• Posted by Konstantin 29.03.2017 No Comments

    The following is an expanded version of an explanatory comment I posted here.

    Alice's Diary

    Alice decided to keep a diary. For that she bought a notebook, and started filling it with lines like:

    1. Bought 5 apples.
    2. Called mom.
    3. Gave Bob $250.
    4. Kissed Carl.
    5. Ate a banana.

    Alice did her best to keep a meticulous account of events, and whenever she had a discussion with friends about something that happened earlier, she would quickly resolve all arguments by taking out the notebook and demonstrating her records. One day she had a dispute with Bob about whether she lent him $250 earlier or not. Unfortunately, Alice did not have her notebook at hand at the time of the dispute, but she promised to bring it tomorrow to prove Bob owed her money.

    Bob really did not want to return the money, so that night he got into Alice's house, found the notebook, found line 132 and carefully replaced it with "132. Kissed Dave". The next day, when Alice opened the notebook, she did not find any records about money being given to Bob, and had to apologize for making a mistake.

    Alice's Blockchain

    A year later Bob's conscience got to him and he confessed his crime to Alice. Alice forgave him, but decided to improve the way she kept the diary, to avoid the risk of forging records in the future. Here's what she came up with. The operating system Linups that she was using had a program named md5sum, which could convert any text to its hash - a strange sequence of 32 characters. Alice did not really understand what the program did with the text, it just seemed to produce a sufficiently random sequence. For example, if you entered "hello" into the program, it would output "b1946ac92492d2347c6235b4d2611184", and if you entered "hello " with a space at the end, the output would be "1a77a8341bddc4b45418f9c30e7102b4".

    Alice scratched her head a bit and invented the following way of making record forging more complicated to people like Bob in the future: after each record she would insert the hash, obtained by feeding the md5sum program with the text of the record and the previous hash. The new diary now looked as follows:

    1. 0000 (the initial hash, let us limit ourselves with just four digits for brevity)
    2. Bought 5 apples.
    3. 4178 (the hash of "0000" and "Bought 5 apples")
    4. Called mom.
    5. 2314 (the hash of "4178" and "Called mom")
    6. Gave Bob $250.
      1010 (the hash of "4492" and "Gave Bob $250")
    7. Kissed Carl.
      8204 (the hash of "1010" and "Kissed Carl")

    Now each record was "confirmed" by a hash. If someone wanted to change the line 132 to something else, they would have to change the corresponding hash (it would not be 1010 anymore). This, in turn, would affect the hash of line 133 (which would not be 8204 anymore), and so on all the way until the end of the diary. In order to change one record Bob would have to rewrite confirmation hashes for all the following diary records, which is fairly time-consuming. This way, hashes "chain" all records together, and what was before a simple journal became now a chain of records or "blocks" - a blockchain.

    Proof-of-Work Blockchain

    Time passed, Alice opened a bank. She still kept her diary, which now included serious banking records like "Gave out a loan" or "Accepted a deposit". Every record was accompanied with a hash to make forging harder. Everything was fine, until one day a guy named Carl took a loan of $1000000. The next night a team of twelve elite Chinese diary hackers (hired by Carl, of course) got into Alice's room, found the journal and substituted in it the line "143313. Gave out a $1000000 loan to Carl" with a new version: "143313. Gave out a $10 loan to Carl". They then quickly recomputed all the necessary hashes for the following records. For a dozen of hackers armed with calculators this did not take too long.

    Fortunately, Alice saw one of the hackers retreating and understood what happened. She needed a more secure system. Her new idea was the following: let us append a number (called "nonce") in brackets to each record, and choose this number so that the confirmation hash for the record would always start with two zeroes. Because hashes are rather unpredictable, the only way to do it is to simply try out different nonce values until one of them results in a proper hash:

    1. 0000
    2. Bought 5 apples (22).
    3. 0042 (the hash of "0000" and "Bought 5 apples (22)")
    4. Called mom (14).
    5. 0089 (the hash of "0042" and "Called mom (14)")
    6. Gave Bob $250 (33).
    7. Kissed Carl (67).
      0093 (the hash of "0001" and "Kissed Carl (67)")

    To confirm each record one now needs to try, on average, about 50 different hashing operations for different nonce values, which makes it 50 times harder to add new records or forge them than previously. Hopefully even a team of hackers wouldn't manage in time. Because each confirmation now requires hard (and somewhat senseless) work, the resulting method is called a proof-of-work system.

    Distributed Blockchain

    Tired of having to search for matching nonces for every record, Alice hired five assistants to help her maintain the journal. Whenever a new record needed to be confirmed, the assistants would start to seek for a suitable nonce in parallel, until one of them completed the job. To motivate the assistants to work faster she allowed them to append the name of the person who found a valid nonce, and promised to give promotions to those who confirmed more records within a year. The journal now looked as follows:

    1. 0000
    2. Bought 5 apples (29, nonce found by Mary).
    3. 0013 (the hash of "0000" and "Bought 5 apples (29, nonce found by Mary)")
    4. Called mom (45, nonce found by Jack).
    5. 0089 (the hash of "0013" and "Called mom (45, nonce found by Jack)")
    6. Gave Bob $250 (08, nonce found by Jack).
    7. Kissed Carl (11, nonce found by Mary).

    A week before Christmas, two assistants came to Alice seeking for a Christmas bonus. Assistant Jack, showed a diary where he confirmed 140 records and Mary confirmed 130, while Mary showed a diary where she, reportedly, confirmed more records than Jack. Each of them was showing Alice a journal with all the valid hashes, but different entries! It turns out that ever since having found out about the promotion the two assistants were working hard to keep their own journals, such that all nonces would have their names. Since they had to maintain the journals individually they had to do all the work confirming records alone rather than splitting it among other assistants. This of course made them so busy that they eventually had to miss some important entries about Alice's bank loans.

    Consequently, Jacks and Mary's "own journals" ended up being shorter than the "real journal", which was, luckily, correctly maintained by the three other assistants. Alice was disappointed, and, of course, did not give neither Jack nor Mary a promotion. "I will only give promotions to assistants who confirm the most records in the valid journal", she said. And the valid journal is the one with the most entries, of course, because the most work has been put into it!

    After this rule has been established, the assistants had no more motivation to cheat by working on their own journal alone - a collective honest effort always produced a longer journal in the end. This rule allowed assistants to work from home and completely without supervision. Alice only needed to check that the journal had the correct hashes in the end when distributing promotions. This way, Alice's blockchain became a distributed blockchain.


    Jack happened to be much more effective finding nonces than Mary and eventually became a Senior Assistant to Alice. He did not need any more promotions. "Could you transfer some of the promotion credits you got from confirming records to me?", Mary asked him one day. "I will pay you $100 for each!". "Wow", Jack thought, "apparently all the confirmations I did still have some value for me now!". They spoke with Alice and invented the following way to make "record confirmation achievements" transferable between parties.

    Whenever an assistant found a matching nonce, they would not simply write their own name to indicate who did it. Instead, they would write their public key. The agreement with Alice was that the corresponding confirmation bonus would belong to whoever owned the matching private key:

    1. 0000
    2. Bought 5 apples (92, confirmation bonus to PubKey61739).
    3. 0032 (the hash of "0000" and "Bought 5 apples (92, confirmation bonus to PubKey61739)")
    4. Called mom (52, confirmation bonus to PubKey55512).
    5. 0056 (the hash of "0032" and "Called mom (52, confirmation bonus to PubKey55512)")
    6. Gave Bob $250 (22, confirmation bonus to PubKey61739).
    7. Kissed Carl (40, confirmation bonus to PubKey55512).

    To transfer confirmation bonuses between parties a special type of record would be added to the same diary. The record would state which confirmation bonus had to be transferred to which new public key owner, and would be signed using the private key of the original confirmation owner to prove it was really his decision:

    1. 0071
    2. Gave Bob $250 (22, confirmation bonus to PubKey6669).
    3. Kissed Carl (40, confirmation bonus to PubKey5551).
    4. TRANSFER BONUS IN RECORD 132 TO OWNER OF PubKey1111, SIGNED BY PrivKey6669. (83, confirmation bonus to PubKey4442).

