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# Publications

Gonçalo M. Correia, Vlad Niculae and André F.T. Martins. In Proceedings of EMNLP, 2019

Comments: Presented as a Talk in Machine Learning Session III.

Attention mechanisms have become ubiquitous in NLP. Recent architectures, notably the Transformer, learn powerful context-aware word representations through layered, multi-headed attention. The multiple heads learn diverse types of word relationships. However, with standard softmax attention, all attention heads are dense, assigning a non-zero weight to all context words. In this work, we introduce the adaptively sparse Transformer, wherein attention heads have flexible, context-dependent sparsity patterns. This sparsity is accomplished by replacing softmax with $\alpha$-entmax: a differentiable generalization of softmax that allows low-scoring words to receive precisely zero weight. Moreover, we derive a method to automatically learn the $\alpha$ parameter -- which controls the shape and sparsity of $\alpha$-entmax -- allowing attention heads to choose between focused or spread-out behavior. Our adaptively sparse Transformer improves interpretability and head diversity when compared to softmax Transformers on machine translation datasets. Findings of the quantitative and qualitative analysis of our approach include that heads in different layers learn different sparsity preferences and tend to be more diverse in their attention distributions than softmax Transformers. Furthermore, at no cost in accuracy, sparsity in attention heads helps to uncover different head specializations.
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@inproceedings{Correia2019b, author = {Correia, Gonçalo M. and Niculae, Vlad and Martins, André F. T.}, booktitle = {Proceedings of EMNLP}, title = {Adaptively Sparse Transformers}, year = {2019}}
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## Unbabel's Submission to the WMT2019 APE Shared Task: BERT-based Encoder-Decoder for Automatic Post-Editing

António V. Lopes, M. Amin Farajian, Gonçalo M. Correia, Jonay Trenous, André F.T. Martins. In Proceedings of WMT, 2019

This paper describes Unbabel's submission to the WMT2019 APE Shared Task for the English-German language pair. Following the recent rise of large, powerful, pre-trained models, we adapt the BERT pretrained model to perform Automatic Post-Editing in an encoder-decoder framework. Analogously to dual-encoder architectures we develop a BERT-based encoder-decoder (BED) model in which a single pretrained BERT encoder receives both the source src and machine translation tgt strings. Furthermore, we explore a conservativeness factor to constrain the APE system to perform fewer edits. As the official results show, when trained on a weighted combination of in-domain and artificial training data, our BED system with the conservativeness penalty improves significantly the translations of a strong Neural Machine Translation system by and in terms of TER and BLEU, respectively.
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@inproceedings{Lopes2019, author = {Lopes, António V. and Farajian, M. Amin and Correia, Gonçalo M. and Trenous, Jonay and Martins, André F. T.}, booktitle = {Proceedings of WMT19}, title = Unbabel's Submission to the WMT2019 APE Shared Task: BERT-based Encoder-Decoder for Automatic Post-Editing, year = {2019} }
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## A Simple and Effective Approach to Automatic Post-Editing with Transfer Learning

Gonçalo M Correia, André FT Martins. In Proceedings of ACL, 2019


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Thermodynamically reversed black holes, firewalls, Casimir effect, null energy condition violations

Scott Aaronson asked a very deep question at Hawking radiation and reversibility about what happens if black hole evolution is reversed thermodynamically. Most of the commenters missed his point entirely. Of course it's overwhelmingly improbable statistically to set up the initial conditions needed to get a thermodynamically reversed black hole evolution. But that's not the point of the question at all. It is theoretically possible to have a thermodynamically reversed black hole evolution as a valid evolving state by CPT symmetry, even if it's improbable for all practical purposes. As it exists as a possible solution, we can ask questions about the properties of such a solution.

In a thermodynamically probable evolution, we can have a massive object with a significant fraction of the black hole's mass fall into a black hole, and the blak hole automatically increases in mass by a huge fraction over a timescale of order of the Schwarzschild radius R. Then, it slowly evaporates away Hawking radiation with a remaining lifetime of order $R^3$. In the thermodynamically reversed version of this event, over a time period of order $R^3$, thermal radiation is fed into a black hole which increases in size very gradually as a result. There is an incredible fine-tuning of the ingoing radiation fed in with fine-tuned higher order multipartite entanglement (this is a theoretical state, not a practical engineering one) so that as a result, over the time period of order $R^3$, no Hawking radiation is emitted from the black hole because of fine-tuned quantum cancellations between the contributions from the black hole, and the contribution from frequency mixing of the radiation fed into the hole. Then, after a time period of order $R^3$, suddenly over a time period of order $R$, the black hole emits as "Hawking radiation" a massive body a huge fraction of its mass, and the black hole's mass decreases accordingly. The area of its event horizon goes down by a huge fraction over a time period of order R. According to Raychaudhuri's optical equation, this requires a significant negative null energy. This is too large to come from Casimir effects, as that would require a time period of order $R^3$ to lose that much black hole mass. So, what is the origin of this huge negative null energy?

Let's look at what happens just outside the black hole before the massive body is ejected. In the semiclassical approximation, the massive body must have come from a firewall ( What are cosmological "firewalls"?) at an exponentially small distance above the event horizon. This exponential factor comes from the very high boost factor and Lorentz contraction. Semiclassically, this firewall couldn't have originated from entangled Hawking pair production just outside the horizon. So, it must have originated from the fine-tuned radiation fed into the hole.

What if a probe is sent during an earlier time to measure this firewall? Unfortunately, this probably can't be done. See, we are conditioning a significant part of the future state by stipulating that a huge massive body is ejected as Hawking radiation. To further specify a probe hovering just above the horizon requires in addition a significant condition on the past. Ordinarily, with no thermodynamic conspiracies, we only have past conditions and no future conditions, and so, there's no problem. However, such fine-tuned conditioning of both the past and future might actually be impossible?

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The question is only getting one point wrong, which is that the Raychoudhuri equation requires the reversed black hole whose horizon shrinks by emitting matter to have negative stress energy.

The reason this last point fails in the reversed case is becuase the horizon character is reversed--- the white hole horizon is a past horizon, it's another extension of the exterior solution. Stuff can come out of the white hole horizon, leading the black hole to shrink.

The past horizon in a white hole can be understood by considering the four quadrants of Minkowski space in Rindler coordinates: in the metric

$$ds^2 = dx^2 - dt^2$$

Has four regions picked out by the origin: I --- (spacelike and to the right of the origin), II-- (spacelike and to the left of the origin) III-(interior of future light cone of origin) IV--- (interior of past light cone of origin).

When you make a horizon by shifting to Rindler coodinates, which are relativistic polar coordinates for the origin, or the coordinates appropriate to a family of accelerated observers that consider each other mutually stationary, you find that the null rays from the horizon, both past and future, become event horizons for the accelerated observers.

In a Schwarzschild metric, near the horizon, the metric becomes Rindler, and therefore the maximal extension interprets each point on the horizon as both a future horizon, into which things can fall, and a past horizon from which things can come out. The past-horizon doesn't make sense near the point of formation of the black hole, and the future horizon doesn't make sense near the point of evaporation. Because collapse in classical GR doesn't have evaporation, people generally said "the past horizon intepretation is unphysical, and the future horizon is physical". But this is not 100% accurate, as with evaporation, both horizons are equally physical for a black hole in equilibrium. The future region, the analog of III, is where infalling perturbations go, and the past region, the analog of IV, is where the past of fluctuations in the Hawking radiation comes from.

This means that in your example, the time-reversed process emits the collapsing star from a past horizon, and this emission shrinks the black hole without any violation of positive energy. Past horizons can only shrink for the same reason that future horizons can only grow. The horizon in thermal equilibrium can only grow only by entropic thermodynamic considerations--- the entropy is unlikely to go down.

So once you accept Susskind's holographic duality, and the complementarity, a static black hole formed long ago has both extensions by different transformation on the physical exterior-measurable variables, it has a past region equally as a future region. It is just a conspiracy for such a black hole to emit stuff non-thermally from it's past region.

This argument is only valid, the nonthermal emissions are only conspiratorial thermodynamically, when the horizon has rung-down to thermal equilibrium. This doesn't happen for near-extremal black holes, whose normal modes are only slightly damped, with a damping time which diverges at extremality. There is no theoretical barrier for non-thermal emissions from such near-extremal black holes, which is because they have a hard time thermalizing.

It is my opinion, not shared by anyone else, that in the extremal case, the matter that falls generally does a full circuit of the interior regions, and comes out of the white hole past region after a finite time. Ignoring the past extension of the black hole is a legacy of pre-evaporation pictures of black holes, and doesn't make sense in the modern AdS/CFT context, where extremal black holes are fully reversible.

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But surely a description of Aaronson's scenario fully in terms of a blavk hole without invoking white holes ought to be possible? –  Bewildered Oct 29 '12 at 13:34
At any rate, if the only way to explain this scenario is by treating the system as a white hole, what if after ejecting the massive body, we now dump a lot of matter into the hole and forget about any further fine-tuned feeding of matter. If it's a white hole, it now has to convert into a black hole, right? –  Bewildered Oct 29 '12 at 13:37
@Bewildered: Black holes and white holes are the same thing--- they are both described in the stationary limit by maximally extended Schwarzschild. They are both black and white away from the initial point of formation or the end point of evaporation, and both extensions (white and black) are from rearranging exterior variables in the quantum description, but in incompatible ways. –  Ron Maimon Oct 29 '12 at 14:14
That's only the case for an eternal Schwarzschild solution. For a hole which formed a finite time in the past, and will end in a finite time in the future, they have different Penrose diagrams. –  Bewildered Oct 29 '12 at 14:34
@Bewildered: They have mutually inconsistent Penrose diagrams. You can't have a past horizon for a black hole which formed, and you can't have a future horizon for a black hole which decays. Nevertheless, the correct classical description includes both extensions away from the two endpoints, and this is something which was only sorted out with Susskind's stuff. –  Ron Maimon Oct 29 '12 at 14:49

The two-state formalism of Aharanov can answer your question. True, in ordinary QM, there's a limit as to how low the Casimir negative null energy can go.

