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Pathfinder - Sacred Geometry calculator

Pathfinder - Sacred Geometry calculator

Sacred geometry is a feat in pathfinder that slows down games quite a bit... I was wondering if anybody knew of a calculator out there for use with this feat.

I could have sworn there was one. Some guy did a Python script but never turned it into an app, however somebody did make a less robustly featured app for Android.

According to various discussions, doing it right is trickier than it looks, but as long as you don't care if it finds ALL possible solutions it's not that bad. Also, apparently Calculating Mind is a waste of a feat and you never need more than 12 ranks in Engineering ever.

Originally Posted by mildly_competent View Post
Yikes. This feat seems incredibly overpowered.
It is. But it also slows down games incredibly.

There used to be a calculator... but it's down now.

It is and it isn't. Unlike Incantantrix and Divine Metamagic, this doesn't let you cast stuff you shouldn't actually be able to cast at all. It just lets you cast it ridiculously cheap. Powerful for sure and probably overdone but not quite as bad as it looks at first glance.

Just because it's less bad than those things doesn't mean it's not still ridiculously bad - it's basically free whatever-metamagic to any spells you want, just capped at your maximum spell level (or L9). That's more Metamagic Song than Divine Metamagic, but pretty much any free metamagic is powerful - and this has no other limits. There's a small risk that the spell fails (though with enough ranks you can probably make any number reliably regardless of rolls - does 12 do it?) I guess which is potentially a big deal, but otherwise adding free Quicken to all your L5-and-below spells is pretty darn amazing even for L20 casters.

To speed up the game, perhaps you could consider adding a Countdown Clock? That would be a) faster, b) better-balanced (you're more likely to fail) and c) more entertaining.

I'm not sure what the best way of calculating this would be, actually. For small-ish numbers of dice you could probably calculate all possible answers just by trying everything, but I think this would start to get a bit unwieldy around 12. If you could show that you definitely never need more than a certain number of ranks, though, you might just get away with it.

You could probably make something which tries combinations much more smartly. For example, at the very least you know that the last operation can't be a multiplication (and I would guess it's probably not going to be a division in most cases). It would be harder then, though, to make sure that you've tried all the possibilities if the more likely things turn out not to work after all.

According to discussion I've seen, you never need more than 12 dice to cover you all the way to the highest possible difficulties with 100% chance of success. Not 100% sure if that's correct, but they seemed to know their stuff, so I'm inclined to believe.

Actually that seems pretty likely.

You only need 4 numbers to make each of the L9 constants, if you can pick them (but you do need at least 4). The lower-level ones are probably the same or easier.

So maybe you need to roll 6, 6, 3 and 1 so that you can do 6x6x3 - 1 = 107. Well, if you don't roll those numbers, you can probably still make them from other numbers. If you get really unlucky you might need as many as five rolls to make a 6 (i.e. roll five 1s - then you can do (1+1+1)x(1+1), worst case scenario) so a very naive upper limit is 20 dice. We can definitely make any of these with 20 rolls, because we can just make those four numbers and then get 107.

But actually it's way better than that.
Firstly, I probably won't need five dice for every number I want to make. You can always make a 1 with just three numbers, for example (the only way you don't already have a 1 or two numbers 1 apart is 2, 4, 6 in which case you can do (2+4)/6 or similar), so in that example I'd be looking at more like 5+5+5?+3 dice, so certainly no more than 18.
Secondly I'll get some overlaps. The only(?) way it takes me 5 dice to make a 6 is if I roll six 1s. Well, if I rolled six 1s, I will just be able to use on of them as my 1, so I won't need more rolls to make that. The worst case for the 6 prevents me from suffering the worst case for my 1, so I'm looking at more like 15 rolls, tops now (maybe fewer, I didn't bother looking at the 3).
Thirdly this is to make one specific expression - but to get a L9 prime, well, I have three different primes I can make, and I just spotted five different four-number expressions which will hit one of those. This gives me way more flexibility - to use the same example again, if in trying to get two 6s I just roll ten 1s, well OK, I'll just make 5x5x4 + 1 instead, because that uses fewer dice.

So yeah, you'd have to work it all through, but 12 seems like a pretty plausible cap - in fact I wouldn't have been surprised if it was lower.

Of course, if you can show that that's true, it means that once you have 12 ranks you don't need a calculator at all because you always win.


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