__________

Related:

• The social construction of Hungarian genius

[ source ]

This talk is arguably the very best introduction to computational complexity I have ever come across. I highly recommend it.

Kissinger at 88 is writing brochures for Kissinger Associates. His last book on China is one such work written by the staff at Kissinger Associates. It is designed to curry favor with the Chinese authorities and nothing else.

I know him personally very well, but he is such a deceptive person; he’s a habitual liar and dissembler. Although I’ve spent a lot of time talking to him, I have no insight on him at all. His book ends with a paean to U.S.-Chinese friendship and how every other country has to fit in. I have to review it for the TLS, but I’ve been delaying it by weeks because I don’t know whether it is a case of senility or utter corruption.

I wonder if Kissinger would consider this a compliment. Anyone can be transparent, but obfuscation requires sophistication.

__________

Source:

David Samuels, Q&A: Edward Luttwak, Tablet Magazine, Sept. 6, 2011.

Here is one of many possible implementations in Haskell:

compose :: [(a -> a)] -> (a -> a)
compose = foldr (.) (\x -> x)

where we use a right-fold to compute the composition. Note that the identity function is defined anonymously. What happens if we apply compose to the empty list? We should obtain the identity function. Let us see if this is the case:

compose [] = foldr (.) (\x -> x) [] = (\x -> x)

Indeed, it is the case: if we apply compose to the empty list, we obtain the identity function. What is the type of this identity function? We load the two-line script above into GHCi and experiment a bit:

*Main> let id = compose []
*Main> :t id
id :: a -> a
*Main> map id [0..9]
[0,1,2,3,4,5,6,7,8,9]
*Main> map id ['a'..'z']
"abcdefghijklmnopqrstuvwxyz"

We thus obtain a polymorphic identity function that merely outputs its input, regardless of the type of the input. So far, so good. What happens if we apply compose to a non-empty list? Here is some more equational reasoning:

compose [f0,f1,f2] = foldr (.) (\x -> x) [f0,f1,f2]
                   = foldr (.) (\x -> x) (f0 : f1 : f2 : [])
                   = f0 . (f1 . (f2 . (\x -> x)))

A couple of days ago, we saw how to compose a function with itself times. This is a special case of composing a list of functions, of course. Using function replicate, we can compose a function f with itself  times in the following manner:

compose (replicate n f) = foldr (.) (\x -> x) (replicate n f)

as suggested by Reid Barton. Let us now repeat the experiment we did perform a couple of days ago, but using the new compose:

*Main> let f =  (\n -> (n+1) `mod` 64)
*Main> let h = compose (replicate 64 f)
*Main> map h [0..63]
[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
 16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
 32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
 48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63]

In the next posts, we will do some interesting things with this compose.

where is the identity function on , i.e., for all . In Haskell, it is easy to create a higher-order function compose that composes an endofunction with itself times:

compose :: (a -> a) -> Integer -> (a -> a)
compose f 0 = (\x -> x)
compose f k = f . (compose f (k-1))

where the identity function is defined anonymously. If you happen to dislike anonymity in your functions, here is another implementation:

compose :: (a -> a) -> Integer -> (a -> a)
compose f 0 = identity
compose f k = f . (compose f (k-1))

identity :: a -> a
identity x = x

There are many other alternatives. Take a look at this Stack Overflow thread. I particularly liked Reid Barton‘s elegant solution using a right-fold:

compose f n = foldr (.) id (replicate n f)

where id = (\x -> x). Beautiful, isn’t it? If this one-liner does not convert you to Haskell, then nothing will.

__________

Example

Let , and let function be defined by

This function corresponds to a cyclic shift. Therefore, we expect to have . Let and . We load the script above into GHCi, and voilà, we have the session:

*Main> let ns = [0..63]
*Main> let f =  (\n -> (n+1) `mod` 64)
*Main> map f ns
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,
 17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,
 33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,
 49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,0]

So far, so good! Let us take a look at and :

*Main> let g = compose f 32
*Main> let h = compose f 64
*Main> map g ns
[32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
 48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,
 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
 16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31]
*Main> map h ns
[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
 16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
 32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
 48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63]

Indeed, as we expected, we have .

__________

Everything in this post is elementary. But now that the elements are solid, we can dare to attempt more sophisticated things. Like what? Like composing lists of (distinct) functions. I am thinking of permutations. I am thinking of perfect shuffles.

Normally I just ignore navel-gazing tech-industry articles like this, but people keep sending it to me, so I guess this guy is famous or something. Michael Arrington posted this article, “Startups Are Hard. So Work More, Cry Less, And Quit All The Whining” which quotes extensively from my 1994 diary.

