trying to make a grid for reactivition where each block is a unique output. since I get a X and Y output, possibly even a Z for rotation, but no idea how to even start with that.

here is a visualization (i didn't fill it in all because I think you get the basic idea): http://localhostr.com/files/jhmQnUl/Screen+shot+2011-11-25+at+10.08.34+AM.jpg

in normal programing language doesn't seem too hard. you set up conditions, if (y1, x1) then [output]. but this would get REALLY messy in pd and seems really inefficient. anyone have suggestions on how to set this up? I basically want a unique output for each of the squares lets say a bang if (x,y) matches. I saw something on tables once, not sure if this is best way to approach this problem.

..the [expr]-object can handle at least easy if-conditions..
..but if the squares are all the same (..in size..), you could quantize the x-y-input (via eg. ), sothat you get the x- & y-sqare-number out.. instead of the x-y-coordinates. Then just compare that "vector" with given ones to bang a desired output...
...hope I got you right...

if (y1, x1) then [output]

[== x] [== y]
| /
| /
[&& ]

make sure you send the (right) y value first

@mod said:

if (y1, x1) then [output]

[== x] [== y]
| /
| /
[&& ]

make sure you send the (right) y value first

this seems like it would be a really messy/repetitive way of getting this done though. it would have 49 variants just for 7 x 7 square. in programing land when I run into that stuff, seems like there is an object that helps organize it.

and even if I do it that way there seems to be a bug. it will match for 2, 2 for example if I do:

[== 2] [== 2]
| /
| /
[&& ]

but if change value of y input and x stays at 2, it still matches and wont throw false output until both values change. seems that it's only when Y stays the same that it will continue to match, if x changes it sends a false or 0.

@lukasz said:

but if change value of y input and x stays at 2, it still matches and wont throw false output until both values change. seems that it's only when Y stays the same that it will continue to match, if x changes it sends a false or 0.

[bondo] works well for getting around that, i find myself using it a lot.

[bondo] works well for getting around that, i find myself using it a lot.

that worked perfect. I'm curious why it is necessary to have bondo? is it that it outputs everything at once instead of individually? documentation doesn't have much on it.

@lukasz said:

there seems to be a bug. it will match for 2, 2 for example if I do:

[== 2] [== 2]
| /
| /
[&& ]

but if change value of y input and x stays at 2, it still matches and wont throw false output until both values change. seems that it's only when Y stays the same that it will continue to match, if x changes it sends a false or 0.

you said this was for reactivision, which i assumed means you will use those marker things, and they will output X,Y,and Z values as a list. Whenever you get an x value, you will also get a y value. I did write 'send the right (y) value first, and if you do that it works as it should. There is no bug.

if you want individual bangs for every single square in your grid, then a simple way is just to add the x value to the y value, multiplied by the length of the x axis.

so, if you have a 6 x 6 grid, like in your diagram, just start x and y at zero, and then do: x + 6y
this numbers each individual box from 0 to 35.

you can then just use [sel 0 1 2 ... ] etc to output bangs for each box and do whatever you want with that.

ok tried to implement this on a larger level and it doesn't seem to work how I thought it would.

http://localhostr.com/files/NOj37hk/Screen+shot+2011-11-28+at+10.04.28+PM.jpg

any idea why 24 is being sent even though it should be matching 38 (or the label (5,2))

http://localhostr.com/files/djL4Rrl/Screen+shot+2011-11-28+at+10.17.40+PM.jpg
this is what I ended up doing, but It seems really Macgyvered haha

[&& ] outputs a 1 if it is true, or a zero if it is false. With your setup, both the 1 and the 0 will bang the message boxes.

you need [sel 1] to just bang from the true state

also, trust me on this...you don't need the [bondo] object. If you delete it, your patch will still work fine.

[unpack 0 0] sends the right output first, and [&& ] needs its right inlet set first. It all matches up. See, there is some logic here.

here's a MUCH simpler thing that does exactly what you want.

in general, there is almost always a way to avoid repetitious patching.

http://www.pdpatchrepo.info/hurleur/simplified.pd

@mod said:

here's a MUCH simpler thing that does exactly what you want.

in general, there is almost always a way to avoid repetitious patching.

awesome. makes sense, yeah I knew it seemed repetitious, was driving me crazy. I will try to implement this. thanks a ton!

@mod said:

here's a MUCH simpler thing that does exactly what you want.

in general, there is almost always a way to avoid repetitious patching.

is f1/f2/f3 etc based on which you connect first to last? lets say I forgot which one I connected first, how would I find which one f1?

$f1 is the left inlet, $f2 is the next one on the right, $f3 is further right after that, etc...

if you send a list, like [10 4 7 (, then $f1 will be 10, $f2 will be 4, etc...