Kite's thoughts on pergens: Difference between revisions

TallKite (talk | contribs)
TallKite (talk | contribs)
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One can explore the universe of possible EI's, and thus possible pergen notations, more easily if using the gedra format, expressing the EI as a combination of A1's, d2's and arrows. Thus vvA1 = [1 0 -2], v<sup>3</sup>m2 = [1 1 -3], etc. Unlike a conventional 2.3.^ monzo, the first two numbers in a A1.d2.^1 monzo are fairly small, and the second number is never negative (since it's an EI). And we can require that the first two numbers be coprime (see the next section). All this facilitates one's search.
One can explore the universe of possible EI's, and thus possible pergen notations, more easily if using the gedra format, expressing the EI as a combination of A1's, d2's and arrows. Thus vvA1 = [1 0 -2], v<sup>3</sup>m2 = [1 1 -3], etc. Unlike a conventional 2.3.^ monzo, the first two numbers in a A1.d2.^1 monzo are fairly small, and the second number is never negative (since it's an EI). And we can require that the first two numbers be coprime (see the next section). All this facilitates one's search.


===Simplifying "doubled" EI's===
===Simplifying a "squared" EI===
Consider an EI of v<sup>3</sup>AA1. AA1 is "doubled" in the sense that AA1 = A1 + A1. The EI's 2.3.^ monzo is [-22 14 -3]. The EI implies a mapping of [(1 2 2) (0 -3 -14)]. The pergen is (P8, P4/3). Here are the [[twin squares]].
Consider an uninflected EI of AA1. AA1 is "squared" in the sense that AA1 = A1 + A1, thus both numbers in its ratio are square numbers. If inflected by an even number of arrows, it would obviously be an invalid EI, for if v<sup>4</sup>AA1 = 0¢, then so does vvA1, and v<sup>4</sup>AA1 could be replaced with vvA1. So the number of arrows must be odd.
Consider an EI of v<sup>3</sup>AA1. The pergen is (P8, P4/3). Here are the [[twin squares]].


<math>
<math>
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\end{array} \right]
\end{array} \right]
</math>
</math>
The two red numbers in the lower left are both even, a direct result of AA1 being a squared ratio. The two red numbers in the upper right must both be even to ensure their rows' dot products with the EI are zero, which is an even number. If we halve all the red numbers and double all the green numbers, all the various row dot products will be unchanged, and the twin squares will remain valid:
<math>
\begin{array} {rrr}
P8 \\
upminor 2nd \\
trud AA1 \\
\end{array}
\left[ \begin{array} {rrr}
1 & 0 & {\color {Green}0} \\
8 & -5 & {\color {Green}2} \\
\hline
{\color {Red}-11} & {\color {Red}7} & -3 \\
\end{array} \right]
\longleftrightarrow
\left[ \begin{array} {rrr}
1 & 2 & {\color {Red}1} \\
0 & -3 & {\color {Red}-7} \\
\hline
{\color {Green}0} & {\color {Green}-2} & -5 \\
\end{array} \right]
</math>