Mega Shark vs Giant Octopus

The Scifi channel have a habit of showing some fairly awful movies and Midori has a habit of recording them - tedious tromps around the woods looking for a Sasquatch that never comes into shot, ghosts of dinosaurs attacking oil drillers in Alaska. Last night she excelled herself. I was too tired to do anything but watch and 20 minutes in it played its trump card. A scene of such ball-shattering awsomnitude that the clip has been watched more than 750,000 times on youtube.

Actually that clip has only been watched 60,000 times but the other clip replaced the movie's soundtrack with a song for no apparent reason so I linked to this version.

Think of how many consecutive failures in quality control had to happen for that to appear on my TV last night.

More importantly, someone was paid to write that scene and in fact the whole film and if that wasn't you then you are losing the game of life!

Computing combinatorials

I was commenting on this fun post about efficiently computing n-choose-c but my attempt to post some code seems to have been eaten by the comment system (I suspect use of square and angle brackets upset it), so I'll post something here instead.

Here's why c always takes integer values and more generally why (k + 1) * (k + 2) * ... * (k + n) is always divisible by factorial(n). There are three ways to show this.

Apologies for the crappy maths typesetting, I should really figure out how to do this nicely.

Proof by hand-waving.

First off, something hand-wavy. One of (k + 1)*(k + 2) is even so 2 divides into that product. One of (k + 1)*(k + 2)*(k + 3) is threeven (not a real word!) so 3 divides into that product. 4 is not so simple. One of (k + 1)*(k + 2)*(k + 3)*(k + 4) is fourven so 4 divides into that product - however maybe the fourven number was also the even number we used to cancel the 2 earlier on so it only has one 2 left for division but if that's the case then since one of the remaining 3 numbers is also even and you can get a 2 from that to make up the 4. 5 works just like 2 and 3. 6 contends with both 2 and 3 but you can resolve that too. This gets messy quickly but essentially, for any prime p, every time you come to the next multiple of p in the divisor you will have already added at least one multiple of p in the k+i product and by the time you hit a multiple of p² you will have already hit a corresponding one in the k+i product too, so you'll still have enough ps in the product overall to cancel all the ps in the divisor. Still a bit hand-wavy but it could be made rigorous, however there is another proof that's easily rigorous but less insightful.

Proof by counting prime factors.

There is a well known formula for finding the power of p in factorial(n). It's

sum(i=1..infinity, [n/pⁱ])

where [] means take the integer part, throwing away any fractions. So if p = 5, and n=30 then this is

[30/5] + [30/25} + [30/125] + [30/625] + ... = 6 + 1 + 0 + 0 + ...

It's all zeros after that, so the end result is 7. So 5⁷ divides factorial(30) and 7 is the highest power of 5 that divides it.

So rewriting the k + i product as factorial(k + n)/factorial(k) we now know that for any prime p, it contains exactly

sum(i=1..infinity, [(n+k)/pⁱ]) - sum(i=1..infinity, [(k)/pⁱ])

powers of p. If this is greater than

sum(i=1..infinity, [(k)/pⁱ])

then we know that factorial(n) divides factorial(k + n)/factorial(k).

Well, [x + y] >= [x] + [y] so

[(n+k)/pⁱ] >= [(n)/pⁱ] + [(k)/pⁱ]

now bring one term over to the left hand side to get

[(n+k)/pⁱ] - [(n)/pⁱ] >= [(k)/pⁱ]

Now sum over i=1..infinity and you get exactly what we needed to show. So for any prime p, there are at least as many powers of p in the k+i product as there are in factorial(n), all the ps in the divisor cancel out and the divisor divides in evenly.

Proof by "it just is".

The final and least insightful way is to note that factorial(n+k)/(factorial(k) * factorial(n)) is n+k choose n and so must be an integer

Code

Here's a python snippet that checks that the division is always even with an assert.

#! /usr/bin/python2.4 import sys def comb(n, k): # k = max(k, n - k) # would make it go faster c = 1 for i in range(k + 1, n + 1): c *= i assert c % (i - k) == 0 c /= i - k return c n = int(sys.argv[1]) for i in range(0, n + 1): print "comb(%d, %d) = %d" % (n, i, comb(n, i))

Give it an n and it will print out comb(n, i) for all i. The assert never fires for me.

As it loops, c gets the values comb(k + 1, 1), comb(k + 2, 2), ..., comb(n, n - k), all of which are integers (and remember that comb(n, n - k) = comb(n, k) so c ends up with the desired value.)

If your lanugage does tail-call optimisation, you could define

comb(n, 0) = 1

comb(n, i) = comb(n - 1, i - 1) * n / (n - k)

and be done. With memoisation you could avoid lots of repeated calculations.

Eddie Hobbs and the car scrappage scheme.

Update:Eddie's reply is at the bottom

Lots of good things in this post but I'm surprised to see positive mention of the car scrappage scheme - "Focus on energy efficiency, green energy and a cash- for- clunkers scheme for low emission cars.". It is neither green nor good for the economy.

