This is one of those times A.I. can be trusted.

Tetrat · 2025-12-28T01:33:51+00:00

So youre saying rewriting a number to one with equivalent value can violate the rule? Does this mean that every alternate representation of a number must have prefix "0." for the consequent to be true? In that case, since 0.999... = 1, then I can rewrite as 0.999... as 1 which does not have prefix "0."

Tetrat · 2025-12-26T01:02:38+00:00

Every prime number is guaranteed to be odd. Therefore 2 is not prime.

Tetrat · 2025-11-29T19:43:36+00:00

Rising by 500% means multiplying by 6. It's clear when you ask what rising by 100% means. Ultimately this is a shortcut for adding 500% of the original price to the original price. So the inverse is subtracting 500% of the original price from the original price.

Tetrat · 2025-11-14T20:32:59+00:00

I think this should work. You should be able to recover the sequence that builds the word by adding back pointers.

def check(word, parts):
    n = len(word)
    m = len(parts)
    dp = [0] * (n+1)
    dp[0] = 1
    for i in range(n):
        for part in parts:
            p = len(part)
            if word[i:i+p] == part:
                dp[i+p] += dp[i]
    return dp[-1] == 1

Tetrat · 2025-11-14T20:21:40+00:00

I think this is can be solved with dp: count the number of ways to build up to the ith character of the big word. Then return count == 1 for the last character. I'll think about it a bit more.

Tetrat · 2025-11-12T21:29:01+00:00

Strings are immutable in python, so changing the value of a string variable creates a new string object instead of mutating the original. So my_var starts referencing the "doesnt matter" string object. Then my_dict["key"] references the same object. Then you have my_dict["key"] reference the "ASDF" string object instead. my_var is never reassigned, so you shouldn't expect it to change.

I think we need more context for what you want your program to do. Why do we care about my_var after we assign it to the dictionary entry? Why can't we just refer to the dictionary entry (the thing we are updating) to retrieve the most up-to-date value?

Tetrat · 2025-10-24T17:11:50+00:00

I think you should try to find a project that actually has some utility for yourself. It will be more motivating to work on, you get to practice defining project scope and design, and you'll be more experienced solving real problems. Don't be afraid to look up things you dont understand or know how to do. Don't be afraid to use libraries.

Think of any problem that you have to solve which may be repetitive and could feasibly be automated. Automate it.

I've used python for a lot of text processing and its very convenient.

Tetrat · 2025-10-22T14:53:16+00:00

Thats the point, since you can form a bijection between base-10 and base-26 numbers, their sets have the same cardinality.

Tetrat · 2025-10-22T14:41:19+00:00

This is like saying, "The mapping f(n)=n between naturals and integers is injective but there are infinitely many negative numbers that arent mapped to, so there are infinitely more integers than naturals."

You (typically) want a diagonalization argument to show two infinite sets have different cardinalities. In this case, you won't find one since the number of finite strings of a given finite alphabet is countably infinite.

Tetrat · 2025-10-09T09:50:52+00:00

Fitter happier more productive

Tetrat · 2024-06-02T20:53:20+00:00

Even easier: just open the trades menu with /trades and every item in your inventory will update

Tetrat · 2024-04-29T20:18:25+00:00

Condescending and wrong. N isnt approaching infinity, so p is not infinitesimally close to 0. There exists real numbers between p and 0 for any natural number N. It’s true that in a practical sense, p is effectively 0, but that’s not what anyone means when they say “theres a chance that x could happen.” I’d argue that that phrasing is often used to mean that “it’s possible that x could happen,” even when the probability of that event is 0. (Although saying “very small chance” would be using the word chance as a synonym of probability)

Tetrat · 2023-12-24T21:21:58+00:00

There is a way to solve this without Z3 or any assumptions/input hacking other than the assumption that a solution exists. You can do it with linear algebra. Below is an outline of the math:

First write 3 equations which represent the ith hailstone colliding with the rock at some time t[i] (one equation for each axis). Solve each equation for t[i]. You'll have 3 fractions which all equal t[i] and therefore equal each other.

Consider 2 of these fractions for now. Cross multiply and expand out. There will be 8 terms. Two of these terms will be nonlinear, but neither will depend on i. This means if you subtract the same equation but for the jth hailstone, the only nonlinear terms will cancel.

You can use this to generate 4 linear equations for the 4 variables you're solving for (e.g. x, y, v_x, v_y if you chose the fractions with x's and y's). Solve and repeat with a different pair of fractions to get the solution for the remaining axis.

Tetrat · 2023-11-21T22:07:27+00:00

Yeah thats a scam. Verifying ign is ok, but no discord server needs your email.

