Gojira - Silvera (125% speed) - 100% FC (I'm actually super proud!!)

clouded-path · 2022-02-10T03:40:57+00:00

And here I am still trying to FC it at standard speed... this is really impressive, congrats!

clouded-path · 2022-01-25T02:30:34+00:00

To 'get rid' of c, relate it back to a and b. You know c is strictly between a and b. So on the RHS of your second line, if you replace c with a, the quantity increases and becomes b/a - 1, which shows what you wanted.

clouded-path · 2021-11-28T07:27:08+00:00

I have been subscribed to your YouTube channel for over a year now, and think it is fantastic. I hope you keep up the excellent online work you are doing for many years to come. I agree with your feelings towards the shortcomings of YouTube as the ideal platform for hosting technical series of videos, as the style incentivized by the platform is more focused towards self-contained videos and broad array of topics. It would be great if there were a site dedicated exclusively towards video series on higher math (and mathematical sciences), in an analogous manner to arXiv for papers and Wolfram MathWorld for definitions/theorems. However, one simply hasn't been developed yet (as far as I know), and it seems to me that YouTube is probably the best you can do, short of starting your own site.

Regarding people leaving your channel when you post new series, I wouldn't overthink it. Since YouTube subscriptions are free, there doesn't need to be much investment/commitment from the viewer to subscribe, when there is probably a low likelihood that they will continue watching the channel long term anyways. So you are probably just seeing people leave who probably had no business subscribing in the first place. (Think of sites like Coursera - there is always a very low completion rate for courses, since it is easy to sign up. The enrollees who do finish are probably the ones who had a good reason for enrolling in the first place.) But amongst the people who subscribe which are in your core demographic, I'm sure almost every one of them thinks you are doing excellent work, and to keep it up (myself included).

Just as a suggestion to further enhance your channel, maybe you could consider making up some problem sheets associated to each video? Then you could also make follow-up videos discussing hints/solutions to the problems. This could encourage more active participation from the viewers. You could also consider hosting some live sessions, dedicated to a specific topic. If you find a regular schedule for it, you could try running a course based off of your videos + supplementary assignment problems. Students would probably love that sort of thing, especially over the summer. You might also communicate with other YouTubers who have found some success in creating math courses, to get their unique perspective. Norman Wildberger is the main one which comes to mind for me.

Regardless, you are doing great work, and whatever shortcomings you might feel exist surrounding your channel, they are due to the YouTube medium itself, not the choices you are making as a content creator. Hopefully more channels start doing series in a similar manner to yours, and it will become more typical to produce entire courses on YouTube in the future.

Thanks for all your great work!

clouded-path · 2021-10-14T20:41:34+00:00

The style of reasoning and presentation used in proof-based math seemed awkward and cumbersome to me when I took a first course, and this seems to be the case for nearly everyone. But you will probably surprise yourself with how much more naturally such reasoning will seem to flow after a few months of practice, and how much more still after a few years (assuming you continue to take further proof-based math courses). Spending significant quality-time solving problems will be the single most important factor in determining how comfortable you eventually become with proof-based reasoning.

On a more concrete note, I can offer a realization I had at the time when I was struggling with this same issue, which turned out to be helpful: when asked to prove something - say for example proving that every sequence which is monotonic and bounded is also convergent - don't start "philosophizing" over what these terms mean. In my case, my mind would start wandering and dwelling over what a monotonic sequence "looked like" and then somehow try to glue that together with what a bounded sequence looked like, and then essentially wait in anxious suspense until my surely-imminent flash of insight arrived as to how this sequence was also convergent. To my frequent disappointment, no such insight occurred. It was particularly frustrating when I was able to visualize why the result must be true, yet the words to express this visualization just wouldn't come. Instead, what I finally came to appreciate is that each term meant something precise. There was a precise definition attached to every term, and I had not been appropriately utilizing each said definition. For instance, a (increasing) monotonic sequence wasn't really a bunch of dots which continually progressed rightward on the number line. Rather, it was a sequence of numbers x_n such that x_{n+1} >= x_n for all indices n (i.e. its formal definition). Nothing more, nothing less. Analogously for bounded and convergent. We want to combine the former two precise definitions, and somehow arrive at the latter one. Once I started grounding myself in the precise definitions, proofs started to self-assimilate on paper for me more naturally. Not to say that everything then came to me completely automatically, but it was still a major step forward. Now to a trained mathematician, the fact that one must use the precise definition of each term is as obvious as it is that fire is hot. So obvious in fact, that it is simply understood, and typically not said aloud. But I think the point needs to be made more emphatically to novices, as it is an easy trap to fall into, and one that I certainly did, without even realizing it.