In What is the physical meaning of weak expectation values?, it was pointed out that the weak expectation value $\langle \chi |A|\psi \rangle/\langle \chi |\psi\rangle$ can be much larger in magnitude than the eigenvalues of A. Similarly, $\langle \chi |\widehat{T}_{vv}(x)|\psi \rangle/\langle \chi |\psi\rangle$ where $\widehat{T}_{vv}(x)$ is the null energy operator can have a much more negative value than permitted by the Casimir effect in ordinary QM where $|\psi\rangle=|\chi\rangle$. What is needed is the case where $|\psi\rangle$ and $|\chi\rangle$ are nearly orthogonal, but not quite.

However, with such huge null energy differences between $\langle \chi |\widehat{T}_{vv}(x)|\chi \rangle$ and $\langle \psi |\widehat{T}_{vv}(x)|\psi \rangle$, backreaction effects will mean both states will have significantly different metrics, and hence, significantly different causal structures, which means the locations of their horizons will differ significantly.

At any rate, don't expect your ejected massive body to come from frequency mixing. If you do your math, you'd find frequency mixing only becomes significant when the wavelength is around the Schwarzschild radius R. The de broglie wavelength of your massive body will be much smaller. This means frequency mixing will be highly suppressed. Your massive body has to come straight from the firewall, and not from Hawking radiation. As another person so aptly put it, it has to be "born in the firewall".

Although come to think about it, if the massive body came from the firewall outside the horizon, why do we need negative null energies at the horizon?

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The comments about the case with conspiracies until a massive body is emitted, followed by dumping in a massive body without conspiracies is interesting. What if it's the same massive body?

Wild wild crazy thought experiment coming:

A conspiracy to feed the hole until a macroscopic sentient observer is pushed out of the hole. This sentient observer tells his story about his life, his conception at the past singularity, and his birth pushed out violently out of the white hole horizon. Then, after telling his fantastic story, his duty is done, and he jumps back into the hole to die. This same observer.

Black hole Penrose diagram version of events: His story is a LIE!!! We conspired to feed the hole until it pushes out this sentient observer, but as fellow conspirators, we planned out all the fine details of what comes out, including the brain memories of this observer, and all the records in his spaceship. They're all implanted false memories. The observer may swear all he wants about the truth of his experiences and memories but we know better, because we implanted them conspiratorially. We'll just nod and wink among ourselves, and exchange knowing glances, even while this observer keeps on protesting. We know better. There is no white hole. Only a black hole. A black hole which emitted a sentient observer with false memories as Hawking radiation coming out from the firewall just outside the horizon, where he was born. Then, this observer jumps back into the hole, passes through the horizon, and dies violently at the future singularity.

White hole Penrose diagrams version of events: It's a white hole, not a black hole. Miraculously, a sentient observer is conceived at the past singularity, and is pushed out through the horizon. He lives to tell his true miraculous life story. Then, he jumps back into the hole. As he approaches the white horizon, he is bombarded by radiation coming out from the white hole as he asymptotically approaches the horizon. He meets his death by this bombardment. Mercifully, this bombardment causes hallucinations in his brain so that just before he is finally destroyed, he hallucinates a future life inside the hole which in his hallucinatory state, he thinks it's a black hole. He doesn't realize he's already dying even as he hallucinates. He doesn't live to tell the story about his hallucination either.

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# Proof Verification: x^n-a^n is divisible by x-a for all n [duplicate]

Proposition: $$x^n-a^n$$ is divisible by $$x-a$$ for all n. Potential proof by induction (which I would like to know whether it is correct):

Suppose n=1. Then $$x^n-a^n=x-a$$, and $$x-a$$ is clearly divisible by $$x-a$$. So the proposition holds for n=1, the base case.

Now suppose the proposition holds for n=k. This means that $$x^k-a^k$$ is divisible by $$x-a$$.

Now we have to prove that the proposition holds for n=k+1. Then it has to be proved that $$x^{k+1}-a^{k+1}$$ is divisible by x-a. $$x^{k+1}-a^{k+1}$$ is the same as $$x^k \cdot x - a^k \cdot a$$, which is equivalent to $$(x^k-a^k) \cdot ax$$. However, it was already established that the proposition holds for n=k, so $$x^k-a^k$$ is divisible by x-a. This renders the expression $$(x^k-a^k) \cdot ax$$ divisible by x-a; when this expression is divided by x-a, only $$ax$$ is left.

I don't know if my proof is correct since I just learned induction.

• Does this answer your question? $a-b$ divides $a^n-b^n$ . Found using Approach0. Note another duplicate is $x-y$ divides $x^n - y^n$ -- prove by mathematical induction. May 7 at 17:12
• You may be interested to know that for a polynomial $p(x)$ it is always true that $p(x)-p(a)$ is divisible by $x-a$. If follows from the statement you are trying to prove, but it can also be shown more briefly using polynomial division. $p(x)-p(a)=q(x)(x-a)+b$ for some polynomial $q(x)$ and some constant remainder $b$. Now set $x=a$ to obtain $b=0$. This conceals the induction (the division algorithm requires it), but is a neat proof and embodies a potentially useful technique. May 7 at 17:15

The step $$x^k\cdot x - a^k \cdot a = (x^k - a^k)\cdot ax$$ is wrong.
A simple counterexample is $$3^2\cdot 3 - 4^2\cdot 4 \neq (3^2 - 4^2)\cdot (3\times4)$$.
To correct the proof, substitute $$x^k = p + a^k$$ with the assumption that $$p$$ is divisible by $$(x-a)$$.
The structure of your proof by induction is correct, however you have an invalid algebraic step. You claim $$x^kx - a^ ka = (x^k -a^k)ax$$, but that is not true. You will need to change your algebraic steps in the proof, but you would have been right except for that part, so you understand induction alright!


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# Positive semidefiniteness of submatrix of positive semidefinite matrix

Assume $A$ is a real symmetric positive semidefinite $k\times k$ matrix. Denote by $A^{(a:b,c:d)}$ the submatrix of $A$ obtained by keeping only the rows between $a$ and $b$, and columns between $c$ and $d$. Now define the $(k-1) \times (k-1)$ matrix defined by

$\tilde{A} = A^{(1:k-1,1:k-1)} - \bigg(A^{(1:k-1,k)}(A^{(k,k)})^{-1}\bigg)A^{(k,k)}\bigg(A^{(1:k-1,k)}(A^{(k,k)})^{-1}\bigg)^{\prime}$.

I'm pretty sure that this matrix is positive semidefinite, but I need some to help to verify that claim rigorously.

## 1 Answer

Writing $$A = \pmatrix{ A_1 & a \\ a^T & c}$$ we have $$\tilde A = A_1 - aa^T c^{-1}.$$ Let $x\in \mathbb R^{k-1}$. Then $$x^T\tilde Ax = x^TA_1x - (x^Ta)(a^Tx) c^{-1}.$$ We want to use the matrix $A$, so let me augment $x$ by a real number $b$ to obtain $$0\le \pmatrix{x \\ b }^TA\pmatrix{x \\ b } = x^TA_1x + 2b a^Tx + b ^2c.$$ Setting $b=-c^{-1} a^Tx$ yields $$0\le \pmatrix{x \\ -c^{-1} a^Tx }^TA\pmatrix{x \\ -c^{-1} a^Tx }=x^TA_1x - (x^Ta)(a^Tx) c^{-1},$$ which shows positive semidefinitness.


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### Spacetime analogies

The elastic sheet model is one of the most used model for telling general relativity. It allows to show as smaller balls "orbit" around the larger ball in the center of the sheet in a way similar to the planets, at least until the balls lose energy due to friction and finally fall into the "gravitational wall".
This way to see the general relativity is directly connected with the embedded diagrams and Flamm paraboloids, the mathematical way to see the spacetime deformations. But this analogy has some problems not only because is inaccurate like all analogies, but also becuase it could be confusing about distorced space and spacetime especially among students. So we can ask: why is the sheet deformed? Because of the weight? This fact implies the use of a circluar argument: usinf gravity to explain gravity! But if the ball isn't in spacetime, where is it?(1)
Despite this objections, I must note that the previous questions have never been exposed in the few activities with the elastic sheet I made, probably because I ever try to explain that the elastic sheet is only an analogy needed to see some particular aspects about gravity. In every case I think that it could be very interesting to use an alternative method to tell spacetime deformation.
Suppose we want to go from Ottawa to Venice by plane. This, which moves relatively close to the planet's surface, will make an arc of circumference to go from the canadian town to the italian one, instead of taking the straight line that connects the two cities directly. To see how, in the presence of gravity, this curve is the most suitable one, we can show three distinct kinds of relativistic curves on the spacetime plane.
In this kind of diagrams, line $a$ is a particle at rest, the line $b$ is a particle moving with constant speed, the line $c$ is a particle moving with constant acceleration. Therefore the arc of circumference between Ottawa and Venice is the curve of an accelerated motion with acceleration equal to \$g£, in the case of the Earth(1).
Further reading: Deviation of Light near the Sun in General Relativity by Christian Magnan: original version, and archived version.
1. Janis, A. I. (2018). On mass, spacetime curvature, and gravity. The Physics Teacher, 56(1), 12-13. doi:10.1119/1.5018679 (4shared


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# Jobs That Use Right Triangle Trigonometry

