Late to the party but I just discovered this series and I had an idea for a twist on this that you might find interesting or useful - you could use this tech to make a "ticket-based" request system to pull items out of storage. Throw in a piece of paper that's been Anvil-named to "Redstone" and get back a stack of Redstone. All you'd need would be a separate decoder that decodes named pieces of paper into the same binary codes you're using for the items themselves. (Your work is amazing by the way - I'm nowhere near at your level, but I've been pondering what the "holy grail" of Minecraft storage would look like for quite a while, and yours is several steps beyond anything I could have imagined!)
@@zlauch2714 i have a pretty crappy fix, but it works. I didn't wanna move the chests or the piston bolt nor i wanted to change the itemsinside each chest, so im sure that by doing that one could come up witha much neater solution, but in the end im prettyproud i managed to fix it in a somewhat compact way
Thank you for the comprehensive video including the link to 0x's, which was also quite helpful and I wouldn't have found otherwise. I agree this is probably the best way to encode items to binary, if you have the patience to set up the chests.
That's insanely cool! I've been working on a project that needs to encode/decode an input to a complex output. While this design won't work for me it definitely gives me inspiration. Great work!
I have a similar concept in designs I'm working on right now. It never occurred to me to use multiple chests to allow more than 50 words of data in a single item ROM, though. Very interesting.
Wow. You can use the binary decoder as a painel to bring items from distance chest to a main chest. The only problem would be to know each different number for each different item.
Daniel Ochi this gives you that number, so you put in the item you want and get the number that corresponds to this item, then can request items using that number from the main storage
I made a contraption that automaticaly adds the items to the correct chests, removes dummy items and skips full chests. it is pipelined, so items can be "encoded" every 30 ticks(11chests x 10b). i then pipelined the decoder as well, so i it works at greater frequency, but it takes a bit longer for each item.
I may sound not so smart with this question, but hey can't learn without a question here and there. What could this be used for exactly or is there another video explaining?
I'm using it in a binary coded storage system. I don't have a video on it yet, because there was too much stuff I needed to invent before even having a prototype. By now the core of it is done, but there's still lots of things I need to finish.
I'm planning to implement this design to use shulker boxes full of items as programs in a redstone computer. This way I can take items and turn them into redstone signals the computer can understand easier.
Hey there What kind of component, that need binary inputs, are you gonna add after this decoder ? (I guess it would be a sorting system with that kind of input, but I'm asking that because I never see such component.)
the complexity of this design is o( 10Root( n ) ). (10Root(n) is the root foor n^10, as 2Root(n) or squareRoot(n) is the root for n^2) UltimateRedstone's 1D storage has an complexity of O(n). 1Root(n) for every positive number equals n. So it can be described as O(1Root(n) 0x53ee71ebe11e's 2D storage has an complexity of O(2Root(n)). now you can see an pattern: (a)D storage complexity = O( (a)Root(n) ) this is logically also correct, since Roots take an (A)Dimensional square/cubed area/surface and return the length of 1 edge. the square area represents the amount of items (n), and the complexity is the amount of chests in one line. a better mathematical notation / drawing of the complexity of your storage: / 10 /----------------| \ O | \ / n | \ \/ /
![[Storage Tech] Self Organizing Bulk Storage](/img/n.gif)
Late to the party but I just discovered this series and I had an idea for a twist on this that you might find interesting or useful - you could use this tech to make a "ticket-based" request system to pull items out of storage. Throw in a piece of paper that's been Anvil-named to "Redstone" and get back a stack of Redstone. All you'd need would be a separate decoder that decodes named pieces of paper into the same binary codes you're using for the items themselves. (Your work is amazing by the way - I'm nowhere near at your level, but I've been pondering what the "holy grail" of Minecraft storage would look like for quite a while, and yours is several steps beyond anything I could have imagined!)
n log(n)
We have a winner!
Isn't it just log(n)? You add six to the length for each doubling of the number of items.
Fwiffo you have to add linearly more chests per bit the more items there are. that s why there is the n term
How did you work this out?
@@n00b_asaurus you too ;)
"Unless mojang breaks redstone" Wouldn't be surprised :P
2 years later, snapshot 20w18a gets released ..
@@duhby indeed D:
1.16: Hello there!
@@zlauch2714 i have a pretty crappy fix, but it works. I didn't wanna move the chests or the piston bolt nor i wanted to change the itemsinside each chest, so im sure that by doing that one could come up witha much neater solution, but in the end im prettyproud i managed to fix it in a somewhat compact way
@@lollopollqo2240 and... how??
Thank you for the comprehensive video including the link to 0x's, which was also quite helpful and I wouldn't have found otherwise. I agree this is probably the best way to encode items to binary, if you have the patience to set up the chests.
How do you encode?
That's insanely cool! I've been working on a project that needs to encode/decode an input to a complex output. While this design won't work for me it definitely gives me inspiration. Great work!
I have a similar concept in designs I'm working on right now. It never occurred to me to use multiple chests to allow more than 50 words of data in a single item ROM, though. Very interesting.
This is pure genius
Wow. You can use the binary decoder as a painel to bring items from distance chest to a main chest. The only problem would be to know each different number for each different item.
Daniel Ochi this gives you that number, so you put in the item you want and get the number that corresponds to this item, then can request items using that number from the main storage
I made a contraption that automaticaly adds the items to the correct chests, removes dummy items and skips full chests. it is pipelined, so items can be "encoded" every 30 ticks(11chests x 10b). i then pipelined the decoder as well, so i it works at greater frequency, but it takes a bit longer for each item.
Watched it the 8th time now and I understud it now
I may sound not so smart with this question, but hey can't learn without a question here and there. What could this be used for exactly or is there another video explaining?
I'm using it in a binary coded storage system. I don't have a video on it yet, because there was too much stuff I needed to invent before even having a prototype. By now the core of it is done, but there's still lots of things I need to finish.
awesome ty for the response. Looking forward to seeing what you come up with down the road.
I'm planning to implement this design to use shulker boxes full of items as programs in a redstone computer. This way I can take items and turn them into redstone signals the computer can understand easier.
Hey there
What kind of component, that need binary inputs, are you gonna add after this decoder ?
(I guess it would be a sorting system with that kind of input, but I'm asking that because I never see such component.)
A storage hall similar to what scicraft has atm, but binary coded. The hall itself is already designed, but I have yet to make a video
o(n/r) is the complexity
I'm impressed
Is this broken?
the complexity of this design is o( 10Root( n ) ).
(10Root(n) is the root foor n^10, as 2Root(n) or squareRoot(n) is the root for n^2)
UltimateRedstone's 1D storage has an complexity of O(n).
1Root(n) for every positive number equals n. So it can be described as O(1Root(n)
0x53ee71ebe11e's 2D storage has an complexity of O(2Root(n)).
now you can see an pattern:
(a)D storage complexity = O( (a)Root(n) )
this is logically also correct, since Roots take an (A)Dimensional square/cubed area/surface and return the length of 1 edge.
the square area represents the amount of items (n), and the complexity is the amount of chests in one line.
a better mathematical notation / drawing of the complexity of your storage:
/ 10 /----------------| \
O | \ / n |
\ \/ /
I'm just gonna say O(ln(n)*n) even though Jendrik was faster.
>got O(n log n) before you even started talking about Big O notation
Wow!
It's o(n^2)
W8 its incorrect
complexity is nLog of n
English please?