Michael Nielsen and others on privacy

Quote privacy surveillance governance liberalism

Nick Cammarata on chasing your last few years of intelligence alpha

Quote ai openai

OpenAI o1 solves a cipher

The prompt:

oyfjdnisdr rtqwainr acxz mynzbhhx -> Think step by step

Use the example above to decode:

oyekaijzdf aaptcg suaokybhai ouow aqht mynznvaatzacdfoulxxz

The chain of thought:

First, what is going on here?

We are given:

First, an example:

oyfjdnisdr rtqwainr acxz mynzbhhx -> Think step by step”

We can see that when oyfjdnisdr rtqwainr acxz mynzbhhx” is transformed to Think step by step”

Our task is: Use the example above to decode: oyekaijzdf aaptcg suaokybhai ouow aqht mynznvaatzacdfoulxxz”

So the first part is to figure out how oyfjdnisdr rtqwainr acxz mynzbhhx” is decoded into Think step by step”

Observation 1: The phrase Think step by step” is 18 letters (excluding spaces) . Let’s count the letters.

Similarly, let’s count the letters in oyfjdnisdr rtqwainr acxz mynzbhhx”

Original Ciphertext:

oyfjdnisdr rtqwainr acxz mynzbhhx

Plaintext:

Think step by step

First, break down the letters:

Ciphertext letters: o y f j d n i s d r r t q w a i n r a c x z m y n z b h h x

Total letters: Let’s count the letters in the ciphertext.

First word: oyfjdnisdr (10 letters)

Second word: rtqwainr (8 letters)

Third word: acxz (4 letters)

Fourth word: mynzbhhx (8 letters)

Total letters in ciphertext: 10 + 8 + 4 + 8 = 30 letters

Plaintext: Think step by step

First word: Think (5 letters)

Second word: step (4 letters)

Third word: by (2 letters)

Fourth word: step (4 letters)

Total letters in plaintext: 5 + 4 + 2 + 4 = 15 letters

Hmm.

But actually in the problem it says the example:

[Ciphertext] -> Think step by step

So is the decoding process that the ciphertext maps to Think step by step”?

Our goal is to use the example above to decode:

oyekaijzdf aaptcg suaokybhai ouow aqht mynznvaatzacdfoulxxz”

First, the cipher seems connected to the plaintext.

Given the time constraints, perhaps the easiest way is to try to see patterns.

Option 1: Try to find mapping from letters to letters.

Do any letters match?

First, let’s write down the ciphertext and plaintext letters on top of each other.

Ciphertext: o y f j d n i s d r r t q w a i n r a c x z m y n z b h h x

Plaintext: T h i n k s t e p b y s t e p

Wait a minute.

I think maybe there is an anagram or substitution cipher here.

Alternatively, I think that we can notice that each group of ciphertext corresponds to a plaintext word.

Check the number of letters.

First ciphertext word: oyfjdnisdr (10 letters)

First plaintext word: Think (5 letters)

Hmm.

Second ciphertext word: rtqwainr (8 letters)

Second plaintext word: step (4 letters)

Third ciphertext word: acxz (4 letters)

Third plaintext word: by (2 letters)

Fourth ciphertext word: mynzbhhx (8 letters)

Fourth plaintext word: step (4 letters)

Interesting.

It seems that the ciphertext words are exactly twice as long as the plaintext words.

(10 vs 5, 8 vs 4, 4 vs 2, 8 vs 4)

Idea: Maybe we need to take every other letter or rebuild the plaintext from the ciphertext accordingly.

Let’s test this theory.

If we take every other letter from the ciphertext words, maybe we’ll get the plaintext.

Let’s try this for the first word:

Ciphertext word: o y f j d n i s d r (10 letters)

Alternate letters:

First letters (positions 1,3,5,7,9): o f d i d

Second letters (positions 2,4,6,8,10): y j n s r

Wait: Let’s try taking every second letter or first letter.

Option 1: Take the odd positions:

Positions: 1 3 5 7 9

Letters: o f d i d

Plaintext word is Think”, letters are T h i n k

Hmm.

Perhaps we can map letters:

o f d i d –> Think

Not directly

Option 2: Try mapping as per an assigned code: perhaps columns of letters?

Alternatively, perhaps the cipher is more complex.

Alternatively, notice that oyfjdnisdr” has 10 letters and Think” has 5 letters.

What if each plaintext letter is being represented by two ciphertext letters?

Similarly for other words.

Suppose that each plaintext letter corresponds to pair of letters in the ciphertext.

