Plate pothos xenoblade 3

This post assumes you are somewhat familiar with self-correcting splitters, as well as having all the upgrades.
*Disclaimer: the information in this post is probably very jumbled, because the 4 efficiencies are heavily interdependent. Furthermore, this post is less specifically about being a guide and more an explanation of how to think about efficiency.
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While playing the game, you may have asked yourself what exactly makes a line more efficient. The general lines are pretty intuitive: use less rollers, have less empty space, make designs somewhat square, use less starters, etc. But there are more imperative ways to determine where exactly to improve a line.
Let's start with the basics: which machines are efficient? Essentially,
(robotic arms aren't even on the list)
What makes a machine efficient is its contribution to creating the item you are crafting. To make a circuit you need 2 circuits and 1 gold. Starters get you those. You also need to make wires. Wire drawers do that. You also need to craft the circuit, hence you need a crafter. So those are necessary machines, and are therefore efficient.
Splitters are slightly more complicated. Essentially, they are only really efficient when splitting basic resources, as they basically become extra starters. In most places you only need one resource, so instead of having two starters outputting 1 copper you use 1 starter and 1 splitter.
Following this logic, triple splitters are the most efficient. Instead of having 3 starters outputting 1, you can have 1 starter and 1 triple splitter, thereby saving a square. The obvious drawback is that the space the triple splitter creates is often hard to use.
Comparison of space required to make 3 diamond wires, regardless of positioning and starter efficiency. The triple splitter design only uses 5 squares instead of 6
But if you have, say, 2 circuits, and are using a splitter to split them into two different crafters, it would be more efficient, if possible, to link the crafters directly to the circuit crafters, and not have to "combine" the two circuits and split them again, thus skipping the splitter.
Rollers obviously do not advance the crafting, and neither do splitters, despite them making things much easier, although on some occasions, explained later, splitters can improve efficiency. Though some designs may require them, transporters are essentially a higher dimension roller, except each one takes 2 squares (an input and an output), so are actually worse for efficiency, in a void.
I'll talk more about sellers later with output efficiency. And understanding timed roller efficiency requires understanding how to use them, which is enough of a topic for its own post, but in general, when splitting basic resources (same reasoning as for splitters), they vary between being slightly worse than regular splitters to slightly better than triple splitters, based on context.
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Space efficiency is, first of all, dependent on machine efficiency. If you're using a lot of rollers, selectors or inefficient splitters that's usually where you should look. But there's also a limit to how much you can make those; it is usually impossible to only have efficient machines. Instead, when you get stuck while optimizing, you should take a look at the number of different space inputs the crafter needs.
Take, for example, the circuit. The crafter has 4 sides, and one side is usually reserved for the output. That leaves 3 sides that you can send items in. Assuming you have no inefficient or inconsistent machines (see self-correcting splitter post for inconsistency), generally speaking, each side can only receive 1 item. You cannot receive 2 wires from 1 side because that would either require you to have "combined" 2 wires by means of, say, a roller, or you have some inconsistent wire splitters.
Cases described above. For efficiency, a circuit generally needs 3 sides for its inputs and 1 for its output
Now why would we care about how many sides a crafter uses? This ties into output efficiency as well. Consider the following design:
Circuit line, 4/s
Starter efficiency aside, this design is okay-ish. The only "wasted efficiency" is the 3 rollers. But the two holes on the side aren't always easy to work with either. The problem is that it is impossible to get rid of these things in this design without compromising on other things. One such compromise would be to add an output, and that works on some designs, but in this one, the two middle squares are still wasted.
4 circuits/s in 2 outputs
One of the possible improvements is using the output side as input. The way to do that is, of course, pointing the crafter into a selector or self-correcting splitter. Generally speaking, you can only self-correct 1 type of item, but selectors can send multiple different items into the same crafter, resulting in it only needing to use 1 side. If said different items are collected together not with inefficient machines but self-correcting splitters, selectors can actually become efficient this way.
Regardless of other efficiencies, this example illustrates the fact that the circuit crafter is only \"using\" one of its 4 sides
There's another exception: if a self-corrected resource is needed in both a first and second stage of a craft, notably in solar panels (1 circuit, 2 gold, 1 diamond):
This type of thing make solar panels surprisingly efficient and fun to design lines for. The diamonds can be in self-correcting splitters, too.
But let me talk more about the term "collecting" from the previous paragraph, which I'd assume most people wouldn't immediately grasp. To understand this, let's talk about one of my more recent designs (in action here), which I've said to be essentially a theoretical maximum:
4.5 server racks/s, single output, theoretical maximum?
This is a very obvious case of "counting sides". Of course, I'd like to be able to connect all the 10 aluminum and 20 plates required directly to the server rack crafter and skip all the rollers. Unfortunately, the game is restricted in 2D; only 4 sides per square. I'd need 4D Assembly Line to make it work, where a hypercube would have 32 sides (edges).
What would then be the least amount of rollers I'd need? We know from before that each plate requires 1 side, so there's 20. We can also figure that the 10 aluminum would need 4 sides at least (3x starter with 3 aluminum + 1x something giving 1 aluminum). I also probably don't want to put any self-correcting splitter shenanigans because of how many items are flowing, and selectors don't seem to immediately improve anything, so I cannot use the output side as any input and have to count 1 extra side. All that sums up to 25.
Then, what would be the most efficient way to make 25? It should be pretty intuitive to figure that adding a straight line of rollers (without bends) would be the best. The direction does not matter. Here's something to illustrate why:
Consider each cutter as a \"useable side\". The two on the left have 10 each, whereas the bended one on the right only has 9
So a straight line of rollers. It's easy to figure out that the crafter starts with 4 sides, and each additional roller adds 2. So to make 25 sides, I'd need 11 rollers (11x2 + 4 = 26 > 25). In other words,
no. sides = 2 * no. rollers + 4
I'd even have an extra side of slack. The rest of the line is just about using triple splitters efficiently, and determining the best way to use the leftover sides; having been put beside machines of separate crafts, each roller no longer provides 2 input sides.
Essentially, the reason I considered that line a theoretical max is because:

• It is not possible to reduce the amount of rollers
• Using selectors by, for example, putting a single aluminum on the line and selecting it into a press later on, actually reduces efficiency since you'd be using 2 sides for 1 plate instead of 1 side/plate
• It is essentially impossible to use self-correcting splitters because of inconsistency and number of items flowing.
• There is already significant use of triple splitters.
• Single output
• It's probable that one or two squares could still be squeezed out, but to make the smallest significant difference (+ 0.05 server rack/s), one would need to add 1 plate/s and 0.5 aluminum/s (essentially 1 aluminum/s) to the line. That would require 1 starter, 1 splitter, 1 press and at least 1 extra roller, and saving 1 starter somewhere (this design uses all of them). This doesn't seem physically possible

* Reminder: selectors can sometimes be efficient because they can significantly reduce the number of sides required in a design. They add the sides like a roller would, and more.
Funnily enough, this seems to be one of the few lines that doesn't benefit from extra outputs/sellers (segue into output efficiency). I normally add an output to my design when I feel like things are getting too crowded, but concretely, it's because theoretically, it adds extra useable sides, and practically, it allows build flexibility, which eventually translates to build efficiency. But it is already inevitable that the half server rack on the right provide a useable path for the other built racks towards the one seller.
Single output vs two outputs side count comparison using the same space: 8 - 10. Conclusion: each additional output grants 2 extra sides, because of s = 2*r + 4 from before
Let us consider the "collecting" from before. If you're gathering items together towards a crafter inefficiently using unnecessary rollers, it stays inefficient. But sometimes, there are surplus sides. Consider the efficient self-correcting circuit line:
3 circuits/s, single output, self-correcting splitters
There's a surplus side on the bottom right, which you could would be using to provide for the rest of the line. In fact, lets discuss circuit making and self-correcting splitters under side counting.
In that post I briefly explained that despite using more space, self-correcting splitters always end up being more efficient, because they automatically gather all outputs to the same point instead of requiring them to be gathered simply. Earlier in this post, I showcased some not great circuit designs and their shortcomings. Then, we've taken a look at the minimum number of sides required to craft a server rack. That concept is basically identical to the way you would decide how to position sellers and how many rollers you would need to gather all the things you craft towards said sellers.
The real reason why self-correcting splitters are so efficient, and so much of an improvement, is that they replace those mandatory rollers required to gather all items with efficient splitters, with no impact on consistency. Consider the design above, and view the three splitters/selector in the middle as the real roller path bringing all circuits together. Furthermore, self-correcting stuff is generally very side-positive; they use less sides than they produce. Remember how straight lines of rollers create more useable sides than bent ones? If we go with a circuit crafter having to use 3 sides as inputs and the fourth as output, then each one would be wasting 2 sides. The best you'd be able to do would be this:
5 circuits/s, single output
This design isn't even that inefficient, actually. It's just that you'll have a very hard time removing those rollers and selector. Comparatively,
6 circuits/s, single output, self-correcting splitters
is essentially the same size and makes 1 more circuit, while being more starter efficient. Which segues into the last topic.
Starter efficiency is normally less of a concern than one may think. Most of the time, if you're lacking starters, it not only means that you've used too many single resource starters, but also that there are significant non starter-related parts of your line you can improve. Use more triple splitters or something. Or find a way to use self-correcting splitters.
When a design feels crowded, consider counting sides. It's pretty much the underlying indicator of building efficiently. And build straight. In that regard, circuits are an exception; building curved also works.
(Too tired to reread and check stuff, if there's anything majorly wrong with information or flow, or something I forgot to explain, please let me know)

Preface, I left sta seven years ago, still visit my folks. Decided to get a drink at tavern, didn’t realize I had to pay to park in an empty lot at 1030 on a Thursday night.

15-1k Free bet.

Top of the First and Harley Winn are the brainchild of a baseball themed, uber flavorful, fun, entire-set-in-2-cards vibe.
The idea is you “step up to the plate” by playing Top of the First on turn one. From there, Harley Winn must “run the bases” to complete a homerun, and thus, complete the quest. The reward is 2 very aggressive tools to close out the game.
Harley “running the bases” must happen in order, and does not all need to occur in the same turn. For example, she can be drawn on Turn 3, and will then stay on “first base” indefinitely. You can then play her on Turn 4, so she will be on “second base” indefinitely.
Keep in mind Harley Winn can’t reach 1st base if she is in your opening hand or mulligan. I added Tradeable to her to avoid this situation and to make reading home possible. I think Tradeable also fits the theme of baseball. She is a clutch player and only shows up precisely when she is needed.
I think the earliest you could theoretically complete the quest is Turn 4 with the coin. She would need to be drawn T1, 2, 3, or 4. You could play her (without the combo), Shadowstep, coin, then trade her into the deck, completing the homerun, all on turn 4. Adding her combo effect reduces the tempo loss if she is played later on.
Also keep in mind that if Harley dies on board, the quest isn’t necessarily over. If she is discovered or resurrected, then bounced to the hand, she is then on 3rd.
Lucille and the Heater are just high tempo, high damage finishing options. Altogether, they represent a whopping 52 damage in total if all 4 weapon hits go face and all 5 echos are played on an empty board. A fair reward for such a crowning achievement!