text:
## The Noisy Channel Coding Theorem

Here’s a charming, easily readable tale of <NAME> and how he came up with information theory:

I hadn’t known his PhD thesis was on genetics! His master’s thesis introduced Boolean logic to circuit design. And as a kid, he once set up a telegraph line to a friend’s house half a mile away.

So, he was perfectly placed to turn information into a mathematical quantity, deeply related to entropy, and prove some now-famous theorems about it.

These theorems set limits on how much information we can transmit through a noisy channel. More excitingly, they say we can cook up coding schemes that let us come as close as we want to this limit, with an arbitrarily low probability of error.

As <NAME> points out, these results are fundamental to the ‘information age’ we live in today:

Can we transmit, say, a high-resolution picture over a telephone line? How long will that take? Is there a best way to do it?

Before Shannon, engineers had no clear answers to these questions. At that time, a wild zoo of technologies was in operation, each with a life of its own—telephone, telegraph, radio, television, radar, and a number of other systems developed during the war. Shannon came up with a unifying, general theory of communication. It didn’t matter whether you transmitted signals using a copper wire, an optical fiber, or a parabolic dish. It didn’t matter if you were transmitting text, voice, or images. Shannon envisioned communication in abstract, mathematical terms; he defined what the once fuzzy concept of “information” meant for communication engineers and proposed a precise way to quantify it. According to him, the information content of any kind of message could be measured in binary digits, or just bits—a name suggested by a colleague at Bell Labs. Shannon took the bit as the fundamental unit in information theory. It was the first time that the term appeared in print.

So, I want to understand Shannon’s theorems and their proofs—especially because they clarify the relation between information and entropy, two concepts I’d like to be an expert on. It’s sort of embarrassing that I don’t already know this stuff! But I thought I’d post some preliminary remarks anyway, in case you too are trying to learn this stuff, or in case you can help me.

There are various different theorems I should learn. For example:

• The source coding theorem says it’s impossible to compress a stream of data to make the average number of bits per symbol in the compressed data less than the Shannon information of the source, without some of the data almost certainly getting lost. However, you can make the number of bits per symbols arbitrarily close to the Shannon entropy with a probability of error as small as you like.

• the noisy channel coding theorem is a generalization to data sent over a noisy channel.

The proof of the noisy channel coding theorem seems not so bad—there’s a sketch of a proof in the Wikipedia article on this theorem. But many theorems have a hard lemma at their heart, and for this one it’s a result in probability theory called the asymptotic equipartition property.

You should not try to dodge the hard lemma at the heart of the theorem you’re trying to understand: there’s a reason it’s there. So what’s the asymptotic equipartition property?

Here’s a somewhat watered-down statement that gets the basic idea across. Suppose you have a method of randomly generating letters—for example, a probability distribution on the set of letters. Suppose you randomly generate a string of $n$ letters, and compute $-(1/n)$ times the logarithm of the probability that you got that string. Then as $n \to \infty$ this number ‘almost surely’ approaches some number $S.$ What’s this number $S$? It’s the entropy of the probability distribution you used to generate those letters!

(Almost surely is probability jargon for ‘with probability 100%’, which is not the same as ‘always’.)

This result is really cool—definitely worth understanding in its own right! It says that while many strings are possible, the ones you’re most likely to see lie in a certain ‘typical set’. The ‘typical’ strings are the ones where when you compute $-(1/n)$ times the log of their probability, the result is close to $S.$ How close? Well, you get to pick that.

The typical strings are not individually the most probable strings! But if you randomly generate a string, it’s very probable that it lies in the typical set. That sounds a bit paradoxical, but if you think about it, you’ll see it’s not. Think of repeatedly flipping a coin that has a 90% chance of landing heads up. The most probable single outcome is that it lands heads up every time. But the typical outcome is that it lands up close to 90% of the time. And, there are lots of ways this can happen. So, if you flip the coin a bunch of times, there’s a very high chance that the outcome is typical.

It’s easy to see how this result is the key to the noisy channel coding theorem. In general, the ‘typical set’ has few elements compared to the whole set of strings with $n$ letters. So, you can make short codes for the strings in this set, and compress your message that way, and this works almost all the time. How much you can compress your message depends on the entropy $S$.

(I said ‘in general’ because there’s one exception: when every $n$-letter string is equally probable, every string is in the typical set. In this very special case, no compression is possible.)

So, we’re seeing the link between information and entropy!

The actual coding schemes that people use are a lot trickier than the simple scheme I’m hinting at here. When you read about them, you see scary things like this:

But presumably they’re faster to implement, hence more practical.

The first coding schemes that come really close to the Shannon limit are the turbo codes. Surprisingly, these codes were developed only in 1993! They’re used in 3G mobile communications and deep space satellite communications.

One key trick is to use, not one decoder, but two. These two decoders keep communicating with each other and improving their guesses about the signal they’re received, until they agree:

This iterative process continues until the two decoders come up with the same hypothesis for the m-bit pattern of the payload, typically in 15 to 18 cycles. An analogy can be drawn between this process and that of solving cross-reference puzzles like crossword or sudoku. Consider a partially completed, possibly garbled crossword puzzle. Two puzzle solvers (decoders) are trying to solve it: one possessing only the “down” clues (parity bits), and the other possessing only the “across” clues. To start, both solvers guess the answers (hypotheses) to their own clues, noting down how confident they are in each letter (payload bit). Then, they compare notes, by exchanging answers and confidence ratings with each other, noticing where and how they differ. Based on this new knowledge, they both come up with updated answers and confidence ratings, repeating the whole process until they converge to the same solution.

This can be seen as “an instance of loopy belief propagation in Bayesian networks.”

By the way, the picture I showed you above is a flowchart of the decoding scheme for a simple turbo code. You can see the two decoders, and maybe the loop where data gets fed back to the decoders.

While I said this picture is “scary”, I actually like it because it’s an example of network theory applied to real-life problems.

### 18 Responses to The Noisy Channel Coding Theorem

1. davidtweed says:

What you’re looking at here is (AIUI) general codes where one is receives all the transmitted “symbols” but there is generally some garbling going on. Another model is the “erasure codes”, where it’s assumed you can “check” (with eg a checksum) each symbol that you receive, so you KNOW if it’s correct or not but you may not receive all the symbols. (Of course these are the ends of a spectrum, eg, erasure could be said to be the case where garbled symbols are “completely” garbled.) In both cases you want to reconstruct an original message from a “redundancy via coding” message that’s gone over the unreliable channel.

The interesting thing is that slightly different mathematics appears to apply best in the different cases, eg, consider “fountain codes” such as described in the last chapter of McKay’s book. There it considers using random graphs to encode, then reconstruct, a message from an arbitrary set of packets that’s just bigger than the minimal amount needed.It would be interesting to see what you could figure out about the representing a “set” of messages (eg, a set of viable genomes) via one of these kind of processes since it might be related to “useful” within species diversity. (It might be useful to have some different genes within a population, but for many — although not all — genetic ailments there is no benefit associated with it. So you want some set, but not all, variants to be “recoverable” from the population of animals in a given species.)

• <NAME> says:

David wrote:

What you’re looking at here is (AIUI) general codes where one is receives all the transmitted “symbols” but there is generally some garbling going on.

That’s right—that’s the model here.

Another model is the “erasure codes”, where it’s assumed you can “check” (with eg a checksum) each symbol that you receive, so you KNOW if it’s correct or not but you may not receive all the symbols.

Hmm! Now you’re making me wonder what’s the most general model of ‘garbled’, ‘partially erased’, or otherwise screwed-up data. Category-theoretic computer scientists and logicians have extremely general theories of data types, which I sort of understand, but I’ve never tried to imagine an extremely general theory of ‘screwed-up’ or ‘damaged’ data.

It seems information theorists are often content to think of data as a string of symbols drawn from a pre-established alphabet. Clearly anything discrete can be encoded in this way, but is this model sufficiently general for thinking about the ways data can be damaged?

It’s easy to imagine a string being ‘garbled’ with some symbols replaced by others, and it’s easy to imagine it being ‘partially deleted’. As you note, the latter can be conceived as a special case of the former if we allow a symbol ‘blank’ and think of deletion as replacing some other symbol by a blank. But are there other fundamentally different ways to damage data… or is there some theorem saying that all other ways can be reduced to ‘garbling’?

And when I say “information theorists are often content to think of data as a string of symbols drawn from a pre-established alphabet” maybe I’m being old-fashioned. Maybe they do more general things I don’t know about. For example, maybe I’m stuck back in the era where data was sent along a fixed ‘channel’, and now I should be thinking about data moving around in ‘packets’… so that one form of ‘garbling’ could be getting a packet that should have gone to someone else, or getting a packet really late, or something like that!

Examples chosen from biology could force us to think about information transmission in even more general ways. Just because you can model data as a string of symbols in some alphabet, doesn’t mean you should. There’s information in the cytoplasm of a human egg cell, which is important for the development of the fetus. But modeling it as a string of symbols seems awkward (while for the genome, it’s not quite so bad).

• <NAME> says:

John said:

Maybe they do more general things I don’t know about.

Having a single communication channel is very simplistic: there is one sender and one receiver which share a channel. For our deeply connected world, this scenario is highly unrealistic! Nowadays, communication is organized in big networks formed by lots of nodes and connections between these. Such a network typically have many senders and many receivers. In such situations, capacity ceases to be a single number; rather, it becomes an entire region in the space of matrices $R_{ij}$, the entries of which stand for the amount of information transmitted from sender $i$ to receiver $j$ when using a certain coding scheme. It’s called the ‘rate region’.

The butterfly network is a beautiful example of how coding in a network can work — and why the internet, with its routing-based approach, is likely to be in a pretty mediocre part of the rate region.

I’m mentioning all this because it’s obviously related to Bayesian networks, and maybe even to other aspects of your work on network theory, I don’t know.

• <NAME> says:

Thanks! More to learn and think about. I’d like to combine network theory with information theory, since green mathematics, if such a thing exists, should involve both.

• Robert says:

DNA copying can also be thought of as data transmission. Besides point transcription and deletion errors, corresponding to those already described, chunks of DNA can get duplicated, or moved around. – ‘genes 1,2,3’ becoming ‘genes 1,2,2,3’ or ‘genes `1,3,2’. I don’t think there’s any equivalent for messages sent over serial channels.

Tangentially, the types of errors you see can indicate how the data is being stored. If ‘riddle’ is misspelt ‘ridlle’, that suggests an underlying representation as r-i-double-d-l-e, with a single error producing r-i-d-double-l-e.

• <NAME> says:

In quantum error correction, people often consider two primary types of errors: bit flips, where $| 0 \rangle$ and $| 1 \rangle$ switch; and sign flips, where $| 0 \rangle$ is unchanged but $| 1 \rangle$ is multiplied by ${-}1$. These are conveniently represented by the Pauli $X$ and $Z$ matrices. There are results saying that as long as you can correct adequately for these discrete sorts of errors, you can correct for all errors in some sense.

I wonder if there are classical analogues of these sorts of result.

Tobias, I agree, it’s really exciting to think about how all this stuff generalizes to topologically interesting situations. It’s entrancing to see linear algebra coming into play in the butterfly network.

• <NAME> says:

Let’s keep talking about all this stuff.

By the way, Jamie, speaking of error correction…

Everyone please remember: on all WordPress blogs, LaTeX is done like this:

$latex E = mc^2$

with the word ‘latex’ directly following the first dollar sign, no space. Double dollar signs don’t work here.

2. romain says:

Noteworthy, the asymptotic equipartition property (AEP) shows that the plug-in estimate of entropy converges to the true entropy in the large sampling limit.

This is interesting with respect to the post “Mathematics of biodiversity 7” to understand the whole concept of bias correction of entropy estimates in the case of a limited sample.

Consider the quantity

$\displaystyle{H_n=-\frac{1}{n} \log p(X_{i_j}^n)}$

where $X_{i_j}^n$ is a sequence of $n$ letters $X_i$ occurring with probabilities $p_i$. And let’s assume there are $m$ different letters..

In the case where the letters are i.i.d., then

$\displaystyle{p(X_{i_j}^n)=\prod_j^n p(X_{i_j})}$

Now, the same $X_i$ may appear many times, say $n_i$, so this formula can be rewritten

$\displaystyle{p(X_i^n)=\prod_i^m p(X_i)^{n_i}}$

so that:

$\displaystyle{H_n=-\sum_i^m \frac{n_i}{n} \log p(X_i)}$

When $N$ is large, then

$\displaystyle{\frac{n_i}{n} \rightarrow p(X_i)}$

QED.

• <NAME> says:

That’s nice! Thanks!

I hope I have enough energy to say more about the asymptotic equipartition property.

3. […] So I’m boning up on the subject by reading Khinchin’s great classic, Mathematical Foundations of Information Theory. In the future I’ll share some findings here, but in the mean time, follow the link to Azimuth and read Baez’s great discussion of the Noisy Channel Coding Theorem. […]

4. arch1 says:

John,

My quick skim of your summary of the hard lemma left me wondering why it is a hard lemma*. Isn’t it just a baby step away from the fact that as n-> infinity, samples of size n look more and more like the underlying distribution from which they are drawn?

