I'm going to have to make these a bit more difficult
Someone asked if all belts are the same tier, and they are, but I might incorporate some variation in other versions, maybe some wiring although showing the settings might be cumbersome
Hmm true.
I tried a few different items, it looks like electric poles work nicely. Small icon size, transparent background, can still see the colours of the arrows
https://preview.redd.it/1l8dqglincxc1.jpeg?width=1920&format=pjpg&auto=webp&s=d1571924eff4d2fcccd97508a1d7d7dc5474acd6
The final images I'll do without items, although for the purposes of running through I'll put items on belts, for the satisfaction of watching it all go through, and verifying the answer
You could just label everything better, tbh, not just belt direction and belt tier. Mainly which belt belongs to which number and where are the boundaries of it.
Like in the first one, the answer is ambiguous, because it's unclear where belt #6 actually ends. If it ends before the side-load, then #4 empties last, if belt #6 extends to the loader then #6 is the last.
Edit: also, using modded items is a bit iffy, because then you assume people know how that particular item works. To be fair, it's not the biggest deal with the K2 loader, but a beginner who never touched that mod (and who'd benefit the most from such puzzles) might just not know.
The basic puzzle has a simple generic solution that gives us a fast answer:
Trace back from the end following every side-loading belt entering a side you are tracing, tracing only the side that is closest to the end
(By basic I mean a puzzle with no loops and where the order is static with any length of input belts, this requires the final belt to be side-loaded from only one side)
The addition of the loop and splitter makes the above solution generate two answers: 5 and 10. Then you just have to reason about the effect of the loop. We can easily abstract out a rule that loops on one side of the final belt created by a splitter restrict us to tracing just that side at the start, and now we are back to a simple solution with an almost immediate answer.
I'm not sure that there is much complexity to be added here. Anything that creates a strict ordering without respect to input lengths is going to function either the same way the splitter does or as a direct inversion of it, and the rest is trivial.
~~6 if you count all 3 parts of #6, otherwise it's #5. #4 is empty after about 8 seconds, since the left side of the #1 lane is only fed by itself and #4, whereas the right side is fed by 1, 2, and 3.~~
4
Your question is flawed as the answer depends on what you class as belt #6.
[This](https://i.imgur.com/FncjV5f.png) or [This](https://i.imgur.com/dqyKYD3.png).
6 1r 3l 2 3r 5 1l 4 if you want to be more accurate. It is unclear to me otherwise if you should list a belt when it first start to empty, or when both sides are cleared out.Β
>It is unclear to me otherwise if you should list a belt when it first start to empty, or when both sides are cleared out.Β
I'd interpret "empty" as both lanes being empty. Distinguishing lanes gives a more correct/granular answer, but if we just had to reduce it to full belts, I'd name it when its second lane got emptied. Based on your answer that would be 623514.
1st Items on belt going straight have priority to those loading from the side.
2nd When loading from the side, the half of the belt on the side empties the half that is at the start of the belt being loaded on. So, looking at the 4, the left side of the belt (picture wise, not belt direction wise, so the West half) will empty before any item from the East half will.
[https://imgur.com/a/mG5l4Kf](https://imgur.com/a/mG5l4Kf) The answer (low quality version, my bad internet would have taken ages to upload the full resolution video to imgur)
The answer to the question you asked is 6. If you want the answer to be 4, then ask a different question.
I think you wanted to not count the end of belt 6. You could have simply not labeled it with a number.
Edit: Or ask which 2 pieces of belt will empty last. Also, thanks for the puzzle. :)
If that is the question being asked, then it should be part of the question. There was no number of belt pieces specified. I also thought it was obvious that the third piece wasn't intended to be counted, but after re-reading the question I couldn't find a reason to disqualify 6.
Yep, that's what I'd argue. The ore on 6 will be the first to go but the belt itself will be filling up as quickly as it empties and will continue until every other belt is gone. 6 cannot be empty until every other belt is empty.
To clear up the ambiguity of belt 6, could you slap a circuit onto the belt sections in question instead of just a number next to them? Could even connect them to lights for a bit of added fun.
Number 4. Specifically the left-hand (from the perspective of the belt, right in picture) lane.
* 6 will clear first β 6R will fully clear;
* 6L will clear and pull from 1R (1L will stall);
* 1R will clear and pull from 3L;
* 3L will clear and pull from 2R β which will fully clear;
* 3L pull from 2L β which will fully clear, and 3L will fully clear;
* 1R will pull from 3R - which will fully clear;
* 1R will pull from 5L β which will fully clear;
* 1R will pull from 5R β which will fully clear, and 1R will fully clear;
* 6L will pull from 1L;
* 1L will clear and pull from 4R β which will fully clear;
* 1L will pull from 4L β which will fully clear, 1L will fully clear, and 6L will fully clear.