    In this example, record 284 transfers bonus for confirming record 132 from whoever it belonged to before (the owner of private key 6669, presumably Jack in our example) to a new party - the owner of private key 1111 (who could be Mary, for example). As it is still a record, there is also a usual bonus for having confirmed it, which went to owner of private key 4442 (who could be John, Carl, Jack, Mary or whoever else - it does not matter here). In effect, record 284 currently describes two different bonuses - one due to transfer, and another for confirmation. These, if necessary, can be further transferred to different parties later using the same procedure.

    Once this system was implemented, it turned out that Alice's assistants and all their friends started actively using the "confirmation bonuses" as a kind of an internal currency, transferring them between each other's public keys, even exchanging for goods and actual money. Note that to buy a "confirmation bonus" one does not need to be Alice's assistant nor register anywhere. One just needs to provide a public key.

    This confirmation bonus trading activity became so prominent that Alice stopped using the diary for her own purposes, and eventually all the records in the diary would only be about "who transferred which confirmation bonus to whom". This idea of a distributed proof-of-work-based blockchain with transferable confirmation bonuses is known as the Bitcoin.

    Smart Contracts

    But wait, we are not done yet. Note how Bitcoin is born from the idea of recording "transfer claims", cryptographically signed by the corresponding private key, into a blockchain-based journal. There is no reason we have to limit ourselves to this particular cryptographic protocol. For example, we could just as well make the following records:

    1. Transfer bonus in record 132 to whoever can provide signatures, corresponding to PubKey1111 AND PubKey3123.

    This would be an example of a collective deposit, which may only be extracted by a pair of collaborating parties. We could generalize further and consider conditions of the form:

    1. Transfer bonus in record 132 to whoever first provides x, such that f(x) = \text{true}.

    Here f(x) could be any predicate describing a "contract". For example, in Bitcoin the contract requires x to be a valid signature, corresponding to a given public key (or several keys). It is thus a "contract", verifying the knowledge of a certain secret (the private key). However, f(x) could just as well be something like:

        \[f(x) = \text{true, if }x = \text{number of bytes in record #42000},\]

    which would be a kind of a "future prediction" contract - it can only be evaluated in the future, once record 42000 becomes available. Alternatively, consider a "puzzle solving contract":

        \[f(x) = \text{true, if }x = \text{valid, machine-verifiable}\]

        \[\qquad\qquad\text{proof of a complex theorem},\]

    Finally, the first part of the contract, namely the phrase "Transfer bonus in record ..." could also be fairly arbitrary. Instead of transferring "bonuses" around we could just as well transfer arbitrary tokens of value:

    1. Whoever first provides x, such that f(x) = \text{true} will be DA BOSS.
    2. x=42 satisifes the condition in record 284.
      Now and forever, John is DA BOSS!

    The value and importance of such arbitrary tokens will, of course, be determined by how they are perceived by the community using the corresponding blockchain. It is not unreasonable to envision situations where being DA BOSS gives certain rights in the society, and having this fact recorded in an automatically-verifiable public record ledger makes it possible to include the this knowledge in various automated systems (e.g. consider a door lock which would only open to whoever is currently known as DA BOSS in the blockchain).

    Honest Computing

    As you see, we can use a distributed blockchain to keep journals, transfer "coins" and implement "smart contracts". These three applications are, however, all consequences of one general, core property. The participants of a distributed blockchain ("assistants" in the Alice example above, or "miners" in Bitcoin-speak) are motivated to precisely follow all rules necessary for confirming the blocks. If the rules say that a valid block is the one where all signatures and hashes are correct, the miners will make sure these indeed are. If the rules say that a valid block is the one where a contract function needs to be executed exactly as specified, the miners will make sure it is the case, etc. They all seek to get their confirmation bonuses, and they will only get them if they participate in building the longest honestly computed chain of blocks.

    Because of that, we can envision blockchain designs where a "block confirmation" requires running arbitrary computational algorithms, provided by the users, and the greedy miners will still execute them exactly as stated. This general idea lies behind the Ethereum blockchain project.

    There is just one place in the description provided above, where miners have some motivational freedom to not be perfectly honest. It is the decision about which records to include in the next block to be confirmed (or which algorithms to execute, if we consider the Ethereum blockchain). Nothing really prevents a miner to refuse to ever confirm a record "John is DA BOSS", ignoring it as if it never existed at all. This problem is overcome in modern blockchains by having users offer additional "tip money" reward for each record included in the confirmed block (or for every algorithmic step executed on the Ethereum blockchain). This aligns the motivation of the network towards maximizing the number of records included, making sure none is lost or ignored. Even if some miners had something against John being DA BOSS, there would probably be enough other participants who would not turn down the opportunity of getting an additional tip.

    Consequently, the whole system is economically incentivised to follow the protocol, and the term "honest computing" seems appropriate to me.

    Now that you know how things work, feel free to transfer all your bitcoins (i.e. block confirmation bonuses for which you know the corresponding private keys) to the address 1JuC76CX4FGo3W3i2Xv7L86Vz4chHHg71m (i.e. a public key, to which I know the corresponding private key).

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  • Posted by Konstantin 28.03.2017 No Comments

    Consider the following question:

    Which of the following two statements is logically true?

    1. All planets of the Solar System orbit the Sun. The Earth orbits the Sun. Consequently, the Earth is a planet of the Solar System.
    2. God is the creator of all things which exist. The Earth exists. Consequently, God created the Earth.

    implicationI've seen this question or variations of it pop up as "provocative" posts in social networks several times. At times they might invite lengthy discussions, where the participants would split into camps - some claim that the first statement is true, because Earth is indeed a planet of the Solar System and God did not create the Earth. Others would laugh at the stupidity of their opponents and argue that, obviously, only the second statement is correct, because it makes a valid logical implication, while the first one does not.

    Not once, however, have I ever seen a proper formal explanation of what is happening here. And although it is fairly trivial (once you know it), I guess it is worth writing up. The root of the problem here is the difference between implication and provability - something I myself remember struggling a bit to understand when I first had to encounter these notions in a course on mathematical logic years ago.

    Indeed, what any textbook on propositional logic will tell you in one of the first chapters that you can write

        \[A \Rightarrow B\]

    to express the statement "A implies B". A chapter or so later you will learn that there is also a possibility to write

        \[A \vdash B\]

    to express a confusingly similar statement, that "B is provable from A". To confirm your confusion, another chapter down the road you should discover, that A \Rightarrow B is the same as \vdash A \Rightarrow B, which, in turn, is logically equivalent to A \vdash B. Therefore, indeed, whenever A \Rightarrow B is true, A \vdash B is true, and vice-versa. Is there a difference between \vdash and \Rightarrow then, and why do we need the two different symbols at all? The "provocative" question above provides an opportunity to illustrate this.

    The spoken language is rather informal, and there can be several ways of formally interpreting the same statement. Both statements in the puzzle are given in the form "A, B, consequently C". Here are at least four different ways to put them formally, which make the two statements true or false in different ways.

    The Pure Logic Interpretation

    Anyone who has enough experience solving logic puzzles would know that both statements should be interpreted as abstract claims about provability (i.e. deducibility):

        \[A, B \vdash C.\]

    As mentioned above, this is equivalent to

        \[(A\,\&\, B) \Rightarrow C.\]


        \[\vdash (A\,\&\, B) \Rightarrow C.\]

    In this interpretation the first statement is wrong and the second is a correct implication.

    The Pragmatic Interpretation

    People who have less experience with math puzzles would often assume that they should not exclude their common sense knowledge from the task. The corresponding formal statement of the problem then becomes the following:

        \[[\text{common knowledge}] \vdash (A\,\&\, B) \Rightarrow C.\]

    In this case both statements become true. The first one is true simply because the consequent C is true on its own, given common knowledge (the Earth is indeed a planet) - the antecedents and provability do not play any role at all. The second is true because it is a valid reasoning, independently of the common knowledge.

    This type of interpretation is used in rhetorical phrases like "If this is true, I am a Dutchman".

    The Overly Strict Interpretation

    Some people may prefer to believe that a logical statement should only be deemed correct if every single part of it is true and logically valid. The two claims must then be interpreted as follows:

        \[([\text{common}] \vdash A)\,\&\, ([\text{common}] \vdash B)\,\&\, (A, B\vdash C).\]

    Here the issue of provability is combined with the question about the truthfulness of the facts used. Both statements are false - the first fails on logic, and the second on facts (assuming that God creating the Earth is not part of common knowledge).

