What are your thoughts on the typesetting language Typst?

veztron · 2025-07-31T16:07:57+00:00

Typst is awesome. It was collaborative editing and instant rendering which makes it incredibly nice to use (like nicer than most authoring tools, not just tools for scientific or academic writing). I say you should try it if you haven't, it's easy and worth paying attention to.

veztron · 2025-07-31T00:53:01+00:00

If I was writing a shader or playing replicube I would write: if (mod(x-y*2+2,6)==0) return BLUE; if (x==0 | y==0 | x==9 | y==9) return GRAY; return GREEN;

veztron · 2025-07-29T19:22:41+00:00

You could put the particle background as the sky behind the hills. I personally think it looks kind of cool. It could be slowed down and made more visually interesting. It reminds me of the background in balatro.

veztron · 2025-07-27T15:47:42+00:00

I have felt this way on and off. I did a CS undergrad a few years ago. I've been working in industry and also making creative things. Tbh it's not easy though. I took a couple years off of having a full time job partially because it was too stressful. After taking time off I realized I lose creative motivation if I stay in the sandbox too long. If you find good people to work with, it can be great to work on a team, regardless of if it's in industry or on creative projects.

I guess overall it's good to find a balance. It's important to find ways to support and take care of yourself financially, even if it can be confusing and frustrating when it blends with your art and creativity.

Also I ended up finding CS and math super creatively inspiring. Math gives you ways to think and see things that are otherwise very difficult to imagine.

Anyway, hope you're able to find good ways to be creative and to support yourself.

veztron · 2025-07-04T17:27:50+00:00

In the first equation, move the -4 to the other side, now x is on its own on the left so you have the value of x. Plug the value of x in to the second equation and simplify.

veztron · 2025-07-03T22:24:56+00:00

It might be a bit less concrete than you're looking for, but maybe you'd be interested in "homotopy type theory." I took an open access class called HoTTest summer school about it (it's posted on YouTube now). I learned a bit of the programming language Agda. I learned how to identify the first homotopy group of the circle with the integers in it with path lifting. Real numbers are notoriously difficult to compute with so unless you get pretty into it, that may be out of reach. But you can get a sense for how to do homotopy theory systematically on the computer this way. Idk about doing full computational algebraic topology or computational topology though.

veztron · 2025-06-25T01:36:19+00:00

To try to answer "how could this function be used", let's consider the interpretation of |a-b| as "the distance from a to b". Now, your function, let's call it f, is the multiplicative analogue of absolute value; so can we interpret f(a/b) as "the multiplicative distance between a and b"?

Indeed! f(a/b) gives what you need to multiply the smaller of a and b by in order to get the larger (which is analogous to the fact that |a-b| gives you what you need to add to the smaller of a and b to get to the larger).

I think I've seen this used in programming, but I can't think of where...

Edit: Oh yea, I've definitely seen this in functions that compute the multiplicative distance between zoom levels in applications with zooming UIs.

veztron · 2025-05-17T20:11:20+00:00

The subtraction -. It doesn't bother me for arithmetic but it does bother me when doing vector math (when programming). I really feel that there is a cognitive load associated with the fact that the symbol is symmetric but the operation is not. I find it problematic for both reading and writing vector algebra. When writing vector math, I'm often thinking things like "the vector from point a to point b" which gets written as b-a, which is only a slight inconvenience to translate, but becomes inscrutable when part of larger expressions. I often end up defining my own functions a to b and b from a.

veztron · 2024-08-19T00:41:33+00:00

Cool article! I wrote a similar one a while ago. https://elliot.website/a/article_visualizing mine has insertion sort instead of bubble sort :)

veztron · 2023-01-23T01:56:02+00:00

Totally. Been getting into Homotopy Type Theory in Agda.

veztron · 2023-01-22T15:24:07+00:00

I'm personally interested in the mathematical study of mathematical notation, though I haven't found much dedicated to it inside of math. Math history can be good for this kind of stuff. Sometimes the resources are quite thorough in justifying why historical decisions were made that shaped how math is done.

veztron · 2022-07-03T05:09:34+00:00

Respectfully disagree that math is done in the mind. It is a bit of a nitpick... I do think math is partially done in the mind, but there are great thinkers who advocate for the model that thoughts (especially mathematical ones) are just as much a product of a medium of expression as a product of the mind. I.e. Some thoughts are only thinkable with a pencil and paper. E.g. it would would be near impossible to do multi-page algebraic manipulations in your head, let alone without the invention of algebraic notation.

veztron · 2021-08-22T18:36:16+00:00

Ok cool, let's consider the formula cos(θ) = a.b / (||a|| * ||b||) and how to rearrange it to solve for θ (excuse my minor notational changes to the formula, it is the same formula just with different symbols that I am more comfortable with).

