I am self learning lisp and thought a nice non-trivial program would be to write a set of standard tree insertion and manipulation routines. I figured it could be done with CONS but wanted to try it with a structure.
I put together one version that worked:
(defstruct treenode data left right)
(defun tree-insert ( value tree )
"Insert data into tree"
(if tree
(if (< value (treenode-data tree))
(setf (treenode-left tree) (tree-insert value (treenode-left tree)))
(setf (treenode-right tree) (tree-insert value (treenode-right tree))))
(setf tree (make-treenode :data value)))
tree)
which rebuilt the tree at every step which seemed computationally inefficient. By inefficient, I mean that I have to use setf every time I do another level of recursion. So I wanted to try a scheme that passed the tree by reference rather than by value so I could make assignments in the subroutine that inserts into a tree.
I cobbled the following together, which does not work (but give me credit for having comments):
(defstruct treenode data left right)
(defun tree-insert ( value tree )
"Insert data value into tree, using pass by reference.
value A datum to insert, in this version has to be a number.
tree The tree passed as a symbol."
(setq tval (symbol-value tree))
(if (eq tval nil)
(set tree (make-treenode :data value)) ; Empty tree. Place data here.
(if (< value (treenode-data tval)) ; Non-empty node. Decide which subtree for insert.
(tree-insert value (treenode-left tval)) ; Left side
(tree-insert value (treenode-right tval)))) ; Right side. This is a stable sort.
nil)
? (setf tr nil)
NIL
? (tree-insert 10 'tr)
NIL
? tr
#S(TREENODE :DATA 10 :LEFT NIL :RIGHT NIL)
?
The initial insert works fine. Passing a symbol the (set tree ...) correctly inserts the structure with left and right porinters nil.
Of course, the problem that follows is that on the recursive call to tree-insert I am not passing a symbol.
That is the hangup. I haven't found a way to refer to a structure slot as a symbol that I can then pass to tree-insert.
I've been looking around for a couple of days and found this interesting comment about the defstruct macro: "defstruct not only defines an access function for each slot, but also arranges for setf to work properly on such access functions, defines a predicate named name-p, defines a constructor function named make-name, and defines a copier function named copy-name. All names of automatically created functions are interned in whatever package is current at the time the defstruct form is processed (see package). Also, all such functions may be declared inline at the discretion of the implementation to improve efficiency; if you do not want some function declared inline, follow the defstruct form with a notinline declaration to override any automatic inline declaration."
So, what could I do to do the magic that setf does? I know I can do assignments to slots with setf, but I haven't gotten setf to work in a function due to lexical scope rules. Maybe like adding automatic functions to allow symbol generating, like (treenode-data-symbol tr)?
Of course, lisp programmers have dealt with binary trees since before my first PDP-8/L. What's the lispy way to do this?
This is an edited question. User Rainer Joswig gave a very fast and concise response. I learned a lot from the example he gave. I was interested in the issue of modifying the tree directly rather than using a return value from a function.
From the comments I've seen here, and the one answer by Rainer Joswig, should I draw the conclusion that the pointer manipulation is computationally low in cost, and that the best lisp approach is to use a function that returns a tree rather than relying on an approach of modifying the argument?
simple version for your inspiration:
(defstruct node a b v)
(defun insert-tree (tree value)
(cond ((null tree)
(setf tree (make-node :v value)))
((> (node-v tree)
value)
(setf (node-a tree)
(insert-tree (node-a tree) value)))
(t
(setf (node-b tree)
(insert-tree (node-b tree) value))))
tree)
using it:
CL-USER 171 > (let ((tree nil))
(loop for i in '(4 7 3 5 9 10 11 8)
do (setf tree (insert-tree tree i)))
(pprint tree)
(values))
#S(NODE :A #S(NODE :A NIL :B NIL :V 3)
:B #S(NODE :A #S(NODE :A NIL :B NIL :V 5)
:B #S(NODE :A #S(NODE :A NIL :B NIL :V 8)
:B #S(NODE :A NIL
:B #S(NODE :A NIL
:B NIL
:V 11)
:V 10)
:V 9)
:V 7)
:V 4)
Now, if wanted to do less setf operations, we could check whether the returned subtree is the same which we passed. This will only not be the case when we create a new node.