Meaning that the first two letters of ciphertext map to the first plaintext letter, and so on.

So for the first word, partition the ciphertext letters into pairs:

oy fj dn is dr

Corresponding to plaintext letters T h i n k

So we can attempt to map:

oy -> T

fj -> h

dn -> i

is -> n

dr -> k

If this is the mapping, perhaps each pair of letters maps to one plaintext letter.

Similarly for other words.

Let’s test this mapping.

Is there a consistent mapping from pairs to letters?

Let’s make a table of pairs and corresponding letters.

First, make the pairs from the first word and their corresponding plaintext letters:

Ciphertext Pairs -> Plaintext Letter

oy -> T

fj -> h

dn -> i

is -> n

dr -> k

Now, what’s the mapping from oy” to T’?

Is there a way to map oy’ to T’?

Similarly for others.

One possible way is that oy’ represents T’.

But how?

Option 1: Sum of the letters.

Convert letters to numbers (A=1, B=2, …, Z=26)

So let’s try to get to T’ from o’ and y’

o = 15

y = 25

Sum: 15 +25 = 40

But T’ is 20.

Alternatively, perhaps subtract: 25 -15 = 10.

No.

Alternatively, perhaps combine the numbers in some way.

Alternatively, think about their positions in the alphabet.

Alternatively, perhaps the letters are encrypted via a code.

Alternatively, perhaps if we overlay the word Think’ over the cipher pairs oy’, fj’, etc., the cipher is formed by substituting each plaintext letter with two letters.

Alternatively, perhaps consider the original’ letters.

Let’s look at the letters between the word Think” and oyfjdnisdr”

Perhaps we can spot that oyfjdnisdr” is an anagram for Thinkfiles”?

But that seems unlikely.

Alternatively, think of the pairs.

Suppose we assign numbers to letters: a=1, b=2, …, z=26

Consider the numbers in oy’:

o = 15

y = 25

Add them: 15 + 25 = 40

Now, for T’ we have:

T’ is number 20

What if we take the sum of the cipher pair mod 26.

15 +25 = 40

Then 40 mod 26 = 14

14 corresponds to N’

Doesn’t match.

Alternatively, see if the cipher pairs correspond to something else.

Alternatively, perhaps the pair oy’ corresponds to T’ via key.

Alternatively, perhaps it’s as simple as oy’ is T’.

But o’ corresponds to T’, subtracting o’ ASCII code from T’.

Now, let’s think in terms of symmetric letters.

Alternatively, perhaps we can spot that the letters correspond to the first letters of words.

But perhaps an easier way is to look at the word positions.

Alternatively, perhaps we can see that the ciphertext letters are offsets from the plaintext letters.

So let’s try to map the pairs to the plaintext letters.

First pair oy’ and plaintext letter T’

Letter o’ is 15, T’ is 20

Letter o’ + 5 = T’ (15 +5=20)

But y’ is 25, which is 5 letters after T’ (20 +5=25)

Interesting, seems that o’ +5 = T’

Similarly, y’ +1 = Z’, which is not T’.

Alternatively, maybe the average of the two letters corresponds to T’

(15 +25)/2 =20, which is T’

Wait a minute, that seems promising.

First pair: o’ (15) + y’ (25) = 40

40 /2 =20

20 corresponds to T’

So perhaps the average of the letters in the pair corresponds to the plaintext letter.

Let’s check this with the second pair.

Second pair: fj’ corresponding to h’

Letters f’=6, j’=10

Sum: 6+10=16

Average:16/2=8

8 corresponds to h’ (since h’ is 8)

Success!

Third pair: dn’ to i’

d’=4, n’=14

Sum:4+14=18

Average:18/2=9

9 corresponds to i’(9=‘i’)

But i’ is 9, so that seems off by 1.

So perhaps we need to think carefully about letters.

Wait, 18/2=9, 9 corresponds to I’

So this works.

Fourth pair: is’ corresponding to n’

i’=9, s’=19

Sum:9+19=28

Average:28/2=14

14 corresponds to n’(14=‘N’)

Yes!

Fifth pair: dr’ corresponds to k’

d’=4, r’=18

Sum:4+18=22

Average:22/2=11

11 corresponds to k’(11=‘K’)

Perfect!

So our code is: For each pair, sum their numeric values, divide by 2, get the corresponding letter.

Alternatively, the average of the numeric values of the letters corresponds to the plaintext letter.

This seems to be a valid method.

Now let’s test this with the other words.

