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April 25, 2012

Nadal’s phenomenal record at Monte Carlo

Filed under: sport — ckrao @ 6:47 am

Rafael Nadal recently won his eighth straight title at Monte Carlo, in the process being the first player to register 20 tournament wins at Masters 1000 level. Unbelievably he is still only 25! Here are some amazing stats about his performance at this tournament – you can see more details of his matches and stats there at the ATP website here.

  • His only loss at the tournament came as a 16 year-old in 2003, against eventual tournament runner-up Guillermo Coria. Nadal was still ranked outside the top 100 at that time. In the previous (second) round of that tournament Nadal defeated the defending French Open champion (and world number 7 at the time) Albert Costa!
  • He matches Guillermo Vilas with 8 as the most titles won at a single tournament in the Open era (Buenos Aires in his case). The women’s record is 12 by Martina Navratilova (Chicago).
  • Nadal is the only player, man or woman,  in the open era to win the same professional singles tournament more than 6 consecutive times. (Graf won the German Open, while Navratilova won Wimbledon and at Chicago 6 consecutive times).
  • He has the record for most consecutive match wins at a tournament: 42 (next is Borg with 41 at Wimbledon). Overall his record at Monte Carlo is 44-1.
  • Top ten players Nadal defeated at the tournament:
    2003: Albert Costa (#7, R32) 7-5 6-3
    2004: Did not play
    2005: Gaston Gaudio (#6, QF) 6-3 6-0, Guillermo Coria (#9, F) 6-3 6-1 0-6 7-5
    2006: Guillermo Coria (#9, QF) 6-2 6-1, Gaston Gaudio (#8, SF) 5-7 6-1 6-1, Roger Federer (#1, F) 6-2 6-7(2) 6-3 7-6(5)
    2007: Roger Federer (#1, F) 6-4 6-4
    2008: David Ferrer (#5, QF) 6-1 7-5, Nikolay Davydenko (#4, SF) 6-3 6-2, Roger Federer (#1, F) 7-5 7-5
    2009: Andy Murray (#4, SF) 6-2 7-6(4), Novak Djokovic  (#3, F) 6-3 2-6 6-1
    2010: highest ranked opponent was Fernando Verdasco (#12) whom he defeated 6-0 6-1 in the final
    2011: Andy Murray (#4, SF) 6-4 2-6 6-1, David Ferrer (#6, F) 6-4 7-5
    2012: Novak Djokovic (#1, F) 6-3 6-1
  • In the last six years (30 matches) he only lost two sets (to Murray and Djokovic, shown above)! In each case he won the deciding set 6-1.
  • In the 2010 edition of the tournament he lost only 14 games!
  • In total he has won a set 6-0 (a bagel)  8 times and 6-1 (a breadstick) 18 times.

Astonishing to think that he never had an off day at this tournament over these 8 years, able to beat some of the best clay court specialists of their time and all-time greats in Federer and Djokovic in five of the finals. Being a Masters 1000 tournament the best players show up there. As Nadal himself put it,

‘To have eight victories, you must be lucky, you have to have no injuries, perfect conditions for eight years in a row. That’s the first thing.’

‘And you have to be playing almost perfect to win eight titles in a row, especially in a Masters. The best in the world always play – you have to win against the best.’

I also liked the praise Djokovic had for Nadal.

‘Fantastic, impressive. The way he’s been treating this sport is a real example of a champion.’

‘I only have nice things to say about him. Every year he comes back and he looks like he’s the first time in this place.’

 

April 21, 2012

Divisibility tips

Filed under: mathematics — ckrao @ 8:17 am

Suppose you want to figure out (without computing aid) if a number n (which is not too large) is prime. Many people are unaware that it suffices to check for prime factors up to \sqrt{n}. This is because factors of n come in pairs: a and n/a. Since we have two numbers multiplying to n the smaller factor must be at most \sqrt{n}:

\displaystyle n = a.(n/a) \geq \left(\min\{a,n/a\}\right). \left(\min\{a,n/a\}\right) = \left(\min\{a,n/a\}\right)^2.

A second tip is that to check whether a prime p is a factor of n, instead of dividing n by p it helps to add or subtract multiples of p to reach a multiple of 10. Then check whether this new number divided by 10 is a multiple of p. The key idea is that we are not interested in the quotient, only that the remainder is non-zero.

For example here is how I would determine whether 2011 is prime. It is possible for each line to be performed mentally. I only use divisibility “tricks” for the cases of 2,3,5,11. It also helps to know that 111 is a multiple of 37. We need to check primes up to 43, since 45^2 = 2025.