    The Oversimplified Interpretation

    Finally, people very unfamiliar with strict logic would sometimes tend to ignore the words "consequently", "therefore" or "then", interpreting them as a kind of an extended synonym for "and". In their minds the two statements could be regarded as follows:

        \[[\text{common}] \vdash A\,\&\, B\,\&\, C.\]

    From this perspective, the first statement becomes true and the second (again, assuming the aspects of creation are not commonly known) is false.

    Although the author of the original question most probably did really assume the "pure logic" interpretation, as is customary for such puzzles, note how much leeway there can be when converting a seemingly simple phrase in English to a formal statement. In particular, observe that questions about provability, where you deliberately have to abstain from relying on common knowledge, may be different from questions about implication, where common sense may be assumed and you can skip the whole "reasoning" part if you know the consequent is true anyway.

    Here is an quiz question to check whether you understood what I meant to explain.

    "The sky is blue, and therefore the Earth is round." True or false?

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  • Posted by Konstantin 21.03.2017 No Comments

    Ever since Erwin Schrödinger described a thought experiment, in which a cat in a sealed box happened to be "both dead and alive at the same time", popular science writers have been relying on it heavily to convey the mysteries of quantum physics to the layman. Unfortunately, instead of providing any useful intuition, this example has instead laid solid base to a whole bunch of misconceptions. Having read or heard something about the strange cat, people would tend to jump to profound conclusions, such as "according to quantum physics, cats can be both dead and alive at the same time" or "the notion of a conscious observer is important in quantum physics". All of these are wrong, as is the image of a cat, who is "both dead and alive at the same time". The corresponding Wikipedia page does not stress this fact well enough, hence I thought the Internet might benefit from a yet another explanatory post.

    The Story of the Cat

    The basic notion in quantum mechanics is a quantum system. Pretty much anything could be modeled as a quantum system, but the most common examples are elementary particles, such as electrons or photons. A quantum system is described by its state. For example, a photon has polarization, which could be vertical or horizontal. Another prominent example of a particle's state is its wave function, which represents its position in space.

    There is nothing special about saying that things have state. For example, we may say that any cat has a "liveness state", because it can be either "dead" or "alive". In quantum mechanics we would denote these basic states using the bra-ket notation as |\mathrm{dead}\rangle and |\mathrm{alive}\rangle. The strange thing about quantum mechanical systems, though, is the fact that quantum states can be combined together to form superpositions. Not only could a photon have a purely vertical polarization \left|\updownarrow\right\rangle or a purely horizontal polarization \left|\leftrightarrow\right\rangle, but it could also be in a superposition of both vertical and horizontal states:

        \[\left|\updownarrow\right\rangle + \left|\leftrightarrow\right\rangle.\]

    This means that if you asked the question "is this photon polarized vertically?", you would get a positive answer with 50% probability - in another 50% of cases the measurement would report the photon as horizontally-polarized. This is not, however, the same kind of uncertainty that you get from flipping a coin. The photon is not either horizontally or vertically polarized. It is both at the same time.

    Amazed by this property of quantum systems, Schrödinger attempted to construct an example, where a domestic cat could be considered to be in the state

        \[|\mathrm{dead}\rangle + |\mathrm{alive}\rangle,\]

    which means being both dead and alive at the same time. The example he came up with, in his own words (citing from Wikipedia), is the following:

    Schrodingers_cat.svgA cat is penned up in a steel chamber, along with the following device (which must be secured against direct interference by the cat): in a Geiger counter, there is a tiny bit of radioactive substance, so small, that perhaps in the course of the hour one of the atoms decays, but also, with equal probability, perhaps none; if it happens, the counter tube discharges and through a relay releases a hammer that shatters a small flask of hydrocyanic acid. If one has left this entire system to itself for an hour, one would say that the cat still lives if meanwhile no atom has decayed. The first atomic decay would have poisoned it.

    The idea is that after an hour of waiting, the radiactive substance must be in the state

        \[|\mathrm{decayed}\rangle + |\text{not decayed}\rangle,\]

    the poison flask should thus be in the state

        \[|\mathrm{broken}\rangle + |\text{not broken}\rangle,\]

    and the cat, consequently, should be

        \[|\mathrm{dead}\rangle + |\mathrm{alive}\rangle.\]

    Correct, right? No.

    The Cat Ensemble

    Superposition, which is being "in both states at once" is not the only type of uncertainty possible in quantum mechanics. There is also the "usual" kind of uncertainty, where a particle is in either of two states, we just do not exactly know which one. For example, if we measure the polarization of a photon, which was originally in the superposition \left|\updownarrow\right\rangle + \left|\leftrightarrow\right\rangle, there is a 50% chance the photon will end up in the state \left|\updownarrow\right\rangle after the measurement, and a 50% chance the resulting state will be \left|\leftrightarrow\right\rangle. If we do the measurement, but do not look at the outcome, we know that the resulting state of the photon must be either of the two options. It is not a superposition anymore. Instead, the corresponding situation is described by a statistical ensemble:

        \[\{\left|\updownarrow\right\rangle: 50\%, \quad\left|\leftrightarrow\right\rangle: 50\%\}.\]

    Although it may seem that the difference between a superposition and a statistical ensemble is a matter of terminology, it is not. The two situations are truly different and can be distinguished experimentally. Essentially, every time a quantum system is measured (which happens, among other things, every time it interacts with a non-quantum system) all the quantum superpositions are "converted" to ensembles - concepts native to the non-quantum world. This process is sometimes referred to as decoherence.

    Now recall the Schrödinger's cat. For the cat to die, a Geiger counter must register a decay event, triggering a killing procedure. The registration within the Geiger counter is effectively an act of measurement, which will, of course, "convert" the superposition state into a statistical ensemble, just like in the case of a photon which we just measured without looking at the outcome. Consequently, the poison flask will never be in a superposition of being "both broken and not". It will be either, just like any non-quantum object should. Similarly, the cat will also end up being either dead or alive - you just cannot know exactly which option it is before you peek into the box. Nothing special or quantum'y about this.

    The Quantum Cat

    "But what gives us the right to claim that the Geiger counter, the flask and the cat in the box are "non-quantum" objects?", an attentive reader might ask here. Could we imagine that everything, including the cat, is a quantum system, so that no actual measurement or decoherence would happen inside the box? Could the cat be "both dead and alive" then?

    Indeed, we could try to model the cat as a quantum system with |\mathrm{dead}\rangle and |\mathrm{alive}\rangle being its basis states. In this case the cat indeed could end up in the state of being both dead and alive. However, this would not be its most exciting capability. Way more suprisingly, we could then kill and revive our cat at will, back and forth, by simply measuring its liveness state appropriately. It is easy to see how this model is unrepresentative of real cats in general, and the worry about them being able to be in superposition is just one of the many inconsistencies. The same goes for the flask and the Geiger counter, which, if considered to be quantum systems, get the magical abilities to "break" and "un-break", "measure" and "un-measure" particles at will. Those would certainly not be a real world flask nor a counter anymore.

    The Cat Multiverse

    There is one way to bring quantum superposition back into the picture, although it requires some rather abstract thinking. There is a theorem in quantum mechanics, which states that any statistical ensemble can be regarded as a partial view of a higher-dimensional superposition. Let us see what this means. Consider a (non-quantum) Schrödinger's cat. As it might be hopefully clear from the explanations above, the cat must be either dead or alive (not both), and we may formally represent this as a statistical ensemble:

        \[\{\left|\text{dead}\right\rangle: 50\%, \quad\left|\text{alive}\right\rangle: 50\%\}.\]

    It turns out that this ensemble is mathematically equivalent in all respects to a superposition state of a higher order:

        \[\left|\text{Universe A}, \text{dead}\right\rangle + \left|\text{Universe B}, \text{alive}\right\rangle,\]

    where "Universe A" and "Universe B" are some abstract, unobservable "states of the world". The situation can be interpreted by imagining two parallel universes: one where the cat is dead and one where it is alive. These universes exist simultaneously in a superposition, and we are present in both of them at the same time, until we open the box. When we do, the universe superposition collapses to a single choice of the two options and we are presented with either a dead, or a live cat.

    Yet, although the universes happen to be in a superposition here, existing both at the same time, the cat itself remains completely ordinary, being either totally dead or fully alive, depending on the chosen universe. The Schrödinger's cat is just a cat, after all.

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  • Posted by Konstantin 23.11.2016 3 Comments

    Basic linear algebra, introductory statistics and some familiarity with core machine learning concepts (such as PCA and linear models) are the prerequisites of this post. Otherwise it will probably make no sense. An abridged version of this text is also posted on Quora.

    Most textbooks on statistics cover covariance right in their first chapters. It is defined as a useful "measure of dependency" between two random variables:

        \[\mathrm{cov}(X,Y) = E[(X - E[X])(Y - E[Y])].\]

    The textbook would usually provide some intuition on why it is defined as it is, prove a couple of properties, such as bilinearity, define the covariance matrix for multiple variables as {\bf\Sigma}_{i,j} = \mathrm{cov}(X_i, X_j), and stop there. Later on the covariance matrix would pop up here and there in seeminly random ways. In one place you would have to take its inverse, in another - compute the eigenvectors, or multiply a vector by it, or do something else for no apparent reason apart from "that's the solution we came up with by solving an optimization task".

    In reality, though, there are some very good and quite intuitive reasons for why the covariance matrix appears in various techniques in one or another way. This post aims to show that, illustrating some curious corners of linear algebra in the process.

    Meet the Normal Distribution

    The best way to truly understand the covariance matrix is to forget the textbook definitions completely and depart from a different point instead. Namely, from the the definition of the multivariate Gaussian distribution:

    We say that the vector \bf x has a normal (or Gaussian) distribution with mean \bf \mu and covariance \bf \Sigma if:

        \[\Pr({\bf x}) =|2\pi{\bf\Sigma}|^{-1/2} \exp\left(-\frac{1}{2}({\bf x} - {\bf\mu})^T{\bf\Sigma}^{-1}({\bf x} - {\bf \mu})\right).\]

    To simplify the math a bit, we will limit ourselves to the centered distribution (i.e. {\bf\mu} = {\bf 0}) and refrain from writing out the normalizing constant |2\pi{\bf\Sigma}|^{-1/2}. Now, the definition of the (centered) multivariate Gaussian looks as follows:

        \[\Pr({\bf x}) \propto \exp\left(-0.5{\bf x}^T{\bf\Sigma}^{-1}{\bf x}\right).\]

    Much simpler, isn't it? Finally, let us define the covariance matrix as nothing else but the parameter of the Gaussian distribution. That's it. You will see where it will lead us in a moment.

    Transforming the Symmetric Gaussian

    Consider a symmetric Gaussian distribution, i.e. the one with {\bf \Sigma = \bf I} (the identity matrix). Let us take a sample from it, which will of course be a symmetric, round cloud of points:

    We know from above that the likelihood of each point in this sample is

    (1)   \[P({\bf x}) \propto \exp(-0.5 {\bf x}^T {\bf x}).\]

    Now let us apply a linear transformation {\bf A} to the points, i.e. let {\bf y} ={\bf Ax}. Suppose that, for the sake of this example, {\bf A} scales the vertical axis by 0.5 and then rotates everything by 30 degrees. We will get the following new cloud of points {\bf y}:

    What is the distribution of {\bf y}? Just substitute {\bf x}={\bf A}^{-1}{\bf y} into (1), to get:

    (2)   \begin{align*} P({\bf y}) &\propto \exp(-0.5 ({\bf A}^{-1}{\bf y})^T({\bf A}^{-1}{\bf y}))\\ &=\exp(-0.5{\bf y}^T({\bf AA}^T)^{-1}{\bf y}) \end{align*}

    This is exactly the Gaussian distribution with covariance {\bf \Sigma} = {\bf AA}^T. The logic works both ways: if we have a Gaussian distribution with covariance \bf \Sigma, we can regard it as a distribution which was obtained by transforming the symmetric Gaussian by some {\bf A}, and we are given {\bf AA}^T.

    More generally, if we have any data, then, when we compute its covariance to be \bf\Sigma, we can say that if our data were Gaussian, then it could have been obtained from a symmetric cloud using some transformation \bf A, and we just estimated the matrix {\bf AA}^T, corresponding to this transformation.

    Note that we do not know the actual \bf A, and it is mathematically totally fair. There can be many different transformations of the symmetric Gaussian which result in the same distribution shape. For example, if \bf A is just a rotation by some angle, the transformation does not affect the shape of the distribution at all. Correspondingly, {\bf AA}^T = {\bf I} for all rotation matrices. When we see a unit covariance matrix we really do not know, whether it is the “originally symmetric” distribution, or a “rotated symmetric distribution”. And we should not really care - those two are identical.

    There is a theorem in linear algebra, which says that any symmetric matrix \bf \Sigma can be represented as:

    (3)   \[{\bf \Sigma} = {\bf VDV}^T,\]

    where {\bf V} is orthogonal (i.e. a rotation) and {\bf D} is diagonal (i.e. a coordinate-wise scaling). If we rewrite it slightly, we will get:

    (4)   \[{\bf \Sigma} = ({\bf VD}^{1/2})({\bf VD}^{1/2})^T = {\bf AA}^T,\]

    where {\bf A} = {\bf VD}^{1/2}. This, in simple words, means that any covariance matrix \bf \Sigma could have been the result of transforming the data using a coordinate-wise scaling {\bf D}^{1/2} followed by a rotation \bf V. Just like in our example with \bf x and \bf y above.

    Principal Component Analysis

    Given the above intuition, PCA already becomes a very obvious technique. Suppose we are given some data. Let us assume (or “pretend”) it came from a normal distribution, and let us ask the following questions:

    1. What could have been the rotation \bf V and scaling {\bf D}^{1/2}, which produced our data from a symmetric cloud?
    2. What were the original, “symmetric-cloud” coordinates \bf x before this transformation was applied.
    3. Which original coordinates were scaled the most by \bf D and thus contribute most to the spread of the data now. Can we only leave those and throw the rest out?

    All of those questions can be answered in a straightforward manner if we just decompose \bf \Sigma into \bf V and \bf D according to (3). But (3) is exactly the eigenvalue decomposition of \bf\Sigma. I’ll leave you to think for just a bit and you’ll see how this observation lets you derive everything there is about PCA and more.

    The Metric Tensor

    Bear me for just a bit more. One way to summarize the observations above is to say that we can (and should) regard {\bf\Sigma}^{-1} as a metric tensor. A metric tensor is just a fancy formal name for a matrix, which summarizes the deformation of space. However, rather than claiming that it in some sense determines a particular transformation \bf A (which it does not, as we saw), we shall say that it affects the way we compute angles and distances in our transformed space.

    Namely, let us redefine, for any two vectors \bf v and \bf w, their inner product as:

    (5)   \[\langle {\bf v}, {\bf w}\rangle_{\Sigma^{-1}} = {\bf v}^T{\bf \Sigma}^{-1}{\bf w}.\]

    To stay consistent we will also need to redefine the norm of any vector as

    (6)   \[|{\bf v}|_{\Sigma^{-1}} = \sqrt{{\bf v}^T{\bf \Sigma}^{-1}{\bf v}},\]

    and the distance between any two vectors as

    (7)   \[|{\bf v}-{\bf w}|_{\Sigma^{-1}} = \sqrt{({\bf v}-{\bf w})^T{\bf \Sigma}^{-1}({\bf v}-{\bf w})}.\]

    Those definitions now describe a kind of a “skewed world” of points. For example, a unit circle (a set of points with “skewed distance” 1 to the center) in this world might look as follows:

    And here is an example of two vectors, which are considered “orthogonal”, a.k.a. “perpendicular” in this strange world:

    Although it may look weird at first, note that the new inner product we defined is actually just the dot product of the “untransformed” originals of the vectors:

    (8)   \[{\bf v}^T{\bf \Sigma}^{-1}{\bf w} = {\bf v}^T({\bf AA}^T)^{-1}{\bf w}=({\bf A}^{-1}{\bf v})^T({\bf A}^{-1}{\bf w}),\]

    The following illustration might shed light on what is actually happening in this \Sigma-“skewed” world. Somehow “deep down inside”, the ellipse thinks of itself as a circle and the two vectors behave as if they were (2,2) and (-2,2).