(defun insert-tree (tree value)
(cond ((null tree)
(setf tree (make-node :v value)))
((> (node-v tree)
value)
(let ((new-tree (insert-tree (node-a tree) value)))
(unless (eql new-tree (node-a tree))
(setf (node-a tree) new-tree))))
(t
(setf (node-b tree)
(insert-tree (node-b tree) value))))
tree)
or with a local macro hiding part of the code:
(defun insert-tree (tree value)
(macrolet ((insert (place call &aux (new-value-sym (gensym "new-value")))
`(let ((,new-value-sym ,call))
(unless (eql ,place ,new-value-sym)
(setf ,place ,new-value-sym)))))
(cond ((null tree)
(setf tree (make-node :v value)))
((> (node-v tree)
value)
(insert (node-a tree) (insert-tree (node-a tree) value)))
(t
(insert (node-b tree) (insert-tree (node-b tree) value))))
tree))
Trying to add an answer from another angle.
In standard Common Lisp structures have a bunch of limitations to make them low-level and efficient to use. Among those limitations:
access to structure slots via slot names is undefined. some implementations do it, others not.
redefining a structure definition has undefined consequences. This means that in some cases, one best restarts Lisp to do that...
The idea behind that: all operations to structures should be able to be inlined and an executing program should not need any further information about structure slots (their names, their memory locations, ...). There would be no dynamic lookup at runtime.
Then Common Lisp in general has this further limitation: it has no first class pointers. There is no mechanism to provide a pointer only pointing directly to the slot of a structure. In some older Lisp dialects this might be possible via a concept of locatives - pointers in those languages. Common Lisp does not support that.
This means practically: to update a slot of a structure, one needs access to the structure and a setter operation.
How do update the slot of a structure?
I can think of two simple ways:
pass the structure, a new value and an indicator what to update -> then dispatch on the indicator and call the right updater
Example
(defun update (s indicator value)
(case indicator
(:a (setf (node-a s) value))
(:b (setf (node-b s) value))))
(update tree :a (make-node :v 100))
Pass a closure, which does the update
Example:
(let ((tree ...))
(flet ((do-something (updater)
(funcall updater (make-node :v 100))))
(do-something (lambda (value) (setf (node-a tree) value) ...)))
With much thanks to Rainer and Will, I understand Common Lisp better, now. The point about not having first class pointers is huge. I don't have to keep looking for that anymore, though I did see a package that implemented refs in my searches.
The key problem in my approach was that I defined an empty tree as nil. Since passing nil doesn't allow any manipulation of the argument, nil being immutable, the algorithm was bound to fail.
Redefining the empty tree as '(nil) allows the program to work.
;; Make list of 5 random numbers.
(setf r5 (loop for i from 1 to 5 collect (random 100)))
;; Initialize tr to empty tree.
;; Empty tree is '(nil). Tree with data is '(data left right),
;; where left and right are either empty tree or tree with data.
(setf tr '(nil))
(defun tree-insert ( value tree )
"Insert data into tree. tree is modified with an insertion."
(if (equal tree '(nil))
(progn ; Empty (sub)tree. Insert value.
(setf (car tree) value)
(setf (cdr tree) (list (list nil)(list nil))))
(progn ; Non-empty subtree.
(if (< value (car tree))
(tree-insert value (second tree)) ; Insert on left.
(tree-insert value (third tree))))) ; Insert on right.
nil)
;; Load tree with the list of random numbers defined above.
(mapc (lambda (val) (tree-insert val tr)) r5)
(defun tree-walk (tree)
"Retrieve keys in sorted order."
(if (car tree)
(progn
(tree-walk (second tree)) ; Left subtree.
(format t " ~d" (car tree))
(tree-walk (third tree))))) ; Right subtree.
;; Walk the tree.
(tree-walk tr)
Example in use:
? (setf r5 (loop for i from 1 to 5 collect (random 100)))
(22 50 76 20 49)
? (setf tr '(nil))
(NIL)
? (mapc (lambda (val) (tree-insert val tr)) r5)
;Compiler warnings :
; In an anonymous lambda form at position 37: Undeclared free variable TR
(22 50 76 20 49)
? tr
(22 (20 (NIL) (NIL)) (50 (49 (NIL) (NIL)) (76 (NIL) (NIL))))
? (tree-walk tr)
20 22 49 50 76
NIL
?
So, several things to make this work. A mutable object has to be passed to the procedure. In this case I redesigned the structure to be a list, either '(nil) for empty, or '(data left right), where left and right are either '(nil) or '(data left right). A list containing nil can be manipulated. However, I had to use car and cdr to access the structure so as to preserve Lisp's pointer that was passed to the procedure.
Another thing I had to do was not use a list constant in the functon definitions. I'm sure knowledgeable people will be in the know about this, and chucle a bit about the opaque error that follow until the issue is understood, but if I had used '((nil)(nil)) rather than (list (list nil)(list nil)) in tree-insert it would not work. Looks like Lisp compiles list shorthand notation to a pointer to an abject in memory which is used on all subsequent calls of the function.
Oh, there is a leftover progn function call in tree-insert. That was from when I wrapped everything with progn to let me add print statements during debugging.
Running timing on the functions was interesting. It's fast! I'll run some timing comparisons to compare the functional reassignment approach vs the search and insert algorithms.
Thanks, again, to the expert comments. Since last contributing I've learned a little about map, loop/collect, and that variables leak out of functions into global space when let isn't used in function definitions. Also wrapping a funciton with a lot of output with (progn ... nil) saves screen space after large data structures are used. I have learned a lot with just this exercise.
Essentially what I am trying to do is take a binary tree with data definition
binary_tree: number | (symbol binary_tree binary_tree)
and create a new version of the tree where each leaf (a number) is replaced with a counter number. I am trying to do this left-to-right and then top-down, so using a breadth first search seems like the obvious choice to visit every node in order. However, my problem is this. I need to accumulate a new binary tree to return it. Is there any possible way to do this since we are visiting each node in order?
So in short if I have a tree defined like this:
(define bt '(foo (bar 26 12) (baz 11 (quux 117 14))))
i need to process and accumulate a new list such that
(define bt '(foo (bar 0 1) (baz 2 (quux 3 4))))
Here is my code:
(define (number-leaves bst)
(define (helper queue counter)
(cond[(non-empty-queue? queue)
(define x (dequeue! queue))
(cond [(number? x)(cons counter (helper queue (+ 1 counter)))]
[(symbol? (car x))(begin (enqueue! queue (car(cdr x)))
(enqueue! queue (car(cdr(cdr x))))
(cons (list(car x)) (helper queue counter)))])]
['()]))
(begin (define q (make-queue))
(enqueue! q bst)
(helper q 0)))
as of now this function returns
(foo bar baz 0 1 2 quux 3 4)
It seems to me that it is impossible to accumulate into a recursive data definition while processing the tree breadth first. What can I do? (NB: car = first and cdr = rest in the EOPL racket dialect)
I think you will need to do two passes.
breath first while making a lookup based on the pairs that hold the leaf and their incremental new value. It's important to use the pairs since the numbers by themselves are not guaranteed unique, thus (eq? 2 2) can be #t even though they are different places.. Comparing the pairs that holds the twos are guaranteed to be only eq? with the very same value.
A standard post order tree traversal where you fetch the new values from the lookup.
The lookup should be a hash for efficiency, but it can be a assoc list for small trees. If you expect the hash to do O(1) when iterating over all the element twice will still only make it O(n).
I have a problem that I need to complete:
We can represent a nonempty binary tree by list (root left_subtree
right_subtree) and the empty tree by the empty list. Each binary
search tree with integer labels can be considered representing a set
of integers. Write a function which, given a set of integers S as a
bst and an integer x, returns both the set of all the integers less
than x and that of all the integers greater than x as bst’s. Use a
pair rather than a list to represent the resulting sets.
I'm very new to Scheme. I've built trees using SML as well as prolog, but can't seem to get a hold of what I need to do for Scheme. Could anyone help me out and guide me towards this goal? Is my tree suppose to just look like this?
