minecraft-java-editionminecraft-redstone
I'd like the output signal to be half the input signal strength.
I was able to use the 'lookup table' approach, which can convert pretty much any input to any output, but is bulky, noisy, somewhat slow, and quite expensive.
Is there a neater, more compact, or faster way to convert input signal strength 0-15 into range 0-8 (or similar, small off-by-one changes allowed, e.g. 0,1->0; 14,15->7).
Bonus: divide by a different number?
The words compact and Redstone don't usually go together :) There could be something simpler than this, probably using some weird combination of water, pistons, and chewing gum, but this is the best I can come up with: a basic delay circuit (from the wiki):
In the screenshot, there is actually a pressure plate on the right, bit hard to see. I prefer not to use buttons, when running for my life. :)
I stumbled on this trying to figure out how to do this, and I found out NONE OF THEM WORK! So I figured it out instead. Place a repeating redstone curcuit going into a command block, and type in the command block this:
/effect @p[r=(number)] 13 1 0
with a comparotor coming out of it.
Then, hook up whatever you want to the comparotor! Now, here's how to know what number to put: the number you put = how many blocks away a player has to be before the redstone comparotor turns on. Also, if you put "m" after r, so "rm" it will work only if the player has LEFT the number of blocks put.
Don't you use ANY () in your command! JUST []
Best Answer
What about this?
When you make your input two units stronger, your output increases by one.
Edit:
This is rather slow, because it takes up to 8 ticks (one for the last repeater plus one for each of the following comparators) to update the output.
Inspired by Fabian Rölling's faster (only one tick) solution, I found out that lining up the repeater's output to adjacent blocks leads to an even more compact solution for the division by two:
Note that the redstone signal can climb up slabs but not down!
Each repeater is reached in two steps more compared to the previous repeater. The last repeater reached is crucial for the result. Because the output blocks of the repeaters are adjacent, the signal's power is reduced by one for each unused repeater passed.
From this pattern, we can derive solutions for a division by any number, for example by three:
Please note that the input to the repeater can come from behind the adjacent solid block. Here each repeater is reached in three steps more compared to the previous one.
The division by five turns out to be very simple: