I managed a guy like that. He was capable of very complex thinking, but he wasn't in love with complexity, he was in love with simplicity. His solutions tended to be of the form, "we can ignore all these things, and just focus on X, and it will provide all the value." He'd notice something and simplify it and the benefit to the company would be measured in multiples of his salary.
Every manager who'd ever directly managed him knew what a treasure he was, but it was often hard for us to convince others of the value of his solutions because they were so simple, and people were convinced that hard problems must have complex solutions. (or else they would have solved them, right?)
He eventually got bored. He retired and joined a seminary.
Money would matter even more than the interpersonal stuff in most cases but on top of it even the managers treasured him so there should've been even less of an issue of communicating value.
Getting bored is totally understandable though given his calibre but that's a separate issue from how the company evaluates performance.
Here is why: I turned off a feature flag in our feature flagging service which saved company 10% infra cost, do you think I can be promoted to Staff+ and lead 50 engineers?
Promotions and/or recognitions in corporate environments works differently.
I don't agree with it, but this is how it works: If what you did feels simple, anyone else can do it as well, why should we promote you for finding such silly mistake or improvement.
I’m not a fan of doing politicking, bjut after much courses on writing and communication, I strongly believe that such simple solutions could have been presented in a way to justify rewards.
There are people that do nothing worthwhile and can find words to justify themselves. If someone brings value, you can find words to earn him recognition.
I mean, we are in the industry where it used to be a standard practice, not so long ago, to deliver daily reports about one's activities while "planking", or throwing a beach ball to another person doing some silly acrobatics...
It should come as no surprise that there's no rigorous assessment protocol for these kinds of things anywhere. Retrospectively, I will admit, that enormous amount of effort and resources are wasted due to bad planning. But it's still not done.
I can imagine that with the field becoming more competitive, eventually, the industry specialists will come together and try to address the problem, but so far and for so long the resources just kept flowing in, the huge waste wasn't really a problem.
I would safely assume that there are no limitations of what mathematicians can do, with one important exception: Andrew, for whom I argued about the mis-uses of Infinity. Andrew is, well, rather famous.
It has become sort of junk food for the brain. Temptations and ads for it everywhere.
Plenty of people are experiencing this nowadays
The idea that no one is being forced to use AI is nonsense
me@localhost:~> bc
d=1; for(i=21; i < 41; i++){d *= i;}; print d; print "\n";
335367096786357081410764800000
n = 1; for(i = 1; i < 21; i++){n *= i;}; print n; print "\n";
2432902008176640000
d/n;
137846528820
* - What it does contain is sine, cosine, exponential, log, arctan, and Bessel J (?!?!?!?!)
let ans = 1
for (let i=1; i<21; ++i) {
ans *= (41 - i)
ans /= i
}
After the i-th iteration of the for loop, ans will contain n!/((n-i)!i!) which is exactly \binom{n}{i}, an integer.
Technically "ans" can grow above the final result in my example, but even that could be fixed if one really wants (e.g. i must divide either ans or n-i, you play a bit with divmod to figure out which division you do first.)
https://www.wilfred.me.uk/blog/2014/10/20/the-fastest-bigint...
- Python's native integer handling, which already has no size limit.
- PLUS part of the Decimal module in Python's stdlib: BC's floats are DECIMAL by default, not binary.
- PLUS an implementation of Bessel's J function, while neglecting Bessel's K.
- Some features for base conversion using `ibase` and `obase`. So, I suppose you can output numbers to base 60. [EDIT: Correction from earlier: ibase is allowed to be at most 16, while POSIX allows for the maximum value of obase to be at least 99, which therefore does allow for formatting output to base 60.]
You can also think of it another way, without using the formula combinations, and only the fact that there are n! permutations of n objects. We can think of this a permutation of 2n items, made up of two groups of n identical items each. Using (2n!) will overcount, due to the fact that each of the "over" steps are identical, and similarly for the "down" group. We have cut down our answer by dividing out all of the repeated sequences. There will be n! redundancies for all the ways we can permute the "over" group and, the same for the "down" group. So this results in (2n!) / (n! * n!), which is exactly equal to 2n choose n. See [1] which explains permutations with repetion this in general. [Note: We pretty much re-derived the formula for combinations!]
[1] https://brilliant.org/wiki/permutations-with-repetition/
So for counting, you can basically think about it as a list of twenty initially empty spots. You first fill it in with your 10 down steps. The remaining 10 spots will then be the ones for the 10 right steps. So really the only choice you have to make is where to place the 10 down steps.
This question boils down to: in how many different ways can you distribute the 10 down steps over the 20 empty spots? That's 20 choose 10.
Off the cuff, notice that the diagonal has n+1 intersection points, and a path that never passes through the diagonal gives a forest via the isomorphism with ballot sequences [0]. Any sequence that does pass below the diagonal can be "rotated" into one that doesn't, and so there are probably n+1 paths in each "path class" on average.
Conversely, this would suggest that all paths contained in just one upper or lower triangle of the square can be counted by the Catalan numbers. Indeed, a 2x2 square has just 2 such paths and (2n C n)/(n+1) = 6/3 = 2.
[0]:https://blog.wilsonb.com/posts/2026-02-27-easy-random-trees....
Anyway, here is the post https://kpatucha.github.io/posts/Dyck-paths-Raneys-lemma/
Perhaps because I was pigeon-holing this as a programming optimization problem.
I wrote about it too! [0]
Also if you help little kids with homework, you'll see that some problems are quite difficult as well and require you to actually think, even if it's problems for 10 year olds.
Two years later, comes a challenge in class... make a formula for summing the integers... well everyone started with 1+2+... I starred with blocks, 1+n, 2+n-1... I had the complete formula in minutes...
That was the very last class for which I was with my peers of that grade... I was put in a HP High Potential class, with a high school algebra book, and although was a bit lonely, was in my element.
The point is- the recognition of the problem, can save huge amounts of time, where as AI can only brute force it, or use a pretrained solution.
this is why I'm not a big fan of "show your work": the "work" is however many years it took to build up my intuition, and often any explanation I could type out for my solution would be a retroactive rationalization. it's still useful, sure -especially for catching your errors, but I place it on the opposite end of the open-fake scale than most people.
of course here the proof is simple: 20 right moves, 20 down moves, any order => of 40 total moves choose any 20 indices to be your down moves => 40 choose 20 is your answer. would that teach you how to solve the next problem though? I'm not so sure.
40 indices, pick 20. Those are East moves, the rest are South moves.
FWIW the generalization of binomial coefficient which allows you to express an n-dimensional solution is called a multinomial coefficient [0]. So in a 3d 20x20x20 box we would have (60 multichoose 20, 20, 20) paths.
Also, the wiki article doesn't mention this but the growth rate of (n multichoose k1, k2, ..., km) as we increase n but fix the ratios p1 = k1 / n, ..., pm = km / n is precisely the Shannon entropy of the categorical distribution with probabilities p1, ..., pm . The wiki article for entropy [1] states the result for the binomial coefficient, which can be written as (n choose k) = (n multichoose k, (n - k)) .
Actually a lot of basic information-theoretic results about entropy and related quantities (e.g. the properties of the Boltzmann distribution/softmax function) can be derived from similar discrete counting problems after taking a large-n limit. I don't have links at the ready but I might edit this comment if I remember places which explain this stuff.
[0] https://en.wikipedia.org/wiki/Multinomial_theorem#Number_of_...
[1] https://en.wikipedia.org/wiki/Entropy_(information_theory)#A...
But I am reminded of how during my engagement 24 years ago, my future father-in-law raised an issue of being able to determine whether they were getting the full amount of sandpaper on large rolls that they were paying for. I was able to simplify the question a bit to one that treated the rolls as if they were simple concentric rolls of a specified thickness and from there could turn it into the good old Gaussian sum formula times 2π to get the length. The engineers working for the company came up with the same solution, but instead of using n(n-1)/2 they did the summation with multiple rows in excel.
I also tried a weird idea involving popcount, but it didn't scale. My approach was to represent each possible path with 0s (don't turn) and 1s (turn), testing the same number of 0s and 1s. However, even with popcount running in O(1) with hardware support, the total number of possible paths made the idea impractical :)
- at zero events: we have one outcome, nothing
- at one event: we have two outcomes, 1X and 1Y
- at two events, each of those two outcomes have two outcomes, but one X and one Y outcome both become XY, so we have 1X^2 2XY 1Y^2.
You can continue combining this way, adding an X and Y to each term (for the two possible outcomes) — and then grouping like terms.
I’m building a grid based game and engine, and I have a game replay format which is not video.
I hit a massive wall with compression, trying to compress unit pathing and was trying to solve a similar solution.
Given an NxN grid, and the 4 cardinal directions (NSEW) you can move in, plus an extra action that makes you move 2 cells instead of 1, and considering you can move 4 cells per second…
What’s the smallest worst-case raw compression artefact you can output for 1 player for a 1 minute game?
It’s an extremely fun problem to solve. I tried:
- encoding changes into bits eg using 2 bits for direction
- movement pattern batching (ie batching 2 moves into 3 bits)
- crowd patterns and movement prediction
- treating movement as a “projectile” and deriving intermediate states
And all sorts of other wild crap that I will write up about on game launch
My game is strictly deterministic, so I get bot movement for free - but the player has agency so I need to capture their deviations
That’s the tricky part! Right now I do capture input (actually just deviations) and can replay whole games, but I think I’m at the limits in terms of compression - talking bytes here not KB
I feel sad because I had forgotten the simple and intuitive construction of choosing “go down” and “go right” directions. When a person learns more advanced mathematics, it is often the case that the person just applies such advanced mathematics by rote without realizing that a solution can be found with more elementary mathematics and more creativity. It reminded me of the time in middle school before derivatives were taught, when my teacher reminded me that using derivatives to solve a problem would receive no credit.
