How to walk an AST with tail recursion in Clojure - clojure

I have an ANTLR3 AST which I need to traverse using a post-order, depth-first traversal which I have implemented as roughly the following Clojure:
(defn walk-tree [^CommonTree node]
(if (zero? (.getChildCount node))
(read-string (.getText node))
(execute-node
(map-action node)
(map walk-tree (.getChildren node)))))))
I would like to convert this to tail recursion using loop...recur, but I haven't been able to figure out how to effectively use an explicit stack to do this since I need a post-order traversal.

Instead of producing a tail recursive solution which traverses the tree and visits each node, you could produce a lazy sequence of the depth first traversal using the tree-seq function and then get the text out of each object in the traversal. Lazy sequences never blow the stack because they store all the state required to produce the next item in the sequence in the heap. They are very often used instead of recursive solutions like this where loop and recur are more diffacult.
I don't know what your tree looks like though a typical answer would look something like this. You would need to play with the "Has Children" "list of children" functions
(map #(.getText %) ;; Has Children? List of Children Input Tree
(tree-seq #(> (.getChildCount #) 0) #(.getChildren %) my-antlr-ast))
If tree-seq does not suit your needs there are other ways to produce a lazy sequence from a tree. Look at the zipper library next.

As you mention, the only way to implement this using tail recursion is to switch to using an explicit stack. One possible approach is to convert the tree structure into a stack structure that is essentially a Reverse Polish notation representation of the tree (using a loop and an intermediate stack to accomplish this). You would then use another loop to traverse the stack and build up your result.
Here's a sample program that I wrote to accomplish this, using the Java code at postorder using tail recursion as an inspiration.
(def op-map {'+ +, '- -, '* *, '/ /})
;; Convert the tree to a linear, postfix notation stack
(defn build-traversal [tree]
(loop [stack [tree] traversal []]
(if (empty? stack)
traversal
(let [e (peek stack)
s (pop stack)]
(if (seq? e)
(recur (into s (rest e))
(conj traversal {:op (first e) :count (count (rest e))}))
(recur s (conj traversal {:arg e})))))))
;; Pop the last n items off the stack, returning a vector with the remaining
;; stack and a list of the last n items in the order they were added to
;; the stack
(defn pop-n [stack n]
(loop [i n s stack t '()]
(if (= i 0)
[s t]
(recur (dec i) (pop s) (conj t (peek s))))))
;; Evaluate the operations in a depth-first manner, using a temporary stack
;; to hold intermediate results.
(defn eval-traversal [traversal]
(loop [op-stack traversal arg-stack []]
(if (empty? op-stack)
(peek arg-stack)
(let [o (peek op-stack)
s (pop op-stack)]
(if-let [a (:arg o)]
(recur s (conj arg-stack a))
(let [[args op-args] (pop-n arg-stack (:count o))]
(recur s (conj args (apply (op-map (:op o)) op-args)))))))))
(defn eval-tree [tree] (-> tree build-traversal eval-traversal))
You can call it as such:
user> (def t '(* (+ 1 2) (- 4 1 2) (/ 6 3)))
#'user/t
user> (eval-tree t)
6
I leave it as an exercise to the reader to convert this to work with a Antlr AST structure ;)

I'm not skilled up on clojure, but I think I understand what you're looking for.
Here's some pseudocode. The stack here in my pseudocode looks like a stateful object, but it's quite feasible to use an immutable one instead. It uses something like O(depth of tree * max children per node) heap.
walk_tree(TreeNode node) {
stack = new Stack<Pair<TreeNode, Boolean>>();
push(Pair(node, True), stack)
walk_tree_aux(stack);
}
walk_tree_aux(Stack<Pair<TreeNode, Boolean>> stack) { -- this should be tail-recursive
if stack is empty, return;
let (topnode, topflag) = pop(stack);
if (topflag is true) {
push Pair(topnode, False) onto stack);
for each child of topnode, in reverse order:
push(Pair(child, True)) onto stack
walk_tree_aux(stack);
} else { -- topflag is false
process(topnode)
walk_tree_aux(stack);
}
}