*That in itself is progress, since my quick skim of Wikipedia’s summary of the hard lemma left me almost clueless as to the content of the hard lemma (it struck me as very vague:-)

• <NAME> says:

I didn’t mean to say the asymptotic equipartition property is extremely hard. However, the rest of the proof looks easy in comparison, so one is inclined to look at this part and say “yuck, that’s some technical fact I’d rather take on faith”. It seems like the pit in the peach. But I was trying to convince everyone that unlike the pit in the peach, it’s highly nutritious, and tasty in its own way.

I stated a watered-down version of the asymptotic equipartition theorem: just for purposes of exposition, I assumed that each letter in the string was drawn independently from the same probability distribution on letters. In other words, I was assuming that they’re ‘independent identically distributed’ random variables. This is clearly too restrictive—it sure ain’t true for English text!

The statement and proof gets a bit harder when we do the full-fledged thing: you can see a proof here for the i.i.d. case and here for the more general case. The more general case uses Lévy’s martingale convergence theorem, Markov’s inequality and Borel-Cantelli lemma. For some people that would be very scary, while others eat such stuff for breakfast.

But anyway, regardless of whether the proof is hard, the result seems both interesting and highly believable: not shocking.

• Sounds like yummy breakfast. But no Cramer large deviation stuff? Last century I ran a seminar on Cramer’s theorem, iterated logarithm, arcsin law etc. Now suddenly it seems Shannon is yummy, too.

• arch1 says:

Thanks for the extra context John.

5. blake561able says:

It is interesting that Weyl was Shannon’s adviser at IAS. The communication between those two must have come close to exceeding these limits on information!

6. L says:

Do you think you can expand on this : “The ‘typical set’ has few elements compared to the whole set of strings with n letters”…? If we start with the uniform distribution of a finite set A, then every string is in the typical set of A^n – they are equiprobable – so I guess the statement needs some qualification, or perhaps I am confused. Does this have something to do with the uniform distribution having the largest entropy?

• <NAME> says:

I haven’t thought about this for about 5 years, but: when the probability distribution on $A$ is uniform, I believe you are right that the ‘typical set’ is all of $A^n$. In this case no compression is possible. For every other probability distribution on $A$, the typical set will become small compared to $A^n$ when $n \to \infty$.

• L says:

Thanks for responding!! For other readers, the quantitative statement is that the number of of $\beta$ typical words (those words $x$ with $| latex |-1/n * log (p(x)) – H(X) | < \beta$ ) can be bounded by $2^{N(H(X) + \beta)}$ (this follows by direct computation from $1 \geq \Sigma_{x Typical} P(x)$). When $H(X) < log |X|$ (which is exactly the non-uniform distribution case) and $\latex \beta$ sufficiently small, $H(X) + \beta – log |X| < 0$, so the ratio of typical words to all words goes to zero as $N$ grows.

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# Prove each equivalence by using the rules for semantic equivalence

Having some issue with some logic - the examples I've been provided with arn't very helpful so I can have no idea where to start. The question is to prove;

           (p ∨ q) ∧ (q ⇒ p) ≡ p

but I have no idea where to start. Attached below are the rules I'm using.

 A ∧ A ≡ A, A ∨ A ≡ A                                         idempotence
A ∧ B ≡ B ∧ A, A ∨ B ≡ B ∨ A                                 commutativity
A ∧ (B ∧ C ) ≡ (A ∧ B) ∧ C , A ∨ (B ∨ C ) ≡ (A ∨ B) ∨ C      associativity
A ∧ (A ∨ B) ≡ A, A ∨ (A ∧ B) ≡ A                             absorption
A ∧ (B ∨ C ) ≡ (A ∧ B) ∨ (A ∧ C )                            distributivity
A ∨ (B ∧ C ) ≡ (A ∨ B) ∧ (A ∨ C )                            distributivity
A ∧ (¬A) ≡ false, A ∨ (¬A) ≡ true                            negation
¬(¬A) ≡ A                                                    double negation
¬(A ∧ B) ≡ (¬A) ∨ (¬B), ¬(A ∨ B) ≡ (¬A) ∧ (¬B)               de Morgan
A ⇒ B ≡ (¬A) ∨ B                                             implication
A ⇔ B ≡ (A ⇒ B) ∧ (B ⇒ A)                                  bi-implication

My first guess is to use the rule of implication first but to be honest, I have no idea where to start.

Would really appreciate any help anyone can provide.

Thanks.

Another solution:

$(p \lor q) \land(q \rightarrow p)$

$(p \lor q) \land(\neg q \lor p)$, implication

$(p \lor q) \land (p \lor \neg q)$, commutativity

$p \lor (q \land \neg q)$, distributivity

$p \lor \text{false}$, negation

$p \lor (p \land \neg p)$, negation again

$p$, absorption

• Ah, that is the trick for removing $\bot$ from a disjunction. Nice. – James Feb 17 '16 at 18:26

A general rule of thumb is always to simplity: Replace $\rightarrow$, push in $\neg$, distribute $\lor$ over $\land$ (and $\land$ over $\lor$)

I omit some parenthesis, all these are justified by associativity. I also distribute on both sides, which uses commutivity implicitly.

$(p \lor q) \land (q \rightarrow p)$

$(p \lor q) \land (\neg q \lor p)$, by implication

$((p \lor q) \land \neg q )\lor ((p \lor q)\land p)$, by distributivity

$(p \land \neg q) \lor (q \land \neg q) \lor (p \land p) \lor (q\land p)$, by distributivity again

$(p \land \neg q) \lor \text{false} \lor p \lor (q\land p)$, by negation and idempotence

$p$, by absorption and the rule $A \lor \text{false} = A$, which you should have.

format: plain


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# Homework Help: Just a small question.

1. Feb 22, 2012

### nesan

1. The problem statement, all variables and given/known data

Sorry, I just have a general question. I know how to calculate velocity of anything relative to the ground, but how do you calculate the velocity of something relative to the air (?) if the velocity of the wind is given.

Please explain, with a small example if possible. Thank you. ^_^

2. What I have so far

For example lets say a plane was flying at a velocity of 10 m/s north and the wind was blowing 5 m/s east. I know the velocity of the plane relative to the ground would be sqrt(10^2 + 5^2)[N x Degree E], but how does this work if you have to find the velocity relative to the air?

Last edited: Feb 22, 2012
2. Feb 22, 2012

### cepheid

Staff Emeritus
I'll use subscripts, p, a, and g for plane, air, and ground respectively. A comma between two of them. such as a,g, means "air relative to ground." So, as you know, it goes:

$$\vec{v}_{p,g} = \vec{v}_{p,a} + \vec{v}_{a,g}$$

The velocity of the plane relative to the ground is equal to the velocity of the plane relative to the air, PLUS the velocity of the air relative to the ground.

