Treating Trees¶

Although any expression in Joy can be considered to describe a tree with the quotes as compound nodes and the non-quote values as leaf nodes, in this page I want to talk about ordered binary trees and how to make and use them.

The basic structure, in a crude type notation, is:

BTree :: [] | [key value BTree BTree]

That says that a BTree is either the empty quote [] or a quote with four items: a key, a value, and two BTrees representing the left and right branches of the tree.

A Function to Traverse this Structure¶

Let's take a crack at writing a function that can recursively iterate or traverse these trees.

Base case []¶

The stopping predicate just has to detect the empty list:

BTree-iter == [not] [E] [R0] [R1] genrec

And since there's nothing at this node, we just pop it:

BTree-iter == [not] [pop] [R0] [R1] genrec

Node case [key value left right]¶

Now we need to figure out R0 and R1:

BTree-iter == [not] [pop] [R0]            [R1] genrec
           == [not] [pop] [R0 [BTree-iter] R1] ifte

Let's look at it in situ:

[key value left right] R0 [BTree-iter] R1

Processing the current node.¶

R0 is almost certainly going to use dup to make a copy of the node and then dip on some function to process the copy with it:

[key value left right] [F] dupdip                 [BTree-iter] R1
[key value left right]  F  [key value left right] [BTree-iter] R1

For example, if we're getting all the keys F would be first:

R0 == [first] dupdip

[key value left right] [first] dupdip                 [BTree-iter] R1
[key value left right]  first  [key value left right] [BTree-iter] R1
key                            [key value left right] [BTree-iter] R1

Recur¶

Now R1 needs to apply [BTree-iter] to left and right. If we drop the key and value from the node using rest twice we are left with an interesting situation:

key [key value left right] [BTree-iter] R1
key [key value left right] [BTree-iter] [rest rest] dip
key [key value left right] rest rest [BTree-iter]
key [left right] [BTree-iter]

Hmm, will step do?

key [left right] [BTree-iter] step
key left BTree-iter [right] [BTree-iter] step
key left-keys [right] [BTree-iter] step
key left-keys right BTree-iter
key left-keys right-keys

Wow. So:

R1 == [rest rest] dip step

Putting it together¶

We have:

BTree-iter == [not] [pop] [[F] dupdip] [[rest rest] dip step] genrec

When I was reading this over I realized rest rest could go in R0:

BTree-iter == [not] [pop] [[F] dupdip rest rest] [step] genrec

(And [step] genrec is such a cool and suggestive combinator!)

Parameterizing the F per-node processing function.¶

[F] BTree-iter == [not] [pop] [[F] dupdip rest rest] [step] genrec

Working backward:

[not] [pop] [[F] dupdip rest rest]            [step] genrec
[not] [pop] [F]       [dupdip rest rest] cons [step] genrec
[F] [not] [pop] roll< [dupdip rest rest] cons [step] genrec

Ergo:

BTree-iter == [not] [pop] roll< [dupdip rest rest] cons [step] genrec

Adding Nodes to the BTree¶

Let's consider adding nodes to a BTree structure.

BTree value key BTree-add == BTree

Adding to an empty node.¶

If the current node is [] then you just return [key value [] []]:

BTree-add == [popop not] [[pop] dipd BTree-new] [R0] [R1] genrec

Where BTree-new is:

value key BTree-new == [key value [] []]

value key swap [[] []] cons cons
key value      [[] []] cons cons
key      [value [] []]      cons
     [key value [] []]

BTree-new == swap [[] []] cons cons

(As an implementation detail, the [[] []] literal used in the definition of BTree-new will be reused to supply the constant tail for all new nodes produced by it. This is one of those cases where you get amortized storage "for free" by using persistent datastructures. Because the tail, which is ((), ((), ())) in Python, is immutable and embedded in the definition body for BTree-new, all new nodes can reuse it as their own tail without fear that some other code somewhere will change it.)

If the current node isn't empty.¶

We now have to derive R0 and R1, consider:

[key_n value_n left right] value key R0 [BTree-add] R1

In this case, there are three possibilites: the key can be greater or less than or equal to the node's key. In two of those cases we will need to apply a copy of BTree-add, so R0 is pretty much out of the picture.