    Getting back to our example with the transformed points, we could now say that the point-cloud \bf y is actually a perfectly round and symmetric cloud “deep down inside”, it just happens to live in a skewed space. The deformation of this space is described by the tensor {\bf\Sigma}^{-1} (which is, as we know, equal to ({\bf AA}^T)^{-1}. The PCA now becomes a method for analyzing the deformation of space, how cool is that.

    The Dual Space

    We are not done yet. There’s one interesting property of “skewed” spaces worth knowing about. Namely, the elements of their dual space have a particular form. No worries, I’ll explain in a second.

    Let us forget the whole skewed space story for a moment, and get back to the usual inner product {\bf w}^T{\bf v}. Think of this inner product as a function f_{\bf w}({\bf v}), which takes a vector \bf v and maps it to a real number, the dot product of \bf v and \bf w. Regard the \bf w here as the parameter (“weight vector”) of the function. If you have done any machine learning at all, you have certainly come across such linear functionals over and over, sometimes in disguise. Now, the set of all possible linear functionals f_{\bf w} is known as the dual space to your “data space”.

    Note that each linear functional is determined uniquely by the parameter vector \bf w, which has the same dimensionality as \bf v, so apparently the dual space is in some sense equivalent to your data space - just the interpretation is different. An element \bf v of your “data space” denotes, well, a data point. An element \bf w of the dual space denotes a function f_{\bf w}, which projects your data points on the direction \bf w (recall that if \bf w is unit-length, {\bf w}^T{\bf v} is exactly the length of the perpendicular projection of \bf v upon the direction \bf w). So, in some sense, if \bf v-s are “vectors”, \bf w-s are “directions, perpendicular to these vectors”. Another way to understand the difference is to note that if, say, the elements of your data points numerically correspond to amounts in kilograms, the elements of \bf w would have to correspond to “units per kilogram”. Still with me?

    Let us now get back to the skewed space. If \bf v are elements of a skewed Euclidean space with the metric tensor {\bf\Sigma}^{-1}, is a function f_{\bf w}({\bf v}) = {\bf w}^T{\bf v} an element of a dual space? Yes, it is, because, after all, it is a linear functional. However, the parameterization of this function is inconvenient, because, due to the skewed tensor, we cannot interpret it as projecting vectors upon \bf w nor can we say that \bf w is an “orthogonal direction” (to a separating hyperplane of a classifier, for example). Because, remember, in the skewed space it is not true that orthogonal vectors satisfy {\bf w}^T{\bf v}=0. Instead, they satisfy {\bf w}^T{\bf \Sigma}^{-1}{\bf v} = 0. Things would therefore look much better if we parameterized our dual space differently. Namely, by considering linear functionals of the form f^{\Sigma^{-1}}_{\bf z}({\bf v}) = {\bf z}^T{\bf \Sigma}^{-1}{\bf v}. The new parameters \bf z could now indeed be interpreted as an “orthogonal direction” and things overall would make more sense.

    However when we work with actual machine learning models, we still prefer to have our functions in the simple form of a dot product, i.e. f_{\bf w}, without any ugly \bf\Sigma-s inside. What happens if we turn a “skewed space” linear functional from its natural representation into a simple inner product?

    (9)   \[f^{\Sigma^{-1}}_{\bf z}({\bf v}) = {\bf z}^T{\bf\Sigma}^{-1}{\bf v} = ({\bf \Sigma}^{-1}{\bf z})^T{\bf v} = f_{\bf w}({\bf v}),\]

    where {\bf w} = {\bf \Sigma}^{-1}{\bf z}. (Note that we can lose the transpose because \bf \Sigma is symmetric).

    What it means, in simple terms, is that when you fit linear models in a skewed space, your resulting weight vectors will always be of the form {\bf \Sigma}^{-1}{\bf z}. Or, in other words, {\bf\Sigma}^{-1} is a transformation, which maps from “skewed perpendiculars” to “true perpendiculars”. Let me show you what this means visually.

    Consider again the two “orthogonal” vectors from the skewed world example above:

    Let us interpret the blue vector as an element of the dual space. That is, it is the \bf z vector in a linear functional {\bf z}^T{\bf\Sigma}^{-1}{\bf v}. The red vector is an element of the “data space”, which would be mapped to 0 by this functional (because the two vectors are “orthogonal”, remember).

    For example, if the blue vector was meant to be a linear classifier, it would have its separating line along the red vector, just like that:

    If we now wanted to use this classifier, we could, of course, work in the “skewed space” and use the expression {\bf z}^T{\bf\Sigma}^{-1}{\bf v} to evaluate the functional. However, why don’t we find the actual normal \bf w to that red separating line so that we wouldn’t need to do an extra matrix multiplication every time we use the function?

    It is not too hard to see that {\bf w}={\bf\Sigma}^{-1}{\bf z} will give us that normal. Here it is, the black arrow:

    Therefore, next time, whenever you see expressions like {\bf w}^T{\bf\Sigma}^{-1}{\bf v} or ({\bf v}-{\bf w})^T{\bf\Sigma}^{-1}({\bf v}-{\bf w}), remember that those are simply inner products and (squared) distances in a skewed space, while {\bf \Sigma}^{-1}{\bf z} is a conversion from a skewed normal to a true normal. Also remember that the “skew” was estimated by pretending that the data were normally-distributed.

    Once you see it, the role of the covariance matrix in some methods like the Fisher’s discriminant or Canonical correlation analysis might become much more obvious.

    The Dual Space Metric Tensor

    “But wait”, you should say here. “You have been talking about expressions like {\bf w}^T{\bf\Sigma}^{-1}{\bf v} all the time, while things like {\bf w}^T{\bf\Sigma}{\bf v} are also quite common in practice. What about those?”

    Hopefully you know enough now to suspect that {\bf w}^T{\bf\Sigma}{\bf v} is again an inner product or a squared norm in some deformed space, just not the “internal data metric space”, that we considered so far. Which space is it? It turns out it is the “internal dual metric space”. That is, whilst the expression {\bf w}^T{\bf\Sigma}^{-1}{\bf v} denoted the “new inner product” between the points, the expression {\bf w}^T{\bf\Sigma}{\bf v} denotes the “new inner product” between the parameter vectors. Let us see why it is so.

    Consider an example again. Suppose that our space transformation \bf A scaled all points by 2 along the x axis. The point (1,0) became (2,0), the point (3, 1) became (6, 1), etc. Think of it as changing the units of measurement - before we measured the x axis in kilograms, and now we measure it in pounds. Consequently, the norm of the point (2,0) according to the new metric, |(2,0)|_{\Sigma^{-1}} will be 1, because 2 pounds is still just 1 kilogram “deep down inside”.

    What should happen to the parameter ("direction") vectors due to this transformation? Can we say that the parameter vector (1,0) also got scaled to (2,0) and that the norm of the parameter vector (2,0) is now therefore also 1? No! Recall that if our initial data denoted kilograms, our dual vectors must have denoted “units per kilogram”. After the transformation they will be denoting “units per pound”, correspondingly. To stay consistent we must therefore convert the parameter vector (”1 unit per kilogram”, 0) to its equivalent (“0.5 units per pound”,0). Consequently, the norm of the parameter vector (0.5,0) in the new metric will be 1 and, by the same logic, the norm of the dual vector (2,0) in the new metric must be 4. You see, the “importance of a parameter/direction” gets scaled inversely to the “importance of data” along that parameter or direction.

    More formally, if {\bf x}'={\bf Ax}, then

    (10)   \begin{align*} f_{\bf w}({\bf x}) &= {\bf w}^T{\bf x} = {\bf w}^T{\bf A}^{-1}{\bf x}'\\ & =(({\bf A}^{-1})^T{\bf w})^T{\bf x}'=f_{({\bf A}^{-1})^T{\bf w}}({\bf x}'). \end{align*}

    This means, that the transformation \bf A of the data points implies the transformation {\bf B}:=({\bf A}^{-1})^T of the dual vectors. The metric tensor for the dual space must thus be:

    (11)   \[({\bf BB}^T)^{-1}=(({\bf A}^{-1})^T{\bf A}^{-1})^{-1}={\bf AA}^T={\bf \Sigma}.\]

    Remember the illustration of the “unit circle” in the {\bf \Sigma}^{-1} metric? This is how the unit circle looks in the corresponding \bf\Sigma metric. It is rotated by the same angle, but it is stretched in the direction where it was squished before.