It is a common experience in mathematical problem solving that the first solution leads to more insight which illuminates a shorter slap-my-forehead solution -- bruised forehead.
Skills do decay, I can't deny that. But even when I was still in school going back and doing an end-of-year problem for a class I took two years ago would have been harder than it was for me at the time... but it would have been easier than the first time if I warmed up properly with a bit of review and practice first, and I mean, not just three minutes glancing over things but taking some serious time for it.
Most of the first 100 problems can be solved without any understanding of the problem, should you so desire.
As stated, the choose(2n,n) solution of course works but as soon as you deviate from a square, things can get more complicated. What if it's a rectangle? An arbitrary shape? One with holes? The dynamic programming solution takes all of this in stride (assuming, of course, that the conditions of only going right and down still hold).
Pascal's triangle is, after all, a dynamic programming solution. It just so happens that there's a "closed form" solution to their entries.
I'm all for clever tricks but I also appreciate much more a solution that generalizes well and gives more insight into a class of problems.
Depending on how well you know binomial coefficients, this can also lead you to the dynamic programming solution. After all, for the rectangle case this is nothing but construction of Pascal's triangle by summing two adjacent numbers in a row to get the number below them. If you recognize this, the solution for paths on grids with holes falls out almost immediately.
return 5 # because mathThe computer programming part of it is just a quick way to develop candidate solutions.
needed to justify viewing this as "arranging down vs right movements" as another comment outlines
And with AI the path of least (initial) effort seems to be to just ask the model to solve it. It might get it wrong and then I'll prompt it again and again. But each individual prompt is fairly low effort on my part. Whereas coming up with the right solution myself might've taken less time but the initial effort is a lot more.
Last year I used to romanticize about building at least 1 thing each month completely by hand without any LLM coding help. The last such project I worked on was 6 months ago so sadly it's not going so well.
I'm sure ~4 yrs ago i would have loved the thought of this. It's so boring. My job is so, so boring.
...
...
Right, right, right, down, down
RRRDD
Well let's actually consider the degenerate cases.
ABCDE
AACDE
AACCE
AACCC
AACCC is isomorphic to RRDDD, which is how we get 10 possible paths to solve the 2*3 grid. We can check this with the binomial theorem: ((2+3) choose 3) = 10.
What's nice about this step by step approach is that it generalizes not just to non-square grids, but to multiple dimensions as well! Imagine trying to get from the top of a 3 by 3 by 3 Rubik's cube to the bottom, how do you do that? Well how many ways are there to rewrite
AAABBBCCC
There is no easy way out, you have to rest but you simply can't stop. Your body will rot, your mind too.
PS: song isn't an ode to the grind culture or how to slave away in an office, as lyrics say "you’ve got to work for yourself - Love yourself, feed yourself".
D[A,B] := number of ways to navigate from grid sized AxB = D[A-1,B]+D[A,B-1]
and the aha moment is realising this is just a binomial coefficient.
Give it too long a rest and you have to go back at full blast for weeks on end to hope to ever achieve past performance.
I am very bad at math and have always been in awe of those who can do it well.
On the other side, my Math ability definitely goes down to Calculus and I definitely forgot most about geometry.
This is high school math.
It's true, if you don't activate this area of your brain often, it's easier to brute force the solution and reach for the easy mechanical calculation. I can feel this when I'm refactoring code. Today, I just have Claude do it for me with a few instructions. Each day, I feel a tiny bit more ignorant about the actual framework's APIs, its abstractions, and its rules. But I still would rather do other things with my time.
As for the problem, luckily for me, this one was easy to derive if you remember factorials, permutations, and remember to account for duplicate patterns
Work out the first few cases by hand (1,2,6,20 in our case) and then look up the sequence on "The On-Line Encyclopedia of Integer Sequences" (OEIS):
https://oeis.org/search?q=1%2C2%2C6%2C20&language=english&go...
At first glance of the question, I had imagined it to be hard but then I read through the solution and other comments to recognize that I had in fact done such a question previously and I had solved it independently during the class if I remember correctly or such classes of problems.
I also agree with the AI and spreadsheets part of thing for what its worth but I can only tell more when I get into job but I have heard such things from my senior brothers.
I feel like there has to be a right balance of complexity though, and for what its worth I think that there are so many other things that one optimizes later on in life with tangential benefits as well with real knowledge about real life use-cases and edge-cases and so much more! I feel like it would be hard to replace with AI as much as (some) people (mostly Marketing) want it to feel so.
I do hope that people don't atrophy their skills though and to solve some coding questions or make projects perhaps as well without LLM by hand if given/having the time. Not everything probably has to be done by the fastest or the most accelerated way as you wouldn't know the destination as it would be found along the way itself. I suppose just like life, so stay safe and have a nice day.