(list value left right)
Yes, that's one way to define a tree. Just so long as you make a proper ADT and use it consistently, it shouldn't matter after you've defined it.
(define (make-tree root left right)
(list root left right))
(define (right tree)
(caddr tree))
(define (left tree)
(cadr tree))
(define (root tree)
(car tree))
(define (empty-tree? tree)
(null? tree))
(define empty-tree '())
(define (leaf? tree)
(and (empty-tree? (left tree)) (empty-tree? right tree)))
(define (split-tree-at tree x)
(let ((less ...)
(more ...))
(cons less more))
(define (in-tree-below-x tree x) ;;assuming a sorted tree
(cond ((empty-tree? tree) (empty-tree))
((>= (root tree) x)
(in-tree-below (left tree) x))
(else (make-tree (root tree)
(left tree)
(in-tree-below-x (right tree) x)))))
Im struggling with this problem for over 2 days now and still somehow I cannot solve it.
I have to write a function in SCHEME that takes a list in a tree and displays items in sorted order.
The way I define trees is '(6 (left... ) (right...))
My function to choose a tree:
(define (tree-sort tree)
(cond ((null? tree) '())
((> (car tree) (cadr tree))
(tree-sort (cadr tree)))
(else
(tree-sort (caddr tree))))
)
So I guess I should also have a function that sorts the most indepth list?
I really dont get it and this is the last time I will ever have to deal with scheme. I have never used stackoverflow so please excuse me if the formating is wrong.
Kindly thank you!
Now that you clarified that the tree is already sorted, then you're looking for an in-order traversal of the tree, which returns a sorted list of the elements - I'm assuming that you're interested in a list as the output, because of the base case shown in the question. Try something like this:
(define (tree-sort tree)
(if (empty-tree? tree)
'()
(append (tree-sort (left-subtree tree))
(list (value tree))
(tree-sort (right-subtree tree)))))
Use the appropriate procedures for testing if the tree is empty and for accessing each node's value, left and right subtrees. The above procedure will return a sorted list with the trees' values.
I am trying to write the algorithm for inorder traversal for a binary tree using RACKET/DR. RACKET
(define (print-records node number)
(cond
[(not (empty? node-left))(print-records (node-left node) number )]
*Do This before moving to next IF clause*
[(not (empty? node-right))(print-records(node-right node) number)]
))
I am trying to follow the following algorithm
InOrder(node)
if node is null return
InOrder(node.left)
Print(node)
InOrder(node.Right)
My problem is that through COND I can execute one expression and it will skip the rest. I tried to add two expressions under one it did not work either e.g ((a)(b)). I also tried to make a helper procedure but that did not work either.
You're using cond in a wrong way. Notice that you have to recursively traverse the left part of the tree, then visit the current node and then recursively traverse the right part of the tree - they're not mutually exclusive alternatives, the three steps need to be performed in precisely that order. Try something like this instead:
(define (print-records node number)
(unless (empty? (node-left node))
(print-records (node-left node) number))
(print (node-value node)) ; replace this line with the actual printing code
(unless (empty? (node-right node))
(print-records (node-right node) number)))
Some explanations:
(unless <condition> <body>) is just shorthand for (cond ((not <condition>) <body>)).
The real work of the traversal is done between the two recursive calls, in this case I wrote (print (node-value node)) as an example, replace that line with the actual printing code for the current node's value.
It's not clear what do you intend to do with the number parameter, as it is it's just being passed around, unused.
Walking a binary-tree is a very general operation. You can make a general procedure and then specialize it with the function to apply to each node.
(define (walker node function)
(unless (empty? node)
(walker (node-left node) function)
(function node)
(walker (node-right node) function)))
Note: it is nice to check for empty? at the beginning of the function.
(define (print-records node number)
(walker node (compose print node-value))) ; ignore number, it seems.
You could also work this as:
(define (walking-with function)
(letrec ((walker (lambda (node)
(unless (empty? node)
(walker (node-left node))
(function node)
(walker (node-right node))))))
walker))
(define print-records-for (walking-with (compose print node-value)))
(print-records-for node)
> ...