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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
In [1]:
from notebook_preamble import D, J, V, define, DefinitionWrapper
In [2]:
define('BTree-iter == [not] [pop] roll< [dupdip rest rest] cons [step] genrec')
In [3]:
J('[] [23] BTree-iter')  #  It doesn't matter what F is as it won't be used.
In [4]:
J('["tommy" 23 [] []] [first] BTree-iter')
'tommy'
In [5]:
J('["tommy" 23 ["richard" 48 [] []] ["jenny" 18 [] []]] [first] BTree-iter')
'tommy' 'richard' 'jenny'
In [6]:
J('["tommy" 23 ["richard" 48 [] []] ["jenny" 18 [] []]] [second] BTree-iter')
23 48 18

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
In [7]:
define('BTree-new == swap [[] []] cons cons')
In [8]:
V('"v" "k" BTree-new')
                . 'v' 'k' BTree-new
            'v' . 'k' BTree-new
        'v' 'k' . BTree-new
        'v' 'k' . swap [[] []] cons cons
        'k' 'v' . [[] []] cons cons
'k' 'v' [[] []] . cons cons
'k' ['v' [] []] . cons
['k' 'v' [] []] . 

(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
In [9]:
define('P == pop roll> pop first')
In [10]:
V('["k" "v" [] []] "vv" "kk" [0] P >')
                              . ['k' 'v' [] []] 'vv' 'kk' [0] P >
              ['k' 'v' [] []] . 'vv' 'kk' [0] P >
         ['k' 'v' [] []] 'vv' . 'kk' [0] P >
    ['k' 'v' [] []] 'vv' 'kk' . [0] P >
['k' 'v' [] []] 'vv' 'kk' [0] . P >
['k' 'v' [] []] 'vv' 'kk' [0] . pop roll> pop first >
    ['k' 'v' [] []] 'vv' 'kk' . roll> pop first >
    'kk' ['k' 'v' [] []] 'vv' . pop first >
         'kk' ['k' 'v' [] []] . first >
                     'kk' 'k' . >
                         True . 

