What does the saving and loans trees look like?

Treat it like two trees, or two branches off the main node.  Left branch is loans, right is savings.

We would expect the two to be highly similar.  When a string of borrowings comes in, the bot expects a similar stream of deposits to arrive. Compare the trees, they should have similar shapes, we can construct a similarity measure.

If the bank were a channel with limited bandwidth, we want savers and borrowers to spend equal time using the channel. It is known to the traders that when the loans are are out of balance, there will be  retro-active rate hike for savers, which is to say the bot allocates more of the channel to savers.

Lending tree failed this part.  If they could have looked at the wo trees separately, the lendng tree bot would have raised loans rates earlier.  The bot would have spotted the string of duplicate loans coming in.

In coding, the code word sent down the channel is foun d by entering the encode tree, and comparing a trade to the left right quant boundaries and  thus moving down the tree to a leaf, collecting the code word on hr way down.  The graph, itself, is the allocation of bandwidth, as measured in he recent past.  Creating the tree is reverse, staring at tyhe leafs,where all the input transactions are listed,the bot find the least probable, and enclosed therm  together, thus building the tree.

You see, the solution has to exist because savings and loan business is simply a queuing problem between savers and borrowers.   Think of it as  Poisson queuing, except you are finite, arrival rates can vary a bit past over some window in the past. The bots, in their maximu entropy computations, naturally keep the queue length such that is exposes the greatest amount of current information.  The larger the bet, the more time on the tree, queue stack up, the bet stream is slowed by the generator bots.

In other words, do not worry; the Lending Tree problem is going away, the Ethereum snaffu is going away.  This problem is contained in a box, packed with minimum redundancy, equipartion tells us there has to be  local, smooth  red/green detector.  Just be clever,  there is a convolution operator that gives you an imbalance measure, and it can be an operator running in an independent thread..