Xerophyte posted:

Looking at issues 17497, 118599 and 28724 in their tracker: this seems to not currently be possible, but going by the discussion in 28724 it's a candidate for maybe 14.10.

Some people in one of the threads worked around it by making sure that each individual executors had a unique tag and then requiring the same tag for each stage of the pipeline to ensure it all ran on the same machine. I don't know Gitlab's build system but presumably that stops you from actually distributing your builds across multiple machines, so sounds like the very bad type of workaround.

Gin_Rummy posted:

Is there a way to layer an image or a div on top of another one as a child without disrupting the contents of the parent div? I basically have groups of square divs with centered spans of text inside of them, and occasionally I am using JS to create an image on top of these divs. The CSS seems pretty inconsistent at placing the image where I want it, and on top of that, it always shoves the text out of alignment. Really, it would be perfect if I could just drop my image on top as its own layer, but is that doable?

pokeyman posted:

position:relative the div then position:absolute the image?

Batterypowered7 posted:

I've been out of college for a while now but decided to take some graduate level courses because my employer is covering 100% of the costs. I'm taking an algorithms course and I'm ashamed to admit that it's kicking my rear end way harder than I expected it to. Would this be the place to ask for some help with an assignment question or is that sort of thing frowned upon? I suspect the solution is much easier than I expect but I'm honestly stumped and could use some help.

Batterypowered7 posted:

I've been out of college for a while now but decided to take some graduate level courses because my employer is covering 100% of the costs. I'm taking an algorithms course and I'm ashamed to admit that it's kicking my rear end way harder than I expected it to. Would this be the place to ask for some help with an assignment question or is that sort of thing frowned upon? I suspect the solution is much easier than I expect but I'm honestly stumped and could use some help.

Volmarias posted:

pokeyman posted:

It's good to be upfront about it being schoolwork so people can tailor their answers, but yeah fire away!

ultrafilter posted:

Tell us what you've tried and we can give more specific hints.

Boba Pearl posted:

I have a website and it's garbage, and every second writing html, css, php, or javascript is bringing me an inch closer to ventilating my own skull. I thought wordpress would make things easier, but instead it makes things juuuust easy enough to create something that looks like poo poo.

https://bobapearlessence.com/
https://bobapearlescence.com/

At the bottom of this page is a "series" box that I pay 200$ a year for, because I don't know how to do what I need to do (write a script that generates first, next, previous, last) buttons.

https://bobapearlessence.com/42-2/

Whenever I use the thing, it makes the buttons line up like this:



I want them to go left to right and be buttons. The thing is, I don't know which part of the code controls that, and when I e-mailed the company I paid for the plugin, they just said it was planned in a future update.

Honestly, I don't want to pay this company every year, 200$, to put buttons on my web comic. I don't know where to go to teach me what I need to know.

Can you point me to where I could find the information I need?

Alternatively, how much do you think it would cost just to pay someone one off to make me the code so I don't have to pay this company every year, 200$? I'm planning to do comics for the next few years, and like, next year I'll have given them 400$, and there has to be a point where I break even just paying a guy to set it up. Or teach me tbh.

Jabor posted:

The problem here explicitly tells you to use dynamic programming, so that's probably where you want to start. Dynamic programming is a specific algorithms technique where you essentially build a table encompassing the entire solution space for a subproblem, use the information in that table to make a new table encompassing the entire solution space for a slightly larger subproblem, and so on until you've built up to and solved the original problem. Keep that table of the entire solution space in mind when you're considering the problem - in a schoolwork setting like this, the reason they're mentioning it is because it's key to solving the problem.

The key here is all about how you break it down into subproblems. Crucially, you want to break it down in a way that having the entire solution space for a subproblem makes it easier (ideally, nearly trivial) to solve the full problem.

So, let's look at the various elements of our problem and see which ones we could conceivably make smaller:

N - While we could make the number of cities smaller and then solve the reduced problem, it's difficult to see how that smaller solution would be useful for solving the full problem.
u, v - While we could change the cities we're looking at, doing so doesn't really make the problem "smaller" in any way. Having this solution also isn't going to help us solve the original problem.
minDays - We're out of other things to try, so hopefully this is it. Can you see how having a table of solutions for (minDays - 1) could be used to solve the full problem?

KillHour posted:

I'm going to spoil this because it gives away the answer but I want to ask about it. Don't look, Batterypowered7


I'm no high-falutin' degree haver, but isn't this just a really complicated way to say "breadth-first search"? This seems like a really simple problem with a well-established answer camouflaged by a bunch of 50 cent words.

Jabor posted:

Not quite - the critical difference with dynamic programming is that the solutions to one subproblem are used to solve multiple subproblems in the next layer up. So once you've solved a subproblem once, you have to keep the answer around to avoid recomputing it the next time it's needed.

You can use recursion-with-memoization instead of explicitly building the table, but that memoization table is effectively the same table you'd have been building had you done it explicitly.

KillHour posted:

I would get an F on that project because my brain would insist that this is obviously a graph theory problem so let me use graph theory to solve it, damnit! Yes, I realize the point is to teach the technique and not actually care about the specific answer. I don't do well in school. Thinking about it, though, the computational complexity should be the same - all you're doing with the table is flattening the graph every iteration, unless I'm missing some nuance there.

Jabor posted:

There's a meet-in-the-middle trick you can do with the dynamic programming solution to replace a linear factor with a log one, which I don't see how you'd accomplish with BFS. That's only something to look at after you've got your head around the basic solution though.

