Wow, theory tells us about the pit boss

The boss will have to annihilate bid o ask, sooner or later.  How? Ley's try a simple, off yhe top of our head:

for bin in the_pit(ignore_singletons,using_dict)
 with len(bin) as count :
 for each singleton in bin
   with bin.valur() as mean
    for each element i  The_pit(compatable_bids,ueing_dict)
,    bin.append(elemnt)
    if count = BinSize : break

Whew, what a bunch of junk, just off the top of my head.

Each one of those for loops is another descent in the pit.  This pit boss code is checking each bin, everywhere, and if it is short of bids, it checks the pits for any singletons that can fit in the bin.

In other words, we know the expected precision, the arrival rate, the bit resolution method, so just Huffman encode to the appropriate bin size everywhere.  The math folkss know what I mean, in his code I am still need to moved and combine bins by probability, the probability that some singleton with a value near this bin value, will arrive.  If the boss can group singletons by value in fixed bins, hen it can probability code the bins, it has a statistically working method to check if a value is equal to a bin.

Those little selector functions in  the iterator?  They work block counts, and values, whatever as they descend the pit.  They can divert the path down with skips and jumps.  A lot of them are lambda functions, I think.