Related

Tracking down a StackOverflow in a Clojure program, contains SSCCE

I am having a hard time tracking down a stackoverflow error produced by my clojure implementation of the Bron-Kerbosch algorithm.
Below is my implementation of the algorithm itself, and here is a link to a SSCCE outlining the issue http://pastebin.com/gtpTT4B4.
(defn Bron-Kerbosch [r p x graph cliques]
;(println (str "R:" r " P:" p " X:" x " Clq:" cliques))
(cond (and (empty? p) (empty? x)) (concat cliques r)
:else
(let [neigh (neighV graph (dec (count p)))]
(loop [loop-clq '(cliques)
loop-cnt (dec (count p))
loop-p '(p)
loop-x '(x)]
(cond (= -1 loop-cnt) loop-clq
:else
(recur (concat loop-clq
(Bron-Kerbosch (concat r loop-cnt)
(concat p neigh)
(filter (set x) neigh)
graph cliques))
(dec loop-cnt)
(filter (set p) loop-cnt)
(concat x loop-cnt)))))))
I would have to assume that the issue obviously lies within one of my two bootstrap conditions (cond (and (empty? p) (empty? x)) and (cond (= -1 loop-cnt) because the function algorithm is recursive.
Though this assumes that I am building the lists x r p correctly. Judging by the output of the commented out print statement (cliques is always printed as an EmptyList) I assume that my list comprehension might also be the issue.
Along the same lines, the other issue I can somewhat see is that I am not actually calling the algorithm properly in the BK-Call function (in the SSCCEE).
My overall question is what is causing this? Though this is somewhat too open, another question that might lead me to my answer is how I might go about using the print statement on the first line.
When the print statement is uncommented it produces the output
R:clojure.lang.LazySeq#2044e0b9 P:clojure.lang.LazySeq#7f9700a5 X:clojure.lang.LazySeq#1 Clq:clojure.lang.PersistentList$EmptyList#1
I assume that if I could see that x r p are at each call I might be able to see where the algorithm is going wrong.
Can anyone point me in the right direction?
EDIT: The neighV function from the SSCCE
(defn neighV [graph nodenum]
(let [ret-list (for [i (range (count graph)) :when (contains? (graph i) nodenum)] i)]
ret-list))
EDIT2: Noisesmith's answers had gotten me closer to the solution and made sense to me. I wrapped all of my concat in doall. After trying to call the function again I was getting "Cannot cast Long to Seq" errors, I figured that these stemmed from trying to concat loop-cnt onto lists in the function
fptests.core> (BK-Call (sanity1))
IllegalArgumentException Don't know how to create ISeq from: java.lang.Long clojure.lang.RT.seqFrom (RT.java:505)
fptests.core> (concat 1 '(2 3))
IllegalArgumentException Don't know how to create ISeq from: java.lang.Long clojure.lang.RT.seqFrom (RT.java:505)
So I then wrapped each loop-cnt in a '() to turn it into a list before it is concat
fptests.core> (concat '(1) '(2 3))
(1 2 3)
Which, after I made all of these changes, I ended back at my stack overflow.. Here is the new Bron-Kerbosch function with all of the edits. I guess I now have the same questions as I did before..
Though the new ones are, did I implement that changes that I should have correctly, does the usage of '() make sense to fix the issue that came up after implementing noisesmith's changes?
(defn Bron-Kerbosch1 [r p x graph cliques]
(cond (and (empty? p) (empty? x)) (doall (concat cliques r))
:else
(let [neigh (neighV graph (dec (count p)))]
(loop [loop-clq '(cliques)
loop-cnt (dec (count p))
loop-p '(p)
loop-x '(x)]
(cond (= -1 loop-cnt) loop-clq
:else
(recur (doall (concat loop-clq
(Bron-Kerbosch1 (doall (concat r '(loop-cnt)))
(doall (concat p neigh))
(filter (set x) neigh)
graph cliques)))
(dec loop-cnt)
(filter (set p) '(loop-cnt))
(doall (concat x '(loop-cnt)))))))))
EDIT3: After patiently waiting for my prn statements to stop (not sure why I put them in when I knew it was in a SO) I found that most if not all statements printed were along the lines of
"R:" (loop-cnt loop-cnt loop-cnt loop-cnt loop-cnt loop-cnt loop-cnt ...)
"P:" (range (count graph) 0 2 3) " X:" () " Clq:" ()
After inspecting this some I realized that I have not been recursively calling the function properly. I have been union'ing items to P instead of removing them. This causes P to continuously grow. This is most likely the cause of my stack overflow. Though there are still some issues. I still am creating a stackoverflow, yet again.
Once I fixed my issue of continuing to union to P my issue is that when I concat loop-cnt it is not, I guess to say, evaluated to a value but it stays as a variable name loop-cnt. I suspect that my stack overflow now lies with my bootstrap condition not being met because it cannot be met if loop-cnt is not evaluated to a number.
So I think my issue now lies with concat loop-cnt to a list as a number and not a variable.
concat is lazy. Recursive calls that build concat on top of concat, without realizing any of the prior layers of laziness, each add to the size of the stack of calls that will be needed to realize the lazy-seq.
Does this concatenation need to be lazy at all? If not, wrap the calls to concat in calls to doall. This will make the concatenation eager, which reduces the size of the call stack needed to realize the final result, and thus eliminating the StackOverflowError.
Also, the correct way to print a lazy-seq and see the contents is prn, you can use pr-str to get the form of the value that pr or prn would use as a string, if needed.
You are misusing quoted lists, I think.
For example, in (defn Bron-Kerbosch1 [ ... ] ... ), '(cliques) evaluates to a list containing the symbol cliques, not to a list containing the argument cliques as its one element. You want (list cliques) for that.
Similar cases abound.