We don't have a car industry, we have a car sales force which funnels money straight out of this economy with a small bit staying locally. Stimulate this and you stimulate the Germany and Japanese economies far more than the Irish.

Scrapping cars that have plenty of life left just to have them replaced with something that is maybe 10% less emitting is going to cause a net increase when you factor in the manufacture of the car. Never mind that fact that some will upgrade their cars to bigger, more-polluting models.

Given all that I'm curious why Eddie thinks it's a good idea. he makes a virtue of not being beholden to vested interests so it's unlikely he's saying it for his mates in the motor trade. There are ways to spend the scrappage money that would produce a greater economic stimulus in this country and produce greater environmental impact per euro too.

I sent Eddia a link to this and he replied quite quickly which is very good of him. He said

Yeah, you’re right to pick me up on it. Technically it’s the wrong thing to do since we don’t have a car manufacturing industry but that is softened somewhat by the emphasis on Band A and B. It is, of course, front-end loading cars sales but the x factor is the sight of 2010 new plates on the road, car rooms with consumers in them and a lift in consumer morale. Very hard to measure I know and a subjective call but morale is pretty important to arrest the spending delay factor caused by deflation. Much will depend on the relative allocation to energy, efficiency etc where we’ll get a proper return so let’s see at teatime.

to which I say, hmmm... maybe. One small snag here is that these newly confident consumers just blew all their money on a new car so although they feel good, they're less likely to do anything as a result...

in reference to: Ten things Brian Linehan must get right (view on Google Sidewiki)

Independent nonsense: negligent doctors

As usual, the Indo provides some useless statistics, this time on malpractice cases. What they tell us is:

A senior house officer -- a grade of junior doctor -- was most likely to be involved (74pc) as against 14pc for registrars and 8pc for consultants.

What they don't tell us is what percentage of patients are dealt with by each type of doctor. So if only 50% of patients were dealt with by junior doctors but they were involved in 74% of malpractice cases then that's a black mark against junior doctors and would be something worth knowing. Elsewhere in the article it's stated that they deal with "the majority" of cases. Without these other figures, the statistics are not just useless, they're possibly quite misleading.

in reference to: http://www.independent.ie/national-news/ae-blunders-responsible-for-one-in-seven-medical-claims-1951185.html (view on Google Sidewiki)

The "open" web awards.

They want me to sign in with a twitter or facebook account, not any of the many other sign-in systems out there. Not very open...

in reference to: Mashable's 2009 Open Web Awards - Votes (view on Google Sidewiki)

Sichuan House - Spicy chilli chicken

Mostly a note to myself. I got in a quick trip to Sichuan House on Sunday with Riona. I ordered Spicy Chilli Chicken (山椒鳥 I think). It was quite tasty but the sauce was not very exciting. I was expecting a different dish actually but now I have no idea what that one is called. This was cubes of chicken but looked a lot like an Irish-style dish (I was assured it wasn't and in fairness it didn't taste like it was, just looked like it).

Looking up Kanji/Hanzi quickly on zhongwen.com

Update: I've turned this into a Google App Engine app - http://zhongwenlookup.appspot.com/

zhongwen.com is an excellent dictionary that shows the decomposition of all Chinese characters and links to the entry for each component. It's the online version of Rick Harbaugh's "Chinese Characters: A Genealogy and Dictionary". I've spent plenty of time looking things up in it while learning Kanji (both online and in my own copy).

The one problem I have is that it's not possible to just paste a character into the search box and get to the entry. You can only search by radical, pronunciation and radical none of which lead directly to a single character. Tonight I finally got bored with that and was about to mail the author to see if he would add a search by character feature. While composing the mail I started poking a bit further and realised that the URL scheme for the site is based on the BIG5 encoding of Chinese characters and so I could just do it myself.

Here's a little bash script that takes characters as arguments and gives you back the zhongwen.com URLs and pass them to a command called browser which, for me, opens them in Firefox. Yes it's ugly. I tried to convert it to Perl but ran into encoding problems that I couldn't be bothered solving.

#! /bin/bash -e if [ -z "$@" ]; then echo Usage: ./harb char1 char2 ... echo echo Will print the zhongwen.com URL for the characters and pass them to browser fi while [ $# -gt 0 ]; do url=$( echo $1| iconv -t big5| \ perl -lne '@c=unpack("CC", $_); print "http://zhongwen.com/d/$c[0]/x$c[1].htm"' ) echo $url browser "$url" shift done

Invoke it as harb 宅 煉 to get the URLs for those 2 characters. Characters must passed as separate arguments (e.g. space separated).

