# Chaitanya's Random Pages

## January 30, 2015

### AB de Villiers’ fastest ODI century

Filed under: cricket,sport — ckrao @ 12:51 pm

Recently we witnessed the fastest century in one-day cricket history, with AB de Villiers coming in during the 39th over at the fall of South Africa’s first wicket (1/247) and proceeding to blast an incredible 149 off 44 balls in just 59 minutes propelling the team to 2/439 (Amla also made 153* and Rossouw 128). The match produced all types of records including many instances of most runs in n consecutive overs, documented in other blog post here.

Here is the ball-by-ball breakdown of his innings followed by a graph of runs versus balls.

4 2 1 | 1 4 6 4 6 | 6 . 2 2 LB | 1 6 6 6 4 | 6 1 4 1 | 1 . 4 4 | . 6 | 4 6 6 6 | LB 6 | 1 | 6 6 4 6 6 2 | 2 . out

We see that three times he hit 28 runs in 5 balls and 26 runs in another space of 5 balls – that’s 110 runs in just 20 balls with 15 6s and 5 4s right there! He was already 82 off 27 balls (7 6s, 7 4s) and then made 63 off his next 13 balls including 9 6s and 2 4s to surge to 145 off 40 balls!! I have never seen such a concentration of 6 hitting. More analysis of his strike rate is at this @dualnoise post.

Here is what he scored (with balls faced) against the four bowlers who were up against him:

Taylor: 30 (13)
Holder: 45 (9)
Russell: 35 (12)
Smith: 39 (10)

Some other amazingly fast centuries in limited over cricket are in this blog post. Here is the breakdown of the previous fastest ODI century (36 balls) by Corey Anderson in this match:

1 | . 4 1 | . 1 | 6 1 . 2 1 | 4 6 1 | 6 6 . 6 . 6 | 1 . . 6 | 6 6 6 6 1 | . 1 | 4 4 1 1 | 6 6 4 1 | 2 4 1 | 6 1 | 2 2 1

## January 26, 2015

### Affine Transformations of Cartesian Coordinates

Filed under: mathematics — ckrao @ 11:43 am

A very common exercise in high school mathematics is to plot transformations of some standard functions. For example, to plot $y = 2\sin (5x + \frac{2\pi}{3}) - 1$ we may start with a standard sine curve and apply the following transformations in turn:

• squeeze it by a factor of 5 in the $x$-direction
• shift it left by $\frac{2\pi}{3}$
• stretch it by a factor of 2 in the $y$-direction
• shift it down by $1$

This leads to the plot shown.

For sine and cosine graphs an alternative is to plot successive peaks/troughs of the curve and interpolate accordingly. For example, to plot $y = 2\sin (5x + \frac{2\pi}{3}) - 1$ we may proceed as follows.

• Since $\sin(x)$ has a peak at $\frac{\pi}{2}$, solve $5x + \frac{2\pi}{3} = \frac{\pi}{2}$ to find $x = -\frac{\pi}{30}$ as a point where there is a peak at $y = 2\times 1 - 1 = 1$. Hence plot the point $(-\frac{\pi}{30},1)$.
• Since the angular frequency is 5, the period is $\frac{2\pi}{5}$ and we may plot successive peaks spaced $\frac{2\pi}{5}$ apart from the point $(-\frac{\pi}{30},1)$.
• Troughs will be equally spaced halfway between the peaks at $y = 2\times (-1) - 1 = -3$ (at $x = -\frac{\pi}{30} + \frac{\pi}{5} + k\frac{2\pi}{5}$). Then join the dots with a sinusoidal curve.
• Additionally $x-$ and $y-$ intercepts may be found by setting $y = 0$ and $x = 0$ respectively. We find that the $x-$intercepts are at $x = \frac{2\pi k}{5} - \frac{\pi}{10}, \frac{2\pi k}{5} + \frac{\pi}{30}\ (k \in \mathbb{Z})$ and $y-$intercept is $y = 2\sin (\frac{2\pi}{3}) - 1 = \sqrt{3}-1$.

The first approach is more generalisable to plotting other functions. Instead of thinking of the graph transforming, we also may consider it as a change of coordinates. For example, if we translate the parabola $y = x^2$ so that its turning point is at $(2,1)$, this is equivalent to keeping the parabola fixed and shifting axes so that the new origin is at $(-2,-1)$ with respect to the old coordinates. This is illustrated below where the black coordinates are modified to the red ones. The parabola has equation $y = x^2$ under the black coordinates and $y - 1 = (x-2)^2$ or $y = (x-2)^2 + 1$ under the red coordinates.

