Better to have more input variables and less output variables

October 15, 2012

 

When trying to describe an invariant map \phi : U \times V \to W, where $U,V$ and $W$ are finite-dimensional spaces,  \phi is “complicated” because its values are vector-valued. It is sometimes useful to look at \phi^{\sharp} : U \times V \times W^{*} \to {\mathbb K}, defined by \phi^{\sharp}(u,v,w^*)=w^*(\phi(u,v)). This map has more variables but takes scalar values.

Vector whose coordinates are pairwise independent

October 8, 2011

Consider the following twelve vectors in \lbrace 0,1 \rbrace^4 :

\begin{array}{l}  V_1=(0,0,0,0),V_2=(0,0,0,1),V_3=(0,0,0,0), \\  V_4=(0,1,0,1),V_5=(0,1,1,0),V_6=(0,1,1,1), \\  V_7=(1,0,0,1),V_8=(1,0,1,0),V_9=(1,0,1,1), \\  V_{10}=(1,1,0,0),V_{11}=(1,1,0,0),V_{12}=(1,1,1,1). \\  \end{array}

Then the multiset V=\lbrace V_k\rbrace_{1 \leq k \leq 12} (note that V_{11}=V_{10}) has the following property : if \overrightarrow{X} is a random vector whose distribution is unform in V, then the coordinates of \overrightarrow{X} are pairwise independent. Moreover, V is minimal with respect to this property. Can V be “explained” rather than just enumerated?

 

Changing the measure

February 27, 2011

Roth’s function g(n) is defined as the largest size of a subset of \lbrace 1,2, \ldots ,n \rbrace without arithmetic progressions. We may “force” the arithmetic progressions into the measure, and instead of taking the usual measure \mu(A)=|A| we may consider \nu(A)=|A|-\begin{displaystyle}\sum_{T\in{\cal T}}c_T1_{T \subseteq A}\end{displaystyle} where \cal T denotes the set of all AP’s in \lbrace 1,2, \ldots ,n \rbrace and 1_{T \subseteq A} equals 1 if T \subseteq A, 0 otherwise, and the c_T are constants. The advantage of doing this is that we now have an optimization problem on all subsets of \lbrace 1,2, \ldots ,n \rbrace instead of the subsets without APs.
For example, we may take all the c_T‘s equal to 1, but the extremal subsets (those which maximize \nu) do not appear to exhibit a very interesting structure. Perhaps there are better choices for the c_T constants.

Decomposing sets without arithmetic progressions

February 23, 2011

Let A be a subset of \lbrace 1,2, \ldots ,8 \rbrace containing no arithmetic progressions. Then one of the following two must hold :

(1) |A \cap \lbrace 3,5 \rbrace | \leq 1 and |A \cap \lbrace 1,2,4,6,7,8 \rbrace | \leq 1.

(2) |A \cap \lbrace 4,6 \rbrace | \leq 1 and |A \cap \lbrace 1,2,3,5,7,8 \rbrace | \leq 1.

Let A be a subset of \lbrace 1,2, \ldots ,10 \rbrace containing no arithmetic progressions. Then one of |A \cap  \lbrace 1,2, \ldots ,7 \rbrace| and |A \cap  \lbrace 4,5, \ldots ,10 \rbrace| is at most $3$.

Let A be a subset of \lbrace 1,2, \ldots ,13 \rbrace containing no arithmetic progressions. Then one of the following two must hold :

(1) |A \cap \lbrace 1,2,3,4,6 \rbrace | \leq 3 and |A \cap \lbrace 5,7,8,9,10,11,12,13 \rbrace | \leq 4.

(2) |A \cap \lbrace 8,10,11,12,13 \rbrace | \leq 3 and |A \cap \lbrace 1,2,3,4,5,6,7,9 \rbrace | \leq 4.

Sets without arithmetic progressions in [1..7]

January 7, 2011

Let A be a subset of \lbrace 1,2, \ldots ,7 \rbrace containing no arithmetic progressions. Then one of the following seven must hold :

(1) A \cap \lbrace 1,4,7 \rbrace=\emptyset.

(2) A \cap \lbrace 3,4,5 \rbrace=\emptyset.

(3) |A \cap T | \leq 1 for any T\in \lbrace \lbrace 1,2,3 \rbrace, \lbrace 1,3,5 \rbrace\rbrace.

(4)  |A \cap T | \leq 1 for any T\in \lbrace \lbrace 3,5,7 \rbrace, \lbrace 5,6,7 \rbrace\rbrace.

(5) |A \cap T | \leq 1 for any T\in \lbrace \lbrace 1,3,5 \rbrace, \lbrace 1,4,7 \rbrace , \lbrace 3,4,5 \rbrace, \lbrace 4,5,6 \rbrace \rbrace.

(6) |A \cap T | \leq 1 for any T\in \lbrace \lbrace 1,4,7 \rbrace, \lbrace 3,5,7 \rbrace , \lbrace 2,3,4 \rbrace, \lbrace 3,4,5 \rbrace \rbrace.

(7) |A \cap T | \leq 1 for any T\in \lbrace \lbrace 2,3,4 \rbrace, \lbrace 3,4,5 \rbrace , \lbrace 4,5,6 \rbrace, \lbrace 2,4,6 \rbrace \rbrace.

 

 

 

Center of gravity of a trapezium

November 14, 2009

It is worth noting that the center of gravity of a trapezium ABCD does not coincide with the center of gravities of the set \lbrace A,B,C,D \rbrace. Indeed, if we consider the special case where the coordinates of A,B,C,D are (-\alpha,0),(\alpha,0),(\beta,h) and (-\beta,h), respectively, the coordinates of the first center of gravity (let’s call it G_1) are (0,\frac{(\alpha+2\beta)h}{3(\alpha+\beta)}) while the coordinates of the second (let’s call it G_2) are (0,\frac{h}{2}). If \alpha>\beta, then G_1 is below G_2. Also, G_1 satisifies (2\alpha+beta)(\overrightarrow{GA}+\overrightarrow{GB})+(\alpha+2\beta)(\overrightarrow{GC}+\overrightarrow{GD})=\overrightarrow{0} while G_2 satisifies \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}.


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