Sacred Geometry and Arithmancy.

[Yirrare]

Since everyone seems to miss it, I would just like to point out the hidden benefit of calculating mind.

[...]Then roll a number of d6s equal to the number of ranks you possess in Knowledge (engineering).[...]

[...]You can use any combination of d6s and d8s that you wish, as long as the number of dice does not exceed the number of ranks you possess in Knowledge (engineering).[...]

I have bolded the relevant parts for clarity. In short, calculating mind gives you the option of which die combination and number of die you wanna use.

But then, to be fair, you will probably want to use as many die as you can along with the elimination method first suggested by DarkSonic1337.

Best Regards
Yirrare

[meschlum]

Further number crunching indicates that Level 1 spells can be done with 5 or more ranks. 4 ranks don't always work: (5, 5, 5, 5) fails, hence the minimum is 5 ranks. Edit: 4 ranks actualy works.

Spoiler: Number crunching. Yum!

Show

To get a Level 2 spell with only 1s, you use 13 = (1 + 1 + 1) * (1 + 1 + 1 + 1) + 1, for 8 dice. So the lower limit is 8 dice.

With 10 or more dice, you have at least two pairs.

If you have a pair of 4s, you have another pair (which gives 1), and you can use 4 * 4 + 1, with four dice, so there are six left, which cancel out.

So you have one 4 at most.

If you have a single 4 and a pair of 2s, you can use 2 * 2 * 4 + 1, and it works.

If you have a single 4 and a single 2, you can use (2 + 1) * 4 + 1, using six dice. The remaining four dice cancel out.

So if you have a single 4, you cannot have any 2s.

If you have a single 4 and one or more 3s, you can use 3 * 4 + 1, with four committed dice (one pair, one 4, one 3), so it works since you have two pairs or more.

So if you have a single 4, you cannot have any 2s or 3s. This leaves nine or more dice with values of 1, 5, or 6.

If you have four or more 1s, you can use (1 + 1 + 1) * 4 + 1, with five or more dice left, so it works. Therefore, you have at most three 1s.

If you have three 1s, you can use (1 + 1 + 1) * 4 + N/N, where N=5 or N=6, since you have six or more dice which are 5s or 6s, so you have at least two pairs. This uses six dice, and the remainder can be canceled.

If you have two 1s, you have seven or more dice that are 5s or 6s, so at least three pairs. (1 + 1 + P/P) * 4 + P/P uses seven dice, leaving one pair. (P-P) = 0, so the remainder cancels out.

If you have one 1, you have at least three pairs of 5s and 6s, so you have a solution using 8 dice. Your remaining dice are either equal and cancel out, or are 5 and 6, and 6-5 = 1, so that cancels out as well.

So if you have a single 4, you can only have 5s or 6s.

If you have four 6s, you can get 6 + 6 + 6/6, which works. So you have three 6s at most.

If you have four 5s, you can get 5 + 5 + 5/5, which works. So you have three 5s at most.

Since you need to have nine or more 5s and 6s, yuo have at least four of either, so you can always cast a level 2 spell with ten or more dice and one or more 4s.


Now we look at what happens if you have no 4s and ten or more dice. In this case, you have at least three pairs.

If you have a pair of 6s or a pair of 5s, you can do 6 + 6 + 1 (or 5 + 5 + 1), using four dice and cancelling the rest.

So you have at most one 5 and one 6. If you have both, 5 + 6 works with the rest cancelling out, so you have at most one die that comes out as a 5 or a 6.

You therefore have nine or more dice that are 1s, 2s, or 3s.

If you have three or more 3s, you can use 3 * (3 + 3) - 1 with five dice (a pair of 2s or a single 1 will contribute the 1)

If you have two 3s, you can use (3 + 3) * 2 + 1 if you have 1s and 2s, or (3 + 3) * 2 + 2/2 if you have only 2s, or (3 + 3) * (1 + 1) + 1 if you have only 1s. So it always works.

If you have one 3, you can use 3 * 2 * 2 + 1 if you have at least two 2s and a 1, or 3 * 2 * 2 + 2/2 if you have only 2s (and a 5 or 6), so you have one 2 at most (and at least eigth 1s). 3 * (1 + 1 + 1 + 1) + 1 uses six dice, so the rest cancels out.

You therefore have nine or more dice that are 1s or 2s, and one that could be 1, 2, 5, or 6.

If you have a 5 or 6, 6 * (1 + 1) + 1 or 5 * (1 + 1) + 1 means you have at most two 1s (so seven or more 2s). 6 * 2 + 2/2 and 5 * 2 + 2/2 mean you have at most two 2s, so there is always a solution if you have a 5 or 6.

Therefore, you have ten or more dice that are 1s or 2s.

With six or more 2s, you have 2 * 2 * 2 *2 + 2/2, and the rest cancels out, so there is always a solution in this case and you have at least five 1s.

With three or more 2s, you have 2 * 2 * (2 + 1) + 1, and the rest cancels out if you have at least two 1s, which is the case.

With two 2s, you have (1 + 1 + 1) * 2 * 2 + 1, using 6 dice and the remaining 1s cancel out.

With one 2, you have a (1 + 1 + 1) * 2 * (1 + 1) + 1, with only 1s otherwise, so it works.

With no 2s, you have only 1s, and there is a solution with eight 1s, the extra 1 cancels out (multiply by 1).

Conclusion: With 10 or more dice, you are always certain of getting a level 2 spell.

It is possible to be certain with as few as 8 dice, but no less.

The absolute minimum dice pool to be certain of a level 1 result is 4 dice (2nd level spells).

So the minimum dice pool for absolute certainty starts out fairly high and rises quickly (to 8 for level 2) before flattening out (at 17 or fewer dice for level 9), meaning a paranoid caster will never bother with low level spells - by the time you're confident that you can cast level 2 spells without failure, you can probably do a lot better.

[grarrrg]

Further number crunching indicates that Level 1 spells can be done with 5 or more ranks. 4 ranks don't always work: (5, 5, 5, 5) fails, hence the minimum is 5 ranks.

(5-5)*5+5=5
Four 5's can get a 5 and pass Level 1

It's late, I ain't checking all of you other stuff....night...

[meschlum]

(5-5)*5+5=5
Four 5's can get a 5 and pass Level 1

It's late, I ain't checking all of you other stuff....night...

Indeed! Which is why it's hard to get absolute lower bounds in many cases. My mistake.

Spoiler: Proof of 4 dice for Level 1

Show

With 4 dice:

If there is a pair and a 3 or a 5, then grarrrg's approach works.

If there are four 1s, then Level 1 is possible

If there are three 1s, then Level 1 is possible (1 + 2, 1 + 4 and 1 +6, or a pair of 1s and a 3 or 5)

If there is a pair of 1s, then the other two dice are 2, 4, or 6

(2, 2): 2 + 2 + 1
(2, 4): 2 + 4 + 1
(2, 6): 2 + 6 - 1
(4, 4): 4 + 4 - 1
(4, 6): 6 - 4 + 1
(6, 6): 6 - 6/(1+1) (requires all 4 dice)

So there is at most one 1.

If there is a 1 and a pair, then the pair cancels out and the remaining options are 2, 4, 6. Since 1 + 2, 1 + 4 and 1 + 6 all work, Level 1 is always possible if there is a 1 and a pair.


If there are four 2s, then 2 + 2 + 2/2 works

If there are three 2s, then we need to check combinations with 2, 4, and 6 (there is a pair of 2s and a 3 or 5 otherwise).

2 + 2 + 2 + 1 works
4 + 2 + 2/2 works
6 - 2 + 2/2 works

If there is a pair of 2s, then the other two dice are 4s, or 6s since Level 1 is always possible with a 1 and a pair of 2s.

(4, 4): 2 + 2 + 4/4
(4, 6): 6 - 4 + 2/2
(6, 6): 2 + 2 + 6/6

So there is a most a single 2.

If there are three or more 3s (or 5s) then there is a pair of 3s (or 5s) and an extra 3 (or 5), so Level 1 is possible.

If there are two 3s, then the other dice are 2s, 4s, or 6s (at most one of each, since a pair and 3 works)

(2, 4): (2 + 4) / 3 + 3
(2, 6): (6 / 2) / 3 * 3
(4, 6): 4 + 6 / 3 - 3

So there is at most a single 3

If there are two 5s, then the following checks apply

(2, 4): (5 - 4) * (5 - 2)
(2, 6): (6 - 5) * (5 - 2)
(4, 6): (4 + 5 + 6) / 5

So there is at most a single 5

If there are four 4s, we use 4 + 4 - 4/4

If there are three 4s, we need to check combinations with 2 and 6

4 + 2 + 4/4
6 - 4 + 4/4

So there are at most two 4s.

If there are two 4s, we need to check combinations with 2 and 6 (since there is a pair, so 1, 3, and 5 all give Level 1 spells), and only one 2 is allowed.

(2, 6): 2 + 6 - 4/4
(6, 6): 4 + 4 - 6/6

So there is at most one 4.

If there are four 6s, then (6 * 6) / (6 + 6) works

If there are three 6s, we need to check with 2 and 4

6 - 6 + 6/2
6 - (6 + 6) / 4

So there are at most two 6s.

If there are two 6s, we need to check combinations with 2 and 4 (since 1, 3, 5 are excluded and there cannot be pairs of 2s or 4s)

(2, 4): 6 + 6 / (2 + 4)

So there is at most one 6.


Therefore, with 4 dice, Level 1 is always possible if there is a pair.

If there are no pairs, the remaining options are:

4 + 3 - 2 * 1
5 + 3 - 2 - 1
6 + 3 - 2 * 1
5 + 4 - 2 * 1
6 + 4 - 2 - 1
6 - 5 + 2 * 1
(5 - 4) * 3 * 1
(6 - 4) * 3 + 1
(6 - 5) * 3 * 1
6 - 5 + 4 * 1
5 + (4 + 2) / 3
6 / (3 * 2) + 4
6 / (3 * 2) * 5
6 / (2 + 4) * 5
(6 - 3) * (5 - 4)

So all 4 dice combinations work for Level 1.

I stand corrected, Level 1 can be done with 4 dice (and there are proofs for 5 and 6+, so it really is 4+ dice).

With three dice, (1, 6, 6) does not work, so the absolute lower limit is 4.

Recap

Level 1: 4+ ranks exactly (level 2+ needed for the feat)
Level 2: 8+ ranks exactly (level 3+ needed to cast)

Level 9: 17+ ranks guaranteed, could be as low as 14 (level 17+ needed to cast)

[GreyBlack]

I am totally the jackass at my table who would use this feat. This feat is WAY OP and should be banned at every table, as it is way too easy to pull off and way too much for way too little.

[Miss Disaster]

Conclusion: With 10 or more dice, you are always certain of getting a level 2 spell.

It may be possible to be certain with as few as 8 dice, but no less.

The absolute minimum dice pool to be certain of a level 1 result is 5 dice (3rd level spells for prepared casters).

So the minimum dice pool for absolute certainty starts out fairly high and rises quickly (to 8-10 for level 2) before flattening out (at 17 or fewer dice for level 9), meaning a paranoid caster will never bother with low level spells - by the time you're confident that you can cast level 2 spells without failure, you can probably do a lot better.

Thanks for number-crunching all that data out, Meschlum. It was an interesting read.

So as a recap to your synopsis that I quoted above .... is there a general guideline for each of the Levels of Spells that has a correlating number total for Knowledge: Engineering ranks in regards to achieving *absolute certainty* with Sacred Geometry?

Ex.
1st Level Spell = 5 Ranks
2nd =
3rd =
4th =
5th =
6th =
7th =
8th =
9th = 17 Ranks

That's 11 Knowledge: Engineering skill ranks between the high and low end ... and 7 spell level to correlate them too. Unless that doesn't work with the stat formula and table spread.

[meschlum]

Thanks for number-crunching all that data out, Meschlum. It was an interesting read.

So as a recap to your synopsis that I quoted above .... is there a general guideline for each of the Levels of Spells that has a correlating number total for Knowledge: Engineering ranks in regards to achieving *absolute certainty* with Sacred Geometry?

Ex.
1st Level Spell = 4 Ranks
2nd =
3rd =
4th =
5th =
6th =
7th =
8th =
9th = 17 Ranks

That's 11 Knowledge: Engineering skill ranks between the high and low end ... and 7 spell level to correlate them too. Unless that doesn't work with the stat formula and table spread.