    Intuitively, the norm (“importance”) of the dual vectors along the directions in which the data was stretched by \bf A becomes proportionally larger (note that the “unit circle” is, on the contrary, “squished” along those directions).

    But the “stretch” of the space deformation in any direction can be measured by the variance of the data. It is therefore not a coincidence that {\bf w}^T{\bf \Sigma}{\bf w} is exactly the variance of the data along the direction \bf w (assuming |{\bf w}|=1).

    The Covariance Estimate

    Once we start viewing the covariance matrix as a transformation-driven metric tensor, many things become clearer, but one thing becomes extremely puzzling: why is the inverse covariance of the data a good estimate for that metric tensor? After all, it is not obvious that {\bf X}^T{\bf X}/n (where \bf X is the data matrix) must be related to the \bf\Sigma in the distribution equation \exp(-0.5{\bf x}^T{\bf\Sigma}^{-1}{\bf x}).

    Here is one possible way to see the connection. Firstly, let us take it for granted that if \bf X is sampled from a symmetric Gaussian, then {\bf X}^T{\bf X}/n is, on average, a unit matrix. This has nothing to do with transformations, but just a consequence of pairwise independence of variables in the symmetric Gaussian.

    Now, consider the transformed data, {\bf Y}={\bf XA}^T (vectors in the data matrix are row-wise, hence the multiplication on the right with a transpose). What is the covariance estimate of \bf Y?

    (12)   \[{\bf Y}^T{\bf Y}/n=({\bf XA}^T)^T{\bf XA}^T/n={\bf A}({\bf X}^T{\bf X}){\bf A}^T/n\approx {\bf AA}^T,\]

    the familiar tensor.

    This is a place where one could see that a covariance matrix may make sense outside the context of a Gaussian distribution, after all. Indeed, if you assume that your data was generated from any distribution P with uncorrelated variables of unit variance and then transformed using some matrix \bf A, the expression {\bf X}^T{\bf X}/n will still be an estimate of {\bf AA}^T, the metric tensor for the corresponding (dual) space deformation.

    However, note that out of all possible initial distributions P, the normal distribution is exactly the one with the maximum entropy, i.e. the “most generic”. Thus, if you base your analysis on the mean and the covariance matrix (which is what you do with PCA, for example), you could just as well assume your data to be normally distributed. In fact, a good rule of thumb is to remember, that whenever you even mention the word "covariance matrix", you are implicitly fitting a Gaussian distribution to your data.

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  • Posted by Konstantin 17.11.2016 No Comments

    Mass on a spring

    Imagine a weight hanging on a spring. Let us pull the weight a bit and release it into motion. What will its motion look like? If you remember some of your high-school physics, you should probably answer that the resulting motion is a simple harmonic oscillation, best described by a sinewave. Although this is a fair answer, it actually misses an interesting property of real-life springs. A property most people don't think much about, because it goes a bit beyond the high school curriculum. This property is best illustrated by

    The Slinky Drop

    The "slinky drop" is a fun little experiment which has got its share of internet fame.

    The Slinky Drop

    The Slinky Drop

    When the top end of a suspended slinky is released, the bottom seems to patiently wait for the top to arrive before starting to fall as well. This looks rather unexpected. After all, we know that things fall down according to a parabola, and we know that springs collapse according to a sinewave, however neither of the two rules seem to apply here. If you browse around, you will see lots of awesome videos demonstrating or explaining this effect. There are news articles, forum discussions, blog posts and even research papers dedicated to the magical slinky. However, most of them are either too sketchy or too complex, and none seem to mention the important general implications, so let me give a shot at another explanation here.

    The Slinky Drop Explained Once More

    Let us start with the classical, "high school" model of a spring. The spring has some length L in the relaxed state, and if we stretch it, making it longer by \Delta L, the two ends of the spring exert a contracting force of k\Delta L. Assume we hold the top of the spring at the vertical coordinate y_{\mathrm{top}}=0 and have it balance out. The lower end will then position at the coordinate y_{\mathrm{bot}} = -(L+mg/k), where the gravity force mg is balanced out exactly by the spring force.

    How would the two ends of the spring behave if we let go off the top now? Here's how:

    The falling spring, version 1

    The horozontal axis here denotes the time, the vertical axis - is the vertical position. The blue curve is the trajectory of the top end of the spring, the green curve - trajectory of the bottom end. The dotted blue line is offset from the blue line by exactly L - the length of the spring in relaxed state.

    Observe that the lower end (the green curve), similarly to the slinky, "waits" for quite a long time for the top to approach before starting to move with discernible velocity. Why is it the case? The trajectory of the lower point can be decomposed in two separate movements. Firstly, the point is trying to fall down due to gravity, following a parabola. Secondly, the point is being affected by string tension and thus follows a cosine trajectory. Here's how the two trajectories look like separately:

    They are surprisingly similar at the start, aren't they? And indeed, the cosine function does resemble a parabola up to o(x^3). Recall the corresponding Taylor expansion:

        \[\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \dots \approx 1 - \frac{x^2}{2}.\]

    If we align the two curves above, we can see how well they match up at the beginning:

    Consequently, the two forces happen to "cancel" each other long enough to leave an impression that the lower end "waits" for the upper for some time. This effect is, however, much more pronounced in the slinky. Why so?

    Because, of course, a single spring is not a good model for the slinky. It is more correct to regard a slinky as a chain of strings. Observe what happens if we model the slinky as a chain of just three simple springs:

    Each curve here is the trajectory of one of the nodes inbetween the three individual springs. We can see that the top two curves behave just like a single spring did - the green node waits a bit for the blue and then starts moving. The red one, however, has to wait longer, until the green node moves sufficiently far away. The bottom, in turn, will only start moving observably when the red node approaches it close enough, which means it has to wait even longer yet - by that time the top has already arrived. If we consider a more detailed model, the movement  of a slinky composed of, say, 9 basic springs, the effect of intermediate nodes "waiting" becomes even more pronounced:

    To make a "mathematically perfect" model of a slinky we have to go to the limit of having infinitely many infinitely small springs. Let's briefly take a look at how that solution looks like.

    The Continuous Slinky

    Let x denote the coordinate of a point on a "relaxed" slinky. For example, in the two discrete models above the slinky had 4 and 10 points, numbered 1,\dots, 4 and 1,\dots, 10 respectively. The continuous slinky will have infinitely many points numbered [0,1].

    Let h(x,t) denote the vertical coordinate of a point x at time t. The acceleration of point x at time t is then, by definition \frac{\partial^2 h(x,t)}{\partial^2 t}, and there are two components affecting it: the gravitational pull -g and the force of the spring.

    The spring force acting on a point x is proportional to the stretch of the spring at that point \frac{\partial h(x,t)}{\partial x}. As each point is affected by the stretch from above and below, we have to consider a difference of the "top" and "bottom" stretches, which is thus the derivative of the stretch, i.e. \frac{\partial^2 h(x,t)}{\partial^2 x}. Consequently, the dynamics of the slinky can be described by the equation:

        \[\frac{\partial^2 h(x,t)}{\partial^2 t} = a\frac{\partial^2 h(x,t)}{\partial^2 x} - g.\]

    where a is some positive constant. Let us denote the second derivatives by h_{tt} and h_{xx}, replace a with v^2 and rearrange to get:

    (1)   \[h_{tt} - v^2 h_{xx} = -g,\]

    which is known as the wave equation. The name stems from the fact that solutions to this equation always resemble "waves" propagating at a constant speed v through some medium. In our case the medium will be the slinky itself. Now it becomes apparent that, indeed, the lower end of the slinky should not move before the wave of disturbance, unleashed by releasing the top end, reaches it. Most of the explanations of the slinky drop seem to refer to that fact. However when it is stated alone, without the wave-equation-model context, it is at best a rather incomplete explanation.

    Given how famous the equation is, it is not too hard to solve it. We'll need to do it twice - first to find the initial configuration of a suspended slinky, then to compute its dynamics when the top is released.

