Ronald Guida wrote:
Thank you, apfelmus.  That was a wonderful explanation; the debit
method in [1] finally makes sense.


A diagram says more than a thousand words :)

My explanation is not entirely faithful to Okasaki, let me elaborate.


In his book, Okasaki calls the process of transferring the debits from
the input

  xs = x1 : x2 : x3 : ... : xn : []
          1    1    1 ... 1    1  0

  ys = y1 : y2 : y3 : ... : ym : []
1 1 1 ... 1 1 0

to the output

  xs ++ ys = x1 : x2 : x3 : ... : xn : y1 : y2 : y3 : ... : ym : []
2 2 2 ... 2 2 1 1 1 ... 1 1 0

"debit inheritance". In other words, the debits of xs and ys (here 1 at
each node) are carried over to xs ++ ys (in addition to the debits
created by ++ itself). In the thesis, he doesn't give it an explicit
name, but discusses this phenomenon in the very last paragraphs of
chapter 3.4 .


The act of relocating debits from child to parent nodes as exemplified with

  xs ++ reverse ys =
     x1 : x2 : x3 : ... : xn : yn : y{n-1} : ... : y1 : []
1 1 1 ... 1 1 n 0 ... 0 0 0

  xs ++ reverse ys =
     x1 : x2 : x3 : ... : xn : yn : y{n-1} : ... : y1 : []
        2    2    2 ... 2    2    0        0 ... 0    0  0

is called "debit passing", but Okasaki doesn't use it earlier than in the chapter "Implicit recursive slowdown". But the example I gave here is useful for understand the scheduled implementation of real time queues. The trick there is to not create a "big" suspension with n debits but to really "physically" distribute them across the data structure

     x1 : x2 : x3 : ... : xn : yn : y{n-1} : ... : y1 : []
        2    2    2 ... 2    2    2        2 ... 2    2  2

and discharge them by forcing a node with every call to  snoc . I say
"physically" because this forcing performs actual work, it does not
simply "mentally" discharge a debit to amortize work that will be done
later. Note that the 2 debits added to each  yi  are an overestimation
here, but the real time queue implementation pays for them nonetheless.


My focus on debit passing in the original explanation might suggest that
debits can only be discharged when actually evaluating the node to which
the debit was assigned. This is not the case, an operation may discharge
any debits, even in parts of the data structure that it doesn't touch.
Of course, it must discharge debits of nodes it does touch.

For instance, in the proof of theorem 3.1 (thesis) for queues, Okasaki
writes "We can restore the invariant by discharging the first (two) debit(s) in the queue" without bothering to analyze which node this will be. So, the front queue might look like

     f1 : f2 : f3 : ... : fn : f{n+1} : f{n+2} : ... : fm : []
        0    0    1 ... 1    1        n        0 ... 0    0  0

and it's one of the nodes that carries one debit, or it could look like

          f2 : f3 : ... : fn : f{n+1} : f{n+2} : ... : fm : []
             0    0 ... 0    0       n-3       0 ... 0    0  0

and it's the node with the large amount of debits. In fact, it's not even immediate that these two are the only possibilities.

However, with the debit passing from my previous post, it's easier to say which node will be discharged. But even then, only tail discharges exactly the debits of nodes it inspects while the snoc operation discharges debits in the untouched front list. Of course, as soon as identifying the nodes becomes tractable, chances are that you can turn it into a real-time data structure.

Another good example are skew heaps from

[2]:
  Chris Okasaki. Fun with binary heap trees.
  in  J. Gibbons, O. de Moor. The Fun of Programming.
  http://www.palgrave.com/PDFs/0333992857.Pdf

Here, the "good" nodes are annotated with one debit. Every join operation discharges O(log n) of them and allocates new ones while walking down the tree, but the "time" to actually walk down the tree is not counted immediately. This is just like (++) walks down the first list and allocates debits without immediately using O(n) time to do that.


Regards,
apfelmus


PS: In a sense, discharging arbitrary debits can still be explained with debit passing: first pass those debits to the top and the discharge them because any operation has to inspect the top.

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