D&D 3E/3.5 v4: Challenge Ratings pdf (3.5 compatible)

CRGreathouse said:
OK, let's start. I'm taking a fairly simple formula, as far as yours go, and making some kind of sense of it. This is the formula in cell D8.

=1+4*(LOG((2^(TRUNC(LOG(D7,2)))),2))+TRUNC((((D7/(2^(TRUNC(LOG(D7,2)))))-1)*4))

Let me break down what I was doing. I'm going to replace CR where it should be, and N where the number of combatants should go.

First part:

=1+4*(LOG((2^(TRUNC(LOG(CR,2)))),2))

What this does, actually, is find the lowest "base 2" starting point from the target CR. I'm not simply taking 2 to the power of log 2-- the TRUNC is important to get the lowest base 2 point. So, for example, if the CR is 18, the LOG2 is 4.169925; truncate this to an even 4, and raise back to the power of 2= 16. The starting off point is 16.

Now for the second part:

+TRUNC((((CR/(2^(TRUNC(LOG(CR,2)))))-1)*4))

What this unholy bastard is doing is finding the sub-divisions between one base-2 plateau and the next. Because double the CR = +4 EL, there are therefore 4 "steps" between each doubling, and furthermore each step is equidistant from the others.

So between CR16 and CR32, there are 4 steps: CR16-19, CR20-23, CR24-27, and CR28-31. What this formula does is find out how far from the next plateau the current CR is by simple division.

Working from the inside out, the LOG2,18 = 4.xx; driven back down to the next lower plateau with the trunc and ^2 operators = 16; 18/16 = 1.125; subtract 1 to get the fractional portion (.125), multiply x4 = .5, and truncated again = 0.

So what's that tell me? That CR18 is "0 steps" away from the CR16 plateau.

If you tried the same formula with CR20-23, the result = 1; with CR24-27, result 2; CR28-31, result 3.

(Also, in a fit of pique, I changed TRUNC to INT -- they work the same if you don't use the optional argument of TRUNC.)

In my mind I read TRUNC as "truncate," which is what the formula is doing. It's just an easier mnemonic device for me.

While this may make sense when making a formula from a chart, there's no reason to come up with a formula that rigidly follows the space constraints on the chart.

I didn't see much value in a spreadsheet that returned different values than UK's pen-and-paper system-- I'd just carry on using the pen and paper version and to hell with the spreadsheet. ;)

Obvious simplification then follows.

=4*(Int(LOG(D7,2)))+1

This is much better. Of course, we can go one step beyond -- there's no reason to round until the end:

=Int(4*LOG(D7,2))+1

Well you know that's not true... order of operations and all that... There's a difference between 4*(INT(2.25)) and INT (4*2.25).

Wulf
 

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Wulf Ratbane said:
I didn't see much value in a spreadsheet that returned different values than UK's pen-and-paper system-- I'd just carry on using the pen and paper version and to hell with the spreadsheet.

Then use the simplification I used *before* my rounding step:

=4*(Int(LOG(D7,2)))+Int(4*(D7/(2^(Int(LOG(D7,2))))))-3

You can substitute TRUNC for INT if you prefer. I've been programming with INT for a decade now, so INT is the obvious one for me.

Wulf Ratbane said:
Well you know that's not true... order of operations and all that... There's a difference between 4*(INT(2.25)) and INT (4*2.25).

I know it changed the value. It just happened to give numbers closer to UK's system, so it looked good to me.
 
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CRGreathouse said:
Then use the simplification I used *before* my rounding step:

=4*(Int(LOG(D7,2)))+Int(4*(D7/(2^(Int(LOG(D7,2))))))-3

Cool, thanks.

I know I got carried away with the parentheses, I just had to have them to get myself through the mental gymnastics (lots of cutting and pasting from one formula to another).

Wulf
 

Wulf Ratbane said:
I know I got carried away with the parentheses, I just had to have them to get myself through the mental gymnastics (lots of cutting and pasting from one formula to another).

I'll try to work on the others over the next few days. Oh, and I could take out another 2 sets of parens in the formula:

=4*Int(LOG(D7,2))+Int(4*D7/(2^(Int(LOG(D7,2)))))-3
 

CRGreathouse said:
I'll try to work on the others over the next few days. Oh, and I could take out another 2 sets of parens in the formula:

=4*Int(LOG(D7,2))+Int(4*D7/(2^(Int(LOG(D7,2)))))-3

Again, thanks.

Just so I am clear-- the formula above is mathematically equivalent to what I had before? You haven't lost any of the "adherence to chart," right?

One thing that might help others would be to set up the formulas with Names and to move them to another part of the sheet, so that the formulas on the main sheet read a little cleaner to the layman.

Actually, I'd love to set them up as a function that might be more cleanly expressed, so the EL function would be:

EL (CR)

the CR function would be

CR (N)

and the adjusted EL function would be

EL (CR, N)

I am looking forward to your further feedback on the CR formula (it's actually more of a mess than the EL one!) and the nasty problem I mentioned to you in our original email (which is almost certainly due to the adherence to the chart).

