La Fée Verte @Absinthe@qoto.org
The green faerie
Joined Aug 2019
La Fée Verte
boosted
🇳🇱
@stateastic I would recommend you have a char with @freemo
La Fée Verte
boosted
@Absinthe I am learning about it and want to shift my career to business analytics. Right now, I am working as sports analyst.
I am hoping to attract them :)
The instance that I am on, is intended to be STEM oriented, and I am pretty sure our admin is a data scientist. What about you?
La Fée Verte
boosted
@Absinthe where are the data science people? Any idea?
Here's a freebie!
This problem was asked by Google.
Given a list of integers S and a target number k, write a function that returns a subset of S that adds up to k. If such a subset cannot be made, then return null.
Integers can appear more than once in the list. You may assume all numbers in the list are positive.
For example, given S = [12, 1, 61, 5, 9, 2] and k = 24, return [12, 9, 2, 1] since it sums up to 24.
@freemo would it help to know that the solution is based on memoization ?
@freemo I said a million, but I meant 10 million.
@freemo They don't tell you that you have to do it in any particular time or space. It is a bad challenge form that part. When you start there are a few tests that you run. Then when you think you are happy and submit it, it runs against more tests. Some of them are over a million long with the numbers being at the beginning and end, and also in the middle. I did put a link in for the actual problem. It has a ui and a working compiling and testing environment. I had the logic solved dead on, but couldn't be certain until I got it efficient enough to go through the big blind tests without timing out. You get very little feedback on a timeout :)
@freemo It was a weird challenge.
Here is the odd part:
[5,3,8,7,8,9,5], 10 would be [3,7]
Because it finished first. Even though [5,5] starts first.
okay, this is not exactly a #toyprogrammingchallenge but I hope you hate this problem as much as I did...
The problem is not just getting the right solution, but getting it in under the running timeout of 12000 ms. Good luck.
La Fée Verte
boosted
Here's a #toyprogrammingchallenge requiring a bit of algebra.
For any four non-collinear points A, B, C, and D in three-dimensional space, there is a unique hyperbolic paraboloid H containing the lines AB, BC, CD, and AD (that is, every point on any of these lines is also on the surface of H). Write a program that accepts Cartesian coordinate triplets representing A, B, C, and D and prints an equation in x, y, and z that is satisfied if and only if <x, y, z> is a point on H.
Your program's output may differ from the examples but should be optimised for human readability. Combine like terms, omit terms equal to zero and avoid unnecessary factors (e.g. prefer "x = y" to "2x + 0 = 4y - 2y" even though both describe the same surface).
Example 1 input:
<0, 1, -1>; <1, 0, 1>; <0, -1, -1>; <-1, 0, 1>
Example 1 output:
z = x² - y²
Example 2 input:
<1, 1, 1>; <1, 0, -1>; <-1, 1, -1>; <-1, 0, 1>
Example 2 output:
2y = xz + 1
Example 3 input:
<0, 1, 1>; <0, 1, -1>; <0, -1, -1>; <0, -1, 1>
Example 3 output:
x = 0
La Fée Verte
boosted
Octave solution & discussion
Pretty straightforward. Might be possible to improve on the factorial one by counting factors, because computers can generally calculate x^n faster than just multiplying it out, so if you figure out that the answer would be 2^a * 3^b * 5^c... it could outpace the naive 1 * 2 * 3 * ... * n method.
That said, I tried such an approach and while it does scale better than linear, the upfront cost is high enough that the breakeven point is way past anything my computer can represent in floating point. So for practical use, this is what I'd put forward.
function term = fibonacci(index)
root = sqrt(5);
term = (((1 + root)/2).^index - ((1 - root)/2).^index) ./ root; end;
function number = factorial(argument)
number = ones(size(argument));
for factor = 1:max(argument)
number(argument >= factor) *= factor; end; end;
Kind of feels like I am phoning this one in, but I really liked this puzzle and hadn't really seen it before. However, if you have already seen this (and it is likely that you have) feel free to just ignore it.
Given a short message containing no more than 10 unique letters
written in the form of a simple equasion. Show the numeric values
for the letters to make it true. No letter that starts a word can
have the value Zero.
For example:
I + LOVE + YOU == DORA
1 + 2784 + 975 == 3760
Here is an interesting that is true in both cases, and will have only one solution:
ONE + TWO + FIVE + NINE + ELEVEN + TWELVE + FIFTY = NINETY
@Koreko well, this hashtag is for the programming challenges that I have been posting. I guess you could try to solve them in bash script if you like. Click on it and scroll all the way down to the beginning, and try the Ninety-Nine bottles one see if you can do it in Bash Script... That would be interesting. :)
La Fée Verte
boosted
Still trying to find a style i like for the double L... Thoughts on this one?
La Fée Verte
boosted
Recursion is almost never the correct solution for any problem in a language that doesn't support tail call optimization.
La Fée Verte
boosted
Octave solution
Should be fairly easy to modify for other bracket types, just add them to the lists at the top and bottom (using the same index for an open bracket and its corresponding close bracket).
function valid = is_balanced(string)
open_brackets = '({[';
close_brackets = ')}]';
expecting = false(size(string));
valid = true;
expected_bracket = [];
for index = 1:numel(string)
if ismember(string(index), open_brackets)
expecting(index) = true;
expected_bracket = strchr(open_brackets, string(index));
elseif ismember(string(index), close_brackets)
if expected_bracket == strchr(close_brackets, string(index))
expecting(find(expecting, 1, 'last')) = false;
expected_bracket = strchr(open_brackets, string(find(expecting, 1, 'last')));
else
valid = false; end; end; end;
if any(expecting)
valid = false; end; end;
