This is the third installment, do read the previous posts. I left some standing questions in the last post, and I am going to answer at least one of them: Can we squeeze an $l_p$ ball´s volume out of a generalized Wallis sieve?. At the same time I will generalize and simplify the proof from the xkcd post. First, lets generalize the Wallis sieve.

In $d$ dimensions, starting with a $p$ sided hypercube and cutting it appropriately (I leave to the reader to draw it), we get the product,

$$A_n^d = \prod_{n=1}^\infty \frac{p^d n^{d-1} \Big(n+\frac{d}{p}\Big)} {(pn+1)^d},$$

which can be rewritten as the limit

$$A_n^d = lim_{n\to\infty} \frac{p^{nd}\Gamma(n+1)^{d-1}\Gamma\Big(n+1+\frac{d}{p}\Big)\Gamma\Big(1+\frac{d}{p}\Big)^d}{p^{nd}\Gamma(1+\frac{d}{p})\Gamma\Big(n+1+\frac{1}{p}\Big)^d}$$

where I have made use of the Euler gamma function recurrent properties (see previous posts).

Note that the $d$ in $A_n^d$ is an index, not an exponent.

The limit can be separated into two factors (after cancelling the $p^{nd}$),

$$A_n^d = lim_{n\to\infty}\Bigg[ \frac{\Gamma(n+1)^{d-1}\Gamma\Big(n+1+\frac{d}{p}\Big)}{\Gamma\Big(n+1+\frac{1}{p}\Big)^d} \Bigg]\Bigg[\frac{\Gamma\Big(1+\frac{1}{p}\Big)^d}{\Gamma(1+\frac{d}{p})}\Bigg]$$.

Remember that $\frac{\Gamma(n+a)}{\Gamma(n)}\sim n^a$, so the first term when $n\to\infty$

$$ \frac{\Gamma(n+1)^{d-1}\Gamma\Big(n+1+\frac{d}{p}\Big)}{\Gamma\Big(n+1+\frac{1}{p}\Big)^d} = \frac{\Gamma(n+1)^{d-1}\Gamma\Big(n+1+\frac{d}{p}\Big)}{\Gamma\Big(n+1+\frac{1}{p}\Big)^{d-1}\Gamma\Big(n+1+\frac{1}{p}\Big)} \sim \frac{n^{\frac{d-1}{p}}}{n^{\frac{d-1}{p}}}\sim 1.$$

So we obtain, finally,

$$A_n^d = lim_{n\to\infty} \frac{\Gamma\Big(1+\frac{1}{p}\Big)^d}{\Gamma(1+\frac{d}{p})} = V_d^p\Big(\frac{1}{2}\Big).$$

The value of the limit is the volume of the $l_p$ ball, with the case of the hypersphere $p=2$ being a particular case.

Another remarkable case happens when $p=1$ and we obtain the volume of the $d$-dimensional cross-polytope, of radius $R=\frac{1}{2},$ which is $\frac{1}{d!},$ . The cross-polytope is the generalization of the octahedron to $n$ dimensions and is the dual of the hypercube we start with.

This formula lets us also interpret the volume of various fat Cantor and other Smith-Cantor-Volterra sets.

The question still standing from last post is: Is there a geometrical interpretation for the intermediate $A_n^d$? And of course, What more can we learn from this relation between $l_p$ and these sets?

# Playing and hacking

## Tuesday, April 26, 2016

## Saturday, April 23, 2016

### Wallis sieve, and lp n-balls II

This post is a continuation of this one. In it I talked about this video by Matt Parker and some

of its consequences for two dimensions. In the video, he also asserted that, for three dimensions, the approach of cutting off pieces of a cube in tha same way that is done in the Wallis sieve but in three dimensions, would give back the volume of a sphere (see the pretty drawings in the post by Evelyn Lamb).

This opened the question of whether this would happen in higher dimensions. Someone in twitter (thanks!) pointed me to this post in the xkcd forum, where they proof this fact. I am going to transcript, dissect and explain this proof. The original ideas are all from the post, and the mistakes all mine.