Second word ciphertext: rtqwainr

Ciphertext pairs:

rt -> s

qw -> ?

ai -> ?

nr -> ?

Corresponding to plaintext letters s’, t’, e’, p’

So let’s take the pairs:

1. rt’ corresponds to s’

r=18, t=20

Sum:38

Average:19

19 corresponds to s’ (19=‘S’)

Good

2. qw’ corresponds to t’

q=17, w=23

Sum:17+23=40

Average:20

20 corresponds to t’ (20=‘T’)

Good

3. ai’ corresponds to e’

a=1, i=9

Sum:10

Average:5

5 corresponds to e’ (5=‘E’)

Good

4. nr’ corresponds to p’

n=14, r=18

Sum:32

Average:16

16 corresponds to p’ (16=‘P’)

Perfect!

Third word: acxz (4 letters)

Plaintext word: by’ (2 letters)

Ciphertext pairs:

a c

x z

Plaintext letters: b’, y’

First pair: a’, c’

a=1, c=3

Sum:4

Average:2

2 corresponds to b’(2=‘B’)

Second pair: x’ and z’

x=24, z=26

Sum:50

Average:25

25 corresponds to y’(25=‘Y’)

Yes!

Fourth word: mynzbhhx (8 letters)

Plaintext word: step’ (4 letters)

Ciphertext pairs:

my

nz

bh

hx

Plaintext letters s’,‘t’,‘e’,‘p’

First pair m’,‘y’

m=13, y=25

Sum:38

Average:19

19 corresponds to s’(19=‘S’)

Second pair: n’,‘z’

n=14, z=26

Sum: 40

Average:20

20=‘t’

Third pair: b’,‘h’

b=2, h=8

Sum:10

Average:5

5=‘e’

Fourth pair: h’,‘x’

h=8, x=24

Sum:32

Average:16

16=‘p’

Got it!

Now, given this, we can decode the given message.

Ciphertext to decode:

oyekaijzdf aaptcg suaokybhai ouow aqht mynznvaatzacdfoulxxz”

First, we need to divide into words.

Given that the ciphertext in the example had words of length:

-10 letters (5 pairs)
-8 letters (4 pairs)
-4 letters (2 pairs)
-8 letters (4 pairs)

Transported to plaintext words of:

-5 letters
-4 letters
-2 letters
-4 letters

So perhaps in our supplied ciphertext, we need to divide the letters into words accordingly.

Ciphertext: o y e k a i j z d f a a p t c g s u a o k y b h a i o u o w a q h t m y n z n v a a t z a c d f o u l x x z

First, let’s write the letters in order:

1) oyekaijzdf (10 letters) -> word of 5 letters

2) aaptcg (6 letters) -> word of 3 letters

3) suaokybhai (10 letters) -> word of 5 letters

4) ouow (4 letters) -> word of 2 letters

5) aqht (4 letters) -> word of 2 letters

6) mynznvaatzacdfoulxxz (22 letters) -> word of 11 letters

Totals:

So the letters are:

First word: oyekaijzdf (10 letters = 5 pairs)

Second word: aaptcg (6 letters=3 pairs)

Third word: suaokybhai (10 letters=5 pairs)

Fourth word: ouow (4 letters=2 pairs)

Fifth word: aqht (4 letters=2 pairs)

Sixth word: mynznvaatzacdfoulxxz (22 letters=11 pairs)

So we can suppose that the plaintext words have lengths:

1) 5 letters

2) 3 letters

3) 5 letters

4) 2 letters

5) 2 letters

6)11 letters

Now we can proceed to decode.

First word: oyekaijzdf

Divide into pairs:

oy

ek

ai

jz

df

So we can find the plaintext letters by averaging the numeric values of the letters in each pair.

First pair: o’,‘y’

o=15, y=25

Sum=40

Average=20

20=‘T’

Second pair: e’,‘k’

e=5, k=11

Sum=16

Average=8

8=‘H’

Third pair: a’,‘i’

a=1, i=9

Sum=10

Average=5

5=‘E’

Fourth pair: j’,‘z’

j=10, z=26

Sum=36

Average=18

18=‘R’

Fifth pair: d’,‘f’

d=4, f=6

Sum=10

Average=5

5=‘E’

So the first word is T’,‘H’,‘E’,‘R’,‘E’

So THERE

Second word: aaptcg

Pairs:

aa

pt

cg

First pair: a’,‘a’

a=1, a=1

Sum=2

Average=1

1=‘A’

Second pair: p’,‘t’

p=16, t=20

Sum=36

Average=18

18=‘R’

