I'm currently learning Clojure and am stuck with list comprehension.
;; https://stackoverflow.com/a/7625207/4110233
(defn gen-primes "Generates an infinite, lazy sequence of prime numbers"
[]
(letfn [(reinsert [table x prime]
(update-in table [(+ prime x)] conj prime))
(primes-step [table d]
(if-let [factors (get table d)]
(recur (reduce #(reinsert %1 d %2) (dissoc table d) factors)
(inc d))
(lazy-seq (cons d (primes-step (assoc table (* d d) (list d))
(inc d))))))]
(primes-step {} 2)))
(defn prime-factors-not-working [x]
(for [y (gen-primes)
:when (= (mod x y) 0)
:while (< y (Math/sqrt x))]
y))
(defn prime-factors-working [x]
(for [y (gen-primes)
:while (< y (Math/sqrt x))
:when (= (mod x y) 0)]
y))
(prime-factors-working 100)
;; ↪ (2 5)
(prime-factors-not-working 100)
;; Goes into infinite loop
(gen-primes) is a lazy sequence of prime numbers. The only difference between the working and not-working sequences is the order of the modifiers while and when. Why does the not-working implementation go into an infinite loop?
The not working variant expands conceptually (but not factually) into this:
(loop [ys (gen-primes)
result []]
(if (seq ys)
(let [y (first ys)]
(if (= (mod x (first ys)) 0) ;; Can be replaced with `(zero? (mod x y))` BTW.
(if (< y (Math/sqrt x))
(recur (next ys) (conj result y))
result)
(recur (next ys) result)))
result))
As you can see, if (mod x (first ys)) is not 0, it will go to the next number - without checking for that <, going forever.
When you exchange :when and :while, the checks in the pseudo-expansion above are also swapped - stopping the iteration once y reaches the square root of x.
the macro expansion of for is sensitive to the order in which you put :when and :while. macroexpanding gives slightly different code.
you can go very far in clojure without relying on complicated macros beyond defn, for is not very common, and this isn't a usecase where it is clearly advantagous over map->filter->take
good expansion: line 28:
(when (< y (Math/sqrt x)) ; XXX
(if (= (mod x y) 0) ; XXX
(do
(chunk-append b__86695 y)
(recur (unchecked-inc i__86694)))
(recur (unchecked-inc i__86694))))
bad expansion: line 28:
(if (= (mod x y) 0) ; XXX
(when (< y (Math/sqrt x)) ; XXX
(do
(chunk-append b__86666 y)
(recur (unchecked-inc i__86665))))
(recur (unchecked-inc i__86665)))
you can learn about the implementation of the for macro by going to it's source code (your editor should have a way for navigating to defs of symbols)
https://github.com/clojure/clojure/blob/master/src/clj/clojure/core.clj#L4654
there is nothing about clojure that requires it's macros to write code in the way you think they should, though in this case it may be a bug in for, it's hard to tell. some use cases may want to be able to when while and when are written.
since this is about learning, and macros are pretty much magic unless you see the code they write out, i think the best way to learn is to figure out how to view the expanded macro forms in your code. this is generally how macros are debugged.
link to topic on macro expansion
how is a macro expanded in clojure?
documentation for expanding macros in cider (emacs clojure editor)
https://docs.cider.mx/cider/debugging/macroexpansion.html
Related
I'm currently learning Racket/Scheme for a course (I'm not sure what's the difference, actually, and I'm not sure if the course covered that). I'm trying a basic example, implementing the Newton method to find a square root of a number; however, I ran into a problem with finding the distance between two numbers.
It seems that for whatever reason, when I'm trying to apply the subtraction operator between two numbers, it returns a list instead.
#lang racket
(define distance
(lambda (x y) (
(print (real? x))
(print (real? y))
(abs (- x y))
)
)
)
(define abs
(lambda x (
(print (list? x))
(if (< x 0) (- x) x)
)
)
)
(distance 2 5)
As you can see, I've added printing of the types of variables to make sure the problem is what I think it is, and the output of all those prints is #t. So:
In calling distance, x and y are both real.
In calling abs, x is a list.
So, the conclusion is that (- x y) returns a list, but why?
I double-checked with the documentation and it seems I'm using the subtraction operator correctly; I've typed (- 2 5) and then (real? (- 2 5)) into the same REPL I'm using to debug my program (Dr. Racket, to be specific), and I'm getting the expected results (-3 and #t, respectively).
Is there any wizard here that can tell me what kind of sorcery is this?
Thanks in advance!
How about this...
(define distance
(lambda (x y)
(print (real? x))
(print (real? y))
(abs (- x y))))
(define abs
(lambda (x) ;; instead of (lambda x ...), we are using (lambda (x) ...) form which is more strict in binding with formals
(print (list? x))
(if (< x 0) (- x) x)))
Read further about various lambda forms and their binding with formals.
I'm trying to edit the current program I have
(define (sumofnumber n)
(if (= n 0)
1
(+ n (sumofnumber (modulo n 2 )))))
so that it returns the sum of an n number of positive squares. For example if you inputted in 3 the program would do 1+4+9 to get 14. I have tried using modulo and other methods but it always goes into an infinite loop.
