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Pigeon Hole Problems
Warmup Problem (for the first day)
Dirichlet’s pigeon hole principle says if we have n pigeons to place in m pigeon holes where the
number of pigeons, n, is greater than the number of pigeon holes, m, then at least one pigeon
hole must contain 2 (or more) pigeons.
Use Dirichlet’s pigeon hole principle to explain why the following is true:
If 10 points are chosen at random from a 1 unit by 1 unit square, then there are two of those
points that are less than 1/2 unit apart:
What does this have to do with “10″? How about “17″ – how close must at least 2 points be if
we choose 17 random points? What about 26 points?
By creating a 3x3 matrix and placing a
point in each box, this leaves one final
point to be placed inside the matrix.
However because there is only 9 boxes,
one box has to have more than one pigeon
hole in it. Using Pythagoreans theorem one
can see that the two units diagonal from
each other are less than half a unit apart
(.47). The same theorem applies to the
other two conditions. For 17, a 4x4 matrix
can be created allowing for 16 pigeon
holes. The 17th and final one also has to fit
into another box with an existing pigeon
hole. Using Pythagoreans theorem again,
the distance between two diagonal points
will be .35. Applying this theory again for 26, a 5 by 5 matrix is drawn and the distance between
two diagonal points will be .28.
Pigeon Hole Principle Problem, number 2
Take 19 distinct integers at random from the sequence
.
Explain why at least 2 of these 19 randomly chosen integers add to 104.
n
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
3n-2
4
7
10
13
16
19
22
25
28
31
34
37
40
43
46
49
55
58
61
64
67
70
73
76
79
82
85
88
91
94
97
100
n
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
3n-2
100
97
94
91
88
85
82
79
76
73
70
67
64
61
58
55
49
46
43
40
37
34
31
28
25
22
19
16
13
10
7
4
sum
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
104
Through creating the table and
summing up both green columns we
can see that all but two pairs of
numbers add up to 104. These numbers
are 1 and 18. These are not part of the
16 sets that add up to 104.
Pigeon hole Principle problem, number 3
Put
consecutive integers:
on slips of paper in a hat.
Draw single slips of paper from the hat at random (do not put them back in the hat).
Show that if you draw at least
slips then the numbers on at least 2 of them differ by exactly n.
Pigeon Hole Principle Problem, number 4
On every square of a 5×5 board there is a single flea.
All the fleas jump to an adjacent square (two squares are adjacent if they share an edge).
Show that after the jump there is an empty square.
If the fleas jump in succession just like the
arrows are drawn, the flea in the top left corner
of the matrix will hop to the right leaving its box
empty. As a result of this, the flea at the bottom
right hand corner will have to jump outside of
the matrix; this leaves one box inside the matrix
empty.
Pigeon hole principle Problem, number 5
5 points are chosen at the nodes of a square lattice (grid).
Show that at least one mid-point of a line joining a pair of these points is also a lattice point.
By using the lattice point formula, we can prove that at least one midpoint of a line is also a lattice point.
A lattice point is located at the intersection of two or more gridlines.
𝑎1 +𝑎2 𝑏1 +𝑏2
Lattice point formula: (
2
,
2
)
Each of the points were randomly placed on the grid.
Using the lattice point formula, I checked each of the
points until at least one point fell on the intersection
of two grid lines. The one point that fell as a
midpoint and on the intersection of two grid lines is
circled in green. The formula was calculated as
follows:
1+1 0+4
(
2
,
2
)= (1, 2); This proves that the lattice
formula works.
Pigeon Hole Principle Problem, number 6
On the complete graph
on 6 vertices, the edges are painted either red or blue.
Show that, no matter how the coloring is done, there is a triangle with all edges of the same color.
This figure has 6 sides and 6 points while a triangle has 3 sides and 3 vertices. In order to find the
amount of possible triangles, wolfram was used and a number of 20 was found through using the
binomial command. Since a triangle has three sides and it can be either red or blue, this means there is 8
possible combinations, (or 2^3), proven by the table below.
1
R
R
R
R
B
B
B
B
Edge
Colors
2
R
R
B
B
R
R
B
B
3
R
B
R
B
R
B
R
B
Now if a vertices has 5 lines coming from it, and they can be either red or blue, this means there are 32
possible combinations, or (2^5). There are two cases that can be disregarded however; (3 blue, 2 red)
will be the same as (3 red, 2 blue) but opposite using the same method implemented as the table
above.
Looking at one vertices with 5 lines coming
from it as it would in the graph K5, It does
not matter what color the yellow line is (red
or blue), because as it crosses through 5
lines it forms 4 triangles and as a result, one
of these four triangles will have all sides of
the same color.