Treating Trees II: ``treestep``
===============================

Let’s consider a tree structure, similar to one described `“Why
functional programming matters” by John
Hughes <https://www.cs.kent.ac.uk/people/staff/dat/miranda/whyfp90.pdf>`__,
that consists of a node value followed by zero or more child trees. (The
asterisk is meant to indicate the `Kleene
star <https://en.wikipedia.org/wiki/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``
-------------------

.. code:: ipython2

    from notebook_preamble import D, J, V, define, DefinitionWrapper

.. code:: ipython2

    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

.. code:: ipython2

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

Running this function on an empty tree value gives zero:

::

      [] [pop 0] [] [sum +] treestep
   ------------------------------------
              0

.. code:: ipython2

    J('[] sumtree')  # Empty tree.


.. parsed-literal::

    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

.. code:: ipython2

    J('[23] sumtree')  # No child trees.


.. parsed-literal::

    23


.. code:: ipython2

    J('[23 []] sumtree')  # Child tree, empty.


.. parsed-literal::

    23


.. code:: ipython2

    J('[23 [2 [4]] [3]] sumtree')  # Non-empty child trees.


.. parsed-literal::

    32


.. code:: ipython2

    J('[23 [2 [8] [9]] [3] [4 []]] sumtree')  # Etc...


.. parsed-literal::

    49


.. code:: ipython2

    J('[23 [2 [8] [9]] [3] [4 []]] [pop 0] [] [cons sum] treestep')  # Alternate "spelling".


.. parsed-literal::

    49


.. code:: ipython2

    J('[23 [2 [8] [9]] [3] [4 []]] [] [pop 23] [cons] treestep')  # Replace each node.


.. parsed-literal::

    [23 [23 [23] [23]] [23] [23 []]]


.. code:: ipython2

    J('[23 [2 [8] [9]] [3] [4 []]] [] [pop 1] [cons] treestep')


.. parsed-literal::

    [1 [1 [1] [1]] [1] [1 []]]


.. code:: ipython2

    J('[23 [2 [8] [9]] [3] [4 []]] [] [pop 1] [cons] treestep sumtree')


.. parsed-literal::

    6


.. code:: ipython2

    J('[23 [2 [8] [9]] [3] [4 []]] [pop 0] [pop 1] [sum +] treestep')  # Combine replace and sum into one function.


.. parsed-literal::

    6


.. code:: ipython2

    J('[4 [3 [] [7]]] [pop 0] [pop 1] [sum +] treestep')  # Combine replace and sum into one function.


.. parsed-literal::

    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:

.. code:: ipython2

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


.. parsed-literal::

    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

.. code:: ipython2

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


.. parsed-literal::

    [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

.. code:: ipython2

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


.. parsed-literal::

    [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

.. code:: ipython2

    J('[["key" "value"] ["left"] ["right"] ] ["B"] ["N"] ["C"] treegrind')


.. parsed-literal::

    ['key' 'value'] 'N' [['left'] ['right']] [[not] ['B'] [uncons ['N'] dip] ['C'] genrec] 'C'


``treegrind`` with ``step``
---------------------------

Iteration through the nodes

.. code:: ipython2

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


.. parsed-literal::

    [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.

.. code:: ipython2

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


.. parsed-literal::

    44


Rebuild the tree using ``map`` (imitating ``treestep``.)

.. code:: ipython2

    J('[[3 0] [[2 0] [] []] [[9 0] [[5 0] [[4 0] [] []] [[8 0] [[6 0] [] [[7 0] [] []]] []]] []]]   [] [[100 +] infra] [map cons] treegrind')


.. parsed-literal::

    [[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.

.. code:: ipython2

    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)

.. code:: ipython2

    J('''\
    
    [[3 13] [[2 12] [] []] [[9 19] [[5 15] [[4 14] [] []] [[8 18] [[6 16] [] [[7 17] [] []]] []]] []]]
    
    [] [5] Tree-get
    
    ''')


.. parsed-literal::

    15


.. code:: ipython2

    J('''\
    
    [[3 13] [[2 12] [] []] [[9 19] [[5 15] [[4 14] [] []] [[8 18] [[6 16] [] [[7 17] [] []]] []]] []]]
    
    [pop "nope"] [25] Tree-get
    
    ''')


.. parsed-literal::

    'nope'