Just as a fun fact, the fact that side lengths have ratios when the angles are the same is fundamental in trigonometry as well, as you use side ratios. 5) Ready, Set, Go Homework: Similarity & Right Triangle Trigonometry 6. In this lesson you will use what you have learned to solve problems. These involve ratios of the lengths of the sides in a right triangle. Triangles on a sphere are also studied, in spherical trigonometry. This study analyzes textbooks from the eighteenth and nineteenth centuries, reviews in contemporary. math 1316 - plane trigonometry In-depth study and applications of trigonometry including definitions, identities, inverse functions, solutions of equations,graphing, and solving triangles. You can type trigonometry into the search form on our Math Beyond School section and you'll find some articles that describe other places for trig. Learn right triangle trigonometry with free interactive flashcards. Graph the six trig functions, apply transformations to their graphs and write trig equations that model periodic phenomena. Any ideas? and if so, how does the profession take use of trigonometry?. Trigonometry frequently uses both measurements, so learn to convert fluidly between them in this theme. 71 degrees (which I expected) But on java it returns: 0. Trigonometry (from Greek trigōnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships between side lengths and angles of triangles. Students move at their own speed through the self-instructional "PACE" workbooks. Sines and cosines are two trig functions that factor heavily into any study of trigonometry; they have their own. The Quiz Mixer allows you to take a prewritten BrainPOP game quiz and either customize/remix it or use it out of the box. In this lesson you will use what you have learned to solve problems. Start studying Unit 7: Right triangles. Since the radius of a unit circle is one, the hypotenuse of these triangles is also one. Use of special Right triangles. I am not thrilled that we have to worry about it in this course, and come at it from the perspective of similar right triangles (instead of unit circle/wrapping function), but that decision is above my paygrade. Develop a graphic organizer or foldable that anchors the special right triangles and right-triangle trigonometry to the Pythagorean Theorem. Prove the Laws of Sines and Cosines and use them to solve problems. Carpenter use it to make cabinets, surveyors use it in their work, you can use what you've learned to measure tall structures. Although that was in 1965-66 I still have a complete understanding of trig and believe that I could make most students understand how it works. Your task is to complete the following activities to practice using the trigonometric ratios to find missing side lengths in triangles. Do the Law of Sines and the Law of Cosines apply to all triangles? Particularly, could you use these laws on right triangles? That is, could you use these laws instead of the Sine=opposite/hypote. We know the formulas pretty well by now. For other math terms see Algebra and Geometry. The only information we have is about the Gandalf triangle, so that's the best place to start. He wants to tie a 40-foot rope to two poles so that he can practice for his upcoming circus show. In the previous set of problems, you were given one side length and one angle. Shop Math: Angles, Triangles & Trigonometry is part four of the Shop Math five-part training series. When applicable students must rationalize the denominator. This assignment continues the work from class today by having students complete right triangle trigonometry problems in context. Apply trigonometry to general triangles 9. An Academy Award winner for sure. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Pythagoras’ theorem and trigonometry are two of those classic topics that pupils revisit year-on-year. Understanding how to translate word problems into mathematical solutions is an essential skill. There are other formula for general triangles such as the Law of sines $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ and the cosine rule $a^2 = b^2 + c^2 - 2 b c \cos A$ where a, b, c are the side lengths and A, B,. Section 6: Trigonometric Functions of Any Angle. First, use the Pythagorean theorem to find the length of the third side: Next, use trigonometric functions to solve for the acute angles: Now you know that sin A = 5 / 8, but you are trying to find out the value of Trigonometry > Trigonometry - Right Triangles Now that you understand inverse trig functions, this opens up a whole new set of problems you can solve. Here is the total set of formulas compiled for students of class XI-XII and those giving various entrance exams based on 10+2 syllabus. Ready, Set, Go Homework: Similarity & Right Triangle Trigonometry 6. Many times, we can use the Pythagorean theorem to find the missing legs or hypotenuse of 45 45 90 triangles. It originates in the study of geometry when we investigate the ratios of sides in similar right triangles, or when we look at the relationship between a chord of a circle and its arc. Use this lesson as a refresher of what trig ratios are and how they work. In practical use, trigonometry is a friend to astronomers who use triangulation to measure the distance between stars. These involve ratios of the lengths of the sides in a right triangle. If you are given the measure of one of the non-right angles and one of the sides, you can find all the values of the right triangle. But its not a good idea. Discover the unit circle and how when combined with triangles they encompass trigonometry. Trigonometry is the branch of mathematics that focuses on measuring lengths and angles of triangles. Suppose that the angle of elevation from a point on level ground to the top of a pole is 40 ∘. Transcript of Real World Applications and Jobs That Use Trigonometry.


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## Averia: The Average Font

This is a clever and interesting idea:  creating a new font by taking the average of all existing fonts.

http://iotic.com/averia/

By overlaying all the small letter a‘s, say, from all the different fonts, one can take a visual average and create a new letter a.  Repeat for the whole alphabet, numerals, and punctuation marks, and voila!, you’ve got Averia.

The idea of taking a visual average may be a bit mysterious, but the author describes a few different approaches in how to combine the images.  Essentially all of the instances of a particular symbol are placed on top of each other, and the the darkest parts of the new image are where the instances intersect the most.  The result is then smoothed over to create a readable letter.

And the font looks pretty nice, if not too exciting.  Just what you might expect from the average font.


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# Exactly when and why can a Turing-machine not solve the halting problem?

I perfectly understand and accept the proof that a Turing-machine cannot solve the halting problem.

Indeed, this is not one of those questions that challenges the proof or result.

However, I feel that there is still something left to be explained ... I am still left wondering exactly why the Halting problem is not solvable. Of course, in the sense that there is a proof, there is a why here ... and yet ... I feel that some important other part of the why is missing.

Let me explain:

First, let's assume we just try to solve the 'empty-tape halting problem' and let's assume that the machines we are interested in only have two symbols: 1 and 0. Now, given some machine, will it halt, when stated on the empty tape (meaning: all 0) or not?

Now, we know that this problem is not Turing-solvable. If it were, we get a logical contradiction. OK, I get it. I have no problem with that at all, and like I said, I can follow the proof and I completely agree with it. I perfectly accept that this halting problem is not solvable.

But suppose I would actually try and give it a go: suppose I would try and solve this halting problem. We know the set of all turing-machines is enumerable, so let's just go through them one by one. Now, presumably this enumeration is such that it starts with relatively 'simple' machines. Indeed, I could first list all the ones with 1 internal state, then all the ones with 2, etc. since for any $$n$$, and with only $$2$$ symbols, there are only finitely many possible machines

Now, for all the machines with $$1$$ state, I can easily predict their behavior. Some halt. Some don't. OK, moving on to the machines with $$2$$ states. With some effort, I can predict the behavior for all of them as well. Cool. On to $$3$$ ... ok, now it becomes more difficult .. but even here I can do it. I know, because people working on the Busy Beaver problem have figured this out. And I believe they figured it out for $$n=4$$ as well ...

Interestingly, these researchers are using computers to help them figure out the halting or non-halting behavior for these relatively 'simple' machines. These computer programs are, in a way, trying to solve the halting problem, at least for very small values of $$n$$. Presumably, these machines 'analyze' and 'break down' the behavior of a machine with $$4$$ states into something that can be demonstrated to halt or not halt. But of course, we know they can't solve it for all $$n$$ ... they can't be perfect. And indeed, for $$n=5$$ the behavior of Turing-machines gets so complicated that human nor machine is able to figure out (yet) whether the machine halts or not.

So ... here is my question: what is it that we're running into that prevents us from figuring out the halting behavior?

The proof of the Halting problem uses self-reference. That is, if a machine could solve the halting, then we can show that thee must be a machine that halts on its own input (i.e. when given its own program, or its own number in some enumeration, or..) if and only if it does not .. a contradiction.

OK, but this is when we have a machine with certain powers ... in a way, a machine that can solve the halting problem is a machine with 'too much' power, which leads to a contradiction.

But, the halting-detection machines used by the Busy Beaver researchers don't have too much power. They have too little power. Currently they can't solve $$n=5$$. OK, so we give them some more power. Maybe at some point they can solve $$n=5$$ ... but they still can't solve $$n=6$$. Maybe we can give them enough power to solve $$n=6$$, or $$n=7$$ or ....

... so my question is: is there some 'special' value of $$n$$, say $$n=m$$ where this has to stop. Where, somehow, the only way to solve $$n=m$$, is by a machine that has 'too much' power? But why would that be? Is it because of some kind of self-reference? Because the only way to solve $$n=m$$ is by a machine that, as it tries to analyze and predict the behavior of some machine with $$m$$ states, cannot break it down to anything 'smaller' than something that requires solving $$n=m$$ itself? Some kind of 'minimal' value not unlike some set of minimum requirements that formal systems need to have in order to apply the Godel construction to them?

One thought I have is that this cannot be: like I said, for any $$n$$, there are only finitely many machines to consider. As such, it is computable; there is some machine that correctly classifies all machines with $$n$$ states as empty-tape halters or non-halters: it takes a machine on the input, goes through its finite list with pre-stored answers, and outputs that answer. There is a machine that does this for $$n=5$$, there is one for $$n=6$$, etc. And, none of those machines have too much power: no contradictions here. They all have too little.

Then again, these machine don't do any explicit analysis of the machines involved ... they just happen to give the right value. So, maybe there is still some value of $$n$$ where the approach of actually trying to analyze and predict the behavior of machine starts to break down for some fundamental, again possibly self-referential, reason?

Or: is it that the analytical approach merely gets harder and harder ... but that there is no 'special' point where it becomes, for some theoretical, fundamental reason, too hard? As such, the contradiction only comes from a machine that can do it for all infinitely many values of $$n$$? Indeed, maybe the problem is that in order to analyze the behavior of all machines with $$n$$ states, we need a machine that has to have more than $$n$$ states ... and so while for every $$n$$, there is a machine $$M$$ that can perform the analysis, the complexity of $$M$$ is greater than any of the machines with $$n$$ states, and hence you'd need another, even more complicated machine $$M'$$ in order to analyze machines with the kind of complexity that $$M$$ has ... thus setting up an infinite regress that you can never complete, i.e. there is no one machine that can 'do it all'?