So, in this case, $v_{p,a}$ is clearly just 10 m/s N. Think about it. If you were in a river that ran east-west, and the current was 5 m/s east, and you just sat there and allowed yourself to be carried along with the current, then your velocity relative to the river would be 0 (because you are moving along with it). Your velocity relative to the bank would be 5 m/s east.

If you then began to swim across the river at a velocity of 10 m/s north, then your velocity relative to the water, instead of being 0, would now be 10 m/s north.

It's the same idea with the plane example.

3. Feb 22, 2012

### nesan

Thank you very much, just what I was looking for. Very helpful. ^_^ Thank you.


text:
## [astro-ph/0509212] Age of High Redshift Objects - a Litmus Test for the Dark Energy Models

 Authors: <NAME>, Abha Dev Abstract: The discovery of the quasar, the APM 08279+5255 at z = 3.91 whose age is 2.1 Gyr has once again led to age crisis''. The noticeable fact about this object is that it cannot be accommodated in a universe with $\Omega_m = 0.27$, currently accepted value of matter density parameter and $\omega = \mathrm{constant}$. In this work, we explore the concordance of various dark energy parameterizations ($w(z)$ models) with the age estimates of the old high redshift objects. It is alarming to note that the quasar cannot be accommodated in any dark energy model even for $\Omega_m = 0.23$, which corresponds to $1 \sigma$ deviation below the best fit value provided by WMAP. There is a need to look for alternative cosmologies or some other dark energy parameterizations which allow the existence of the high redshift objects. [PDF]  [PS]  [BibTex]  [Bookmark]

Discussion related to specific recent arXiv papers
<NAME>
Posts: 59
Joined: July 19 2005
Affiliation: Published independent

### [astro-ph/0509212] Age of High Redshift Objects - a Litmus T

Any comments on these extraordinary claims?

<NAME>
Posts: 87
Joined: February 24 2005
Affiliation: Institute of Astronomy, Nicolaus Copernicus University
Contact:

### Re: [astro-ph/0509212] Age of High Redshift Objects - a Litm

<NAME> wrote:Any comments on these extraordinary claims?
The title is not so extraordinary: this is a good type of test. But the interesting question is how precise the 2.1Gyr estimate is.

If this estimate were 2.1 \pm 0.1 Gyr (no h dependence) including random and systematic error, then it would start to become significantly (several sigma) in contradiction with the concordance values of the local cosmological parameters (e.g. 71,0.27,0.73,-1.0).

If it's 2.1 \pm 0.5 Gyr (no h dependence) then it's not yet precise enough.

From a 5 minute scan of Hamann & Ferland (1993) + Komossa & Hasinger (2002) it is not obvious to me that we have 2.1 \pm 0.1 Gyr (no h dependence), but maybe that's simply because 5 minutes is not enough. If you are interested, then please read the papers, understand them, and then you can summarise for us. :)

<NAME>
Posts: 59
Joined: July 19 2005
Affiliation: Published independent

### [astro-ph/0509212] Age of High Redshift Objects - a Litmus T

Yes, the age is determined from the Fe/O and Fe/Ne relative abundances and it takes time to form Fe in the observed quantities; Fe/O ~ 3. It would be essential to firm up on the error bars of this age determination for APM08279+5255, is there any more recent information about this?

Jain & Dev seem to have pulled the value of 2.1 Gyr out of thin air, with no error bars at all. Is this a generally accepted value for that system or have they just 'made it up'?

<NAME>
Posts: 59
Joined: July 19 2005
Affiliation: Published independent

### [astro-ph/0509212] Age of High Redshift Objects - a Litmus T

In answer to my last question I found http://www.ingentaconnect.com/content/b ... 4/art00002
"Cosmological implications of APM 08279+5255, an old quasar at z= 3.91" Alcaniz J.S.; L<NAME>.; <NAME>, MNRAS, Volume 340, Number 4, April 2003, pp. L39-L42(1)
The existence of old high-redshift objects provides an important tool for constraining the expanding age of the Universe and the formation epoch of the first objects. In a recent paper, Hasinger, Schartel & Komossa reported the discovery of the quasar APM 08279 + 5255 at redshift z= 3.91 with an extremely high iron abundance, and estimated age of 2-3 Gyr. By assuming the lower limit for this age estimate and the latest measurements of the Hubble parameter as given by the HST key project, we study some cosmological implications from the existence of this object. In particular, we derive new limits on the dark matter and vacuum energy contribution. Our analysis is also extended to quintessence scenarios in which the dark energy is parametrized by a smooth component with an equation of state px= x x (-1 x < 0) . For flat models with a relic cosmological constant we show that the vacuum energy density parameter is constrained to be 0.78 , a result that is marginally compatible with recent observations from type Ia supernovae (SNe Ia) and cosmic microwave background (CMB). For quintessence scenarios the same analysis restricts the cosmic parameter to x -0.22 . Limits on a possible first epoch of quasar formation are also briefly discussed. The existence of this object pushes the formation era back to extremely high redshifts.
(emphasis mine)

So it looks as if Jain and Dev we quoting the lower end of that age range, with a possible lower error of -0.1 Gyr. May their conclusions therefore be of some significance?


text:
# Resource[v13+] Unreal Time System 1.1.1

#### -FL-

##### Pokémon Island Creator
Member
-FL- submitted a new resource:

Unreal Time System - Time passes like Minecraft and Zelda: Ocarina of Time.

This script makes the time in game uses its own clock that only pass when you are in game instead of using real time (like Minecraft and Zelda: Ocarina of Time).

Tested on Essentials v13, v17.2, v18.1, v19.1 and v20. If this script isn't working on latest Essentials version, please...

#### TheDizzy1

##### Novice
Member
I’m starting work on my game again, and I absolutely love this and can’t wait to start it back up again. As I’m thinking of scenery, is it possible to set time of day events while using this amazing resource?

#### TechSkylander1518

##### Wiki Dweeb
Member
I’m starting work on my game again, and I absolutely love this and can’t wait to start it back up again. As I’m thinking of scenery, is it possible to set time of day events while using this amazing resource?
FL's got some instructions for this at the top of the Script.rb file -
# same thing, in fact, EXTRA_SECONDS/EXTRA_DAYS call these methods.
#
# 3. 'UnrealTime.advance_to(16,17,18)' advance the time to a fixed time of day,
# 16:17:18 on this example.

#### corvo00

##### Rookie
Member
Dosn't work in 20.1? I installed the script above main and dosn't work.

Sorry for my english

#### TechSkylander1518

##### Wiki Dweeb
Member
Dosn't work in 20.1? I installed the script above main and dosn't work.

Sorry for my english
You've got to describe the problem beyond just "it doesn't work".