He’s trying to make the point that the only path to success in the software industry is to work insane hours, sleep under your desk, and give up your one and only youth, and if you don’t do that, you’re a pussy. He’s using my words to try and back up that thesis.

I hate this, because it’s not true, and it’s disingenuous.

What is true is that for a VC’s business model to work, it’s necessary for you to give up your life in order for him to become richer.

Follow the fucking money. When a VC tells you what’s good for you, check your wallet, then count your fingers.

He’s telling you the story of, “If you bust your ass and don’t sleep, you’ll get rich” because the only way that people in his line of work get richer is if young, poorly-socialized, naive geniuses believe that story! Without those coat-tails to ride, VCs might have to work for a living. Once that kid burns out, they’ll just slot a new one in.

I did make a bunch of money by winning the Netscape Startup Lottery, it’s true. So did most of the early engineers. But the people who made 100x as much as the engineers did? I can tell you for a fact that none of them slept under their desk. If you look at a list of financially successful people from the software industry, I’ll bet you get a very different view of what kind of sleep habits and office hours are successful than the one presented here.

So if your goal is to enrich the Arringtons of the world while maybe, if you win the lottery, scooping some of the groundscore that they overlooked, then by all means, bust your ass while the bankers and speculators cheer you on.

Instead of that, I recommend that you do what you love because you love doing it. If that means long hours, fantastic. If that means leaving the office by 6pm every day for your underwater basket-weaving class, also fantastic.

__________

The words that Zawinski italicized for emphasis I have chosen to present in bold, due to the fact that the whole quoted text is already italicized. Amusingly, Zawinski’s blog post caused quite a bit of “fear and (self-) loathing” at Hacker News ;-)

__________

Source:

Jamie Zawinski, Watch a VC use my name to sell a con, November 28, 2011.

Indeed, it is. Let us introduce a function , defined by , where and . By construction, the positive zero of function is . Do note that if is odd, then is not a zero of . Recall that Newton’s method uses the recurrence relation , where function is defined by

where is the first derivative of . Hence, we obtain

Thus, we have the generalized Heron’s method

If we make , we obtain the standard Heron’s method, as we expected. The following Haskell script implements this generalized Heron’s method using arbitrary-precision rational numbers (included in the Data.Ratio library) to represent and the sequence of approximations :

import Data.Ratio

-- define generalized Heron map
g :: Integer -> Rational -> Rational -> Rational
g n y x = (1 % n) * ((fromIntegral (n-1) * x) + (y / (x^(n-1))))

-- initial approximation
x0 :: Rational
x0 = 1 % 1

-- list of approximations of the n-th root of y
roots :: Integer -> Rational -> [Rational]
roots n y | n  > 1 = iterate (g n y) x0
          | n <= 1 = error "Invalid value of n!!"

We load this script into GHCi. Let us now perform some numerical experiments.

__________

Experiment #1

We would like to compute a rational approximation of . We have and . Here is a very brief GHCi session:

*Main Data.Ratio> take 6 (roots 2 2)
[1 % 1,3 % 2,17 % 12,577 % 408,665857 % 470832,
 886731088897 % 627013566048]

We obtain the following rational approximation after five iterations

which is the same approximation I obtained twice before.

__________

Experiment #2

We would like to compute a rational approximation of . We have and . Here is a very brief GHCi session:

*Main Data.Ratio> take 7 (roots 2 5)
[1 % 1,3 % 1,7 % 3,47 % 21,2207 % 987,4870847 % 2178309,
 23725150497407 % 10610209857723]

We obtain the following rational approximation after six iterations

How good is this approximation? Let us check:

*Main Data.Ratio> (23725150497407 % 10610209857723)^2 - 5
4 % 112576553224922323902744729
*Main Data.Ratio> 4 / 112576553224922323902744729
3.5531377408652764e-26

Not bad! We could now obtain a rational approximation of the famous golden ratio , as follows:

*Main Data.Ratio> let xs = roots 2 5
*Main Data.Ratio> let phis = map (/2) (map (+1) xs)
*Main Data.Ratio> take 7 phis
[1 % 1,2 % 1,5 % 3,34 % 21,1597 % 987,3524578 % 2178309,
 17167680177565 % 10610209857723]

We obtain the following rational approximation after six iterations

How good is this approximation? Let us recall that the golden ratio is one of the solutions of the equation . Hence, we can check how close to zero is:

*Main Data.Ratio> let phi6 = 17167680177565 % 10610209857723
*Main Data.Ratio> phi6^2 - phi6 - 1
1 % 112576553224922323902744729
*Main Data.Ratio> 1 / 112576553224922323902744729
8.882844352163191e-27

I would say that is fairly close.

__________

Experiment #3

We now would like to compute a rational approximation of . We have and . Here is a very brief GHCi session:

*Main Data.Ratio> take 6 (roots 3 10)
[1 % 1,4 % 1,23 % 8,4909 % 2116,55223315303 % 25495981298,
83759169926117983945469262167029 % 38876457805393768546966848104041]

We obtain the following rational approximation after five iterations

Is this approximation any good? Let us check:

*Main Data.Ratio> let x5 = (roots 3 10) !! 5 
*Main Data.Ratio> x5^3
5876207108075025254603649658428809744645894398375582704
87986732898174481543266076676033790365389 % 58757060813
2677736272189042625165443146909261971794220978628909441
43919908404893015241356540921
*Main Data.Ratio> 5876207108075025254603649658428809744
6458943983755827048798673289817448154326607667603379036
5389 / 587570608132677736272189042625165443146909261971
79422097862890944143919908404893015241356540921
10.000852709004354

This is somewhat disappointing. The rational approximation is a ratio of two enormously long integers, but we do not have such a small error. Note that . We can do some quick error analysis. Let . The 1st order Taylor approximation of around evaluated at is

and, since , we obtain the estimate of the error

In other words, the output deviation should be divided by the slope to obtain an estimate of the magnitude of the approximation error . Let us use GHCi yet once again:

*Main Data.Ratio> let error = (x5^3 - 10) / (3 * x5^2)
*Main Data.Ratio> error
1670089160826306272530773923851043922672595525468316978
5941152245094153072382174540074985393 % 272741620880842
8590518540423731512512814773109630109912907276686930689
12248623218095571481624481
*Main Data.Ratio> 1670089160826306272530773923851043922
6725955254683169785941152245094153072382174540074985393
 / 2727416208808428590518540423731512512814773109630109
91290727668693068912248623218095571481624481
6.123338108179484e-5

This is still quite a huge error, I would say.

__________

Bonus Experiment

Let us go crazy and attempt to compute a rational approximation of . We have and . We expect the rational numbers to quickly become ratios of astronomically long integers. Therefore, we will not be showing any rational approximations on GHCi. Here is something scary:

*Main Data.Ratio> let x3 = (roots 100 100) !! 3
*Main Data.Ratio> let error = (x3^100 - 100) / (100 * x3^99)
*Main Data.Ratio> let error_float = fromIntegral (numerator error) 
                                  / fromIntegral (denominator error)
*Main Data.Ratio> error_float

After some 20 minutes without an output, I decided to abort this experiment. If you want to show x3, you can, but you will see thousands of digits. Thousands! You have been warned.

Let be an approximation of . Hence, we have that  is the geometric mean of and . If we replace the geometric mean of these two numbers by their arithmetic mean, we hopefully obtain a “better” approximation , as follows

If all goes well, then will be “sufficiently close” to for a “sufficiently large” integer . This algorithm is the famous Heron’s method, devised by Heron of Alexandria some 2000 years ago. Some call it the Babylonian method, which suggests that the algorithm may be some 4000 years old. It’s not merely vintage, it’s antique.

Nonetheless, this is not a post on archaeology. We don’t want to sit around and talk about Heron’s method, we want to actually implement it. What programming language should we use? Let us use Haskell. How are we going to represent and the sequence of approximations ?

__________

Heron’s method using floating-point arithmetic

Though millions of people have already implemented Heron’s method using floating-point arithmetic (I implemented it in C in late 2000), one more implementation would do no harm:

y :: Floating a => a
y = 2

-- define Heron map
g :: Floating a => a -> a
g x = 0.5 * (x + y/x)

-- initial approximation
x0:: Floating a => a
x0 = 1

-- list of approximations
xs :: Floating a => [a]
xs = iterate g x0

Let us load this script into GHCi. Here’s a GHCi session:

*Main> take 6 xs
[1.0,1.5,1.4166666666666665,1.4142156862745097,
 1.4142135623746899,1.414213562373095]
*Main> let x5 = xs !! 5
*Main> x5**2 - 2
-4.440892098500626e-16

We thus have the following approximation after five iterations

__________

Heron’s method using rational arithmetic

Let us now use rational numbers of arbitrary precision. The following script uses the Data.Ratio library:

import Data.Ratio

y :: Rational
y = 2

-- define Heron map
g :: Rational -> Rational
g x = 0.5 * (x + y/x)

-- initial approximation
x0:: Rational
x0 = 1 % 1

-- list of approximations
xs :: [Rational]
xs = iterate g x0

Let us load this script into GHCi. Here’s a GHCi session:

*Main> take 6 xs
[1 % 1,3 % 2,17 % 12,577 % 408,665857 % 470832,
 886731088897 % 627013566048]
*Main> let x5 = xs !! 5
*Main> x5*x5 - 2
1 % 393146012008229658338304
*Main> 1 / 393146012008229658338304
2.5435842395854372e-24

Note that the error is eight orders of magnitude smaller than in the previous case (where we used finite-precision floating-point numbers). We pay for more accuracy with more computation. Thus, after five iterations, we have the following rational approximation

This is probably not news to you, my dear reader. Last April, I did, in fact, implement Heron’s method under the disguise of Newton’s method (see the appendix).