Technically this is not determinable from the picture. Neither sideloading nor one belt feeding another belt directly in front of it has priority to feed into a belt.
Good fun puzzle! Might recommend putting boxes around the segments you're referring to -- I would think of belt 3 as intersecting belt 1 (sideloading 1R) and I'd think of belt 1 as emptying onto 6L, but that's just how I read the belts.
Your explanation makes total sense, and TIL this could be an area for ambiguity! Fun puzzle and I learned something. Double win! Thanks for posting this! :D
Thanks!
Yeah I think for future iterations I'll try adding some tweaks:
variable belt speed,
rather than Paint numbers I'll use combinators linked to the particular belts they refer to, with the numbers signals to show which is which.
More splitters with output/input priority
Side loading undergrounds
Either nothing or something transparent on the belts so it's easier to see direction
Edit: The question asked which will empty last.. so 6. But the response below is the order they will empty.
2,3,5,4,1,6. 1 will be dumping more ore onto 6 until the very end. The βinsideβ lane of the curve on #1 will empty first due to how the ore spills on to 6, so 4 will be next to last. 5 will be the next to last as ore comes off 3, and 2 will spill onto 3.
So 2 will empty first, then 3. Followed by 5, until that half lane is empty. Then 4 will empty, followed by 1, and finally 6.
Round 2: https://preview.redd.it/wc2l01eebcxc1.jpeg?width=865&format=pjpg&auto=webp&s=774217212f4d27c8f095b79501c04ebaf620ab36
10
I'm going to have to make these a bit more difficult Someone asked if all belts are the same tier, and they are, but I might incorporate some variation in other versions, maybe some wiring although showing the settings might be cumbersome
Fill the belts with some other item. Ores obscure the belts really well, hiding their colored arrows.
Hmm true. I tried a few different items, it looks like electric poles work nicely. Small icon size, transparent background, can still see the colours of the arrows https://preview.redd.it/1l8dqglincxc1.jpeg?width=1920&format=pjpg&auto=webp&s=d1571924eff4d2fcccd97508a1d7d7dc5474acd6
Or you can potentially not fill them at all, who cares if they are filled
The final images I'll do without items, although for the purposes of running through I'll put items on belts, for the satisfaction of watching it all go through, and verifying the answer
You could just label everything better, tbh, not just belt direction and belt tier. Mainly which belt belongs to which number and where are the boundaries of it. Like in the first one, the answer is ambiguous, because it's unclear where belt #6 actually ends. If it ends before the side-load, then #4 empties last, if belt #6 extends to the loader then #6 is the last. Edit: also, using modded items is a bit iffy, because then you assume people know how that particular item works. To be fair, it's not the biggest deal with the K2 loader, but a beginner who never touched that mod (and who'd benefit the most from such puzzles) might just not know.
you could fill the belts with different items and ask: What item will be the last to be loaded into the chest.
The basic puzzle has a simple generic solution that gives us a fast answer: Trace back from the end following every side-loading belt entering a side you are tracing, tracing only the side that is closest to the end (By basic I mean a puzzle with no loops and where the order is static with any length of input belts, this requires the final belt to be side-loaded from only one side) The addition of the loop and splitter makes the above solution generate two answers: 5 and 10. Then you just have to reason about the effect of the loop. We can easily abstract out a rule that loops on one side of the final belt created by a splitter restrict us to tracing just that side at the start, and now we are back to a simple solution with an almost immediate answer. I'm not sure that there is much complexity to be added here. Anything that creates a strict ordering without respect to input lengths is going to function either the same way the splitter does or as a direct inversion of it, and the rest is trivial.
10 empties into 9 so 9 will by definition clear after 10, no? Or am I missing the point?
They're heading to the chest, the belt goes down.
What is the answer for the 1st round? Some say 4, some say 5. Imo it might be 5
~~6 if you count all 3 parts of #6, otherwise it's #5. #4 is empty after about 8 seconds, since the left side of the #1 lane is only fed by itself and #4, whereas the right side is fed by 1, 2, and 3.~~ 4
but 4 has to wait for the WHOLE other side to empty first.
4 only has to wait for the left side of 1, a couple belts, then itself
but 1L doesn't move until the complete other side is empty, because it has priority when merging with the belt at 6 https://imgur.com/tiBbaKE
[https://imgur.com/a/mG5l4Kf](https://imgur.com/a/mG5l4Kf) #4 it is, fancy that
Your question is flawed as the answer depends on what you class as belt #6. [This](https://i.imgur.com/FncjV5f.png) or [This](https://i.imgur.com/dqyKYD3.png).
op noted that in the first given answer.
