Treating Trees II: treestep

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

tree = [] | [node tree*]

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

tree [B] [N] [C] treestep

If the current tree node is empty then just execute B:

   [] [B] [N] [C] treestep
---------------------------
   []  B

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

(Later on we’ll experiment with making map part of C so you can use other combinators.)

Derive the recursive function.

We can begin to derive it by finding the ifte stage that genrec will produce.

K == [not] [B] [R0]   [R1] genrec
  == [not] [B] [R0 [K] R1] ifte

So we just have to derive J:

J == R0 [K] R1

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*] uncons [N] dip
node [tree*]        [N] dip
node N [tree*]

Next, map K over the 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 == uncons [N] dip [K] map C

Plug it in and convert to genrec:

K == [not] [B] [J                       ] ifte
  == [not] [B] [uncons [N] dip [K] map C] ifte
  == [not] [B] [uncons [N] dip]   [map C] genrec

Extract the givens to parameterize the program.

Working backwards:

[not] [B]          [uncons [N] dip]                  [map C] genrec
[B] [not] swap     [uncons [N] dip]                  [map C] genrec
[B]                [uncons [N] dip] [[not] swap] dip [map C] genrec
                                    ^^^^^^^^^^^^^^^^
[B] [[N] dip]      [uncons] swoncat [[not] swap] dip [map C] genrec
[B] [N] [dip] cons [uncons] swoncat [[not] swap] dip [map C] genrec
        ^^^^^^^^^^^^^^^^^^^^^^^^^^^

Extract a couple of auxiliary definitions:

TS.0 == [[not] swap] dip
TS.1 == [dip] cons [uncons] swoncat
[B] [N] TS.1 TS.0 [map C]                         genrec
[B] [N]           [map C]         [TS.1 TS.0] dip genrec
[B] [N] [C]         [map] swoncat [TS.1 TS.0] dip genrec

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

(alternate) Extract the givens to parameterize the program.

Working backwards:

[not] [B]           [uncons [N] dip]            [map C] genrec
[not] [B] [N]       [dip] cons [uncons] swoncat [map C] genrec
[B] [N] [not] roll> [dip] cons [uncons] swoncat [map C] genrec
        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Define treestep

from notebook_preamble import D, J, V, define, DefinitionWrapper
DefinitionWrapper.add_definitions('''

    _treestep_0 == [[not] swap] dip
    _treestep_1 == [dip] cons [uncons] swoncat
    treegrind == [_treestep_1 _treestep_0] dip genrec
    treestep == [map] swoncat treegrind

''', D)

Examples

Consider trees, the nodes of which are integers. We can find the sum of all nodes in a tree with this function:

sumtree == [pop 0] [] [sum +] treestep
define('sumtree == [pop 0] [] [sum +] treestep')

Running this function on an empty tree value gives zero:

   [] [pop 0] [] [sum +] treestep
------------------------------------
           0
J('[] sumtree')  # Empty tree.
0

Running it on a non-empty node:

[n tree*]  [pop 0] [] [sum +] treestep
n [tree*] [[pop 0] [] [sum +] treestep] map sum +
n [ ... ]                                   sum +
n m                                             +
n+m
J('[23] sumtree')  # No child trees.
23
J('[23 []] sumtree')  # Child tree, empty.
23
J('[23 [2 [4]] [3]] sumtree')  # Non-empty child trees.
32
J('[23 [2 [8] [9]] [3] [4 []]] sumtree')  # Etc...
49
J('[23 [2 [8] [9]] [3] [4 []]] [pop 0] [] [cons sum] treestep')  # Alternate "spelling".
49
J('[23 [2 [8] [9]] [3] [4 []]] [] [pop 23] [cons] treestep')  # Replace each node.
[23 [23 [23] [23]] [23] [23 []]]
J('[23 [2 [8] [9]] [3] [4 []]] [] [pop 1] [cons] treestep')
[1 [1 [1] [1]] [1] [1 []]]
J('[23 [2 [8] [9]] [3] [4 []]] [] [pop 1] [cons] treestep sumtree')
6
J('[23 [2 [8] [9]] [3] [4 []]] [pop 0] [pop 1] [sum +] treestep')  # Combine replace and sum into one function.
6
J('[4 [3 [] [7]]] [pop 0] [pop 1] [sum +] treestep')  # Combine replace and sum into one function.
3

Redefining the Ordered Binary Tree in terms of treestep.