#### corvo00

##### Rookie
Member
You've got to describe the problem beyond just "it doesn't work".
I get this when I open the game

#### TechSkylander1518

##### <NAME>
Member
I get this when I open the game
I'm not getting this error when I try it. Have you removed anything from the script? That error message is pointing to line 170, but the script doesn't reference pbGraphicsUpdate until line 271. And you're getting this error message because PokeBattle_Scene isn't a thing in v20, so there's no pbGraphicsUpdate to alias, but FL definitely accounted for that with the line PokeBattle_Scene = Battle::Scene if !defined?(PokeBattle_Scene).

#### corvo00

##### Rookie
Member
I'm not getting this error when I try it. Have you removed anything from the script? That error message is pointing to line 170, but the script doesn't reference pbGraphicsUpdate until line 271. And you're getting this error message because PokeBattle_Scene isn't a thing in v20, so there's no pbGraphicsUpdate to alias, but FL definitely accounted for that with the line PokeBattle_Scene = Battle::Scene if !defined?(PokeBattle_Scene).
I'm not getting this error when I try it. Have you removed anything from the script? That error message is pointing to line 170, but the script doesn't reference pbGraphicsUpdate until line 271. And you're getting this error message because PokeBattle_Scene isn't a thing in v20, so there's no pbGraphicsUpdate to alias, but FL definitely accounted for that with the line PokeBattle_Scene = Battle::Scene if !defined?(PokeBattle_Scene).
Ups! I had deleted a line unintentionally, what a fool... Thank you for responding and sorry for the inconvenience.

#### Westrah

##### Novice
Member
Is there a way to make this script work with Apricorns so that they respawn after "24 hours" of this scripts time?

#### -FL-

##### Pokémon Island Creator
Member
Is there a way to make this script work with Apricorns so that they respawn after "24 hours" of this scripts time?
An Apricorn system who works without this scripts should work with this script automatically.

#### Taiga

##### Rookie
Member
I kind of struggle to set the date to a specific time. I put in
Ruby:
pbGetTimeNow.year(2028)
pbGetTimeNow.mon(8)
pbGetTimeNow.day(1)
pbGetTimeNow.hour(8)
but it just gives me an error message.

#### TechSkylander1518

##### Wiki Dweeb
Member
I kind of struggle to set the date to a specific time. I put in
Ruby:
pbGetTimeNow.year(2028)
pbGetTimeNow.mon(8)
pbGetTimeNow.day(1)
pbGetTimeNow.hour(8)
but it just gives me an error message.
"An error message" could be anything, you need to share the full thing. But I'm guessing that the error is coming from the fact that pbGetTimeNow is supposed to return the current time, not set it. You should be using these commands -

#### SirWeibrot

##### Novice
Member
Hi, I just noticed that a caught Pokemon shows its date received based on Time.local, rather than the actual date it was received. Is there a way to change/fix this?

#### -FL-

##### Pokémon Island Creator
Member
Hi, I just noticed that a caught Pokemon shows its date received based on Time.local, rather than the actual date it was received. Is there a way to change/fix this?
The default Essentials use 'pbGetTimeNow' for pokémon caught date (search for 'pbGetTimeNow' on class Pokemon, should be a '@timeReceived = pbGetTimeNow.to_i'), so this may be some external script changing it.

If you want to change others 'Time.now' to use Unreal Time (or vice-versa):
Script said:
# If you wish to some parts still use real time like the Trainer Card start time
# and Pokémon Trainer Memo, just change 'pbGetTimeNow' to 'Time.now' in their
# scripts.
Do the reverse for the opposite effect if you want.


text:
# FxPaul

Math in finance or vice versa

## Maximum Likelihood Estimation of Stochastic Process Parameters

Maximum Likelihood estimation (MLE) is a method of parameter estimations of statistical model. The base idea is to establish joint density probability for observations and to maximize its value by model’s parameters. To say it differently, we are looking for the most probable explanation of observed data.

### Problem setup

Assume that we’re given a one-dimensional stochastic process:
$dS_t = \mu dt + \sigma dW_t$
where $\mu$ and $\sigma$ are some functions of arguments $\theta$.

We observe this process by measuring $\latex S_i(t_i)$ where $i=1..N$. For sake of simplicity assume that observations are equidistant in time, i.e. $\Delta t = t_{i-1} - t_i = const$.

So, let’s estimate parameters.

Written by fxpaul

November 2, 2011 at 08:00


text:
## 'Triangle Incircle Iteration' printed from http://nrich.maths.org/

If the angles in the first triangle are $a$, $b$ and $c$ prove that the angles in the second triangle are given (in degrees) by
$f(x) = (90 - x/2)$
where $x$ takes the values $a$, $b$ and $c$. Choose some triangles, investigate this iteration numerically and try to give reasons for what happens.


text:
It’s now easier than ever to write major modes in Emacs. If you haven’t written a major mode in a while, or you’re just starting out, here are my three top tips:

## Use regexp-opt

As of Emacs 24, regexp-opt takes a 'symbols option. You should write your font-lock keywords like this:

This has two advantages. By whitelisting keywords, users can quickly spot mistakes when editing:

This also prevents a classic bug where Emacs highlights substrings that happen to be keywords:

## (Don’t) use company

Company is excellent, and I highly recommend it. However, not all Emacsers use company. You don’t need to force company on your users.

Instead, you can use completion-at-point-functions. Your completion functionality will work in stock Emacs, and company users will benefit too through company-capf.

Ah, but what about all the extra annotations you can supply to company? We can have our cake and eat it too:

## Test with assess

Historically, it’s been rather awkward to test major modes. Many authors didn’t bother.

That’s all changed with the release of assess. Assess provides great assertions with readable error messages.

For example, here’s a simple indentation test from cask-mode:

Highlighting is particularly helped by assess:

If this test fails, we get a helpful message describing which faces were actually used:

These tips are all new things I’ve learnt writing a new major mode for Cask files. If you’re just getting started with Emacs development, check out adding a new language to Emacs. Finally, if I’ve missed your favourite tip, leave a comment on the /r/emacs discussion!


text:
# How do you find all zeros with multiplicities of f(x)=-17x^3+5x^2+34x-10?