First, remember the basic Wallis product formula,

$$\frac{\pi}{4}=\prod_{n=1}^{\infty}\frac{4n(n+1)}{(2n+1)^2}=lim_{n\to\infty} \frac{4^n n! (n+1)!}{(2n+1)!!^2}.$$

Remember that the doble factorial is the product $n!! = n(n-2)(n-4)\cdots~$ which stops when the terms would cease to be positive, i.e. with $⌈n/2⌉$ terms. An important property of the double factorial is that it can be written in terms of Euler gamma function,

$$\Gamma\Big(n+\frac{1}{2}\Big) = \frac{(2n-1)!!\sqrt{\pi}}{2^n}.$$

Remember also that the volume of an $l_p$ hyperball is

$$V_d^p(R) = \frac{(2\Gamma(\frac{1}{p} + 1)R)^d}{\Gamma(\frac{d}{p} + 1)}.$$

So the, taking into account $\Gamma\big(\frac{3}{2}\big)=\frac{\sqrt{\pi}}{2}$, the volume of an d-hypersphere, which is the $l_2$ hyperball is

$$V_d^2(R) = \frac{R^d\pi^{d/2}}{\Gamma(\frac{d}{2} + 1)}.$$

The Wallis product can be written,

$$\frac{\pi}{4}=\prod_{n=1}^{\infty}\Bigg[1-\frac{1}{(2n+1)^2}\Bigg],$$

which is convergent, and we can move around terms (if we are careful), because the series

$$\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2},$$

is absolutely convergent and we can apply the test for product convergence.

The general product for d dimensions for the Wallis sieve turns out to be

$$\prod_{n=1}^\infty\frac{2^dn^{d-1}(n+\frac{d}{2})}{(2n+1)^d}.$$

For even $d$ we can write this product as

$$lim_{n\to\infty} \frac{2^{nd}(n!)^{d-1}(n+\frac{d}{2})!}{(\frac{d}{2})!((2n+1)!!)^d}.$$

What we have done here is that the factors coming from $n+\frac{d}{2}$ have been expanded into

$\frac{(n+\frac{d}{2})!}{(\frac{d}{2})!}$. This is the tricky part where we have assumed $

d$ to be even. For odd $d$ we can either rewrite in terms of $k$, i.e. $d=2k+1$, and follow a similar approach or be more general and use the gamma function.

The trick to calculate the limit is to separate it into the product of two parts, one of which can be identified as a Wallis product, between square brackets,

$$lim_{n\to\infty} A_n^d = lim_{n\to\infty} \Bigg[\frac{4^{n-1} n! (n-1)!}{(2n-1)!!}\Bigg]^{d/2}\Bigg(\frac{2^dn^{d/2}(n+\frac{d}{2})!}{(\frac{d}{2})!n!(2n+1)^d}\Bigg),$$

$$lim_{n\to\infty} A_n^d =\Big(\frac{\pi}{2}\Big)^{d/2}\Bigg(\frac{2^dn^{d/2}(n+\frac{d}{2})!}{(\frac{d}{2})!n!(2n+1)^d}\Bigg).$$

The last part is to show that the right term converges to $\Big(\frac{d}{2}\Big)!$.

We show it by parts, first, when $n \to \infty$

$$\frac{\Big(n+\frac{d}{2}\Big)!}{n!}=(n+1)(n+2)\cdots(n+\frac{d}{2}) \sim n^{d/2},$$

so the limit can be rewritten as

$$lim_{n\to\infty} A_n^d =\Big(\frac{\pi}{2}\Big)^{d/2}\Bigg(\frac{(2n)^d}{(\frac{d}{2})!(2n+1)^d}\Bigg).$$.

