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This article was excerpted from Machine Learning with TensorFlow . Before jumping into , you should first familiarize yourself with how to use the tools. This article covers some essential advantages of TensorFlow, to convince you it’s the machine learning library of choice. machine learning algorithms As a thought experiment, let’s imagine what happens when we write Python code without a handy computing library. It’ll be like using a new smartphone without installing any extra apps. The phone still works, but you’d be more productive if you had the right apps. prices_. Another vector of size 100 called_ amounts Consider the following… You’re a business owner tracking the flow of sales. You want to calculate your revenue from selling your products. Your inventory consists of 100 different products, and you represent each price in a vector called represents the inventory count of each item. You can write the following chunk of Python code shown in listing 1 to calculate the revenue of selling all products. Keep in mind that this code doesn’t import any libraries. That’s a lot of code just to calculate the inner-product of two vectors (also known as ). Imagine how much code would be required for something more complicated, such as solving linear equations or computing the distance between two vectors. dot product By installing the TensorFlow library, you also install a well-known and robust Python library called NumPy, which facilitates mathematical manipulation in Python. Using Python without its libraries (e.g. NumPy and TensorFlow) is like using a camera without autofocus: you gain more flexibility, but you can easily make careless mistakes. It’s already pretty easy to make mistakes in machine learning, so let’s keep our camera on auto-focus and use TensorFlow to help automate some tedious software development. Listing 2 shows how to concisely write the same inner-product using NumPy. Python is a succinct language. Fortunately for you, that means you won’t see pages and pages of cryptic code. On the other hand, the brevity of the Python language implies that a lot is happening behind each line of code, which you should familiarize yourself with carefully as you work. Detailed documentation about various functions for the Python and C++ APIs for TensorFlow are available at . By the way… https://www.tensorflow.org/api_docs/index.html This article is geared toward using TensorFlow for computations, because machine learning relies on mathematical formulations. After going through the examples and code listings, you’ll be able to use TensorFlow for some arbitrary tasks, such as computing statistics on big data. The focus here will entirely be about how to use TensorFlow, as opposed to machine learning in general. Machine learning algorithms require a large amount of mathematical operations. Often, an algorithm boils down to a composition of simple functions iterated until convergence. Sure, you might use any standard programming language to perform these computations, but the secret to both manageable and performant code is the use of a well-written library. That sounds like a gentle start, right? Without further ado, let’s write our first TensorFlow code! Ensuring TensorFlow works First, we need to ensure that everything is working correctly. Check the oil level in your car, repair the blown fuse in your basement, and ensure that your credit balance is zero. Just kidding, I’m talking about TensorFlow. Go ahead and create a new file called for our first piece of code. Import TensorFlow by running the following script: test.py import tensorflow as tf This single import prepares TensorFlow for your bidding. If the Python interpreter doesn’t complain, then we’re ready to start using TensorFlow! A common cause of error at this step is if you installed the GPU version and the library fails to search for CUDA drivers. Remember, if you compiled the library with CUDA, you need to update your environment variables with the path to CUDA. Check the CUDA instructions on TensorFlow. (See for further information). Having technical difficulty? https://www.tensorflow.org/versions/master/get_started/os_setup.html#optional-linux-enable-gpu-support Sticking with TensorFlow conventions The TensorFlow library is usually imported with the qualified name. Generally, qualifying TensorFlow with is a good idea to remain consistent with other developers and open-source TensorFlow projects. You may choose not to qualify it or change the qualification name, but then successfully reusing other people’s snippets of TensorFlow code in your own projects will be an involved process. tf tf Representing tensors Now that we know how to import TensorFlow into a Python source file, let’s start using it! A convenient way to describe an object in the real world is by listing out its properties, or features. For example, you can describe a car by its color, model, engine type, and mileage. An ordered list of some features is called a and that’s exactly what we’ll represent in TensorFlow code. feature vector, Feature vectors are one of the most useful devices in machine learning because of their simplicity (they’re lists of numbers). Each data item typically consists of a feature vector and a good dataset has thousands, if not thousands, of these feature vectors. No doubt, you’ll often be dealing with more than one vector at a time. A concisely represents a list of vectors, where each column of a matrix is a feature vector. matrix The syntax to represent matrices in TensorFlow is a vector of vectors, each of the same length. Figure 1 is an example of a matrix with two rows and three columns, such as [[1, 2, 3], [4, 5, 6]]. Notice, this is a vector containing two elements, and each element corresponds to a row of the matrix. Figure 1. The matrix in the lower half of the diagram is a visualization from its compact code notation in the upper half of the diagram. This form of notation is a common paradigm in most scientific computing libraries. We access an element in a matrix by specifying its row and column indices. For example, the first row and first column indicate the first top-left element. Sometimes it’s convenient to use more than two indices, such as when referencing a pixel in a color image not only by its row and column, but also its red/green/blue channel. A is a generalization of a matrix that specifies an element by an arbitrary number of indices. tensor Suppose an elementary school enforces assigned seating to its students. You’re the principal, and you’re terrible with names. Luckily, each classroom has a grid of seats, where you can easily nickname a student by his or her row and column index. Example of a tensor… There are multiple classrooms, so you can’t say “Good morning 4,10! Keep up the good work.” You need to also specify the classroom, “Hi 4,10 from classroom 2.” Unlike a matrix, which needs only two indices to specify an element, the students in this school need three numbers. They’re all a part of a rank three tensor! The syntax for tensors is even more nested vectors. For example, a 2-by-3-by-2 tensor is [[[1,2], [3,4], [5,6]], [[7,8], [9,10], [11,12]]], which can be thought of as two matrices, each of size 3-by-2. Consequently, we say this tensor has a of 3. In general, the rank of a tensor is the number of indices required to specify an element. Machine learning algorithms in TensorFlow act on Tensors, and it’s important to understand how to use them. rank Figure 2. This tensor can be thought of as multiple matrices stacked on top of each other. To specify an element, you must indicate the row and column, as well as which matrix is being accessed. Therefore, the rank of this tensor is three. It’s easy to get lost in the many ways to represent a tensor. Intuitively, each of the following three lines of code in Listing 3 is trying to represent the same 2-by-2 matrix. This matrix represents two features vectors of two dimensions each. It could, for example, represent two people’s ratings of two movies. Each person, indexed by the row of the matrix, assigns a number to describe his or her review of the movie, indexed by the column. Run the code to see how to generate a matrix in TensorFlow. The first variable ( ) is a list, the second variable ( ) is an from the NumPy library, and the last variable ( ) is TensorFlow’s object. All operators in TensorFlow, such as , are designed to operate on tensor objects. A convenient function we can sprinkle anywhere to make sure that we’re dealing with tensors, as opposed to the other types, is . Most functions in the TensorFlow library already perform this function (redundantly), even if you forget to. Using is optional, but I show it here because it helps demystify the implicit type system being handled across the library. The aforementioned listing 3 produces the following output three times: m1 m2 ndarray m3 Tensor neg tf.convert_to_tensor( … ) tf.convert_to_tensor( … ) <class ‘tensorflow.python.framework.ops.Tensor’> Let’s take another look at defining tensors in code. After importing the TensorFlow library, we can use the constant operator as follows in Listing 4. Running listing 4 produces the following output: Tensor( “Const:0”,shape=TensorShape([Dimension(1),Dimension(2)]),dtype=float32 ) Tensor( “Const_1:0”,shape=TensorShape([Dimension(2),Dimension(1)]),dtype=int32 ) Tensor( “Const_2:0”,shape=TensorShape([Dimension(2),Dimension(3),Dimension(2)]),dtype=int32 ) As you can see from the output, each tensor is represented by the aptly named object. Each object has a unique label ( ), a dimension ( ) to define its structure, and data type ( ) to specify the kind of values we’ll manipulate. Because we didn’t explicitly provide a name, the library automatically generated them: “Const:0”, “Const_1:0”, and “Const_2:0”. Tensor Tensor name shape dtype Tensor types Notice that each of the elements of end with a decimal point. The decimal point tells Python that the data type of the elements isn’t an integer, but instead a float. We can pass in explicit values. Much like NumPy arrays, tensors take on a data type that specifies the kind of values we’ll manipulate in that tensor. matrix1 dtype TensorFlow also comes with a few convenient constructors for some simple tensors. For example, creates a tensor with all values initialized at zero of a specific shape. Similarly, creates a tensor of a specific shape with all values initialized at one. The shape argument is a one-dimensional (1D) tensor of type (a list of integers) describing the dimensions of the tensor. tf.zeros(shape) tf.ones(shape) int32 Creating operators Now that we have a few starting tensors ready to use, we can apply more interesting operators, such as addition or multiplication. Consider each row of a matrix representing the transaction of money to (positive value) and from (negative value) another person. Negating the matrix is a way to represent the transaction history of the other person’s flow of money. Let’s start simple and run the negation op (short for operation) on our tensor from listing 4. Negating a matrix turns the positive numbers into negative numbers of the same magnitude, and vice versa. matrix1 Negation is one of the simplest operations. As shown in listing 5, negation takes only one tensor as input, and produces a tensor with every element negated — now, try running the code yourself. If you master how to define negation, it’ll provide a stepping stone to generalize that skill to all other TensorFlow operations. an operation, such as negation, is different from it. Aside… Defining running Listing 5 generates the following output: Tensor(“Neg:0”, shape=(1, 2), dtype=int32) Useful TensorFlow operators The official documentation carefully lays out all available math ops: . https://www.tensorflow.org/api_docs/Python/math_ops.html Some specific examples of commonly used operators include: tf.add(x, y)Add two tensors of the same type, x + y tf.sub(x, y)Subtract tensors of the same type, x — y tf.mul(x, y)Multiply two tensors element-wise tf.pow(x, y)Take the element-wise power of x to y tf.exp(x)Equivalent to pow(e, x), where e is Euler’s number (2.718…) tf.sqrt(x)Equivalent to pow(x, 0.5) tf.div(x, y)Take the element-wise division of x and y tf.truediv(x, y)Same as tf.div, except casts the arguments as a float tf.floordiv(x, y)Same as truediv, except rounds down the final answer into an integer tf.mod(x, y)Takes the element-wise remainder from division Use the TensorFlow operators we’ve learned to produce the Gaussian Distribution (also known as Normal Distribution). See Figure 3 for a hint. For reference, you can find the probability density of the normal distribution online: Exercise… https://en.wikipedia.org/wiki/Normal_distribution. Most mathematical expressions such as “*”, “-“, “+”, etc. are shortcuts for their TensorFlow equivalent, for the sake of brevity. The Gaussian function includes many operations, and it’s cleaner to use some short-hand notations as follows: from math import pi mean = 1.0sigma = 0.0 (tf.exp(tf.neg(tf.pow(x — mean, 2.0) /(2.0 * tf.pow(sigma, 2.0) ))) *(1.0 / (sigma * tf.sqrt(2.0 * pi) ))) Figure 3. The graph represents the operations needed to produce a Gaussian distribution. The links between the nodes represent how data flows from one operation to the next. The operations themselves are simple, but complexity arises in how they intertwine. As you can see, TensorFlow algorithms are easy to visualize. They can be described by flowcharts. The technical (and more correct) term for the flowchart is a . Every arrow in a flowchart is called the of the graph. In addition, every state of the flowchart is called a . graph edge node Executing operators with sessions A session is an environment of a software system that describes how the lines of code should run. In TensorFlow, a session sets up how the hardware devices (such as CPU and GPU) talk to each other. That way, you can design your machine learning algorithm without worrying about micro-managing the hardware that it runs on. Of course, you can later configure the session to change its behavior without changing a line of the machine learning code. To execute an operation and retrieve its calculated value, TensorFlow requires a session. Only a registered session may fill the values of a Tensor object. To do so, you must create a session class using and tell it to run an operator (listing 6). The result will be a value you can later use for further computations. tf.Session() Congratulations! You have just written your first full TensorFlow code. Although all it does is negate a matrix to produce [[-1, -2]], the core overhead and framework are just the same as everything else in TensorFlow. Session configurations You can also pass options to tf.Session. For example, TensorFlow automatically determines the best way to assign a GPU or CPU device to an operation, depending on what is available. We can pass an additional option, , when creating a Session, as shown in listing 7. log_device_placements=True This outputs info about which CPU/GPU devices are used in the session for each operation. For example, running listing 6 results in traces of output like the following to show which device was used to run the negation op: Neg: /job:localhost/replica:0/task:0/cpu:0 Sessions are essential in TensorFlow code. You need to call a session to actually “run” the math. Figure 4 maps out how the different components on TensorFlow interact with the machine learning pipeline. A session not only runs a graph operation, but can also take placeholders, variables, and constants as input. We’ve used constants so far, but in later sections we’ll start using variables and placeholders. Here’s a quick overview of these three types of values. : A value that is unassigned, but will be initialized by the session wherever it is run. Placeholder A value that can change, such a parameter of a machine learning model. Variable: A value that does not change, such as hyper-parameters or settings. Constant: Figure 4. The session dictates how the hardware will be used to most efficiently process the graph. When the session starts, it assigns the CPU and GPU devices to each of the nodes. After processing, the session outputs data in a usable format, such as a NumPy array. A session optionally may be fed placeholders, variables, and constants. Hungry for more? That’s it for now, I hope that you have successfully acquainted yourself with some of the basic workings of TensorFlow. If this article has left you ravenous for more delicious TensorFlow tidbits, please go and see this for more information (and a ). download the first chapter of Machine Learning with TensorFlow Slideshare presentation discount code