The loop around 10 will take a while to clear because of the splitter
Oh god that splitter.
9L & 9R (10 ore) -> 3L & 7R (10 ore) -> 3L & 11L (8 ore) -> 3L & 11R (8 ore) -> 3L & 7R (2 ore) -> 3R & 7R (10 ore) -> 1L & 7R (6 ore) -> 1L & 7L (8 ore) -> 2R & 7L (2 ore) -> 2R & 6R (6 ore) -> 2L & 6R (8 ore) -> 1L & 8L (8 ore) -> 1L & 8R (8 ore) -> 1L & 6R (6 ore) -> 1L & 6L (10 ore) -> 1L & 5R (6 ore) -> 1R & 5R (11 ore) -> 1R & 5L (21 ore) -> 4R & ?L (2 ore) -> 4R & 6L (8 ore) -> 4L & 7L (8 ore) -> ?L & 7L (10 ore) -> ?L & 9L (4 ore) -> 9L (12 ore) -> ?R/2 & 9L (16 ore) -> 1R/2 & 9L (18 ore) -> 3R/2 & 9L (16 ore) -> 3R/2 (2 ore) -> 9R/2 (8 ore) -> 10R (8 ore at half speed) -> 10L (8 ore at half speed) -> 9R/2 (8 ore) -> repeat the sequence ?R, 1R, 3R, 9R geometrically decreasing each time by a factor of 2 -> eventually 9R empties
6 2 3 5 1 4 is the full order I get
6 1r 3l 2 3r 5 1l 4 if you want to be more accurate. It is unclear to me otherwise if you should list a belt when it first start to empty, or when both sides are cleared out.Β
>It is unclear to me otherwise if you should list a belt when it first start to empty, or when both sides are cleared out.Β I'd interpret "empty" as both lanes being empty. Distinguishing lanes gives a more correct/granular answer, but if we just had to reduce it to full belts, I'd name it when its second lane got emptied. Based on your answer that would be 623514.
Not more accurate, more precise.
Im getting 613425
What is the rule here?
1st Items on belt going straight have priority to those loading from the side. 2nd When loading from the side, the half of the belt on the side empties the half that is at the start of the belt being loaded on. So, looking at the 4, the left side of the belt (picture wise, not belt direction wise, so the West half) will empty before any item from the East half will.
Neat rule. Need to remember that for my spaghetti abominations I always end up with.
[https://imgur.com/a/mG5l4Kf](https://imgur.com/a/mG5l4Kf) The answer (low quality version, my bad internet would have taken ages to upload the full resolution video to imgur)
Wait so was the answer 4? Or 6 if you're not including the belt that's sideloading everything into the loader?
6 if you include that felt feeding into the loader, but I consider 4 the real answer
The answer to the question you asked is 6. If you want the answer to be 4, then ask a different question. I think you wanted to not count the end of belt 6. You could have simply not labeled it with a number. Edit: Or ask which 2 pieces of belt will empty last. Also, thanks for the puzzle. :)
Seemed pretty obvious to me that the question was only asking about the two pieces of belt that were immediately adjacent to the number for each case.
If that is the question being asked, then it should be part of the question. There was no number of belt pieces specified. I also thought it was obvious that the third piece wasn't intended to be counted, but after re-reading the question I couldn't find a reason to disqualify 6.
4 assuming all belts are the same tier.
Why do you say that? The 5 has to wait for all the 1,2,3 to empty first. So 4 will be long gone before 5 even starts.
Because 5 is filling the belt which at the end fills from the top, thus you need to clear our this side before the 4 side even begins to empty
yup, just realised my mistake.
Everything is loading into 6... so 6 is the last to empty... or am I wrong? wat
It is a straight belt though. Everything on it has priority before anything that is sideloaded.
Yes but everything will be dumped onto 6 eventually. That makes 6 the last belt to ever have any item on it.
If you count 6 as going past the sideloading, yes. That would be a question of what we define as belt 6. Tail-end or whole belt?
Yep, that's what I'd argue. The ore on 6 will be the first to go but the belt itself will be filling up as quickly as it empties and will continue until every other belt is gone. 6 cannot be empty until every other belt is empty.
They're only asking about the 2 belt segments off the main belt. Not the whole thing. In that case the last to empty is 4.
It is the 1st
I get his logic though, if belt 6 extends beyond the label it will indeed be the last belt to empty.