Finally, for big enough $n$, we can use approximate the second term as,

$$\frac{(2n)^d}{(\frac{d}{2})!(2n+1)^d}\sim\frac{1}{(\frac{d}{2})!}=\frac{1}{\Gamma(\frac{d}{2}+1)},$$

so

$$lim_{n\to\infty} A_n^d = \frac{(\frac{\pi}{4})^{d/2}}{\Gamma(\frac{d}{2} + 1)}=V_d^2\Big(\frac{1}{2}\Big).$$

So, now that we have the general formula for $d$ dimensions, the question still stands, can we interpret $A_n^d$ in geometrical terms as we did for $d=2$? What about $l_p$ for $p\neq2$? Stay tuned.

of its consequences for two dimensions. In the video, he also asserted that, for three dimensions, the approach of cutting off pieces of a cube in tha same way that is done in the Wallis sieve but in three dimensions, would give back the volume of a sphere (see the pretty drawings in the post by Evelyn Lamb).

This opened the question of whether this would happen in higher dimensions. Someone in twitter (thanks!) pointed me to this post in the xkcd forum, where they proof this fact. I am going to transcript, dissect and explain this proof. The original ideas are all from the post, and the mistakes all mine.

First, remember the basic Wallis product formula,

$$\frac{\pi}{4}=\prod_{n=1}^{\infty}\frac{4n(n+1)}{(2n+1)^2}=lim_{n\to\infty} \frac{4^n n! (n+1)!}{(2n+1)!!^2}.$$

Remember that the doble factorial is the product $n!! = n(n-2)(n-4)\cdots~$ which stops when the terms would cease to be positive, i.e. with $⌈n/2⌉$ terms. An important property of the double factorial is that it can be written in terms of Euler gamma function,

$$\Gamma\Big(n+\frac{1}{2}\Big) = \frac{(2n-1)!!\sqrt{\pi}}{2^n}.$$

Remember also that the volume of an $l_p$ hyperball is

$$V_d^p(R) = \frac{(2\Gamma(\frac{1}{p} + 1)R)^d}{\Gamma(\frac{d}{p} + 1)}.$$

So the, taking into account $\Gamma\big(\frac{3}{2}\big)=\frac{\sqrt{\pi}}{2}$, the volume of an d-hypersphere, which is the $l_2$ hyperball is

$$V_d^2(R) = \frac{R^d\pi^{d/2}}{\Gamma(\frac{d}{2} + 1)}.$$

The Wallis product can be written,

$$\frac{\pi}{4}=\prod_{n=1}^{\infty}\Bigg[1-\frac{1}{(2n+1)^2}\Bigg],$$

which is convergent, and we can move around terms (if we are careful), because the series

$$\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2},$$

is absolutely convergent and we can apply the test for product convergence.

The general product for d dimensions for the Wallis sieve turns out to be

$$\prod_{n=1}^\infty\frac{2^dn^{d-1}(n+\frac{d}{2})}{(2n+1)^d}.$$

For even $d$ we can write this product as

$$lim_{n\to\infty} \frac{2^{nd}(n!)^{d-1}(n+\frac{d}{2})!}{(\frac{d}{2})!((2n+1)!!)^d}.$$

What we have done here is that the factors coming from $n+\frac{d}{2}$ have been expanded into

$\frac{(n+\frac{d}{2})!}{(\frac{d}{2})!}$. This is the tricky part where we have assumed $

d$ to be even. For odd $d$ we can either rewrite in terms of $k$, i.e. $d=2k+1$, and follow a similar approach or be more general and use the gamma function.

The trick to calculate the limit is to separate it into the product of two parts, one of which can be identified as a Wallis product, between square brackets,

$$lim_{n\to\infty} A_n^d = lim_{n\to\infty} \Bigg[\frac{4^{n-1} n! (n-1)!}{(2n-1)!!}\Bigg]^{d/2}\Bigg(\frac{2^dn^{d/2}(n+\frac{d}{2})!}{(\frac{d}{2})!n!(2n+1)^d}\Bigg),$$

$$lim_{n\to\infty} A_n^d =\Big(\frac{\pi}{2}\Big)^{d/2}\Bigg(\frac{2^dn^{d/2}(n+\frac{d}{2})!}{(\frac{d}{2})!n!(2n+1)^d}\Bigg).$$

The last part is to show that the right term converges to $\Big(\frac{d}{2}\Big)!$.