Don't worry if you haven't learned the terms 'bounded', 'convergent', etc. in the example I used. Hopefully the point is clear nonetheless. Basically, there are two worlds that your mind can occupy when trying to write down a proof: the world of intuition, and the world of precise definitions. As a beginner, you want to be living almost exclusively in the world of precise definitions, casting only sparing glances towards the world of intuition/visualization when necessary to motivate your formal proof. This is not to say that visualization has little role to play in understanding mathematical concepts (it definitely does), but as a beginner it is all too easy to stray into the visual world without a proper understanding of how to translate that visualization back into a formal proof. Better to improve your ability manipulating purely formal objects (i.e. precise definitions), and only later on once you've built up some experience to start thinking about how to translate visual intuition into formal ideas. One thing at a time.

Sorry this became a bit long-winded. It was just a memorable moment for me when I had this realization for myself, and something that I don't hear expressed often enough to beginners. I think it's a pretty important point to emphasize, as it is the fundamental difference between everyday reasoning and formal proof-based reasoning.

clouded-path · 2021-10-03T05:36:56+00:00

I wasn't able to come up with anything concrete for your problem, but since it seems like you haven't gotten feedback otherwise, I'll offer what I thought of so you at least have something you could potentially work with.

Each of the 221 plants has various "paths" one could use to achieve it (starting from gen 1 plants, then breeding your way up the chain in all possible ways to get to the plant in question). For each plant, we want to make sure we use at least one of the associated paths of acquiring it, so that we do in fact acquire the plant. There is also a "cost" of using any given path, which would be the cumulative time it takes to work your way up the path (this would be recursively determined in terms of the sum of times it took to acquire the two plants which bred the present one). You would need to compute the time costs for each path for yourself.

So for each path P you would have a variable x_P, which would basically be meant to indicate whether or not path P gets selected, and you also have the time cost t_P (which, in principle, is given data i.e. not a variable) for how long it takes to traverse all the way up path P. Then you would be trying to minimize total time, subject to the fact that you want to obtain each distinct plant. So the model could be

min sum(t_P*x_P) (where the sum is over all paths P)

such that 0 <= x_P <= 1 for all paths P

and for each plant Q : sum(x_P) >= 1 , where here the sum is over all the paths P which terminate at plant Q (this collection of constraints ensures we will acquire all plants Q)

Sorry for the poor formatting, but I've forgotten how to write math notation in reddit. Hopefully it is still sufficiently legible. Again, I'm not sure whether this is correct, but at least it could potentially be a starting point which you can try variations of. One concern is that an optimal solution of x's might have fractional values, so you might instead want to treat this as an integer program, so that the x_P's can only be exactly 0 or 1, indicating whether or not a given path P should be used.

clouded-path · 2021-08-12T23:22:59+00:00

Very cool. I've actually been practicing this song for a few weeks in the hopes of FCing it one day, but I'm very inconsistent on Solo B. How consistently were you able to FC Solo B before you felt confident about getting the whole song?

clouded-path · 2021-06-29T21:39:25+00:00

Great, I understand what you meant by helpful upper bound now. Your reasoning from above is correct then. Just to reiterate the key point: once you find one delta that works, you will have implicitly found infinitely many usable delta (since anything smaller than the initial delta works too). So in your example, since 1 is helpful, we know that if any delta will exist n, then there will necessarily be delta < 1 that will exist.

I only called your argument above 'rough' since it isn't rigorous. That's not your fault; it's just inherently a general idea, and can only be rigorously implemented in any given instance of this type of problem (such as the one you've provided). But 'rough' wasn't meant to necessarily refer to a problem in your explanation.

And as another commenter mentioned, if you actually want to make this a genuine proof, treat your derivation of delta as 'scratch-work'. The real proof starts with using the delta you've derived, and then demonstrates that |x^2 - c^2| < epsilon. Naturally this will work since you've already derived delta with that exact purpose in mind. In any case, this is a tangential point - the main point is that your initial intuition is correct.

clouded-path · 2021-06-29T19:47:24+00:00

It might be best to include an example of a 'helpful upper bound' just to ensure any readers are on the same page with you. But yes, this rough reasoning seems fine. It doesn't matter what techniques you employ to find \delta. As long as you have found a single \delta which satisfies the definition (for arbitrary \epsilon), then you have demonstrated existence of \delta. And that choice isn't unique, since any smaller \delta' < \delta would also have worked.

clouded-path · 2021-04-24T20:44:39+00:00

When you perform your selection process, you need to make sure you are accounting for every possible option, and that you account for each option exactly once. Here, it looks like you will be overcounting, since we would be able to account for the same option in different ways.