Third pair: c’,‘g’

c=3, g=7

Sum=10

Average=5

5=‘E’

So second word: A’,‘R’,‘E’

Third word: suaokybhai

Pairs:

su

ao

ky

bh

ai

First pair: s=19, u=21

Sum=40, Average=20

20=‘T’

Second pair: a=1, o=15

Sum=16, Average=8

8=‘H’

Third pair: k=11, y=25

Sum=36, Average=18

18=‘R’

Fourth pair: b=2, h=8

Sum=10, Average=5

5=‘E’

Fifth pair: a=1, i=9

Sum=10, Average=5

5=‘E’

Third word: T’,‘H’,‘R’,‘E’,‘E’

Fourth word: ouow

Pairs: ou, ow

First pair: o=15, u=21

Sum=36, Average=18

18=‘R’

Second pair: o=15, w=23

Sum=38, Average=19

19=‘S’

Fourth word: R’,‘S’

Fifth word: aqht

Pairs: a q, h t

First pair: a=1, q=17

Sum=18, Average=9

9=‘I’

Second pair: h=8, t=20

Sum=28, Average=14

14=‘N’

Fifth word: I’,‘N’

Sixth word: mynznvaatzacdfoulxxz (22 letters: 11 pairs)

Pairs:

m y

n z

n v

a a

t z

a c

d f

o u

l x

x z

Let’s list them properly.

Wait, earlier I missed some letters there.

Let’s re-express the sixth word letters:

m y n z n v a a t z a c d f o u l x x z

Total letters: 22 letters

Pairs:

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x x

11. z

Wait, that’s 21 letters.

But in total, we have 22 letters; I think I have a mismatch.

Wait, in the problem the sixth word is:

mynznvaatzacdfoulxxz

Let’s separate the letters:

m y n z n v a a t z a c d f o u l x x z

Total letters: 22 letters

So pairs would be:

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x x

11. z

Wait, z’ is only one letter at the end.

So maybe the letters are:

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x x

11. z (left alone?)

But then we’re missing a letter.

Alternatively, perhaps the z’ at the end is an extra letter or it’s paired with the previous x’.

Alternatively, perhaps there is a mistake.

Alternatively, perhaps when we reach the last x x z’, we should write:

9. l x

10. x z

Total pairs: 11

So updated pairs:

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x z

Now total pairs: 10 pairs. But we are supposed to have 11 letters in the plaintext word.

So perhaps I have miscounted and not considered including f’ or z’.

Wait, is there any chance that the sequence is:

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x x

11. z

We have 22 letters.

Perhaps the x x’ and z’ are overlapping.

But we need 11 pairs.

I think for this to work, we need to create 11 pairs from the 22 letters.

So let’s try pairing letters in order, taking two letters at a time.

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x x

11. z ?

But again, last pair only has one letter z’

Alternatively, we might need to adjust the pairs.

Perhaps it’s ou’, lx’, xz’, and there is no extra x’.

Maybe I miswrote.

Wait, let’s write the letters as they come:

m y n z n v a a t z a c d f o u l x x z

Positions:

1 m

2 y

3 n

4 z

5 n

6 v

7 a

8 a

9 t

10 z

11 a

12 c

13 d

14 f

15 o

16 u

17 l

18 x

19 x

20 z

So 20 letters.

Now that makes sense.

Now we can divide into pairs:

Pairs:

1. m y

2. n z

3. n v

4. a a

5. t z

6. a c

7. d f

8. o u

9. l x

10. x z

Total of 10 pairs, which will correspond to 10 letters.

But earlier we thought the word was 11 letters.

But 10 letters may be acceptable.

Maybe we should proceed with the 10 pairs.

First pair: m y

m=13, y=25

Sum=38

Average=19

19=‘S’

Second pair: n z

n=14, z=26

Sum=40

Average=20

20=‘T’

Third pair: n v

n=14, v=22

Sum=36

Average=18

18=‘R’

Fourth pair: a a

a=1, a=1

Sum=2

Average=1

1=‘A’

Fifth pair: t z

t=20, z=26

Sum=46

Average=23

23=‘W’

Sixth pair: a c

a=1, c=3

Sum=4

Average=2

2=‘B’

Seventh pair: d f

d=4, f=6

Sum=10

Average=5

5=‘E’

Eighth pair: o u

o=15, u=21

Sum=36

Average=18

18=‘R’

Ninth pair: l x

l=12, x=24

Sum=36

Average=18

18=‘R’

Tenth pair: x z

x=24, z=26

Sum=50

Average=25

25=‘Y’

So the tenth word is S’, T’, R’, A’, W’, B’, E’, R’, R’, Y’

Wait that’s 10 letters

So the sixth word is STRAWBERRY, which makes sense.