review my beginners clojure reverse function [closed]

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I am just starting out in my clojure journey and I wonder if anybody can point out my beginners mistakes in my function below that simply reverses a list. I know that there is already a reverse function so this is purely for learning.
(defn rev
([l]
(if (= (count l) 1) l (rev '() l)))
([l orig]
(if (= (count orig) 0)
l
(rev (conj l (first orig)) (rest orig)))))
In my defence, it does work but what I am finding myself doing a lot in clojure is overloading the arguments list to take into account when I need a working list like in this example where I conj new items onto.
First, it's a really good idea to look at the existing implementation of reverse function first:
(defn reverse [coll]
(reduce conj () coll))
The main difference from your code is that existing implementation of reverse uses higher-order function reduce. It's a good practice to use higher-order functions instead of recursion wherever possible.
--
But let's assume that your goal is to learn recursion. Here is how I would've written it:
(defn rev
([coll]
(rev coll ()))
([coll acc]
(if-let [[h & ts] (seq coll)]
(recur ts (conj acc h))
acc)))
Let's have a closer look at my code.
First, I'm using if-let and seq to check that coll is a non-empty collection.
Then I'm using destructuring to get the first element of the given collection and the rest of it.
In other words, my construction
(if-let [[h & ts] (seq coll)]
(recur ts (conj acc h))
acc)
could be rewritten with if, let, first and rest:
(if-not (empty? coll)
(let [h (first coll)
ts (rest coll)]
(recur ts (conj acc h)))
acc)
which is almost what you wrote yourself.
The last important thing is that I'm using recur instead of calling rev directly.
It allow clojure compiler to perform tail recursion optimization.
You should also consider using loop instead of creating an overloaded function, unless you want to make two-arguments form public:
(defn rev [coll]
(loop [coll coll
acc ()]
(if-let [[h & ts] (seq coll)]
(recur ts (conj acc h))
acc)))
--
So, there is a lot of things to improve in your code. But I can see only three real mistakes there:
you should use recur;
there is no point in checking (= (count l) 1);
you should use empty? instead of (= (count orig) 0).
Here I fixed these two mistakes in your code:
(defn rev
([l]
(rev '() l))
([l orig]
(if (empty? orig)
l
(recur (conj l (first orig)) (rest orig)))))
If you don't want to expose the function with other arities, define and employ them locally:
(defn rev [l]
((fn rev2 [l orig]
(if (empty? orig)
l
(rev2 (conj (first orig) l) (rest orig))))
() l))
You may find it easier to use let or letfn:
(defn rev [l]
(letfn [(rev2 [l orig]
(if (empty? orig)
l
(rev2 (conj l (first orig)) (rest orig))))]
(rev2 () l)))
And, by the way, ...
Don't count the sequence if you just want to know whether there is
anything in it. Use seq or empty?.
Get rid of the special case of 1 element sequences: it isn't special.
() doesn't need quoting.
And, of course, we can and ought to use recur instead of the recursive call to rev2, to avoid blowing the stack on long sequences:
(defn rev [l]
(letfn [(rev2 [l orig]
(if (empty? orig)
l
(recur (conj l (first orig)) (rest orig))))]
(rev2 () l)))