Getting to Know TensorFlow

MATHEMATICAL MEMORIES: NEWTON'S BINOMIAL THEOREM

Prompted by the quick of the vaccine against COVID-19, blockchains are taking over the world with giant steps. From business to governmental institutions, the of enterprises that finally went in favor of blockchain never ceases to impress. In the view of increasing trust in Bitcoin, the world’s largest decentralized currency, blockchain may catch even more attention, even if not all are convinced by its straightforward benefits just yet. roll-out number authorities Despite the advantages on both industrial and , blockchains still remain a nascent technological trend not fully deprived of its short-offs. Such, for instance, is the fragmented nature of the blockchain market, that does not provide possibilities for a full-blown scale-up. societal levels Another one is a way too rapid pace of change, which is sure to leave behind many of the dominant projects in a few years’ time. This leaves entrepreneurs extremely cautious regarding their choice of blockchains, as their choices rely on the whole future of the project. In order to build something long-lasting, the businessmen need a guarantee that this technological trend will withstand the trial of time and won’t go into oblivion each time as the new, better-equipped blockchain rises on the horizon. An Interview With The CEO and Co-founder of Bifrost In view of all this, integrating the technological foundation that merges all existing blockchain into one. The platform will support multichain DApp development mainly tailored to DeFi applications for the time being. the Bifrost platform I was excited by the opportunity to exchange opinions with Dohyun Pak, CEO and Co-founder of Bifrost. I prepared a set of intriguing questions. We tried to envisage how the year ahead of blockchain will look like, while also delving deeper into the functionality of the technology, which makes many entrepreneurs scratch their heads. Edward Moon: Hi, pleased to host you as a speaker. Could you please share with me your outlook on 2021, and what do you think is in store for blockchain technology? I believe we may be in the due process of critical changes on an industrial scale. Among them, I believe that the multichain technology will prove itself to be paramount in the current year of 2021. Until now, different blockchain protocols have functioned independently and demonstrated that there was no solution in allowing users to use various blockchains simultaneously. Multichain will present opportunities to increase the scalability of the blockchain ecosystem, provide interoperability between a myriad of blockchains and give flexibility to developers and users when using DApps or DeFi. Dohyun Pak: Edward Moon: Lately, blockchain has proved itself as an indispensable part of healthcare during the latest round of vaccine distribution; what other industries could take immediate advantage of blockchain, and how covid-19 could spur their adoption rate? Finance is an industry that can immediately take advantage of blockchain. I see DeFi (Decentralized Finance) as the future of finance because DeFi can solve problems that traditional finance cannot solve. There are 5 features that make DeFi the future of finance. Dohyun Pak: Firstly, DeFi is a service for all. To use DeFi, all people need are smartphones and the Internet. Secondly, DeFi is a service without borders. This means that anyone can use DeFi anywhere in the world. Thirdly, DeFi is a service that has strengthened security because it is based on blockchain technology in which anyone can check the transactions at any time. Fourthly, DeFi is a service without government. Interest rates are determined by the supply and demand of the market. Lastly, DeFi is a service with plug & play. To use DeFi, people do not have to go through massive amounts of documents, but instead, go through simple, easy steps. DeFi has these characteristics because it is built on top of blockchain technology. Before the spread of COVID-19, people went to banks to create accounts or directly meet the tellers to use certain financial services. As COVID-19 spread around the world, most financial services started to conduct through mobile phones, and fintech companies are leading changes in finance. DeFi will improve financial services from fintech using blockchain. For example, not only will it strengthen security, but it also reduces transaction costs because DeFi has no need for middlemen. Edward Moon: How do you feel about the progress made in the DeFi space so far? What would you say is the main obstacle on its way to becoming the global mass movement, and what supportive steps could be taken for it to become one? In my opinion, DeFi is the last stage of the financial revolution. Although DeFi will not replace all the traditional finance, services that require security and lowering transaction fees will be improved by DeFi. Dohyun Pak: I am excited to see the rapid growth of DeFi; however, this is just a start. There are limitations to current DeFi services. The biggest limitation is that DeFi services are restricted under a single blockchain protocol so they cannot use features of different ones. For example, DeFi services under Ethereum cannot interact with other services on different networks. In order for DeFi services to truly become the next generation of finance, it would have to adopt a multichain technology to lower gas fees and allow interactions among different blockchain protocols. Edward Moon: How far can the popularity of the multi-protocol platform go, and would other players be prompted to follow your example? There has been a buzz on projects that connect different blockchain networks such as Polkadot and Cosmos. However, no actual project has proven such a feat of multichain technology. So, we developed BiFi to prove this, and if multichain technology operates with the flow, BiFi and Bifrost will then most likely be mainstream. Dohyun Pak: The demand for a multichain middleware platform will increase as the blockchain and DApp industry grows. Currently, DeFi users are paying high gas fees because most of DeFi services are operating on top of a single blockchain. If gas fees continue to rise, then no one will use DeFi. Restrictions under one blockchain are not just a problem for DeFi, but also other types of DApps. Bifrost is a solution that can solve these issues of DeFi and DAppshave. As a multichain middleware platform, Bifrost will not only connect different blockchain networks but also give choices for developers to choose different blockchain networks for their own DApps. Bifrost will be the paradigm of multichain. I saw that your platform came up with two native tokens – BFC and BiFi. How would you describe the difference between them, and in which way is the value of one interlinked with the other? BFC is the currency of the DApps in the Bifrost Multichain Ecosystem. Developers pay BFC for using the multichain middleware to develop and operate their DApps. BiFi is the governance token of BiFi service. As BiFi is the first DApp to be powered by Bifrost’s multichain technology, BiFi will need to pay BFC to Bifrost as a multichain service fee. Edward Moon: With more than 35 existing partners, how far are you planning to take your blockchain network? Is there any plan of how you want your project to look on the scale of 2-5 years? First, I would like to clarify that Bifrost does not have a blockchain of its own, nor is it an exclusive ecosystem. It is a multichain middleware platform that connects different blockchains. Our current goal is to develop our own DeFi project utilizing the power of Bifrost, and support the project of our partners, to ultimately provide seamless interactions with the greater blockchain ecosystem. Dohyun Pak: In Q4 2021, we plan to release the Bifrost Suite, a development platform tailored for DeFi. Bifrost will attempt to create a decentralized infrastructure that connects the fragmented markets. Using Bifrost, DApps, and DeFi projects will be able to choose their target blockchains that allow for seamless connections among all blockchains.

"I See DeFi as The Future of Finance" — Dohyun Pak, CEO and Co-founder of Bifrost