[R0] == []

A predicate to compare keys.¶

The first thing we need to do is compare the the key we're adding to see if it is greater than the node key and branch accordingly, although in this case it's easier to write a destructive predicate and then use ifte to apply it nullary:

[key_n value_n left right] value key [BTree-add] R1
[key_n value_n left right] value key [BTree-add] [P >] [T] [E] ifte

[key_n value_n left right] value key [BTree-add] P                   >
[key_n value_n left right] value key [BTree-add] pop roll> pop first >
[key_n value_n left right] value key                 roll> pop first >
key [key_n value_n left right] value                 roll> pop first >
key key_n                                                            >
Boolean

P > == pop roll> pop first >
P < == pop roll> pop first <
P   == pop roll> pop first

If the key we're adding is greater than the node's key.¶

Here the parantheses are meant to signify that the right-hand side (RHS) is not literal, the code in the parentheses is meant to have been evaluated:

[key_n value_n left right] value key [BTree-add] T == [key_n value_n left (BTree-add key value right)]

Use infra on K.¶

So how do we do this? We know we're going to want to use infra on some function K that has the key and value to work with, as well as the quoted copy of BTree-add to apply somehow:

right left value_n key_n value key [BTree-add] K
    ...
right value key BTree-add left value_n key_n

Pretty easy:

right left value_n key_n value key [BTree-add] cons cons dipdd
right left value_n key_n [value key BTree-add]           dipdd
right value key BTree-add left value_n key_n

So:

K == cons cons dipdd

And:

[key_n value_n left right] [value key [BTree-add] K] infra

Derive T.¶

So now we're at getting from this to this:

[key_n value_n left right]  value key [BTree-add] T
    ...
[key_n value_n left right] [value key [BTree-add] K] infra

And so T is just:

value key [BTree-add] T == [value key [BTree-add] K]                infra
                      T == [                      K] cons cons cons infra

If the key we're adding is less than the node's key.¶

This is very very similar to the above:

[key_n value_n left right] value key [BTree-add] E
[key_n value_n left right] value key [BTree-add] [P <] [Te] [Ee] ifte

In this case Te works that same as T but on the left child tree instead of the right, so the only difference is that it must use dipd instead of dipdd:

Te == [cons cons dipd] cons cons cons infra

This suggests an alternate factorization:

ccons == cons cons
T == [ccons dipdd] ccons cons infra
Te == [ccons dipd] ccons cons infra

But whatever.

Else the keys must be equal.¶

This means we must find:

[key_n value_n left right] value key [BTree-add] Ee
    ...
[key value left right]

This is another easy one:

Ee == pop swap roll< rest rest cons cons

[key_n value_n left right] value key [BTree-add] pop swap roll< rest rest cons cons
[key_n value_n left right] value key                 swap roll< rest rest cons cons
[key_n value_n left right] key value                      roll< rest rest cons cons
key value [key_n value_n left right]                            rest rest cons cons
key value [              left right]                                      cons cons
          [key   value   left right]

Now we can define BTree-add¶

BTree-add == [popop not] [[pop] dipd BTree-new] [] [[P >] [T] [E] ifte] genrec

Putting it all together:

BTree-new == swap [[] []] cons cons
P == pop roll> pop first
T == [cons cons dipdd] cons cons cons infra
Te == [cons cons dipd] cons cons cons infra
Ee == pop swap roll< rest rest cons cons
E == [P <] [Te] [Ee] ifte

BTree-add == [popop not] [[pop] dipd BTree-new] [] [[P >] [T] [E] ifte] genrec

We can use this to make a set-like datastructure by just setting values to e.g. 0 and ignoring them. It's set-like in that duplicate items added to it will only occur once within it, and we can query it in $O(\log_2 N)$ time.