Here's my browser script, while I'm at it

#! /bin/bash if firefox -remote 'ping()'; then firefox -remote "openURL($1,new-tab)" else echo no ping firefox "$1" & fi

1889 Kanji Characters

It's been slow going for a while now but I'm 153 away from finishing the book. I would have said that's just a few weeks away but it seems that every week now something comes up and I end up add only 10 or so kanji. The current chapter is a mixed bag of kanji that don't really fit into the scheme at all and is supposedly the most difficult chapter. I hope I can finish the whole thing before December. I'd like to be able to say I cracked it in less than a year. At this point there is no way I'm giving up!

Some stats - I've done 22451 repetitions, that means I've written that many kanji. So on average I've written each one about 12 times which is not so bad really. I have an 88.2% success rate on "mature cards" which is also OK but I find recently that cards from 6 months ago are coming up and I'm flummoxed. I think Anki is a bit too aggressive in increasing the interval between repetitions. I could tune that but at this point it's mostly working and I don't want to screw with it. I've been doing this for for 319 days. On 160 of them I added new kanji, on 159 of them I added no new kanji.

Finally a graph:

Hyperbinary numbers

Update: The proof is over here.

I've been following along the a discussion on The Math Less Travelled about hyperbinary numbers (here, here and here). I'm posting this here because it's a bit too big to fit in the margins of the other blog (although I have found no remarkable proof). Since I have very little time to work on this (and have already given myself a few nights with insufficient sleep!) I thought I'd do a brain-dump of what I did in case someone else can finish it. Apologies for the awful formatting, I don't post much maths so have no nice styles for it and no time to find them.

Quick summary

h(n) is the number of ways of writing n as a sum of powers of 2 where each power of 2 can only occur 0, 1 or 2 times. See the articles for more details but the key facts are (I'll refer to them by these numbers):

1. h(2n) = h(n) + h(n - 1)
2. h(2n + 1) = h(n)

From those you can get several relationships, like h(k * 2^n - 1) = h(k-1). Other interesting stuff is in those posts. The main question is how to find the inverse of h(n). That is, given k, what values of n have h(n) = k. It turns out that each even solution to that leads to an infinite number of odd solutions and all odd solutions can be traced back to an even solution. So the question then is what are the even solutions - the "primary occurrences"? My contribution so far has been to notice that there seem to be φ(k) even solutions for a given k although I am still stumped on how to prove that.

Another useful things is that h(2n + 1) = h(n) means that all the odd positions are just repeats of earlier values. Applying fact 1 to the right hand side of fact 2 gets you h(2n) = h(2n + 1) + h(2n - 1). So all the even positions are just the sum of the values that come immediately before and after.

Finally, h(2^n - 1) = 1 and h(2^n) = n + 1 and from that you can get h(2^n + 1) = n

New stuff

The last bits got me thinking about h(2^n + k). It's fairly easy to figure that out for more and more values of k. For h(2^n + 2) just apply fact 1 to get that as the sum of 2 things we already know:

h(2^n + 2) = h(2^(n-1) + 1) + h(2^(n-1)) = (n-1) + (n-1) + 1 # note this line for discussion below. = 2n - 1.

Looking at the table of values primary occurrences (here) you can see that all of the odd entries do indeed have 2^n + 2. In fact after a bit, it's the lowest value for each odd entry.

What about 1? 2*1 - 1 = 1 but if n = 1 the line commented above is not valid because h(2^n + 1) = n is only valid for n > 1.

To get a bit further, consider h(2n) = h(2n + 1) + h(2n - 1) from above. Moving things around a little that becomes h(2n + 1) = h(2n) - h(2n - 1). So we can get the next odd entry by subtracting the previous odd entry from the previous even entry. So h(2^n + 3) = n - 1. Since 2^n + 3 is odd this doesn't tell us anything about primary occurrences but it's still useful as all odd entries are duplicated later on.

Now we can get h(2^n + 4) = h(2^(n-1) + 2) + h(2^(n-1) + 1) and again we know what they are = (2n - 1) + n = 3n - 1. And low and behold, all the entries (except 2) in the primary occurrences table of the form 3n - 1, include have 2^n + 4 (5 -> 12, 8 -> 20, 11 -> 36, ...). What about 2? 2 = 3*1 -1 but if n=1 then the relation above breaks because the 2n - 1 rule is only valid for n > 1.

Continuing on in the same vein you get the first 100:

h(2^n + 0) = 1n + 1 h(2^n + 2) = 2n - 1 h(2^n + 4) = 3n - 4 h(2^n + 6) = 3n - 5 h(2^n + 8) = 4n - 9 h(2^n + 10) = 5n - 12 h(2^n + 12) = 5n - 13 h(2^n + 14) = 4n - 11 h(2^n + 16) = 5n - 16 h(2^n + 18) = 7n - 23 h(2^n + 20) = 8n - 27 h(2^n + 22) = 7n - 24 h(2^n + 24) = 7n - 25 h(2^n + 26) = 8n - 29 h(2^n + 28) = 7n - 26 h(2^n + 30) = 5n - 19 h(2^n + 32) = 6n - 25 h(2^n + 34) = 9n - 38 h(2^n + 36) = 11n - 47 h(2^n + 38) = 10n - 43 h(2^n + 40) = 11n - 48 h(2^n + 42) = 13n - 57 h(2^n + 44) = 12n - 53 h(2^n + 46) = 9n - 40 h(2^n + 48) = 9n - 41 h(2^n + 50) = 12n - 55 h(2^n + 52) = 13n - 60 h(2^n + 54) = 11n - 51 h(2^n + 56) = 10n - 47 h(2^n + 58) = 11n - 52 h(2^n + 60) = 9n - 43 h(2^n + 62) = 6n - 29 h(2^n + 64) = 7n - 36 h(2^n + 66) = 11n - 57 h(2^n + 68) = 14n - 73 h(2^n + 70) = 13n - 68 h(2^n + 72) = 15n - 79 h(2^n + 74) = 18n - 95 h(2^n + 76) = 17n - 90 h(2^n + 78) = 13n - 69 h(2^n + 80) = 14n - 75 h(2^n + 82) = 19n - 102 h(2^n + 84) = 21n - 113 h(2^n + 86) = 18n - 97 h(2^n + 88) = 17n - 92 h(2^n + 90) = 19n - 103 h(2^n + 92) = 16n - 87 h(2^n + 94) = 11n - 60 h(2^n + 96) = 11n - 61 h(2^n + 98) = 16n - 89 h(2^n + 100) = 19n - 106

With the 2^n + k only being valid for n such that 2^n > k.

Obviously calculating these by hand gets a bit boring. To write some code to do it I had to play with f(n, m) = h(2^n + m) and it breaks down into 2 cases for m even and odd.

For the odd case, let m = 2k + 1. So

f(n, m) = h(2^n + m) = h(2^n + 2k + 1) = h(2n + 1) = h(2^n + 2k) - h(2^n + 2k - 1)

For the even case, let m = 2k. So

f(n, m) = h(2^n + m) = h(2^n + 2k) = h(2(2^(n-1) + k)) = h(2^(n-1) + k) + h(2^(n-1) + k -1) (applying fact 1) = h(2^(n-1) + m/2) + h(2^(n-1) + m/2 -1) = f(n - 1, m/2) + f(n -1, m/2 - 1)

Turning that into python and adding a couple of base cases gives the following code (it tries to also cope with 2^n - m but there's a bug somewhere). #! /usr/bin/python2.4 import sys class ANPlusB(object): def __init__(self, a, b): self.a = a self.b = b def Sub(self, n_plus): return ANPlusB(self.a, self.b + self.a * n_plus) def Eval(self, n): return self.a * n + self.b def __str__(self): if self.b < 0: s = "-" b = - self.b else: s = "+" b = self.b return "% 4sn %s % 4s" % (self.a, s, b) def __add__(self, other): return ANPlusB(self.a + other.a, self.b + other.b) def __sub__(self, other): return ANPlusB(self.a - other.a, self.b - other.b) TABLE = {-2: ANPlusB(1, -1), -1: ANPlusB(0, 1), 0: ANPlusB(1, 1), 1: ANPlusB(1, 0), } biggest = 1 smallest = -1 def F(n_plus, k): """Find the a*n +b for 2^(n + n_plus) + k .""" # Note something is broken for negaitve n_plus, going below -10 # causes infinite recursion. global smallest, biggest if k not in TABLE: if k < smallest: val = _FSmaller(k) smallest = k else: val = _FBigger(k) biggest = k TABLE[k] = val val = TABLE[k] return val.Sub(n_plus) def _FBigger(k): if k % 2: return F(0, k - 1) - F(0, k - 2) else: return F(-1, k / 2) + F(-1, (k / 2) - 1) def _FSmaller(k): if k % 2: return F(0, k + 2) - F(0, k + 1) else: return F(-1, k / 2) + F(-1, (k / 2) - 1) def main(): min, max = [int(x) for x in sys.argv[1:]] for i in range(min, max+1, 2): print "h(2^n + % 4d) = %s" % (i, F(0, i)) if __name__ == "__main__": main()

You can run it as ./hyperbinary.py 0 100 or to get the first 50 even entries. Looking at the output it seems that h(2^n + m) = h(m) * n + c. I haven't proved it formally but it should be easy enough, just show that the coefficient of n obeys the fact 1 and fact 2. I didn't find a good relation between m or h(m) and the c above (the best I can spot is that int(c/h(m)) is always log_2(m) and when m is 2^k c=k*h(m) + 1). Nor did I relate any of this to φ although at least now there appears to be some modulo arithmetic going on.

My contribution to the Lisbon debate.

This time around I have neither the time nor the energy to do very much. However I would really love to have a wind-driven rotating sign to express my opinion on the level of the debate. The kind that often stand outside shops for ice cream etc. Since that isn't going to happen I tried my hand at an animated GIF:

I'm a 35 year old man trapped in the body of...