• @MattSamuel The OP is aware of this, as they state in their question. – Noah Schweber Dec 26 '18 at 22:02
• @NoahSchweber Thanks Noah ... you are actually just the person I was hoping would see this post ... am I making sense? Do you see what I'm getting at? – Bram28 Dec 26 '18 at 22:02
• The issue is that having more "power" will not solve the problem. No matter how powerful your machine is it cannot solve a busy beaver state even for low n. All the turing machine can do is continue running and running and even after 1000000000000 steps we cannot truely be sure it doesn't halt without a pattern, and many of these busy beaver states don't exhibit any obvious pattern. Only an infinitely powerful turing machine (infinite time turing machine) can solve the problem. – Matthew Liu Dec 26 '18 at 22:27
• Defining how a program $P$ halts is easy : there exists a finite integer $n$ such that running $P$ for $n$ steps yields the $0$ state. Defining what properties of $P$ guaranty that $n$ doesn't exist is exactly the purpose of our logic/mathematical theories (PA, ZFC...). That there exists two different integers such that $P$ is in the same state after $n$ and $m$ steps is one of those properties. But when looking at theories formalizing the arithmetic, we get much more, and we reach the incompleteness problem : we have no way to guaranty our theory is consistent. – reuns Dec 26 '18 at 22:39
• @Bram28 Still won't matter how much the machine is able to "reason". Fundamentally speaking reasoning is no different from calculation speed. Our human brains are no different from machines other than the fact that our brains are still much more powerful but we aren't made out of metal so we make more mistakes. – Matthew Liu Dec 26 '18 at 23:14

I think the following portion from your question is most important:

But, the halting-detection machines used by the Busy Beaver researchers don't have too much power. They have too little power. Currently they can't solve $$n=5$$. OK, so we give them some more power. Maybe at some point they can solve $$n=5$$ ... but they still can't solve $$n=6$$. Maybe we can give them enough power to solve $$n=6$$, or $$n=7$$

or ....

... so my question is: is there some 'special' value of $$n$$, say $$n=m$$ where this has to stop. Where, somehow, the only way to solve $$n=m$$, is by a machine that has 'too much' power? But why would that be?

The solution to solving $$\Sigma(5)$$ isn't simply giving Turing machines "more power." The reason we don't know $$\Sigma(5)$$ right now is because there are 5-state Turing machines which mathematicians believe will never halt, but have not been able to prove will never halt. The problem is not as simple as just enumerating through all of the 5-state Turing machines since once you've done that, you still need to figure out if the Turing machine halts or not, which, as you know, is not a trivial problem. We've been able to do this for 4-state Turing machines, but we don't know yet if we can do this for 5-state Turing machines because there may very well be 5-state Turing machines which we can never prove to be halting/non-halting within the system of classical mathematics (that is, ZFC).

Now, you've asked what is the magic number is: What is the magic number $$n=m$$ such that we will never be able to solve for $$\Sigma(n)$$? As I've said above, that magic number could very well be $$n=5$$, but that hasn't been proven yet. However, mathematicians have proven that $$\Sigma(1919)$$ is indeed un-knowable within ZFC. Before explaining this, I think it might be helpful to quote your question again:

Then again, these machine don't do any explicit analysis of the machines involved ... they just happen to give the right value. So, maybe there is still some value of $$n$$ where the approach of actually trying to analyze and predict the behavior of machine starts to break down for some fundamental, again possibly self-referential, reason?

My answer to this question is yes, there is a 1919-state Turing machine such that trying to predict if the machine halts would be fundamentally unsolvable within our system of mathematics. See, the way mathematicians proved $$\Sigma(1919)$$ is unsolvable is by constructing a 1919-state Turing machine $$M$$ which halts if there is a contradiction within ZFC and never halts if ZFC is consistent. However, there's no way to prove ZFC is consistent using the axioms of ZFC because of Godel's Second Incompleteness Theorems. This means we can never use the ZFC axioms of mathematics to prove machine $$M$$ ever halts or not because doing so would constitute a proof that ZFC is consistent. Thus, mathematicians can't predict if machine $$M$$ halts or not because of Godel's Incompleteness Theorem, which means that the busy-beaver problem for 1919-state Turing machines is unsolvable.

I hope this gives you some more insight into why $$\Sigma(n)$$ is solvable for small values of $$n$$ but unsolvable for larger values of $$n$$. Anyway, I am certainly not an expert in theory of computation, so if someone would like to add an alternate explanation/clarifying comments to my answer, feel free. Thanks!

• Someone can find a theory T that is "obviously consistent" and does prove an explicit upper bound for $\Sigma(1919)$ ? If T,ZFC both are consistent then T is stronger than ZFC since it will prove ZFC is consistent. – reuns Dec 26 '18 at 23:03
• @Bram28 Yes, if there is a contradiction in ZFC, this Turing machine will find it and halt. On the other hand, if the Turing machine halts, then it has found a contradiction in ZFC and ZFC is inconsistent. Therefore, this is a biconditional statement. – Noble Mushtak Dec 26 '18 at 23:06
• @Bram28 And continuing Noble's comment, this equivalence - appropriately phrased in the language of arithmetic - is provable in PA, and indeed much less. – Noah Schweber Dec 26 '18 at 23:18
• @NobleMushtak: NBG is known to be equiconsistent with ZFC, so looking for an inconsistency of either of them is basically the same task. (I don't know for sure, but I could imagine the optimizations necessary for cramming it into 1919 states involve much more drastic reformulations of the system anyway). NF versus ZFC would be a different matter ... – Henning Makholm Dec 27 '18 at 14:52
• @Bram28 : Currently 1919 is special. But, in 2016, this special number was 7918. (I know the paper I cite had led to work condensing the machine to fewer states and that there was success. I do not know if the 1919 result follows from that chain of work or possibly a different chain of work.) So, tomorrow, someone could show that the TM can be condensed to a 500 state machine, or to a 5 state machine? This is currently unknown. It may be unknowable how small a machine does this computation -- what is the intrinsic entropy of a computation? – Eric Towers Dec 27 '18 at 21:25

Since, as you observe, any finite amount of the halting problem - that is, any set of the form $$H\upharpoonright s:=\{x - is computable, there isn't any particular sharp impossibility point. There are some interesting "phase transitions" which appear relevant - e.g. at a certain point we hit our first universal machine - but I don't know of any which have any claim to be the point where the halting problem becomes noncomputable.

On the other hand, as you also observe the way in which the $$H\upharpoonright s$$s are computable is non-uniform (otherwise, the whole halting problem would be computable). So we can try to measure this "ongoing complexity." There are two, to my mind, natural approaches:

• Given $$n$$, how up up the "hierarchy of theories" - from fragments of PA, to fragments of $$Z_2$$, to fragments of ZFC, to ZFC + large cardinals - do we have to go to get a theory which can decide whether each of the first $$n$$ Turing machines halts on input $$0$$?

• Given $$n$$, how complicated is the finite string consisting of the first $$n$$ bits of the characteristic function of the halting problem (call this string "$$\eta_n$$")?

Of these two approaches, the first has some draw which the second lacks, but it's also far more vague and limited. The second winds up leading to a very rich theory, namely the theory of Kolmogorov complexity (and its attendant notions, like algorithmic randomness), and also partially subsumes the former question. So I think that's my answer to your question: ultimately you won't find a sharp transition point, but the study of the dynamic behavior of the complexity of the halting problem is quite rewarding.

• Thanks Noah! ... a little over my head .. though I like your "phase transitions" notion .... from one of the other answers I gather that indeed for some number of states we "transit" into the consistency of ZFC ... which is certainly 'special' real-life-practicing-mathematics in some sense ... but maybe not 'special' in the sense I had in mind? – Bram28 Dec 26 '18 at 23:17
• So thinking about this a little more .... "transitioning" into the consistency-of-ZFC space is pretty 'special' for sure ... and yet it doesn't feel 'special' enough to me: The proof of the unsolvability of the Halting problem seems so deeply logical that I was looking for a likewise more deeply logical barrier that we'd run into as we contemplate more and more complicated TMs ... the fact that ZFC plays such a prominent role in mathematics makes it 'special' ... but I feel not 'special' enough. So, maybe it's back to the thought that there is no deeply logical 'special' point. – Bram28 Dec 27 '18 at 1:46
• I think your first sentence is a bit misleading. Any set $H\restriction s$ is finite and hence computable, but that doesn't mean that we can actually compute it, because we don't know what program computes this particular finite set because we don't know what set it is. There are indeed concrete finite values of $s$ like 1919 for which we definitely cannot prove that it is equal to anything in particular. This is an "impossibility point" by any reasonable measure. (If you are working in a different system than ZFC the point may be elsewhere, but it will be there regardless.) – Mario Carneiro Dec 28 '18 at 6:35
• @MarioCarneiro I disagree - we can compute it, we just don't know what program will do the job. More substantively, each specific theory hits a wall somewhere, but I don't see any reason to argue that any one theory perfectly captures the "right" situation (re: "any reasonable measure"); so the right picture is still a "sliding scale." – Noah Schweber Dec 28 '18 at 17:59

For example, you can construct a Turing machine (I don't know how many states you need, but it is a finite number) that searches for a counterexample to Goldbach's conjecture, i.e. an even number $$> 2$$ that is not the sum of two primes, going through the even numbers one by one; for even number $$n > 2$$ it checks each $$k$$ from $$2$$ to $$n/2$$; if $$k$$ is prime and $$n-k$$ is prime it goes to the next $$n$$, but if it gets through all the $$k$$ it halts. Thus this Turing machine will halt if and only if Goldbach's conjecture is false. In order to decide whether it will halt, your analysis will need to decide Goldbach's conjecture.

And when you're done with that one, there are lots of other conjectures you could check with a Turing machine.