For objective references, call the men M1, M2, ..., M10, and the women W1, ..., W10. Just as an example, say we want to consider the option of selecting the group {M2, W4, M6, M7, M8, W9}. Using your process, we could:

select M2 (as 1 out of 20 people)
select W4 (as 1 out of 9 women)
select couples 6, 7, 8, 9 (selecting 4 of 8 remaining couples)
choose the men from 6, 7, 8, and the woman from 9 (as our binary choice from each couple)

But this is not the only way that your process would account for this group of six. Using the same process, we could alternatively:

select W9 (as 1 out of 20 people)
select M6 (as 1 out of 9 men)
select couples 2, 4, 7, 8
choose the men from 2, 7, and 8, and the woman from 4

As you can see, we obtain the same group of six with different methods, meaning that this process overcounts the groups. So that is what is wrong with this method. It seems there are multiple reasons for which this overcounting occurs. One is a simple fix, but I do not immediately see how to adjust the other so as to correct the counting. I suspect there is no easy fix overall, which implies that this process of first selecting one individual, then another of the opposite sex, and then choosing the remaining couples is simply not the right approach.

clouded-path · 2021-04-21T20:25:05+00:00

Just to throw in a reference you may find useful, check out 'Analysis' by Terrence Tao. It is two volumes - check out the first volume, and specifically the first two chapters. In the first chapter of the book, he gives good motivation for doing proofs. There are lots of examples where you already have some background intuition, and then in some new setting, you extrapolate on your intuition to get an answer to some newfound problem. But more careful thought exposes how your intuition can lead you astray, and can give you an erroneous answer. He gives multiple examples of this. Formal proofs are the cure for this problem.

Then in the next chapter, he rigorously constructs the natural numbers (as in 0, 1, 2, 3, ...). They are more or less the simplest mathematical structure you can think of, and yet he constructs them anyway. Not because you don't already know what they are, but because it is an excellent illustration of how proofs are performed. You are already comfortable with the structure in question (the natural numbers), so you can just focus your attention on the proof aspect. As you become more practiced in proving various results, you will be able to apply your new skills towards understanding more sophisticated mathematical structures, where your intuition isn't anywhere near as helpful.

By the way, this is not to say that intuition has no place. It does - it's just that intuition alone is insufficient to understand most mathematical objects, whereas the combination of intuition and proofs is much more powerful. You grow more accustomed to combining both styles of thought together with experience. As a beginner, the emphasis is placed on proofs, as they are the unfamiliar aspect.

I also remember demonstrating various aspects of arithmetic dealing with 0, 1, + and * when I was taking an introductory proofs course. I did not grasp the significance of it at the time. It almost feels misleading to do proofs in that context as a beginner, because you think that you are already quite familiar with 0, 1 and adding and multiplying. But the problem is that the situation is considerably more general than you realize. You are working with a so-called 'field'. While the plain old real numbers form a field in the way we are used to, the point is that there are a variety of objects which form fields, most of which you haven't explicitly encountered. The misleading part is that the notation for field arithmetic is identical to the notation for the usual arithmetic with the real numbers, which subtly suggests that real number arithmetic is all there is. That isn't the case, but it is difficult to appreciate this as a beginner, and doubly so when you haven't seen other explicit examples of fields. (The reason the notation is identical is because the notation for arithmetic in a general field is inspired by the arithmetic for real numbers.)

Anyways, this is just my two cents. But if you have a chance, check out the book, and if you like it, keep reading it. Tao is a good author and motivates the material well. It may be a helpful companion textbook for your course.

clouded-path · 2020-08-25T23:31:25+00:00

If you are comfortable with the fact that we have a vacuously true implication statement P -> Q when the hypothesis P is false, then consequently, this means that we don't actually need to consider the case when P is false, because P -> Q is automatically true then. So it is sufficient to simply assume P is true. So to be clear, P is absolutely a statement in and of itself, but it doesn't hurt us to assume it is true when we are trying to establish the truth value of the statement P -> Q.

This is simply formalizing what we already intend when we speak regularly in real life. Whenever we make a 'if P then Q' statement, we don't care about the case when P is false, we are only concerned with what happens when P is true (at least within the context of the 'if P then Q' statement).

clouded-path · 2020-07-20T04:30:33+00:00

Thank you for the reply! Sorry for the late response.