There are certain things, at certain times, that I start to acutely crave. These cravings are programmed behaviors that always execute in the same mechanical, linear way. I’d like this to not happen. So I spent a bunch of time closely observing how these programs execute. And developing approaches that help rewrite them. I have been able to completely and permanently get myself un-addicted to social media. To arguing with people online. To playing poker. I am close to doing the same for sugar, netflix and videogames. Why hack yourself? There is a lot of (uninformed) talk about how Instagram/Facebook is hacking you. Usually accompanied by outrage at reality. Or denial, despite overwhelming evidence. A more constructive approach is to embrace reality for what it is. And embrace yourself for what you are — a biorobot, following a set of programs. Find ways to observe how the programs execute. Hack yourself in service of your own goals. The way Instagram is trying to hack you in service of theirs. What do I mean by “cravings”? Basically “ ” programs that execute at certain moments in time when something I know is bad for me becomes intensely desirable and hard to resist. I think “addictions” is a very similar concept. What I crave, when: There are 3 things I currently crave: to eat , to watch , and to play . sugar Netflix videogames This only happens in evenings when certain triggers arise and willpower is lower. Why do I want to get rid of these? Sugar is very bad for you. In any amount. For many reasons. This is an undisputed fact in the world today. Just like “cigarettes are bad for you.” Netflix/videogames still take 1–2 hours a day that I would like to spend on more useful types of rest. Time with friends, meditation, reading, sleeping, listening to podcasts, going to the spa etc. If I were able to replace the former with the latter permanently it would be a big win for my life. More generally I enjoy personal growth. I want to reprogram and change myself more easily. It is clear that the ability to reprogram habits is a “muscle.” It can itself be trained. So every time I change a habit there are second-order consequences. Reprogramming other habits becomes easier. That last bit is extremely valuable. What are the symptoms of a of a craving program executing? My cravings are a consistent, precise program. It always executes in the same way: . Usually eating dinner, winding down before sleep, or simply coming home in the evening. I eliminated the more obvious triggers like having bad food around. Situational triggers above cannot be easily removed. Trigger that launches the program . My mind starts constructing scenarios of me going and ordering some dessert. Booting up Steam and playing a game. This seems to be a hallmark of various desires. e.g. if I meditate for a long time, I start to fantasize about standing up and getting red of the pain in my legs. I was discussing this with . It appears that this is a very common way such programs execute. The motor centers of the brain actually switch on. Signal the body to act out the program physically. Fantasizing about physically acting the craving out Paul Conti, my therapist . As in “ ” or “ ” or “ ” It is hilarious how actively my mind searches for rationalizations here. Inventing rationalizations for why I should do it I am in this hotel with a michelin star restaurant, I can have dessert from there, it is a unique dessert that i can’t get elsewhere I just had a 2 hour workout in the gym, Peter Attia says this is when glycogen reserves are depleted and sugar is less harmful because it does not stay in the bloodstream and gets taken up into muscle/liver fast so I should eat it now my genius co-founder plays video games! fucking Elon Musk plays video games, I want to be like that guy so I should play them! If you read such (highly recommended) books as The Elephant in The Brain , The Happiness Hypothesis or The Righteous Mind , you will remember that this is something we do all the fucking time. First decide what we want to be the conclusion. Then look for arguments to support it. We think we are being rational independent thinkers. In fact we are robots executing confirmation-bias programs. . I feel upset at the very idea that after a 2 hour gym workout I don’t let myself have some small session of watching Rick and Morty. Just 20 minutes! (even though of course it never is 20 minutes). A feeling of acute emotional annoyance at arguments against acting on the craving Craving programs are highly dependent on whether I acted on them the day before. As a result, I have multi-day streaks of relapsing or resisting. If for the last several days I have given in to craving programs, they are intensely strong and usually win. Unless there is some strong external factor driving me to resist. If for the last several days I have resisted, the craving programs are weaker or disappear. Until there is some strong external factor driving me to give in and relapse. As a result I tend to have multi-week periods of relapsing intermixed with multi-week periods of resisting. What “strong factors” drive switches from one streak to another? Things that make me relapse and start a new streak of acting on the craving program: — stress, jetlag, flu etc. things that acutely deplete willpower — e.g. i am in an amazing restaurant with a unique dessert and it becomes an excuse to relapse. or on LSD food tastes . A piece of chocolate on acid is a symphony of the best taste imaginable. It is actually a shame to not eat it. Once I do, I relapse. powerful situational cues spectacularly fucking good On an unrelated note, it is a miracle that LSD is not addictive. Things that make me resist and start a new streak of ignoring the craving programs: — e.g. I just spent time with my grandma (who sadly has Alzheimers) and also just spoke with my doc about how sugar appears to be a major contributor to Alzheimers. or I just read this . It compellingly argues that if you decide to eat that sugar today, you will always make the same choice and thus eat it forever. And you will also have lower willpower all your life on all other dimensions. Which i find . powerful situational cues Peter Attia very complicated but wonderful paper on the breakdown of willpower truly fucking unacceptable . Observing the cravings for 5 minutes makes them disappear. although in the moment cravings seem eternal and extremely important. they just vanish. meditation So what is the conclusion of all this? How do we hack cravings? My most successful craving hack — social media I used to be extremely addicted to Facebook. I fixed it by banning it for a long time. Edited the mac hosts file, changed passwords/2FA and gave them to other people, deleted the apps etc. For the first couple weeks it felt almost painful. I would constantly start typing “facebook” into the browser and be annoyed when it would not resolve. I also just wasn’t sure what to do with all this newfound free time I have. After a year of not using it, I felt zero attachment to it. I feel similarly about Instagram etc. I used a similar set of hacks to reduce reactivity to arguing with people online and in comments to my articles. It becomes easier and easier. Same with an addiction to online poker I used to have. More general thoughts on hacking cravings You have to observe yourself. Notice patterns in your own behavior. They might be different from mine. But the patterns are there. Again: you are a hackable, programmable biorobot constructed by evolution. You are just executing programs. Embrace the reality of what you are. Instagram is hacking you to get what they want. So you must hack yourself to get what you want. For patterns similar to mine, the following things work: . My home screen on my phone is designed to minimize bad triggers (messengers, instagram etc.) and maximize good triggers (books, podcasts). Unfortunately it isn’t possible to filter the world for sugar the way I can filter my devices for Instagram. Remove triggers that can be removed It is a very obvious program that you can observe getting executed. Meditate on it for 5 minutes once symptoms of its execution appear. It will go away. Learn to notice the start of a craving. . Delete accounts, delete games, tell hotels to remove sweets from your rooms, ban yourself from online casinos. Put artificial barriers to acting on the cravings in place . Sometimes people say “ ” Truly fucking stupid idea. You are a different person in a jetlagged sick state vs. a healthy well-rested state. If you try to just rationalize from your state now to your future state, you will miserably fail. Do not deny the reality of how your brain executes its programs with fantasies about being an independent free-willed entity. Observe past situational behavior as a guide to future behavior oh so you just have to be extra cautious when you are jetlagged with the flu to not do these things! . Example crazy idea: have some paintings of cancer cells on the kitchen wall along with an article about the (that sugar preferentially feeds cancer cells). Here the main challenge are everyday logistics of life. In my case, I travel a lot and live in hotel rooms, so there isn’t a way to pre-decorate them with weird art. Observe what cues drive resistance. Surround yourself with these cues Warburg effect . As in start that 3 michelin star dinner by saying to the waiter and the table “ ” Then if you give in afterwards, the waiter / your friends see that you are a douchebag weak hypocrite. That usually is a powerful motivator to not give in. Recognize situations that drive relapses. Preemptively act to prevent them Do not bring bread for me. I also do not eat sugar, pls replace my dessert with some cheese/nuts/berries. I guess I should also remove all bad food and just leave some veggies around when I take LSD. That becomes hard because other people object. Again: everyday logistics of life make some of this challenging. So come up with ideas that work in the context of your life. And of course observe and iterate! There is a ton of upside to learning to rewrite your programs. Especially in enhancing the meta-program of rewriting programs. Happy growth and self-hacking, fellow biorobots! Some of my other articles below. If you like them — pls share and follow me, your feedback helps me write more :) _Other deep-dive articles by Serge:_hackernoon.com How to biohack your intelligence — with everything from sex to modafinil to MDMA _Other deep-dive articles by Serge:_hackernoon.com I’m 32 and spent $200k on biohacking. Became calmer, thinner, extroverted, healthier & happier. _Other deep-dive articles by Serge:_hackernoon.com How I set and execute 50-year goals.