• Thanks! I am guessing we can program something like this with a Turing-machines with close to 100 states ... so that gives us some measure (or at least a kind of minimum) of 'how difficult' it is to figure out the halting behavior of machines with that number of states. One of the other answers does something similar with the consistency of ZFC, which has been shown to take 1919 states at the most. Interesting! – Bram28 Dec 26 '18 at 23:13
• Goldbach's conjecture is usually an example of what the halting problem is not. We don't know a proof for a conjecture but it may eventually be found (or we might find a proof that it isn't true), it is not an intrinsically unsolvable problem (we have a diagonalization proof that you can't build a machine that can solve the halting problem) – pqnet Dec 29 '18 at 7:25

A potential Busy Beaver has three possibilities:

1. It is easy to show it stops
2. It is easy to show it never stops
3. It is difficult to show either case

Number 1 either stops quickly, or it has a repeating pattern with an eventual flaw that causes it to stop.

Number 2 has a repeating pattern and never has a flaw, causing it to continue forever.

Number 3 is the interesting case. Its behaviour is chaotic. It might have short-term patterns, but it has no long-term patterns. Its future states can be predicted a short way without actually running the machine. There comes a certain point where its behaviour can no longer be predicted without running it. At this point you need a machine capable of emulating a Turning machine but faster. However, you will reach the same problem with this hypothetical new machine too, as long as it has finite power, which all real-world machines have.

If you improve your analysis of Busy Beavers, you can decide whether certain candidates are actually case 1 or case 2. We can think of it as a spectrum of behaviour with stopping at 0, running forever at 2, and undecidability at 1. To start with we know that 0 to 0.5 will stop (case 1) and 1.5 to 2 will run forever (case 2), with 0.5 to 1.5 being undecidable without running it (case 3). We improve our understanding of Busy Beavers. Now we know that 0 to 0.95 will stop and 1.05 to 2 will run forever, with 0.95 to 1.05 being undecidable.

At some point, there's no way to predict without running the machine whether it will halt or not. The only way to determine its behaviour is to run it, and if it stops, it usually gives you no insight you can use for other potential Busy Beavers. If it doesn't stop after $$10^6$$ cycles, you can try $$10^7$$, then $$10^8$$, and so on. At some point you give up without knowing whether it will stop if given enough time.

The problem is similar to drawing a Mandelbrot set. Certain points "escape" to infinity quickly, others settle into a repeating pattern quickly. The points on the boundary of the Mandelbrot set are difficult to predict either way without calculating them. There are methods to make the decision easier, but there will always be chaotic behaviour between the two easily detectable outcomes that can't be predicted.

Hopefully I've answered "why". Then "when" will be determined by your understanding of the specific problem you're trying to solve and the power of the machine you're using. Over time we can eat into the chaos and make it decidable, but it grows much faster than we can ever solve.

• So basically a universal computer is a kind of "useful discrete chaos", then? Interesting, but seems to make sense on a certain level. – The_Sympathizer Dec 27 '18 at 10:34

I would like to offer an alternative way of thinking about the halting problem, which helped me understanding better why the halting problem is non computable, or better, why in general there are functions which are non computable.

In his famous paper on Computability, Turing mentions that he is going to prove that there are real numbers which are not computable. Computable numbers are defined as those whose decimals are calculable by finite means, or in other words, decimals that can be computed by a machine.

He also mentions that it is equally easy to define and investigate computable functions instead of computable numbers and that is what I would like to show. I will report in short the lecture of the link I have posted already (https://www.youtube.com/watch?v=9JeIG_CsgvI&index=14&list=FLIknGRIW8gX2yutAMeHVEKw) because I think it is worth it: it is indeed the first part of a lecture which proves Goedel first Incompleteness Theorem. Credits go to "UC Davis Academics" of course.

Let's define a function $$f$$ from non negative integers to the set $$\{0,1\}$$. We let $$Q$$ be the set of all such functions. It is clear that $$Q$$ is infinite (we wil prove indeed it is uncountable infinite essentially).

Also a function $$f$$ in $$Q$$ is defined as computable if there is a computer program $$P$$, (loosely speaking a Turing Machine), that can take any non negative integer $$x$$ and output $$f(x)$$. We add the constraints that $$P$$ must always terminate in finite time and $$P$$ must be correct, in other words, output the correct value of $$f$$ for all non negative integers.

We let $$A$$ be all the functions in $$Q$$ which are computable. We can show that there exists a function in $$Q$$ that is not in $$A$$, i.e. there exist uncomputable functions.

We define a program as a series of finite statements over some finite alphabet $$\alpha$$, in other words it can be thought as a single finite string. Suppose that the language $$L$$ we are using to express our program is Turing complete, i.e. it can be used to simulate any Turing machine.

We can start enumerating in order of length the strings expressible in $$\alpha$$. Strings of the same length are taken based on an alphabetical order that can be defined arbitrarily in $$\alpha$$.

You could indeed write a program $$T$$ to enumerate all those strings expressible in $$\alpha$$, so for any string $$s$$ expressible in $$\alpha$$, $$T$$ in finite time will generate $$s$$.

Therefore you have a list $$Z$$ of strings in $$\alpha$$ ordered by length. Some of those strings in $$Z$$ will be legitimate programs in our chosen programming language $$L$$. Indeed all possible programs will be in $$Z$$, and in particular the programs that compute functions in $$A$$ must all be there (by definition of computable function) and they are ordered in $$Z$$.

Let's call $$H$$ this ordered list of functions in $$A$$, $$\{f_{1}, f_{2},...\}$$ . Now applying the diagonalization method, by defining $$g(x)=1-f_{i}(i)$$ It is easy to see that $$g$$ is in $$Q$$, however $$g$$ is not computable since it is not in $$A$$, and so we are done.


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# error in trapezoidal approximation Santa Fe Springs, California

Use $K\le 3.6$ (or even $2+\pi$). Show Answer If you have found a typo or mistake on a page them please contact me and let me know of the typo/mistake. Close × Select Your Country Choose your country to get translated content where available and see local events and offers. Or if you did want accurate estimates of errors that small, mightn't the additional errors introduced by the finite differencing make that difficult?

Consider a single subinterval $[a,b]$ and let $A[f]$, $T[f]$, and $M[f]$ represent the exact area integral of $f$ on $[a,b]$, the trapezoidal estimate, and the midpoint estimate, respectively. However, I got some strange number. An animation showing how the trapezoidal rule approximation improves with more strips. Also, when I first started this site I did try to help as many as I could and quickly found that for a small group of people I was becoming a

United States Patents Trademarks Privacy Policy Preventing Piracy Terms of Use © 1994-2016 The MathWorks, Inc. Apply Today MATLAB Academy New to MATLAB? Wird verarbeitet... It is clear that, if there's no inflection point in the interval, then, if the trapezoid of the midpoint rule is an overestimate of the integral, the trapezoid of the trapezoidal

The function f(x) (in blue) is approximated by a linear function (in red). Wird geladen... calculus sequences-and-series share|cite|improve this question asked Dec 12 '13 at 5:20 citelao 12315 2 The constant $k$ is not necessarily the same. We get $$f''(x)=-x\cos x-\sin x-\sin x=-(2\sin x+x\cos x).$$ Now in principle, to find the best value of $K$, we should find the maximum of the absolute value of the second derivative.

Nächstes Video Error Estimates (Midpoint Rule, Trapezoid Rule, Simpson's Rule) - Dauer: 9:37 BriTheMathGuy 635 Aufrufe 9:37 Maximum Error in Trapezoidal Rule & Simpson's Rule READ DESCRIPTION - Dauer: 20:13 ProfRobBob In general, three techniques are used in the analysis of error:[6] Fourier series Residue calculus Euler–Maclaurin summation formula:[7][8] An asymptotic error estimate for N → ∞ is given by error = Instead, the experimental error would be contained in the uncertainty of the fitted curve (assuming the fit is correctly weighted). You will find several of these in the File Exchange.

Combining this with the previous estimate gives us ((f(b+(b-a))-f(b))-(f(a)-f(a-(b-a))))*(b-a)/24for the estimated error within the single interval from a to b. I also have quite a few duties in my department that keep me quite busy at times. It is especially true for some exponents and occasionally a "double prime" 2nd derivative notation will look like a "single prime". My data is discrete.

Show Answer There are a variety of ways to download pdf versions of the material on the site. You can access the Site Map Page from the Misc Links Menu or from the link at the bottom of every page. Autoplay Wenn Autoplay aktiviert ist, wird die Wiedergabe automatisch mit einem der aktuellen Videovorschläge fortgesetzt. Melde dich an, um unangemessene Inhalte zu melden.

Wird geladen... I won't go into that now.You will find the matlab function 'diff' useful in calculating the above second difference. Trapezoid Rule For this rule we will do the same set up as for the Midpoint Rule.  We will break up the interval  into n subintervals of width, Then on The question is whether this residual error is larger or smaller in magnitude than the error of the trapezoidal rule.

Then Example #1 [Using Flash] [Using Java] [The Trapezoidal Rule approximation was calculated in Example #1 of this page.] Example #2 [Using Flash] [Using Java] [The Trapezoidal Rule approximation Asking client for discount on tickets to amusement park Passport check for Hidden City Ticketing where only final destination is international How to answer my boss's question about my ex-coworker's current Show Answer If the equations are overlapping the text (they are probably all shifted downwards from where they should be) then you are probably using Internet Explorer 10 or Internet Explorer Related Content Join the 15-year community celebration.

Simpson's rule is another member of the same family, and in general has faster convergence than the trapezoidal rule for functions which are twice continuously differentiable, though not in all specific There are some formulas available here http://cmd.inp.nsk.su/old/cmd2/manuals/cernlib/shortwrups/node88.html, http://wwwasdoc.web.cern.ch/wwwasdoc/shortwrupsdir/d108/top.htmlMy solution was to implement the trapz() algorithm by hand, and to manually take care of the error propagation at each step. I am attempting to find a way around this but it is a function of the program that I use to convert the source documents to web pages and so I'm Your cache administrator is webmaster.

