Targil 5 – double counting.
1. In one country, there are 5 big and 19 small cities. The country is divided into 9
regions. Each big city is connected by bus to at least 14 cities, while each small
city is connected by bus to at most 3 cities (each bus goes in both directions).
Show that there exists a region in which no two cities are connected by bus.
Solution. Estimate the connections with 19 small cities in two ways:
(a) Each small city is connected to at most 3 big ones, totally at most 57
connections.
(b) Each big city is connected to 14 cities at least, only 4 of those may be the
other big cities, so at least 10 of connected cities are small, which is at least
50 connections.
We see that (b) is just slightly greater then (a), while (a) counts each (b)-type
connection of big and small cities once, and each connection of two small cities
twice. So twice the number of connections between 2 small cities is at most 7.
Therefore there can be at most 3 connections between pairs of small cities.
Since there are 9 regions, there’s a region that contains neither those 3 connections
between small cities nor any of the 5 big cities. Therefore, in that regions there are
only small cities which are not connected to each other.
2. a. Show that each map on a sphere has a country with less than 6 neighbors, and
conclude that each map can be painted in 6 colors, so that countries having
common border of positive length will be of different colors.
b. Show that each map on a sphere can be painted in 5 colors.
c. What is the maximal number of necessary colors for a map on a torus?
Solution. a. As in the famous Euler formula, denote: F number of faces, E number
of edges, V number of vertices.
Count the edge-vertex incidence relations X two ways: it is exactly 2E, and at least
3V, since each vertex is at least on 3 edges. Hence 2E/3 = X/3 ≥ V .
Now count the number Y of face-edge incidence relations: it is again 2E, since
each edge is of 2 faces, and at least 6F if we assume that each country has at least 6
neighbors, therefore E/3 = Y/6 ≥ F .
If we sum up the two inequalities we get E ≥ V + F, therefore the Euler expression
V – E + F is not positive, but it is equal to 2 by Euler formula.
Exercise. Find a map on sphere which is a counter-example to Euler formula and
complete the proof for those cases too (hint in the end of the solution).
Remark. This statement (and Euler formula itself) can be proved by doublecounting of angles in either Euclidean, spherical or hyperbolic geometries, but then
an extra explanation is required that we can make all the borders straight line.
For example in Euclidean geometry: total sum of all angles is 360 degrees times
number of vertexes, so the average angle is at most 120 degrees, so the average
number of vertexes in the polygon is at most 6, but if you do it carefully you make
it less than 6.
Theorem about 6 colors is proved by induction over the number of countries.
Since there’s a country of less than 6 neighbors, if one of the neighbors would
temporary take that country, we would be able to paint it into 6 colors by induction
assumption, and when that country gains independence again, we can choose one
of 6 colors which is different from all of its neighbors.
Remark. As for b’ - we can’t prove that there’s a country of less than 5 neighbors:
a dodecahedron is just one counter-example, and there’s a lot of others, so a more
subtle approach is required.
b. If the map has a country of less than 5 neighbors, we can use the induction. If
not, consider a country C with just 5 neighbors (it can’t be that all countries have at
least 6 neighbors, as we have proven before).
It cannot happen that all 5 neighbors of C are connected to each other, since a full
graph with 5 vertexes, K5, is not a planar graph. So A and B, some two neighbors
of C, are not connected.
Temporary unite A, B, C into one country. Since number of countries is smaller,
now the map can be painted in 5 colors. Then make C a separate unpainted
country. It has 5 neighbors, out of those 2 have the same color, so its neighbors
have only 4 colors, so we can choose for C one of the 5 colors different from all its
neighbors. A and B already have a color, and it doesn’t make a problem since they
are not neighbors.
c. Every map on torus can be painted into 7 colors, and there are
maps that require all 7 colors. An example of a map which requires
7 colors is on the picture. It is cut out of periodic hexagonal pattern
of 7 colors, each color touches every other color. When we glue the
corresponding opposite sides of the parallelogram, we get a map of 7 countries on
the torus, where each country is a neighbor of each.
As in a’, double count and Euler formula prove that there is a country with no
more than 6 neighbors, and after that we can apply induction.
Euler formula on torus is V – E + F = 0.
By double-counting we get: 2E/3 ≥ V and E/3 > 2E/7 ≥ F (if each country has at
least 7 neighbors).
So E > V + F and Euler expression is negative when it should be 0.
Hint to the exercise: consider countries with holes.
3. There are l unit vectors in n-dimensional space which are pair-wise orthogonal,
and the orthogonal projection of each vector to a given k-dimensional subspace is
longer than ε. Show that l ≤ k / ε2 .