We show it by parts, first, when $n \to \infty$

$$\frac{\Big(n+\frac{d}{2}\Big)!}{n!}=(n+1)(n+2)\cdots(n+\frac{d}{2}) \sim n^{d/2},$$

so the limit can be rewritten as

$$lim_{n\to\infty} A_n^d =\Big(\frac{\pi}{2}\Big)^{d/2}\Bigg(\frac{(2n)^d}{(\frac{d}{2})!(2n+1)^d}\Bigg).$$.

Finally, for big enough $n$, we can use approximate the second term as,

$$\frac{(2n)^d}{(\frac{d}{2})!(2n+1)^d}\sim\frac{1}{(\frac{d}{2})!}=\frac{1}{\Gamma(\frac{d}{2}+1)},$$

so

$$lim_{n\to\infty} A_n^d = \frac{(\frac{\pi}{4})^{d/2}}{\Gamma(\frac{d}{2} + 1)}=V_d^2\Big(\frac{1}{2}\Big).$$

So, now that we have the general formula for $d$ dimensions, the question still stands, can we interpret $A_n^d$ in geometrical terms as we did for $d=2$? What about $l_p$ for $p\neq2$? Stay tuned.

## Wednesday, April 20, 2016

### Wallis sieve, and lp n-balls

I heard about the Wallis sieve the first time in this video by Matt Parker, which is fascinating. Instantly I recognized the pattern. I thought the relation between the Wallis sieve and the formula for the volume of an lp n-ball would be trivial and well-known, it turns out it is neither. Later, I read the blog post in scientific american by Evelyn Lamb and I still thought it would be easy to relate both. Finally, I sat down and did the work and found that the result is not only surprising, but (at least to me), completely non-obvious and has the potential to be very interesting.

The volume of an $l_p$ n-ball, a generalized ball is of radius $R$ is, (for more details and the calculation and history of the formula, see Xiafu Wang's paper),

$$V_d^p(R) = \frac{(2\Gamma(\frac{1}{p} + 1)R)^d}{\Gamma(\frac{d}{p} + 1)}.$$

Note that the ball for $l_2$ is an hypersphere of dimension $d$.

This explains the relation between the Euler gamma function and $\pi$, one of my favorites,

$$\Gamma\Big(\frac{3}{2}\Big) = \frac{\sqrt{\pi}}{2}.$$

At the same time, the gamma function is a generalization of the factorial and satisfies all sorts of recursive formulas similar to the Wallis sieve.

I will refer you to Evelyn Lamb's post for a detailed introduction, but the Wallis sieve can be easily written as a limit using the gamma formula,

$$\frac{\pi}{2} = \prod_{n=1}^{\infty}\Bigg[\frac{(2n)^2}{(2n-1)(2n+1)}\Bigg] = \frac{2\cdot 2\cdot 4\cdot 6\cdot 6\ldots}{1\cdot 2\cdot 2\cdot 5\cdot 5\cdot 7\ldots},$$

which can be rewritten as a limit,

$$\lim_{n\to\infty} \frac{2^{4n}}{n{{2n}\choose{n}}^2} = \pi \lim_{n\to\infty} \frac{n \Gamma(n)^2}{\Gamma(\frac{1}{2}+n)^2} = \pi.$$

We can then apply the gamma duplication formula,

$$\Gamma(z)\Gamma(z+\frac{1}{2})= 2^{1-2z}\sqrt{\pi}\Gamma(2z),$$

and the functional relation,

$$\Gamma(z+1) = z\Gamma(z),$$

to rewrite again the limit,

$$\pi = \lim_{n\to\infty} n\Bigg[\frac{\Gamma(n)^2}{\Gamma(2n)2^{1-2n}}\Bigg]^2 = \lim_{n\to\infty} \frac{1}{n}\Bigg[\frac{\Gamma(1+n)^2 2^{2n}}{\Gamma(2n+1)}\Bigg]^2,$$

so

$$\pi = \lim_{n\to\infty}\frac{V_2^{\frac{1}{n}}(2^n)^2}{4n}.$$

This is, to say the least, surprising. Instead of hyperspheres and a trivial relationship, we get something which looks like an astroid (the image comes from Wikipedia).