Guy Koren Guy Koren (view profile) 1 question 0 answers 0 accepted answers Reputation: 0 on 4 Jan 2013 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/57737#comment_120703 Thanks a lot.. The question is to obtain the uncertainty in the integrated area given the uncertainty in each of the data points.There are a couple of approaches one could take. Wird geladen... How do I download pdf versions of the pages?

Melde dich bei YouTube an, damit dein Feedback gezählt wird. Passport check for Hidden City Ticketing where only final destination is international Define a hammer in Pathfinder Draw an asterisk triangle Dictionaries are ordered in Python 3.6 more hot questions question Can Homeowners insurance be cancelled for non-removal of tree debris? In the "Add this website" box Internet Explorer should already have filled in "lamar.edu" for you, if not fill that in.

The above relation obviously holds for the functions $f(x)=1$ and $f(x)=x$. If you use the trapezoidal approximation, (f(a)+f(b))/2*(b-a), to approximate the integral of a quadratic function f(x) from a to b (which is what 'trapz' does,) it can be shown that the Hence, for an estimate of this error you need to obtain a good approximation to your data's second derivative within each of your trapezoid intervals.If your data is very accurate and Learn more MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi Learn more Discover what MATLAB® can do for your career.

Download Page - This will take you to a page where you can download a pdf version of the content on the site. You will be presented with a variety of links for pdf files associated with the page you are on. Let me know what page you are on and just what you feel the typo/mistake is.


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RFC 4279, section 5.2, says this about identity hints: In the absence of an application profile specification specifying otherwise, servers SHOULD NOT provide an identity hint and clients MUST ignore the identity hint field. Applications that do use this field MUST specify its contents, how the value is chosen by the TLS server, and what the TLS client is ...

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Blockchains seem to be a common buzzword these days. And more often than not it is used by people, who don't understand the actual concept in detail. For example, that blockchains are based on assumptions about the distribution of processing power. And when you use it outside the context of bitcoins, you still need an incentive for many people to contribute, ...

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I would just like to know on what basis they can say it is PQC without any NP problem reduction. I believe that the point you're making is assuming that a Quantum Computer could solve any problem in NP that's not actually NP-complete quickly (or, at least, in polynomial time). That's not known to be true; Quantum Computers would be able to solve some ...

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Let's approach the question from this direction. PAKE tries to solve this problem: Alice has a password $PW_{a}$, and Bob has a password $PW_{b}$; consider a protocol where Alice and Bob exchange messages, and generate encryption keys $E_{a}, E_{b}$ where $E_{a} = E_{b}$ if $PW_a = PW_b$ (and unrelated if $PW_a \ne PW_b$); if they're the same, Alice and Bob ...

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How does one verify a key revocation? As Jon Callas already stated: you simply don’t. In case a different wording helps, here’s a quote related to the exact same question… https://lists.gnupg.org/pipermail/gnupg-users/2014-February/049100.html … On 02/19/2014 11:55 AM, Hauke Laging wrote: Am Di 18.02.2014, 23:19:33 schrieb Tadas ...

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Assertion : The K_(a) value o...

# Assertion : The K_(a) value of alpha-amino acids is very low. Reason : alpha-Amino acids have zwitterionic structure.

Answer

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If both assertion and reason are correct and reason is correct explanation for assertion.If both assertion and reason are correct but reason is not correct explanation for assertion.If assertion is correct but reason is incorrect.If assertion and reason are both incorrect.

Answer :
B
Solution :
Correct explanation. In the Zwitterion structure, overset(+)(N)H_(3) group and not COOH group acts as an acid. Therefore, it is a very weak acid.


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# Given a RGB color x, how to find the most contrasting color y? [duplicate]

Possible Duplicate:
Making a symbol appear on any colour

I have to mark a certain item in a way that will make it stick-out in the background. I need it to be surrounded with the color that contrasts the background as much as possible so it will pop out and easily noticeable by the player.

Lets say I know the background is color 'x', how do I find 'y' such that it will be very contrasting to 'x' and easy to notice in a background where 'x' is a dominant color?

I first thought about inverting color 'x' and then I noticed that when 'x' is a medium shade of gray, if I invert 'x' to get 'y', then 'y' is also a medium shade of gray which does not work.

• The "most contrasting color" is often going to be really ugly, is that really what you want, or do you just want a sufficiently contrasting color? – user744 Oct 8 '12 at 13:28
• If you can address the concern you mention that would be exquisite. I was thinking of lowering the bar and thought it would be difficult to answer. It would be much better if it was sufficiently contrasting while taking into account looks. – AturSams Oct 8 '12 at 13:52
• A-ha, I knew I had answered this question before, I just couldn't find it on my answers list searching for "color". UKians... – user744 Oct 8 '12 at 14:20
• I decided to take time testing both answers. :) – AturSams Oct 8 '12 at 19:53
• Not to diss the great answers below, but - have you considered just using an outline (or in your case, an additional outline)? e.g. take a look how the text here is clearly visible over both light and dark backgrounds, due to it having an outline in a color that strongly contrasts with the text color. Will also make your coloring choice consistent, which is a plus imo. Just a thought. And if you need to really make it stick out, you can animate that outline! – Oak Oct 8 '12 at 22:02

A quick and easy way - though not 100% precise one - is to consider just the five extreme points white, black, red, green and blue.

First, let's transform RGB into linear space. Officially this is usually done by this formula (assuming the source data is in sRGB, which is the default for most graphic card operations on 8-bit data and nearly every image you see displayed on the web):

... where a = 0.055 for C = {R, G, B}

Or, if you just want a quick approximation, use Clinear = Csrgb2.2.

## Black or White?

You can then calculate two helpful values, luminance Y and saturation S. Useful formulas (for our linear-space RGB values) are:

Y = 0.2126 R + 0.7152 G + 0.0722 B

S = (max(R, G, B) - min(R, G, B)) / max(R, G, B)

The second one is the calculation from the HSV colour space.

If the saturation is low enough (pick any value you like; something between 0.3 and 0.5 would work fine), check the luminance; if that's > 0.5, your contrasting colour is black, else if it's < 0.5 the colour is white. For exactly Y = 0.5, both work.

## Red, Green or Blue?

For colours with high saturation, you can calculate a hue H and decide according to it. Since I'm lazy, I'll just copy the calculation from Wikipedia (hue is named H2):

If your hue is between 60° and 180°, your colour is green, if it's between 180° and 300°, it's blue, else it's red. Extend it to yellow, turquoise and magenta for more variation, or just use the opposite of the just-calculated hue at maximum saturation and lightness.

• +1 This is an amazingly good answer, the main reason I picked Snake's was simplicity, I don't fully understand the part with the (alpha) (Beta) (H2) (C2).. – AturSams Oct 8 '12 at 14:52
• @ArthurWulfWhite Ignore the C2, and both apha and beta are just used to calculate the hue (H2) - that's all you need. They represent the coordinates in which the colour you're using is on the 2D plane below. – Martin Sojka Oct 8 '12 at 15:02
• This is really very cool stuff. So H2 is basically the Hue value? I really love applied mathematics and this seems like a shining example to me. – AturSams Oct 8 '12 at 17:26
• In the first part, how do I calculate the sRGB from regular RGB and what is a value? – AturSams Oct 14 '12 at 23:32
• @ArthurWulfWhite: You don't calculate it; "sRGB" is a color space, and you can assume nearly every image you find is already in this color space unless specifically defined otherwise. Quick starting points for further reading: Wikipedia W3C – Martin Sojka Oct 15 '12 at 4:59

The most contrasting color would be the color that is as far as possible from color X. It's easy to get it this way (assuming 0,0,0=black and 1,1,1=white -- floating point colors):

y = rgb_color(
x.r > 0.5 ? 0 : 1,
x.g > 0.5 ? 0 : 1,
x.b > 0.5 ? 0 : 1
);

The result is quite ugly though, so you might want to consider a few more things:

1) generate only black / white colors:

v = ( x.r + x.g + x.b ) / 3 > 0.5 ? 0 : 1;
y = rgb_color( v, v, v );

2) change the values "0" and "1" to something less sharp

3) work in the HSV color space:

y = hsv_color(
( x.h + 0.5 ) % 1,
x.s > 0.5 ? 0 : 1,
x.v > 0.5 ? 0 : 1
);

• That is exactly what I just thought myself and as Joe Wreschnig and you mentioned it is not very pretty. It does answer the question perfectly. Working in hsv sounds interesting, however, I fear that both 0 and 1 are red in hsv so for hue I would have to do a (x+0.5)%1.0 kind of thing. and also I will have to keep value slightly above and bellow 0 and 1 to make it less sharp. – AturSams Oct 8 '12 at 14:05
• @ArthurWulfWhite You're right about hue, I fixed it in my answer. – snake5 Oct 8 '12 at 14:15
• What you first propose is not even the most contrasting color, RGB is not perceptually uniform nor split along what are thought to be the opponent colors in our visual processing. Change your recommendation from HSV to HSL and your third option is an okay answer, though. But Martin's is waaaay better. – user744 Oct 8 '12 at 14:15
• @JoeWreschnig: Given that I use way more calculations and don't derive many more benefits in terms of results out of them (aside from correct luminance values and thus better B&W/colour threshold), I'm not sure "better" is the right word. More extensible and more easily tweakable, maybe. :) – Martin Sojka Oct 8 '12 at 14:29
• @snake5 I really like this answer the best, it is very simple and concise. I learned a lot from Martin Sojka's answer about converting bases too. I will pick this as the accepted answer cause it is well thought out and a quick fix :) – AturSams Oct 8 '12 at 14:54

Just to distill Martin Sojka's excellent answer into something simple to apply, here's how to decide whether black or white text would have higher contrast on a given background color (R, G, B) in the sRGB color space:

const float gamma = 2.2;
float L = 0.2126 * pow( R, gamma )
+ 0.7152 * pow( G, gamma )
+ 0.0722 * pow( B, gamma );

boolean use_black = ( L > pow( 0.5, gamma ) );

This assumes that R, G and B are floating-point numbers ranging from 0.0 to 1.0. If what you have is, say, integers from 0 to 255, convert them to floats and divide them by 255.

(I would not suggest using colored text, both because the human eye has much poorer spatial resolution for color than for luminance, and also because combinations of highly saturated complementary colors tend to be irritating to look at.)

Note that, if L is close to 0.52.2 ≈ 0.2, small changes in the background color could cause the most contrasting text color to flip from black to white or vice versa. To avoid this happening too frequently, you could save the previous text color and only change it if L moves too far from the threshold:

if ( L > 0.3 ) {
use_black = true;
} else if ( L < 0.1 ) {
use_black = false;
} else {
// keep previous text color
}

Ps. If you wanted an even quicker approximation, you could round the exponent gamma down to 2.0, allowing you to replace the pow() with a simple multiplication:

float L = 0.2126 * R*R + 0.7152 * G*G + 0.0722 * B*B;

This approximate formula is still within ±0.05 of the correct luminance calculated with the official piecewise formula given by Martin (or with the gamma = 2.2 approximation used above, which itself is within ±0.01 of the official formula), and so more than close enough for this purpose. Besides, you can mostly compensate for the error simply by adjusting the threshold slightly.

Finally, note that, if the background has a very busy texture, it's possible that no single color may stand out from it very effectively. In such cases, you may want to consider using combinations of colors, such as black text with a white outline.

Ps. Earlier versions of this answer suggested using the uncorrected threshold L > 0.5. However, this suggestion ignores the fact that the gamma correction used in the RGB color space approximates the perceptual non-linearity of the human eye, so that the shade of gray that looks about halfway between black and white is indeed close to RGB(0.5, 0.5, 0.5). Thus, to get an accurate measure of perceptual luminance, we really need to apply a final inverse gamma correction to the linear luminance L — or, equivalently, just apply gamma to the thresholds as well.

This perceptual non-linearity also explains why just linearly mixing the uncorrected RGB components into a luma value, and basing the decision on that, also works reasonably well:

float luma = 0.2126 * R + 0.7152 * G + 0.0722 * B;
boolean use_black = ( luma > 0.5 );

This formula does tend to overestimate the brightness of highly saturated colors; for example, for pure green it gives luma = 0.7152, whereas properly applying gamma correction would give pow(L, gamma) ≈ 0.478. Fortunately, for saturated colors, either black or white will provide a decent contrast anyway.

Since there is some subjectivity to any answer here, I'll offer one that is easy to understand and easy to compute. Whether it suits you and your needs is up to you.

Imagine color space represented by R, G, and B values each on the interval [0,1]. Imagine the RGB value being a vector in color space, where that vector is confined to a cube. For any RGB value in that space, the most contrasting value would be geometrically as far away from that value while still being on that cube.

Offset the cube, so it is centered at the origin, with new values r, g, and b on the interval [-0.5, 0.5]. Do the by subtracting 0.5 from all of the original RGB values. Now compute the sign of each component of rgb and multiply by -0.5. This will put you at one of the vertices of the offset color space cube. Add 0.5 to each value to restore the values to the original color space cube.

It depends how you invert the color, the standard way of doing so is something like:

newColor = new Color(255-oldColor.r, 255-oldColor.g, 255-oldColor.b);

Which means that something grey like (128,128,128) will turn into something very grey (127, 127, 127). You may want to try inverting the hue instead. Convert your color to HSV mode and invert the hue color.hue = 360-color.hue.

You could additionally invert the saturation.

Ultimately, it sounds like you're the one choosing the original colors here. So just choose colors that invert nicely.

Another option would be to check for colors that won't change much. Something like:

if((255-oldColor.r)-oldColor.r < 10)
newColor.r = 0;

• If I convert the hue of a gray color, it would stay gray. – AturSams Oct 8 '12 at 13:49


text:
# Centroid of a Beam Section

Not Reviewed
bary =
Tags:
Rating
ID
MichaelBartmess.Centroid of a Beam Section
UUID
c374e102-ec14-11e6-9770-bc764e2038f2

The Centroid of a Beam Section calculator computes the y-centroid (y being vertical axis of an I-beam's cross-section).

INSTRUCTIONS: Choose units and enter the following:

• (D1) Top width
• (W1) Top thickness
• (W2) Vertical thickness
• (D2) Interior height
• (D3) Bottom width
• (W3) Bottom thickness

y-centroid (y): The calculator return the centroid in millimeters.  However, this can be automatically converted to compatible units via the pull-down menu.

#### The Math / Science

The Centroid of a Beam Section calculator computes the y-centroid (y being vertical axis of an I-beam's cross-section). The centroid is the center of mass of the beam's cross-sectional sections.  The centroid is used to compute the moment of inertia of the beam, which is used in analysis of the beam's bending stress under weight, beam shear, and beam deflection or other analysis questions.

We split the cross-section into into three segments, each segment having a nice rectangular symmetry. We then calculate the area and y-centroid of each of the three segments and compute the entire centroid as:

bary = sum(A_i*y_i)/(sum A_i)

The three segments are shown in the figure below, where A_i = w_i *D_i:

And the y-centroids of the segments are given as:

• y_1 = w_3 +D_2 + w_1/2
• y_2 = w_3 +D_2/2
• y_3 = w_3/2


text:
# Forum: The Enigmatic Mr. Putin: Taking the Pulse of Russia's President and Politics

May 1, 2007

Website

### Full Text

Who is Mr. Putin?” The question reverberated in world capitals when Boris Yeltsin called a press conference on August 9, 1999, to introduce Vladimir Vladimirovich Putin as his choice for prime minister of Russia and as his heir when a presidential election rolled around the next summer. Yeltsin accelerated the timetable by resigning on December 31 and decreeing his protégé acting president. On March 26, 2000, Putin won a popular mandate in his own right.

We normally size up a new leader by looking for guidance to his or her biography. In this regard, the Putin story, as lived from his birth in Leningrad (now St. Petersburg) in 1952, presented several problems.

For one, he was a late bloomer who had left few tracks in his earlier jobs. Only in 1996 did he relocate to Moscow and only in 1997 did he catch Yeltsin’s attention, one year before being appointed director of the FSB, the successor to the Soviet Union’s security and intelligence agency, the KGB. Putin had not previously headed any organization and had not once run for elective office. And he never aspired to be Russia’s top leader: he was astonished when Yeltsin broached the idea.

Another issue had to do with Putin’s having spent the bulk of his career in the foreign intelligence wing of the KGB, five years of it at the KGB station on Angelikastrasse in Dresden, East Germany. Espionage as a line of work is cloaked in secrecy and oftentimes in deception. It selects for individuals who have the aptitude for these arts and it hones them through training. I have met Putin twice, in hospitable group settings. He shakes hands firmly and makes eye contact, yet blushes slightly and rocks back on his heels as he greets you. The impression telegraphed is of someone disinclined by nature and vocation to give away much about himself.

Extrapolation from his past in 1999 or 2000 would have brought a further discordance—mixed signals. The Putin dossier testified to the KGB entanglement, on the one hand. But on the other, it spoke of a legal education, a credential he shared with Mikhail Gorbachev, and an association with Anatolii Sobchak, one of the better-known Russian democrats of the perestroika era. Sobchak was Putin’s professor at Leningrad State University in the 1970s, and from 1991 to 1996, when Sobchak was the elected mayor of St. Petersburg, Putin was his right-hand man. The reason Putin and his family migrated to Moscow in 1996 was that he was persona non grata in his hometown once his patron lost a bitter reelection battle. When Sobchak then came under attack for alleged but never proven corruption, Putin arranged at considerable personal risk to have him spirited out of the country to Paris in 1997. (Sobchak returned to Russia in 1999 and died in early 2000.)

Once president, Putin soon pursued a public line that, whatever the private yearnings and calculations behind it, set the tone for his administration. Its logic can be summarized in a pair of oft-repeated phrases: ukrepleniye gosudarstva (“strengthening of the state”) and upravlyayemaya demokratiya (“managed democracy”). It obviously was closer in spirit to the demimonde of the security services than to a Sobchak or Gorbachev.

Yeltsin in his day transformed Russia by loosening the apron strings of state control over economic activity, decentralizing, and tolerating criticism in the mass media and a plenitude of organized points of view in the political sphere. The gestalt was closer to what Thomas Carothers of the Carnegie Endowment for International Peace dubs “feckless pluralism” than it was to a textbook democracy. But the liberalizing trend was unmistakable. Yeltsin was content to achieve his breakthrough and leave the next generation to perfect the system. The flip side of the coin was government’s loss of steering power and capacity, faults that were spectacularly on display in two successive crises late in his second term: the financial emergency of August 1998, in which many banks failed and the treasury defaulted on its debts; and the launching of a second war over the breakaway republic of Chechnya in August-September 1999, after Chechen fighters raided a neighboring Russian section. It was the Caucasus war that provided Putin with the golden chance to display his leadership qualities in the winter of 1999-2000.

Right after taking the oath of office, Putin took action on a series of fronts to rectify what he saw as the overindulgence of the Yeltsin era, without ever denigrating Yeltsin personally. Having ousted Chechnya’s suicidal separatist government, he used his army to grind down the guerrillas and refused to negotiate with their remnants. He overlaid seven “federal districts” on the 89 provinces of Russia, each with a Kremlin plenipotentiary, and removed the provincial governors from the upper house of parliament. In 2005 he pushed through the abolition of the direct election of governors, who were henceforth to be nominated by the president and merely confirmed by the provincial legislature.