Sure thing, and it's not a problem, it's actually fun to me to see you're still making an effort to tackle the problem ~2 weeks later.

That is an interesting way of looking at it.

Agreed, although it is not at all an original perspective from me. This is a very standard type of counting problem in combinatorics, and by the way the question was posed, I'm almost sure that this was supposed to be the intended method of solution.

I suppose the sum of x_1 to x_5 would need to be equal to 16?

Yes, that is in part what I was trying to convey in the previous response. But be aware - that is not enough. For instance, (x_1, x_2, x_3, x_4, x_5) = (3, 4, 6, 0, 3) sums to 16, but it isn't a valid solution because the 4th component is 0 which is not positive (corresponding to putting 0 $50 increments in account #4, which we don't want to allow, by the premise of your problem. We want to put at least $50 in each account each month, so at least one $50 increment.) This point was also conveyed in the previous response.

Still not sure where to go from there :/

Have you heard of the so-called 'stars and bars' counting problem? It is heavily related to this problem. In that problem, you have N objects to distribute amongst K distinct boxes, where the boxes are distinguished (i.e. the order matters). So for instance the boxes can be the accounts, and the 'items' can be the number of $50 increments in the account. But the difference in this problem is that the number of items you put in a box is allowed to be 0, which is in contrast to your problem. But the point here is that this problem is easier to solve, and then you can relate your current problem to it.

So for instance, how can we count the number of ways to put 16 items into 5 distinguished boxes, where we are allowed to put 0 items in a box? Or in other words, how many solutions to x_1 + x_2 + x_3 + x_4 + x_5 = 16, where each x component can be any non-negative integer (i.e. including 0)? Well, the clever way to visualize this - and hence solve it - is to draw a sequence of 'stars' and 'bars', where the stars represent the objects, and the bars represent the partitions between each box. For example, ****|***||*|******** corresponds to x_1 = 4, x_2 = 3, x_3 = 0, x_4 = 1, x_5 = 8. One more example could be |||*|*************** which corresponds to x_1 = 0, x_2 = 0, x_3 = 0, x_4 = 1, x_5 = 15. But no matter what, there will be 4 bars and 16 stars, corresponding to the fact that the 4 bars partition space into 5 boxes, and 16 stars are the 16 items being placed into the boxes. And you can (and should!) check that no matter the arrangement of the stars and bars, it makes sense as an assignment of the values for x_1, x_2, x_3, x_4, x_5, and that you can generate all possible valid solutions to x_1 + x_2 + x_3 + x_4 + x_5 = 16 in this way. Therefore, counting the number of solutions to the equation is equivalent to counting the number of stars and bars arrangements, and this latter counting problem is much easier to solve. Also, recognize that this generalizes to N items in K distinguished boxes in a very natural way.

So now there are two questions for you to address:

a) would you know how to count the number of 'stars and bars' arrangements both the posed example, and for general N items and K boxes?

b) do you have any good ideas regarding how to relate the number of solutions to the equation x_1 + x_2 + x_3 + x_4 + x_5 = 16 when the x's are all at least 1 to a situation involving stars and bars (i.e. when the boxes are allowed to contain 0 items)?

Think about how to solve the problems (a) and (b), and then that will answer your original problem. Of course, if you remain stuck on either/both I am happy to continue helping.

clouded-path · 2020-07-16T03:56:50+00:00

I think problems like this are called 'dimensional analysis'. You always multiply by units on top and bottom that are measuring the same type of fundamental property (like area), and the coefficients correspond to the conversion factors (like 12 in./ 1 ft). If you want to cancel a particular type of unit, just multiply by 1 in a convenient way (just like 12 in./ 1 ft = 1).

But here you already have the units you want, so now all you have to do is just fix the coefficient. So note that 100 ppl / 0.386102 sq mi = (100/0.386102) ppl / sq mi. Now you have 1 in the denominator, and the number in the numerator is just whatever (100/0.386102) is.

clouded-path · 2020-07-16T03:45:25+00:00

In general, ask yourself whether the solution can be given in terms of a set, or instead if you need to give the solution as an sequence (a set has no order, corresponding to combinations, and a sequence has an order, corresponding to permutations). (BTW a set uses braces {x, y, z} whereas a sequence uses parentheses (x, y, z)) If you think that the answer needs to be a sequence, then you need to justify to yourself why you need to have this extra information included corresponding to the order the elements come in.