And with that we can write a little program to remove duplicate items from a list.

cmp combinator¶

Instead of all this mucking about with nested ifte let's just go whole hog and define cmp which takes two values and three quoted programs on the stack and runs one of the three depending on the results of comparing the two values:

   a b [G] [E] [L] cmp
------------------------- a > b
        G

   a b [G] [E] [L] cmp
------------------------- a = b
            E

   a b [G] [E] [L] cmp
------------------------- a < b
                L

We need a new non-destructive predicate P:

[key_n value_n left right] value key [BTree-add] P
[key_n value_n left right] value key [BTree-add] over [Q] nullary
[key_n value_n left right] value key [BTree-add] key  [Q] nullary
[key_n value_n left right] value key [BTree-add] key   Q
[key_n value_n left right] value key [BTree-add] key   popop popop first
[key_n value_n left right] value key                         popop first
[key_n value_n left right]                                         first
 key_n
[key_n value_n left right] value key [BTree-add] key  [Q] nullary
[key_n value_n left right] value key [BTree-add] key key_n

P == over [popop popop first] nullary

Here are the definitions again, pruned and renamed in some cases:

BTree-new == swap [[] []] cons cons
P == over [popop popop first] nullary
T> == [cons cons dipdd] cons cons cons infra
T< == [cons cons dipd] cons cons cons infra
E == pop swap roll< rest rest cons cons

Using cmp to simplify our code above at R1:

[key_n value_n left right] value key [BTree-add] R1
[key_n value_n left right] value key [BTree-add] P [T>] [E] [T<] cmp

The line above becomes one of the three lines below:

[key_n value_n left right] value key [BTree-add] T>
[key_n value_n left right] value key [BTree-add] E
[key_n value_n left right] value key [BTree-add] T<

The definition is a little longer but, I think, more elegant and easier to understand:

BTree-add == [popop not] [[pop] dipd BTree-new] [] [P [T>] [E] [T<] cmp] genrec

Factoring and naming¶

It may seem silly, but a big part of programming in Forth (and therefore in Joy) is the idea of small, highly-factored definitions. If you choose names carefully the resulting definitions can take on a semantic role.

get-node-key == popop popop first
remove-key-and-value-from-node == rest rest
pack-key-and-value == cons cons
prep-new-key-and-value == pop swap roll<
pack-and-apply == [pack-key-and-value] swoncat cons pack-key-and-value infra

BTree-new == swap [[] []] pack-key-and-value
P == over [get-node-key] nullary
T> == [dipdd] pack-and-apply
T< == [dipd] pack-and-apply
E == prep-new-key-and-value remove-key-and-value-from-node pack-key-and-value

A Version of BTree-iter that does In-Order Traversal¶

If you look back to the non-empty case of the BTree-iter function we can design a varient that first processes the left child, then the current node, then the right child. This will allow us to traverse the tree in sort order.

BTree-iter-order == [not] [pop] [R0 [BTree-iter] R1] ifte

To define R0 and R1 it helps to look at them as they will appear when they run:

[key value left right] R0 [BTree-iter-order] R1

Process the left child.¶

Staring at this for a bit suggests dup third to start:

[key value left right] R0        [BTree-iter-order] R1
[key value left right] dup third [BTree-iter-order] R1
[key value left right] left      [BTree-iter-order] R1

Now maybe:

[key value left right] left [BTree-iter-order] [cons dip] dupdip
[key value left right] left [BTree-iter-order] cons dip [BTree-iter-order]
[key value left right] [left BTree-iter-order]      dip [BTree-iter-order]
left BTree-iter-order [key value left right]            [BTree-iter-order]

Process the current node.¶

So far, so good. Now we need to process the current node's values:

left BTree-iter-order [key value left right] [BTree-iter-order] [[F] dupdip] dip
left BTree-iter-order [key value left right] [F] dupdip [BTree-iter-order]
left BTree-iter-order [key value left right] F [key value left right] [BTree-iter-order]

If F needs items from the stack below the left stuff it should have cons'd them before beginning maybe? For functions like first it works fine as-is.

left BTree-iter-order [key value left right] first [key value left right] [BTree-iter-order]
left BTree-iter-order key [key value left right] [BTree-iter-order]

Process the right child.¶

First ditch the rest of the node and get the right child:

left BTree-iter-order key [key value left right] [BTree-iter-order] [rest rest rest first] dip
left BTree-iter-order key right [BTree-iter-order]