Jul 14, 2017

The zeros are $x = \frac{5}{17} , x = \sqrt{2} , x = - \sqrt{2}$

#### Explanation:

$f \left(x\right) = - 17 {x}^{3} + 5 {x}^{2} + 34 x - 10$ or

$f \left(x\right) = - 17 {x}^{3} + 34 x + 5 {x}^{2} - 10$ or

$f \left(x\right) = - 17 x \left({x}^{2} - 2\right) + 5 \left({x}^{2} - 2\right)$ or

$f \left(x\right) = \left({x}^{2} - 2\right) \left(- 17 x + 5\right)$ or

$f \left(x\right) = \left(x + \sqrt{2}\right) \left(x - \sqrt{2}\right) \left(- 17 x + 5\right)$

$f \left(x\right) = 0$ When $\left(x + \sqrt{2}\right) = 0 \mathmr{and} x = - \sqrt{2}$ ,

$\left(x - \sqrt{2}\right) = 0 \mathmr{and} x = \sqrt{2}$ and $\left(- 17 x + 5\right) = 0 \mathmr{and} 17 x = 5 \mathmr{and} x = \frac{5}{17}$

The zeros are $x = \frac{5}{17} , x = \sqrt{2} , x = - \sqrt{2}$ [Ans]


text:
• A
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Regular version of the site
Of all publications in the section: 8
Sort:
by name
by year
Article
<NAME>., <NAME>., <NAME>. Russian Journal of Nonlinear Dynamics. 2021. Vol. 17. No. 1. P. 5-21.

We suggest a universal map capable of recovering the behavior of a wide range of dynamical systems given by ODEs. The map is built as an artificial neural network whose weights encode a modeled system. We assume that ODEs are known and prepare training datasets using the equations directly without computing numerical time series. Parameter variations are taken into account in the course of training so that the network model captures bifurcation scenarios of the modeled system. The theoretical benefit from this approach is that the universal model admits applying common mathematical methods without needing to develop a unique theory for each particular dynamical equations. From the practical point of view the developed method can be considered as an alternative numerical method for solving dynamical ODEs suitable for running on contemporary neural network specific hardware. We consider the Lorenz system, the Rцssler system and also the Hindmarch – Rose model. For these three examples the network model is created and its dynamics is compared with ordinary numerical solutions. A high similarity is observed for visual images of attractors, power spectra, bifurcation diagrams and Lyapunov exponents.

Article
<NAME>., <NAME>. Russian Journal of Nonlinear Dynamics. 2020. Vol. 16. No. 2. P. 259-273.

As is well known, many small celestial bodies are of a rather complex shape. Therefore, the study of the dynamics of a spacecraft in their vicinity, based on terms up to the second order of smallness in the expansion of the potential of attraction, seems to be insufficient for an adequate description of the observed dynamical effects related, for example, to positioning of the libration points.In this paper, such effects are demonstrated for spacecraft dynamics in the vicinity of the asteroid (2063) Bacchus. The libration points are computed for various approximations of the gravitational potential. The results of this computation are compared with similar results obtained before for the so-called Sludsky – Werner – Scheeres potential. The dependence of the structure of the regions of possible motions on approximation of the gravitational potential is also studied.

Article
<NAME>., <NAME>., <NAME>. Russian Journal of Nonlinear Dynamics. 2021. Vol. 17. No. 2. P. 165-174.

We consider the system of two coupled one-dimensional parabola maps. It is well known that the parabola map is the simplest map that can exhibit chaotic dynamics, chaos in this map appears through an infinite cascade of period-doubling bifurcations. For two coupled parabola maps we focus on studying attractors of two types: those which resemble the well-known discrete Lorenz-like attractors and those which are similar to the discrete Shilnikov attractors. We describe and illustrate the scenarios of occurrence of chaotic attractors of both types.

Article
<NAME>., <NAME>., <NAME>. Russian Journal of Nonlinear Dynamics. 2021. Vol. 17. No. 3. P. 321-334.

The present paper gives a partial answer to Smale’s question which diagrams can correspond to (A,B)-diffeomorphisms. Model diffeomorphisms of the two-dimensional torus derived by “Smale surgery” are considered, and necessary and sufficient conditions for their topological conjugacy are found. Also, a class G of (A,B)-diffeomorphisms on surfaces which are the connected sum of the model diffeomorphisms is introduced. Diffeomorphisms of the class G realize any connected Hasse diagrams (abstract Smale graph). Examples of diffeomorphisms from G with isomorphic labeled Smale diagrams which are not ambiently Ω-conjugated are constructed. Moreover, a subset G∗ ⊂ G of diffeomorphisms for which the isomorphism class of labeled Smale diagrams is a complete invariant of the ambient Ω-conjugacy is singled out.

Article
<NAME>., <NAME>., <NAME>. et al. Russian Journal of Nonlinear Dynamics. 2019. Vol. 15. No. 2. P. 199-211.

We consider the class $G$ of gradient-like orientation preserving diffeomorphisms of the 2-sphere with saddles of negative orientation type. We show that the for every diffeomorphism $f\in G$ every saddle point is fixed. We show that there are exactly three equivalence classes (up to topological conjugacy) $G=G_1\cup G_2\cup G_3$ where a diffeomorphism $f_1\in G_1$ has exactly one saddle and three nodes (one fixed source and two periodic sinks); a diffeomorphism $f_2\in G_2$ has exactly two saddles and four nodes (two periodic sources and two periodic sinks) and a diffeomorphism $f_3\in G_3$ is topologically conjugate to a diffeomorphism $f_1^{-1}$. The main result is the proof that every diffeomorphism $f\in G$ can be connected to the source-sink'' diffeomorphism by a stable arc and this arc contains at most finitely many points of period-doubling bifurcations.

Article
Nozdrinova E. Russian Journal of Nonlinear Dynamics. 2018. Vol. 14. No. 4. P. 543-551.

The problem of the existence of a simple arc connecting two structurally stable systems on a closed manifold is included in the list of the fifty most important problems of dynamical systems. This problem was solved by <NAME> and <NAME> for Morse-Smale flows on an arbitrary closed manifold in 1980. As follows from the works of <NAME>, <NAME>, <NAME>, <NAME>, <NAME>, for the Morse-Smale cascades, obstructions to the existence of such an arc exist on closed manifolds of any dimension. In these works, necessary and sufficient conditions for belonging to the same simple isotopic class for gradient-like diffeomorphisms on a surface or a three-dimensional sphere were found. This article is the next step in this direction. Namely, the author has established that all orientation-reversing diffeomorphisms of a circle are in one component of a simple connection, whereas the simple isotopy class of an orientation-preserving transformation of a circle is completely determined by the rotation number of Poincare.


text:
## mathdummy4 Group Title i just need help with the formula here...find the number of ways in which each committee can be selected. 3 people from a group of 5 2 years ago 2 years ago

1. vinrar

5C3

2. lgbasallote

the formula would be combinations $nCr = \frac{n!}{r!(n-r)!}$ does that help?

