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ID: f-p8-decay-renorm
Type: f

H4. Accumulate as the baseline does, then set the position's total support to what the arriving
evidence justifies, rather than to a constant.

THE RULE. Let s be lsa_sum256 of the accumulated vector, and let the target t be the LSA sum --
lsa_add folded, not the maximum -- of the supports of the applications that fired at this position
on this time step, which is the correct combination when the applications come from disjoint
observations, per the second free parameter in #f-p8's derivation. Subtract lsa_sub(s, t) from every
one of the 256 entries, floored at 0. When nothing fired, t is 0 and the position decays to the
floor. When s is at or below t nothing is subtracted.

WHAT A UNIFORM SUBTRACTION CAN AND CANNOT DO, which is what this alternative rests on. Subtracting a
constant d from every entry of a position's vector divides every count it represents by 2^d. That
leaves the softmax distribution EXACTLY unchanged and changes only the total number of observations
the position claims to rest on. So a renormalisation is never a statement about WHICH byte the
position favours; it is only ever a statement about HOW MUCH EVIDENCE the position represents.

WHAT TO EXPECT, stated before the run so the measurement can contradict it: because the subtraction
is uniform it cannot reorder the ES, so every effect this alternative has arrives through the 0
floor and through what the next time step accumulates on top of a rescaled row. Its interest is not
the settled bytes but the CONFIDENCE the settled state reports -- it is the only alternative on this
axis whose vectors mean the same thing at time step 5 and at time step 100.

HISTORY.
Generation 5, A3's idea (#f-p8-renorm) restated, and the restatement is a correction. A3
renormalised to a sum of 255 and was read as imposing an invariant on the distribution, which is
both the thing #p8v2_gen3 ruled out ("there is no constant sum ever in an ES") and a thing the
operation cannot do. MJC on this generation's choice C: "Renormalization was never understood
correctly by the model... 'along the same lines' is fine if you interpret it like a good probability
theorist."
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