Chaitanya's Random Pages

February 29, 2012

Expensive bowling in ODIs

Filed under: cricket,sport — ckrao @ 8:39 pm

Fascinating that just a few days after I blogged about economical bowling performances in cricket ODIs, we see the most expensive analysis in terms of economy rate (qualification: 5 overs or more). I found about it at Andy Zaltzman’s twitter feed here.

The list of expensive analyses can be found here. Earlier in the month a bowler (Brian Vitori of Zimbabwe) had conceded 100+ runs for only the fourth time in an ODI (9-0-105-1 against New Zealand at Napier).

Lasith Malinga bowled 7.4 overs for 96 runs (and one wicket) as India managed to chase down a victory target of 321 in just 36.4 overs. It was the fifth highest run rate for a 300+ score according to this.

Here is Malinga’s ball-by-ball breakdown. After 14 balls he had only conceded 13, then he conceded 83 off his next 32 balls!

1 1 1 0 0 2 | 1 2 1 0 4 0 | 0 0 4 1 4 6 | 4 x 2W 0 4 1 4 | 4 1 0 4 1 W 4lb | 1 1 4 1 1 0 | 2 6 4 4 4 4 | 1 1 4 4

(W = wide, x = wicket, lb = legbye)

For economical analyses I pointed out those that happened in recent times. The following lists the most expensive analyses that occurred prior to 2000.

  • Saurav Ganguly (India) bowled 5 overs for 62 against Pakistan at Toronto in 1998.
  • Rajab Ali (Kenya) bowled 6 overs for 67 against SL at Kandy in 1996 as SL posted a massive score of 5/398 (the highest at the time) in the World Cup.
  • Ravi Shastri (India) bowled 7 overs for 77 against the West Indies at Jamshedpur in 1983. Viv Richards scored 149 off 99 balls in the game.
  • Greg Matthews (Australia) bowled 5 overs for 54 against India at Delhi in 1986.
  • Heath Davis (New Zealand) bowled 5 overs for 54 against India at Bangalore in 1997.

February 28, 2012

Finite differences for polynomial extrapolation

Filed under: mathematics — ckrao @ 9:41 pm

If I am given the values of a polynomial at some consecutive integers and I want to find its value at the next integer, how might I proceed? For example, if I know p(0) = 1, p(1) = 3, p(2) = 11, p(3) = 28, p(4) = 57, how would I find p(5)?

The easiest way I know is to set up a finite difference table. Write the polynomial values in a row, then underneath each row write the difference of the two numbers above it. Continue until all the numbers in a row are equal (or the row below it contains 0s).

For example:

Then we extrapolate by starting at the bottom row and writing the next equal number (in this case 6), then proceeding upwards by writing the sum of the number to its left and below it. In our example we obtain the following.

Hence we conclude that p(5) = 101. Notice that we did not need to find p(x) for general x (it turns out to be p(x) = (x^3 + 3x^2 + 2)/2 in this example).

The above principle works based on the fact that if p(x) is a polynomial, then p(x+1) - p(x) reduces its degree by 1. By doing this successively the degree eventually becomes 0. Constant polynomials are uniquely determined by one value so by extrapolating this and propagating the result upwards, we can extrapolate any degree of polynomial in a unique manner provided we have a complete table.

What if we wanted to find p(10) in the above example? Would we have to resort to finding p(x) by solving a system of linear equations in its coefficients? That is, do we need to set p(x) = ax^3 + bx^2 + cx + d, substitute known values and then solve for a, b, c, d? In this post I want to show how to proceed in a different way.

Firstly, consider what occurs if our original row were taken from a diagonal of Pascal’s triangle (i.e. binomial coefficients). For example, here in the first row we show numbers of the form \binom{n}{3} = n(n-1)(n-2)/6.

We find the second row has numbers of the form \binom{n}{2} and the third row has numbers of the form \binom{n}{1}. In general if the first row has numbers of the form \binom{n}{k} (where k is fixed and n varies), then the i‘th row has numbers of the form \binom{n}{k-i+1}. This is a consequence of the fundamental identity

\displaystyle \binom{n+1}{k} - \binom{n}{k} = \binom{n}{k-1}.

Given this, it makes sense to try to write our polynomial as a linear combination of binomial coefficients (\binom{n}{k}) rather than as a linear combination of monomials (x^n).