3. mathdummy4

so well thats what i did...5!/(2!-3!)...and the answer was wrong

4. lgbasallote

that's because it's $\frac{5!}{3!2!}$not minus

5. amistre64

5P3 ---- 3!

6. amistre64

5.4.3 5.4 ----- = --- = 20/2 3.2.1 2

7. mathdummy4

oo okay oops!...thanks:)


text:
# Is it incorrect to say that "passenger jets like to cruise at higher altitudes because there's less drag"?

I hear it stated quite often that aircraft like to fly high because there's less drag at higher altitudes.

$$\text{Range} = \frac{V}{g} \frac{1}{\text{SFC}} \frac{L}{D} \ln \left( \frac{W_\text{initial}}{W_\text{final}} \right) \; .$$

Aside from SFC, it seems that none of the components of the range equation are altitude dependent. Lift/Drag should remain constant assuming the lift and drag coefficients remain constant. Therefore, is it incorrect to say that there is a benefit to range flying at a higher altitude due to decreased drag?

Another question is, does the lift/drag ratio of an aircraft increase somehow at higher altitudes?

There's less drag at the same speed (and other things being equal, aside from altitude).

This is due to lower air density at higher altitude, which linearly affects drag. But just the same, it affects lift! You get less drag, but less lift too. If you want to fly level, you can't afford that: lift must be equal weight.

So what do you do to restore lift? You either fly faster, or increase angle of attack (AoA), or both. Either way, drag will roughly be the same.

Flying faster with the same drag is the benefit. However, sooner or later you'll hit the speed limit. (Most commonly for jets, this will be the Mach limit). From then on, you can only increase AoA.

Changing the AoA affects the L/D ratio. It may increase or decrease, depending on how the wing was designed. There is a certain AoA which will be the 'best'. If the jet was optimised for cruise, it will fly close to that angle.

Does it make any sense to fly even higher, where you have to increase AoA past the best L/D angle? It might, but not much. Primarily because you might squeeze a little better fuel consumption in colder air (that SFC component), so the optimum may be slightly higher. But flying at a higher-than-optimal AoA may be unstable by speed. (Which is manageable, but is another difficulty you don't want).

At some point, you will exhaust either the AoA, getting dangerously close to stall (while, remember, being at the speed limit at the same time - the so called 'coffin corner'), or thrust: it will keep reducing in thinner air, until it becomes lower than drag. That's your ceiling.

• The first sentence could be clarified by stating less drag at the same indicated airspeed. (Because TAS is higher...) Nov 18, 2020 at 0:51
• Ah yes, this all makes sense. To clarify, will an altitude be reached where the optimal cruise Mach for an aircraft will no longer provide enough lift at maximum L/D? Can this altitude be above the troposphere, and can flight planning put the plane above the troposphere to extend range? Thanks, this question has been on my mind. Nov 18, 2020 at 1:14
• @MichaelHall don't you mean less drag at the same TAS? The drag at the same indicated a/s should be the same, the dynamic pressure being the same. Nov 18, 2020 at 1:33
• @RoryMcDonald most airliners already cruise above the troposphere except in the tropical lats where the trop is closer to 40000 ft. At mid latitudes it's in the low-mid 30s, and in the north in the 20s, and as low as the teens in the arctic in winter. Nov 18, 2020 at 1:38
• @John K, no... I meant what I said, but probably not clear. Trying to point out that for a constant drag at cruise IAS there is a benefit to higher TAS for the same IAS at altitude. Nov 18, 2020 at 21:35


text:
# rotation to quaternion matrix handeness

I've understand that quaternions do not have handness but rotation matricies derived from unit quaternions does.

The following formula is given by wikipedia for quaternion to rotation matrix conversion :

Given the unit quaternion $q = w + xi + yj + zk$ , the equivalent left-handed (Post-Multiplied) 3×3 rotation matrix is $$Q = \begin{bmatrix} 1 - 2 y^2 - 2 z^2 & 2 x y - 2 z w & 2 x z + 2 y w \\ 2 x y + 2 z w & 1 - 2 x^2 - 2 z^2 & 2 y z - 2 x w \\ 2 x z - 2 y w & 2 y z + 2 x w & 1 - 2 x^2 - 2 y^2 \end{bmatrix} .$$

As mentioned, this formula is relative to a left-handed coordinate frame. What's the right-handed counterpart ?

Best : ) ,

• What have you tried? The difference between conventions is just a matter of quite straight forward calculations. – skyking Jan 29 '16 at 10:41
• I've a bug in my software and i want to search for possible candidates : i want to be 100% sure that i'm not missing something with those handness conventions. – jcolafrancesco Jan 29 '16 at 10:52
• Then it would maybe be a point to post what conventions you're using (fx your coordinate system, the way you apply rotation matrixes and the way you apply quaternions) and what formulas you're using. If nothing else it would show that you've done some own effort. – skyking Jan 29 '16 at 11:00

Hint:

Given the unit quaternion $q = w + x \vec i + y\vec j + z \vec k=w+ \vec v$, the matrix $Q$ results from the representation of a rotation of a vector $\vec p=p_x \vec i+p_y \vec j+ p_z \vec k$ as: $$Q(\vec p)= q \vec p q^{-1}$$ The resulting rotation is a rotation of angle $\theta$ around an axis oriented by a versor $\vec u$ such that:

$$\cos \frac{\theta}{2}=\frac{w}{|q|} \qquad \sin \frac{\theta}{2}=\frac{|\vec v|}{|q|}$$ and $$\vec u=\frac{\vec v}{|\vec v|}$$

This is a counter-clockwise rotation $R_{\vec u, \theta}$ around the axis $\vec u$.

You can find the clockwise rotation around the same axis changing the angle to $-\theta$, or inverting the orientation of the versor $\vec u$ (note that if you perform all the two transformation the rotation remain the same).

What does this means for the quaternion $q$ ?

• I'm i right if i say that right-handed version is just the transpose of the left-handed ? – jcolafrancesco Jan 29 '16 at 13:31
• Change the sigh of $theta$ or the sigh of $\vec u$ is the same as take the conjugate of $q$, i.e. $\bar q=w-xi-yj-zk$ and this gives a matrix $Q'$ that is the transpose of $Q$... so : Yes! you are right. – <NAME> Jan 29 '16 at 13:39


text:
# Lesson 7

Exploring the Area of a Circle

### Lesson Narrative

This lesson is the first of two lessons that develop the formula for the area of a circle. Students start by estimating the area inside different circles, deepening their understanding of the concept of area as the number of unit squares that cover a region, and discovering that area (unlike circumference) is not proportional to diameter.