Those of you who emailed me for the excel sheet, check your email.

Wulf
 

Wulf Ratbane said:
Just so I am clear-- the formula above is mathematically equivalent to what I had before? You haven't lost any of the "adherence to chart," right?

That's right. The formula adheres to the original inaccuracies of the chart.

Wulf Ratbane said:
Actually, I'd love to set them up as a function that might be more cleanly expressed, so the EL function would be:

EL (CR)

the CR function would be

CR (N)

and the adjusted EL function would be

EL (CR, N)

With VBA functions? It looks neater, but then you have the whole permissions thing to deal with...

Wulf Ratbane said:
I am looking forward to your further feedback on the CR formula (it's actually more of a mess than the EL one!) and the nasty problem I mentioned to you in our original email (which is almost certainly due to the adherence to the chart).

I'm working on it.
 


Hi Upper_Krust!

Well, I've beenusing your PDF for some time now and I've found it much better than SS in so many ways that I actually sold my copy :]

I'm a big psionic fan and now WotC has released a new book, the XPH (expanded psionics book).

I was wondering how would you rate some of the racial abilities that are present in that book?
 

Otakkun said:
Hi Upper_Krust!

Hey there Otakkun! :)

Otakkun said:
Well, I've beenusing your PDF for some time now and I've found it much better than SS in so many ways that I actually sold my copy :]

He he! :D

Well I'm flattered, but more importantly I'm glad you liked it and are getting use from it. ;)

Otakkun said:
I'm a big psionic fan and now WotC has released a new book, the XPH (expanded psionics book).

I was wondering how would you rate some of the racial abilities that are present in that book?

I have to be honest I haven't bought that book as yet, nor have I seen it over here yet. So I am really in the dark as regards the racial abilities.

If you briefly outline one or two examples here (or email me) then I would be happy to look them over and give you my opinion.
 

Just a recap about my XP system: it's based on the idea that doubling the CR of a creature increases its EL by +4, and that increase corresponds to a quadrupling of XP awarded. That idea I lifted directly from the tables for Challenging Challenge Ratings and Encountering Encounter Levels.

I wondered what kind of function had the property that doubling the input quadrupled the output. I realized that f(x)=ax^2 fit the bill. It turns out that the particular function we want is a=75, though a=90 or a=100 would also probably work.

It's easy to verify that for f(x)=75x^2, f(2x)=4*f(x). Just plug in a few numbers. f(8)=4,800, and f(4)=1,200. In other words, f(8)=4*f(4). More generally,
Code:
If f(x) = ax^2, then
   f(2x)= a(2x)^2
        = a*2x*2x
        = 4ax*x
        = 4ax^2
        = 4f(x)

A little creative rounding makes some nice looking tables. For CR of 2 or higher, round down to the nearest 100.

Code:
CR          XP
1           75 
2           300
3           675 ===> 600
4           1,200
5           1,875 => 1,800
etc.

A party of four 1st level characters needs 13.333 CR 2 encounters to get to 2nd level (because of the standard array they are CR 2, so a CR 1 encounter would be too easy). That's 13.3333 encounters worth 300 xp (before being divided 4 ways) or 1000 xp each.

Four 2nd level PCs need 13.333 CR 3 encounters to get to 3rd level, which is 2,000 more xp each, for a total of 3,000 xp. So the first few levels are:

Code:
Level      XP
1          0
2          1,000
3          3,000
4          7,000
5          13,000

To extend the table, calculate 100*level^3, and round up to the nearest 1000.

Code:
Level      XP
2          800 ====> 1,000
3          2,700 ==> 3,000
4          6,400 ==> 7,000
5          12,500 => 13,000
6          21,600 => 22,000
7          34,300 => 35,000
8          51,200 => 52,000
etc.

At higher levels the influence of rounding becomes less important. The difference in xp between levels becomes equal to 100*[(x+1)^3-(x^3)], or 100*(3x^2+3x+1). If x is small compared to x^2, you could treat this as 300*x^2. Four characters require a total of four times as much xp, so 1200*x^2 xp at level x. This is exactly 16 times the xp awarded for an encounter whose CR is equal to x. A little slower than the standard progression, but it is likely that characters will take on CRs high enough that 13 encounters will be enough for them to advance.

To see this, note that a fair encounter for a character of level x tends to be higher than x+1. Better stats than the standard array, more equipment, or the fact that 1 character level = +1.15 CR (at least if the silver rule is ignored) all make the characters more capable. So a fair encounter will not be one whose CR is equal to x, but a little higher. Higher CR encounters mean more xp, which mean fewer encounters per level. The exact number is hard to estimate, but could easily fall in the desired 12 to 14 range.

Since the standard treasure for a character is given by Upper Krust as being (level x level x level) x 100 gp, it turns out the treasure value of an encounter should be about equal to the xp value of that encounter. I regard that as being a nice coincidence.
 

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