectively. If such a complete covering would have existed, then we must have had which is not possible. Therefore we are done.
A rectangle can be covered by rectangles if and only if
Solution(from A. Engel-P.S.S)
We have,if then our covering is obvious.
So we consider the case when Now colour the board accordingly as in our previous diagram using the colours There are squares of each of the colours and squares of each of the colours . The horizontal tiles each covering one square of each colour, after being placed, we will have squares of each of the colours and of each of the colours Thus must divide their difference.
We can also state this for the space, whereas if an block can be covered with bricks we have
An art gallery is the shape of an -gon. Find the minimum number of watchmen needed to guard the gallery.
This is known as the Art Gallery problem.
We join non-intersecting diagonals to form disjoint triangles. Now we colour the vertices of each triangle with distinct colours, so that we can make a proper colouring of each of the triangles (recall this term and prove it using induction). Now, the colour used least number of times will be our required number, which is equal to
This solution or theorem is known as Chvátal’s art gallery theorem.
For further research you can visit the wikipedia link I gave above.
I will post more problems soon, I have to leave now. I am posting some Exercises and post more within 5-8 hours.
1. Consider the following figure, where 14 cities are connected by roads.
Find a path passing through each city exactly once.
In case of graphs, a path or walk means a proper way of starting from a vertex of a graph and walking along the edges, to reach a final point is called a path. Thus a path can be written as
A path is said to be simple if , and a closed path if I will come back to this topic again in future when discussing about graphs.
2. Every point in the space is coloured with exactly one of the colours Red, Green, or Blue. The sets consist of the lengths of those segments in space with both endpoints red, green, and blue respectively. Show that at least one of these sets contain all non-negative integers.
Experiment with an equilateral triangle
About a month ago, I was researching with colouring problems that are solved using the Pigeonhole principle and can be proved elegantly. However, I have only dealt with planar figures and there colourings only, so I hope to present my little (and flop) experiment that I made about a month ago.
But before that, I want you to get used to some problems on colourings that require the PHP to be proved elegantly.
The PHP states that if balls are put into boxed, we must have a box with more than ball. This can be generalised as follows:
1. If we put balls into boxes, we must have a box with more than balls.
2. If the average of positive numbers is then at least one of the numbers is greater than or equal to As a consequence we must have at least one of the numbers less than or equal to
However, our main point of concern is the first and most elementary form of PHP. In order to prove these theorems, we can assume the converse, and show that it leads to a contradiction. The proof is left to the readers.
Now, we might be given problems on colourings like:
If the plane is coloured with two colours, then there exists at least one line with its endpoints and its midpoint of the same colour.
After drawing a diagram it is obvious. Since there are infinitely many points in the plane, we can get two points of the same colour always in the plane.
Denote them by where signifies red and signifies blue. (say). Now let us see what its midpoint is. If is , we are done. So WLOG assume to be blue. Now, we produce the line to both sides at a length of Now we get a new line If , we are done since in both cases we get a line
So both of are blue and we have a new segment of our required type. Hence proved.
If all points of the plane are coloured red or blue, show that there exists an equilateral triangle with three vertices of the same colour.
We must have a line with its endpoints and its midpoint of the same colour, as according to E1.
Let be a line segment with blue. Draw an equilateral triangle on this, forcing to be in colour. Now lets take the midpoints of respectively. Then we again have forced to become (else we get and as the required triangles). In this case too we are forced to get as our required triangle with all vertices red in colour. Hence we are done.
(My inspiration to this experiment: Indian National MO 2008) If all the lattice points of the plane are coloured with three colours: (standing for blue, red, green respectively) show that we have at least one right triangle with three vertices of different colours; given that
Consider the lattice points on the lines and other than If any one of them , say is coloured green, then we have a right triangle with and as vertices, all having different colours.
If not, then the lattice points on and are all red or blue. Now consider three different cases:
Suppose the point is blue. Consider a blue point in the plane. suppose If it’s projection on the axis is red, then are the required vertices.
If is blue, then we can consider the vertices are the vertices of the required RAT.
If then the points form the required RAT.
A point on the line is red. The logic is similar to case I.
Suppose all the lattice points on the line are red and all the points on the line are blue.
Consider a green point where
Consider an isosceles RAT with such that the hypotenuse is a part of the x- axis. Let intersect in . Then is a red point and is a blue point. Hence is a desired right triangle.
So now that we all know how I have been inspired to perform an experiment, why not lets come to the point?
Here; I thought about a small trick. The problem I thought about was to prove that:
There exists a right triangle with three vertices of the same colour, if we have the plane coloured with red and blue.
According to our above problem, there must exist an equilateral triangle with all its 3 vertices of the same colour. Refer to figure 1. Let the triangle be and let be the midpoint of segment But, as we see, we must have , because if were to be red, we could have found a triangle or satisfying our requirements. Now join and
As in our figure, let and be the midpoints of and segments respectively.Consider right triangle If would have been red, we would have done; since is also an altitude of equilateral So is forced again.
Now let us join ; and let it intersect at If were to be blue, we could have found or to fulfill our requirements. So again we see that has to be red in colour.
Now, if we would have been done. But is equilateral, so why not find out about this angle? Well, we can let the triangle have equal sides of length units, so that Also . Note that from we have
So we get therefore we have
which is not -, how can we proceed?
No, never give up hope. We now try to walk a bit farther towards our goal. Let us consider as in Figure 2. Note that if any point on would have been red, we would have been done. So it is completely forced that all points on except are blue in colour.
What else can we say? we want a right triangle, am I correct? So why not try to make one? Let us take a point on which satisfies so we must have . How funny, now we have three points in a row that are red. But have we done yet?
The answer is unfortunately “no”. We still have a bit to go, but we can feel it that we are near to our goal.
OK, last try. Refer to Figure 3.
Draw a line parallel to such that is on . Again we have forced . Now we can almost finish our proof if proceeded a bit more. Let us draw a right angle with on We can show that such a point exists since and due to the acuteness property we force to be in the same side of with respect to
Wow, we have proved earlier that and in this case we see that is our required triangle ; and we are done.
A really nice problem that I made, isn’t it? The solution might be lengthy and not elegant, but still, it is an interesting solution, all the same.
(To be updated)