Tree = [] | [[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

Traversal

[key value] first [left right] [K] map i
key [value]       [left right] [K] map i
key               [left right] [K] map i
key               [lkey rkey ]         i
key                lkey rkey

This doesn’t quite work:

J('[[3 0] [[2 0] [][]] [[9 0] [[5 0] [[4 0] [][]] [[8 0] [[6 0] [] [[7 0] [][]]][]]][]]] ["B"] [first] [i] treestep')
3 'B' 'B'

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:

[] [first] [flatten cons] treestep
J('[[3 0] [[2 0] [] []] [[9 0] [[5 0] [[4 0] [] []] [[8 0] [[6 0] [] [[7 0] [] []]] []]] []]]   [] [first] [flatten cons] treestep')
[3 2 9 5 4 8 6 7]

There we go.

In-order traversal

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
J('[[3 0] [[2 0] [] []] [[9 0] [[5 0] [[4 0] [] []] [[8 0] [[6 0] [] [[7 0] [] []]] []]] []]]   [] [uncons pop] [i roll< swons concat] treestep')
[2 3 4 5 6 7 8 9]

With treegrind?

The treegrind function doesn’t include the map combinator, so the [C] function must arrange to use some combinator on the quoted recursive copy [K]. With this function, the pattern for processing a non-empty node is:

node N [tree*] [K] C

Plugging in our BTree structure:

[key value] N [left right] [K] C
J('[["key" "value"] ["left"] ["right"] ] ["B"] ["N"] ["C"] treegrind')
['key' 'value'] 'N' [['left'] ['right']] [[not] ['B'] [uncons ['N'] dip] ['C'] genrec] 'C'

treegrind with step

Iteration through the nodes

J('[[3 0] [[2 0] [] []] [[9 0] [[5 0] [[4 0] [] []] [[8 0] [[6 0] [] [[7 0] [] []]] []]] []]]   [pop] ["N"] [step] treegrind')
[3 0] 'N' [2 0] 'N' [9 0] 'N' [5 0] 'N' [4 0] 'N' [8 0] 'N' [6 0] 'N' [7 0] 'N'

Sum the nodes’ keys.

J('0 [[3 0] [[2 0] [] []] [[9 0] [[5 0] [[4 0] [] []] [[8 0] [[6 0] [] [[7 0] [] []]] []]] []]]   [pop] [first +] [step] treegrind')
44

Rebuild the tree using map (imitating treestep.)

J('[[3 0] [[2 0] [] []] [[9 0] [[5 0] [[4 0] [] []] [[8 0] [[6 0] [] [[7 0] [] []]] []]] []]]   [] [[100 +] infra] [map cons] treegrind')
[[103 0] [[102 0] [] []] [[109 0] [[105 0] [[104 0] [] []] [[108 0] [[106 0] [] [[107 0] [] []]] []]] []]]

Do we have the flexibility to reimplement Tree-get?

I think we do:

[B] [N] [C] treegrind

We’ll start by saying that the base-case (the key is not in the tree) is user defined, and the per-node function is just the query key literal:

[B] [query_key] [C] treegrind

This means we just have to define C from:

[key value] query_key [left right] [K] C

Let’s try cmp:

C == P [T>] [E] [T<] cmp

[key value] query_key [left right] [K] P [T>] [E] [T<] cmp

The predicate P

Seems pretty easy (we must preserve the value in case the keys are equal):

[key value] query_key [left right] [K] P
[key value] query_key [left right] [K] roll<
[key value] [left right] [K] query_key       [roll< uncons swap] dip

[key value] [left right] [K] roll< uncons swap query_key
[left right] [K] [key value]       uncons swap query_key
[left right] [K] key [value]              swap query_key
[left right] [K] [value] key                   query_key

P == roll< [roll< uncons swap] dip

(Possibly with a swap at the end? Or just swap T< and T>.)

So now:

[left right] [K] [value] key query_key [T>] [E] [T<] cmp

Becomes one of these three:

[left right] [K] [value] T>
[left right] [K] [value] E
[left right] [K] [value] T<

E

Easy.

E == roll> popop first

T< and T>

T< == pop [first] dip i
T> == pop [second] dip i

Putting it together

T> == pop [first] dip i
T< == pop [second] dip i
E == roll> popop first
P == roll< [roll< uncons swap] dip

Tree-get == [P [T>] [E] [T<] cmp] treegrind

To me, that seems simpler than the genrec version.

DefinitionWrapper.add_definitions('''

    T> == pop [first] dip i
    T< == pop [second] dip i
    E == roll> popop first
    P == roll< [roll< uncons swap] dip

    Tree-get == [P [T>] [E] [T<] cmp] treegrind

''', D)
J('''\

[[3 13] [[2 12] [] []] [[9 19] [[5 15] [[4 14] [] []] [[8 18] [[6 16] [] [[7 17] [] []]] []]] []]]

[] [5] Tree-get

''')
15
J('''\

[[3 13] [[2 12] [] []] [[9 19] [[5 15] [[4 14] [] []] [[8 18] [[6 16] [] [[7 17] [] []]] []]] []]]

[pop "nope"] [25] Tree-get

''')
'nope'