So it is the limit of the square of the volume of this figure as it collapses upon itself, its inner radius getting smaller while the outer radio grows. This result is bizarre and not at all trivial.

The formula can be generalized (I will play with this the next time I have some free time) to higher dimensions. The video talks about this, but I have not written their formula down. Also, it will be interesting if fat Cantor sets can be written in terms of hyperballs too.

I have the conjecture which it will be related with the taxicab measure astroid ball, whatever that is.

Edit: fixed a missing n in the denominator in the limit.

The volume of an $l_p$ n-ball, a generalized ball is of radius $R$ is, (for more details and the calculation and history of the formula, see Xiafu Wang's paper),

$$V_d^p(R) = \frac{(2\Gamma(\frac{1}{p} + 1)R)^d}{\Gamma(\frac{d}{p} + 1)}.$$

Note that the ball for $l_2$ is an hypersphere of dimension $d$.

This explains the relation between the Euler gamma function and $\pi$, one of my favorites,

$$\Gamma\Big(\frac{3}{2}\Big) = \frac{\sqrt{\pi}}{2}.$$

At the same time, the gamma function is a generalization of the factorial and satisfies all sorts of recursive formulas similar to the Wallis sieve.

I will refer you to Evelyn Lamb's post for a detailed introduction, but the Wallis sieve can be easily written as a limit using the gamma formula,

$$\frac{\pi}{2} = \prod_{n=1}^{\infty}\Bigg[\frac{(2n)^2}{(2n-1)(2n+1)}\Bigg] = \frac{2\cdot 2\cdot 4\cdot 6\cdot 6\ldots}{1\cdot 2\cdot 2\cdot 5\cdot 5\cdot 7\ldots},$$

which can be rewritten as a limit,

$$\lim_{n\to\infty} \frac{2^{4n}}{n{{2n}\choose{n}}^2} = \pi \lim_{n\to\infty} \frac{n \Gamma(n)^2}{\Gamma(\frac{1}{2}+n)^2} = \pi.$$

We can then apply the gamma duplication formula,

$$\Gamma(z)\Gamma(z+\frac{1}{2})= 2^{1-2z}\sqrt{\pi}\Gamma(2z),$$

and the functional relation,

$$\Gamma(z+1) = z\Gamma(z),$$

to rewrite again the limit,

$$\pi = \lim_{n\to\infty} n\Bigg[\frac{\Gamma(n)^2}{\Gamma(2n)2^{1-2n}}\Bigg]^2 = \lim_{n\to\infty} \frac{1}{n}\Bigg[\frac{\Gamma(1+n)^2 2^{2n}}{\Gamma(2n+1)}\Bigg]^2,$$

so

$$\pi = \lim_{n\to\infty}\frac{V_2^{\frac{1}{n}}(2^n)^2}{4n}.$$

This is, to say the least, surprising. Instead of hyperspheres and a trivial relationship, we get something which looks like an astroid (the image comes from Wikipedia).

So it is the limit of the square of the volume of this figure as it collapses upon itself, its inner radius getting smaller while the outer radio grows. This result is bizarre and not at all trivial.

The formula can be generalized (I will play with this the next time I have some free time) to higher dimensions. The video talks about this, but I have not written their formula down. Also, it will be interesting if fat Cantor sets can be written in terms of hyperballs too.

I have the conjecture which it will be related with the taxicab measure astroid ball, whatever that is.

Edit: fixed a missing n in the denominator in the limit.

## Tuesday, April 12, 2016

### Swap without temporary space

There is a really bad technical interview question which appears now and then.

It goes like this:

It gets better, because the question is voided in some programming languages with tuple literals or multiple assignment. For example in go, the solution is trivial,

a, b = b, a

And you are done with it. Even better, the compiler may generate a swap of registers, which is as efficient as it gets.

