This works fine as well. More to digest. Thanks.
On 19 Aug, 20:06, Benny Tsai wrote:
> This is one way to do it functionally, though it's a bit more verbose.
>
> "get-nodes" performs a BFS walk of the tree between two nodes, returning the
> set of visited nodes:
>
> (defn get-nodes [from to]
>
It works great. Thanks a lot. I had a feeling concat and reduce were
needed; I just couldn't figure out how.
On 19 Aug, 22:10, Justin Kramer wrote:
> Oops, renamed the function: get-edges = candidates->edges.
>
> Justin
>
>
>
>
>
>
>
> On Friday, August 19, 2011 4:03:27 PM UTC-4, Ulrik Sandberg w
Oops, renamed the function: get-edges = candidates->edges.
Justin
On Friday, August 19, 2011 4:03:27 PM UTC-4, Ulrik Sandberg wrote:
>
> And get-edges?
>
> On 19 Aug, 20:52, Justin Kramer wrote:
> > Here's another way, which constructs a sequence of edges using
> candidates,
> > which are th
And get-edges?
On 19 Aug, 20:52, Justin Kramer wrote:
> Here's another way, which constructs a sequence of edges using candidates,
> which are then fed into reduce to build an adjacency list.
>
> (defn candidates->edges [candidates from to]
> (when-let [kids (seq (candidates from to))]
> (c
Here's another way, which constructs a sequence of edges using candidates,
which are then fed into reduce to build an adjacency list.
(defn candidates->edges [candidates from to]
(when-let [kids (seq (candidates from to))]
(concat (for [k kids] [from k])
(mapcat #(get-edges cand
Probably a good idea for "get-nodes" to handle potential cycles:
(defn get-nodes [from to]
(loop [visited #{}
[x & xs] [from]]
(if-not x
visited
(let [new-visited (conj visited x)
neighbors (candidates x to)
new-queue (remove new-visited (concat x
This is one way to do it functionally, though it's a bit more verbose.
"get-nodes" performs a BFS walk of the tree between two nodes, returning the
set of visited nodes:
(defn get-nodes [from to]
(loop [queue [from]
visited #{}]
(let [[x & xs] queue]
(if-not x
visite
Sorry for being unclear.
I would in fact like to _create_ an adjacency list, given a start and
end key, and the candidates function I mentioned. However, I have
problems with coming up with a functional algorithm. I have a stateful
one that works, but it's very ugly.
(defn build-tree [from to m]
I'm not sure if I'm addressing your needs exactly, but if you want to turn
an adjacency list into a nested structure, here's one way:
(def adj {:a [:b] :b [:c :d] :c [:e] :d [:e]})
(defn build-tree [adj from to]
{:name from
:children (when (not= from to)
(map #(build-tree adj
Thanks. That looks very interesting.
> ;; adjacency list
> {:a [:b] :b [:c :d] :c [:e] :d [:e]}
For some reason, I have trouble constructing this recursively. I seem
to always end up with something like this:
user> (build-tree :a :e)
(:a (:b (:c (:e)) (:d
(:e
(defn build-tree [from to]
(c
There are many ways one could model a tree/graph:
;; collection of edges
[[:a :b] [:b :c] [:b :d] [:c :e] [:d :e]]
;; adjacency list
{:a [:b] :b [:c :d] :c [:e] :d [:e]}
If you're looking for a pre-made solution, the loom graph library
(https://github.com/jkk/loom) may work:
(ns example
(:us
I have data that I would like to represent as a non-binary tree, or perhaps
a graph, but I'm not sure how to do that. The sample data looks like this:
A->B
B->C,D
C->E
D->E
Here the number of subnodes from B are just two, but they may be any number,
which is the source of my
HB writes:
> I'm trying to implement searching algorithm for binary tree.
>
> (defrecord Node [data left-child right-child])
>
> (defrecord Tree [root])
>
> (defn find-node [tree data]
> "Search for a node"
> (let [root (:root tree)]
> (l
On Tue, Feb 22, 2011 at 8:37 PM, HB wrote:
> Hi,
>
> I'm trying to implement searching algorithm for binary tree.
> Here is my Java implementation:
>
> public Node findNode(Integer data) {
> Node current = root;
> while (!data.equals(current.getDat
Hi,
I'm trying to implement searching algorithm for binary tree.
Here is my Java implementation:
public Node findNode(Integer data) {
Node current = root;
while (!data.equals(current.getData())) {
if (data < current.getData())
can move to the next row however the column index has to be either
> increased by one or decreased by one(and the new cell is empty).
>
>
> If you had a balanced binary tree, you could just take log_2(N) where N is
> the number of nodes. :) But since you don't you have to do so
On Jun 30, 2009, at 3:50 AM, Nicolas Oury wrote:
>
> If you use depth/size a lot, it might be faster to store them in the
> nodes. (it is pseudo code)
>
> (defstruct bintree :head :left :right :size :depth)
>
> (defn make-tree [head left right]
> (struct bintree head left right
>
ed by one or decreased by one(and the new cell is
> > empty).
>
>
> If you had a balanced binary tree, you could just take log_2(N) where
> N is the number of nodes. :) But since you don't you have to do
> something like get the max depth of the left branch and t
m one cell of any
> vector I can move to the next row however the column index has to
> be either increased by one or decreased by one(and the new cell is
> empty).
If you had a balanced binary tree, you could just take log_2(N) where
N is the number of nodes. :) But since you
Hi,
Am 30.06.2009 um 09:05 schrieb Emeka:
I have a BinaryTree Nodes that I want to resolve it's depth and the
BinaryTree may be unbalanced. How do I do this? Can't just figure it
out on my own?
It is in the form of vector of vectors, and from one cell of any
vector I can move to the next r
Hello All,
I have a BinaryTree Nodes that I want to resolve it's depth and the
BinaryTree may be unbalanced. How do I do this? Can't just figure it out on
my own?
It is in the form of vector of vectors, and from one cell of any vector I
can move to the next row however the column index has to be
On Tue, Nov 18, 2008 at 4:55 PM, mb <[EMAIL PROTECTED]> wrote:
[...]
> I don't know, whether this is more idiomatic Clojure code, but
> it works...
[...]
What revision is that? On r1099 I got:
java.lang.IllegalArgumentException: Don't know how to create ISeq
from: Symbol (NO_SOURCE_FILE:31)
so
Hi,
blindly copying code is usually not a good way to learn a new
language
I don't know, whether this is more idiomatic Clojure code, but
it works...
(defn build-tree
[item depth]
(when (< 0 depth)
(let [i (* 2 item)
d (dec depth)]
[item (build-tree (dec i) d) (build
On Tue, Nov 18, 2008 at 5:58 AM, Simon Brooke <[EMAIL PROTECTED]> wrote:
>
> However Giraud uses the Common
> LISP ASH (arithmetic shift) function, and, if there's a built-in
> function in Clojure, I did not find it;
find-doc is your friend in this case:
user=> (find-doc "shift")
---
As a learning exercise and also to continue to investigate Clojure
performance I've roughly translated the Alioth binary-tree benchmark
into Clojure. I chose the binary-tree simply because it's the first of
the Alioth benchmarks in alphabetical collation; I may do others
later.
I&#
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