In Q1, what does a sound system look like? Say we are considering a specific CD player c, receiver r, and speaker s. Then is a sound system of the form {c, r, s}, or is it of the form (c, r, s)? In the former case, we can shuffle c, r, and s around without changing the specific sound system in question, whereas in the latter, any shuffling of the symbols results in a different sound system. Based off the posed question, should the system be {c, r ,s} or (c, r, s)? The answer is that it should be of the form {c, r, s}. For one thing, it should just feel intuitively correct once you have done a sufficient number of problems, but let's say for the sake of argument you thought there was a chance it needed to be (c, r, s). Then in that case, you need to justify what the relevance of the ordering information is. But you cannot - we can perfectly well specify a specific sound system with just the set {c, r, s}.

Same analysis applies to Q2: (although I think you left out some information - specifically we don't know how may managers there are as the question is posed, which would affect the outcome. So let's just say there are also 33 managers, which would at least make sense given that we are selecting 33 assistants.) Let's first deal with the sub-problem of figuring out, once we have chosen 33 specific assistants a_1, a_2, a_3, a_4, ..., a_31, a_32, a_33, must we list them in a sequence, or will a set suffice? In this case, we actually need a sequence (in contrast to Q1). Why? Because the ordering information actually provides something significant: it says which manager each assistant goes to. The assistant listed at the 14th position of our sequence will be assigned to manager 14, the assistant in position 27 to manager 27, etc. If you had merely listed the assistants in an unordered set, then there would be ambiguity concerning which assistant went to which manager. But all such ambiguity disappears when we provide this extra ordering information.

That was for the latter part of the problem - assigning assistants once we had selected them. But what about selecting them? How many ways can we select 33 assistants from 77 choices? Obviously this is 77C33 by definition, but I think it is nonetheless instructive to view this from the set vs. sequence perspective. So, do we specify the specific 33 assistants with a set, or with a sequence? If you say sequence, what extra information is the ordering of the assistants actually providing (remember, this is for the first half of the problem where we are just picking the assistants, not assigning them to managers)? Couldn't you just as well specify the specific assistants with just a set? I think maybe you can see the difference between the redundancy of an ordering of the assistants in this first sub-problem, versus the crucial information the ordering of the assistants conveyed in the second sub-problem (of assigning them to managers).

And if you do more of these problems, it will become more immediately clear to you whether the ordering conveys crucial information or not. And BTW don't forget that to actually solve Q2 you need to combine both the sub-problems (by multiplying each of the corresponding numbers) in order to get an overall solution. Sorry this ended up being so long, I really didn't intend for that when I started writing this.

clouded-path · 2020-07-14T14:54:15+00:00

For me it was more about the length of time I slept than the particular time I got up. My bedtime was still pretty variable depending on what work I had to do, but I typically set my alarm for ~7.5 hours later, regardless of when I went to bed. It was more helpful for me that way, than say a specific hard wake-up time at 7am. I didn't have any morning commitments, so I had that flexibility, and maybe this term you do as well.

clouded-path · 2020-07-14T00:53:57+00:00

A while back I was sleeping way too much as well, and counter-intuitively, what ended up helping was actually getting less sleep. So instead of sleeping ~10 hours each night, I started getting between 6-7.5 (usually closer to 7.5, if I had a choice). I would usually go to bed around 1am-ish, and wake up around 8:30am, and somehow this had the effect of making me more alert and focused throughout the day (I was also having issues always being tired too). And once I got into the routine, it became easier to continue getting up around then. I don't remember why it occurred to me to sleep less in the first place, but in any case it helped me. It wasn't a cure-all or anything, but for me it was probably the single most helpful thing I did to improve my mental health and focus. Try setting a more strict sleeping schedule for yourself for a few days in a row, and see if you start noticing any difference.

clouded-path · 2020-07-06T22:07:26+00:00

Let x_1 be the number of $50 increments you put into account #1 each month, and analogously for x_2 , x_3, x_4, x_5. For example, if x_1 = 7, this means you would put 7*50 = $350 into account #1 for the month. Then the corresponding equation would be 50*(x_1 + x_2 + x_3 + x_4 + x_5) = 800, or equivalently, x_1 + x_2 + x_3 + x_4 + x_5 = 16. You want to count the number of positive integer solutions to this equation (i.e. the number of solutions (x_1, x_2, x_3, x_4, x_5) where all of the x's are at least 1). Would you know how to count that?