Embrace being a programmable biorobot, hack your own cravings.

Imagine yourself as the proud owner of a souvenir shop inside the Spot-On Change Hotel. When closing the cash register, you notice an excess of 8 euros. It seems that a guest didn’t receive the exact change. This could tarnish the hotel’s reputation. To avoid this, you decide to solve the mystery of the incorrect change. Upon opening the shop’s cash system, you discover that the error occurred on two different accounts! You devise a plan: to visit each room and ask the guests what change they received in the shop until you find the two with the wrong bill. Arriving on the first floor, a guest reports receiving 4 euros in change. You calculate that you need to find a number that, when added to the 4 euros, results in 8. Moving to the second floor, the guest’s change is 5 euros. 4 + 5 results in 9, so it can’t be this one… Fifth floor, sixth… Tenth, NO, NO, NO! Since you didn’t find the value on the first attempt, you go down to the second floor and start asking all the guests again about their change so that you can compare it with the value from the next starting floor, exhausting, isn’t it? In computer science, this search method is called brute force, an extremely slow search method that works by comparing one item with the following ones in the array. This consumes a lot of time and is not efficient, imagine going up and down all the floors of the Spot-On Change Hotel several times! This is not practical for you and also not practical for the computer. If you’re curious about how this going up and down the floors would be in a software program, it would look like this in JavaScript: twoSum(nums, target) {
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] + nums[j] === target) {
        return [i, j];
      }
    }
  }
} After assessing the situation calmly, you realize that there’s a more efficient way to find the two guests with the wrong change. You refine your initial mathematical intuition that 9 euros equals x + y. Applying a bit of arithmetic, you realize that: x + y = 9 is the same as saying that y = 9 — x, and that makes all the difference when increasing the efficiency of your search. Another thing that changed in your strategy is that now you will take a notepad to write down the values that the guests tell you, so you don’t have to ask the same value twice for the guests, and you don’t have to spend the day going up and down stairs. (What a terrible time for the elevator to break!). With the new tools in hand, you head to the first floor, where a guest informs you that they have 5 euros in change. You record this as {5: 0}, indicating a value of 5 at position 0. Substituting x with 5 in the equation, you calculate y = 8–5, resulting in y = 3. Since your notepad is still empty, you don’t have the number 3 recorded, which would be the complementary number to 5 to reach the result 8. You then write down the number 5 on your notepad (remember to always write down the number verified at the moment and not its complement; you will only use the complement to compare it with the target number that you have written down). You proceed to the second guest, who mentions receiving 2 euros in change. Applying the formula again: y = 8–2 => y = 6, you check your record, where you don’t find any number 6. Thus, you add the 2 to your record, which now stands as {5: 0, 2: 1}. Continuing the search, the next guest reports receiving 3 euros in change. You calculate again: y = 8–3 => y = 5. Bingo! You have a 5 recorded on your notepad! The guest from floor 0 is complementary to the one from floor 2! This approach is known as a hash table, and it’s much faster and efficient, both for you and the computer. In JavaScript, this would be implemented as follows: twoSum(nums, target) {
  const myTable = {};
  for (let i = 0; i < nums.length; i++) {
    const complementaryNumber = target - nums[i];
    if (complementaryNumber in myTable) {
      return [i, myTable[complementaryNumber]];
    }
    myTable[nums[i]] = i;
  }
} This is an easy problem from Leetcode that only becomes easy after you understand the concepts behind it. I recommend that you try to solve the problem on your own and spend some time trying to build intuition behind it, create your own analogy, review, and try to write pseudocode before moving on to the solution. I hope my analogy has helped you better understand the concepts behind the two sum challenge. See you next time! Imagine yourself as the proud owner of a souvenir shop inside the Spot-On Change Hotel. When closing the cash register, you notice an excess of 8 euros. It seems that a guest didn’t receive the exact change. This could tarnish the hotel’s reputation. To avoid this, you decide to solve the mystery of the incorrect change. Upon opening the shop’s cash system, you discover that the error occurred on two different accounts! You devise a plan: to visit each room and ask the guests what change they received in the shop until you find the two with the wrong bill. Arriving on the first floor, a guest reports receiving 4 euros in change. You calculate that you need to find a number that, when added to the 4 euros, results in 8. Moving to the second floor, the guest’s change is 5 euros. 4 + 5 results in 9, so it can’t be this one… Fifth floor, sixth… Tenth, NO, NO, NO! Since you didn’t find the value on the first attempt, you go down to the second floor and start asking all the guests again about their change so that you can compare it with the value from the next starting floor, exhausting, isn’t it? In computer science, this search method is called brute force, an extremely slow search method that works by comparing one item with the following ones in the array. This consumes a lot of time and is not efficient, imagine going up and down all the floors of the Spot-On Change Hotel several times! This is not practical for you and also not practical for the computer. If you’re curious about how this going up and down the floors would be in a software program, it would look like this in JavaScript: twoSum(nums, target) {
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] + nums[j] === target) {
        return [i, j];
      }
    }
  }
} twoSum(nums, target) {
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] + nums[j] === target) {
        return [i, j];
      }
    }
  }
} After assessing the situation calmly, you realize that there’s a more efficient way to find the two guests with the wrong change. You refine your initial mathematical intuition that 9 euros equals x + y. Applying a bit of arithmetic, you realize that: x + y = 9 is the same as saying that y = 9 — x, and that makes all the difference when increasing the efficiency of your search. Another thing that changed in your strategy is that now you will take a notepad to write down the values that the guests tell you, so you don’t have to ask the same value twice for the guests, and you don’t have to spend the day going up and down stairs. (What a terrible time for the elevator to break!). With the new tools in hand, you head to the first floor, where a guest informs you that they have 5 euros in change. You record this as {5: 0}, indicating a value of 5 at position 0. Substituting x with 5 in the equation, you calculate y = 8–5, resulting in y = 3. Since your notepad is still empty, you don’t have the number 3 recorded, which would be the complementary number to 5 to reach the result 8. You then write down the number 5 on your notepad (remember to always write down the number verified at the moment and not its complement; you will only use the complement to compare it with the target number that you have written down). You proceed to the second guest, who mentions receiving 2 euros in change. Applying the formula again: y = 8–2 => y = 6, you check your record, where you don’t find any number 6. Thus, you add the 2 to your record, which now stands as {5: 0, 2: 1}. Continuing the search, the next guest reports receiving 3 euros in change. You calculate again: y = 8–3 => y = 5. Bingo! You have a 5 recorded on your notepad! The guest from floor 0 is complementary to the one from floor 2! This approach is known as a hash table, and it’s much faster and efficient, both for you and the computer. In JavaScript, this would be implemented as follows: twoSum(nums, target) {
  const myTable = {};
  for (let i = 0; i < nums.length; i++) {
    const complementaryNumber = target - nums[i];
    if (complementaryNumber in myTable) {
      return [i, myTable[complementaryNumber]];
    }
    myTable[nums[i]] = i;
  }
} twoSum(nums, target) {
  const myTable = {};
  for (let i = 0; i < nums.length; i++) {
    const complementaryNumber = target - nums[i];
    if (complementaryNumber in myTable) {
      return [i, myTable[complementaryNumber]];
    }
    myTable[nums[i]] = i;
  }
} This is an easy problem from Leetcode that only becomes easy after you understand the concepts behind it. I recommend that you try to solve the problem on your own and spend some time trying to build intuition behind it, create your own analogy, review, and try to write pseudocode before moving on to the solution. Leetcode I hope my analogy has helped you better understand the concepts behind the two sum challenge. See you next time!

Leetcode: Two-sum an Intuitive Approach

লিটকোড: দুই-সমষ্টি একটি স্বজ্ঞাত পদ্ধতি

Leetcode: Zweisummen, ein intuitiver Ansatz

Leetcode: dos sumas y un enfoque intuitivo

Leetcode : une approche intuitive à deux sommes

लेटकोड: दो-योग एक सहज दृष्टिकोण

Leetcode: 2 つの合計を直感的に計算するアプローチ

Leetcode: 직관적인 접근 방식의 2합

Leetcode: duas somas e uma abordagem intuitiva

Leetcode: интуитивный подход с двумя суммами

Leetcode: İki Toplamlı Sezgisel Bir Yaklaşım

Leetcode: Tổng hợp hai cách tiếp cận trực quan

Leetcode：二和的直观方法

Today, nearly all companies in the IT industry use AB testing to evaluate the value of a feature. They test all changes in applications and websites, allowing someone to determine which feature is the winner every day. As a result, numerous discussions revolve around experiments and AB tests, with a common misconception being the prohibition of applying t-tests for abnormal distributions. Let's delve into this topic and find the correct answer! Problem Formulation Let’s look at a familiar example Imagine a common experiment with different paywalls that suggest a purchase to a customer. We have two variants that show two different and independent groups of users. The decision will be made by a conversion rate to purchase. Every user was assigned to a control or test group randomly, so both samples are independent. Also, every user is assigned 1 or 0 meaning they either made a purchase or not. Thus, our samples have a bi-nominal distribution. In this case, the conversion rate to purchase equals the sample mean. So, we will compare p1n1 and p2n2. Is it possible to use a t-test here or not? Let’s generalize the problem We’re still considering the two independent distributed samples X1 and X2. We know that they have abnormal distributions with finite expected values μ1 and μ2 and finite variances σ1 and σ2. Now, we want to check a hypothesis about the equality of means: Theoretical Proof Let’s consider the Welch’s t-test in a case of different variances. When the variances of both samples are the same, then the Student’s and Welch’s t-tests have similar results. Thus, let’s consider a statistic: Can it be compared with critical values of Student’s distribution? By the central limit theorem, if n1,2 → ∞ Therefore, Also, per Slutsky’s theorem, this convergence in probability is the convergence in distribution. Thus, the constructed statistic t ∼ N(μ1 −μ2,1). Under the null hypothesis t ∼ N(0,1), which is asymptotically equivalent to the Student’s distribution. Empirical Proof Finally, we proved that a t-test can estimate the experiment with two variants of paywall. Also, it is worth mentioning that only two conditions were used for that example: n→∞
and finite of moments for distribution of random value. Let’s try to simulate a pair of distributions to check the prohibition to apply a t-test for them! Bernoulli data Bernoulli data is a widely used case for the experimentation process. A metric like CTR, any type of conversion rate, retention rate, etc., is often chosen as a target metric in an experiment. First, let’s simulate two samples of the Bernoulli distribution with the same parameters. After that, we started increasing the parameter for the second sample. For all samples, we calculated the p-value using Student’s t-test. We see that the p-value drops to zero almost immediately. import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

p_diff_list = []
p_value_list = []

N = 10000
p_1 = 0.01
p_2 = p_1

while p_2 <= 1:
  p_diff_list.append(p_2-p_1)
  x_1 = np.random.binomial(1, p_1, N)
  x_2 = np.random.binomial(1, p_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_list.append(p_value)
  p_2 += 0.01

fig, ax = plt.subplots(figsize=(10,4))

ax.plot(p_diff_list, p_value_list)
ax.set(xlabel='p_2 - p_1', ylabel='p-value',
       title='p-value changes for different Bernoulli distributions')

fig.tight_layout()
plt.show() Now, let’s simulate 100,000 times two samples with the same parameters and 100,000 times two samples with different parameters. After that, we can check the first and second-type errors. Indeed, the t-test is a powerful criterion for metrics with parameters like in our simulations—the first type error is around 5%, and the power is around 90%! import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

n_attempts = 100000
N = 10000

# False Negative calculation
p_value_eq_list = []
p = 0.01

i = 1
while i <= n_attempts:
  x_1 = np.random.binomial(1, p, N)
  x_2 = np.random.binomial(1, p, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_eq_list.append(p_value)
  i += 1

False_Negative = sum([i < 0.05 for i in p_value_eq_list])/len(p_value_eq_list)

# False Positive calculation
p_value_noneq_list = []
p_1 = 0.01
p_2 = 0.015

i = 1
while i <= n_attempts:
  x_1 = np.random.binomial(1, p_1, N)
  x_2 = np.random.binomial(1, p_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_noneq_list.append(p_value)
  i += 1

False_Positive = 1 - sum([i < 0.05 for i in p_value_noneq_list])/len(p_value_noneq_list) Exponential data Now, let’s consider the exponential data prevalent in the real world, too. Two cases of exponential sample comparisons are shown below. In the first case, samples with the same parameters are considered. The p-value corresponds to reality; it says that samples are similar. In the right part, there is a case with two samples with different parameters and a p-value lower than 0.05. import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

# Generate similar samples
N = 10000
beta = 3

xs_1 = np.random.exponential(beta, N)
xs_2 = np.random.exponential(beta, N)

_, p_value_s = stats.ttest_ind(xs_1, xs_2, equal_var=False)

# Generate different samples
N = 10000
beta_1 = 3
beta_2 = 10

xd_1 = np.random.exponential(beta_1, N)
xd_2 = np.random.exponential(beta_2, N)