  • It ends in 1, so it is divisible by neither 2 nor 5.
  • Its digit sum is 4 so it is not divisible by 3.
  • 2011 – 21 = 1990, 199 is not divisible by 7 since 210 (11 away) is.
  • 2 + 1 \neq 0 + 1 (summing alternating digits), so 2011 is not divisible by 11.
  • 2011 is divisible by 13 iff 2011-13 = 1998 is, but 1998 = 2 \times 999 = 2 \times 111 \times 9 does not have 13 as a factor (111 is 19 away from 130).
  • 2011 – 51 = 1960, but 196 is not divisible by 17 since 170 (26 away) is.
  • 2011 + 19 = 2030, but 203 is not divisble by 19 since 190 (13 away) is.
  • 2300 – 2011 = 289, 289 – 230 = 59, which is not divisible by 23, hence 2011 is not divisible by 23.
  • 2011 + 29 = 2040, 204 is not divisible by 29 since 203 = 7 \times (30 - 1) is.
  • 2011 – 31 = 1980, 198 is not divisible by 31 since 186 (12 away) is.
  • 2011 is not divisible by 37 since 1998 (13 away) as found above is 2 \times 111 \times 9 = 2 \times 37 \times 3 \times 9 is.
  • 2011 – 41 = 1970, 197 is not divisible by 41 since 205 is.
  • 2011 + 129 = 2140, 214 is not divisible by 43 since 215 is.

Hence 2011 is prime! You might want to try a similar exercise to show 2003 is also prime.

Finally, note that if the original number is less than 1072, then one only needs to check primes up to 23, and remember the following special cases involving higher factors:

  • 841 = 29^2
  • 899 = 30^2 - 1 = (30 + 1)(30-1)
  • 961 = 31^2

How does 1073 factorise? If one has a suspicion that it has large factors, one can find a square just larger and then if one gets lucky use the difference of squares formula. Indeed here,

\displaystyle 1073 = 1089 - 16 = 33^3 - 4^2 = (33 - 4)(33 + 4) = 29 \times 37.

April 16, 2012

Federer’s latest winning streak

Filed under: sport — ckrao @ 12:11 pm

Roger Federer has compiled a most impressive list of winning streaks, including the following.

  • consecutive matches won on grass: 65
  • consecutive matches won on hard court: 56
  • consecutive wins against top 10 players: 26
  • consecutive wins in North America: 55
  • consecutive finals won: 24
  • consecutive matches won in grand slam matches against players outside the top 5: 124

A streak which just ended for him was a 77-match winning streak against players ranked outside the top 20. On either side of these wins were losses to former world number ones in Lleyton Hewitt and Andy Roddick. He had 19 other losses during this time but they were all to players in the top 20 at the time. I looked here and found two other similar winning streaks for him:

  •  A 55-match winning streak against outside-top-20 players in 2008-9 (losses to Ivo Karlovic and Julien Benneteau on either side).
  • A 78-match winning streak against outside-top-20 players in 2005-6 (losses to Richard Gasquet and Andy Murray on either side). Only one of his six losses during that time was not against Rafael Nadal!

 

April 10, 2012

Integers equal to the sum of three cubes

Filed under: mathematics — ckrao @ 11:32 am

One easy-to-state open problem in number theory is whether every integer not of the form 9k \pm 4 is the sum of three cubes. Some solutions are easy to find, while others are notoriously difficult.

For example 29 can be written as 3^3 + 1^3 + 1^3, or even 4^3 - 3^3 - 2^3.

It is clear that 9k \pm 4 cannot be written as the sum of three cubes since x^3 \equiv -1, 0 \text{ or } 1 \mod 9.

There also exist infinite families of solutions in certain cases (though these need not represent all solutions). For example,

\displaystyle (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 = 1

\displaystyle (6t^3 + 1)^3 - (6t^3 - 1)^3 - (6t^2)^3 = 2

For a long time it was not known whether 30 can be written as the sum of cubes. The following is the “simplest” way this can be done, found as recently as 1999.

\begin{array}{lcl} 30 &=& (2,220,422,932)^3 + (-2,218,888,517)^3 + (-283,059,965)^3\\ &=& 10,947,302,325,566,084,787,191,541,568 - 10,924,622,727,902,378,924,946,084,413 - 22,679,597,663,705,862,245,457,125 \end{array}

It is still not known whether 33, 42 or 74 can be written as the sum of three cubes (edit: it is now known that 74 can be – see [5])! In these cases no solution has been found when any of the cubes is less than (10^{14})^3 in magnitude [4].

One of the computational algorithms to find solutions to a^3 + b^3 + c^3 = n where n is relatively small, proposed by Elkies, involves converting the equation to (-a/c)^3 + (-b/c)^3 \approx 1 and then considering rational points close to the curve y = \sqrt[3]{1-x^3}, x \in [0, 1/\sqrt[3]{2}]. This curve is covered by small parallelograms and the problem is converted to finding lattice points in a pyramid using basis reduction followed by the Fincke-Pohst algorithm [4].

References

[1] http://www.math.niu.edu/~rusin/known-math/96/3cubes

[2] Hisanori Mishima, Chapter 4. n=x^3+y^3+z^3

[3] D.J. Bernstein, http://cr.yp.to/threecubes.html

[4] A.-S. Elsenhans and J. Jahnel, New sums of three cubes, Math. Comp. 78 (2009), 1227–1230, available here.

[5] Sander G. Huisman, Newer sums of three cubes, http://arxiv.org/abs/1604.07746, April 2016.

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