Returning to our previous example, we wish to evaluate p(10) given p(0) = 1, p(1) = 3, p(2) = 11, p(3) = 28, p(4) = 57. Notice how the entire finite difference table can be generated from its first diagonal (running NW-SE). For example,

\begin{array}{lcl} 28 &=& 11 + 17\\ &=& (3 + 8) + (8 + 9)\\ &=& 3 + 2\times 8 + 9\\ &=& (1+2) + 2(2 + 6) + (6 + 3)\\ &=& 1 + 3.2 + 3.6 + 3.\end{array}

This shows that p(3) = 28 can be written in terms of the first diagonal (1, 2, 6, 3) with the binomial coefficients 1,3,3,1. Another way of thinking of this is to see how many ways each number in the first diagonal can trickle across the table. In the above example, the number 6 can trickle up to p(3) in \binom{3}{2} ways: it needs to go up twice and across once in three total moves (up-up-across, up-across-up or across-up-up). This should hopefully explain how the binomial coefficients enter the picture. In general if we need to go across p places and up q places, this can be done in \binom{p+q}{p} ways.

Now if we wish to find p(10), we need to go 10 spaces across from p(0) = 1, nine spaces across and one space up fromĀ  2 = p(1)-p(0), eight spaces across and two spaces up from 6, and seven spaces across and three spaces up from 3. We conclude that we may write

\displaystyle p(10) = 1\binom{10}{0} + 2\binom{10}{1} + 6\binom{10}{2} + 3\binom{10}{3}.

Hence p(10) = 1 + 2\times 10 + 6\times (10.9)/2 + 3\times (10.9.8)/6 = 1 + 20 + 270 + 360 = 651. In general, if the first diagonal of the table is a_0, a_1, \ldots, a_n and a_0= p(0), then

\displaystyle p(x) = \sum_{i=0}^n a_i \binom{x}{i}.

There is actually a strong link here between this formula and the Taylor series expansion, forming the basis for umbral calculus. See [2] for more details.


[1] Jim Loy’s pages, Finite Difference

[2] Weisstein, Eric W. “Newton’s Forward Difference Formula.” From MathWorld–A Wolfram Web Resource.

February 25, 2012

Economical bowling in ODIs

Filed under: cricket,sport — ckrao @ 12:58 am

A couple of months ago Saeed Ajmal (Pakistan) produced a spell of 7 overs, 4 maidens, 2 wickets for 6 runs in a one-day international against Bangladesh. I wondered how often in recent times it has happened that a bowler has been so economical in ODIs, especially with the trend towards higher run-rates (2006 saw the first 400+ team score and now it has occurred ten times). I used CricInfo’s Statsguru to find out. It turns out that there have been 31 instances of a bowler bowling at least 7 overs with an economy rate less than 1 run per over, with a fair number of those occuring in the 2000s, though mostly against the weaker teams.

The list can be seen here.

The most economical was 4 wickets for 3 runs in 10 overs by Phil Simmons (WI) against Pakistan! Only two of the 31 performances have been five wicket hauls:

  • Muralitharan against NZ at Sharjah in 2002 (10-3-9-5) (New Zealand still posted 218 and won the match!)
  • Sunil Joshi against South Africa at Nairobi in 1999 (10-6-6-5)

Below are some other notable performances of “recent” times.

  • Curtly Ambrose (West Indies) had 10-5-5-1 against Sri Lanka at Sharjah in 1999. His overs were bowled in a single spell.
  • Andre Botha (Ireland) had 8-4-5-2 against Pakistan at Kingston in 2007 (during the World Cup), a game which Pakistan famously lost. This was the last time we saw such an economical bowling display in an ODI against a non-minnow (non-Irish to be more specific) side.
  • Glenn McGrath (Australia) had 10-4-8-4 against India at Sydney in 2000. This was McGrath at the peak of his powers.
  • Aasif Karim (Kenya) had 8.2-6-7-3 against Australia at Durbin in 2003 (during the World Cup Super Sixes stage as a 39 year-old). He had 5 taken off his last two balls so at one point his figures were 8-6-2-3!
  • Trevor Gripper (Zim) had 7-4-6-0 against West Indies at Kandy in 2001. Interestingly he only played 8 ODIs.

February 13, 2012

Decomposition of a 3x3x3 cube

Filed under: mathematics — ckrao @ 9:14 pm

Here is a decomposition of a 3 by 3 by 3 cube that I learnt recently: it can be divided into six 2x2x1 blocks and three unit cubes. The following diagram attempts to show this in layers, where the unit cubes are in red and the 2x2x1 blocks are each shown in a different colour.

A question for the interested reader: do all such decompositions necessarily have the unit cubes appearing along a diagonal of the 3x3x3 cube?

Create a free website or blog at

%d bloggers like this: