Help with Math

I need some assistance with a formula for a macro in my game, because I am not very mathematically inclined. I need a formula that when it pulls the value of attribute @{X}, X having a value of either 0 or a positive integer, that it produces 0 if X was 0, or 1 otherwise. Since this is for a macro, it has to be something which fits inside an inline roll. Anyone with math skills able to help me on this?

Any idea what the possible range of the attribute is? This should be 0 or 1: [[ (((@{X}/1000)+.49999)d1) ]] Hopefully Brian will be by soon, he's brilliant at these things...

Here's a little modulo trick that works for positive integers of any size: [[ ((@{X}+2) % (@{X}+1)) ]] (Edit: formatting.)

Simon S. said: Here's a little modulo trick that works for positive integers of any size: [[ ((@{X}+2) % (@{X}+1)) ]] How about the reverse? (Return 0 if it's above 0, return 1 if it's 0)

Mark said: Simon S. said: Here's a little modulo trick that works for positive integers of any size: [[ ((@{X}+2) % (@{X}+1)) ]] How about the reverse? (Return 0 if it's above 0, return 1 if it's 0) That is clever! I didn't realize we had access to modulo? Could reverse with: [[ ((((@{X}+2) % (@{X}+1))-1)*-1) ]]

Travis M. said: I need some assistance with a formula for a macro in my game, because I am not very mathematically inclined. I need a formula that when it pulls the value of attribute @{X}, X having a value of either 0 or a positive integer, that it produces 0 if X was 0, or 1 otherwise. Since this is for a macro, it has to be something which fits inside an inline roll. Anyone with math skills able to help me on this? [[((@{X} + 1) - abs(@{X} - 1)) / 2]] That will produce min(@{X}, 1) . With the guarantee that X is a non-negative integer, the result will either be 0 or 1. Mark said: Simon S. said: Here's a little modulo trick that works for positive integers of any size: [[ ((@{X}+2) % (@{X}+1)) ]] How about the reverse? (Return 0 if it's above 0, return 1 if it's 0) If you have a formula that produces 0 in one case and 1 in all other cases, you can flip that result by subtracting your formula from 1. 1 - 0 produces 1, and 1 - 1 produces 0.

Brian said: Mark said: Simon S. said: Here's a little modulo trick that works for positive integers of any size: [[ ((@{X}+2) % (@{X}+1)) ]] How about the reverse? (Return 0 if it's above 0, return 1 if it's 0) If you have a formula that produces 0 in one case and 1 in all other cases, you can flip that result by subtracting your formula from 1. 1 - 0 produces 1, and 1 - 1 produces 0. The above part is key. I have 0, 2, 6, 3.5, etc. I'm just optimizing a formula at this point though. Aaron's reverse is decent, but it doesn't shorten my original formula. I was hoping to find a short version like Simon posted.

Yeah, like I said, Brian is brilliant at this stuff. =D

Thanks for the help guys! Actually, after posting this I walked away from my computer and bumped into a friend, who I asked about this problem. He managed to make a shorter one, although it's not as useful in as many situations as your guys' work: @{y}=ceil(@{x}/(@{x}+1)) Works fine in math-only rolls, and when doing things like [[1d20+@{y}]] (which is what I was using it for), but it doesn't seem to work for things like [[(@{y})d6]]. I'll keep the ones you guys gave me saved for when I inevitably need to do a roll like that, though. Will come in handy when/if I get around to making a sheet that's also a fully automated character builder...

Travis , you cannot use "ceil" for the "x" portion of xDy unless you put it in an inline roll. Most people get around this by using +0.5. The dice roller automatically rounds (0.5 up, 0.4999 down) the "x" portion of xDy. Example: (3/2+0.5)d6 = 2d6

Mark said: Brian said: Mark said: Simon S. said: Here's a little modulo trick that works for positive integers of any size: [[ ((@{X}+2) % (@{X}+1)) ]] How about the reverse? (Return 0 if it's above 0, return 1 if it's 0) If you have a formula that produces 0 in one case and 1 in all other cases, you can flip that result by subtracting your formula from 1. 1 - 0 produces 1, and 1 - 1 produces 0. The above part is key. I have 0, 2, 6, 3.5, etc. I'm just optimizing a formula at this point though. Aaron's reverse is decent, but it doesn't shorten my original formula. I was hoping to find a short version like Simon posted. I count three solutions in this thread which will result in 0 for X=0 and 1 for X>0. In all of those cases, (1 - @{solution}) will produce 1 for X=0 and 0 for X>0. Travis M. said: Works fine in math-only rolls, and when doing things like [[1d20+@{y}]] (which is what I was using it for), but it doesn't seem to work for things like [[(@{y})d6]]. I'll keep the ones you guys gave me saved for when I inevitably need to do a roll like that, though. Will come in handy when/if I get around to making a sheet that's also a fully automated character builder... You cannot use any functions for calculating the number of dice, unfortunately. Simon's solution ought to work, though.