... a 35 year old man. Basically I have no excuse or no reason to plead a special case. Well, my 2 small kids seem to sleep less than some others but not outside the ranges given in baby books. I just can't do things like I could before. I used to play a lot of soccer which made me quite fit (self-praise is no praise). When I played again a few weeks ago I was in a steaming heap after 10 minutes. Some of it is lack of exercise some is just getting older. I hardly drink any more because I just hate losing time to hangovers and I don't have to drink very much to get one anymore. It was nice be able to abuse my body and not pay the price in my teens and twenties (self-abuse is no abuse!).

What's the point? Nothing much. Just when I changing my socks I was playing with the phrase "An X trapped in the body of a Y" and I realised that if I claimed to be different on the inside and the outside I'd be kidding myself.

To vastly over-generalise and ignore changing life-circumstances, as people age they stop doing some things. I don't think they stop because they have lost the desire to do them but because they have lost the ability to do them. Of course that may then impact the desire or simply cause the focus to shift to new things which can be done but I think you don't stop wanting to do anything simply because you're older. Which probably explains the childishness of many people (myself included).

In Soviet Canada, healthcare system pays for your problems.

Their nearest neighbour makes it very hard to find a way to exclusively make fun of Canada...

Smalltalk's become() in python

Posting this here because blogger's sucky comment system won't let me post formatted code in a comment on this post. This is an implementation of Smalltalk's become() in python. It will not work on strings, int etc. but otherwise is identical

#! /usr/bin/python def become(a, b): swap(a, b, "__class__") swap(a, b, "__dict__") def swap(a, b, attr): tmp = getattr(a, attr) setattr(a, attr, getattr(b, attr)) setattr(b, attr, tmp) class Animal(object): def __init__(self, name): self.name = name class Dog(Animal): def Cry(self): print "%s says woof" % self.name class Sheep(Animal): def Cry(self): print "%s says baa" % self.name a = Dog("rover") b = Sheep("shaun") c = a a.Cry() c.Cry() become(a, b) a.Cry() c.Cry()

This prints

rover says woof rover says woof shaun says baa shaun says baa

Note how both a and c are now sheep. To do the same thing in perl you monkey around with bless and resetting all the hash entries, it's a bit uglier.

Answer to 1 of the 2 puzzles and a new puzzle.

I take Nicole's comment to mean she gives up, although Nicole, it turns out that your comment has a deep relation to this problem and raises a new question.

To find out on average how many items are left unmoved by a random permutation, you can consider every permutation and how many are left unmoved by each, add them all up and divide by the total number of permutations. To avoid mathsy notations I'm going to write a tiny program that calculates this total.

total = 0 # Check what every permutation does to every item, if it leaves it # in the same place then increase the total by 1 for p in perms: for i in 1..N: if p(i) == i: total = total + 1

Now, here's the key to this problem - swap the loops. The final answer stays the same we just end up adding things up in a different order.

total = 0 # Check what every permutation does to every item, if it leaves it # in the same place then increase the total by 1 for i in 1..N: for p in perms: if p(i) == i: total = total + 1

So now what we're doing is looping over the items and counting how many permutations leave that item in place and that's something we can work out. The number of permutations that fix a given item is (N-1)! (where ! is factorial) because when you fix 1 item, you're left with N-1 other items which you move around however you like so it's just the number of permutations of N-1 items which is (N-1)!. So now the code can be just

total = 0 # Check what every permutation does to every item, if it leaves it # in the same place then increase the total by 1 for i in 1..N: total = total + (N-1)!

which is just adding (N-1)! N times. That's N x (N-1)! which is just N!. Now we have to divide that by the total number of permutations which is also N! to get an average of 1.

You can also get to this answer by drawing a grid with all the permutations across the top and all the items down the side. You put a tick in a cell in the grid if the permutation on the top leaves the item on the left unmoved. Now count the ticks. Doing it column by column is hard but doing it row by row is easy because there will be (N-1)! ticks in each row.

So the conclusion is that if you do a perfect shuffle of a deck of cards, on average 1 card will end up in the same places as before.

So where does Nicole's comment come in? I knew it was Ralph Wiggum who said it but I couldn't remember the context turns out. I found it here. It's

  Lisa: Hey Ralph, want to come with me and Alison to play "Anagrams"?
Alison: We take proper names and rearrange the letters to form a
        description of that person.
 Ralph: My cat's breath smells like cat food.

and what we've just shown is that given a random anagram, you'd expect 1 letter on average to end up in the same place - well kind of, that's only true if there are no repeated letters. I hadn't thought of this in terms of anagrams before and that adds a nice twist in terms of repeated letters. Thanks Nicole!

So the new puzzle is - given a string written in scrabble tiles (it can have repeated letters), if you mess up the tiles and make a new string at random, how many letters do you expect to be in the same place as before?

No code has no bugs.

So the more no code you have, the more no bugs you have.

Pat, "the cope" Gallagher: dishonest or just an idiot?

