- Action sequence: split the numerator, cycle one of the summands, and re-merge.
- Precondition: subterm matching \( \cyc \frac{\sum_i ?x_i}{?y} \).
- Triggers: unclear.
- Progress signs: cycling exposes a shared factor.

However, the challenge with writing the primal trick is that the pattern appears often in proof steps. Do we need to try all \(\mathrm{nCk}(n,n/2)\) possibilities at every proof step? Do we need to continue each possibility until the progress sign is determined to be absent? Accumulated wisdom suggests that it is hard to write down appropriate heuristics for when to try, how to group, and when to give up on the path.

Also, is it even its own trick? Grouping, splitting, cycling, merging are all in the action space already. We will consider this question again later on.

It is not immediately obvious how to generate many variants of the puzzle. In the case of this problem, the problem of finding the \( h \) looked like this: \begin{align*} (a-b) g(a,b,c) &= (a-b) \cyc (a^2 + ab) \\ &= a^3 + ab^2 + ac^2 + a^2b + abc + a^2 c - ba^2 - b^3 - bc^2 - b^2a - b^2c - bca & \text{expand} \\ &= a^3 + ac^2 + a^2 c - b^3 - bc^2 - b^2c & \text{cancel} \\ &= a^3 + ac^2 + a^2 c - b^3 - bc^2 - b^2c + (c^3 - c^3) & \text{add0 (magic)} \\ &= (c^3 + a^3 + ac^2 + a^2 c) - (c^3 + b^3 - bc^2 - b^2c) & \text{group (magic)}\\ &= (c^2+a^2)(c+a) - (b^2+c^2)(b+c) & \text{factor} \\ &= h(a,b,c) - h(b,c,a) \end{align*} This seems as amazing as the primal trick, and does not seem like a recipe that would easily yield a large amount of data.

Arguably the primal trick is

- Action sequence: massage products of sums (with some fractions) so everything cancels after Holder's.
- Precondition: product of sums, upto some "noise"
- Triggers: I can only speculate.
- Progress signs: cancellation after Holder.

What about the dual trick? Arguably it is: \begin{align*} n & = \sum_{i=1}^n 1 \\ & = \sum_{i=1}^n \prod_{j=1}^m f_{ij}(x) f_{ij}(x)^{-1} \\ & = \sum_{i=1}^n \prod_{k=1}^K f_{ik}(x) & \text{group (arbitrarily)} \\ & = \sum_{i=1}^n \prod_{k=1}^K (f_{ik}(x)^{K})^{1/K} & \text{group (arbitrarily)} \\ & = \sum_{i=1}^n \prod_{k=1}^K h_{ik}(x)^{1/K} & \text{define} \\ & \leq \prod_{k=1}^K \left( \sum_{i=1}^n h_{ik}(x) \right)^{1/K} & \text{holder} \\ & = \left( \prod_{k=1}^K \sum_{i=1}^n h_{ik}(x) \right)^{1/K} \end{align*} and then obfuscate the \( \sum_{i=1}^n h_{ik}(x) \) , e.g. \( \cyc (a+b) \to \cyc 2a \to 2 \cyc a \).

In contrast to the first dual trick discussed above, this process is unconstrained. As long as the obfuscation is reasonably interesting, it seems this dual trick can produce an infinite number of puzzles with solutions.