As another example suppose we take a unit circle and stretch it by a factor of $2$ in the $x$ direction and a factor of $3$ in the $y$ direction. This is the equivalent of changing scale so that the $x$-axis is squeezed by $2$ and the $y$-axis is squeezed by $3$.

Under this stretching of the circle or squeezing of axes, the unit circle equation $x^2 + y^2 = 1$ transforms to that of the ellipse $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$.

More generally, by stretching a Cartesian graph $f(x,y) = 0$ by $a$ in the $x$-direction and $b$ in the $y$-direction, then shifting it along the vector $(h,k)$, we obtain the equation

$\displaystyle f\left(\frac{x-h}{a}, \frac{y-k}{b}\right) = 0.$

This uses the fact that $\frac{x-h}{a}$ and $\frac{y-k}{b}$ are the inverses of $ax+h$ and $by+k$ respectively. Note that if $|a|$ or $|b|$ are less than 1, the stretch becomes a squeezing of the graph, while $a < 0$ or $b < 0$ correspond to a reflection in the $x$ or $y$ axes.

We can extend this idea to the rotation of a graph. Suppose for example we wish to rotate the hyperbola $y = 1/x$ by 45 degrees anti-clockwise. This is equivalent to a rotation of the axes by 45 degrees clockwise and the matrix corresponding to this linear transform is

$\displaystyle \left[ \begin{array} {c} x' \\ y' \end{array} \right] = \frac{1}{\sqrt{2}}\left[ \begin{array}{rr} 1 & 1\\ -1 & 1 \end{array} \right] \left[ \begin{array} {c} x \\ y \end{array} \right] = \left[ \begin{array} {c} \frac{x+y}{\sqrt{2}} \\ \frac{-x+y}{\sqrt{2}}\end{array} \right].$

(Here the columns of the change of basis matrix correspond to where the basis vectors (1,0) and (0,1) map to under a 45 degree clockwise rotation.)

In other words we replace $x'$ with $(x+y)/\sqrt{2}$ and $y'$ with $(-x+y)/\sqrt{2}$ in the equation $y' = 1/x'$ and obtain $(-x+y)/\sqrt{2} = 1/( (x+y)/\sqrt{2})$ or $y^2 - x^2 = 2$.

Here is the same transformation applied to the parabola $y = x^2$ to obtain $\frac{-x+y}{\sqrt{2}} = \frac{(x+y)^2}{2}$ or $x^2 + y^2 + 2xy + \sqrt{2}(x-y) = 0$:

If a graph is affinely transformed (by an invertible map) so that $(1,0)$ maps to $(a,b)$ and $(0,1)$ maps to $(c,d)$ followed by a shift along the vector $(h,k)$, then this is equivalent to the coordinates shifting by $(-h,-k)$ and then transforming under the inverse mapping $\displaystyle \left[ \begin{array}{cc} a & c\\ b & d\end{array} \right]^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -c\\ -b & a\end{array} \right]$:

\displaystyle \boxed{\begin{aligned} x' &= (x-h) \frac{d}{ad-bc} - (y-k) \frac{c}{ad-bd}\\ y' &= -(x-h) \frac{b}{ad-bc} + (y-k) \frac{a}{ad-bd}\end{aligned}}

Here are some special cases of this formula:

• rotation of the graph by $\theta$ anti-clockwise: $\displaystyle x' = x\cos \theta + y \sin \theta, y' = - x \sin \theta + y\cos \theta$ (the example $\theta = \pi/4$ was done above)
• reflection of the graph in $y = x$: $x' = y, y' = x$
• reflection of the graph in $y = mx$ where $m = \tan \theta$: verify that $a = \cos 2\theta, b = \sin 2\theta, c = \sin 2\theta, d = -\cos 2\theta$ so $\displaystyle x' = x \cos 2\theta + y \sin 2\theta, y' = x \sin 2\theta - y \cos 2\theta$
• reflection of the graph in $y = mx + c$ where $m = \tan \theta$: this is equivalent to a reflection in the line $y = mx$ followed by a shift along the vector $(h,k) = (-c \sin 2\theta, c (1+\cos 2 \theta) )$ so
\displaystyle \begin{aligned} x' &= (x+c \sin 2\theta)\cos 2\theta + (y-c(1+\cos 2\theta))\sin 2\theta\\ &= x \cos 2\theta + (y-c)\sin 2\theta,\\ y' &= (x+c \sin 2\theta)\sin 2\theta - (y-c(1+\cos 2\theta))\cos 2\theta\\ &= x \sin 2\theta - (y-c) \cos 2\theta + c \end{aligned}
• reflection of the graph in $y = x + c$ (special instance of the previous case with $\sin 2\theta = 1, \cos 2\theta = 0$): $x'= y - c, y' = x+c$

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