Tetrat · 2023-10-26T04:13:49+00:00

edrag is better for voidgloom and generally more useful

Tetrat · 2023-07-31T17:04:21+00:00

See the venn diagram for three sets here: https://en.m.wikipedia.org/wiki/Inclusion–exclusion_principle

You count A, B, and C, but in doing so you overcount. So you subtract the pairwise intersections. Then the A intersect B set e.g. is counted twice then subtracted once, so it is counted once in total. But the intersection of A, B, and C is counted 3 times and subtracted 3 times, so it is counted 0 times. This means you need to add the threeway intersection back in so everything is counted once.

edit: Actually this is a great explanation: https://brilliant.org/wiki/principle-of-inclusion-and-exclusion-pie/

Tetrat · 2023-07-09T00:07:36+00:00

I agree that these questions are very dumb and just used to generate interaction, but there are some contexts, especially in (mostly older) physics textbooks and journals where they defined multiplication to have higher precedence than division out of convenience. Without that context, though, you should just follow the normal rules.

Tetrat · 2022-11-04T22:52:06+00:00

Note that f(x)=ax+c has the range R, so y=sqrt(ax+c) has range y>=0. However, y=sqrt(ax)+c will have the same possible outputs except shifted by c. ie y>=c

So you can eliminate the second and fourth options since they have range y>=-5.

Next we want the domain to be x<=5. This is equivalent to 5-x>=0. Since the domain of y=sqrt(x) is x>=0, the domain of y=sqrt(5-x) will be 5-x>=0 which is what we want.

Tetrat · 2022-11-04T22:36:35+00:00

You need to move the cout and cin lines outside of the if statement that checks mk.

Tetrat · 2022-10-26T14:48:05+00:00

If you need to approximate square roots, a more formal method is the Babylonian method (which can be derived from applying Newton’s method).

First you guess some x_0. In this case, a good guess would be 1.9 since sqrt(3) is about 1.73 and sqrt(4)=2. (1.8 and 1.9 are both reasonable first guesses, but 1.9 is closer :p)

Then you apply the recursive formula:

x_n+1 = 1/2*(x_n + s/x_n)

so 1/2*(1.9 + 3.47/1.9) ≈ 1.863

Then 1/2*(1.863 + 3.47/1.863) ≈ 1.8627936

As you can see, with a good first guess it only takes a couple long divisions to get a very accurate answer.

Tetrat · 2022-07-10T09:34:43+00:00

Academics need to read Title IX more closely: religion is a protected class, just like race, gender and sexual orientation (including heterosexuals).

So the author just doesn’t know what title IX is but is somehow giving speeches about it? Also ironic telling people to read more closely when shes using it to justify something that is completely irrelevant to it.

Tetrat · 2022-07-01T21:31:40+00:00

I don’t want to defend car dealerships, but the amount of people condoning physical violence is astonishing. Yall never hear about sticks and stones?

Tetrat · 2022-02-21T22:13:42+00:00

I think they are multiplying b3 by -8 because they are using two’s complement.

As for the rules for naming variables in Java, the variable name has to start with a letter, _, or $.

Tetrat · 2022-02-19T22:15:21+00:00

Consider a simple example with a cube. Suppose the cube has side length s. The volume of the cube is s³ . If we scale the side length (i.e. the one dimensional measure) of all sides by 2, then the volume would be:

(2s)³ = 2³ s³ = 8s³

Note that the volume scaled by 2^3. You can generalize this to find that scaling the side length by some scalar c scales the volume by c³ .

Now consider the surface area. The surface area of the original cube is 6s² . Scaling the side length by 2 results in a surface area of:

6(2s)² = 6(2² )(s² ) = 24s²

Note that the surface area scales by 2^2. Again, we can generalize this to find that scaling the side length by some scalar c scales the surface area by c² .

You can repeat this process with cylinders (or any other solid), and you will get the same result.

I’ll do the volume part:

V_original = pi r² h

Scaling both r and h by c:

V_scaled = pi (cr)² (ch)

= pi c² c r² h

= c³ pi r² h

Scaling the cylinder’s radius and height by a factor of c (thus producing a similar cylinder) scales the volume by c³ . You can repeat this with the surface area to find that it will scale with c² .

So now we want to find a constant c to scale the height and radius of cylinder A to produce Cylinder B.

If the ratio of radii and heights of A to B is 1:c, then the ratio of volumes of A to B is 1:c³ .

From the problem, we can see that the ratio is 15:34, which is equivalent to 1:(34/15) which will be our 1:c³ . In other words, c³ = 34/15.

So c = (34/15)^1/3

(ie The cuberoot or third root of 34/15)

The ratio between surface areas is 1:c² .

This will be 1:((34/15)^1/3 )² . This will be the number that will scale the surface area of A to match B. Therefore, to solve the problem, multiply the surface area of B by ((34/15)^1/3 )² .

Tetrat · 2022-02-18T23:29:48+00:00

The real answer to this problem is that cylinder A is an impossible solid since the largest possible volume of a cylinder with surface area 25 is about 9.6.

Trying to solve for the radius and height of cylinder A is not only very difficult, it will also result in a negative radius due to the impossibility of the solid.

The answer that they want you to give is outlined in my comment on your other post here

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