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
In [11]:
define('K == cons cons dipdd')
define('T == [K] cons cons cons infra')
In [12]:
V('"r" "l" "v" "k" "vv" "kk" [0] K')
                              . 'r' 'l' 'v' 'k' 'vv' 'kk' [0] K
                          'r' . 'l' 'v' 'k' 'vv' 'kk' [0] K
                      'r' 'l' . 'v' 'k' 'vv' 'kk' [0] K
                  'r' 'l' 'v' . 'k' 'vv' 'kk' [0] K
              'r' 'l' 'v' 'k' . 'vv' 'kk' [0] K
         'r' 'l' 'v' 'k' 'vv' . 'kk' [0] K
    'r' 'l' 'v' 'k' 'vv' 'kk' . [0] K
'r' 'l' 'v' 'k' 'vv' 'kk' [0] . K
'r' 'l' 'v' 'k' 'vv' 'kk' [0] . cons cons dipdd
'r' 'l' 'v' 'k' 'vv' ['kk' 0] . cons dipdd
'r' 'l' 'v' 'k' ['vv' 'kk' 0] . dipdd
                          'r' . 'vv' 'kk' 0 'l' 'v' 'k'
                     'r' 'vv' . 'kk' 0 'l' 'v' 'k'
                'r' 'vv' 'kk' . 0 'l' 'v' 'k'
              'r' 'vv' 'kk' 0 . 'l' 'v' 'k'
          'r' 'vv' 'kk' 0 'l' . 'v' 'k'
      'r' 'vv' 'kk' 0 'l' 'v' . 'k'
  'r' 'vv' 'kk' 0 'l' 'v' 'k' . 
In [13]:
V('["k" "v" "l" "r"] "vv" "kk" [0] T')
                                    . ['k' 'v' 'l' 'r'] 'vv' 'kk' [0] T
                  ['k' 'v' 'l' 'r'] . 'vv' 'kk' [0] T
             ['k' 'v' 'l' 'r'] 'vv' . 'kk' [0] T
        ['k' 'v' 'l' 'r'] 'vv' 'kk' . [0] T
    ['k' 'v' 'l' 'r'] 'vv' 'kk' [0] . T
    ['k' 'v' 'l' 'r'] 'vv' 'kk' [0] . [K] cons cons cons infra
['k' 'v' 'l' 'r'] 'vv' 'kk' [0] [K] . cons cons cons infra
['k' 'v' 'l' 'r'] 'vv' 'kk' [[0] K] . cons cons infra
['k' 'v' 'l' 'r'] 'vv' ['kk' [0] K] . cons infra
['k' 'v' 'l' 'r'] ['vv' 'kk' [0] K] . infra
                    'r' 'l' 'v' 'k' . 'vv' 'kk' [0] K [] swaack
               'r' 'l' 'v' 'k' 'vv' . 'kk' [0] K [] swaack
          'r' 'l' 'v' 'k' 'vv' 'kk' . [0] K [] swaack
      'r' 'l' 'v' 'k' 'vv' 'kk' [0] . K [] swaack
      'r' 'l' 'v' 'k' 'vv' 'kk' [0] . cons cons dipdd [] swaack
      'r' 'l' 'v' 'k' 'vv' ['kk' 0] . cons dipdd [] swaack
      'r' 'l' 'v' 'k' ['vv' 'kk' 0] . dipdd [] swaack
                                'r' . 'vv' 'kk' 0 'l' 'v' 'k' [] swaack
                           'r' 'vv' . 'kk' 0 'l' 'v' 'k' [] swaack
                      'r' 'vv' 'kk' . 0 'l' 'v' 'k' [] swaack
                    'r' 'vv' 'kk' 0 . 'l' 'v' 'k' [] swaack
                'r' 'vv' 'kk' 0 'l' . 'v' 'k' [] swaack
            'r' 'vv' 'kk' 0 'l' 'v' . 'k' [] swaack
        'r' 'vv' 'kk' 0 'l' 'v' 'k' . [] swaack
     'r' 'vv' 'kk' 0 'l' 'v' 'k' [] . swaack
      ['k' 'v' 'l' 0 'kk' 'vv' 'r'] . 

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.

In [14]:
define('Te == [cons cons dipd] cons cons cons infra')
In [15]:
V('["k" "v" "l" "r"] "vv" "kk" [0] Te')
                                                 . ['k' 'v' 'l' 'r'] 'vv' 'kk' [0] Te
                               ['k' 'v' 'l' 'r'] . 'vv' 'kk' [0] Te
                          ['k' 'v' 'l' 'r'] 'vv' . 'kk' [0] Te
                     ['k' 'v' 'l' 'r'] 'vv' 'kk' . [0] Te
                 ['k' 'v' 'l' 'r'] 'vv' 'kk' [0] . Te
                 ['k' 'v' 'l' 'r'] 'vv' 'kk' [0] . [cons cons dipd] cons cons cons infra
['k' 'v' 'l' 'r'] 'vv' 'kk' [0] [cons cons dipd] . cons cons cons infra
['k' 'v' 'l' 'r'] 'vv' 'kk' [[0] cons cons dipd] . cons cons infra
['k' 'v' 'l' 'r'] 'vv' ['kk' [0] cons cons dipd] . cons infra
['k' 'v' 'l' 'r'] ['vv' 'kk' [0] cons cons dipd] . infra
                                 'r' 'l' 'v' 'k' . 'vv' 'kk' [0] cons cons dipd [] swaack
                            'r' 'l' 'v' 'k' 'vv' . 'kk' [0] cons cons dipd [] swaack
                       'r' 'l' 'v' 'k' 'vv' 'kk' . [0] cons cons dipd [] swaack
                   'r' 'l' 'v' 'k' 'vv' 'kk' [0] . cons cons dipd [] swaack
                   'r' 'l' 'v' 'k' 'vv' ['kk' 0] . cons dipd [] swaack
                   'r' 'l' 'v' 'k' ['vv' 'kk' 0] . dipd [] swaack
                                         'r' 'l' . 'vv' 'kk' 0 'v' 'k' [] swaack
                                    'r' 'l' 'vv' . 'kk' 0 'v' 'k' [] swaack
                               'r' 'l' 'vv' 'kk' . 0 'v' 'k' [] swaack
                             'r' 'l' 'vv' 'kk' 0 . 'v' 'k' [] swaack
                         'r' 'l' 'vv' 'kk' 0 'v' . 'k' [] swaack
                     'r' 'l' 'vv' 'kk' 0 'v' 'k' . [] swaack
                  'r' 'l' 'vv' 'kk' 0 'v' 'k' [] . swaack
                   ['k' 'v' 0 'kk' 'vv' 'l' 'r'] . 

