Hi I am new to Racket using it for a binary tree structure.
Using the following structure
(define-struct human(age hight))
I have created the following object/variable/human
(define James(make-human 10 50))
If I have a node in a binary tree structure
(define-struct node (left human right))
How can I compare a different object's hight (say Michael) with James given that James is within the node, so for an example:
(define (insert-human-into-tree human node)
(cond
[(empty? node)(make-node empty human empty)]
[(<= human-hight( **node-human-hight**))
I need to know how to access the hight field of the human object which will be within the node ( node-human-hight).
Use the accessor procedures in the struct, for example:
(define-struct human(age hight))
(define james (make-human 10 50))
(define-struct node (left human right))
(define anode (make-node null james null))
; access the human in the node, and then the height of that human
(human-hight (node-human anode))
=> 50
... And it's spelled "height", not "hight". So, to answer the question, the comparison would look like this:
(<= (human-hight (node-human node)) (human-hight human))
Your insert-human-into-tree is going to need a bit more beyond the comparison to actually insert a new node. It will look something like this (which also answers your prime question):
(define (insert-human-into-tree human node)
(if (empty? node)
(make-node empty human empty)
(let ((other (node-human node)))
(cond ((< (human-height other) (human-height human))
(insert-human-into-tree human (node-left node))) ; go left
((> (human-height other) (human-height human))
(insert-human-into-tree human (node-right node))) ; go right
(else ;; same height
...)))))
This will get you started.
Related
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.
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.
How would I count how many nodes a tree has?
;;; A Binary is one of:
;;; - Number
;;; - (make-node BT BT)
(define-struct node (left right))
(define tree1 (make-node (make-node 10 9)
(make-node 3
(make-node 1 5))))
(define (how-many? nd)
(cond
[(number? nd)....]
[(node? n)
(..... (how-many? (node-left nd))
(how-many? (node-right nd)))]))
so for tree1 I should get
(check-expect (how-many? tree1) 5)
I think I got the template right. If it's a number, you should return 1. But if it's a node, what type of function should I put in the dotted lines?
This is somewhat similar to how you'd process a list. The difference? that now you'll go to the "left" and to the "right" of each node, instead of going just to the "rest". Other than that, the same ideas apply:
(define (how-many? nd)
(cond
[(number? nd) <???>] ; if it's a leaf, then how many numbers does it have?
[(node? nd)
(<???> (how-many? (node-left nd)) ; how do we combine the results?
(how-many? (node-right nd)))]))
You got the base case right, as I can see in the question. The recursive case is simple! remember, how did we combine the results when we were adding the elements in a list? here's the same thing, it's just that we combine the results from the left and from the right subtrees.
I have this code:
(define tree `(A (B (C)) (D (E)) (C (E))))
(define (prog1 graph)
(let ([seen `()])
(define (sub g)
(cond
[(member (car g) seen) `()]
[else
(set! seen (cons (car g) seen))
(cond
[(null? (cdr g)) (list (car g))]
[else
(cons (car g) (map sub (cdr g)))])]))
(delete `() (sub graph))))
(define delete
(lambda (x y)
(if (null? y )
`()
(if (eqv? (car y) x)
(delete x (cdr y))
(cons (car y) (delete x (cdr y)))))))
It prints a connected graph where all the nodes appear once.
Running (prog1 tree)
prints: (A (B (C)) (D (E)))
I have looked at various depth-first searches in lisp (something which is similar to what I'm trying to do) and they appear to be much more elegant to this, some using iterative approaches. I am aware that the program isn't very efficient (on huge trees it runs pretty slow) so how would I go about improving the efficiency of this code?
Thanks, James
The member procedure performs an O(n) lookup on lists every time it's invoked. That's not what we want for quickly testing set membership, for that you should use a data structure providing an O(1) complexity for both adding elements and testing element membership in a collection, ideally a Set data structure or in its place a Hash Table. For example, in Racket try replacing these lines (or use the default hash table implementation in your Scheme interpreter):
(let ([seen `()]) => (let ([seen (make-hash)])
[(member (car g) seen) `()] => [(hash-has-key? seen (car g)) '()]
(set! seen (cons (car g) seen)) => (hash-set! seen (car g) 'ok)
Also, in general you want to use quotes in your code: '() instead of quasiquotes: `(), see the links to understand the difference and when it's appropriate to use quasiquoting.
Finally, you can use the built-in remove procedure, there's no need to implement your own delete.
In most cases the bottleneck in this code will not be the tree traversal, but the member lookup. The complexity of your function seems to be roughly O(M*N), where M is the number of distinct nodes and N is the number of total nodes. The reason why M goes into this as a factor is because you're looking up nodes in a linear list, which takes time proportional to the length of the list (which in your case is proportional to the number of distinct nodes).
The way to get rid of the M is to use a more efficient data structure for the lookup. R6RS defines hash tables, for instance.