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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