Then, of course, we just need i to run BTree-iter-order on the right side:

left BTree-iter-order key right [BTree-iter-order] i
left BTree-iter-order key right BTree-iter-order

Defining BTree-iter-order¶

The result is a little awkward:

R1 == [cons dip] dupdip [[F] dupdip] dip [rest rest rest first] dip i

Let's do a little semantic factoring:

fourth == rest rest rest first

proc_left == [cons dip] dupdip
proc_current == [[F] dupdip] dip
proc_right == [fourth] dip i

BTree-iter-order == [not] [pop] [dup third] [proc_left proc_current proc_right] genrec

Now we can sort sequences.

Getting values by key¶

Let's derive a function that accepts a tree and a key and returns the value associated with that key.

   tree key BTree-get
------------------------
        value

The base case []¶

As before, the stopping predicate just has to detect the empty list:

BTree-get == [pop not] [E] [R0] [R1] genrec

But what do we do if the key isn't in the tree? In Python we might raise a KeyError but I'd like to avoid exceptions in Joy if possible, and here I think it's possible. (Division by zero is an example of where I think it's probably better to let Python crash Joy. Sometimes the machinery fails and you have to "stop the line", methinks.)

Let's pass the buck to the caller by making the base case a given, you have to decide for yourself what [E] should be.

   tree key [E] BTree-get
---------------------------- key in tree
           value

   tree key [E] BTree-get
---------------------------- key not in tree
         tree key E

Now we define:

BTree-get == [pop not] swap [R0] [R1] genrec

Note that this BTree-get creates a slightly different function than itself and that function does the actual recursion. This kind of higher-level programming is unusual in most languages but natural in Joy.

tree key [E] [pop not] swap [R0] [R1] genrec
tree key [pop not] [E] [R0] [R1] genrec

The anonymous specialized recursive function that will do the real work.

[pop not] [E] [R0] [R1] genrec

Node case [key value left right]¶

Now we need to figure out R0 and R1:

[key value left right] key R0 [BTree-get] R1

We want to compare the search key with the key in the node, and if they are the same return the value and if they differ then recurse on one of the child nodes. So it's very similar to the above funtion, with [R0] == [] and R1 == P [T>] [E] [T<] cmp:

[key value left right] key [BTree-get] P [T>] [E] [T<] cmp

So:

get-node-key == pop popop first
P == over [get-node-key] nullary

The only difference is that get-node-key does one less pop because there's no value to discard. Now we have to derive the branches:

[key_n value_n left right] key [BTree-get] T>
[key_n value_n left right] key [BTree-get] E
[key_n value_n left right] key [BTree-get] T<

The cases of T> and T< are similar to above but instead of using infra we have to discard the rest of the structure:

[key_n value_n left right] key [BTree-get] T> == right key BTree-get
[key_n value_n left right] key [BTree-get] T< == left key BTree-get

So:

T> == [fourth] dipd i
T< == [third] dipd i

E.g.:

[key_n value_n left right]        key [BTree-get] [fourth] dipd i
[key_n value_n left right] fourth key [BTree-get]               i
                    right         key [BTree-get]               i
                    right         key  BTree-get

And:

[key_n value_n left right] key [BTree-get] E == value_n

E == popop second

So:

fourth == rest rest rest first
get-node-key == pop popop first
P == over [get-node-key] nullary
T> == [fourth] dipd i
T< == [third] dipd i
E == popop second

BTree-get == [pop not] swap [] [P [T>] [E] [T<] cmp] genrec

BTree-delete¶

Now let's write a function that can return a tree datastructure with a key, value pair deleted:

   tree key BTree-delete
---------------------------
       tree

If the key is not in tree it just returns the tree unchanged.