Clojure Zipper of nested Maps repressing a TRIE

How can I create a Clojure zipper for a TRIE, represented by nested maps, were the keys are the letters.?
Something like this:
{\b {\a {\n {\a {\n {\a {'$ '$}}}}}} \a {\n {\a {'$ '$}}}}
Represents a trie with 2 words 'banana' and 'ana'. (If necessary , its possible to make some changes here in maps..)
I've tried to pass map? vals assoc as the 3 functions to the zipper,respectively.
But it doesnt seem to work..
What 3 functions should I use?
And how the insert-into-trie would look like based on the zipper ?
map? vals #(zipmap (keys %1) %2) would do but doesn't support insertion/removal of children (since children are only values, you don't know which key to remove/add).
The map-zipper below does support insertion/removal because nodes are [k v] pairs (except the root which is a map).
(defn map-zipper [m]
(z/zipper
(fn [x] (or (map? x) (map? (nth x 1))))
(fn [x] (seq (if (map? x) x (nth x 1))))
(fn [x children]
(if (map? x)
(into {} children)
(assoc x 1 (into {} children))))
m))
The solution proposed by #cgrant is great, but implicitly describes a tree where all branches and leaf nodes have an associated value (the key in the dictionary) except for the root node that is just a branch without a value.
So, the tree {"/" nil}, is not a tree with a single leaf node, but a tree with an anonymous root branch and a single leaf node with value /.
In practice, this means that every traversal of the tree has to first execute a (zip/down t) in order to descend the root node.
An alternative solution is to explicitly model the root inside the map, that is, only create zippers from maps with a single key at the root. For example: {"/" {"etc/" {"hosts" nil}}}
The zipper can then be implemented with:
(defn map-zipper [map-or-pair]
"Define a zipper data-structure to navigate trees represented as nested dictionaries."
(if (or (and (map? map-or-pair) (= 1 (count map-or-pair))) (and (= 2 (count map-or-pair))))
(let [pair (if (map? map-or-pair) (first (seq map-or-pair)) map-or-pair)]
(zip/zipper
(fn [x] (map? (nth x 1)))
(fn [x] (seq (nth x 1)))
(fn [x children] (assoc x 1 (into {} children)))
pair))
(throw (Exception. "Input must be a map with a single root node or a pair."))))

Weird behaviour binding in loop recursion

I'm learning Clojure, and I'm trying to solve the Problem 31: Write a function which packs consecutive duplicates into sub-lists.
(= (__ [1 1 2 1 1 1 3 3]) '((1 1) (2) (1 1 1) (3 3)))
I know I can solve this using identity, and in a functional way, but I want to solve it using recursion, because I've not well established this idea in my brain.
My solution would be this:
(defn packing [lista]
(loop [[fst snd :as all] lista mem [] tmp '(fst)]
(print "all is " all "\n\n") ;; something is wrong; it always is an empty list
(if (seq? all)
(if (= fst snd)
(recur (rest all) mem (cons snd tmp))
(recur (rest all) (conj mem tmp) (list snd)))
(seq mem))))
My idea is a recursive loop always taking the first 2 items and comparing. If they are the same number, I include this inside a temporary list tmp; if they're different, I include my temporary list inside men. (This is my final list; a better name would be final_list.)
Because it compares the first 2 items, but at the same time it needs a recursive loop only bypassing the first item, I named the entire list all.
I don't know if the logic is good but inclusive if this was wrong I'm not sure why when I print.
(print "all is " all "\n\n") I receive an empty list
A few points:
'(fst) creates a list containing a symbol fst, not the value of fst, this is one of the reasons to prefer using vectors, e.g., [fst]
you should avoid assuming the input will not be empty
you can use conj for both lists and vectors
destructuring is nestable
(defn packing [coll]
(loop [[x & [y :as more] :as all] coll
result []
same '()]
(if all
(if (= x y)
(recur more result (conj same x))
(recur more (conj result (conj same x)) '()))
result)))
in your code all isn't empty..only happen than is an infinite loop and you always see a empty list...in the firsts lines you can see than it works like expected..
the mistake is in (seq? all) because a empty list is a seq too... try (seq? '()) and return true...then you do a empty loop
you need change this for (empty? all) your code would be
other mistake is '(fst) because it return the simbol fst and not the value...change it for (list fst)
(defn badpacking [lista]
(loop [[fst snd :as all] lista mem [] tmp (list fst)]
(if (empty? all)
(seq mem)
(if (= fst snd)
(recur (rest all) mem (cons snd tmp))
(recur (rest all) (conj mem tmp) (list snd))))))