In any case, I was chatting about this question with a friend and I remembered some ideas I thought I had read somewhere, maybe in

somewhere else. After checking, apparently, I hadn't read it in any of them, so maybe I have come up with them myself. In any case, the gist of it is, if you are ever asked this question, you can use matrices to go completely overboard with the answer.

So, say you want to swap two variables and you want to do it without temporary storage. One of the classic ways to do this is,

a = a + b

b = a - b

a = a - b

So how can we describe this in terms of matrices?

Well, each of the assignments is actually the multiplication of the vector

$\begin{bmatrix}a\\ b\end{bmatrix}$ by a matrix and as long as the matrix determinant is not zero, you

don't lose any information.

For example, the first assignment may be written in math,

$a' = a + b$

$b' = 0 + b$

or in matrix form:

$$\begin{bmatrix}a'\\ b'\end{bmatrix} = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}a\\ b\end{bmatrix}$$

So, the three matrices describing the previous assignments are,

$$M = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$

$$N = \begin{bmatrix}1 & 0\\1 & -1\end{bmatrix}$$

$$R = \begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}$$

The multiplication of these matrices (be careful, the order has to be right)

$$RNM = \begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\1 & -1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$

which is a reverse identity i.e. a swap.

This already works (even if it overflows). In all truth any N factors of the

reverse identity do the trick. You may even rescale them, for example, multiply the first by 2 and the second by 1/2, if you are in floating point, for more obscurity. Or use reciprocals for integers (another trick from

We can go even further and work in $GF2$, i.e. binary bit by bit operations.

In this space, the addition is the xor (^) and each number is its own inverse,

so the above equation can be written,

a = a ^ b

b = a ^ b

a = a ^ b

You can also write this code in terms of factors of the reverse identity with binary matrices.

The three assignments with the xor is probably what the (now completely stunned) interviewer was aiming for.

It goes like this:

**“how would you exchange the value of two variables without****using temporary space?**” Whenever I hear about this question I cringe. It is one of those questions that does not measure anything other than: have you seen this before? or did you read Hacker's Delight?, which, by the way, I wholeheartedly recommend. And you may be a fine developer and human being and be just unlucky enough to not have seen this trick before.It gets better, because the question is voided in some programming languages with tuple literals or multiple assignment. For example in go, the solution is trivial,

a, b = b, a

And you are done with it. Even better, the compiler may generate a swap of registers, which is as efficient as it gets.

In any case, I was chatting about this question with a friend and I remembered some ideas I thought I had read somewhere, maybe in

*Hacker's Delight*, maybesomewhere else. After checking, apparently, I hadn't read it in any of them, so maybe I have come up with them myself. In any case, the gist of it is, if you are ever asked this question, you can use matrices to go completely overboard with the answer.

So, say you want to swap two variables and you want to do it without temporary storage. One of the classic ways to do this is,

a = a + b

b = a - b

a = a - b

So how can we describe this in terms of matrices?

Well, each of the assignments is actually the multiplication of the vector

$\begin{bmatrix}a\\ b\end{bmatrix}$ by a matrix and as long as the matrix determinant is not zero, you

don't lose any information.

For example, the first assignment may be written in math,

$a' = a + b$

$b' = 0 + b$

or in matrix form:

$$\begin{bmatrix}a'\\ b'\end{bmatrix} = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}a\\ b\end{bmatrix}$$

So, the three matrices describing the previous assignments are,

$$M = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$

$$N = \begin{bmatrix}1 & 0\\1 & -1\end{bmatrix}$$

$$R = \begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}$$

The multiplication of these matrices (be careful, the order has to be right)

$$RNM = \begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\1 & -1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$

which is a reverse identity i.e. a swap.

This already works (even if it overflows). In all truth any N factors of the

reverse identity do the trick. You may even rescale them, for example, multiply the first by 2 and the second by 1/2, if you are in floating point, for more obscurity. Or use reciprocals for integers (another trick from

*Hacker's Delight*).We can go even further and work in $GF2$, i.e. binary bit by bit operations.