So:

BTree-Delete == [pop not] swap [R0] [R1] genrec

             [Er] BTree-delete
-------------------------------------
   [pop not] [Er] [R0] [R1] genrec

[n_key n_value left right] [BTree-get] 
[n_key n_value left right] [BTree-get] E
[n_key n_value left right] [BTree-get] T<

Now we get to figure out the recursive case:

w/ D == [pop not] [Er] [R0] [R1] genrec

[node_key node_value left right] key R0                  [D] R1
[node_key node_value left right] key over first swap dup [D] R1
[node_key node_value left right] node_key key key        [D] R1

And then:

[node_key node_value left right] node_key key key [D] R1
[node_key node_value left right] node_key key key [D] cons roll> [T>] [E] [T<] cmp
[node_key node_value left right] node_key key [key D]      roll> [T>] [E] [T<] cmp
[node_key node_value left right] [key D] node_key key            [T>] [E] [T<] cmp

Now this:;

[node_key node_value left right] [key D] node_key key [T>] [E] [T<] cmp

Becomes one of these three:;

[node_key node_value left right] [key D] T>
[node_key node_value left right] [key D] E
[node_key node_value left right] [key D] T<

Greater than case and less than case¶

   [node_key node_value left right] [key D] T>
-------------------------------------------------
   [node_key node_value left key D right]

First:

right left       node_value node_key [key D] dipd
right left key D node_value node_key
right left'      node_value node_key

Ergo:

[node_key node_value left right] [key D] [dipd] cons infra

So:

T> == [dipd] cons infra
T< == [dipdd] cons infra

The else case¶

[node_key node_value left right] [key D] E

We have to handle three cases, so let's use cond.

The first two cases are symmetrical, if we only have one non-empty child node return it.

E == [
    [[pop third not] pop fourth]
    [[pop fourth not] pop third]
    [default]
] cond

(If both child nodes are empty return an empty node.)

The initial structure of the default function:

default == [E'] cons infra

[node_key node_value left right] [key D] default
[node_key node_value left right] [key D] [E'] cons infra
[node_key node_value left right] [[key D] E']      infra

right left node_value node_key [key D] E'

If both child nodes are non-empty, we find the highest node in our lower sub-tree, take its key and value to replace (delete) our own, then get rid of it by recursively calling delete() on our lower sub-node with our new key.

(We could also find the lowest node in our higher sub-tree and take its key and value and delete it. I only implemented one of these two symmetrical options. Over a lot of deletions this might make the tree more unbalanced. Oh well.)

First things first, we no longer need this node's key and value:

right left node_value node_key [key D] roll> popop E''
right left [key D] node_value node_key       popop E''
right left [key D]                                 E''

Then we have to we find the highest (right-most) node in our lower (left) sub-tree:

right left [key D] E''

Ditch the key:

right left [key D] rest E'''
right left     [D]      E'''

Find the right-most node:

right left        [D] [dup W] dip E''''
right left dup  W [D]             E''''
right left left W [D]             E''''

Consider:

left W

We know left is not empty:

[L_key L_value L_left L_right] W

We want to keep extracting the right node as long as it is not empty:

left [P] [B] while W'

The predicate:

[L_key L_value L_left L_right] P
[L_key L_value L_left L_right] fourth
                      L_right

(This has a bug, can run on [] so must be guarded:

if_not_empty == [] swap [] ifte
?fourth == [fourth] if_not_empty
W.rightmost == [?fourth] [fourth] while

The body is also fourth:

left [fourth] [fourth] while W'
rightest                     W'

We know rightest is not empty:

[R_key R_value R_left R_right] W'
[R_key R_value R_left R_right] uncons uncons pop
R_key [R_value R_left R_right]        uncons pop
R_key R_value [R_left R_right]               pop
R_key R_value

So:

W == [fourth] [fourth] while uncons uncons pop

And:

right left left W        [D] E''''
right left R_key R_value [D] E''''

Final stretch. We want to end up with something like:

right left [R_key D] i R_value R_key
right left  R_key D    R_value R_key
right left'            R_value R_key

If we adjust our definition of W to include over at the end:

W == [fourth] [fourth] while uncons uncons pop over

That will give us:

right left R_key R_value R_key [D] E''''

right left         R_key R_value R_key [D] cons dipdd E'''''
right left         R_key R_value [R_key D]      dipdd E'''''
right left R_key D R_key R_value                      E'''''
right left'        R_key R_value                      E'''''
right left'        R_key R_value                      swap
right left' R_value R_key

