Difference between revisions of "Kakeya problem"

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Define a '''Kakeya set''' to be a subset A of <math>[3]^n \equiv ({\Bbb Z}/3{\Bbb Z})^n</math> that contains an [[algebraic line]] in every direction.  Thus, for every <math>r \in ({\Bbb Z}/3{\Bbb Z})^n</math>, there exists <math>a \in ({\Bbb Z}/3{\Bbb Z})^n</math> such that a, a+r, a+2r all lie in A.  Let <math>k_n</math> be the smallest size of a Kakeya set in n dimensions.
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Define a '''Kakeya set''' to be a subset <math>A</math> of <math>[3]^r\equiv{\mathbb F}_3^r</math> that contains an [[algebraic line]] in every direction; that is, for every <math>d\in{\mathbb F}_3^r</math>, there exists <math>a\in{\mathbb F}_3^r</math> such that <math>a,a+d,a+2d</math> all lie in <math>A</math>.  Let <math>k_r</math> be the smallest size of a Kakeya set in <math>{\mathbb F}_3^r</math>.
  
  
Clearly, we have <math>k_1=3</math>, and it is easy to see that <math>k_2=7</math>. Using a computer, I also found <math>k_3=13</math> and <math>k_4\le 27</math>. I suspect that, indeed, <math>k_4=27</math> holds (meaning that in <math>{\mathbb F}_3^4</math> one cannot get away with just 26 elements), and I am very curious to know whether <math>k_5=53</math>: notice the pattern in
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Clearly, we have <math>k_1=3</math>, and it is easy to see that <math>k_2=7</math>. Using a computer, I also found <math>k_3=13</math> and <math>k_4\le 27</math>. I suspect that, indeed, <math>k_4=27</math> holds (meaning that in <math>{\mathbb F}_3^4</math> one cannot get away with just <math>26</math> elements), and I am very curious to know whether <math>k_5=53</math>: notice the pattern in
  
 
:<math>3,7,13,27,53,\ldots</math>
 
:<math>3,7,13,27,53,\ldots</math>
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:<math>k_r\le 2^{r+1}-1</math>:
 
:<math>k_r\le 2^{r+1}-1</math>:
  
the former since for each <math>d\in {\mathbb F}_3^r\setminus\{0\}</math> there are at least three ordered pairs of elements of a Kakeya set with difference d, the latter due to the fact that the set of all vectors in <math>{\mathbb F}_3^r</math> such that at least one of the numbers 1 and 2 is missing among their coordinates is a Kakeya set. (I actually can improve the lower bound to something like <math>k_r\gg 3^{0.51r}</math>.) Also, we have the trivial inequalities
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the former since for each <math>d\in {\mathbb F}_3^r\setminus\{0\}</math> there are at least three ordered pairs of elements of a Kakeya set with difference <math>d</math>, the latter due to the fact that the set of all vectors in <math>{\mathbb F}_3^r</math> such that at least one of the numbers <math>1</math> and <math>2</math> is missing among their coordinates is a Kakeya set. (I actually can improve the lower bound to something like <math>k_r\gg 3^{0.51r}</math>.) Also, we have the trivial inequalities
  
 
:<math>k_r\le k_{r+1}\le 3k_r</math>;
 
:<math>k_r\le k_{r+1}\le 3k_r</math>;

Revision as of 12:45, 14 March 2009

Define a Kakeya set to be a subset [math]A[/math] of [math][3]^r\equiv{\mathbb F}_3^r[/math] that contains an algebraic line in every direction; that is, for every [math]d\in{\mathbb F}_3^r[/math], there exists [math]a\in{\mathbb F}_3^r[/math] such that [math]a,a+d,a+2d[/math] all lie in [math]A[/math]. Let [math]k_r[/math] be the smallest size of a Kakeya set in [math]{\mathbb F}_3^r[/math].


Clearly, we have [math]k_1=3[/math], and it is easy to see that [math]k_2=7[/math]. Using a computer, I also found [math]k_3=13[/math] and [math]k_4\le 27[/math]. I suspect that, indeed, [math]k_4=27[/math] holds (meaning that in [math]{\mathbb F}_3^4[/math] one cannot get away with just [math]26[/math] elements), and I am very curious to know whether [math]k_5=53[/math]: notice the pattern in

[math]3,7,13,27,53,\ldots[/math]

As to the general estimates, we have

[math]k_r(k_r-1)\ge 3(3^r-1)[/math]

and, on the other hand,

[math]k_r\le 2^{r+1}-1[/math]:

the former since for each [math]d\in {\mathbb F}_3^r\setminus\{0\}[/math] there are at least three ordered pairs of elements of a Kakeya set with difference [math]d[/math], the latter due to the fact that the set of all vectors in [math]{\mathbb F}_3^r[/math] such that at least one of the numbers [math]1[/math] and [math]2[/math] is missing among their coordinates is a Kakeya set. (I actually can improve the lower bound to something like [math]k_r\gg 3^{0.51r}[/math].) Also, we have the trivial inequalities

[math]k_r\le k_{r+1}\le 3k_r[/math];

can the upper bound be strengthened to [math]k_{r+1}\le 2k_r+1[/math]?

Dvir, Kopparty, Saraf, and Sudan showed that [math]k_r \geq 3^r / 2^r[/math].

The Cartesian product of two Kakeya sets is another Kakeya set; this implies that [math]k_{r+s} \leq k_r k_s[/math].