Jabor posted:

Possibly you can start by computing the absolute cheapest way from each A to B regardless of days? Perhaps even using BFS for that, and logging each step into the same table that you'll use for the dynamic programming stage that takes into account the minimum days requirement so you don't need to repeat that work.

KillHour posted:

You saying that made me realize the difference - you're computing the cheapest option to get from everywhere to everywhere else in n days, not just starting from the place you are currently. So if you need to get from A to B in 4 days, by day 2, you can add up every combination of A => x and x => B to figure out the cheapest overall, and then that generalizes out to, as you said, log time. Here's the one thing I don't think works with that, though, but maybe you have the answer - what if the cheapest amount possible actually takes 5 days?

rjmccall posted:

What's the longest such a trip could take?

KillHour posted:

Holy poo poo you're getting ripped off. I'm not a web / front end developer, but this is for sure something you could either pay someone a modest amount of money to do or figure out yourself in a few days of futzing with it.

First, making a button: The way to make a link look like a button is to stick in an anchor tag with CSS styling made to look like a button. Here's a website with some examples of cool looking buttons that are links: https://fdossena.com/index.php?p=html5cool/buttons/i.frag

I spent 15 minutes futzing around with CSS in the browser to give an example of what you could do just with styling (and your style sheets could probably use a once-over by a developer because they're a bit of a mess, no offense).



Now for your expensive plugin problem. The "hard" part is writing a small piece of JS to figure out what the correct link is and replace the href field (the thing that tells the link where to go) with the correct destination. This isn't actually hard but I can't tell you how to do it without some digging because it depends on how your site is laid out.

Boba Pearl posted:

So this does exactly what I want, but I don't know where to put the code from Stack Exchange, or how exactly it works, would I put it in the single post file in the theme?

https://wordpress.stackexchange.com/questions/149826/display-posts-from-the-same-category-using-next-previous-post-link

Wait actually I found something here:

https://developer.wordpress.org/reference/functions/next_post_link/

This would do it right? Where do I put that code? In the post template?

Then I can use this to get the first and last link, and use that to make the First and Last buttons, then I can just line them up across the page, and just rip out the plugin I was using for this.

https://wordpress.stackexchange.com/questions/363822/how-can-you-get-first-post-last-post-and-post-count-in-a-category

Doom Mathematic posted:

I think "Design a Divide and Conquer algorithm" is a massive mislead because you can do this in O(N) in a single pass over the ballots with no division or conquering (and constant storage).

Xerophyte posted:

Possibly I'm just saying that because I definitely did not arrive at that algorithm.

Batterypowered7 posted:

The TAs specified that you'd receive no points if you didn't use a Divide and Conquer approach, so this is what I arrived at (using a little bit of the idea of the clever algorithm):

code:

function wrapper(N):

	//prepare our container of voters 1 through n
	container
	for i = 1 to N
		insert i into container

	return getMaxVote(container)[1]

//list, array, whatever
function getMaxVote(container):

	//if the container only has one element
	if len(container) == 1:
		//that voter's candidate gets one vote
		return [container[1], 1]

	k = floor(len(container)/2)

	left = getMaxVote(container[1:k]) //slice of 1 through k inclusive
	right = getMaxVote(container[k+1:len(container)]) //slice of k through n

	//if both left and right voted for the same person
	if Agree(left[1], right[1]):
		//return the voter on the left and sum their votes
		return [left[1], left[2] + right[2]]

	if !Agree(left[1], right[1]):
		//if left's candidate has more votes
		if left[2] > right[2]:
			return left
		
		//if right's candidate has more votes
		else if left[2] < right[2]:
			return right

		//if they have the same number of votes, just return left with one fewer votes
		else:
			return [left[1], left[2] - 1]

I tested the above in Python with a variety of inputs and it seems to work. Running time is O{n log n).

I also found a solution for a similar problem that might do it in O(n) time, but I wasn't confident when running it by hand, so I don't plan to submit it as my answer:

code:

GetMajorityElementLinear(a[1...n])
	//Input: Array a of objects
	//Output: Majority element of a

	if n = 2:
		if a[1] = a[2] return a[1]
	else return NO-MAJORITY-ELEMENT

	Create a temporary array temp
	for i = 1 to n:
		if a[i] = a[i+1]:
			Insert a[i] into temp
		i = i+1

	return GetMajorityElementLinear(temp)

The step that reads "if a = a[i + 1]" would be "if Agree(a,a[i+1]". What I was unsure about was that "i = i + 1" step. Is the person that wrote the algorithm just advancing i by one, or are they advancing it by an [i]additional one (since the for loop is guaranteed to advance it by one already), so that we're comparing 1 to 2, 3 to 4, etc? I [i]think it's the latter, but then it only really works when n is a power of 2, since as soon as n is odd we miss a comparison with the nth member. If we're not skipping any indices, then it just feels about as bad as doing two nested for loops (even though n decreases with each recursion).

Batterypowered7 posted:

Also, the implementation I wasn't quite sure about had a "sanity check" function to make sure whoever you returned did indeed have n/2 + 1 votes or more, but the problem specifically states that one candidate got that many votes, so I omitted the step where we check. From what I've seen on the discussion boards, most people who have solved it have done so with n log n time complexity, so that's what I'm going to settle with. I might ask the TAs independently about this other solution after the assignment period is over.