Guess only work with integers, specially for the floor function that is going to give you an integer at the end everytime.
Not my idea, learned it somewhere while doing college in an statistics class. The idea is that the exponential function grow really fast, so small difference on variables become extreme difference on the exponential, then the log function reverse the exponential, but because it grew more for the biggest variable it reverts to the max variable making the other variables the decimal part (this is why you need the floor function). I think is cool because works for any number of variables, unlike mathematician 2 who only work for 2 variables (maybe it can be generalized for more variables but I don’t think can be done).
For a min fuction it can be use ceiling(-ln(e^-x + e^-y))
to be fair it does seem to work for any two numbers where one is >1. As lim x,y–> inf ln(e^x+e^y) <= lim x,y --> inf ln(2 e^(max(x,y))) = max(x,y) + ln(2).
I think is cool because works for any number of variables
using the same proof as before we can see that: lim,x_i -->inf ln(sum_{i/in I} e^(x_i)) <= ln(|I|) +max{x_i | i /in I}.
So it would only work for at most [base of your log, so e<3 for ln] variables.
I don’t have a mathematical proof, but doing some experimental tests on excel, using multiple (more than 3) numers and using negative numbers (including only negative numbers) it works perfectly every time.
Try (100,100,100,100,100,101) or 50 ones and a two, should result in 102 and 4 as a max respectively. I tried using less numbers, but the less numbers you use, the higher the values (to be exact less off a deviation(%-difference) between the values, resulting in higher numbers) have to be and wolframAlpha does not like 10^100 values so I stopped trying.
I do think the upper bound on that page is wrong thought. Incedentally in the article itself only the lower bound is prooven, but in its sources this paper prooves what I did in my comment before as well:
for the upper bound it has max +log(n) . (Section 2, eq 4) This lets us construct an example (see reply to your other comment) to disproove the notion about beeing able to calculate the max for many integers.
I just remembered where I learned about that function, in this course on convex optimization that unfortunately I never had the opportunity to finishing it but is really good.
Where’s the Julia programmer that hits every one of these with @benchmark and then works for six hours to shave three nanoseconds off of the fastest one?
GNU C. Also works with Clang. Avoids evaluating the arguments multiple times. The optimizer will convert the branch into a conditional move, if it doesn’t I’d replace the ternary with the “bit hacker 2” version.
And then your customer changes their mind. Instead of two numbers, they will now input three numbers. How easy will it be for you to change your code?
And then the customer changes their mind. Instead of three numbers, they will now input any series of numbers. How easy will it be for you to change your code? And why didn’t you already do this is the previous step?
And then the customer changes their mind. Instead of any set of numbers, they will now input numbers and text. How easy will it be to change your code?
And then the customer changes their mind. They now have no idea of what they’re sending you or if they’re even sending you anything. Nevermind the code now, you already did that in the previous step, right? How easy will it be to explain what you’re invoicing them for?
If only you’d done the most bloated and well documented function first, you could have saved yourself the time and your client for the money.
I mean this is the kind of shit my boss would argue about. Why pay for the first attempts that didn’t work… blabla bla. He always ends up paying but it’s always such a hassle.
Otherwise, realistically, I’m prob the worst of all worlds … the procrastinator waiting/hoping to be the pair programmer that has hopefully remembered to just be the thief.
Simple, really. Abs(x-y) is the difference between the two numbers, absolute, so positive value. So, adding abs(x-y) to the smaller of the two numbers turns it into the bigger number. Plus the bigger number, now you have 2 times the bigger number
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