In his letter to the Indo calling for a "yes" to Lisbon, Pat defends our membership of the Euro. I think there are upsides and downsides to it but for Pat even the downsides are upsides. He happily points out that

The eurozone has ensured that billions of euro have been made available to Irish banks via the operation of the European Central Bank.

and that

As a member of the eurozone, Ireland has benefited from very low interest rates, notwithstanding the very difficult economic situation that faces us.

Surely he must know that these 2 things are the root cause of our massive bubble. The IMF and even Brian Lenihan have said as much. So if he knows this, then to list them as upsides of eurozone membership is basically swearing that black is white and up is down. There is of course that odd "notwithstanding the very difficult economic situation that faces us" which seems to be an attempt to cover all angles - as if to say "yes I know this is what fucked up the country but still it was great craic at the time."

Of course maybe he truly disagrees with me and thinks that the oversupply of cheap credit was an entirely positive thing and we should be thankful for it. This would certainly be compatible with FF's policy over the last decade, right up the point where the shit hit the fan.

2 fun maths puzzles

I guess they're fun because I figured them out! If there's anyone actually reading this, post your solutions in the comments. If I get any responses I'll post solutions and if I don't, I might not.

1. On average, given a random permutation of N items, how many elements do we expect to stay put? E.g. if N=3 there are only 6 perumtations, 1 leaves all 3 in place, 3 leave 1 in place and 2 shift all of them (left or right). So that's (1*3 + 3*1) / 6 = 1.
2. On average, given a random function from 1..N->1..N which may or may not be injective or surjective (that is there may be X and Y such that f(X) = F(Y) and there maybe be a Z such that f(X) != Z for any X), how many elements do we expect to stay put - f(X)=X? E.g. if N=2 there are 4 possible functions, f(X) = X, f(X) = 1, f(X) = 2 and f(X) = 3-X. These fix 2, 1, 1 and 0 elements respectively. So on average that's (2+1+1)/4 = 1 but what about larger N?

Movie: Blood, the Last Vampire

I went to see it tonight. It was mostly OK, some good action scenes but often things were moving too quickly to really be able to appreciate it. I found myself wanting to watch Ong Bak,the super-awesome Thai movie starring Tony Jaa. It's super-awesome because there's no wire-work in it. I dislike wire fighting. If you're going to do it with wires, you may as well just do it with camera tricks or CGI and then even I could do it - not awesome at all. The stuff Tony Jaa does in Ong Bak is just fantastic and - more importantly - surprising and often funny.

Aaaaaanyway. Blood is essentially Blade but instead of Wesley Snipes the half-vampire vampire killer is a Japanese schoolgirl. What could possibly go wrong? Mostly nothing goes particularly wrong or right. Add in an unsatisfying ending and you've got "meh".

I would have been annoyed if I hadn't managed to get to it but I'll have forgotten about it in a month.

Movie: A tale of two sisters

I saw this Korean movie years ago but my trip to the cinema tonight reminded me of it because there is just a single sudden fright in the movie and everyone in the cinema leapt, popcorn flew and a few seconds later everyone was nervously giggling and apologising to the complete stranger next to them. A great movie. It turns out it was remade as The Uninvited which was out a few months ago. From the reviews I can find, they made a pig's mickey of it. Ah well.

Mouth water chicken - yummy but dangerous

After my movie trip I went to Sichuan House and ate the spiciest food I've eaten for some time. I came home and ate ice cream in the vain hope that that will neutralise some of the spicy before it makes its way out again. On the menu as "口水鸡 - steamed chicken in chili sauce" (first in the list of salads if you're looking for it), it literally translates as "mouth water chicken". It's a half a cold chicken, drenched in spicy, oily sauce with peanuts, sugar, spring onions and other stuff. It smells exactly like the smell you get in a real Sichuan restaurant - it's nice to know what I was smelling way back when. It's damn tasty and I brought half of it home on the back of the bike! Not sure if I'll eat it myself. I'm going to a BBQ tomorrow, I might bring it along for others to sample. In the meantime, I think I'll go put a pack of wet-wipes in the fridge...

P.S. Ow! Jesus! I rubbed my eye! Chili, chili, chili! Ow!

Movie: drag me to hell

I just saw Drag Me To Hell and I had a great time. It had the scariest old woman I ever saw in a movie. It had the scariest lace handkerchief I ever saw it a movie. It had lots of out-of-seat jumping. It had gumming (not a typo) and goats (also not a typo) and it made me laugh out very loud - just once, mostly it wasn't trying to be funny. I'm really glad I saw it. There's how's that for elevated expectations leading to certain disappointment?

Afterwards I ate some lovely Chinese food.

Too much syntactic sugar...

... gave Perl syntactic diabetes.

1362 Kanji Characters

1362 might not sound like significant number but that is 2/3 of the way through the 2042 characters of Remembering the Kanji. It has become a bit of a slow slog but the progress is fairly steady. I thought I'd get more done while Midori and the kids were away but so far I haven't done anything amazing. In fact 54 in the last 7 days is less than 10 per day - I was occasionally able to beat that with the family around. It's way off the pace from my first 4 weeks but I had 2 weeks of rest before that and I was excited and motivated by the discovery of a sure-fire method of learning the kanji.

It's nice having a fixed goal of 2042. This covers the standard high school kanji and some extras that are too useful to leave out or necessary for the method. It also allows me to set motivational targets. Without a fixed goal, I could set targets of doubling my kanji count and meet the pretty easily to being with. Going from 128 to 256 in about a week was possible. After that, doubling gets harder and harder and doubling 600 to 1200 took a very long time because of lack of free time. But with a fixed target, now I'm on the down slope. The next target is 3/4 (1531), then 4/5 (1633) and so on.

The first hint of trouble begins when I finish number 2002. That's when I go across the 49/50 mark. The trouble is that I also cross the 50/51 mark which makes kanji number 2002 twice as exciting, that might not seem like trouble but more is to come. Kanji 2003 only crosses 51/62 and Kanji 2009 only crosses the 60/61 mark but from there on in every kanji crosses at least 2 marks. Kanji 2041 crosses 1020 marks! All the way from 1020/1021 to 2040/2041. But the real problem is the last kanji. When I complete that I will cross an infinite number of motivational markers, presumably resulting in infinite excitement, that can't be a good thing.

To avoid danger to myself, my family and small animals near by I have already ordered the follow on volume which covers a further 1000 useful but less common kanji. I will deploy that as soon as I feel the excitement is getting too much. Mr Heisig sure is a devious man.

The real reason for bailing out Anglo Irish bank.

I like Gene Kerrigan's articles in the Sunday Indo. He had 2 recently: "Servile surrender sowed seed of doom" and "Saving economy while people go down" which deal with our governments obsession with saving Anglo Irish Bank and the others at the expense of absolutely everything else. Why this is happening is a really important question. He makes a lot of sense but he seems to think that the govt are simply mesmerised by the bankers, "deferential and submissive" just as they were in the past to the church. He makes it sound like the govt is just in awe of these guys and willing to do whatever they request - the same way some politicians seem to be in awe of the "free market" and espousers free market ideology. I actually don't think it's that simple (I doubt Gene Kerrigan does either).

Mr Kerrigan says that we should let Anglo go to the wall and let the bond holders take the hit. The argument is that they gambled, they knew the risks and they lost - there is no reason for the Irish government to cover their bet. The problem here is that when they placed their bet, they were ostensibly betting on regulated a bank in a modern, Western, open democracy. As such, on paper, their risk was low and their rates of returns would have been low too (I haven't checked the rates...). On paper, such a bank, regulated by a well run EU member, should never go bust. The problem is that there was no such regulation, worse still there may even have been collusion between the bank and the regulator. For the government to then say "sorry boys you lost, thanks for playing" raises some problems. These bond holders would certainly not be happy with that and would presumably draw even more attention to the misdeeds of the bank and the regulator, possibly even taking legal actions (I have no clue what recourse they would have) and at the very least making it very hard for AIB, BOI and the other to raise capital as they were and still are under the same crappy regulatory system.

The whole thing is a bit like cheating at poker in a gentleman's casino. Ireland sat at the poker table and won hand after hand. It won because it was cheating - it didn't regulate its banks properly. Come the end of the and by now everyone knows Ireland was cheating and Ireland realises that they know. Of course "gentlemen" don't like to bring up this sort of thing in public, it's impolite. Ireland has 2 choices:

1. it can pretend it didn't cheat, leave the casino with its winnings - in this case it will be black-balled and no one will ever play poker with Ireland again. The genetlemen might then abandon their usual decorum, call out the cheating directly and look for their money back.
2. it can say "what a splendid evening we've all had" and leave its winnings on deposit in the casino's safe, then come back and play a losing game every night until it has lost enough to repay it's cheating debt plus interest.

Gene Kerrigan is calling for #1, this government is doing for #2.

The good news is that finance is not the same as poker. There is a third option. We can do #1 and also avoid being black-balled and banned from every table if we can convincingly promise never to cheat again. We must acknowledge the wrong doing, punish the wrong doers and put in place legislation and regulators with real powers. They must not be mates of the politicians or the bankers and they must be willing to remain adversarial. No more jobs for the boys. Without this step, #1 would be extremely dangerous for all Irish banks. That still leaves us open to attempts to recoup our ill gotten gains but the worst that can happen is we end up paying them all back, just as in #2 except.

The bad news is that Fianna Fáil certainly don't have the balls to do this. They are much happier to take #2 whereby everybody remains gentlemanly, nobody loses face and the tax payer picks up the bill (to a large extent it was the tax payer who benefited but it will not be those who benefited most who end up paying back the most). I don't believe that any of the other parties have the balls to try it either or to even contemplate it, they don't want to rock the boat and they have all shown that jobs for the is what its all about.