My pleasure! Math is fun (to me).

(Note that Level 1 can be done with 4 ranks, as revealed by number crunching above - low levels are painful).

(Also note that Level 9 may be possible with fewer than 17 ranks (as few as 14), but the proof (if there is one) becomes harder).

Short answer: no. If there were a way to easily procedurally generate prime numbers, a lot of the current paradigms for password security would be overthrown.

Slightly longer answer: for the limited range of prime factors that are used by the feat, it's possible to do the math. The issue is doing it in a sane amount of time.


I can work on filling in the table, though!

Spoiler: Proof of 8 dice for Level 2

Show

Level 2, revised. We know that Level 2 can be done with 10 or more ranks, and at least 8 are needed. Let's see if eight or more is possible.

If you roll two 6s, two 5s, or two 4s, and have a pair, you get Level 2 with 4 dice (5 + 5 + 1, 6 + 6 + 1, 4 * 4 + 1). The remaining four or more dice cancel out.

If you roll two 6s, two 5s, or two 4s and have no other pairs, the remaining six dice are (1, 2, 3, 4, 5, 6). You can use the 1, and have five dice left, which cancel out.

So you have at most one 4, one 5, and one 6.

If you roll a 5 and a 6, you have six dice left which cancel out and 5 + 6 works.

So you have at most one (5 or 6)

If you roll a 4 and a 6, you have six dice left with values in the 1 to 3 range, so there is a double, and you can get 4 * 4 + 1 with four dice. Your remaining four or more dice cancel out, so you can't have a 4 and a 6.

If you roll a 4 and a 5, the remaining dice are in the 1 to 3 range. If you also roll a 3, 4 * 5 - 3 works (and you have 5 dice left, which cancel out). If you also roll a 2, 4 + 5 + 2 works (and you have 5 dice left, whcih cancel out). If you have six 1s, 4 + 5 + 1 + 1 works.

So one die can be 4, 5, or 6 (or 1, 2, 3), and the remaining seven must be 1s, 2s, or 3s.

If you have two 3s and a 2, the extra dice cancel out.

If you have two 3s and two 1s, the extra dice cancel out.

So if you have two 3s, you can have at most one 1 and the other dice are 3s (or one die at 4, 5, 6)

With seven 3s, you have 3 * 3 + 3 + 3/3 with five dice, and a pair of 3s left, so (3-3) cancels the rest.

With six 3s and a 1, you have 3 * 3 + 3 + 1 with four dice, and the rest cancels out.

So you have at most one 3.

If you have a 3 and a 4, you have six dice with 1s and 2s, so you have a pair. 3 * 4 + 1 works.

If you have a 3 and a 5, you have six dice with 1s and 2s, so you have at least two pairs 3 * 5 + 1 + 1 works, leaving you with two dice worth 1 or 2. If they are identical, they cancel out. If they are 1 and 2, then 2-1 = 1, and that cancels as well.

If you have a 3 and a 6, you have six dice with 1s and 2s, so you have a pair. 3 * 6 - 1 works.

So you have at most one 'wild' die which can be worth 3, 4, 5, or 6, and the remaining are 1s and 2s.

If the wild die is a 4, you have 4 * 2 * 2 + 2/2 and 4 * (1 + 1 + 1) + 1, so there are at most three 1s and three 2s and three dice are not used to get the total. These unused dice are 1s or 2s, so there is a pair, and they cancel out. Since there are at least seven 1s and 2s, a Level 2 spell can always be cast if a 4 is rolled.

If the wild die is a 3, you have 3 * 2 * 2 + 2/2, 3 * (1 + 1 + 1 + 1) + 1, and 3 * 2 * 2 + 1, so there are at most four 1s and three 2s. If there are 1s and 2s, there is at most a single 2 (since 4 dice are used in 3 * 2 * 2 + 1, so the rest cancels out). So if there are 1s and 2s, there must be at least six 1s, and the extra 1s cancel out - so there must be either 1s only or 2s only. In either case, there are at least two dice left, which are identical, so they cancel out as we have a pair.

If the wild die is a 5, you have 5 * 2 + 2/2 with four dice, so there are at most two 2s. You also have 5 * (1 + 1) + 1, so there are at most two 1s. Since there must be at least seven 1s and 2s, a Level 2 spell is always possible in this case.

If the wild die is a 6, you have 6 * 2 + 2/2 and 6 * (1 + 1) + 1, so again a Level spell is always possible.

Therefore, you need all eight dice to be 1s and 2s.

Since the dice are 1s and 2s, any two dice will cancel out (either equal or multiply by 2 - 1).

If you have six or more 2s, 2 * 2 * 2 * 2 + 2/2 is a solution, and the remaining two dice cancel out. So you have at most five 2s, and at least three 1s.

If you have four or more 2s, 2 * 2 * 2 * 2 + 1 works, and you have enough 1s to make it possible. So you have at most three 2s, and at least five 1s.

With three 2s, you get 2 * 2 * (2 + 1) + 1, which works (the remaining 1s cancel out).

With two 2s, you get 2 * 2 * (1 + 1 + 1) + 1, which works (6 dice).

With one 2, you get 2 * (1 + 1) * (1 + 1 + 1) + 1, which works (7 dice).

With no 2s, you have eight 1s, which work (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1

So Level 2 is possible with eight dice, and not with seven (try seven 1s)

The new table is:

1st Level Spell = 4 Ranks (exact)
2nd = 8 Ranks (exact)
3rd =
4th =
5th =
6th =
7th =
8th =
9th = 17 Ranks (could be as few as 14)

So there is a rather steep initial curve, then 9 Ranks cover 7 Levels. And it's not impossible that some intermediate spell levels could require more dice than the high spell levels!

I've got proofs that Level 3 works with 11 dice, but it could be as low as 9. That's for later, though.

[grarrrg]

My pleasure! Math is fun (to me).
[STUFF]

Might I recommend Spoiler Tags? The length is getting a bit much.
My poor, poor scroll wheel!

[meschlum]

Might I recommend Spoiler Tags? The length is getting a bit much.
My poor, poor scroll wheel!

But then how can I inflict math on hapless passers-by?

Done anyway.

Spoiler: Level 3 with 9 dice

Show

If we have a 6 and a (3, 4, or 5), there are 7 or more dice left, so there is a pair. 6 * 5 - 1, 6 * 4 - 1 and 6 * 3 + 1 all work for Level 3.

So if there is a 6, there are only 1s, 2s, and 6s.

If there are five or more 6s, 6 + 6 + 6 + 6/6 is a solution, with at least 4 dice left, so it cancels out. Therefore, there are four 6s at most.

If there are three 6s and a pair, 6 + 6 + 6 + 1 is a solution. With four 6s at most, there are at least five dice that are 1s or 2s, so there are at least two pairs. One pair cancels out the extra, so there is a solution.

If there are one or two 6s, there are at least seven dice that are 1s or 2s. Since 6 * (2 + 1) + 1 is a four die solution (and the remaining five dice cancel out), there is at most one 1 if there are any 2s.

With a 6 and a 1, there are at least six 2s if any 2s are present. 6 * (2 + 2/2) + 1 uses five dice, and the remaining four or more cancel out.

With a 6 and seven or more 1s, 6 * (1 + 1 + 1) + 1 uses five dice, and the remaining four or more cancel out.

So if there are one or two 6s, the other seven or more dice are 2s. 6 * (2 + 2) - 2/2 uses five dice, and the remaining four or more cancel out.

Therefore, a Level 3 spell is always possible with nine dice if one is a 6.

If we have a 5 and a 4, there are seven dice left, so there is a pair, 5 * 4 - 1 is a solution, and uses four dice or less, so the remainder cancels out.

If we have a 5 and a (3 or 5), 5 * (3 + 1) - 1 and 5 * (5 - 1) - 1 are solutions, so there is one 1 at most, and one pair at most among the seven dice.

If there is one pair among seven dice, then (1, 2, 3, 4, and 5) are present, and one is a triplet. This means that the pair can create a 1, and the other 1 can be used as well. With five dice in all (2 for the pair, one for the 1), the remaining four dice can be cancelled, so there is a solution if there is a 5 and a 3 or two or more 5s.

If we have a single 5 and the remaining dice are 1s and 2s, 5 * 2 * 2 - 2/2 is a 5 die solution and the rest cancels out, so there are three 2s at most. 5 * (1 + 1) * (1 + 1) - 1 is a solution as well, with six dice. Since the remaining dice are 1s or 2s, there is a pair among them, and they cancel out. Thus, there are four 1s at most. This gives seven dice other than the 5 at most, and we use nine dice in all, so there is always a solution.

Therefore, with nine dice, a Level 3 spell is always possible if there is a 5 or a 6.

If there are two or more 4s, 4 * (4 + 1) - 1 is a solution. Since we have seven dice spread across 1, 2, 3, and 4, there are at least two pairs, so the solution is possible, using six dice in all. The solution uses five dice if there is at least one 1, allowing the rest to cancel, so the six die solution must have no 1s. This leaves seven dice spread across 2, 3, and 4. If the remaining three or more dice after the solution include a pair, the remainder cancels out and there is a solution. Otherwise, there are exactly three remaining dice (9 dice total) and the remaining dice are 2, 3, and 4, with a solution of 19. If we add 4 * (3 - 2), we get a solution of 23, which is also valid.

Therefore, there is at most a single 4 and the eight or more other dice are 1s, 2s, and 3s.

If there is a 4, 3, 2, 1, then 4 * (3 + 2) - 1 is a four die solution and the remainder cancel out. Therefore, if there is a 4, all three other dice values do not appear.

If there is a 4, 3, 2, then 4 * (3 + 2) - 1 is a five die solution (using one pair) and the remainder cancel out. With a 4, 3, 2, there are at least six other dice spread across 2 and 3, so there is a pair and the solution exists.

If there is a 4, 3, 1, then 4 * (3 + 1) - 1 is a five die solution (using one pair). Again, the pair is possible and the remainder cancels out.

If there is a 4, 2, 1, then 4 * (2 + 1 + 1) - 1 is a seven die solution using two pairs. This leaves at least two dice which are 1s and 2s - if there is a pair, the remainder canels out. Otherwise, there are exactly two dice left, which are 1 and 2, and we can multiply the result by (2-1), so the solution exists.

Therefore, if there is a 4, all the other eight or more dice have the same value.

3 * 3 * 3 - 4 is a four die solution and the remainder cancel out
4 * (2 * 2 + 2/2) - 2/2 is a seven die solution, so there are at least two 2s left, which give 2-2 and cancel the remainder
4 * (1 + 1 + 1 + 1 + 1) - 1 is a seven die solution, and all the remaining dice are 1s, so we can multiply and keep the solution.

Therefore, there is a solution with nine or more dice if any of the dice are 4, 5, or 6.

3 * 3 * 2 + 1 is a solution where the remainder cancel out, so if there are at least two 3s, there cannot be 1s and 2s together.

(3 + 2) * (3 + 1) - 1 is a solution where the remainder cancel out, so if there are at least two 3s and some 1s, there is a single 1 at most.

(3 + 2) * (3 + 2) - 2 is a solution where the remainder cancel out, so if there are at least two 3s and some 2s, there is a single 2 at most.

So if there are two 3s, there are at least eight 3s and one (1 or 2).

3 * 3 * (3 - 3/3) + 3/3 uses seven 3s. If the remaining dice are 3s, the solution exists. If the remaining dice are a 2 and a 3, multiply by 3-2 and a solution exists, so if there are two or more 3s, a solution exists unless there is a 1 as well.

3 * 3 * (3 - 1) + 3/3 uses five 3s and a 1, so the remaining three or more dice are all 3s, and cancel out.

Therefore, there is at most a single 3 with nine or more dice, and eight dice that are 1s or 2s.

3 * (2 + 1) * 2 + 1 is a solution where the remainder cancel out, so if there is a 3 and 1s and 2s, there is either a single 1 or a single 2.

With a single 2, 3 * (2 + 1) * (1 + 1) + 1 is a solution, and the remaining dice are all 1s, so they cancel out.

With a single 1, 3 * (2 + 1) * 2 + 2/2 is a solution using six dice, so there are at least three 2s remaining, and 2-2 cancels them all.

So if there is a 3, all remaining dice are 1s or all remaining dice are 2s.

3 * 2 * 2 * 2 - 2/2 is a six die solution, and the remaining three or more 2s can be cancelled with 2-2
3 * (1 + 1) * (1 + 1 + 1) + 1 is a seven die solution, and the remaining two or more 1s cancel out.