So:

E' == roll> popop E''

E'' == rest E'''

E''' == [dup W] dip E''''

E'''' == cons dipdd swap

Substituting:

W == [fourth] [fourth] while uncons uncons pop over
E' == roll> popop rest [dup W] dip cons dipdd swap
E == [
    [[pop third not] pop fourth]
    [[pop fourth not] pop third]
    [[E'] cons infra]
] cond

Minor rearrangement:

W == dup [fourth] [fourth] while uncons uncons pop over
E' == roll> popop rest [W] dip cons dipdd swap
E == [
    [[pop third not] pop fourth]
    [[pop fourth not] pop third]
    [[E'] cons infra]
] cond

Refactoring¶

W.rightmost == [fourth] [fourth] while
W.unpack == uncons uncons pop
E.clear_stuff == roll> popop rest
E.delete == cons dipdd
W == dup W.rightmost W.unpack over
E.0 == E.clear_stuff [W] dip E.delete swap
E == [
    [[pop third not] pop fourth]
    [[pop fourth not] pop third]
    [[E.0] cons infra]
] cond
T> == [dipd] cons infra
T< == [dipdd] cons infra
R0 == over first swap dup
R1 == cons roll> [T>] [E] [T<] cmp
BTree-Delete == [pop not] swap [R0] [R1] genrec

By the standards of the code I've written so far, this is a huge Joy program.

One bug, I forgot to put not in the first two clauses of the cond.

The behavior of the [Er] function should maybe be different: either just silently fail, or maybe implement some sort of function that can grab the pending expression up to a sentinel value or something, allowing for a kind of "except"-ish control-flow?

Then, once we have add, get, and delete we can see about abstracting them.

Tree with node and list of trees.¶

Let's consider a tree structure, similar to one described "Why functional programming matters" by John Hughes, that consists of a node value and a sequence of zero or more child trees. (The asterisk is meant to indicate the Kleene star.)

tree = [] | [node [tree*]]

treestep¶

In the spirit of step we are going to define a combinator treestep which expects a tree and three additional items: a base-case value z, and two quoted programs [C] and [N].

tree z [C] [N] treestep

If the current tree node is empty then just leave z on the stack in lieu:

   [] z [C] [N] treestep
---------------------------
      z

Otherwise, evaluate N on the node value, map the whole function (abbreviated here as k) over the child trees recursively, and then combine the result with C.

   [node [tree*]] z [C] [N] treestep
--------------------------------------- w/ K == z [C] [N] treestep
       node N [tree*] [K] map C

Derive the recursive form.¶

Since this is a recursive function, we can begin to derive it by finding the ifte stage that genrec will produce. The predicate and base-case functions are trivial, so we just have to derive J.

K == [not] [pop z] [J] ifte

The behavior of J is to accept a (non-empty) tree node and arrive at the desired outcome.

       [node [tree*]] J
------------------------------
   node N [tree*] [K] map C

So J will have some form like:

J == .. [N] .. [K] .. [C] ..

Let's dive in. First, unquote the node and dip N.

[node [tree*]] i [N] dip
 node [tree*]    [N] dip
node N [tree*]

Next, map K over teh child trees and combine with C.

node N [tree*] [K] map C
node N [tree*] [K] map C
node N [K.tree*]       C

So:

J == i [N] dip [K] map C

Plug it in and convert to genrec:

K == [not] [pop z] [i [N] dip [K] map C] ifte
K == [not] [pop z] [i [N] dip]   [map C] genrec