    In the beginning the slinky must satisfy h_t(x, t) = 0 (because it is not moving at all), h(0, t) = 0 (because the top end is located at coordinate 0), and h_x(1, t) = 0 (because there is no stretch at the bottom). Combining this with (1) and searching for a polynomial solution, we get:

        \[h(x, t) = h_0(x) = \frac{g}{2v^2}x(x-2).\]

    Next, we release the slinky, hence the conditions h_t(x,t)=0 and h(0,t)=0 disappear and we may use the d'Alembert's formula with reflected boundaries to get the solution:

        \[h(x,t) = \frac{1}{2}(\phi(x-vt) + \phi(x+vt)) - \frac{gt^2}{2},\]

        \[\text{ where }\phi(x) = h_0(\mathrm{mod}(x, 2)).\]

    Here's how the solution looks like visually:

    Notice how the part of the slinky to which the wave has not arrived yet, stays completely fixed in place. Here are the trajectories of 4 equally-spaced points on the slinky:

    Note how, quite surprisingly, all points of the slinky are actually moving with a constant speed, changing it abruptly at certain moments. Somewhat magically, the mean of all these piecewise-linear trajectories (i.e. the trajectory of the center of mass of the slinky) is still a smooth parabola, just as it should be:

    The Secret of Spring Motion

    Now let us come back to where we started. Imagine a weight on a spring. What will its motion be like? Obviously, any real-life spring is, just like the slinky, best modeled not as a Hooke's simple spring, but rather via the wave equation. Which means that when you let go off the weight, the weight will send a deformation wave, which will move along the spring back and forth, affecting the pure sinewave movement you might be expecting from the simple Hooke's law. Watch closely:

    Here is how the movement of the individual nodes looks like:

    The fat red line is the trajectory of the weight, and it is certainly not a sinewave. It is a curve inbetween the piecewise-linear "sawtooth" (which is the limit case when the weight is zero) and the true sinusoid (which is the limit case when the mass of the spring is zero). Here's how the zero-weight case looks like:

    And this is the other extreme - the massless spring:

    These observations can be summarized into the following obviously-sounding conclusion: the basic Hooke's law applies exactly only to the the massless spring. Any real spring has a mass and thus forms an oscillation wave traveling back and forth along its length, which will interfere with the weight's simple harmonic oscillation, making it "less simple and harmonic". Luckily, if the mass of the weight is large enough, this interference is negligible.

    And that is, in my opinion, one of the interesting, yet often overlooked aspects of spring motion.

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  • Posted by Konstantin 11.11.2016 No Comments

    A question on Quora reminded me that I wanted to post this explanation here every time I got a chance to teach SVMs and Kernel methods, but I never found the time. The post expects basic knowledge of those topics from the reader.

    Introductory Background

    The concept of kernel methods is probably one of the coolest tricks in machine learning. With most machine learning research nowadays being centered around neural networks, they have gone somewhat out of fashion recently, but I suspect they will strike back one day in some way or another.

    The idea of a kernel method starts with the curious observation that if you take a dot product of two vectors, x^T y, and square it, the result can be regarded as a dot product of two "feature vectors", where the features are all pairwise products of the original inputs:

        \begin{align*} &k(x,y) = (x^Ty)^2 = (\sum_i x_iy_i)^2 = \sum_{i,j} x_ix_jy_iy_j \\ & = \langle (x_1x_1, x_1x_2,\dots,x_ix_j,\dots,x_nx_n), (y_1y_1,\dots,y_iy_j,\dots,y_ny_n)\rangle \end{align*}

    Analogously, if you raise x^Ty to the third power, you are essentially computing a dot product within a space of all possible three-way products of your inputs, and so on, without ever actually having to see those features explicitly.

    If you now take any linear model (e.g. linear regression, linear classification, PCA, etc) it turns out you can replace the "real" dot product in its formulation model with such a kernel function, and this will magically convert your model into a linear model with nonlinear features (e.g. pairwise or triple products). As those features are never explicitly computed, there is no problem if there were millions or billions of them.

    Consider, for example, plain old linear regression: f(x) = w^Tx + b. We can "kernelize" it by first representing w as a linear combination of the data points (this is called a dual representation):

        \[f(x) = \left(\sum_i \alpha_i x_i\right)^T x + b = \sum_i \alpha_i (x_i^T x) + b,\]

    and then swapping all the dot products x_i^T x with a custom kernel function:

        \[f(x) = \sum_i \alpha_i k(x_i,x) + b.\]

    If we now substitute k(x,y) = (x^T y)^2 here, our model becomes a second degree polynomial regression. If k(x,y) = (x^T y)^5 it is the fifth degree polynomial regression, etc. It's like magic, you plug in different functions and things just work.

    It turns out that there are lots of valid choices for the kernel function k, and, of course, the Gaussian function is one of these choices:

        \[k(x, y) = \exp\left(-\frac{|x - y|^2}{2\sigma^2}\right).\]

    It is not too surprising - the Gaussian function tends to pop up everywhere, after all, but it is not obvious what "implicit features" it should represent when viewed as a kernel function. Most textbooks do not seem to cover this question in sufficient detail, usually, so let me do it here.

    The Gaussian Kernel

    To see the meaning of the Gaussian kernel we need to understand the couple of ways in which any kernel functions can be combined. We saw before that raising a linear kernel to the power i makes a kernel with a feature space, which includes all i-wise products. Now let us examine what happens if we add two or more kernel functions. Consider k(x,y) = x^Ty + (x^Ty)^2, for example. It is not hard to see that it corresponds to an inner product of feature vectors of the form

        \[(x_1, x_2, \dots, x_n, \quad x_1x_1, x_1x_2,\dots,x_ix_j,\dots, x_n x_n),\]

    i.e. the concatenation of degree-1 (untransformed) features, and degree-2 (pairwise product) features.

    Multiplying a kernel function with a constant c is also meaningful. It corresponds to scaling the corresponding features by \sqrt{c}. For example, k(x,y) = 4x^Ty = (2x)^T(2y).

    Still with me? Great, now let us combine the tricks above and consider the following kernel:

        \[k(x,y) = 1 + x^Ty + \frac{(x^Ty)^2}{2} + \frac{(x^Ty)^3}{6}.\]

    Apparently, it is a kernel which corresponds to a feature mapping, which concatenates a constant feature, all original features, all pairwise products scaled down by \sqrt{2} and all triple products scaled down by \sqrt{6}.

    Looks impressive, right? Let us continue and add more members to this kernel, so that it would contain all four-wise, five-wise, and so on up to infinity-wise products of input features. We shall choose the scaling coefficients for each term carefully, so that the resulting infinite sum would resemble a familiar expression:

        \[\sum_{i=0}^\infty \frac{(x^Ty)^i}{i!} = \exp(x^Ty).\]

    We can conclude here that k(x,y) = \exp(x^Ty) is a valid kernel function, which corresponds to a feature space, which includes products of input features of any degree, up to infinity.

    But we are not done yet. Suppose that we decide to normalize the inputs before applying our linear model. That is, we want to convert each vector x to \frac{x}{|x|} before feeding it to the model. This is quite often a smart idea, which improves generalization. It turns out we can do this “data normalization” without really touching the data points themselves, but by only tuning the kernel instead.

    Consider again the linear kernel k(x,y) = x^Ty. If we apply it to normalized vectors, we get

        \[k\left(\frac{x}{|x|}, \frac{y}{|y|}\right) = \left(\frac{x}{|x|}\right)^T\left(\frac{y}{|y|}\right) = \frac{x^Ty}{|x||y|} = \frac{k(x,y)}{\sqrt{k(x,x)k(y,y)}}.\]

    With some reflection you will see that the latter expression would normalize the features for any kernel.

    Let us see what happens if we apply this kernel normalization to the “infinite polynomial” (i.e. exponential) kernel we just derived:

        \begin{align*} \frac{\exp(x^Ty)}{\sqrt{\exp(x^Tx)\exp(y^Ty)}} &= \frac{\exp(x^Ty)}{\exp(|x|^2/2)\exp(|y|^2/2)}  \\ &= \exp\left(-\frac{1}{2}(|x|^2+|y|^2-2x^Ty)\right) \\ &= \exp\left(-\frac{|x-y|^2}{2}\right). \end{align*}

    Voilà, the Gaussian kernel. Well, it still lacks \sigma^2 in the denominator but by now you hopefully see that adding it is equivalent to scaling the inputs by 1/\sigma

    To conclude: the Gaussian kernel is a normalized polynomial kernel of infinite degree (where feature products if i-th degree are scaled down by \sqrt{i!} before normalization). Simple, right?