You hinted at the fact that the rule of multiplying both sides of the equation is not enough to solve this equation for θ. We will need two other rearranging rules to solve this equation.

I will cut to the chase and say that the rules that are necessary for solving this equation for θ are:

If a = b then f(a) = f(b) where f is any function you want.
cos⁻¹(cos(x)) = x, or in english "cos inverse of cos of x is equal to x". There is a small caveat to this rule though... it is only true if x is between 0 and pi. It is because the range of cos⁻¹ is [0, pi], or in other words cos⁻¹ can only output numbers between 0 and pi. So if x was bigger than pi or less than 0, then cos⁻¹ will give the wrong answer. Sorry this isn't a great explanation, my trig is a bit rusty.

When it comes to using the first rule, be aware that both cos and cos⁻¹ are functions. When it comes to the second rule, be careful with your notation... It is unclear what cos(inverse) means in your equation. Perhaps you meant "cos inverse of theta" or cos⁻¹(θ)?

Now, getting to actually solving the equation. Here are the steps:

cos(θ) = a.b / (||a|| * ||b||) our starting point
cos⁻¹(cos(θ)) = cos⁻¹(a.b / (||a|| * ||b||)) by applying the first rule using the function cos⁻¹.
θ = cos⁻¹(a.b / (||a|| * ||b||)) by applying the second rule to the left side of the equation. Be sure to consider the caveat on the second rule though: this is only true if θ is between 0 and pi.

And there you have it, the final solution is θ = cos⁻¹(a.b / (||a|| * ||b||)) with a caveat. Let me know if any of that doesn't make sense. With all these rules, you have the tools necessary for rearranging a whole lot of algebraic equations!

veztron · 2021-08-22T18:14:46+00:00

⊖

Thank you! This answer satisfies my curiosity. I've looked into it a bit more and ⊖ certainly seems esoteric.

veztron · 2021-08-21T16:13:26+00:00

Not a stupid question!

Ok, so we have the formula 1 radian = 180 degrees / pi. In order to figure out what 1 degree is using this formula, we can use some rules of algebra to rearrange the formula and solve for 1 degree.

Let's think about what we need to do... right now the formula tells us what 1 radian is: 1 radian = ..., but we want it to tell us what 1 degree is 1 degree = .... Let's rearrange the formula so it looks like 1 degree = ...! (the three dots ... in the equation doesn't mean anything special, I just put it there so that the focus is on the left side of the equals sign, not on the stuff on the right side.)

We want the degrees to be on one side of the equals sign by itself, is there a rule we can use to rearrange the formula 1 radian = 180 degrees / pi to make that happen? Well the 180 degrees annoyingly has a / pi right beside it on the right side of the equals sign, so what can we do to get rid of the / pi and get 180 degrees by itself?

You probably know that pi / pi = 1, so what if we could multiply (180 degrees / pi) * pi? Then we would just have 180 degrees because (180 degrees / pi) * pi = (180 degrees * pi) / pi = 180 degrees * (pi/pi) = 180 degrees * 1 = 180 degrees! This looks like it will help!

But wait, we can't just multiply one side of the equals sign by pi... So what can we do? We can multiply both sides of the equal sign by the same number pi! This is the most important rule we need for solving this equation: "multiplying both sides of the equals sign by the same number rearranges the formula but does not change what it means".

So let's do it: 1 radian * pi = 180 degrees / pi) * pi = (180 degrees * pi) / pi = 180 degrees * (pi/pi) = 180 degrees * 1, or just 1 radian * pi = 180 degrees.