In this space, the addition is the xor (^) and each number is its own inverse,

so the above equation can be written,

a = a ^ b

b = a ^ b

a = a ^ b

You can also write this code in terms of factors of the reverse identity with binary matrices.

The three assignments with the xor is probably what the (now completely stunned) interviewer was aiming for.

## Tuesday, March 27, 2012

### Taking measures and drawing them

I have spent the last couple of months measuring nix, the operating system I am working on. Measuring is hard. Quoting Feynman:

"The first principle is that you must not fool yourself, and

**you are the easiest person to fool**."When you are measuring complex systems and the amd64 machine we are measuring is quite complex, it is important to separate the variables that can influence the results. Each possible variable needs to be modified separately in a different experiment. Each experiment has to be made many, many times. Then, we calculate some statistics and we are done.

Except...

Trying to modify each variable separately is not easy. In many cases, they are not independent. Also, some of them may not be under your control (network traffic, for example). Still, that is the subject for another post. In this one we are interested in calculating simple statistics and drawing them, which should always be the first approximation to a new problem.

**Mean and standard deviation**

We were taught how to interpret experiment results at school, by calculating the average and the standard deviation. It is very possible, being a programmer, that you think that this is trivial and to do it yourself in your benchmark. Well, it is not difficult, but it takes some care. It is easy to do it wrong (I know from experience).

First, the average needs to be calculated using an online algorithm. An online algorithm for the average (normally called a running average) just takes one sample at a time generates an average of what you have seen already.

Why is this important?, why can't I just add all the samples and divide by N?

Because, if you take a lot of samples (which you should) and add them, you will overflow the variable you use as accumulator.

There is also the question of numerical stability. The representation the computer uses for floating point is not equally precise across all the range. You have to be careful with the operations you do, so that they do not amplify the errors added by the representation. This consideration is specially important for calculating the standard deviation. If you calculate the standard deviation using the classical offline algorithm you will end up subtracting very big numbers to obtain a small one, one of the worst things to do. It is very easy when taking the square root to

**end up with an imaginary standard deviation**, because of this problem.The solution is to use an online algorithm to calculate both the average and the standard deviation. One which is carefully designed not to be unstable.

**Donald Knuth's Art of Computer Programming**is the right resource for this kind of algorithms, and fair enough, we can find one in page 232, Vol 2, 3rd Edition.A simple description of the algorithm can be found here:

The gist of it is:

Initialize M1 = x1 and S1 = 0.

For subsequent x's, use the recurrence formulas

Mk = Mk-1+ (xk - Mk-1)/k

Sk = Sk-1 + (xk - Mk-1)*(xk - Mk).

For 2 ≤ k ≤ n, the kth estimate of the variance is s2 = Sk/(k - 1).

This is an example of this algorithm written in awk I use for files with the results of time, which look like this:

1.34u 34.5s 25.78r

1.34u 33.5s 23.78r...

So, we already know how to calculate the standard deviation and the average and that is all, we can draw an error bar diagram and we are done. Not quite. Without going deep into statistics, more can be done as first approximation to interpreting a group of samples.

**Whiskers**

The standard deviation tells you nothing about how the samples are really distributed. Yes, it can tell you if they are spread out but not how asymmetric or skewed the distribution is.

If the distribution is not gaussian or worse, not symmetrical, and many real distributions are not, the standard deviation is a bad estimation of how spread the the error is. It is better not to make any assumptions.

I prefer to use a

**box plot or whisker plot**. A whiskers diagram draws a box spanning the first and third**quartile**(Q1 and Q3) with the**median**(the second quartile or Q2) in the middle and two whiskers depicting the error (what the whiskers represents varies among versions of the whisker diagram).First of all, what is a quartile? The median is the sample that divides the samples in two equal sets (samples bigger than the median and samples smaller):

A quartile is the median of each of the sets which the median creates by dividing the original set in two. This can be applied recursively. If the number of samples of any of this sets is even, the median is defined as the average of the two central samples. For example for the set:

1 3 4 6 7 8

the median (or second quartile Q2) is (4+6)/2 = 5.