So Level 3 spells can be done with nine dice if any come up 3, 4, 5, or 6.

2 * 2 * 2 * 2 + 2 + 2/2 uses seven dice, and the two or mor remaining are 1s and 2s, which cancel out (2-2, 1-1 or multiply by 2-1 if there are exactly two remaining). Therefore, there are at most six 2s and at least three 1s.

(2 + 1) * (2 + 1) * 2 + 1 uses six dice, and the remaining three or more dice are 1s and 2s, and so include a pair and cancel out. This means that there are at most two 2s, and at least seven 1s.

(2 + 1) * (2 + 1) * (1 + 1) + 1 uses seven dice, and the remaining dice must be 1s since the 2s have been used, so they cancel out. Thus, there is at most a single 2 and at least eight 1s

(2 + 1) * (1 + 1 + 1) * (1 + 1) + 1 uses eight dice, and the remaining dice must be 1s since the 2 has been used. Again, the solution works, and there cannot be any 2s

(1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) + 1 uses nine dice, and the remaining dice must be 1s since no others are allowed.

Therefore, Level 3 spells can always be cast with 9 or more dice, and cannot always be cast with eight or less (try eight 1s)

Level 1: 4 or more Ranks (exact)
Level 2: 8 or more Ranks (exact)
Level 3: 9 or more Ranks (exact)
Level 4:
Level 5:
Level 6:
Level 7:
Level 8:
Level 9: 17 or more Ranks (could be as low as 14)

I expect that Level 6 and Level 8 will be problematic, because there are no convenient shortcuts to get close to the associated primes.

[Miss Disaster]

Level 1: 4 or more Ranks (exact)
Level 2: 8 or more Ranks (exact)
Level 3: 9 or more Ranks (exact)
Level 4:
Level 5:
Level 6:
Level 7:
Level 8:
Level 9: 17 or more Ranks (could be as low as 14)

I expect that Level 6 and Level 8 will be problematic, because there are no convenient shortcuts to get close to the associated primes.

More interesting insight, Meschlum. Thanks again.

Your quoted text above is intriguing in your claim that the impromptu table doesn't actually have consistent and predictable progression (where you mention that 6th & 8th level gets complicated). Makes you wonder if the designer(s) of the feat were aware of that arithmetic quirk.

Still, whether this feat makes it to my gaming table or not, I do appreciate that it has forced me a few others in my group to bone-up on our admittedly rusty math and statistics knowledge.

[meschlum]

More interesting insight, Meschlum. Thanks again.

My pleasure!

Your quoted text above is intriguing in your claim that the impromptu table doesn't actually have consistent and predictable progression (where you mention that 6th & 8th level gets complicated). Makes you wonder if the designer(s) of the feat were aware of that arithmetic quirk.

I'm fairly certain they were not, since the series used is just three consecutive primes for each spell level, rather than anything clever. Besides, the pattern of required ranks goes against what I'd expect from sensible design - you want low level boosts to be easy, so people benefit from the feat early on with minor Metamagic, and high level boosts to be difficult, so there is a reason not to cast all your magic missiles maximized and quickened. And the trend of having low incerases for higher level effects goes against that.

Spoiler: Let's try Level 4 with 11 dice (the minimum)

Show

If you have two or more 6s, you have nine dice left, so there is a pair. 6 * 6 + 1 works, so there is one 6 at most (four dice used, so the remaining seven or more cancel out).

If you have two or more 5s, you have nine dice left, so there are two pairs 5 * (5 + 1) + 1 works, so there is one 5 at most (six dice used, so the remaining five or more cancel out).

If you have a 5 and a 6, you have nine dice left, so there is a pair. 5 * 6 + 1 works, so one die can come up as a 5 or a 6 (four dice used, so the remaining seven or more cancel out).

If you have three fours, you have eight dice left, so there is a pair. 4 * (4 + 4) - 1 works, so there are two fours at most (five dice used, so the remaining six or more cancel out).

If you have a 4 and a 6, you have nine dice left (two pairs). 6 * (4 + 1) + 1 works, with six dice used and the remaining five or more cancel out.

If you have a 4 and a 5, you have nine dice left, all of which are 1s to 4s. This means three pairs, and (4 + 1) * (5 + 1) + 1 works, using eight dice. The remaining three dice are all in the 1 to 4 range. If there is a double, they cancel out, so the following options exist:

1, 2, 3: add 3 - 2 - 1
1, 2, 4: multiply by (4 - 2 - 1)
1, 3, 4: add 4 - 3 - 1
2, 3, 4: multiply by 2 + 3 - 4

Therefore, if you have any 4s you cannot have a 5 or 6.

If you have two 4s, you have nine dice left, which are all 1s to 3s.

If you have two 4s and a 2, you have eight dice left (one pair). 4 * 4 * 2 - 1 works, so you can't have a 2 if there are two 4s (five dice used, so the remaining six or more cancel out), and the other dice are all 1s or 3s.

If you have two 4s and a 3, you have eight dice left which are 1s or 3s, so three pairs. 4 * (4 + 3) + 3 is a solution, using four dice (and the remainder cancel out), so there is a single 3 at most, and at least seven 1s. (4 + 1 + 1) * (4 + 1) + 1 uses six dice (the remaining five cancel out) and four 1s, so it is a solution if there are seven or more 1s.

If you have two 4s and only 1s, (4 + 1 + 1) * (4 + 1) + 1 works, and there are at least nine 1s, with the remainder canceling out.

So there is at most a single 4, which is not compatible with having a 5 or 6, so one die can be 4, 5, or 6 and the others are all 1s, 2s, and 3s

If there are three or more 3s, there are eight dice left, one of which can be 4, 5, or 6, so there are two pairs among the seven dice coming up 1, 2, and 3. 3 * 3 * (3 + 1) + 1 is a solution, using seven dice. This leaves four dice, which cancel out. Therefore, there are two 3s at most.

If there are exactly two 3s, there are nine or more dice left, one of which can be 4, 5, or 6. This leaves eight or more dice that can be only 1s or 2s, guaranteeing three pairs.

If there are two 3s and two or more 2s, there are seven or more dice left, of which six are 1s or 2s, and so include a pair. 3 * 3 * (2 + 2) + 1 is a solution with six dice, and the rest cancel out. So if there are two 3s, there is at most a single 2, one 4, 5, or 6, and the other seven or more are 1s.

With two 3s and seven 1s, we have 3 * 3 * (1 + 1 + 1 + 1) + 1 using five 1s, which leaves two 1s to cancel out the rest via 1-1. Therefore, there cannot be two 3s.

If there is one 3, there are nine or more dice that must be 1s or 2s (and one that can be anything but a 3). 3 * 2 * 2 * (2 + 2/2) + 2/2 uses eight dice and seven 2s, so there must be at least three 1s. 3 * 2 * 2 * (2 + 1) + 1 is a solution with less than three 1s, so there must be two 2s at most, and at least seven 1s.

With one 3 and two 2s, we get 3 * 2 * 2 * (1 + 1 + 1) + 1, leaving three or more 1s to cancel the rest. So there is one 2 at most (and eight 1s or more).

With one 3 and one 2, we get 3 * 2 * (1 + 1) * (1 + 1 + 1) + 1, using six 1s. Since there are at least eight 1s, the remaining two or more cancel out via 1-1.

With one 3 and nine or more 1s, we get 3 * (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1, using eight 1s. This leaves at least one 1 and a die that must be 1, 4, 5, or 6.

If the wild die is a 1, they cancel out.

If the wild die is a 4, 3 * 4 * (1 + 1 + 1) + 1 leaves five or more 1s to cancel the rest.

If the wild die is a 5, 3 * (5 + 1) * (1 + 1) + 1 leaves five or more 1s to cancel the rest.

If the wild die is a 6, 3 * 6 * (1 + 1) + 1 leaves six or more 1s to cancel the rest.

So there cannot be any 3s, and we have ten dice that are 1s or 2s, as well as a wild die which is 1, 2, 4, 5, or 6.

If the wild die is a 4, then 4 * (1 + 1) * (1 + 1) * (1 + 1) - 1 is an eight die solution and the three remaining dice are 1s or 2s, so there is a pair and they cancel out. Therefore, there are six 1s or less and at least four 2s. If there is a 4 and three 2s, there are seven dice left (1s or 2s), so there is at least one pair, and we have 4 * 2 * 2 * 2 - 1 as a solution, using six dice and three 2s as well as a pair. The remaining dice cancel out, so the wild die cannot be a 4.

If the wild die is a 5, then 5 * (1 + 1) * (1 + 1 + 1) + 1 is a seven die solution and the remainder cancels out. This means there are at most five 1s, and at least five 2s. If there is a 5 and four 2s, there are six dice left (1s and 2s), so there is at least one pair, and we have 5 * 2 * (2 + 2/2) + 1 as a seven die solution. The remainder cancels out, so the wild die cannot be a 5.

If the wild die is a 6, then 6 * (1 + 1 + 1 + 1 + 1) + 1 a seven die solution and the remainder cancels out. This means there are at most five 1s, and at least five 2s. If there is a 6 and four 2s, there are six dice left (1s and 2s), so there is at least one pair, and we have 6 * 2 * (2 + 2/2) + 1 as a seven die solution. The remainder cancels out, so the wild die cannot be a 6.

Thus all eleven dice must be 1s or 2s.

If there are seven or more 2s, 2 * 2 * 2 * 2 * 2 - 2/2 is a solution, leaving four dice that cancel out. Thus, there are at most six 2s, and at least four 1s.

If there are five or more 2s, 2 * 2 * 2 * 2 * 2 - 1 is a six die solution, and the remainder cancels out. This leaves at most four 2s, and at least seven 1s.

If there are two or more 2s, 2 * 2 * (1 + 1) * (1 + 1) * (1 + 1) - 1 is a nine die solution with seven 1s (which is required). The remaining two dice are 1s or 2s, and cancel out (either as a pair or multiply by 2-1)

If there is a single 2, 2 * (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) - 1 is a ten die solution, with only 1s left, so the remainder cancels out. Thus, there are no 2s.

With eleven 1s, (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) - 1 is a solution.

Therefore, there is always a solution for level 4 with eleven dice, and there may not be with ten dice (all 1s).

Level 1: 4 (exact)
Level 2: 8 (exact)
Level 3: 9 (exact)
Level 4: 11 (exact)
Level 5:
Level 6:
Level 7:
Level 8:
Level 9: 17 (could be as low as 14)

It looks like for high levels the solution with all 1s is also the lower limit (my solution for 9 was done before I'd done this a lot - note that it's not true for Level 1), but that's partly because so far I've been lucky to have two convenient products close to the selected primes (12 and 18 for level 2, 18 and 24 for level 3, 32 and 36 for level 4). Level 5 has 48 and 54 so it should work, but level 6 only has 60 (which isn't that good), level 7 has only 72, level 8 is 96, and level 9 is 108. These aren't the only convenient products (100 is good for level 9 too), but it does suggest that the solution with only 1s may not be the 'hardest' one, so more dice can be required.

[Sith_Happens]

our admittedly rusty math and statistics knowledge.

I don't see where statistics comes into it. Meschlum is currently determining the minimums for guaranteed success. If he were looking for the minimums that provide, say, a 99.9% chance of success, that would involve statistics. I also suspect that most of those minimums are at least several ranks lower.

[grarrrg]

Minimum all 1's
For the sake of space and sanity, a "3" is actually "1+1+1"
Level 1: 3 (3 = 3)
Level 2: 8 (11 = 3*3+2)
Level 3: 9 (19 = 3*3*2+1)
Level 4: 11 (37 = 3*3*2*2+1)
Level 5: 12 (53 = 3*3*3*2-1)
Level 6: 13 (61 = 5*3*2*2+1 (holy cow! a FIVE!))
Level 7: 13 (73 = 3*3*2*2*2+1)
Level 8: 14 (83 = 3*3*3*3+2)
Level 9: 14 (107 = 3*3*3*2*2-1)

I agree that level 6 might be tricky, mainly due to me having to resort to a dirty, icky 5 to get to it.
Also, level 8 might be easier than you think. It winds up having the largest "spread" of all of them (i.e. level 1 ranges from 3 -to- 7, has a 5 point spread, level 8 ranges from 83 -to- 97, has a 15 point spread, the rest are all 7, 9 or 11). So while there is no 'easy' jump-in point, there are more potential jump-in points.

"Hail Mary" minimum (absolute lowest needed for a _possible_ success, however unlikely)
Level 1: 1 (3 = 3)
Level 2: 2 (11 = 6+5)
Level 3: 3 (19 = 6*3+1)
Level 4: 3 (37 = 6*6+1)
Level 5: 4 (43 = 6*6+6+1)
Level 6: 4 (61 = 6*5*2+1)
Level 7: 4 (73 = 6*6*2+1)
Level 8: 4 (89 = 6*5*3-1)
Level 9: 4 (107 = 6*6*3-1)
A couple of these go down a notch if using d8's with Calculating Mind (level 6 drops to 3 > 8*8-3)

[Khosan]

What you can also do to make it easier is break it up into sets of...2-3 or so.

At 2 or 3 dice, there are 21 or 56 possible combinations respectively you can get, which is pretty easy to solve for. You'd just need to list out every possible outcome from any possible combination of dice (and the equation that gets you that, but I figure the DM would stop asking by the umpteenth time). It should work pretty well, saving a lot of time and pretty much allowing you turn a great big pile of rolls into something meaningful.

I guess as an example:

[rollv]17d6[/rollv] - Apparently that doesn't work any more

I'll try to solve for a 9th level spell. Might take me a bit. I don't have a table of that information in front of me and it's 2AM so I'm not making it.

EDIT: http://invisiblecastle.com/roller/view/4592457/

So 3 1s, 5 2s, 3 3s, 2 4s, 1 5 and 3 6s. Looking to manage 101, 103 or 107.

1, 2, 2 gives me any value between 0 and 6, it's like a wildcard (and, I assume, a pile of 3-number combinations do the same), and I'll take the 6. Multiply by 6, multiply by 3 and we're up at 108. Subtract 1 and then clear the rest. Done.

EDIT: Actually, better would be to find those 'wildcard' number combinations that can solve into anything. They pretty much free you up to do whatever. After that, all you need is a free odd number to get you to the prime.

[Hecuba]

Some thoughts on doing this in a timely fashion.

3 1s, 5 2s, 3 3s, 2 4s, 1 5 and 3 6s. Looking to manage 101, 103 or 107

(borrowing you're rolls as an example).

Subtract a small number from the top prime target to get something that factors well, then negate the remaining dice. If you can't negate, you can likely just subtract & move down 1 target.

107-2 is 105. Thus

107=105+2=2+7*3*5=2+(3+4)*3*5

That leaves us

6-6+2*3-6+4-2-2+2/1-1-1=0

It won't always work, but it often will. And it will be much faster than working from the dice (since you can know the convenient decompositions of the target primes in advance).

[meschlum]

While there are benefits to processing a few dice at a time, there are no hard and fast rules for the number of dice you need to get a given value (it's the point of the exercise), so you're likely to wildly overestimate the number of dice needed if you break down the requirements that way. Using pairs to get 1 and 0 is the really critical part, as is knowing that four or more dice can always cancel out.

And you can figure out decompositions just by looking at the calculations I perform. Not always optimal, but guaranteed to work (other solutions exist too, usually) - the overall method is "Get the highest dice, see how you can combine them to come close, then tinker with the rest".

Spoiler: Let's see Level 5 with 12 dice

Show

If you have two 6s and a 4, (6 + 6) * 4 - 1 works, using the known dice and a pair. The rest cancels out.

If you have two 6s and a 2, (6 + 2) * 6 - 1 works, with the rest canceling out.

If you have two 6s and a 5, (5 + 2) * 6 + 1 works, with the rest canceling out.

If you have two 6s and a 3, (6 + 3) * 6 - 1 works, with the rest canceling out.

So if you have two 6s, all dice are 1s or 6s. With three or more 1s, you have (6 + 1 + 1) * 6 - 1, using five dice and the remaining seven dice cancel out. Therefore, there are at most two 1s and at least ten 6s. If the 1s are replaced with pairs of 6s in the equation, eight 6s are used, which is possible since there are at least ten. This leaves four or more dice, so the rest cancel out.

Therefore, there is at most a single 6.

If there are two 4s and a 3, 4 * 4 * 3 - 1 is a solution, using a pair and the rest cancels out.

If there are three 4s, 4 * 4 * (4 - 1) - 1 is a solution, using two pairs (possible with nine dice), and the rest cancels out.

If there are two 4s and a 2, 4 * 4 * (2 + 1) - 1 is a solution, with the rest canceling out.

If there are two 4s and a 6, 6 * (4 + 4) - 1 is a solution using one pair, and the rest cancels.

If there are two 4s and a 5, (5 + 1) * (4 + 4) - 1 is a solution using two pairs, and the rest cancels.

So if there are two 4s, the other dice are 1s, leading to ten or more 1s. 4 * 4 * (1 + 1 + 1) - 1 is a solution using six dice in this case.

Therefore, there is at most a single 4.

If there are two 3s and a 6, 3 * 3 * 6 - 1 is a solution using one pair, and the rest cancels out.

If there are two 3s and a 5, 3 * 3 * (5 + 1) - 1 is a solution using two pairs, and the rest cancels out.

If there are two 3s and a 4, 3 * 4 * (3 + 1) - 1 is a solution using two pairs, and the rest cancels out.

So if there are two or more 3s, all the other dice are 1s, 2s, or 3s.

From 3 * 3 * 3 * 2 - 1, using 3s and 2s as well as a pair (as there are 8 dice left after the 3s and 2s), we know the other dice cannot have 3s and 2s together.

If there is at least one extra 3, all the dice are 3s and 1s, 3 * 3 * 3 * (1 + 1) - 1 means there are two 1s at most, so ten or more 3s. (3 + 3) * 3 * 3 - 3/3 uses six dice, so the rest cancel out.

So there are two 3s at most, and the other dice are 1s and 2s. 3 * 3 * (2 + 2 + 2) - 2/2 is a seven die solution (canceling the rest), so there are four 2s at most, hence at least six 1s. 3 * 3 * (1 + 1 + 1) * (1 + 1) - 1 uses six 1s, and so is possible, using eight dice. The remaining four cancel out.

Thus there is at most a single 3.

If there is a 6, a 2, and a 5, then 6 * (2 + 5) + 1 is a solution. Thus, if there is a 6, there are either 1s and 5s or 1s and 2s besides the other wild die.

If there is a 6 and a 5, 6 * (5 + 1 + 1) + 1 is a solution, so there are two 1s at most. This means that there are at least eight 5s (6, 5, 1, 1, and a wild die are set, the remaining are 5s). (6 + 5) * 5 - (5 + 5) / 5 is a solution, using six dice and five 5s, so it works and the rest cancel out. Thus if there is a 6, all the other (non wild) dice are 1s or 2s.

6 * 2 * 2 * 2 - 2/2 is a solution with five 2s and six dice in all, so there are at most four 2s and at least six 1s (plus the wild die).

6 * (1 + 1 + 1 + 1) * (1 + 1) - 1 is a solution with six 6s and a pair, for nine dice in all. After using a 6 and six 1s, there are four non-wild dice left, which are all 1s or 2s, so there is a pair among them. The three or more dice left are 1s, 2s, and a wild die (which is not 6). If the wild die is 1 or 2, there is a pair and the remainder cancels out. If the wild die is a 4, it can replace four of the 1s, which leaves a pair of 1s to cancel the rest. If the wild die is a 3, it an replace three of the 1s, with the same result. Thus, there is always a solution when one of the wild dice is a 6, so the wild dice are either 3 or 4.

We therefore have ten dice that are 1, 2, or 5 and two wild ones that can be anything but 6 (but with a maximum of one 3 and one 4).

If there is a 3 and a 4, 3 * 4 * (2 + 2) - 2/2 means that there are at most three 2s. Replacing the 2s with 1 + 1 (and the 2/2 with a single 1) means that there are at most four 1s. Therefore, there are at least three 5s. 3 * 4 * (5 - 5/5) - 1 is a solution using one pair, for a total of seven dice. The remaining dice can cancel, so there cannot be a 3 and a 4, meaning there is a single 'wild' die which can be 3 or 4.

With three or more 5s, 5 * (5 + 5 - 1) - 2 is a solution, so there must be only 5s and 2s or only 5s and 1s.

5 * (5 + 5 - 5/5) - (5 + 5) / 5 is a solution using eight 5s, so there are at most seven 5s, and there are 2s or 1s. The remaining four or more dice cancel out.

With three or more 5s and three or more 2s, 5 * (5 + 5 - 2/2) - 2 is a solution, so if there are three or more 5s, there are two 2s at most. In this case, as there are eleven non wild dice, this means there are at least nine 5s, wich is enough for the solution above.

Similar reasoning can be applied with three or more 5s and three or more 1s, using 5 * (5 + 5 - 1) - 1 - 1.

Thus there are two 5s at most.

We have (5 + 2) * (5 + 2/2) + 2/2 as a solution, so if there are two 5s, there are at most four 2s. This means there are at least five 1s. Since (5 + 1 + 1) * (5 + 1) + 1 is a solution using four 1s, there is always a solution with two 5s.

Therefore, there is a single 5 at most.

If there is a 4 and a 5 (a 3 and a 5), then (4 + 5) * (1 + 1) * (1 + 1 + 1) - 1 (or (3 + 5) * (1 + 1 + 1) * (1 + 1) - 1)) is a solution using six 1s. Thus there are at most five 1s, and at least five 2s.

With a 4 and a 5, we have (4 + 2) * (5 + 2) + 2/2 as a six die solution with four 2s, so it is possibile, and there cannot be a 4 and a 5.

With a 3 and a 5, we have (3 + 5) * (2 + 2 + 2) - 2/2 as a seven die solution with five 2s, so it is possible and there cannot be a 3 and a 5.

Thus, there is a single wild die that can be 3, 4, or 5 and the other eleven dice are 1s and 2s.

From (5 + 2) * (2 + 2 + 2) + 2/2, we see that there are at most five 2s if there is a 5, hence at least six 1s.

A 5 and two or more 2s (and six or more 1s) is solved by (5 + 1 + 1) * (2 + 2 + 1 + 1) + 1, using eight dice, so the rest cancels out. Hence if there is a 5, there is at most a single 2 and the other dice are 1s (at least ten).

With 5 * (1 + 1 + 1) * (1 + 1 + 1) - 1 - 1, we have a nine die solution using eight 1s. This leaves at least two 1s, which cancel the rest via 1-1.

So the wild die cannot be 5.

From 4 * 2 * (2 + 2 + 2) - 2/2, we see that there are at most five 2s if there is a 4, so at least six 1s.

A 4 and two or more 2s (and six or more 1s) is solved by 4 * 2 * (2 + 1 + 1 + 1 + 1) - 1, using eight dice. Thus there is a single 2 at most and ten or more 1s if there is a 4.

With 4 * (1 + 1 + 1) * (1 + 1 + 1 + 1) - 1, we have a nine die solution using eight 1s, so there is a pair of 1s left to cancel the remainder.

So the wild die cannot be 4.

From 3 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) - 1, we get a ten die solution using nine 1s. The remaining dice are 1s and 2s, and will cancel out if there is a pair. If there is no pair, multiplying by 2-1 will cancel the rest. Thus there are eight 1s at most, and at least three 2s.

With 3 * 2 * 2 * 2 * 2 - 2/2, we get a solution using six 2s (and seven dice in all). Therefore, there are at most five 2s, and at least six 1s.

We get 3 * 2 * 2 * 2 * (1 + 1) - 1 using three 2s and three 1s, which satisfies the conditions above. Therefore there is always a solution when there is a 3.

So there is no wild die, and all dice must be 1s or 2s.

2 * 2 * 2 * 2 * (2 + 2/2) - 2/2 is a nine die solution. The remainder or three or more dice is 1s and 2s, so there is at least a pair to cancel it. Thus, there are at most eight 2s and at least four 1s.

With four or more 2s and at least four 1s, we get 2 * 2 * 2 * 2 * (1 + 1 + 1) - 1 is an eight die solution (the remaining four or more dice cancel out), so there must be three or less 2s (and nine or more 1s).

With two or more 2s and at least nine 1s, we get 2 * 2 * (1 + 1 + 1 + 1) * (1 + 1 + 1) - 1 as a ten die solution. The other two or more dice are 1s and 2s. If there is a pair, they cancel out. If there is none, then multiply by 2-1 to cancel the remaining two dice.

With one 2 and at least eleven 1s, we get 2 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) - 1 as an eleven die solution. The remainingoe or more dice are 1s, so they cancel via multiplication

With twelve 1s, there is a solution as well.

Level 1: 4 dice (exact)
Level 2: 8 dice (exact)
Level 3: 9 dice (exact)
Level 4: 11 dice (exact)
Level 5: 12 dice (exact)
Level 6:
Level 7:
Level 8:
Level 9: 17 dice (may be as few as 14)

[meschlum]

So... time to see how hard things get for Bards and the like.

Spoiler: Poking at Level 6 with 13 dice

Show

With three 6s, you have 6 * (6 + 6 - 1) + 1, using two pairs (allowed as you have ten 'free' dice) and seven dice in all, so the remainder cancels out. Thus you have two 6s at most.

With a 6 and a 5, you have 6 * 5 * (1 + 1) - 1, using three pairs (which your eleven 'free' dice allow) and eight dice in all, so the remainder cancels out.

With two 6s and a 4, you have 6 * (6 + 4) + 1, using one pair and five dice in all.

With two 6s and a 3, you have 6 * (6 + 3 + 1) + 1, using two pairs (possible as there are ten 'free' dice) and seven dice, so it cancels out.

So if you have two 6s, the other dice are only 1s and 2s.

From 6 * (6 - 2/2) * 2 + 2/2, using seven dice (so the rest cancels), we see that there are four 2s at most, and at least seven 1s.

With 6 * (6 - 1) * (1 + 1) + 1 using six dice and four 1s, we see that all cases with wo 6s are valid (as are all cases with one 6 and a 5).

If there are three 5s, there are no 6s, and all the dice are 1s to 5s, so the remaining ten dice include at least three pairs. (5 + 1) * (5 + 5 + 1) + 1 is a solution using nine dice, and the rest cancels out as there are thirteen or more dice in all.

So there are two 5s at most.

If there are two 5s and a 4, all the ten or more other dice are 1s to 4s, allowing three pairs. (5 + 5) * (4 + 1 + 1) + 1 is a solution using all three pairs and nine dice in all. The remaining four or more dice cancel out.

If there are two 5s and a 3, he same reasoning applies, using (5 + 5) * 3 * (1 + 1) + 1 instead.

So if there are two 5s, all other dice must be 1s or 2s.

We have 5 * 2 * 2 * (2 + 2/2) + 2/2 as an eight die solution (the rest cancels), so there are six 2s at most, and at least six 1s.

We have 5 * 2 * (1 + 1) * (1 + 1 + 1) + 1 as an eight de solution using six 1s and a single 2, so there cannot be a 2 if there is a 5.

We have 5 * (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1 as a nine die solution, with the rest canceling out (and all equal to 1 anyway).

So there is one 5 at most, and a single 'wild' die that can be equal to 5 or 6.

If there are three 4s, we have 4 * (4 - 1) * (4 + 1) + 1 as a solution, using nine dice. With nine 'free' dice (and a wild die) after setting three 4s, all in the 1 to 4 range, we have at least three pairs, so the solution is possible.

Therefore, there are two 4s at most.

Two 4s and a 6 give 4 * (4 - 1) * (6 - 1) + 1, using three pairs and nine dice. The remainder cancels out, and there are at least three pairs in ten dice ranging from 1 to 4, so it's possible.

Two 4s and a 5 give 4 * (4 - 1)* 5 + 1, using two pairs and seven dice, so if there are two 4s the other dice cannot be 5

Two 4s and a 3 give 4 * (4 + 1)* 3 + 1, using two pairs and seven dice, so if there are two 4s the other dice cannot be 3

So if there are two 4s, the other dice must be 1s or 2s.

We have 4 * (2 * 2 * 2 * 2 - 2/2) + 2/2 as a nine die solution with eight 2s, so there are at most seven 2s and at least five 1s.

We have 4 * (2 * 2 * 2 * (1 + 1) - 1) + 1 as an eight die solution with four 1s, so there are at most two 2s and at least ten 1s.

We have 4 * (1 + 1 + 1 + 1 + 1) * (1 + 1 + 1) + 1 as a ten die solution with nine 1s, and the remaining three or more dice are all 1s and 2s, so they include a pair, which cancels the rest.

Therefore, there is at most a single 4, and (apart from the 'wild' die) all other dice are 1s to 3s.

If there is a 4, a 3, and a 6, we have 4 * (6 - 1) * 3 + 1 as a solution using seven dice and two pairs (which is possible since there are ten dice in the 1 to 3 range). Therefore, with a 4 and a 6 (or a 4 and a 5, replacing the 6-1 with a 5), there cannot be any 3s.

If there is a 4, a 2, and a 6, we have 6 * 2 * (4 + 1) + 1 as a solution following the same criteria as above. Replacing the 6 with a 5 + 1 requires three pairs, which is possible since there are ten dice in the 1 to 2 range.

So if there is a 4 and a 6, or a 4 and a 5, the other dice must be 1s. 5 * 4 * (1 + 1 + 1) + 1 and (6 + 4) * (1 + 1 + 1) * (1 + 1) + 1 are solutions, meaning that 4 is on the 'wild' die only.

So we have twelve dice in the 1 to 3 range (guaranteeing five or more pairs), and one wild die which can be 1 to 6.

We have (3 + 3) * (3 * 3 + 1) + 1 as a solution using four 3s and two pairs (possible with eight dice in the 1 to 3 range), for eight dice in all. Thus there are three 3s at most.

With three 3s, we have nine or more 1s and 2s. (3 + 3) * (3 * 2 * 2 - 2) + 2/2 is a solution using five 2s and eight dice, so there are at most four 2s and five or more 1s. (3 + 3) * (3 * (1 + 1 + 1) + 1) + 1 is a solution using five 1s and eight dice, so there is always a solution if there are three 3s exactly.

With two 3s, we have ten or more 1s and 2s. (3 + 3) * (2 * 2 * 2 + 2) + 2/2 is a solution with six 2s and eight dice, so there are at most five 2s and five or more 1s.

We have (3 + 3) * (2 * 2 * (1 + 1) + 1 + 1) + 1 as a solution using five 1s, two 2s, and nine dice in all, so the rest cancels out and there is at most a single 2, so nine or more 1s.

We have (3 + 3) * ((1 + 1 + 1) * (1 + 1 + 1) + 1) + 1 as a solution using eight 1s and ten dice in all. The remaining dice include at least one 1, and the other two can be 1s, 2s, or the wild die. If there is a pair, the remainder cancels out, so the three dice must be 1, 2, and 3 to 6.

If the wild die is a 3, there are three 3s, and there is a solution.

If the wild die is a 4, we have (3 + 3) * (4 + 1) * (1 + 1) + 1 as a solution with seven dice and four 1s, so it is valid.

If the wild die is a 5, we have the same as the above, with 5 instead of 4+1

If the wild die is a 6 we have the same as the above, with 6-1 instead of 4+1

So the remaining three dice either cancel out or there is a solution using the wild die instead.

Therefore, there is a single 3 at most, and the other dice are 1s or 2s (besides the wild die).

With a 3 and a 4, we have 3 * 4 * (2 + 2 + 2/2) + 2/2, so there are five 2s at most (eight dice used, so the rest cancels), and at least six 1s. Six 1s can be used to replace the 2s in the solution exactly, so there is always a solution in this case.

With a 3 and a 6, we have 3 * (6 - 2/2) * (2 + 2) + 2/2, so there are five 2s at most, and at least six 1s. Again, the six 2s can be replaced with six 1s in the specified solution, so there is a solution in this case. If we replace the 6 with a 5 * 1 (or 5 * 2/2), the same formula applies.

Therefore, the wild die can be 3 to 6, and the twelve other dice are 1s and 2s.

From 3 * 2 * 2 * (2 + 2 + 2/2) + 2/2, we have a nine die solution using eight 2s, so there are seven 2s at most and at least five 1s.

With 3 * 2 * 2 * (2 + 1 + 1 + 1) + 1, we have an eight die solution using four 1s, so there are two 2s at most and at least ten 1s.

With 3 * (1 + 1 + 1 + 1 + 1) * (1 + 1 + 1 + 1) + 1, we have an eleven die solution using ten 1s, and the remaining two or more dice are either a pair (and cancel out) or 1 and 2, and we can multiply the result by 2-1.

So the wild die cannot be a 3.

With 4 * (2 + 2/2) * (2 + 2 + 2/2) + 2/2, we have a ten die solution using nine 2s. The three or more remaining dice are 1s or 2s, so include a pair and cancel out. Thus we have at most eight 2s and at least four 1s.

With 4 * (2 + 1) * (2 + 2 + 1) + 1, we have a seven die solution using three 2s and four 1s. Thus there are at most two 2s, and at least ten 1s.

With (4 + 1) * (1 + 1 + 1) * (1 + 1 + 1 + 1) + 1, we have a ten die solution using nine 1s. The remaining three or more dice are 1s or 2s, and so include a pair and cancel out.

So the wild die cannot be a 4.

With 5 * 2 * 2 * (2 + 2/2) + 2/2, we have an eight die solution using seven 2s. Thus there are at most six 2s, and at least six 1s.

With 5 * 2 * (1 + 1) * (1 + 1 + 1) + 1, we have an eight die solution using six 1s and a 2. Thus there are no 2s, and at least twelve 1s. The 2 can be replaced with two 1s, giving a nine die solution with eight 1s, so it's allowed.

So the wild die cannot be a 5.

With 6 * (2 + 2 + 2 + 2 + 2) + 2/2, we have an eight die solution using seven 2s. Thus there are at most six 2s, and at least six 1s.

With 6 * 2 * (1 + 1 + 1 + 1 + 1) + 1, we have an eight die solution using six 1s and a single 2. Thus there must be at least twelve 1s. We can replace the 2 with 1+1, for a nine die solution using a 6 and eight 1s.

So there is no wild die, and all dice must be 1s and 2s.

With 2 * 2 * (2 * 2 * 2 * 2 - 2/2) + 2/2, we have a ten die solution. The three or more remaining dice are 1s and 2s, so there is a pair and they cancel out. Thus there are at most nine 2s and at least four 1s.

With 2 * 2 * (2 * 2 * 2 * (1 + 1) - 1) + 1, we have a nine die solution using four 1s and five 2s. Thus there are at most four 2s, and at least nine 1s.

With 2 * 2 * (2 + 2 + 1) * (1 + 1 + 1) + 1, we have a nine die solution with four 2s, so there are three 2s at most (and at least ten 1s).

With 2 * 2 * (2 + 1) * (1 + 1 + 1 + 1 + 1) + 1, we have a ten die solution with three 2s exactly, and the remaining dice cancel out a they are all 1s. So there are two 2s at most.

With 2 * 2 * (1 + 1 + 1) * (1 + 1 + 1 + 1 + 1) + 1, we have an eleven die solution. Again, the 1s cancel out. The 2s can be replaced with 1+1, adding one die to the solution size. Since the remaining dice are all 1s, they cancel out anyway.

So 13 dice work out.

Level 1: 4 (exact)
Level 2: 8 (exact)
Level 3: 9 (exact)
Level 4: 11 (exact)
Level 5: 12 (exact)
Level 6: 13 (exact)
Level 7:
Level 8:
Level 9: 17 (could be as low as 14)

It's definitely looking like practice and the range of numbers we have mean that 14 dice are all you'll need at level 9.

Spontaneous casters (including level 6 limits like bards) are about 1 level behind at low levels, and then catch up or outgrow their limits (bards before sorcerers). Prepared casters are more like 2 levels behind up to level 12, and then catch up over the course of (probably) two levels. A rather traditional caster progression, granting lots of power at high levels for some inconvenience at low levels.

[meschlum]

There are numbers to crunch!

Spoiler: Level 7, 13 dice or more, let's count the ways

Show

With two 6s, we have 6 * 6 * (1 + 1) + 1 as a solution, using three pairs (which is possible as there are eleven free dice) and eight dice, so the rest cancels out.

There is at most a single 6.

With three 4s, we have 4 * 4 * (4 + 1) - 1 as a solution, using two pairs (possible as there are ten free dice) and seven dice, so there are at most two 4s.

Two 4s and a 5 is a solution by the equation above (using a single pair)

Two 4s and a 6 is a solution by the equation above (with 6 - 1 instead of 4 + 1)

So if there are two 4s, all other dice are 1s to 3s.