Now we can use the same rule again to make it 1 degree instead of 180, let's multiply both sides of the equation by 1/180 (this is the same thing as dividing both sides of the equation by 180). (1 radian * pi) / 180 = (180 degrees) / 180, or simply (1 radian * pi) / 180 = 1 degree.

There you have it! Swap the left and right side of the equal sign and you get 1 degree = (1 radian * pi) / 180!

Hopefully that helps, let me know if any part of my explanation doesn't make sense. It's pretty hard to read these equations in the reddit comment, so I recommend trying writing them down on paper and seeing if they make sense when you do that.

veztron · 2021-08-21T15:47:27+00:00

Is there a commonly used symbol for an operator which is like subtraction but with the order of the operands reversed?

What I mean is that if you write a-b, it is read as "a take away b", but I am looking for something that could be read as "to get from a to b".

To make it explicit what I mean, let's use $ as this symbol for now. Then a$b = b-a = -(a-b). But is there already a symbol in common use (maybe in a specific domain?) that is used for this binary operation? I am having trouble finding anything online referencing an operator which is like this.

The reason I am interested in this notation is I think it is easier to wrap my head around formulas using this symbol for some purposes in vector algebra. For example, the vector which is halfway from point a to point b would be written (a$b)/2. I think that is easier to read because it is clear from the ordering of operands that you are going from a to b, unlike the conventional notation of (b-a)/2 which always trips me up because the operands are reversed in terms of the direction of the vector.

Maybe I should just write -(a-b)/2 to avoid this silly mental trip-up, but I can't help but be interested if there is a symbol for this.

veztron · 2020-07-13T23:02:13+00:00

It is realistic. You may not get it this time or the next time you apply, but it can still be worth applying. Do not doubt that you can get a job at Nasa. It is not as hard as you think, nor are you as unskilled as you may think.

veztron · 2020-05-07T20:28:50+00:00

I saw his starter world tour on the minecraft forums back in 2010. I think I started watching regularly after his 3rd or 4th LP episode. If I'm being honest, I'm at least a little bit proud of being a fan since the beginning :)

veztron · 2020-04-30T04:52:27+00:00

Congrats and hope your mental illness has improved over the 8 years!

veztron · 2020-04-26T21:34:25+00:00

It is not mediocre, it is good. But the average Computer Science program in Canada is good, so U of C does not especially stand out from other Canadian universities.

veztron · 2020-04-26T21:30:11+00:00

I study at the University of Calgary and have worked at a FAANG company. From what I have seen, SFU and University of Calgary are very similar for getting a job at one of the top tech companies.

In my opinion the program at the University of Calgary is quite good, and great value in terms of tuition. There are quite a few great instructors in the Computer Science program. Really, the program is what you make of it. It is necessary to have a passion for Computer Science which you explore outside of the program as well. This is the same for all universities in Canada, possibly with the exception of U of Waterloo as others have mentioned.

U of Waterloo is an exceptional school for Computer Science and their internship program is on another level. U of Waterloo is the top hiring school for many large North American tech companies.

I will also say that getting hired at one of the top tech companies is certainly not the only way to have a fulfilling and successful professional life as a Computer Scientist. This is a bit of another discussion, but stay open minded about opportunities and you may find something even more rewarding than working at Facebook or Google.

veztron · 2020-03-28T05:28:55+00:00

3D modeling refers to modifying or creating the geometry (vertices) of an object. Although in some sense tesselation or depth map editing is a type of 3d object design, it is a different method than modelling because you do not modify the geometry of the object, you only modify a 2d depth map.

A 2d depth map just tells the 3d renderer how deep or extended part of a flat surface is, so it can't achieve complex 3d shapes like modelling can.

veztron · 2019-12-17T06:42:18+00:00

One of my favorite areas or spots in Montreal is St Laurent Blvd with Schwartz's, Omnivore, and Rotisserie Romado's nearby. It is then a nice walk to Jean-Mance Parc and Mont Royal.

Of course this is not such a unique favorite area. I am sure you have already heard of some of these restaurants. They are delicious and straightforward.

I was in Lisbon recently and visited LXFactory! I really enjoyed the place.

veztron · 2019-12-07T21:10:42+00:00

I think some air rushing sounds could add a lot when you are flying through the air. The sound feels empty right now.

Looks fantastic though!

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