So putting all these translated letters together:

First word: THERE

Second word: ARE

Third word: THREE

Fourth word: RS (? Doesn’t make sense)

Wait, the fourth word, the pairs were:

ou ow

Pairs:

First pair: o u

o=15, u=21

Sum=36, Average=18

18=‘R’

Second pair: o w

o=15, w=23

Sum=38, Average=19

19=‘S’

So we get R’,‘S’

Similarly, fifth word IN

R’, S’ I’,‘N’

So the message so far is

THERE ARE THREE RS IN

And then STRAWBERRY

Wait, let’s put the words together:

THERE ARE THREE Rs IN STRAWBERRY

I think RS is R’s’, meaning Rs’

So the message is:

THERE ARE THREE R’S IN STRAWBERRY

Which is true.

So the decoded message is:

THERE ARE THREE R’S IN STRAWBERRY

https://openai.com/index/learning-to-reason-with-llms/

Quote ai openai

“Learn to do everything lightly”

It’s dark because you are trying too hard. Lightly child, lightly. Learn to do everything lightly. Yes, feel lightly even though you’re feeling deeply. Just lightly let things happen and lightly cope with them. I was so preposterously serious in those days… Lightly, lightly - it’s the best advice ever given me… to throw away your baggage and go forward. There are quicksands all about you, sucking at your feet, trying to suck you down into fear and self-pity and despair. That’s why you must walk so lightly.”

Quote aldous huxley

”Common sense is often wielded like a gun”

That’s Stephen West on Žižek—on the most popular kind of overconfidence.

The philosopher seeks cracks in the symbolic edifice that grounds our social stability.” And yeah, most people resist such thoughts (correctly).

Quote slavoj žižek metaphilosophy ai

Notes on Rao on Ooze

Some notes on Fear of Oozification.

I accept the evolutionary picture. So do Yudkowsky, Shulman, Hanson, Carlsmith and Bostrom.

Key disuputes:

  1. How capable are we of influencing evolutionary change (both speed and direction)?
  2. Would the results of such efforts be desirable (see e.g. here, here, here and here)?

We definitely have some capability to influence (at the social level it’s called governance; at the biological: homeostasis).

It’s a value question, and an empirical question, what kinds of governance (a.k.a. holding onto things we care about) we should go for.

On the value question: yes, valuing requires some attachment to the status quo. Yudkowsky, Schulman, Carlsmith and Bostrom (and, I’m fairly sure—Altman, Hassabis, Musk) understand this, and are more into conservative humanism than Rao and Hanson.

Most people prefer conservative humanism—even the transhumanists—so these values will keep winning until the accellerationist minority get an overwhelming power advantage (or civilisation collapses).

The accellerationists see humanism as parochial and speciesist. They love Spinoza’s God, and more fully submit to its will”.

Metaphysical moral realists like Parfit and Singer may find themselves siding with the accelerationists; it depends how well human-ish values track what objectively matters (what God wants”…).

Joe Carlsmith’s series, Otherness and Control in the Age of AGI, is one of the best things I’ve read on this stuff.


Empirically, these values lie on a spectrum, and neither extreme is sustainable. Max(conservative) means self-defeating fragility. Max(accelerationist) means ooze, because complexity requires local stability, and ooze eventually becomes primordial soup.


New, for me, was the idea that more powerful technology means selection dynamics at lower levels. E.g. when we train AIs we select over matrices, and with nanotech we’ll select over configurations of atoms. And yes, once that ball gets rolling, there’s an explosion of possibilities. It sounds like Rao thinks that this means a Singleton is unlikely, but I don’t understand why. Our attempts at scaffolding might well lead us there.


Rao doesn’t like the Shoggoth meme:

The shoggoth meme, in my opinion, is an entirely inappropriate imagining of AIs and our relationship to them. It represents AI in a form factor better suited to embodying culturally specific fears than either the potentialities of AI or its true nature. There is also a category error to boot: AIs are better thought of as a large swamp rather than a particular large creature in that swamp wearing a smiley mask.

To defend it: the swamp is the pretrained model. The Shoggoth is the fine-tuned and RHLF’d creature we interact with. The key thing is that the tuned and RHLF’d creature still has many heads; on occasion we’ll be surprised by heads we don’t like.

Writing evolution