Next, they investigate how the area of a circle compares to the area of a square that has side lengths equal to the circle’s radius. Students may choose tools strategically from their geometry toolkits to help them make these comparisons (MP5). Students find an approximate formula: the area of a circle is a little bigger than $$3r^2$$, and they check their earlier estimates with this formula. At this point, it is a reasonable guess that the exact formula is $$A=\pi r^2$$, but the next lesson will focus on using informal dissection arguments to establish this formula.

When we say “area of a circle” we technically mean “area of the region enclosed by a circle.” However, “area of a circle” is the phrase most commonly used.

### Learning Goals

Teacher Facing

• Create a table and a graph that represent the relationship between the diameter and area of circles of various sizes, and justify (using words and other representations) that this relationship is not proportional.
• Estimate the area of a circle on a grid by decomposing and approximating it with polygons.

### Student Facing

Let’s investigate the areas of circles.

### Required Preparation

For the first activity, you will need the Estimating Areas of Circles blackline master. Prepare 1 copy for every 12 students. (Each group of 2 students gets one of the pages.)

For the second activity, make sure students have access to their geometry toolkits, especially tracing paper and scissors, if they so choose (but try not to influence students' choices about what tools to use).

### Student Facing

• If I know a circle’s radius or diameter, I can find an approximation for its area.
• I know whether or not the relationship between the diameter and area of a circle is proportional and can explain how I know.

Building Towards

### Glossary Entries

• area of a circle

If the radius of a circle is $$r$$ units, then the area of the circle is $$\pi r^2$$ square units.

For example, a circle has radius 3 inches. Its area is $$\pi 3^2$$ square inches, or $$9\pi$$ square inches, which is approximately 28.3 square inches.

### Print Formatted Materials

For access, consult one of our IM Certified Partners.


text:
# Variation of scalar curvature upon frame deformation

Following the textbook "Ideas and Methods of Supersymmetry and Supergravity" by <NAME> and <NAME> (p38 - 40):

If we consider the a frame deformation in our vierbein induced by a symmetric rank 2 Lorentz tensor $H$:

$$e^m_a \rightarrow e_a^m + H_a^b e^m_b$$

One can easily show: $$g_{mn} = e^a_m e^b_n \eta_{ab}$$ $$\implies \delta g_{mn} = -2 e_m^a e_n^b H_{ab}$$ $$\implies H_{ab} = - \frac{1}{2} e^m_a e^n_b \delta g_{mn}$$

The next goal would be to identity how the curvature varies with respect to the variation in the metric so that various invariants could be identified (p40-41), however upon attempting to find the variation in the scalar curvature:

$$\delta R = 2 \nabla^c \nabla_c H^a_a - 2 \nabla^a \nabla^b H_{ab} + 2 H^{ab}R_{ab}$$ $$\implies \delta R = - \nabla^c \nabla_c (\eta^{ab} e_a^{m}e_b^{n} \delta g_{mn}) + \nabla^a \nabla^b (e_a^{m}e_b^{n} \delta g_{mn}) - (e_a^{m}e_b^{n} \delta g_{mn})R_{ab}$$

By using the compatibility of the vielbein with the covariant derivative:

$$\delta R = - \eta^{ab} e_a^{m}e_b^{n} \nabla^c \nabla_c \delta g_{mn} + e_a^{m}e_b^{n} \nabla^a \nabla^b \delta g_{mn} - e_a^{m}e_b^{n} \delta g_{mn}R_{ab}$$

Which does not seem to be of the correct form. Is there anything further I could try or have I approached this incorrectly?

A useful equation might be $$g^{\mu\nu} \delta R_{\mu\nu} = (g_{\mu\nu}\nabla^2 -\nabla_\mu \nabla_\nu) \delta g^{\mu\nu},$$ but may be you are already using this?
I don't think that you have the compatibility correct, though. The connection is defined by specifying the covariant derivative of the basis vectors of $T(M)$. In the case of a vielbein frame ${\bf e}_a$, the covariant derivative of the vector ${\bf e}_a$ is written as $$\nabla_\mu{\bf e}_a = {\bf e}_b {\omega^b}_{a\mu}$$ which one can write in coordinate-frame components (i.e. where ${\bf e}_a=e_a^\mu \partial_\mu$) as $$\nabla_\mu e^\nu_a \equiv (\nabla_\mu{\bf e}_a)^\nu= e^\nu_b {\omega^b}_{a\mu}.$$ The covariant derivative of the vierbein basis is therefore not zero. I have seen people argue for passing vierbeins through covariant derivatives by writing this last equation as $$\partial_\mu e^\nu_a + {\Gamma^\nu}_{\lambda\mu} e^\lambda_a - e^\nu_b {\omega^b}_{a\mu}=0$$ and thinking of this as a kind of "generalized" covariant derivative being zero. In doing this they are making the mistake of imagining that the the "$a$'' in $e_a^\mu$ is an index rather than a label telling us which frame vector ${\bf e}_a$ is. It's perhaps a useful mnemonic, but is also kind of schizophrenic, as they are attempting to work simultaneously with a vielbein frame and a co-ordinate basis for the tangent space $T(M)$. It makes no mathematical sense to interpret the definition of the frame connection ${\omega^b}_{a\mu}$ that way. Certainly the expression $$\partial_\mu e^\nu_a + {\Gamma^\nu}_{\lambda\mu} e^\lambda_a - e^\nu_b {\omega^b}_{a\mu}$$ is not the $\nu$-th component of the covariant derivative $\nabla_\mu {\bf e}_a$! Treating it as if it were will inevitably lead to confusion, and this is what I suspect is happening in your calculation.
Another useful formula for the variation of the tosion-free spin connection under a change of vielbein frame is $$(\delta \omega_{ij\mu}) e^\mu_k =-\frac 12\left\{( \eta_{ib}( \nabla_j [e^{*b}_\alpha \delta e^\alpha_k]- \nabla_k [e^{*b}_\alpha \delta e^\alpha_j]) +\eta_{jb}( \nabla_k [e^{*b}_\alpha \delta e^\alpha_i]- \nabla_i [e^{*b}_\alpha \delta e^\alpha_k]) -\eta_{kb}( \nabla_i [e^{*b}_\alpha \delta e^\alpha_j]- \nabla_j [e^{*b}_\alpha \delta e^\alpha_i])\right\}.$$ where I think my $\eta_{ib}e^{*b}_\alpha \delta e^\alpha_j= {\bf e}_i\cdot \delta{\bf e}_j$ are the same thing as your $H_{ij}$
• @raptakem.Yes eq vanished. we had a power outage as I was ediing it. I don't think torsion-free is relevent. The problem is the multiple roles of the "$a$" in $e_a^\mu$. I'll amend my answer to discuss this. – mike stone Jun 11 '18 at 12:35