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]
In [16]:
define('Ee == pop swap roll< rest rest cons cons')
In [17]:
V('["k" "v" "l" "r"] "vv" "k" [0] Ee')
                               . ['k' 'v' 'l' 'r'] 'vv' 'k' [0] Ee
             ['k' 'v' 'l' 'r'] . 'vv' 'k' [0] Ee
        ['k' 'v' 'l' 'r'] 'vv' . 'k' [0] Ee
    ['k' 'v' 'l' 'r'] 'vv' 'k' . [0] Ee
['k' 'v' 'l' 'r'] 'vv' 'k' [0] . Ee
['k' 'v' 'l' 'r'] 'vv' 'k' [0] . pop swap roll< rest rest cons cons
    ['k' 'v' 'l' 'r'] 'vv' 'k' . swap roll< rest rest cons cons
    ['k' 'v' 'l' 'r'] 'k' 'vv' . roll< rest rest cons cons
    'k' 'vv' ['k' 'v' 'l' 'r'] . rest rest cons cons
        'k' 'vv' ['v' 'l' 'r'] . rest cons cons
            'k' 'vv' ['l' 'r'] . cons cons
            'k' ['vv' 'l' 'r'] . cons
            ['k' 'vv' 'l' 'r'] . 
In [18]:
define('E == [P <] [Te] [Ee] ifte')

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
In [19]:
define('BTree-add == [popop not] [[pop] dipd BTree-new] [] [[P >] [T] [E] ifte] genrec')
In [20]:
J('[] 23 "b" BTree-add')  # Initial
['b' 23 [] []]
In [21]:
J('["b" 23 [] []] 88 "c" BTree-add')  # Less than
['b' 23 [] ['c' 88 [] []]]
In [22]:
J('["b" 23 [] []] 88 "a" BTree-add')  # Greater than
['b' 23 ['a' 88 [] []] []]
In [23]:
J('["b" 23 [] []] 88 "b" BTree-add')  # Equal to
['b' 88 [] []]
In [24]:
J('[] 23 "a" BTree-add 88 "b" BTree-add 44 "c" BTree-add')  # Series.
['a' 23 [] ['b' 88 [] ['c' 44 [] []]]]

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.

In [25]:
J('[] [3 9 5 2 8 6 7 8 4] [0 swap BTree-add] step')
[3 0 [2 0 [] []] [9 0 [5 0 [4 0 [] []] [8 0 [6 0 [] [7 0 [] []]] []]] []]]
In [26]:
define('to_set == [] swap [0 swap BTree-add] step')
In [27]:
J('[3 9 5 2 8 6 7 8 4] to_set')
[3 0 [2 0 [] []] [9 0 [5 0 [4 0 [] []] [8 0 [6 0 [] [7 0 [] []]] []]] []]]

