I am working on a problem which requires finding all paths between two nodes in a directed graph. The graph may have cycles.
Notice that this particular implementation approach is an iterative DFS.
Several approaches I've considered are as follows -
BFS does not appear to have a way to neatly manage this kind of pathing relationships between nodes.
I don't see an easy mechanism for a DFS recursive algorithm to pass up the path when the terminating node is found. (Likely enough it could be done, if I implement a Maybe monad kind of thing).
Creating a GRAPH-PARENT routine. That would add a decent amount of churn (& bugs) in the existing code.
Abstractly, what needs to happen is a tree needs to be generated, with the start node as root, and all leafs are the terminating nodes. Each path from leaf to root is a legal path. That is what a recursive DFS would trace out.
I'm reasonably sure it can be done here, but I don't see exactly how to do it.
I've defined a protocol for this algorithm where GRAPH-EQUAL and GRAPH-NEXT can be defined for arbitrary objects.
The debug node type is a SEARCH-NODE, and it has the data accessor SEARCH-NODE-DATA.
(defun all-paths (start end)
(let ((stack (list start))
(mark-list (list start)) ;I've chosen to hold marking information local to all-paths, instead of marking the objects themselves.
(all-path-list '())) ; Not used yet, using debug statements to think about the problem
(do () ;; intializing no variables
;; While Stack still has elements
((not stack))
(let ((item (pop stack)))
;; I'm looking at the item.
(format t "I: ~a~%" (search-node-data item))
(cond ((graph-equal item end)
(format t "*Q: ~a~%" (loop for var in stack collect (search-node-data var)))
;;Unmark the terminal node so we can view it it next time.
(setf mark-list (remove item mark-list))))
(loop for next in (graph-next item)
do
(cond ((not (in next mark-list :test #'graph-equal))
;; mark the node
(push next mark-list)
;;Put it on the stack
(push next stack))))))))
See A Very General Method for Computing Shortest Paths for an algorithm that can return all paths in a graph (even when there are cycles) as regular expressions over the alphabet of edges in finite time (assuming a finite graph).
You need to pass the path list (mark-list) along with the nodes, since that is part of the state. I've renamed it path in this code:
(defun all-paths (start end)
(let ((stack (list '(start (start)))) ; list of (node path) elements
(all-path-list '()))
(do ()
((not stack))
(let ((item (pop stack)))
(let ((node (first item))
(path (second item)))
(format t "I: ~a~%" (search-node-data node))
(cond ((graph-equal node end)
(format t "*Q: ~a~%"
(loop for var in path
collect (search-node-data var)))))
(loop for next in (graph-next node)
do
(cond ((not (in next path :test #'graph-equal))
(push (list next (cons next path)) stack)))))))))
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.
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 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)
> ...
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.
Im trying to implement Depth First Search in Scheme, but I can only get it to partially work.
This is my code:
(define (depth-first-search graph node neighbour path dest)
(cond ((null? neighbour) #f)
((equal? node dest) path)
((member (car neighbour) path) (depth-first-search graph node (cdr neighbour) path dest))
((memq (car neighbour) (car graph)) (depth-first-search (cdr graph) (car neighbour) (memq (car neighbour) (car graph)) (append path (list (car neighbour))) dest))
(else depth-first-search (cdr graph) path dest)))
And this is my graph, data structure:
(define complete-graph
'((a b c d e)
(b a c)
(c a b f)
(d a e h)
(e a d)
(f c g i)
(g f h i j)
(h d g j)
(i f g j)
(j g h i)))
This is how I call the procedure:
(depth-first-search complete-graph (caar complete-graph) (cdar complete-graph) (list (caar complete-graph)) 'd)
To procedure shoud return the full path frome the start-node to the dest(ination) as a list, but it only seems to work with some start and destination nodes. If I start with 'a and 'c, it returns the right list '(a b c), but if I try with 'a and 'd, I get #f in return. So it probably is something wrong with the backtracking in the algorithm. But I have looked at the code for too long, and I really can't find the problem..
Assuming node 'a has children '(b c d e), node 'b has children '(a c), node ... First you need a function that expands a node to its children.
(define (expand graph node)
(let ((c (assq node graph)))
(if c (cdr c) '())))
Second: you have to remember all visited nodes. In general hat is different from the path (maybe it does not matter in this example). Third: you need to remember all nodes you want to visit (result from the node expansion process). So define a helper function
(define (dfs* graph visited border path dest)
If there are no nodes left to visit, then no road exist.
(cond ((null? border) #f)
If the first element in border is equal to our destination, then we are happy
((eq? (car border) dest) (cons (car border) path))
Lets check all visited nodes. If the first node in border was visited before then proceed without node expansion
((memq (car border) visited)
(dfs* graph visited (cdr border) path dest))
Otherwise expand the first node of border
(else (dfs* graph
(cons (car border) visited)
(append (expand graph (car border)) (cdr border))
(cons (car border) path)
dest))))
Call that helper function with starting values for visited, border and path:
(define (dfs graph src dst)
(dfs* graph '() (list src) '() dst)
For breath first search: append the expanded node at the end of border
Edit:
a) visited and path are the same, you can drop one of them
b) the path is returned in reverse order
c) The procedure is not correct, path contains all visited nodes. But a post processing of the result of dfs* will do the job.
You don't want to change the graph as you do the depth-first search, just the current node. If you want to do things purely functionally, have your function look like:
(define (depth-first-search graph node dest path)
(let dfs ((node node) (path path))
(let ((recur (lambda (node) (dfs node (cons node path)))))
; Write code here
; Recursive calls should use recur, not dfs or depth-first-search
...)))
(returning a path or #f as the result). You can use ormap (in Racket or SRFI-1) to iterate through all neighbors of a node, returning the first value that is not #f.