Extract the givens to parameterize the program.¶

[not] [pop z] [i [N] dip]   [map C] genrec

[not] [pop z]                   [i [N] dip] [map C] genrec
[not] [z]         [pop] swoncat [i [N] dip] [map C] genrec
[not]  z     unit [pop] swoncat [i [N] dip] [map C] genrec
z [not] swap unit [pop] swoncat [i [N] dip] [map C] genrec
  \  .........TS0............./
   \/
z TS0 [i [N] dip]                       [map C] genrec
z     [i [N] dip]             [TS0] dip [map C] genrec
z       [[N] dip] [i] swoncat [TS0] dip [map C] genrec
z  [N] [dip] cons [i] swoncat [TS0] dip [map C] genrec
       \  ......TS1........./
        \/
z [N] TS1 [TS0] dip [map C]                      genrec
z [N]               [map C]  [TS1 [TS0] dip] dip genrec
z [N] [C]      [map] swoncat [TS1 [TS0] dip] dip genrec
z [C] [N] swap [map] swoncat [TS1 [TS0] dip] dip genrec

The givens are all to the left so we have our definition.

Define treestep¶

     TS0 == [not] swap unit [pop] swoncat
     TS1 == [dip] cons [i] swoncat
treestep == swap [map] swoncat [TS1 [TS0] dip] dip genrec

   [] 0 [C] [N] treestep
---------------------------
      0


      [n [tree*]] 0 [sum +] [] treestep
   --------------------------------------------------
       n [tree*] [0 [sum +] [] treestep] map sum +

A slight modification.¶

Let's simplify the tree datastructure definition slightly by just letting the children be the rest of the tree:

tree = [] | [node tree*]

The J function changes slightly.

        [node tree*] J
------------------------------
   node N [tree*] [K] map C


[node tree*] uncons [N] dip [K] map C
node [tree*]        [N] dip [K] map C
node N [tree*]              [K] map C
node N [tree*]              [K] map C
node N [K.tree*]                    C

J == uncons [N] dip [K] map C

K == [not] [pop z] [uncons [N] dip] [map C] genrec

I think these trees seem a little easier to read.

Redefining our BTree in terms of this form.¶

BTree = [] | [[key value] left right]

What kind of functions can we write for this with our treestep? The pattern for processing a non-empty node is:

node N [tree*] [K] map C

Plugging in our BTree structure:

[key value] N [left right] [K] map C


[key value] uncons pop [left right] [K] map i
key [value]        pop [left right] [K] map i
key                    [left right] [K] map i
key                    [lkey rkey ]         i
key                     lkey rkey

Doesn't work because map extracts the first item of whatever its mapped function produces. We have to return a list, rather than depositing our results directly on the stack.

[key value] N     [left right] [K] map C

[key value] first [left right] [K] map flatten cons
key               [left right] [K] map flatten cons
key               [[lk] [rk] ]         flatten cons
key               [ lk   rk  ]                 cons
                  [key  lk   rk  ]

So:

[] [flatten cons] [first] treestep

There we go.

In-order traversal with treestep.¶

From here:

key [[lk] [rk]] C
key [[lk] [rk]] i
key  [lk] [rk] roll<
[lk] [rk] key swons concat
[lk] [key rk]       concat
[lk   key rk]

So:

[] [i roll< swons concat] [first] treestep

Miscellaneous Crap¶

Toy with it.¶

Let's reexamine:

[key value left right] R0 [BTree-iter-order] R1
    ...
left BTree-iter-order key value F right BTree-iter-order


[key value left right] unstack swap
 key value left right               swap
 key value right left

key value right left [BTree-iter-order] [cons dipdd] dupdip
key value right left [BTree-iter-order] cons dipdd [BTree-iter-order]
key value right [left BTree-iter-order]      dipdd [BTree-iter-order]
left BTree-iter-order key value right              [BTree-iter-order]

left BTree-iter-order key value   right [F] dip [BTree-iter-order]
left BTree-iter-order key value F right         [BTree-iter-order] i
left BTree-iter-order key value F right          BTree-iter-order

So:

R0 == unstack swap
R1 == [cons dipdd [F] dip] dupdip i

[key value left right] R0                [BTree-iter-order] R1
[key value left right] unstack swap [BTree-iter-order] [cons dipdd [F] dip] dupdip i
 key value right left                    [BTree-iter-order] [cons dipdd [F] dip] dupdip i

 key value right left [BTree-iter-order] cons dipdd [F] dip [BTree-iter-order] i
 key value right [left BTree-iter-order]      dipdd [F] dip [BTree-iter-order] i
 left BTree-iter-order key value   right            [F] dip [BTree-iter-order] i
 left BTree-iter-order key value F right                    [BTree-iter-order] i
 left BTree-iter-order key value F right                     BTree-iter-order