Solution. For any set of less than n vectors, we can find a unit vector orthogonal to
them all. So, we can complete our l unit vectors to an orthonormal system of n
vectors. Choose a rotated coordinate system such that the mentioned k-dimensional
space would contain the first k coordinate axis.
Now write down each vector in coordinates as a column, one after another, first the
l given vectors and then the rest, and you get an orthogonal matrix.
Within this matrix, consider a sub-block of first k rows in the first l columns.
Apply double-counting to the sum of squares of the numbers in that sub-block.
Each of l first vectors in projection to the first k coordinates is of length ε2 at least,
so the sum is at least lε2. On the other hand, sum of squares in each row (even the
whole row of any orthogonal matrix, not to talk of the sub-block), is at most 1, so
the total sum of squares in the sub-block is at most k.
Conclusion: lε2 ≤ k. QED.
4. In a table there are N columns and M rows, N > M.
Some cells are marked by stars, and in each column there’s at least one star.
Show that there is a star for which there are less stars in its column than in its row.
[Another way to formulate essentially the same problem: the books in the library
were rearranged, so that for each book we have more books on the same shelf with
it than before; show that now there is an empty shelf].
Solution. Draw two tables, A and B, of the same size as the original table.
The cells that correspond to the empty cells of A or B will still be empty.
The cells that correspond to the marked cells of the original table will contain
numbers according to the following rule:
If in the original table the row of the original cell contains k stars, the number that
in table A will replace each star of that row is 1/k.
If in the original table the column of the original cell has l stars, the number that in
table B will replace each star of that column is 1/l.
Sum of all numbers in B is equal N, the number of columns, since each column
had a star and sum in each is 1. Sum of numbers in A is at most M, the number of
rows, which is less than N. So for some cell, the corresponding number in A is
smaller than in B, so for that cell 1/k > 1/l, and l > k, so for that star number of stars
in the same row is bigger than the number of stars in the same column.
Remark. A combinatorial solution that doesn’t use double counting can be devised
for that problem, but neither as short nor as elegant, see for instance
http://taharut.org/Solutions/Ikarit/T11/Autumn/S_11_A_O_I6.doc
Of course, we could have allowed to mark more than one star in each cell, then
there would be more than one number in both A and B table, but the proof would
remain
the
same.
The story about books: if the rows of the table are shelves before the reordering,
the columns of the table are the shelves after the reordering, and the stars are the
books, we get again the same problem.
5*. Show that in a group of 50 people there are two that have an even number of
common friends (maybe 0), assuming that friendship is symmetric.
First solution. Assume there’s someone (denote him Yossi) who has an odd
number of friend. Consider the subgroup of friends of Yossi. If there’s anyone with
even number of friends inside that subgroup, we got someone who actually has
even number of friends with Yossi, QED.
If not, then we have that between odd number of people, each having odd number
of friends in that group, so total number of pairs “someone and his friend” is odd,
but it cannot be since it is equal to twice the number of friendships in that group.
So, it remains to consider the case where each person among the 50 has even
number of friends. Then consider someone called Kobi. The rest of the crowd can
be divided into two parts: even number of the friends of Kobi, denote them F, and
odd number of the others, that group will be denoted as O.
If someone in has even number of friends in then he has an even number of the
common friends with Kobi and we are done. So, it remains to assume that each
person in has an odd number of friends in F. In particular, each person in F has
even number of friends: odd number in F, Kobi himself, and even number from O.
So the number of friendships between F and O is even number of even numbers,
and that is even.
On the other hand, each person in O has odd number of friends in F, and O is of
odd order, so the number of friendships between F and O is an odd number of odd
numbers, and that is odd.
That is a contradiction: the number of friendship connections between O and F is
both even and odd.
Second solution. Consider the incidence matrix A: each number in column i row j
is 1 if people i and j are friends and 0 otherwise. It can be considered in two ways:
either as a matrix over integers or as a matrix with coefficients in the {0,1} field of
two elements; we shall use the field of two elements.
Let v be a vector all coordinates of which are 1.
Like we have shown in the beginning of the previous solution, WLOG we can
assume that each person has even number of friends. That means, each row/column
is orthogonal to v, since the sum in each row is even, so Av = 0.
Consider A2: On the cell (i , j) for i ≠ j You will have the number of common
friends of i and j which is always 1 unless we get what we want, and on diagonal
cell You get the number of friends of someone mod 2, which is 0.
So we know the A2 matrix precisely, and A2v = v ≠ 0, which contradicts Av = 0.