_, p_value_d = stats.ttest_ind(xd_1, xd_2, equal_var=False)

n_bins = 100
fig, axs = plt.subplots(1, 2, figsize=(12,4))

axs[0].hist([xs_1, xs_2], n_bins, histtype='bar')
axs[0].set_title('Exponential similar samples')
axs[0].annotate(f'p-value = {p_value_s:.2f}', xy=(.6, .8), xycoords=axs[0].transAxes)

axs[1].hist([xd_1, xd_2], n_bins, histtype='bar')
axs[1].set_title('Exponential different samples')
axs[1].annotate(f'p-value = {p_value_d:.2f}', xy=(1.8, .8), xycoords=axs[0].transAxes)

fig.tight_layout()
plt.show() Now, we do the same calculations as with Bernoulli data for evaluating the first and second-type errors. The results are the same—the criteria are very powerful. import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

n_attempts = 100000
N = 10000

# False Negative calculation
p_value_eq_list = []
beta = 1

i = 1
while i <= n_attempts:
  x_1 = np.random.exponential(beta, N)
  x_2 = np.random.exponential(beta, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_eq_list.append(p_value)
  i += 1

False_Negative = sum([i < 0.05 for i in p_value_eq_list])/len(p_value_eq_list)

# False Positive calculation
p_value_noneq_list = []
beta_1 = 1
beta_2 = 1.05

i = 1
while i <= n_attempts:
  x_1 = np.random.exponential(beta_1, N)
  x_2 = np.random.exponential(beta_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_noneq_list.append(p_value)
  i += 1

False_Positive = 1 - sum([i < 0.05 for i in p_value_noneq_list])/len(p_value_noneq_list) Conclusion Now we know the answer to the question about a prohibition on applying a t-test for abnormal distribution—yes, we can apply it to most distributions! Today, nearly all companies in the IT industry use AB testing to evaluate the value of a feature. They test all changes in applications and websites, allowing someone to determine which feature is the winner every day. As a result, numerous discussions revolve around experiments and AB tests, with a common misconception being the prohibition of applying t-tests for abnormal distributions. Let's delve into this topic and find the correct answer! Problem Formulation Let’s look at a familiar example Imagine a common experiment with different paywalls that suggest a purchase to a customer. We have two variants that show two different and independent groups of users. The decision will be made by a conversion rate to purchase. Every user was assigned to a control or test group randomly, so both samples are independent. Also, every user is assigned 1 or 0 meaning they either made a purchase or not. Thus, our samples have a bi-nominal distribution. In this case, the conversion rate to purchase equals the sample mean. So, we will compare p1n1 and p2n2. Is it possible to use a t-test here or not? Is it possible to use a t-test here or not? Let’s generalize the problem We’re still considering the two independent distributed samples X1 and X2. We know that they have abnormal distributions with finite expected values μ1 and μ2 and finite variances σ1 and σ2. Now, we want to check a hypothesis about the equality of means: Theoretical Proof Let’s consider the Welch’s t-test in a case of different variances. When the variances of both samples are the same, then the Student’s and Welch’s t-tests have similar results. Thus, let’s consider a statistic: Can it be compared with critical values of Student’s distribution? By the central limit theorem, if n1,2 → ∞ Therefore, Also, per Slutsky’s theorem, this convergence in probability is the convergence in distribution. Thus, the constructed statistic t ∼ N(μ1 −μ2,1). Under the null hypothesis t ∼ N(0,1), which is asymptotically equivalent to the Student’s distribution. Empirical Proof Finally, we proved that a t-test can estimate the experiment with two variants of paywall. Also, it is worth mentioning that only two conditions were used for that example: n→∞ and finite of moments for distribution of random value. n→∞ and finite of moments for distribution of random value. Let’s try to simulate a pair of distributions to check the prohibition to apply a t-test for them! Bernoulli data Bernoulli data is a widely used case for the experimentation process. A metric like CTR, any type of conversion rate, retention rate, etc., is often chosen as a target metric in an experiment. First, let’s simulate two samples of the Bernoulli distribution with the same parameters. After that, we started increasing the parameter for the second sample. For all samples, we calculated the p-value using Student’s t-test. We see that the p-value drops to zero almost immediately. import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

p_diff_list = []
p_value_list = []

N = 10000
p_1 = 0.01
p_2 = p_1

while p_2 <= 1:
  p_diff_list.append(p_2-p_1)
  x_1 = np.random.binomial(1, p_1, N)
  x_2 = np.random.binomial(1, p_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_list.append(p_value)
  p_2 += 0.01

fig, ax = plt.subplots(figsize=(10,4))

ax.plot(p_diff_list, p_value_list)
ax.set(xlabel='p_2 - p_1', ylabel='p-value',
       title='p-value changes for different Bernoulli distributions')

fig.tight_layout()
plt.show() import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

p_diff_list = []
p_value_list = []

N = 10000
p_1 = 0.01
p_2 = p_1

while p_2 <= 1:
  p_diff_list.append(p_2-p_1)
  x_1 = np.random.binomial(1, p_1, N)
  x_2 = np.random.binomial(1, p_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_list.append(p_value)
  p_2 += 0.01

fig, ax = plt.subplots(figsize=(10,4))

ax.plot(p_diff_list, p_value_list)
ax.set(xlabel='p_2 - p_1', ylabel='p-value',
       title='p-value changes for different Bernoulli distributions')

fig.tight_layout()
plt.show() Now, let’s simulate 100,000 times two samples with the same parameters and 100,000 times two samples with different parameters. After that, we can check the first and second-type errors. Indeed, the t-test is a powerful criterion for metrics with parameters like in our simulations—the first type error is around 5%, and the power is around 90%! import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

n_attempts = 100000
N = 10000

# False Negative calculation
p_value_eq_list = []
p = 0.01

i = 1
while i <= n_attempts:
  x_1 = np.random.binomial(1, p, N)
  x_2 = np.random.binomial(1, p, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_eq_list.append(p_value)
  i += 1

False_Negative = sum([i < 0.05 for i in p_value_eq_list])/len(p_value_eq_list)

# False Positive calculation
p_value_noneq_list = []
p_1 = 0.01
p_2 = 0.015

i = 1
while i <= n_attempts:
  x_1 = np.random.binomial(1, p_1, N)
  x_2 = np.random.binomial(1, p_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_noneq_list.append(p_value)
  i += 1

False_Positive = 1 - sum([i < 0.05 for i in p_value_noneq_list])/len(p_value_noneq_list) import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

n_attempts = 100000
N = 10000

# False Negative calculation
p_value_eq_list = []
p = 0.01

i = 1
while i <= n_attempts:
  x_1 = np.random.binomial(1, p, N)
  x_2 = np.random.binomial(1, p, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_eq_list.append(p_value)
  i += 1

False_Negative = sum([i < 0.05 for i in p_value_eq_list])/len(p_value_eq_list)

# False Positive calculation
p_value_noneq_list = []
p_1 = 0.01
p_2 = 0.015

i = 1
while i <= n_attempts:
  x_1 = np.random.binomial(1, p_1, N)
  x_2 = np.random.binomial(1, p_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_noneq_list.append(p_value)
  i += 1

False_Positive = 1 - sum([i < 0.05 for i in p_value_noneq_list])/len(p_value_noneq_list) Exponential data Now, let’s consider the exponential data prevalent in the real world, too. Two cases of exponential sample comparisons are shown below. In the first case, samples with the same parameters are considered. The p-value corresponds to reality; it says that samples are similar. In the right part, there is a case with two samples with different parameters and a p-value lower than 0.05. import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

# Generate similar samples
N = 10000
beta = 3

xs_1 = np.random.exponential(beta, N)
xs_2 = np.random.exponential(beta, N)

_, p_value_s = stats.ttest_ind(xs_1, xs_2, equal_var=False)

# Generate different samples
N = 10000
beta_1 = 3
beta_2 = 10

xd_1 = np.random.exponential(beta_1, N)
xd_2 = np.random.exponential(beta_2, N)

_, p_value_d = stats.ttest_ind(xd_1, xd_2, equal_var=False)

n_bins = 100
fig, axs = plt.subplots(1, 2, figsize=(12,4))

axs[0].hist([xs_1, xs_2], n_bins, histtype='bar')
axs[0].set_title('Exponential similar samples')
axs[0].annotate(f'p-value = {p_value_s:.2f}', xy=(.6, .8), xycoords=axs[0].transAxes)

axs[1].hist([xd_1, xd_2], n_bins, histtype='bar')
axs[1].set_title('Exponential different samples')
axs[1].annotate(f'p-value = {p_value_d:.2f}', xy=(1.8, .8), xycoords=axs[0].transAxes)

fig.tight_layout()
plt.show() import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