    An Example

    The derivations above may look somewhat theoretic if not "magical", so let us work through a couple of numeric examples. Suppose our original vectors are one-dimensional (that is, real numbers), and let x = 1, y = 2. The value of the Gaussian kernel k(x, y) for these inputs is:

        \[k(x, y) = \exp(-0.5|1-2|^2) \approx 0.6065306597...\]

    Let us see whether we can obtain the same value as a simple dot product of normalized polynomial feature vectors of a high degree. For that, we first need to compute the corresponding unnormalized feature representation:

        \[\phi'(x) = \left(1, x, \frac{x^2}{\sqrt{2!}}, \frac{x^3}{\sqrt{3!}}, \frac{x^4}{\sqrt{4!}}, \frac{x^5}{\sqrt{5!}}\dots\right).\]

    As our inputs are rather small in magnitude, we can hope that the feature sequence quickly approaches zero, so we don't really have to work with infinite vectors. Indeed, here is how the feature sequences look like:

    ----\phi'(1) = (1, 1, 0.707, 0.408, 0.204, 0.091, 0.037, 0.014, 0.005, 0.002, 0.001, 0.000, 0.000, ...)

    ----\phi'(2) = (1, 2, 2.828, 3.266, 3.266, 2.921, 2.385, 1.803, 1.275, 0.850, 0.538, 0.324, 0.187, 0.104, 0.055, 0.029, 0.014, 0.007, 0.003, 0.002, 0.001, ...)

    Let us limit ourselves to just these first 21 features. To obtain the final Gaussian kernel feature representations \phi we just need to normalize:

    ----\phi(1) =\frac{\phi'(1)}{|\phi'(1)|} = \frac{\phi'(1)}{1.649} = (0.607, 0.607, 0.429, 0.248, 0.124, 0.055, 0.023, 0.009, 0.003, 0.001, 0.000, ...)

    ----\phi(2) =\frac{\phi'(2)}{|\phi'(2)|} = \frac{\phi'(2)}{7.389} = (0.135, 0.271, 0.383, 0.442, 0.442, 0.395, 0.323, 0.244, 0.173, 0.115, 0.073, 0.044, 0.025, 0.014, 0.008, 0.004, 0.002, 0.001, 0.000, ...)

    Finally, we compute the simple dot product of these two vectors:

        \[\scriptstyle\phi(1)^T\phi(2) = 0.607\cdot 0.135 + 0.607\cdot 0.271 + \dots = {\bf 0.6065306}602....\]

    In boldface are the decimal digits, which match the value of \exp(-0.5|1-2|^2). The discrepancy is probably more due to lack of floating-point precision rather than to our approximation.

    A 2D Example

    The one-dimensional example might have seemed somewhat too simplistic, so let us also go through a two-dimensional case. Here our unnormalized feature representation is the following:

        \begin{align*} \scriptscriptstyle\phi'(&\scriptscriptstyle x_1, x_2) = \left(1, x_1, \frac{x_1x_1}{\sqrt{2!}}, \frac{x_1x_2}{\sqrt{2!}}, \frac{x_2x_1}{\sqrt{2!}}, \frac{x_2x_2}{\sqrt{2!}}, \right. \\ &\scriptscriptstyle \left.\frac{x_1x_1x_1}{\sqrt{3!}}, \frac{x_1x_1x_2}{\sqrt{3!}}, \frac{x_1x_2x_1}{\sqrt{3!}}, \frac{x_1x_2x_2}{\sqrt{3!}}, \frac{x_2x_1x_2}{\sqrt{3!}},  \frac{x_2x_2x_1}{\sqrt{3!}}, \dots\right).\\ \end{align*}

    This looks pretty heavy, and we didn't even finish writing out the third degree products. If we wanted to continue all the way up to degree 20, we would end up with a vector with 2097151 elements!

    Note that many products are repeated, however (e.g. x_1x_2 = x_2x_1), hence these are not really all different features. Let us try to pack them more efficiently. As you'll see in a moment, this will open up a much nicer perspective on the feature vector in general.

    Basic combinatorics will tell us, that each feature of the form \frac{x_1^a x_2^b}{\sqrt{n!}} must be repeated exactly \frac{n!}{a!b!} times in our current feature vector. Thus, instead of repeating it, we could replace it with a single feature, scaled by \sqrt{\frac{n!}{a!b!}}. "Why the square root?" you might ask here. Because when combining a repeated feature we must preserve the overall vector norm. Consider a vector (v, v, v), for example. Its norm is \sqrt{v^2 + v^2 + v^2} = \sqrt{3}|v|, exactly the same as the norm of the single-element vector (\sqrt{3}v).

    As we do this scaling, each feature gets converted to a nice symmetric form:

        \[\sqrt{\frac{n!}{a!b!}}\frac{x_1^a x_2^b}{\sqrt{n!}} = \frac{x_1^a x_2^b}{\sqrt{a!b!}} = \frac{x^a}{\sqrt{a!}}\frac{x^b}{\sqrt{b!}}.\]

    This means that we can compute the 2-dimensional feature vector by first expanding each parameter into a vector of powers, like we did in the previous example, and then taking all their pairwise products. This way, if we wanted to limit ourselves with maximum degree 20, we would only have to deal with 21^2 = 231 features instead of 2097151. Nice!

    Here is a new view of the unnormalized feature vector up to degree 3:

        \[\phi'_3(x_1, x_2) = \scriptstyle\left(1, x_1, x_2, \frac{x_1^2}{\sqrt{2!}}, \frac{x_1x_2}{\sqrt{1!1!}}, \frac{x^2}{\sqrt{2!}}, \frac{x_1^3}{\sqrt{3!}}, \frac{x_1^2x_2}{\sqrt{2!1!}}, \frac{x_1x_2^2}{\sqrt{1!2!}}, \frac{x_2^3}{\sqrt{3!}}\right).\]

    Let us limit ourselves to this degree-3 example and let x = (0.7, 0.2), y = (0.1, 0.4) (if we picked larger values, we would need to expand our feature vectors to a higher degree to get a reasonable approximation of the Gaussian kernel). Now:

    ----\phi'_3(0.7, 0.2) = (1, 0.7, 0.2, 0.346, 0.140, 0.028, 0.140, 0.069, 0.020, 0.003),

    ----\phi'_3(0.1, 0.4) = (1, 0.1, 0.4, 0.007, 0.040, 0.113, 0.000, 0.003, 0.011, 0.026).

    After normalization:

    ----\phi_3(0.7, 0.2) = (0.768, 0.538, 0.154, 0.266, 0.108, 0.022, 0.108, 0.053, 0.015, 0.003),

    ----\phi_3(0.1, 0.4) = (0.919, 0.092, 0.367, 0.006, 0.037, 0.104, 0.000, 0.003, 0.010, 0.024).

    The dot product of these vectors is 0.8196\dots, what about the exact Gaussian kernel value?

        \[\exp(-0.5|x-y|^2) = 0.{\bf 81}87\dots.\]

    Close enough. Of course, this could be done for any number of dimensions, so let us conclude the post with the new observation we made:

    The features of the unnormalized d-dimensional Gaussian kernel are:

        \[\phi({\bf x})_{\bf a} = \prod_{i = 1}^d \frac{x_i^{a_i}}{\sqrt{a_i!}},\]

    where {\bf a} = (a_1, \dots, a_d) \in \mathbb{N}_0^d.

    The Gaussian kernel is just the normalized version of that, and we know that the norm to divide by is \sqrt{\exp({\bf x}^T{\bf x})}. Thus we may also state the following:

    The features of the d-dimensional Gaussian kernel are:

        \[\phi({\bf x})_{\bf a} = \exp(-0.5|{\bf x}|^2)\prod_{i = 1}^d \frac{x_i^{a_i}}{\sqrt{a_i!}},\]

    where {\bf a} = (a_1, \dots, a_d) \in \mathbb{N}_0^d.

    That's it, now you have seen the soul of the Gaussian kernel.

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