Problems
At the elections to the High Government every voter who comes, votes for himself (if he is a candidate) and for those candidates who are his friends. The forecast of the media service of the mayor’s office is considered good if it correctly predicts the number of votes at least for one of the candidates, and not considered good otherwise. Prove that for any forecast voters can show up at the elections in such a way that this forecast will not be considered good.
Draw a Sperner’s coloring for the following \(3\)-dimensional simplex. The blue segments are visible, the grey ones are inside the tetrahedron. The point \(F\) is on the face \(ABC\), point \(E\) is on the face \(BCD\), point \(G\) is on the face \(ACD\) and the point \(H\) is on the face \(ABD\).
Prove Sperner’s lemma in dimension \(1\), namely on a line.
The simplex in this case is just a segment, the triangulation is
subdivision of the segment into multiple small segments, and the
conditions of a Sperner’s coloring are the following:
There are only two colors;
The opposite ends of the main segment are colored differently;
Then one needs to prove that there exists a small segment with two ends colored in different colors. In particular there is an odd number of such small segments.
We want to prove Monsky’s theorem as a corollary of Sperner’s lemma:
it is not possible to dissect a square into an odd number of triangles
of equal area. After scaling we can consider the square with
coordinates: \((0,0),(0,1),(1,0),(1,1)\), which we want to
cut into \(n\) triangles with area
\(\frac{1}{n}\) each for an odd \(n\). Consider the following coloring of all
the points with rational coordinates \((\frac{p}{q},\frac{r}{s})\) inside the
square:
We look at the powers of \(2\) in the
fractions \((\frac{p}{q},\frac{r}{s})\), first of all
the numbers \(p,q\) are coprime, and
thus only one of them is divisible by \(2\), same with \(r,s\). Then the following possibilities
might occur:
Neither \(q\) nor \(s\) is divisible by \(2\). In this case we color the point red.
\(\frac{r}{s}\) is divisible by a larger or equal power of \(2\) than \(\frac{p}{q}\) and \(p\) is not divisible by \(2\). In this case we color the point blue.
\(\frac{p}{q}\) is divisible by a strictly larger power of \(2\) than \(\frac{r}{s}\) and \(r\) is not divisible by \(2\). In this case we color the point green.
Under an assumption (which you do not have to prove) that the area of any rainbow triangle is at least \(\frac{1}{2}\) prove the Monsky’s theorem.
Prove the general version of Sperner’s lemma: Consider an \(n\)-dimensional simplex \(\mathcal{A} = A_1A_2...A_{n+1}\). Strictly speaking a simplex is a convex linear combination of \(n+1\) points in general position (when \(k\) points are never in one subspace of dimension \(k-1\)). One can view it as an \(n\)-dimensional tetrahedron or a body spanned over vertices \((0,0,...,0), (1,0,0,...,0), (0,1,0,0...,0), ... (0,0,...0,1)\). \[\mathcal{A} = \{\sum_{i=1}^{n+1}a_i(0,0,...,1,...,0), \,\,\, a_i \geq 0, \,\,\,\, \sum_{i=1}^{n+1}a_i = 1\}.\]
A simplicial subdivision of an \(n\)-dimensional simplex \(\mathcal{A}\) is a partition of \(\mathcal{A}\) into small simplices (cells)
of the same dimension, such that any two cells are either disjoint, or
they share a full face of a certain dimension.
Define a Sperner’s coloring of a simplicial subdivision as an assignment
of \(n+1\) colors to the vertices of
the subdivision, so that the vertices of \(\mathcal{A}\) receive all different colors,
and points on each face of \(\mathcal{A}\) use only the colors of the
vertices defining the respective face of \(\mathcal{A}\).
Prove that every Sperner’s coloring of any subdivision of an \(n\)-dimensional simplex contains an odd
number of rainbow simplexes, namely whose vertices are colored using all
\(n+1\) colors.
Draw Sperner’s coloring for the following triangulation. Try to avoid rainbow triangles at all costs.
Consider an \(n\)-dimensional simplex \(\mathcal{A} = A_1A_2...A_{n+1}\), namely a body spanned over vertices \((0,0,...,0), (1,0,0,...,0), (0,1,0,0...,0), ... (0,0,...0,1)\). \[\mathcal{A} = \{\sum_{i=0}^{n}a_i(0,0,...,1,...,0), \,\,\, a_i \geq 0, \,\,\,\, \sum_{i=1}^{n+1}a_i = 1\}.\] Where next to \(a_i\) there is a point with coordinate where \(1\) is in \(i\)-th place. The point \((0,0,...,0)\) belongs to the simplex as well.
A simplicial subdivision of an \(n\)-dimensional simplex \(\mathcal{A}\) is a partition of \(\mathcal{A}\) into small simplices (cells)
of the same dimension, such that any two cells are either disjoint, or
they share a full face of a certain dimension.
Define a Sperner’s coloring of a simplicial subdivision as an assignment
of \(n+1\) colors to the vertices of
the subdivision, so that the vertices of \(\mathcal{A}\) receive all different colors,
and points on each face of \(\mathcal{A}\) use only the colors of the
vertices defining the respective face of \(\mathcal{A}\).
Consider a simplicial subdivision given by pairwise connected middles of
all the segments in the original simplex. Assign the numbers \(0,1,2...,n\) to the subdivision vertices in
such a way as to conduct a Sperner’s coloring in such a way that you
will have only one rainbow simplex.
Consider an equilateral triangle \(ABC\). Parallel to each side, five equally spaced segments are drawn across the triangle so that \(ABC\) is subdivided into \(36\) smaller equilateral triangles. Vertices \(A,B\) and \(C\) are painted red, blue and green, respectively in counterclockwise order. Alice and Bob take turns painting each vertex of the smaller triangles either red, green, or blue with the following restrictions: the segment \(AB\) can only have red or green, the segment \(BC\) can only have green and blue, and the segment \(AC\) can only have blue and red. The interior vertices can be drawn freely. Once all vertices have been painted, Alice gets a point for every smaller triangle whose vertices have the three colors (red, green, and blue) appearing in the counterclockwise direction. Bob gets a point for every smaller triangle whose vertices have the three colors but in the clockwise direction. Who wins?
There are \(100\) people in a room, and each person has at least one friend in the room. Prove that amongst them there are two people with the same number of friends in the room (we don’t count being friends with oneself).
Can we obtain the polynomial \(h(x)=x\) by adding, subtracting, or multiplying the polynomials \(f(x)=x^2+x\) and \(g(x)=x^2+2\)?