__________

Heron’s method using integer arithmetic

A rational number  can be expressed as a fraction , where and are integers and . Note that can (obviously?) be represented as a pair of integers . Since Haskell has arbitrary-precision integers, we can implement Heron’s method using pairs of arbitrary-precision integers to represent and the sequence of approximations .

Let and . The recurrence relation

can thus be rewritten in the following form

We then have two coupled recurrence relations

where , ,  and are integers. Here is a Haskell script that implements this vector recurrence relation:

a, b :: Integer
a = 2
b = 1

-- define Heron map
g :: (Integer,Integer) -> (Integer,Integer)
g (p,q) = (b * p^2 + a * q^2, 2 * b * p * q)

-- initial approximation
p0, q0 :: Integer
p0 = 1
q0 = 1

-- list of approximations
xs :: [(Integer,Integer)]
xs = iterate g (p0,q0)

Let us load this script into GHCi. Here’s a GHCi session:

*Main> take 6 xs
[(1,1),(3,2),(17,12),(577,408),(665857,470832),
 (886731088897,627013566048)]
*Main> let x5 = xs !! 5
*Main> x5
(886731088897,627013566048)

Hence, after five iterations, we obtain the very same rational approximation we obtained in the previous case (where we used rational numbers of arbitrary-precision)

It should be noted, however, that this implementation using arbitrary-precision integers is different in one very important aspect from the one using arbitrary-precision rational numbers. What is the difference? Let GHCi answer this question:

*Main> import Data.Ratio
*Main Data.Ratio> 2 % 2
1 % 1
*Main Data.Ratio> 4 % 4
1 % 1
*Main Data.Ratio> 3 % 6
1 % 2

The main difference between the two implementations is that when we use arbitrary-precision rational numbers, there is automatic reduction of every fraction to its irreducible form, which requires computing the greatest common divisor (GCD) of the numerator and denominator of each rational number at each iteration. This has a computational cost, of course, but it also has the benefit of avoiding reducible fractions (which are inefficient representations of rational numbers). When we use pairs of arbitrary-precision integers, we do not compute the GCD (although we could, if we wanted to), but we may have to work with pairs of unnecessarily long integers (that correspond to reducible fractions), which is potentially dangerous.

For example, suppose that we replace the first three lines in the script above with the following three lines:

a, b :: Integer
a = 2000
b = 1000

We run the script again. Here’s yet another GHCi session:

 
*Main> take 6 xs
[(1,1),(3000,2000),(17000000000,12000000000),
(577000000000000000000000,408000000000000000000000),
(665857000000000000000000000000000000000000000000000,
470832000000000000000000000000000000000000000000000),
(8867310888970000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000,
62701356604800000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000)]
*Main> let x5 = xs !! 5
*Main> x5
(8867310888970000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000,
62701356604800000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000)

Frankly, I am shocked! Multiplying and by results in and being multiplied by . I may have made a mistake counting all those zeros, but it’s a lot more zeros than I expected.

__________

Appendix: from Heron to Newton

Interestingly, Heron’s method can be derived from the no less famous Newton’s method. Let us introduce a function , defined by . By construction, the zeros of function are and . Recall that Newton’s method uses the recurrence relation , where function is defined by

where is the first derivative of . Hence, we obtain the following

which is the recurrence relation used in Heron’s method.

The whole modern world has divided itself into Conservatives and Progressives. The business of Progressives is to go on making mistakes. The business of the Conservatives is to prevent the mistakes being corrected. Even when the revolutionist might himself repent of his revolution, the traditionalist is already defending it as part of his tradition. Thus we have the two great types—the advanced person who rushes us into ruin, and the retrospective person who admires the ruins. He admires them especially by moonlight, not to say moonshine. Each new blunder of the progressive or prig becomes instantly a legend of immemorial antiquity for the snob. This is called the balance, or mutual check, in our Constitution.

This passage is taken from an article entitled The Blunders of Our Parties that was published on April 19, 1924 in the now-defunct The Illustrated London News.

__________

Source:

Gilbert Keith Chesterton, The Collected Works of G. K. Chesterton, Volume XXXIII: The Illustrated London News 1923-1925, Ignatius Press, San Francisco, 1990.