Clojure: index of a value in a list or other collection

How do I get the index of any of the elements on a list of strings as so:
(list "a" "b" "c")
For example, (function "a") would have to return 0, (function "b") 1, (function "c") 2 and so on.
and... will it be better to use any other type of collection if dealing with a very long list of data?
Christian Berg's answer is fine. Also it is possible to just fall back on Java's indexOf method of class String:
(.indexOf (appl­y str (list­ "a" "b" "c"))­ "c")
; => 2
Of course, this will only work with lists (or more general, seqs) of strings (of length 1) or characters.
A more general approach would be:
(defn index-of [e coll] (first (keep-indexed #(if (= e %2) %1) coll)))
More idiomatic would be to lazily return all indexes and only ask for the ones you need:
(defn indexes-of [e coll] (keep-indexed #(if (= e %2) %1) coll))
(first (indexes-of "a" (list "a" "a" "b"))) ;; => 0
I'm not sure I understand your question. Do you want the nth letter of each of the strings in a list? That could be accomplished like this:
(map #(nth % 1) (list "abc" "def" "ghi"))
The result is:
(\b \e \h)
Update
After reading your comment on my initial answer, I assume your question is "How do I find the index (position) of a search string in a list?"
One possibility is to search for the string from the beginning of the list and count all the entries you have to skip:
(defn index-of [item coll]
(count (take-while (partial not= item) coll)))
Example: (index-of "b" (list "a" "b" "c")) returns 1.
If you have to do a lot of look-ups, it might be more efficient to construct a hash-map of all strings and their indices:
(def my-list (list "a" "b" "c"))
(def index-map (zipmap my-list (range)))
(index-map "b") ;; returns 1
Note that with the above definitions, when there are duplicate entries in the list index-of will return the first index, while index-map will return the last.
You can use the Java .indexOf method reliably for strings and vectors, but not for lists. This solution should work for all collections, I think:
(defn index-of
"Clojure doesn't have an index-of function. The Java .indexOf method
works reliably for vectors and strings, but not for lists. This solution
works for all three."
[item coll]
(let [v (if
(or (vector? coll) (string? coll))
coll
(apply vector coll))]
(.indexOf coll item)))
Do you mean, how do you get the nth element of a list?
For example, if you want to get the 2nd element on the list (with zero-based index):
(nth (list "a" "b" "c") 2)
yields
"c"
Cat-skinning is fun. Here's a low-level approach.
(defn index-of
([item coll]
(index-of item coll 0))
([item coll from-idx]
(loop [idx from-idx coll (seq (drop from-idx coll))]
(if coll
(if (= item (first coll))
idx
(recur (inc idx) (next coll)))
-1))))
This is a Lispy answer, I suspect those expert in Clojure could do it better:
(defn position
"Returns the position of elt in this list, or nil if not present"
([list elt n]
(cond
(empty? list) nil
(= (first list) elt) n
true (position (rest list) elt (inc n))))
([list elt]
(position list elt 0)))
You seem to want to use the nth function.
From the docs for that function:
clojure.core/nth
([coll index] [coll index not-found])
Returns the value at the index. get returns nil if index out of
bounds, nth throws an exception unless not-found is supplied. nth
also works for strings, Java arrays, regex Matchers and Lists, and,
in O(n) time, for sequences.
That last clause means that in practice, nth is slower for elements "farther off" in sequences, with no guarantee to work quicker for collections that in principle support faster access (~ O(n)) to indexed elements. For (clojure) sequences, this makes sense; the clojure seq API is based on the linked-list API and in a linked list, you can only access the nth item by traversing every item before it. Keeping that restriction is what makes concrete list implementations interchangeable with lazy sequences.
Clojure collection access functions are generally designed this way; functions that do have significantly better access times on specific collections have separate names and cannot be used "by accident" on slower collections.
As an example of a collection type that supports fast "random" access to items, clojure vectors are callable; (vector-collection index-number) yields the item at index index-number - and note that clojure seqs are not callable.
I know this question has been answered a million time but here is a recursive solution that leverages deconstructing.
(defn index-of-coll
([coll elm]
(index-of-coll coll elm 0))
([[first & rest :as coll] elm idx]
(cond (empty? coll) -1
(= first elm) idx
:else (recur rest elm (inc idx)))))
(defn index-of [item items]
(or (last (first (filter (fn [x] (= (first x) item))
(map list items (range (count items))))))
-1))
seems to work - but I only have like three items in my list