With two 4s and two 3s, we have (4 + 4) * 3 * 3 + 1 using six dice, so there is one 3 at most.

With two 4s and one 3, all other dice are 1s and 2s. (4 + 4) * 3 * (2 + 2/2) + 2/2 uses eight dice, so there are at most four 2s and at least six 1s. (4 + 4) * 3 * (1 + 1 + 1) + 1 uses seven dice and four 1s, so there is always a solution with two 4s and a 3.

So if there are two 4s, the other dice are 1s and 2s.

We have 4 * 4 * (2 + 2 + 2/2) - 2/2 as an eight die solution, so there are at most five 2s, and at least six 1s. 4 * 4 * (1 + 1 + 1 + 1 + 1) - 1 uses six 1s and eight dice, so there is always a solution if there are two 4s.

With a 4 and a 6, we have 4 * 6 * 3 + 1 as a five die solution (including a pair), so there cannot be a 3. 4 * 6 * (2 + 1) + 1 is a seven die solution (including two pairs, possible as there are ten free dice), so there are no 2s.

With a 4 and a 6, the other eleven or more dice must be 1s or 5s. If there is a 5, there are four pairs, and we have 5 * 4 * (6 - 1 - 1) - 1 as a solution, using nine dice. The remaining four cancel out, so we have 4, 6, and 1s. 4 * 6 * (1 + 1 + 1) + 1 is a six die solution.

Therefore, there cannot be a 4 and a 6, so there is a single 'wild' die that can be 4 or 6, and the other dice are 1, 2, 3, or 5.

If there is one 3 and two 5s, we have 5 * (3 + 5) * (1 + 1) - 1 as a solution using three pairs (we have nine free non wild dice, so three pairs are possible as non wild dice have only four possible values) with nine dice in all, so the remaining four or more cancel out. So if there are two or more 5s, other non wild dice are 1, 2, or 5.

If there are four or more 5s, we have (5 * 5 - 5) * (5 - 1) - 1 as an eight die solution using two pairs. With eight non wild dice, two pairs are possible, so we have three 5s at most.

With exactly three 5s, we have (5 * 5 - 5) * (2 + 2) - 1 as a six die solution using a pair (there are eight non wild dice worth 1, 2, or 5, so one pair is possible). Therfore, there is a single 2 at most, and at least nine 1s. We have (5 * 5 - 5) * (1 + 1 + 1 + 1) - 1 as an eight die solution using five 1s, so there is always a solution with exactly three 5s.

With exactly two 5s, the remaining eleven or more dice are 1s and 2s (ensuring five pairs). We have (5 * 5 - 1) * (2 + 1) + 1 using three pairs and nine dice, so there cannot be a 2. Replacing the2 with two 1s, and the pairs with 1s as well, gives a seven die solution.

Therefore, there is at most a single 5, and other non wild dice are 1, 2, or 3.

If there is a 4 and a 5, we have 4 * 5 * (3 + 1) - 1 as a solution using two pairs and seven dice (there are ten free dice, so two pairs are possible). 4 * 5 * (2 + 2) - 1 is a solution as well, using six dice, so there is at most a single 2 and no 3s if there is a 4 and a 5. This means there are at least ten 1s, and 4 * 5 * (1 + 1 + 1 + 1) - 1 is a solution.

If there is a 6 and a 5, we have 6 * (5 + 1) * 3 + 1 as a solution using two pairs and seven dice again. Thus there cannot be any 3s in this case. 6 * (5 + 1) * (2 + 1) + 1 uses three pairs, drawn from ten free dice worth 1 or 2, which is possible. So with a 6 and a 5, the other dice are 1s, and we have 6 * (5 + 1) * (1 + 1 + 1) + 1 as a seven die solution.

Therefore, the wild die can be 4, 5, or 6 and the other dice are all 1s, 2s, and 3s.

If there are four 3s, 3 * 3* (3 * 3 - 1) + 1 uses eight dice and two pairs, so it is a solution. Therefore, there are at most three 3s.

If there are two or more 3s, 3 * 3 * 2 * 2 * 2 + 1 uses seven dice and one pair, so it is a solution. Therefore, if there are two or more 3s, there are at most two 2s. With three or fewer 3s and two 2s at most, this leaves seven 1s at least (plus the wild die).

With two 3s and seven 1s, we have 3 * 3 * (1 + 1) * (1 + 1) * (1 + 1) + 1 as a nine die solution, so the rest cancels out.

Therefore, there is a single 3 at most.

If there is a 3 and a 4, we have 3 * 4 * (2 + 2 + 2) + 1 as a seven die solution (one pair), so there are two 2s at most, and at least nine 1s. The solution can be reproduced by replacing 2 + 2 + 2 with (1 + 1) * (1 + 1 + 1), using six 1s in all, so there is always a solution.

If there is a 3 and a 5, we have 3 * (2 + 2) * (5 + 1) + 1 as an eight die solution (two pairs), so there is at most a single 2 and at least ten 1s. The solution can be reproduced by replacing 2 + 2 with 1 + 1 + 1 + 1, using six 1s in all, so there is always a solution.

If there is a 3 and a 6, we have 3 * 6 * (2 + 2) + 1 as a six die solution (one pair). As above, we can replace the 2s with 1s and get a solution as well.

Therefore, 3 can be present on the wild die only.

We have twelve dice that can be 1s or 2s, and one wild die which can be 1 to 6.

If there is a 6, we have 6 * (2 + 2) * (2 + 1) + 1 as an eight die solution using two pairs. There are therefore two 2s or less, and at least ten 1s. The three 2s in the solution can thus be replaced with six 1s for a nine die solution using eight 1s, and the rest cancels out.

If there is a 5, we have 5 * (2 + 2 + 2 + 2) - 1 as a seven die solution using one pair. There are therefore three 2s or less, and at least nine 1s. The four 2s in the solution can be replaced with eight 1s for a ten die solution using nine 1s. The remaining three dice are 1s or 2s, so there is a pair and they cancel out.

If there is a 4, we have 4 * (2 + 2 + 2) * (2 + 1) + 1 as a nine die solution using two pairs. There are therefore three 2s or less, and at least nine 1s. 4 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) + 1 is ten die solution using nine 1s, and the remaining three dice are 1s or 2s, so there is a pair and they cancel out.

If there is a 3, we have 3 * (2 + 1) * 2 * 2 * 2 + 1 as a nine die solution using two pairs. There are therefore three 2s or less, and at least nine 1s.

If there is a 2, we have 3 * (1 + 1 + 1) * 2 * (1 + 1) * (1 + 1) + 1 as a ten die solution using eight 1s, and the remaining three dice include at least one pai as they are 1s and 2s.

If there is a 3, all the other dice are therefore 1s, and 3 * (1 + 1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) + 1 is an eleven die solution. The remaining dice are 1s, so they cancel out.

Therefore, all the dice are 1s and 2s.

If there are five 2s, (2 + 1) * (2 + 1) * 2 * 2 * 2 + 1 uses eleven dice and three pairs, which is possible as there are eight dice which are equal to 1 or 2. The remaining two or more dice either have a pair and cancel out, or are 1 and 2, and multiplying by 2-1 cancels them. Therefore, there are at most four 2s, and at least nine 1s.

We can replace the 2s and pairs with 1s, to get (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) * 2 * 2 + 1 with eleven dice and nine 1s. Again the remaining dice cancel out as they must be 1s or 2s, and there is at most a single 2 and at least twelve 1s.

2 * (1 + 1) * (1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) + 1 is a twelve die solution, and the remaining dice are 1s, so they cancel out.

There is a thirteen die solution using only 1s, so level 7 works as expected.

Level 1: 4
Level 2: 8
Level 3: 9
Level 4: 11
Level 5: 12
Level 6: 13
Level 7: 13
Level 8:
Level 9: 17 (probably 14)

An inverted approach, seeing the maximum number of 1s for each case, could work too, but it looks somewhat harder to extend to an arbitrary number of dice (if it works for X dice, you want to be sure that it also works for X + 1, X + 2, etc.)

[Snowbluff]

Nice work. This is great.

[meschlum]

Almost done. Time to try a two for one wall of spoilered text, shall we?

Spoiler: You roll 14 dice and what do you get?

Show

If you have a 4, 5, and 6, then you have eleven other dice, which guarantees three pairs. (4 - 1) * (5 + 1) * 6 - 1 is a nine die solution for Level 9, and 6 * 4 * (5 - 1) + 1 is a seven die solution for level 8. In both cases, the extra dice cancel out, so there is always a solution if you roll, 4, 5, and 6.

If you have two 4s and one 6, the level 8 solution requires even less dice, and 4 * (4 + 1) * (6 - 1) + 1 is a nine die solution for level 9.

If you have one 4 and two 6s, (6 + 6) * 4 * (1 + 1) + 1 is a nine die solution for level 8, and 6 * 6 * (4 - 1) - 1 is a seven die solution for level 9.

So if you have 4s and 6s, there is at most one of each, and no 5s.

The remaining twelve dice are 1s, 2s, and 3s, guaranteeing five pairs. 4 * 6 * (1 + 1 + 1 + 1) + 1 is a twelve die solution for level 8. If the remaining two or more dice are pairs or one apart (1, 2 or 2, 3), then they cancel out. If they are 1 and 3, then the (1 + 1 + 1 + 1) term can be replaced with (1 + 3), which leaves four pairs to cancel out the rest. If there is a 3, 6 * 3 * (4 + 1 + 1) - 1 is a nine die solution for level 9. Without 3s, there are twelve dice that are 1s and 2s, and (6 - 1) * (4 + 1) * 2 * 2 + 1 uses two 2s and three pairs (possible with ten dice that are 1s and 2s) for a total of ten dice. So there is at most a single 2, meaning there are at least eleven 1s, and the solution above is possible with seven 1s, with a total of nine dice.

So there cannot be 4s and 6s together.

Two 5s and one 6 give (6 - 1) * 5 * (5 - 1) + 1, a nine die solution for level 9, and 6 * (5 - 1) * (5 - 1) + 1, a nine die solution for level 8.

Two 6s and one 5 give 6 * 6 * (5 - 1 - 1) - 1, a nine die solution for level 9, and 6 * 6 * 5 / (1 + 1) - 1 is a nine die solution for level 8.

So if you have 5s and 6s, there is at most one of each, and no 4s.

The remaining twelve dice are 1s, 2s, and 3s, guaranteeing five pairs. (5 + 1) * 6 * (1 + 1 + 1) - 1 is a twelve die solution for level 9, and the remaining dice cancel out unless they are 3 and 1, in which case the (1 + 1 + 1) can be replaced by 3, leaving three pairs to cancel the rest. 5 * 6 * (1 + 1 + 1) - 1 is a ten die solution for level 8, and the rest cancels out.

So there cannot be 5s and 6s together, meaning that if there are one or more 6s, the other dice are 1s, 2s, and 3s.

Two 4s and one 5 give 4 * (4 + 1) * 5 +1 for level 9, and 4 * 4 *(5 + 1) + 1 for level 8.

Two 5s and one 4 give 4 * 5 * 5 + 1 for level 9, and 4 * (5 - 1) * (5 + 1) + 1 for level 9, a nine die solution.

So if you have 4s and 5s, there is at most one of each, and no 6s.

The remaining twelve dice are 1s, 2s, and 3s, guaranteeing five pairs. If there is a 3 and a 2 (so ten free dice, and four pairs), then 4 * 5 * (3 + 2) + 1 is a six die solution for level 9, and 4 * 2 * (5 + 1) * (3 - 1) + 1 is a solution for level 8 that uses three pairs and ten dice, and the other four cancel out. So the remaining twelve dice are 1s and 2s or 1s and 3s (and there are at least five pairs). If there is a 3, then 4 * 5 * (3 + 1 + 1) + 1 is a nine die solution for level 9, and 4 * (3 + 1) * (5 + 1) + 1 is a nine die solution for level 8. Thus the remaining twelve dice are 1s and 2s. If there is a 2, then 4 * 5 * (2 + 1 + 1 + 1) + 1 uses eleven dice as a solution for level 9, and the remaining three dice include a pair and so cancel out. 4 * 2 * (1 + 1) * (5 + 1) + 1 is an eleven die solution for level 8, and therefore works, using the same reasoning. Thus we have 4, 5, and twelve 1s. Replacing the 2s in the formula for using a 2, and replacing the pairs with 1s, gives eight dice for level 9 and level 8 solutions.

So there cannot be 4s and 5s together, meaning that we need to review cases with 1s, 2s, 3s, and 4s, cases with 1s, 2s, 3s, and 5s, and cases with 1s, 2s, 3s, and 6s.

If there are two 6s, there are twelve other dice, so at least four pairs, and we have 6 * 6 * (1 + 1 + 1) - 1 as a ten die solution for level 9, so there is at most a single 6. With level 8, we use 6 * (6 + 1) * (1 + 1) - 1 to get the same result.

If there are two 5s and a 3, we have 5 * 5 * (3 + 1) + 1 as a seven die solution for level 9, and 5 * (5 + 1) * 3 - 1 for level 8. So if there are two or more 5s, there are no 3s. Two 5s and a 2 gives 5 * 5 * (2 + 1 + 1) + 1 for level 9, and 5 * (5 + 1) * (2 + 1) - 1 for level 8, both as nine die solutions using three pairs. So if there are two or more 5s, there are only 5s and 1s, which means there are at least five pairs among the remaining twelve dice. Replacing the 2s in the previous equations with 1+1 (two pairs) gives twelve die solutions using five pairs. If the remaing two dice are a pair, they cancel out. Otherwise, we have a 5 and a 1, so there are at least three 5s and one 1 (and four pairs). In this case, 5 * 5 * (5 - 1) + 1 is a six die solution for level 9, and (5 + 5 - 1) * 5 * (1 + 1) - 1 uses ten dice and three pairs for level 8.

Therefore, there is at most a single 5.

With two 4s, there are twelve other dice and at least four pairs, so 4 * 4 * (1 + 1 + 1) + 1 is a ten die solution for level 8.

If there are two 4s and one 3 there are eleven other dice (at least four pairs), and we have 4 * 3 * (4 * (1 + 1) + 1) - 1 as an eleven die solution for level 9. The remaining dice cancel out if there are pairs, so they are three values from 1, 2, 3, 4. If there is a 2 remaining, it can replace the (1 + 1) in the equation, leaving two pairs to cancel everything. So the remaining dice are 1, 3, 4, and 4-3-1 is zero. Therefore, there are no 3s for level 9 if there are two 4s.

If there are three 4s, we have (4 + 1) * (4 + 1) * 4 + 1 as a nine die solution for level 9, so there are two 4s at most.

If there are two 4s and two 2s, we have (4 + 1) * (4 + 1) * 2 * 2 + 1 as a ten die solution for level 9, so there is one 2 at most, and there are at least eleven 1s. These can replace the 2s and pairs in the solution, for a nine die solution using seven 1s, so there is always a solution if there are two or more 4s.

Therefore, there is a single 'wild' die that can be equal to 4, 5, or 6, and the other thirteen dice are all 1s, 2s, or 3s.

3 * 3 * 3 * (3 + 1) - 1 is a solution for level 9 with four 3s and two pairs, so eight dice in all. Therefore, there are at most three 3s for level 9. 3 * 3 * (3 * 3 + 1) - 1 is a solution for level 8 with the same number of 3s, so the same restriction applies.

With three 3s, there are eleven other dice (with three possible values), which allows at least four pairs, and we have 3 * 3 * 3 * 4 - 1, 3 * 3 * 3 * (5 - 1) - 1 and (3 + 3) * 3 * 6 - 1 as six and eight die solutions for level 9 if there is a 4, 5, or 6, so for level 9 all dice must be 1s, 2s, and 3s if there are three 3s. Level 8 can be dealt with as 3 * (3 * 3 - 1) * 4 + 1, 3 * (3 + 5) * (3 + 1) + 1, and 3 * (3 + 6 - 1) * (3 + 1) + 1 for eight and ten die solutions. Therefore, if there are three 3s, the other dice must be 1s and 2s. 3 * 3 * 3 * 2 * (1 + 1) - 1 uses three pairs and ten dice for level 9, so level 9 has only 1s - and replacing the 2 and pairs with 1s gives a solution. (3 * 3 + 1) * 3 * (2 + 1) - 1 uses three pairs and ten dice for level 8, so the same reasoning applies.

There are therefore two 3s at most.

With two 3s, if the wild die is a 4, 5, or 6, we have eleven dice that are 1s or 2s, so five pairs. 3 * 3 * 4 * (1 + 1 + 1) - 1 is an eleven die solution for level 9, and the remaining three dice are 1s or 2s, so there is a pair and they cancel out. (3 + 1) * 3 * 4 * (1 + 1) + 1 is also an eleven die solution for level 8, so it works as well. 3 * 3 * (5 + 1) * (1 + 1) - 1 is another eleven die solution for level 9 and 3 * (3 * 5 + 1) * (1 + 1) + 1 is an eleven die solution for level 8. 3 * 3 * 6 * (1 + 1) - 1 is a nine die solution for level 9, and 6 * (3 + 3 + 1) * (1 + 1) - 1 is an eleven die solution for level 8.

So if there are two 3s, the twelve or more other dice are 1s and 2s, for at least five pairs (and two or more remaining dice will cancel out).

We have 3 * 3 * (2 + 1) * (2 + 2) - 1 as a nine die solution for level 9, so there are at most two 2s if there are two 3s. Level 8 is done via (3 + 1) * (2 + 2) * 2 * 3 + 1, for the same criterion. With two 3s and two 2s at most, we have at least ten 1s, and the 2s and pairs in the earlier solutions can be replaced by these 1s, eight in all. This gives ten die solutions, so there is always a solution if there are two 3s.

With one 3 and the wild die at 4, 5, or 6, we have twelve dice that are 1s or 2s, so five pairs. 3 * 4 * 2 * 2 * (1 + 1) + 1 is a ten die solution for level 8, and 3 * 4 * (2 + 1) * (2 + 1) - 1 is its level 9 analog. Replacing the 4 with 5-1 gives a twelve die solution in either case, and the remaining two dice are 1s and 2s, so they cancel out (either a pair or 2-1). (3 + 1) * 6 * 2 * 2 + 1 is an eight die solution for level 8, and (3 + 6) * 2 * 2 * (1 + 1 + 1) - 1 is a twelve die solution for level 9 (the remaining two dice cancel out as they are 1s and 2s).

So the wild die includes 3, 4, 5, and 6, and the other thirteen dice are all 1s or 2s (six pairs).

2 * 2 * 2 * 2 * 2 * (2 + 1) + 1 is a ten die solution for level 8, using two pairs. Thus level 8 has five 2s at most, and at least eight 1s.

2 * 2 * 2 * (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1 is an eleven die solution for level 8, using eight 1s. The remaining dice are 1s, 2s, and the wild die, so they cancel out unless they are 1, 2, and the wild die. If the wild die is 3, 3-2-1 cancels out. With 4, muiplying by 4-2-1 works. With 5, we can replace (1 + 1) * (1 + 1 + 1) with 5 + 1, so there are multiple 1s left and they cancel out. The same is done if the wild die is 6.

Therefore, there are at most two 2s, and at least eleven 1s. We can replace one of the 2s in the solution above with (1 + 1), for a twelve die solution using two 2s and ten 1s, and the remaining dice are a 1 and the wild die. If it is 1 or 2, it cancels out directly. If it is 3, it can replace the (1 + 1 + 1), and the rest cancels out. Similar replacements are possible for the other wild die values.

So there is at most one 2, and twelve 1s. Another 2 can be replaced, leaving a thirteen die solution with one 2 and the wild die. Again, replacements are possible for all the wild die values.

There is a solution for level 8 with fourteen 1s.

For level 9, we have (2 + 2 + 2) * (2 + 2 + 2) * (2 + 1) - 1 as an eleven die solution using two pairs. A wild die of 3, 4, or 6 can be replaced to cancel the remainder. If the remaining dice are 1, 2, and 5, we can replace three 2s with 1 + 5, which gives a pair of 2s to cancel the rest. So there are at most six 2s, and at least seven 1s.

With seven 1s, we can use 2 * (2 + 2 + 2) * (1 + 1 + 1) * (1 + 1 + 1) - 1, as an eleven die solution using seven 1s. Again, a wild die of 3, 4, or 6 can be replaced, and 2 + 2 + 2 can be replaced by 1+ 5. So there are at most three 2s, and at least ten 1s.

With ten 1s, we can use 2 * 2 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) - 1, a twelve die solution using ten 1s. The remaining dice are 1s, 2s, or the wild values. 1s, 2s, and pairs cancel out. Again, 3, 4, and 6 can be replaced, so the options for the remaining dice are 1, 5 and 2, 5. The 5 can replace 2 * (1 + 1 + 1) with 5 + 1, freeing a 2 and two 1s, so there is a pair to cancel the remainder. Thus, there is a single 2 at most, and at least twelve 1s.

This gives 2 * (1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) - 1, a thirteen die solution. The remaining die is the wild die. A value of 1 cancels out, 2 replaces (1 + 1) and the pair of 1s cancel out, 3, 4, and 6 can also be cancelled, and 5 can be replaced as well.

Thus there are no 2s in the solution, and there is a fourteen die solution for level 9.

Level 1: 4 dice
Level 2: 8 dice
Level 3: 9 dice
Level 4: 11 dice
Level 5: 12 dice
Level 6: 13 dice
Level 7: 13 dice
Level 8: 14 dice
Level 9: 14 dice

Using more than one prime as the target is critical to getting the results this low. Beyond that, the general outcome is that you need lots of dice only if you roll a lot of 1s and 2s. Get a few 4s, 5s, and 6s, and you're pretty much set with (possibly much) lower dice pools.

Sadly, this ends my math frenzy. Maybe Pathfinder will come up with some other silly numeric manipulation?

[Serafina]

Congratulations, you just made Sacred Geometry usable - a player can just point to your results, treat the minimum number of dice as a minimum rank requirement and just skip actually doing all the math since there is a proven solution anyway.

Which is a good thing since it will prevent huge slowdowns, and a bad thing since the feat is still overpowered. Oh Paizo

If you want more math, you could analyze the follow-up feat that allows using D8s

[grarrrg]

Level 1: 4 dice
Level 2: 8 dice
Level 3: 9 dice
Level 4: 11 dice
Level 5: 12 dice
Level 6: 13 dice
Level 7: 13 dice
Level 8: 14 dice
Level 9: 14 dice

This exactly matches the "all 1's" results.
What a bunch of wasted time! You could have just gone with the all 1's numbers and been done!

But seriously.
Low levels the feat isn't that good, mainly due to lack of options/spell-levels "Woo, I can get a Free +1 Meta on a Cantrip!"
But once you hit level level 13+ everything auto-succeeds making you grossly uber-powerful.

Good job Paizo...good job...you just made Casters better. Again. Because we all know they needed help. Especially at high levels.

[Kantolin]

Well, also remember: This is just the 100% success rate.

You can get a 'you are extremely unlikely to fail' success rate from just having six or seven dice. Even with five you're unlikely to fail, especially a +1.

[Snowbluff]

Congratulations, you just made Sacred Geometry usable - a player can just point to your results, treat the minimum number of dice as a minimum rank requirement and just skip actually doing all the math since there is a proven solution anyway.

Which is a good thing since it will prevent huge slowdowns, and a bad thing since the feat is still overpowered. Oh Paizo

Yeah, you lose the fun and slowdown for power and faster pacing. Which is a good trade off. Like I said earlier, I wish their was a way to make this feat functional while still requiring a puzzle.

I suggest Echoing and Quicken as MM for this. Good action economy and longevity for your caster. However, you can take this multiple times for greater effect.

[grarrrg]

so long as you have four or more dice, you can generate a zero

Was thinking about the "4+ dice go bye-bye" thing today.
It's actually _3_+ dice go bye-bye.

While you do need 4 (or more) to guarantee a 0 or a 1, 3 dice will guaranteed get you a 0, a 1, or combine to make any fourth number you add.
Example:
4, 5, 6 will not get a 0 or a 1, BUT
4, 5, 6, 1 > (6+4)/5 - 1 = 1
4, 5, 6, 2 > 5 - 6*2/4 = 2
4, 5, 6, 3 > 6*4/3 - 5 = 3
4, 5, 6, 4 > (6*4 - 4)/5 = 4
4, 5, 6, 5 > (6-4)*5 - 5 = 5
4, 5, 6, 6 > (6*5 - 6)/4 = 6

I've ran the numbers, and this works for all combinations of 3 dice (will post if needed, or if I get bored later).
If 3 dice do not reduce to a 0 or a 1, they WILL let you "convert" any other number into itself.