Putin moved with equal dispatch to clip the wings of the “oligarchs,” the upstart moguls who tried to play kingmaker in the late 1990s. Vladimir Gusinskii, owner of the commercial television network NTV, was accused of financial abuses and fled to the United States and Israel in 2000. The flamboyant Boris Berezovskii, who among other things had called the shots at ORT, the main state-owned television outlet, suffered the same fate forthwith and took up residence in London. The heaviest blow was struck in October 2003, when government agents arrested Mikhail Khodorkovskii, the wealthiest man in Russia; he was convicted of fraud in May 2005 and sentenced to nine years in a Siberian prison camp. The main assets of Yukos, his oil company, were redistributed to government-owned firms, paving the way for a widening of the state role in the petroleum sector. The Russian natural-gas industry, the world’s largest, had never been privatized and remained the preserve of a state concern, Gazprom; in the oil industry, second to Saudi Arabia in output, government equity rose to about 30 percent.

The reassertion of clout was no less pronounced in politics. The elimination of Gusinskii and Berezovskii was followed by the imposition of a uniformly pro-regime line in the television news. In the electoral realm, Putin gave his blessing to a “party of power” named United Russia, designed to represent the ruling group at all levels. In the 2003 parliamentary election, it captured almost 40 percent of the countrywide vote and two-thirds of the seats in the State Duma, the lower house of the national legislature. Provincial governors were now expected to join United Russia as a token of their loyalty. Other Russian parties have been subjected to elaborate and intrusive regulation, while barriers to entry by newcomers have been stiffened. In 2005 the Putin government instituted parallel restrictions on nongovernmental organizations of every stripe.

The final application of Putinism is to foreign policy. Russia with him at the helm has been a more demanding partner in dealings with the other world powers. In relations with the United States, new strains have emerged in Putin’s second term (which began eight months before George W. Bush’s second term), and old ones have been exacerbated. In its former imperium in Eurasia, with Russia pushier and testier than was the case under Yeltsin, spats over energy pricing and transport, military bases, and incorporation into the Western alliance have been regular occurrences. Even President Aleksandr Lukashenko of Belarus, the most pro-Russian of all the ex–Soviet republics, is barely on speaking terms with Moscow.

In pondering these developments, it is essential to keep in mind their provenance, their boundaries, and the forces nurturing them. It would be a gross oversimplification to say they sprang from the head of one man. Putin’s previous experiences and personal preferences, as noted, were quite complex. Like any effective leader, he responds to circumstances and empowers subordinates to flesh out shared goals. Having recruited many subordinates with backgrounds akin to his, he receives a steady diet of advice that reinforces his own predilections, or aspects of them.

He was motivated to formulate his program, aside from his anxiety about the weakness of the state, by a natural desire to differentiate himself from his predecessor. Boris Yeltsin, do not forget, handpicked him as heir. He did so because he detected in Putin qualities he lacked and judged him competent not only to maintain key gains of democratization but to correct its excesses. Putin, needless to say, has excelled more in the latter task than in the former. Moreover, he has improvised along the road. This applies, for example, to the landmark decision to put an end to the election of governors. That followed a horrendous attack by Chechen terrorists on a secondary school in Beslan, North Ossetia, on September 1, 2004, which resulted in the deaths of about 350 people, half of them children. The Beslan tragedy emboldened Putin to extend his grip and biased both rank-and-file and elite Russian opinion in his favor.

Although the authoritarian drift is beyond denial, it has had its limits. United Russia is a coalition of opportunists and hangers-on, not a purposive or disciplined force. The Putin team sets priority targets: television, political parties and elections, and uppity businessmen. Outside that zone, the vise has been tightened less or, on many points, not at all. For instance, there has been no attempt to encumber interpersonal communication or access to information networks. Cellphone penetration has surged to 60 percent countrywide and to 80 or 90 percent in the big cities, and 25 million Russians are estimated to have surfed the Internet unimpeded in 2006, in contrast to China, where the government screens websites. Nor has there been serious infringement on intellectual and cultural life, which by all indicators has been rebounding from a low point during the economic hard times after 1991. In the vast majority of court cases that are nonpolitical, the judiciary functions better than in the 1990s, and jury trials have been made compulsory for the most severe crimes. Economically, private capital still predominates, despite state inroads in the energy field. As for the business class, it has made far more money during Putin’s reign than it did during Yeltsin’s. Forbes magazine registered not a single Russian billionaire in 1999 or 2000, in the aftermath of the 1998 crash. In 2006 it listed 33, the third highest total in the world (behind the United States and Germany), and their net worth had soared from $91 billion to$172 billion during the preceding year.

Putin’s hybrid regime has drawn sustenance from three underlying forces. The first and foremost has been economic. High oil and gas prices, combined with the long-term benefits of the Yeltsin economic reforms and with investment-friendly decisions since 2000, have brought Russia a sustained boom; this comes as an immense relief after the decade of contraction and hardship under Gorbachev and Yeltsin, during which the economy shrank by more than 50 percent—worse than the Great Depression in the United States. The World Bank reckons that gross domestic product (GDP) increased by an average of 6.2 percent annually in 2001 through 2006; the real disposable income of the population surged by 11.2 percent per year from 2002 to 2006. Among other boons, the recovery has filled the coffers and greased the wheels of the state machine. The federal government has a primary budget surplus and a huge currency stabilization fund and has poured billions into paying down the national debt, rebuilding the military, and social services.

Economic strength has fed political strength. Putin’s aura, however, goes beyond GDP and family incomes. His approval ratings have been flat at the 70 to 80 percent level since his inauguration, remarkably impervious to failures and embarrassments. He won reelection with 71 percent of the vote in March 2004, or 18 percentage points more than he polled four years before. Electoral fraud accounts at best for a tiny fraction of this showing. Rightly or wrongly, the average citizen lauds Putin for his businesslike style and for bringing the country order and self-respect. His reputation sails miles above that of his policies: polls consistently show large minorities or pluralities doubting specific decisions on, say, inflation or corruption while keeping the faith in the architect of those policies.

The international environment has been the third big prop for the president and his system of power. Russians appreciate their nation’s resurgence in world politics, and the leader they credit for this revival. Opinionmakers attuned to Putin share two deep grievances that in turn resonate with the mass audience. Toward the West, there is rancor at what they view as the indifference to Russia’s plight in the first postcommunist decade, when the country was at a low ebb and NATO, overriding protests from Yeltsin, expanded into eastern Europe. This resentment dovetails with anger at American unilateralism under President Bush. Whatever lingering receptivity the Russian elite had to U.S. moralizing about peace and democracy was done in by the invasion and occupation of Iraq. Russia’s indignation over American plans to base missile-defense units near its borders are but a first taste of what is likely to come in the arms-control domain. Toward the former Soviet republics, Putin’s Russia betrays a welter of emotions, but prime among them are nostalgia for empire, a sour resentment at perceived ingratitude, and a determination to block further incursions by third parties. On all these points, Putin is if anything more moderate than the average member of the apparatus he heads.

The enigmatic Mr. Putin insists that he is about to leave center stage, and so Russians and Russia watchers are forced to think of the future. The second consecutive term allowed by the 1993 constitution expires next spring, and Putin has repeatedly declared his opposition to a constitutional amendment enabling him to serve longer. The authorities have already scheduled the presidential election for March 9, 2008. Putin has said he will recommend a candidate during the campaign, and the odds surely favor voters going along with his preference, as they did with Putin himself eight years before. Kremlin strategists have coined a new slogan for the pre-election interval: defense of Russia’s suverennaya demokratiya, or “sovereign democracy”—in other words, maintenance of the Putinist status quo.

Whoever prevails in the short term—the current betting is on either of two first deputy prime ministers, Sergei Ivanov and Dmitrii Medvedev, but dark horses lurk—the Russian political environment dictates that we will not know much more in advance about the victor’s inner world than we did about Putin’s in 1999-2000.

Looking five or so years out, the enigmatic Mr. X will have one continuity-enhancing variable to cope with that Putin did not have: namely, the presence on the scene of a relatively young, healthy, and admired predecessor (Yeltsin was elderly, unhealthy, and unloved). Putin has hinted that he will play a significant role after he steps down. This could be an informal one, not unlike that of Deng Xiaoping as China’s senior statesman in the 1980s and 1990s, or a more formalized role, such as chairman of United Russia or conceivably prime minister (Putin could hold this office without constitutional amendment). But the trouble with looking over another’s shoulder, as the parents of teenagers inevitably discover, is that it does not guarantee compliance—and may indeed be counterproductive.

Fact is, Russia’s incoming leader, with an eye to self-assertion and to his place in history, will have the same incentives to separate from the outgoing leader that Vladimir Putin had. In theory, Mr. X could, of course, change the country for the worse by intensifying the illiberal tendencies of the Putin years. Alternatively, he could nudge it in the opposite, more liberal direction.

We should not dismiss the possibility of a more liberal course. The combination of economic dynamism and political inertia is not always a viable one, as the democratization of countries such as Spain, South Korea, and Taiwan goes to show. Significant segments of the Russian elite, while mostly respectful of Putin’s contribution to stability, chafe at the prominence of ex-KGB functionaries, at the infantile state of the news media, and at the curbs on political competition, and harbor a fear that, without more ironclad protections for individual rights and private property, they, too, are in danger of going the way of Khodorkovskii. In personal terms, Mr. X will have difficulty equaling Putin’s rapport with the grassroots, and slavish adherence to the Putin program would make the new man look all the paler by comparison. Then there is the international factor. It will not be decisive on its own, but in combination with domestic factors it may make a difference. We will have a modest chance to influence Russia’s developmental choices if, this time around, we can imagine a place for it in the global community in which its worst instincts are restrained and its best instincts encouraged.

Last updated on 01/02/2014