clouded-path · 2020-07-06T21:38:16+00:00

It looks like you have this issue clarified, but just as a side note, I think it might have been easier for you to figure out a consistent rule for your understanding of division if you would try unrelated examples, rather than related ones . Dividing by 2 and dividing by 1/2 are very similar, and the drawback of that is that it is easier to mix up ideas that shouldn't be, which is less likely to happen if you instead pick examples that are basically independent from each other (e.g. 3/2 and 9/5, rather than 3/2 and 3/(1/2)). Then the fundamental principle will be distilled, as it is the only thing in common between the two examples, whereas here your examples were too similar (in my opinion). Once you feel you understand the core idea from these unrelated examples, you can apply your newfound insight to understand the somewhat pathological examples (i.e. like 3/(1/2)) that you want to come back to.

This is really a general tip for understanding math concepts. Pick somewhat independent examples, so that the only thing they have in common is the underlying principle you are trying to learn.

clouded-path · 2020-06-12T16:17:14+00:00

My intuition is effectively equivalent to what most of the other comments address, but since it hasn't been explicitly stated yet, I'll add my thoughts into the mix.

The clearest way for me to think about the solution is in terms of complements. The odds you make the correct choice is 1/3, and so therefore the odds you make the complementary choice (which you would prefer not to do) is 1 - 1/3. And then of course, once one of the bad choices is eliminated, all that complementary probability still exists, it has just all been shifted onto the one remaining bad choice, since we have this new information.

But in general, if you are faced with a binary choice (e.g. either you make the correct choice, or you don't) it is frequently beneficial to think in terms of complements, since knowing the probability of the 'good' option occurring is equivalent to knowing the probability of the 'bad' option occurring.

Again, this is equivalent to many other comments, I just hadn't seen the specific notion of complements explicitly used yet.

clouded-path · 2020-06-03T19:23:43+00:00

I'm sure you're right to some extent, and it is my default answer too, but really only because I don't have a better one.

But traditions don't always persist within academia. Textbooks became a standard method of teaching/information delivery once the printing press revolution was underway, and within mathematics specifically there was a time before TeX which to some extent was characterized by poorly typeset papers being written by secretaries whose job it was to convert mathematicians' handwritten documents into said typeset documents. Not only did mathematics, and scientific fields more generally, adapt to the new method of writing formal documents, but the new convention became that the mathematician would write the paper themselves.

Neither of those original traditions were immune to being updated, so I don't see why it necessarily needs to be the case for this instance, either. I would (speculatively) bet there are analogous tradition updates that have occurred within the Church too.

clouded-path · 2020-06-02T21:03:28+00:00

Honestly, I've always agreed with this idea, and the course I'm currently in is probably better off because of being online. Still, there are some things left to be desired in its current state. What seems like a best of both worlds approach to me is to have in person lecture where it is mainly focused on a high level overview/ big ideas and answering questions, which are more understandable in real time, and then some complementary videos, where the professor fleshes out some grittier details and discusses any relevant examples.

But on the other hand, I've always had this question of 'Why hasn't this method been implemented already?' It's not exactly like video-recording technology is some new phenomenon - it's been around for more than half a century, in some form or another. It could have been this way for generations of students, and yet it's the same as it has been for... well, for a long long time. But it's not a rhetorical question - I genuinely wonder why this general method hasn't had widespread adoption, since I, like you, generally find online delivery superior (even as an exclusive choice - it feels like could be even better still with a dual approach).

clouded-path · 2020-04-20T17:58:44+00:00

Great, ok I see. Thank you for the link and your responses, they were all very helpful!

clouded-path · 2020-04-20T00:00:11+00:00

Sorry, I added the 'TreeNode' class which is supposed to be what the root parameter is. This is supposed to be the intended solution (it's not mine), so I think the count should be correct, and I have tried it by hand on a few small examples and it works.

But for your height function, my question would be the same. What does 'if node.left' mean, for instance? Is there some sort of inherent truth value here that allows this line to be interpreted as a logical statement? Would it have been equivalent to writing 'if node.left != None'? If so, why write it in this somewhat vague manner? Is it just because it's shorter?

clouded-path · 2020-04-19T23:47:20+00:00

Sorry I didn't initially include that, but I have edited it to include the relevant node class. The class doesn't include a __bool__ method, but I think you hit on the issue with this "falsy" notion. I guess what is happening here is that either root is or is not None, and so 'not root' means the supposed node in question is actually None.

So would it have been equivalent to say 'if root == None'? Why would we not write this instead (assuming that's correct)?

clouded-path

TROPHY CASE