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Draft of CubicSymbol #194
Draft of CubicSymbol #194
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Could you possibly return
(CubicSymbol, EisensteinInteger)
and make corresponding changes toextractPrimaryContributions
andcubicHelper
? This will be more natural representation thanInteger
andMod 3
, I think.I suggest to pattern match on possible values of
remainder
and return corresponding cubic symbols directly.There was a problem hiding this comment.
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This is probably better
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I changed it but I am disappointed by the result. The code looks messier and it runs more slowly. I thought about it but I could not find any better way of implementing these changes. If you have any suggestions to improve it, let me know.
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That's surprising, I'll take a look next week. I should revisit
M.NT.Q.EisensteinIntegers
to check that all operations are as efficient as possible.In the meantime I have three suggestions:
stimes
(which appears to be awkward) we can actually define our own, supporting zeros and negatives. This basically brings us back to your very first design, sorry for the digression.symbolToNum :: CubicSymbol -> EisensteinInteger
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I did this. In the new commit, the timings went down to the previous level.
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It seems we can do this "table lookup" more explicit, pattern-matching by all possible values of
remainder
. Pattern-matching provides an elegant way to definesymb
(equal tostimes powerUnit Omega
), and then one can choose eithere * symbolToNum symb
or-e * symbolToNum
, depending on which of them is equivalent to 1 modulo 3.There was a problem hiding this comment.
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As I understand it, this method is not going to work. My understanding is that you want to define
factor = e * symbolToNum symb
orfactor = -e * symbolToNum symb
depending on which one is equivalent to 1 mod 3. In general, I don't think that either of them need be equivalent to 1 mod 3. For instance, suppose that I find that(1 + ω) * e A.rem 3 = 1
. Then its associated cubic symbol is ω. Then you see that bothe * ω
and-e * ω
cannot have the above property. The issue is that I need to knowpowerUnit
modulo 6 rather that modulo 3. One solution could be to pattern match onremainder
and define bothfactor
andsymb
at the same time. Let me know what you think.There was a problem hiding this comment.
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I have taken this approach in the latest commit. I would like to make the pattern matching more readable but the program does not compile if I use ω when I list the cases. Also, it gives me a warning for redundant pattern matching which I don't agree with.