The Blue-Eyed Logician

14Feb/120

Learn over fifty percent of Swedish vocabulary in under one day

This is a word frequency list based on a Swedish corpus containing around 45 million words collected from publicly available sources by a web crawler. The corpus I used was collected and analysed by Hans Christensen. Additional information, as well as corpora for many languages, can be found on his site (HC Corpora).

Theoretically, to understand over half of the words used in Swedish you need not learn more than 135 words. If you wish to learn an additional 5% of Swedish vocabulary you need to cram around 100 additional words. How much cramming to gain knowledge of 70% of the vocabulary? Well, another 1000 words.

Rank	Cumulative %	Swedish	English	Notes
1	2,82	och	and
2	5,4	att	to
3	7,87	det	it, that
4	10,14	i	in
5	12,11	på	on
6	14,09	är	is, are
7	15,88	jag	I
8	17,55	en	one, indefinite article
9	19,19	som	who, that
10	20,4	med	with
11	21,6	för	for
12	22,75	inte	not
13	23,84	har	have	ha: present
14	24,79	till	to, till
15	25,64	av	of, by
16	26,47	om	about
17	27,29	så	so
18	28,07	den	it, that
19	28,78	men	but
20	29,45	de	they
21	30,09	ett	one, indefinite article
22	30,7	vi	we
23	31,29	man	man, one
24	31,83	var	where, every
25	32,32	nu	now
26	32,78	kan	may
27	33,22	ska	will, shall
28	33,62	han	he
29	34	när	when
30	34,38	sig	yourself, oneself
31	34,74	du	you
32	35,09	mig	me, myself
33	35,39	från	from
34	35,68	eller	or
35	35,95	vad	what, how
36	36,22	bara	just, merely
37	36,48	här	here
38	36,74	då	when, then
39	36,99	lite	little
40	37,24	vara	product, ware, be
41	37,49	hon	she
42	37,74	bra	good
43	37,99	där	there
44	38,23	alla	everybody, all
45	38,47	kommer	come	komma: present
46	38,71	får	may, sheep
47	38,94	hur	how
48	39,17	blir	become, get	bli: present
49	39,39	ju	"the", of course
50	39,61	ut	out
51	39,83	min	my, facial expression
52	40,04	efter	after
53	40,25	hade	had	ha: imperfect
54	40,46	skulle	would, should
55	40,66	ha	have
56	40,86	vill	will, want	vilja: present
57	41,07	mycket	much, very
58	41,26	också	also
59	41,45	upp	up
60	41,64	in	into, in
61	41,81	än	still, than
62	41,99	idag	today
63	42,16	säger	say	säga: present
64	42,33	finns	exist, there is	finnas: present
65	42,49	mer	more
66	42,66	få	get, must, may, few
67	42,82	år	year
68	42,97	över	over
69	43,12	andra	second, others
70	43,27	bli	become, get
71	43,42	något	something, anything
72	43,57	mot	toward, against, versus
73	43,71	går	go, walk	gå: present
74	43,85	under	below, under, miracle	wonder
75	43,99	någon	someone, anyone
76	44,13	sin	its, one's
77	44,27	allt	everything, all
78	44,41	gör	do, make	göra: present
79	44,55	fick	get, must, may, few	få: imperfect
80	44,68	måste	must, have to
81	44,81	kanske	possibly, perhaps
82	44,94	göra	do, make
83	45,07	ta	take, choose, persist
84	45,2	många	many
85	45,32	sedan	then, since
86	45,45	även	even, also
87	45,57	helt	absolutely, completely, whole
88	45,7	blev	became, got, turned into	bli: imperfect
89	45,82	se	see
90	45,94	utan	without
91	46,06	två	two
92	46,17	vid	by, next to, wide
93	46,29	detta	that, this
94	46,41	ja	yes
95	46,52	varit	product, ware, be	vara: supine
96	46,63	hela	complete, whole	hel: plural
97	46,74	vet	know	veta: present
98	46,85	dag	day
99	46,95	sen	then, late
100	47,05	ser	see	se: preesens
101	47,16	igen	again
102	47,26	nya	new	ny: plural
103	47,36	ni	you	y'all
104	47,46	gå	go, walk
105	47,56	just	just, quite recently
106	47,66	oss	us, ourselves
107	47,75	några	some, any, a few
108	47,85	väl	well, accurately
109	47,95	tror	think, believe	tro: present
110	48,04	ingen	no
111	48,14	själv	self, oneself
112	48,23	kom	came	komma: imperfect
113	48,32	första	first
114	48,42	dem	them
115	48,51	nog	probably, surely, enough
116	48,6	mitt	my, middle, center
117	48,68	tycker	think, fancy	tycka: present
118	48,77	dig	you	thee
119	48,85	fram	forward, in front
120	48,93	o	o
121	49,01	sina	one's, their
122	49,09	hem	home
123	49,17	aldrig	never
124	49,24	rätt	right, correct
125	49,32	Sverige	Sweden
126	49,39	tar	take, choose, persist	ta: present
127	49,47	innan	before
128	49,54	riktigt	quite, correctly
129	49,62	gick	go, walk	gå: imperfekct
130	49,69	tack	thanks
131	49,76	annat	other, else
132	49,83	samma	same
133	49,91	varför	why
134	49,98	bättre	better
135	50,05	dom	they, conviction
136	50,12	del	share, part
137	50,18	snart	soon
138	50,25	fått	get, must, may, few	få: supine
139	50,32	alltid	always, forever
140	50,39	ändå	still, nevertheless
141	50,46	mina	my
142	50,52	tid	time, period
143	50,59	åt	to
144	50,66	ner	down
145	50,72	gång	passage, path
146	50,79	denna	this, that
147	50,85	känns	know, sense	basic form: känna
148	50,91	redan	already
149	50,98	genom	through, across
150	51,04	sitt	one's, their