The first quartile Q1 is extracted from the lower set

1 3

and would be (1+3)/2=2 and Q3 would be 7.5.

There are other algorithms where the lowest and highest samples and the median are not used when calculating Q1 and Q3, this is method 1 in the wikipedia entry.

I normally use method 1, but there is not much difference in practice.

Going back to whiskers, what do they represent? Well, it depends on the version of the whiskers diagram. First, there is something called IQR (inter quartile range) which is Q3 - Q1. For many versions of the whiskers diagram, the whiskers represent Q1 - 1.5*IQR and Q3 + 1.5*IQR (I will call this version

**kind A**). The other most popular version is that they represent the maximum and the minimum of the whole data (I will call this version**kind B**). I prefer the second one because, again, it does not make any assumptions about the data.Anything outside of the whiskers (if there is anything) is called an outlier and sometimes they are drawn in the diagram and sometimes they are not.

In any case, a whisker diagram is very good because from a glance one can deduce

- How skewed the data is by looking at the relative position of the median
- How spreaded the data can be by looking at the whiskers
- If two groups of samples are clearly different by seeing where the boxes and the whiskers lay relatively to one other.

So, the next question is how to draw this diagrams given the samples.

As an example, see this little script which calculates the values of the box for numbers in the first column of a diagram of

**kind A**. I use it with grap sometimes, when I need complete control.What if I want to have a very pretty diagram with most of the work done automatically by the machine? In this case, I use R with ggplot2.

First, you need to install the R programming language. This can be done downloading some binary package for your operating system from the web page. This includes (among other things) two programs of interest to us:

- R, a text interpreter and

- Rscript, which is R but can be used as a interpreter for #! in scripts

Why Rscript is needed and R does not behave like any other interpreter, is beyond me, but it is how it is.

In any case, after that, next install ggplot2.

Fire R from the command line and run install.package like this:

> install.packages("ggplot2")

select the mirror and exit R and you are done.

You are ready to create the first plot. For this, first you need data. I am going to use as an example this file, which has a header and a body and looks something like this:

`, Ncores, Time, Scheduler `

`X0, 10, 3.84, amp`

`X0, 10, 3.89, amp`

`The first line is the header, telling me what do the samples stand for. The first column identifies samples to be grouped together, for example all rows with X0 stand for a group of samples that should go in one box. We are ready to write a script to generate the graph.`

`We open a file and write this, which can also be downloaded here:`

- #!/usr/bin/Rscript
- data <-read.csv("data")
- dframe = data[, c("Ncores", "Time","Scheduler")]
- colnames(dframe) = c("Ncores", "Time","Scheduler")
- library(ggplot2)
- ggplot(dframe, aes(x=factor(Ncores), y=Time, fill=Scheduler)) + geom_boxplot(outlier.size=0, size=0.2)
- ggsave(file="data.pdf")

Line 1 is the to make it a script, remember to use Rscript instead of R.

Line 3 selects which columns of data we are going to represent and line 6 gives them names.

Then we include ggplot2 (line 8) and we plot calling ggplot and geom_boxplot.

The second parameter is the aesthetics, which are parameters related to how the graph is displayed. Fill, for example, gives the color for the background of the box, which is chosen automatically to be a different one for each possible value. geom_boxplot parameter describes how outliers should be drawn (I am not depicting them) and ggsave writes the plot to a pdf file.

This scripts plots a whiskers plot of

**kind A,**which is also called tukey diagram in honor to its creator (who also coined the term bit and the most used FFT algorithm).If the meaning of the whiskers need to be changed, this can be more involved. An example can be seen in this script, where the result is not

**kind A**or kind ,**kind B,**but a custom made one.We are finally ready to contemplate the result in all its glory (a pdf scalable version can be found here). Of course, this graph is quite baroque, but then again it expresses a lot of information.

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