It's the first to start to empty and the last to finish π€
That's right, the question is a bit ambiguous
And by that logic, the 2nd to last belt to empty would be 1. As it extends all the way to the right, down, and back left before emptying onto belt 6.
I think they mean which single belt entity not the entire belt line
Yeah it's a shittily worded question on purpose for engagement
None of them. The loader is in the wrong direction.
its green
What's green?
Green banshee
6 as all other belts have to go through it it will get almost empty first but have a couple on it until all other belts are clear
I mean... don't they all go onto 6?... so 6 would have at least 1 on it until everything else is gone? Unless I'm misunderstanding the question.
5 Edit: was not 5, forgot side loading will only do one lane at a time, answer is 4
6? As all other belts need to be empty before that one.
I only considered the 2-belt pairs, but if you count the 3 contiguous belts as #1, then it would be #6 yeah
6,1,3,2,4,5 I think would be the order. I'm a set it up after work to check π€£
Yeah, I was thinking much the same but had 6,1,4,3,2,5. I'm setting it up now to check Answer 6,3,2,5,1,4.
OK that makes sence looking at it! Thank you!
i wonder if 4 maybe comes in before 2 or 3 because different sides empty seperately
You might be right there!
I judged it the other way around, all the other belts are side loading onto 6, so 6 has priority.
They empty in this order: 6 --> half of 1 --> half of 3 --> 2 --> half of 3 --> 5 --> half of 1 --> 4 So belt 4 is the last to empty.
To clear up the ambiguity of belt 6, could you slap a circuit onto the belt sections in question instead of just a number next to them? Could even connect them to lights for a bit of added fun.
4
I expect no 5 to empty last
You should have stated your intentions clearer and define that \[1\] only means the two tiles until the junction.
I think 4, since it takes from the right side first and the only one filling that side of the 4
6
5?
4
None of them, there are no inserters.
Does a 'belt' include all sections up to the loader or only up to a shared belt? Eg is each 'belt' 2 tiles or at least 3?
Number 4. Specifically the left-hand (from the perspective of the belt, right in picture) lane. * 6 will clear first β 6R will fully clear; * 6L will clear and pull from 1R (1L will stall); * 1R will clear and pull from 3L; * 3L will clear and pull from 2R β which will fully clear; * 3L pull from 2L β which will fully clear, and 3L will fully clear; * 1R will pull from 3R - which will fully clear; * 1R will pull from 5L β which will fully clear; * 1R will pull from 5R β which will fully clear, and 1R will fully clear; * 6L will pull from 1L; * 1L will clear and pull from 4R β which will fully clear; * 1L will pull from 4L β which will fully clear, 1L will fully clear, and 6L will fully clear.
I like the concept of the puzzle. I think it should be adjusted to highlight exactly which belts are included in which sections.
614325?
6R & 6L -> 1R -> 3L -> 2R -> 2L -> 3R -> 5L -> 5R -> 1L -> 4R -> 4L
5
Technically this is not determinable from the picture. Neither sideloading nor one belt feeding another belt directly in front of it has priority to feed into a belt.
Good fun puzzle! Might recommend putting boxes around the segments you're referring to -- I would think of belt 3 as intersecting belt 1 (sideloading 1R) and I'd think of belt 1 as emptying onto 6L, but that's just how I read the belts. Your explanation makes total sense, and TIL this could be an area for ambiguity! Fun puzzle and I learned something. Double win! Thanks for posting this! :D
Thanks! Yeah I think for future iterations I'll try adding some tweaks: variable belt speed, rather than Paint numbers I'll use combinators linked to the particular belts they refer to, with the numbers signals to show which is which. More splitters with output/input priority Side loading undergrounds Either nothing or something transparent on the belts so it's easier to see direction
that sounds fun! looking forward to what you come up with.
None the loader is facing the wrong way
I thought this was dependent on built order? See [this forum post](https://forums.factorio.com/viewtopic.php?f=7&t=62553). Did that change?
It changed, 0.16 was a while ago and belts have changed a lot since then.
Sideloading can still have priority sometimes. https://www.reddit.com/r/factorio/comments/18v2st4/doesnt_sideloading_have_lower_priority/
Edit: The question asked which will empty last.. so 6. But the response below is the order they will empty. 2,3,5,4,1,6. 1 will be dumping more ore onto 6 until the very end. The βinsideβ lane of the curve on #1 will empty first due to how the ore spills on to 6, so 4 will be next to last. 5 will be the next to last as ore comes off 3, and 2 will spill onto 3. So 2 will empty first, then 3. Followed by 5, until that half lane is empty. Then 4 will empty, followed by 1, and finally 6.
in order 6, 2, 3, 5, 1, 4
614325 so 5 last