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

In [28]:
define('unique == [to_set [first] BTree-iter] cons run')
In [29]:
J('[3 9 3 5 2 9 8 8 8 6 2 7 8 4 3] unique')  # Filter duplicate items.
[7 6 8 4 5 9 2 3]

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
In [30]:
from joy.library import FunctionWrapper
from joy.utils.stack import concat
from notebook_preamble import D


@FunctionWrapper
def cmp_(stack, expression, dictionary):
    '''
    cmp 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
    '''
    L, (E, (G, (b, (a, stack)))) = stack
    expression = concat(G if a > b else L if a < b else E, expression)
    return stack, expression, dictionary


D['cmp'] = cmp_
In [31]:
from joy.library import FunctionWrapper, S_ifte


@FunctionWrapper
def cond(stack, expression, dictionary):
  '''
  like a case statement; works by rewriting into a chain of ifte.

  [..[[Bi] Ti]..[D]] -> ...


        [[[B0] T0] [[B1] T1] [D]] cond
  -----------------------------------------
     [B0] [T0] [[B1] [T1] [D] ifte] ifte

  '''
  conditions, stack = stack
  if conditions:
    expression = _cond(conditions, expression)
    try:
      # Attempt to preload the args to first ifte.
      (P, (T, (E, expression))) = expression
    except ValueError:
      # If, for any reason, the argument to cond should happen to contain
      # only the default clause then this optimization will fail.
      pass
    else:
      stack = (E, (T, (P, stack)))
  return stack, expression, dictionary


def _cond(conditions, expression):
  (clause, rest) = conditions
  if not rest:  # clause is [D]
    return clause
  P, T = clause
  return (P, (T, (_cond(rest, ()), (S_ifte, expression))))



D['cond'] = cond
In [32]:
J("1 0 ['G'] ['E'] ['L'] cmp")
'G'
In [33]:
J("1 1 ['G'] ['E'] ['L'] cmp")
'E'
In [34]:
J("0 1 ['G'] ['E'] ['L'] cmp")
'L'
In [35]:
from joy.library import DefinitionWrapper


DefinitionWrapper.add_definitions('''

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

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

''', D)
In [36]:
J('[] 23 "b" BTree-add')  # Initial
['b' 23 [] []]
In [37]:
J('["b" 23 [] []] 88 "c" BTree-add')  # Less than
['b' 23 [] ['c' 88 [] []]]
In [38]:
J('["b" 23 [] []] 88 "a" BTree-add')  # Greater than
['b' 23 ['a' 88 [] []] []]
In [39]:
J('["b" 23 [] []] 88 "b" BTree-add')  # Equal to
['b' 88 [] []]
In [40]:
J('[] 23 "a" BTree-add 88 "b" BTree-add 44 "c" BTree-add')  # Series.
['a' 23 [] ['b' 88 [] ['c' 44 [] []]]]

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.

In [41]:
define('BTree-iter-order == [not] [pop] [dup third] [[cons dip] dupdip [[first] dupdip] dip [rest rest rest first] dip i] genrec')
In [42]:
J('[3 9 5 2 8 6 7 8 4] to_set BTree-iter-order')
2 3 4 5 6 7 8 9

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
In [43]:
# I don't want to deal with name conflicts with the above so I'm inlining everything here.
# The original Joy system has "hide" which is a meta-command which allows you to use named
# definitions that are only in scope for a given definition.  I don't want to implement
# that (yet) so...


define('''
BTree-get == [pop not] swap [] [
  over [pop popop first] nullary
  [[rest rest rest first] dipd i]
  [popop second]
  [[third] dipd i]
  cmp
  ] genrec
''')
In [44]:
J('[] "gary" [popop "err"] BTree-get')
'err'
In [45]:
J('["gary" 23 [] []] "gary" [popop "err"] BTree-get')
23
In [46]:
J('''

    [] [[0 'a'] [1 'b'] [2 'c']] [i BTree-add] step

    'c' [popop 'not found'] BTree-get

''')
2

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
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