# Generate similar samples
N = 10000
beta = 3

xs_1 = np.random.exponential(beta, N)
xs_2 = np.random.exponential(beta, N)

_, p_value_s = stats.ttest_ind(xs_1, xs_2, equal_var=False)

# Generate different samples
N = 10000
beta_1 = 3
beta_2 = 10

xd_1 = np.random.exponential(beta_1, N)
xd_2 = np.random.exponential(beta_2, N)

_, p_value_d = stats.ttest_ind(xd_1, xd_2, equal_var=False)

n_bins = 100
fig, axs = plt.subplots(1, 2, figsize=(12,4))

axs[0].hist([xs_1, xs_2], n_bins, histtype='bar')
axs[0].set_title('Exponential similar samples')
axs[0].annotate(f'p-value = {p_value_s:.2f}', xy=(.6, .8), xycoords=axs[0].transAxes)

axs[1].hist([xd_1, xd_2], n_bins, histtype='bar')
axs[1].set_title('Exponential different samples')
axs[1].annotate(f'p-value = {p_value_d:.2f}', xy=(1.8, .8), xycoords=axs[0].transAxes)

fig.tight_layout()
plt.show() Now, we do the same calculations as with Bernoulli data for evaluating the first and second-type errors. The results are the same—the criteria are very powerful. import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

n_attempts = 100000
N = 10000

# False Negative calculation
p_value_eq_list = []
beta = 1

i = 1
while i <= n_attempts:
  x_1 = np.random.exponential(beta, N)
  x_2 = np.random.exponential(beta, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_eq_list.append(p_value)
  i += 1

False_Negative = sum([i < 0.05 for i in p_value_eq_list])/len(p_value_eq_list)

# False Positive calculation
p_value_noneq_list = []
beta_1 = 1
beta_2 = 1.05

i = 1
while i <= n_attempts:
  x_1 = np.random.exponential(beta_1, N)
  x_2 = np.random.exponential(beta_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_noneq_list.append(p_value)
  i += 1

False_Positive = 1 - sum([i < 0.05 for i in p_value_noneq_list])/len(p_value_noneq_list) import numpy as np
from scipy import stats

import matplotlib.pyplot as plt

n_attempts = 100000
N = 10000

# False Negative calculation
p_value_eq_list = []
beta = 1

i = 1
while i <= n_attempts:
  x_1 = np.random.exponential(beta, N)
  x_2 = np.random.exponential(beta, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_eq_list.append(p_value)
  i += 1

False_Negative = sum([i < 0.05 for i in p_value_eq_list])/len(p_value_eq_list)

# False Positive calculation
p_value_noneq_list = []
beta_1 = 1
beta_2 = 1.05

i = 1
while i <= n_attempts:
  x_1 = np.random.exponential(beta_1, N)
  x_2 = np.random.exponential(beta_2, N)
  _, p_value = stats.ttest_ind(x_1, x_2, equal_var=False)
  p_value_noneq_list.append(p_value)
  i += 1

False_Positive = 1 - sum([i < 0.05 for i in p_value_noneq_list])/len(p_value_noneq_list) Conclusion Now we know the answer to the question about a prohibition on applying a t-test for abnormal distribution—yes, we can apply it to most distributions!

Using T-tests for Abnormal Data in AB Testing

This paper is available on arxiv under CC 4.0 license.
Authors:
(1) Sam Bowyer, Equal contribution, Department of Mathematics and sam.bowyer@bristol.ac.uk;
(2)  Thomas Heap, Equal contribution, Department of Computer Science University of Bristol and thomas.heap@bristol.ac.uk;
(3) Laurence Aitchison, Department of Computer Science University of Bristol and laurence.aitchison@bristol.ac.uk. Table of Links Abstract & Introduction
Related work
Background
Methods
Experiments
"Conclusion, Limitations, and References"
Derivations
Algorithms
Experimental Datasets And Model Experimental Datasets And Models MovieLens Dataset The MovieLens100K1 (Harper and Konstan 2015) dataset contains 100k ratings of N=1682 films from among M=943 users. The original user ratings run from 0 to 5. Following Geffner and Domke (2022), we binarise the ratings into just likes/dislikes, by taking user-ratings of {0, 1, 2, 3} as a binarised-rating of 0 (dislikes) and user-ratings of {4, 5} as a binarised-rating of 1 (likes). We assume binarised ratings of 0 for films which users have not previously rated. The probabilistic graphical model is given in Fig 3. We use n to index films, and m to index users. Each film has a known feature vector, xn, while each user has a latent weight-vector, zm, of the same length, describing whether or not they like any given feature. There are 18 features, indicating which genre tags the film has (Action, Adventure, Animation, Childrens,...). Each film may have more than one tag. The probability of the user liking a film is given by taking the dot-product of the film’s feature vector with the latent weight-vector, and applying a sigmoid (σ(·)) (final line) Additionally, we have latent vectors for the global mean, µ, and variance, ψ, of the weight vectors. We use a random subset of N = 20 films and M = 450 users for our experiment to ensure high levels of uncertainty. We use equally sized but disjoint subset of users held aside for calculation of the predictive log-likelihood. In Fig. 1c,d we consider the variance and MSE of estimators of zm. Bus Delay Datase In our second experiment, we model the length of delays to New York school bus journeys,2 working with a dataset supplied by the City of New York (DOE 2023). Our goal is to predict the length of a delay, based on the year, y, borough, b, bus company, c and journey type, j. Specifically, our data includes the years 2015 − 2022 inclusive and covers the five New York boroughs (Brooklyn, Manhatten, The Bronx, Queens, Staten Island) as well as some surrounding areas (Nassau County, New Jersey, Connecticut, Rockland County, Westchester). There are 57 bus companies, and 6 different journey types (e.g. pre-K/elementary school route, general education AM/PM route etc.) We take I = 60 delayed buses in each borough and year, and take Y = 3 years and B = 3 boroughs. We then split along the I dimension to get two equally sized train and test sets. Thus, each delay is uniquely identified by the year, y, the borough, b, and the index, i, giving delayybi The delays are recorded as an integer number of minutes and we discard any entries greater than 130 minutes. We have a hierarchical latent variable model describing the impact of each of the features (year, borough, bus company and journey type) on the length of delays. Specifically, the integer delay is modelled with a Negative Binomial distribution, with fixed total count of 130, as we discard entries with delays longer than this. The expected delay length is controlled by a logits latent variable, logitsybi, with one logits for each delayed bus. The logits is a sum of three terms: one for the borough and year jointly, one for the bus company and one for the journey type. Each of these three terms is themselves a latent variable that must be inferred. First, we have a term for the year and borough, YearBoroughWeightyb, which has a hierarchical prior. Specifically, we begin by sampling a global mean and variance, GlobalMean and GlobalVariance. Then for each year, we use GlobalMean and GlobalVariance to sample a mean for each year, YearMeany. Additionally, we sample a variance for each borough, BoroughVarianceb. Then we sample a YearBoroughWeightyb from a Gaussian distribution with a year-dependent mean, YearMeany, and a borough-dependent variance, BoroughVarianceb. Next, the weights for the bus company and journey type are very similar. Specifically, we have one latent weight for each bus company, CompanyWeightc, with c ∈ {1, . . . , 57}, and for each journey type, JourneyTypeWeightj, with j ∈ {1, . . . , 6}. We have a table identifying the bus company, bybi, and journey type, jybi, for each delayed bus journey (remember that a particular delayed bus journey is uniquely identified by the year, y, borough, b and index i). In logitsybi we use these tables to pick out the right company and journey type weight for that particular delayed bus journey, CompanyWeightcybi and JourneyTypeWeight cybi. The final generative model is, and the graphical model is given in Fig. 4 Experiment Details We ran all experiments on an 80GB NVIDIA A100 GPU and averaged the results over three different dataset splits generated using different random seeds. On each dataset split, the MovieLens experiments required a maximum 26.9 GB of memory at any one time. For the NYC Bus Breakdown experiments this value was 0.9 GB. For each experiment using a single dataset  split, we used global IS, MP IS, VI and HMC methods to obtain (for each value of K in the IS methods, or each iteration of VI/HMC) 1000 values of the ELBO, predictive log-likelihood (on a test set that was disjoint from the training set), and posterior mean estimates for each latent variable in the model. These 1000 values were averaged for the ELBO and predictive log-likelihood, whilst the 1000 mean estimates were used to calculate mean squared error or variance per variable, based on whether the observed data had been sampled from the model itself—in which case the true variable values were known—or not. We present error bars in the top rows of Fig. 1 and Fig. 2 representing the standard deviation over the three dataset splits. For VI, we optimize our approximate posterior using Adam with learning rates of 10−2, 10−3 and 10−4 (we found that learning rates faster than 10−2 were unstable), and ran 100 optimization steps for both for the MovieLens and Bus Breakdown models. For HMC, we ran the No U-Turn Sampler on our GPU using JAX (Bradbury et al. 2018) to obtain 10 consecutive samples after a single step with PyMC’s (Salvatier, Wiecki, and Fonnesbeck 2016) pre-tuning and with all other parameters set to their default values. We present results of the first 5 samples (not including the tuning sample) in order to align as best we can with the small time-scale of the IS methods. This paper is available on arxiv under CC 4.0 license. Authors: (1) Sam Bowyer, Equal contribution, Department of Mathematics and sam.bowyer@bristol.ac.uk; (2)  Thomas Heap, Equal contribution, Department of Computer Science University of Bristol and thomas.heap@bristol.ac.uk; (3) Laurence Aitchison, Department of Computer Science University of Bristol and laurence.aitchison@bristol.ac.uk. This paper is available on arxiv under CC 4.0 license. Authors: Authors: (1) Sam Bowyer, Equal contribution, Department of Mathematics and sam.bowyer@bristol.ac.uk; (2)  Thomas Heap, Equal contribution, Department of Computer Science University of Bristol and thomas.heap@bristol.ac.uk; (3) Laurence Aitchison, Department of Computer Science University of Bristol and laurence.aitchison@bristol.ac.uk. Table of Links Abstract & Introduction Related work Background Methods Experiments "Conclusion, Limitations, and References" Derivations Algorithms Experimental Datasets And Model Abstract & Introduction Abstract & Introduction Related work Related work Background Background Methods Methods Experiments Experiments "Conclusion, Limitations, and References" "Conclusion, Limitations, and References" Derivations Derivations Algorithms Algorithms Experimental Datasets And Model Experimental Datasets And Model Experimental Datasets And Models MovieLens Dataset The MovieLens100K1 (Harper and Konstan 2015) dataset contains 100k ratings of N=1682 films from among M=943 users. The original user ratings run from 0 to 5. Following Geffner and Domke (2022), we binarise the ratings into just likes/dislikes, by taking user-ratings of {0, 1, 2, 3} as a binarised-rating of 0 (dislikes) and user-ratings of {4, 5} as a binarised-rating of 1 (likes). We assume binarised ratings of 0 for films which users have not previously rated. The probabilistic graphical model is given in Fig 3. We use n to index films, and m to index users. Each film has a known feature vector, xn, while each user has a latent weight-vector, zm, of the same length, describing whether or not they like any given feature. There are 18 features, indicating which genre tags the film has (Action, Adventure, Animation, Childrens,...). Each film may have more than one tag. The probability of the user liking a film is given by taking the dot-product of the film’s feature vector with the latent weight-vector, and applying a sigmoid (σ(·)) (final line) Additionally, we have latent vectors for the global mean, µ, and variance, ψ, of the weight vectors. We use a random subset of N = 20 films and M = 450 users for our experiment to ensure high levels of uncertainty. We use equally sized but disjoint subset of users held aside for calculation of the predictive log-likelihood. In Fig. 1c,d we consider the variance and MSE of estimators of zm. Bus Delay Datase In our second experiment, we model the length of delays to New York school bus journeys,2 working with a dataset supplied by the City of New York (DOE 2023). Our goal is to predict the length of a delay, based on the year, y, borough, b, bus company, c and journey type, j. Specifically, our data includes the years 2015 − 2022 inclusive and covers the five New York boroughs (Brooklyn, Manhatten, The Bronx, Queens, Staten Island) as well as some surrounding areas (Nassau County, New Jersey, Connecticut, Rockland County, Westchester). There are 57 bus companies, and 6 different journey types (e.g. pre-K/elementary school route, general education AM/PM route etc.) We take I = 60 delayed buses in each borough and year, and take Y = 3 years and B = 3 boroughs. We then split along the I dimension to get two equally sized train and test sets. Thus, each delay is uniquely identified by the year, y, the borough, b, and the index, i, giving delayybi The delays are recorded as an integer number of minutes and we discard any entries greater than 130 minutes. We have a hierarchical latent variable model describing the impact of each of the features (year, borough, bus company and journey type) on the length of delays. Specifically, the integer delay is modelled with a Negative Binomial distribution, with fixed total count of 130, as we discard entries with delays longer than this. The expected delay length is controlled by a logits latent variable, logitsybi, with one logits for each delayed bus. The logits is a sum of three terms: one for the borough and year jointly, one for the bus company and one for the journey type. Each of these three terms is themselves a latent variable that must be inferred. First, we have a term for the year and borough, YearBoroughWeightyb, which has a hierarchical prior. Specifically, we begin by sampling a global mean and variance, GlobalMean and GlobalVariance. Then for each year, we use GlobalMean and GlobalVariance to sample a mean for each year, YearMeany. Additionally, we sample a variance for each borough, BoroughVarianceb. Then we sample a YearBoroughWeightyb from a Gaussian distribution with a year-dependent mean, YearMeany, and a borough-dependent variance, BoroughVarianceb. Next, the weights for the bus company and journey type are very similar. Specifically, we have one latent weight for each bus company, CompanyWeightc, with c ∈ {1, . . . , 57}, and for each journey type, JourneyTypeWeightj, with j ∈ {1, . . . , 6}. We have a table identifying the bus company, bybi, and journey type, jybi, for each delayed bus journey (remember that a particular delayed bus journey is uniquely identified by the year, y, borough, b and index i). In logitsybi we use these tables to pick out the right company and journey type weight for that particular delayed bus journey, CompanyWeightcybi and JourneyTypeWeight cybi. The final generative model is, and the graphical model is given in Fig. 4 Experiment Details We ran all experiments on an 80GB NVIDIA A100 GPU and averaged the results over three different dataset splits generated using different random seeds. On each dataset split, the MovieLens experiments required a maximum 26.9 GB of memory at any one time. For the NYC Bus Breakdown experiments this value was 0.9 GB. For each experiment using a single dataset  split, we used global IS, MP IS, VI and HMC methods to obtain (for each value of K in the IS methods, or each iteration of VI/HMC) 1000 values of the ELBO, predictive log-likelihood (on a test set that was disjoint from the training set), and posterior mean estimates for each latent variable in the model. These 1000 values were averaged for the ELBO and predictive log-likelihood, whilst the 1000 mean estimates were used to calculate mean squared error or variance per variable, based on whether the observed data had been sampled from the model itself—in which case the true variable values were known—or not. We present error bars in the top rows of Fig. 1 and Fig. 2 representing the standard deviation over the three dataset splits. For VI, we optimize our approximate posterior using Adam with learning rates of 10−2, 10−3 and 10−4 (we found that learning rates faster than 10−2 were unstable), and ran 100 optimization steps for both for the MovieLens and Bus Breakdown models. For HMC, we ran the No U-Turn Sampler on our GPU using JAX (Bradbury et al. 2018) to obtain 10 consecutive samples after a single step with PyMC’s (Salvatier, Wiecki, and Fonnesbeck 2016) pre-tuning and with all other parameters set to their default values. We present results of the first 5 samples (not including the tuning sample) in order to align as best we can with the small time-scale of the IS methods.

Using Autodiff to Estimate Posterior Moments, Marginals and Samples: Experimental Datasets and Model

This paper is available on arxiv under CC 4.0 license. Authors: (1) Jeffrey Ouyang-Zhang, UT Austin (2) Daniel J. Diaz, UT Austin (3) Adam R. Klivans, UT Austin (4) Philipp Krähenbühl, UT Austin Table of Links Abstract & Introduction Related Works Preliminary Methodology Results Limitations Conclusion, Acknowledgements, & References Experimental Setting & Additional Results 7 Conclusion We present a method that efficiently scales thermodynamic stability prediction from single mutations to higher-order mutations. Our key insight is that the effects of mutations on the same protein are correlated. Thus, for a target protein, it suffices to run a deep backbone once and decode the effect of all mutations simultaneously using a shallow decoder. With the AlphaFold model as our backbone, our method outperforms existing methods on a variety of single and multiple mutation benchmarks. 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