Spoiler: EDIT: BORED LATER!

Show

Start with all Unique 3-die combinations (56 unique), then remove any with a duplicate number, as these easily reduce to 0 (20 non-doubles). Finally, remove all the combinations that reduce to 0 or 1. This leaves 5 combinations:
1, 2, 5
1, 2, 6
3, 4, 5
3, 5, 6
4, 5, 6
We already proved 4 ,5, 6 in the example

1, 2, 5, 1 > 5-2-1-1 = 1
1, 2, 5, 2 > 5-2-2+1 = 2
1, 2, 5, 3 > 5-3+2-1 = 3
1, 2, 5, 4 > 5-4+2+1 = 4
1, 2, 5, 5 > (5*2-5)*1 = 5
1, 2, 5, 6 > (5+1)*2-6 = 6

1, 2, 6, 1 > 6/2-1-1 = 1
1, 2, 6, 2 > (6-2-2)*1 = 2
1, 2, 6, 3 > 6/3+2-1 = 3
1, 2, 6, 4 > (6-4+2)*1 = 4
1, 2, 6, 5 > (6-1)*2-5 = 5
1, 2, 6, 6 > (6*2-6)*1 = 6

3, 4, 5, 1 > 4+3-5-1 = 1
3, 4, 5, 2 > 5+3-4-2 = 2
3, 4, 5, 3 > 5*3-4*3 = 3
3, 4, 5, 4 > (5-3)*4-4 = 4
3, 4, 5, 5 > 5*4-5*3 = 5
3, 4, 5, 6 > 5+4+3-6 = 6

3, 5, 6, 1 > 5+3-6-1 = 1
3, 5, 6, 2 > 6+3-5-2 = 2
3, 5, 6, 3 > (5*3-6)/3 = 3
3, 5, 6, 4 > 6+5-4-3 = 4
3, 5, 6, 5 > 6*5/3-5 = 5
3, 5, 6, 6 > 6+6/(5-3) = 6

Apologizes if I typed a wrong symbol and missed it.

[meschlum]

This exactly matches the "all 1's" results.
What a bunch of wasted time! You could have just gone with the all 1's numbers and been done!

Not quite: at the dangerously overpowered levels, you are a bit behind: you only need 3 dice to succeed with "all 1s" for a level 1 spell.

But seriously.
Low levels the feat isn't that good, mainly due to lack of options/spell-levels "Woo, I can get a Free +1 Meta on a Cantrip!"
But once you hit level level 13+ everything auto-succeeds making you grossly uber-powerful.

Good job Paizo...good job...you just made Casters better. Again. Because we all know they needed help. Especially at high levels.

Obviously, since it's harder to get auto-successes at low levels than high ones!


Nice trick on the 3-dice bit! My solutions are somewhat brute force and had a certain amount of learning involved to get the most effective approaches (plus I experimented a bit), but refinements are always nice.


In return, an insight into the workings of the d8 option: it's never better, since the player can always have all the d8s come up in the 1-6 range, so the solutions using d6s are lower bounds, and the d8 solutions could be worse.

For instance, with d2s, there is always a 3 die solution for level 1 - but it goes up to four dice with d6s.

Spoiler: Playing with d8s

Show

First, let's see what we can cancel out. The answer is 4 dice to get 0 and 1.

If all the dice come up in the 1 to 6 range, you need four (or three) to cancel everything.

Otherwise...

Doubles are excluded, of course.

If one of the dice is a 1, none of the others can be adjacent. (e.g. 1, 4, 5)
If one of the dice is a 2, none of the others can be 2 apart. (e.g. 2, 5, 7)
If one of the dice is a 3, none of the others can be 3 apart. (e.g. 3, 4, 7)

At least one die must be 7 or 8

8 / 4 - 2 = 0 removes some options.

If there are two pairs of numbers that are adjacent, or two apart, then they cancel out (e.g. (7 - 5) - (4 - 2))

If the lowest die is a 5, the four dice are adjacent, so they cancel out.

If the lowest die is a 4, and the highest is a 7, there are two adjacent pairs. So if the lowest die is a 4, the highest is an 8.

4, 5, 6, 8: 4 + 8 - 5 - 6 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 4.
4, 6, 7, 8: 6 + 7 - 4 - 8 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 4.

If the lowest die is a 3, none of the other dice can be 3 apart (4, 7 or 5, 8).
3, 4, 6, 8: 3 + 8 - 4 - 6 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 3.
3, 5, 6, 7: 5 + 6 - 3 - 7 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 3.

If the lowest die is a 2, none of the other dice can be 2 apart (3, 5; 4, 6; 5, 7; or 6, 8), and we also cannot have 4, 8.

2, 3, 4, 7: rule of 3 (4, 7)
2, 3, 4, 8: 2, 4, 8 present
2, 3, 5, 7: rule of 2 (5, 7)
2, 3, 5, 8: rule of 3 (5, 8)
2, 3, 6, 7: two adjacent pairs
2, 3, 6, 8: rule of 2 (6, 8)
2, 3, 7, 8: two adjacent pairs
2, 4, 5, 7: rule of 2 (5, 7)
2, 4, 5, 8: 2, 4, 8 present
2, 4, 6, 7: rule of 2 (4, 6)
2, 4, 6, 8: 2, 4, 8 present
2, 4, 7, 8: 2, 4, 8 present
2, 5, 6, 7: rule of 2 (5, 7)
2, 5, 6, 8: rule of 2 (6, 8)
2, 5, 7, 8: rule of 2 (5, 7)
2, 6, 7, 8: rule of 2 (6, 8)

So the lowest die must be a 1, limiting options further.

1, 2, 4, 7: 7 - 4 - 2 - 1 = 0
1, 2, 5, 8: 8 - 5 - 2 - 1 = 0
1, 3, 5, 7: (7 - 5) - (3 - 1) = 0
1, 3, 6, 8: (8 - 6) - (3 - 1) = 0
1, 4, 6, 8: 4 + 6 - 1 - 8 = 1. If another die is added, there is a pair, adjacent numbers, or 1, 2, 4, 6, 8, and 2, 4, 8 cancels out.

Therefore, with 4 dice, it is always possible to cancel out the d8s.


Now let's see what fate awaits an arithmancer trying to use four d8s and a Level 1 spell.

There must be at least one 7 or one 8 (or some of both), as solutions with 6s and less are known.

If there are two 8s, two 7s, or a 7 and an 8, thy can be compressed to 1.

This leaves the following options:

1, 1: 1 + 1 + 1
1, 2: 1 + 2 * 1
1, 3: 1 + 3 + 1
1, 4: 1 + 4 * 1
1, 5: 1 + 5 + 1
1, 6: 1 + 6 * 1
1, 7: 1 + 7 - 1
1, 8: 1 * 7 * 1
2, 2: 2 + 2 + 1
2, 3: 2 + 3 * 1
2, 4: 2 + 4 + 1
2, 5: 2 + 5 * 1
2, 6: 2 + 6 - 1
2, 7: 7 - 2 * 1
2, 8: 8 - 2 + 1
3, 3: 3 + 3 + 1
3, 4: 3 + 4 * 1
3, 5: 3 + 5 - 1
3, 6: 6 - 3 * 1
3, 7: 7 - 3 + 1
3, 8: 8 - 3 * 1
4, 4: 4 + 4 - 1
4, 5: no solutions with 1, 4, 5
4, 6: 6 - 4 + 1
4, 7: 7 - 4 * 1
4, 8: 8 - 4 + 1
5, 5: no solutions with 1, 5, 5
5, 6: no solutions with 1, 5, 6
5, 7: 7 - 5 + 1
5, 8: 8 - 5 * 1
6, 6: no solutions with 1, 6, 6
6, 7: no solutions with 1, 6, 7
6, 8: 8 - 6 + 1
7, 8: no solutions with 1, 7, 8
8, 8: no solutions with 1, 8, 8

With two 8s, if either of the other dice is a 7, we have (8 - 8) * X + 7 as a solution, so the remaining case is four 8s, which is solved by (8 + 8 + 8) / 8.
With 7 and 8, we have 7, 7, 7, 8 or 7, 7, 8, 8 or 7, 8, 8, 8. All of these are solved by taking 0 * X + 7, where the 0 is created with a pair.
With 6 and 7, we have 6, 7, 8, 8 (solved as above), 6, 7, 7, 8 (solved by converting 7, 7 into a 1), and 6, 7, 7, 7 (solved as above).
With 6 and 6, we have 6, 6, 8, 8 (solved via 8 - 6 / (8 - 6)), 6, 6, 7, 8 (solved as above), and 6, 6, 7, 7 (solved as above).
With 5 and 6, we have 5, 6, 8, 8 (solved as above), 5, 6, 7, 8 (solved by converting 6, 7 into a 1), and 5, 6, 7, 7 (solved as above)
With 5 and 5, we have 5, 5, 8, 8 (solved as above), 5, 5, 7, 8 (solved as above), and 5, 5, 7, 7 (solved as above)
With 4 and 5, we have 4, 5, 8, 8 (solved as above), 4, 5, 7, 8 (solved by 4 * 5 - 7 - 8), and 4, 5, 7, 7 (solved as above)

So there is at most a single die that is 7 or 8, and the other three are between 1 and 6.

If the wild die is a 7 and the other three dice are equal to 0 or 1, there is a solution. The remaining cases are:

1, 2, 5, 7: 7 + 2 - 1 - 5
1, 2, 6, 7: 7 + 1 - 6/2
1, 3, 5, 7: 7 + 3 - 5 * 1
3, 4, 5, 7: 7 - (3 + 5) / 4
3, 5, 6, 7: 7 + 6 - 3 - 5
4, 5, 6, 7: 7 - (4 + 6) / 5

So there is always a solution if there is a 7.

The wild die must therefore be an 8, and the other dice are 1s to 6s.

If there is a 6, 4, or 2, there is a solution if the other two dice are equal to 1 or 3 (and 5 if the die is not a 4). Pairs give 1, so the remaining cases are:

1, 5: 8 - (5 - 4) * 1
1, 6: 8 + 4 - 6 + 1
2, 4: 8 - 4 + 6 / 2, other cases have a pair (2, 4, 4, 8 and 2, 2, 4, 8)
3, 5: 8 - 5 + 6 / 3, 5 * (8 - 3 - 4), 3 * (8 - 5 - 2)
4, 6: see solutions for 2, 4

So all the other dice must be odd, and their sum ranges from 3 to 15.

3: 8 - 3
5: 8 - 5
7: to be reviewed
9: to be reviewed
11: 11 - 8
13: 13 - 8
15: 15 - 8

The sums to be reviewed can be obtained as follows:

1, 1, 5: 8 - 5 + 1 - 1
1, 3, 3: 8 - 1 + 3 - 3
1, 3, 5: 8 + 3 - 5 - 1
3, 3, 3: 8 - 3 + 3 - 3

So there is always a solution if there is an 8.

So with d8s, we have 4 dice for Level 1. No improvement at the certainty level.

[Sith_Happens]

Now that the minimum ranks for 100% success seem to have all been found, how about the minimum ranks for 99% success? I think a 1/100 fizzle chance is a more than acceptable risk for free metamagic under most circumstances.

[OldTrees1]

Now that the minimum ranks for 100% success seem to have all been found, how about the minimum ranks for 99% success? I think a 1/100 fizzle chance is a more than acceptable risk for free metamagic under most circumstances.

Something close to this can be determined by checking the probability for (Minimum dice - 1 die)

[Sith_Happens]

Something close to this can be determined by checking the probability for (Minimum dice - 1 die)

Cool, maybe someone who knows how to do that (meschlum?) can even make a table of success chances for assorted target spell level and ranks combinations.

[meschlum]

Eh. With Level 2 or more (where "all 1s" is the worst case scenario), it's plausible that all 1s is the only case involving that many dice. Of course, the odds of rolling all 1s go below 1% as soon as you hit three dice, so it doesn't help much. Plus, there may be cases with other combinations that get close to (or at) the limit, so it's not guaranteed.

The thing with any approach but 100% success is that you need to work out all the combinations, and that is painful.


Still, a low level bonus:

Level 1 fails with 1, 4, 5 (among others), which occurs with probability 1/36 > 1%. Therefore, to have a 99% chance of success, you need more than 3 dice (so 4 dice). Happy?