Filed under: Linguistics, Swedish No Comments

30Jan/120

A Geometric Probability Problem

Michael Lugo posted an interesting geometric probability problem on his blog God Plays Dice:

Here's a cute problem (from Robert M. Young, Excursions in Calculus, p. 244): "What is the average straight line distance between two points on a sphere of radius 1?"

Solution

You can find Michael's solution here. My approach was a bit different so here's how I did it.

Geometric Probability

Without loss of generality we can fix one of the points to be (1,0,0). Take the fixed point as P and the centre of the sphere as O. To find the distance between P and a general point Q on the surface we will use the Pythagorean theorem. Let angle PÔQ = θ. The distance PQ is given by $\sqrt{(1-\cos \theta)^2+\sin^2 \theta} = 2 \sin \frac{\theta}{2}.$

The distance PQ is the same, $2 \sin \frac{\theta}{2}$ , for all points that are situated on a circle at the angle θ. The circle has radius sin θ.

Hence the average distance between two points is $\frac{1}{4 \pi} \int^{\pi}_0 ( 2 \pi \sin \theta) ( 2 \sin \frac{\theta}{2} ) d \theta = \frac{4}{3}.$

A sphere of radius r

How about the average straight line distance between two points on a sphere of radius n.

The distance PQ is $2 r \sin \frac{\theta}{2}$ and is the same for all points on a circle at the angle θ. The circle has radius r sin θ. Hence the average distance between two points is $\frac{4 r}{3}.$

Here's a question for you!

What is the average straight line distance between two points on an unit n-sphere?

Tagged as: n-sphere, unit sphere No Comments

27Jan/120

How to Draw the Tesseract using Sets

The tesseract is to the cube as the cube is to the square, in other words it is the four-dimensional hypercube. The topic for today is how would you go about drawing (a projection of) it on a piece of paper. Of course the tesseract can be constructed in a number of ways, but in my opinion this method is particularly good for deepening one's understanding of higher dimensions.

Setting the Scene

To begin with, let's define what a mathematician means by a set. Basically, a set is a collection of objects. These objects are known as the elements of the set and may be anything from people to two digit squares, as long as they are well defined. This means that it must be always possible to decide from the definition whether or not an object belong to the set. Another thing one needs to know is the definition of a subset. In essence, if every element of a set A also belongs to a set B, then A is a subset of B. A mathematician would write A ⊆ B (A is contained in B).

For example, the subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, and ∅. The symbol ∅ stands for the empty set.

Now let's take a look what would happen if we took all subsets of the set {1,2,3} and connected each subset with an edge to a subset that has one element less if one is a subset of the other.

Hooray, it looks like a cube!

Easy as 0, 1, 2, 3...

Let's try out the same thing with sets {1,2}, {1}, and ∅.

The subset graph of the set {1,2} turns out to be a square.

The subset graph of the set {1} looks like a line.

Needless to add, the subset graph of the empty set is a single point.

Hyperleap

Now then, to draw a projection of a tesseract – you guessed it – take the set {1,2,3,4} and draw a subset graph of it. Ladies and gentlemen, I present to you a two-dimensional projection of the tesseract.

More on the Subject of Four-dimensional Space

Tagged as: 4d, four-dimensional, subset graph, tesseract No Comments

24Jan/122

The Blue-Eyed Islanders

There is an island of 1000 people, 100 of whom have blue eyes, and 900 of whom have brown. There are no mirrors, nor any other reflective surfaces on the island, and the local religion forbids all discussion of eye-colour. Furthermore, anyone who discovers their own eye-colour must commit suicide that same day.

One day, an explorer lands on the island, and is invited to speak before the whole population. The explorer, unaware of the local customs, commits a faux pas stating "How pleasant it’s to see another pair of blue-eyes, after all these months at sea". The explorer doesn't direct his words to anyone in particular.

The question is: What happens next?

Assume that everyone on the island follows their religion unerringly and that all islanders are hyper logical; if there is some way by which someone can deduce their eye-colour, they will do so instantly.

Well, what happens?

To get an idea of how the puzzle works let’s start by first considering a much simpler case where there exists only one blue-eyed islander, call him Jack.

Jack knows from the explorer’s words that there exists at least one blue-eyed islander. Since he can see none, Jack correctly concludes that he must be the blue-eyed islander and commits suicide on the first day.

Now suppose that instead of one there are two blue-eyed islanders, Johnny and Hugh. Johnny can see Hugh and so knows that there exists at least one blue-eyed islander. However, on the second day, Johnny sees that Hugh hasn’t committed suicide, and therefor deduces that Hugh must also see a blue-eyed islander. Since Johnny can see only one, he concludes that he must be the blue-eyed islander that Hugh sees. They both commit suicide on the second day.

What about the brown-eyed islanders? After realizing what has happened, they will all commit suicide on the next day (assuming they knew that all islanders have either brown of blue eyes).

So the general statement is: If there are blue-eyed islanders, they will all commit suicide on the n^th day. And all brown-eyed islanders will commit suicide on the (n+1)^th day.

The answer to the original puzzle therefore is: The blue-eyed islanders will commit suicide on the 100^th day and the brown-eyed will do the same on the 101^st day.

Proof by Induction

Proof by induction consists of two simple steps.

Step 1. Showing that the statement holds when n is equal to the lowest value in question.
Step 2. Showing that if the statement holds for some n, then the statement also holds when n+1 is substituted for n.

We've already worked out the first step. In the case that there is only one blue-eyed islander he will commit suicide on the first day.

Now then, if we suppose that n is larger than 1. Each blue-eyed islander will reason along the lines: “If I don't have blue eyes, there will only be n-1 blue-eyed islanders, and they will all commit suicide n-1 days after the explorer’s blunder”. When n-1 days pass no-one commits suicide as none of the blue-eyed islanders yet has proof for themselves being blue-eyed. However, after nobody commits suicide on the (n-1)^th day, each blue-eyed islanders has to conclude that they themselves must be blue-eyed, and so they will all commit suicide on the n^th day.

Quod erat demonstrandum.

But... – That can’t be right!

There exists an argument against the solution I’ve offered. The argument states that, although the explanation works in the case that there exists only one blue-eyed islander, it will not do so when more than one blue-eyed islander are present because the explorer doesn’t tell them anything they don’t already know. That is to say, everyone on the island already knows that there exists at least one blue-eyed islander.

However this is not true! The explorer introduces new information.

To understand this, first consider the following example. There is a group of students sitting in a class room; each one may individually notice that the teacher isn’t wearing any trousers, but they don’t know whether the others have also noticed. Now then, if the teacher mentions the fact the information becomes common knowledge, also known as second-order knowledge. In other words, everyone in the class room knows that everyone knows that the teacher isn’t wearing trousers.

So in a more general sort of way: If everyone in a group of people knows X, then X is said to be first-order knowledge. If everyone knows that everyone knows X, this is to say if everyone knows that X is first-order knowledge, then X is considered to be second-order knowledge. Generally, X is (n+1)^th-order knowledge, if everyone knows that it’s n^th-order knowledge.

When there exists only one blue-eyed islander, the explorer directly increases the stock of first-order knowledge.

When there are two blue-eyed islanders, Johnny and Hugh, everyone knows that there exists at least one blue-eyed islander, and so much is first order knowledge. However, it isn’t second-order knowledge until the explorer speaks. This is because Johnny doesn’t know if Hugh knows whether there exist any blue-eyed islanders, and likewise.

Too long, didn’t read

The blue-eyed islanders commit suicide on the 100^th day and the brown-eyed islanders commit suicide on the 101^st day.

Too short, I want to read more

Randall has written a very carefully worded version of the problem, which can be found here (xkcd.com/blue_eyes.html).

Tagged as: induction 2 Comments

As I Was Saying...

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The Blue-Eyed Logician No, no, you're not thinking; you're just being logical. ~Niels Bohr