Incidentally, I don't believe the bond holders were dummies either. I imagine they knew how poorly regulated the bank was. However knowing that the government would feel forced to save them, they could make their bets based on the imaginary state of regulation, not the real state of it.

Quick review: aldi and lidl tins of tomato

More of a note to myself than a review. I make a tomato dip that is a cheap and easy version of a friends more lovely but more labour intensive tomato dip. It's great with a cook-it-yourself ciabatta from Aldi. You just get a tin of chopped tomatoes (I have been using Aldi chunk chopped with herbs), add a teaspoon or two of pesto (I have been using Lidl's) maybe some extra olive oil if you like and a bit of salt and pepper.

So the point of this post is to note that when I made this with Lidl's "Nostia" chopped tomatoes it was nowhere near as nice. Theirs seems much more watery. I haven't tried it Roma. I doubt I will since they are 3 times the price of Aldi's and unlikely to be 3 times tastier!

Quick review: graco stadium duo double buggy

About 9 months ago, we bought this for about 170euro after having our second child and we're pretty happy with it. It's quite heavy and big but so much that Midori can't lift it into the boot of the car (it just fits cross-wise in the boot of a Prius). The kids are comfortable in it, Seán can sleep in the back section when its flattened out. There's plenty of space underneath for carry shopping etc and it's surprisingly maneuverable for it's size. I think Midori even went to town on the bus with it once but I wouldn't try that myself unless I knew the bus would be very empty.

Overall a thumbs up, especially at the price.

making dvds from video files

Lest I forget again, DeVeDe did a decent job of putting an flv file that I had downloaded onto a DVD for playback on any TV and it's apt-gettable on ubuntu.

A fun shell idiom.

To find what's in filea that isn't in fileb, I have always done

cat filea fileb fileb | sort | uniq -u

similar things give you union, intersection, symetric difference etc. Recently I shortened it to

cat filea fileb{,} | sort | uniq -u

which I thought was good fun.

Quality reporting from the Indo

From this article on ESRI models of the economy (my emphasis added):

"It would be wrong to call that a forecast. It is more of an assumption, because migration is so hard to predict."
With employment officially defined as more than one hour's work a week, the numbers on the live register could reach 500,000 on these forecasts.

1021 Kanji characters

Yay! I just got to character number 1021 - half way - in the super book Remembering the Kanji. I can write all of these characters from memory. Actually my success rate is 89.8% but that's fine, it's remaining stable as I add more characters. Striving for 100% success would be far more work for only 10% more reward. The tricky characters will eventually stick.

You can see from the progress graph that certain periods have been more productive than others - 600 in the first 4 weeks, 16 weeks for the next 400! What a difference having kids around makes. I accelerated again recently and hope to finish in about 3 months, although the last 2 weeks were pretty poor due to everyone around in my house being ill in a variety of sticky, gooey ways!

It would have been a lot harder without kanji.koohii.com for stories when I couldn't think of good ones myself and Anki for making efficient learning just a matter of sitting down every night and clicking the buttons. Also thanks to the inspiring and entertaining Khatzumoto-san at All Japanese All the Time for documenting that it can be done and that it's a worthwhile step along the road to Japanese proficiency.

Now to try get in another few characters before Seán - asleep on the couch beside me - wakes up!

Infantile coprophagia and numerical synthaesia

I watched a very interesting documentary a few years back when Google video first started. It was about a Daniel Tammet, a semi-autistic savant who learned to speak icelandic in a week, could memorise thousand of items and perform incredible calculations. He "suffered" from synthaesia - a condition where the senses interfere with each other. Some people can "see" sounds or "smell" colours etc. Daniel Tammet could see numbers, they had particular shapes and colours and it was consistent - a week later, the same number would have the same shape and colour.

So what's that got to do with infantile coprophagia? The participants shall remain nameless to protect them but a couple of weeks ago (let the record show that it was Tue March 25th 2009) my youngest child was left alone briefly with a nappy loaded full of my eldest child's poo. When my wife returned, she found that not only had he tucked in to a tasty treat and was smiling a joyous brown smile but he had decorated the floor, the toys, the books and anything else he could get to.

So while I have no reason to believe that he will be an autistic savant, he does know what number 2 tastes like...

Notes from lidl trip

Mostly for my records so I can remember to buy it again:

1. Twinner smoked peppered salamai. V nice.
2. Smoked peppered salmon. V nice. (peppered brie from Aldi, also v. nice but I know that without making notes).
3. Grafenwalder beer. Meh.

The powers of oneechan

"Oneechan" (お姉ちゃん) means "big sister" in Japanese and Ríona now understands that she is a big sister. Many things can be explained by this. Her ability to walk along the edge of a wall or to climb up something or to eat by herself are all because she is "Oneechan"

As usual though, she doesn't think this means what we think she thinks it means. Yesterday, watching Bambi, we got to the scene where he fights the dogs and saves his girlfriend. Ríona announced "お姉ちゃんですからね" - "that's because he's a big sister". Not much you can do but agree.