BTree-iter-order == [not] [pop] [unstack swap] [[cons dipdd [F] dip] dupdip i] genrec

Refactor cons cons¶

cons2 == cons cons

Refactoring:

BTree-new == swap [[] []] cons2
T == [cons2 dipdd] cons2 cons infra
Te == [cons2 dipd] cons2 cons infra
Ee == pop swap roll< rest rest cons2

It's used a lot because it's tied to the fact that there are two "data items" in each node. This point to a more general factorization that would render a combinator that could work for other geometries of trees.

A General Form for Trees¶

A general form for tree data with N children per node:

[[data] [child0] ... [childN-1]]

Suggests a general form of recursive iterator, but I have to go walk the dogs at the mo'.

For a given structure, you would have a structure of operator functions and sort of merge them and run them, possibly in a different order (pre- post- in- y'know). The Cn functions could all be the same and use the step trick if the children nodes are all of the right kind. If they are heterogeneous then we need a way to get the different Cn into the structure in the right order. If I understand correctly, the "Bananas..." paper shows how to do this automatically from a type description. They present, if I have it right, a tiny machine that accepts some sort of algebraic data type description and returns a function that can recusre over it, I think.

   [data.. [c0] [c1] ... [cN]] [F C0 C1 ... CN] infil
--------------------------------------------------------
   data F [c0] C0 [c1] C1 ... [cN] CN

Just make [F] a parameter.¶

We can generalize to a sort of pure form:

BTree-iter == [not] [pop] [[F]]            [R1] genrec
           == [not] [pop] [[F] [BTree-iter] R1] ifte

Putting [F] to the left as a given:

 [F] unit [not] [pop] roll< [R1] genrec
[[F]]     [not] [pop] roll< [R1] genrec
          [not] [pop] [[F]] [R1] genrec

Let's us define a parameterized form:

BTree-iter == unit [not] [pop] roll< [R1] genrec

So in the general case of non-empty nodes:

[key value left right] [F] [BTree-iter] R1

We just define R1 to do whatever it has to to process the node. For example:

[key value left right] [F] [BTree-iter] R1
    ...
key value F   left BTree-iter   right BTree-iter
left BTree-iter   key value F   right BTree-iter
left BTree-iter   right BTree-iter   key value F

Pre-, ??-, post-order traversals.

[key value  left right] uncons uncons
 key value [left right]

For pre- and post-order we can use the step trick:

[left right] [BTree-iter] step
    ...
left BTree-iter right BTree-iter

We worked out one scheme for ?in-order? traversal above, but maybe we can do better?

[key value left right]              [F] [BTree-iter] [unstack] dipd
[key value left right] unstack [F] [BTree-iter]
 key value left right               [F] [BTree-iter]

key value left right [F] [BTree-iter] R1.1

Hmm...

key value left right              [F] [BTree-iter] tuck
key value left right [BTree-iter] [F] [BTree-iter] 


[key value left right]                          [F] [BTree-iter] [unstack [roll>] dip] dipd
[key value left right] unstack [roll>] dip [F] [BTree-iter]
 key value left right               [roll>] dip [F] [BTree-iter]
 key value left roll> right                     [F] [BTree-iter]
 left key value right                           [F] [BTree-iter]

left            key value   right              [F] [BTree-iter] tuck foo
left            key value   right [BTree-iter] [F] [BTree-iter] foo
    ...
left BTree-iter key value F right  BTree-iter

We could just let [R1] be a parameter too, for maximum flexibility.

Automatically deriving the recursion combinator for a data type?¶

If I understand it correctly, the "Bananas..." paper talks about a way to build the processor function automatically from the description of the type. I think if we came up with an elegant way for the Joy code to express that, it would be cool. In Joypy the definitions can be circular because lookup happens at evaluation, not parsing. E.g.:

A == ... B ...
B == ... A ...

That's fine. Circular datastructures can't be made though.