The base case is incorrect (the square of zero is zero), and so is the recursive step (why are you taking the modulo?) and the actual operation (where are you squaring the value?). This is how the procedure should look like:
(define (sum-of-squares n)
(if (= n 0)
0
(+ (* n n)
(sum-of-squares (- n 1)))))
A definition using composition rather than recursion. Read the comments from bottom to top for the procedural logic:
(define (sum-of-squares n)
(foldl + ; sum the list
0
(map (lambda(x)(* x x)) ; square each number in list
(map (lambda(x)(+ x 1)) ; correct for range yielding 0...(n - 1)
(range n))))) ; get a list of numbers bounded by n
I provide this because you are well on your way to understanding the idiom of recursion. Composition is another of Racket's idioms worth exploring and often covered after recursion in educational contexts.
Sometimes I find composition easier to apply to a problem than recursion. Other times, I don't.
You're not squaring anything, so there's no reason to expect that to be a sum of squares.
Write down how you got 1 + 4 + 9 with n = 3 (^ is exponentiation):
1^2 + 2^2 + 3^2
This is
(sum-of-squares 2) + 3^2
or
(sum-of-squares (- 3 1)) + 3^2
that is,
(sum-of-squares (- n 1)) + n^2
Notice that modulo does not occur anywhere, nor do you add n to anything.
(And the square of 0 is 0 , not 1.)
You can break the problem into small chunks.
1. Create a list of numbers from 1 to n
2. Map a square function over list to square each number
3. Apply + to add all the numbers in squared list
(define (sum-of-number n)
(apply + (map (lambda (x) (* x x)) (sequence->list (in-range 1 (+ n 1))))))
> (sum-of-number 3)
14
This is the perfect opportunity for using the transducers technique.
Calculating the sum of a list is a fold. Map and filter are folds, too. Composing several folds together in a nested fashion, as in (sum...(filter...(map...sqr...))), leads to multiple (here, three) list traversals.
But when the nested folds are fused, their reducing functions combine in a nested fashion, giving us a one-traversal fold instead, with the one combined reducer function:
(define (((mapping f) kons) x acc) (kons (f x) acc)) ; the "mapping" transducer
(define (((filtering p) kons) x acc) (if (p x) (kons x acc) acc)) ; the "filtering" one
(define (sum-of-positive-squares n)
(foldl ((compose (mapping sqr) ; ((mapping sqr)
(filtering (lambda (x) (> x 0)))) ; ((filtering {> _ 0})
+) 0 (range (+ 1 n)))) ; +))
; > (sum-of-positive-squares 3)
; 14
Of course ((compose f g) x) is the same as (f (g x)). The combined / "composed" (pun intended) reducer function is created just by substituting the arguments into the definitions, as
((mapping sqr) ((filtering {> _ 0}) +))
=
( (lambda (kons)
(lambda (x acc) (kons (sqr x) acc)))
((filtering {> _ 0}) +))
=
(lambda (x acc)
( ((filtering {> _ 0}) +)
(sqr x) acc))
=
(lambda (x acc)
( ( (lambda (kons)
(lambda (x acc) (if ({> _ 0} x) (kons x acc) acc)))
+)
(sqr x) acc))
=
(lambda (x acc)
( (lambda (x acc) (if (> x 0) (+ x acc) acc))
(sqr x) acc))
=
(lambda (x acc)
(let ([x (sqr x)] [acc acc])
(if (> x 0) (+ x acc) acc)))
which looks almost as something a programmer would write. As an exercise,
((filtering {> _ 0}) ((mapping sqr) +))
=
( (lambda (kons)
(lambda (x acc) (if ({> _ 0} x) (kons x acc) acc)))
((mapping sqr) +))
=
(lambda (x acc)
(if (> x 0) (((mapping sqr) +) x acc) acc))
=
(lambda (x acc)
(if (> x 0) (+ (sqr x) acc) acc))
So instead of writing the fused reducer function definitions ourselves, which as every human activity is error-prone, we can compose these reducer functions from more atomic "transformations" nay transducers.
Works in DrRacket.
I wrote the code below for game I am working on. But it seems a little slow. If you have not checked the code yet, it's the A* search/pathfinding algorithm. It takes about 100-600 ms for a 100x100 grid, depending on the heuristic used (and consequently the number of tiles visited).
There are no reflection warnings. However, I suspect boxing might be an issue. But I don't know how to get rid of boxing in this case, because the computation is split among several functions. Also, I save tiles/coordinates as vectors of two numbers, like this: [x y]. But then the numbers will be boxed, right? A typical piece of code, if you don't want to read through it all, is: (def add-pos (partial mapv + pos)) where pos is the aforementioned kind of two-number vector. There are sereval of places where the numbers are manipulated in a way similar to add-pos above, and put back in a vector afterwards. Is there any way to optimize code like this? Any other tips is welcome too, performance-related or other.
EDIT: Thinking some more about it, I came up with a few follow-up questions: Can a Clojure function ever return primitives? Can a Clojure function ever take primitives (without any boxing)? Can I put primitives in a type/record without boxing?
(ns game.server.pathfinding
(:use game.utils)
(:require [clojure.math.numeric-tower :as math]
[game.math :as gmath]
[clojure.data.priority-map :as pm]))
(defn walkable? [x]
(and x (= 1 x)))
(defn point->tile
([p] (apply point->tile p))
([x y] [(int x) (int y)]))
(defn get-tile [m v]
"Gets the type of the tile at the point v in
the grid m. v is a point in R^2, not grid indices."
(get-in m (point->tile v)))
(defn integer-points
"Given an equation: x = start + t * step, returns a list of the
values for t that make x an integer between start and stop,
or nil if there is no such value for t."
[start stop step]
(if-not (zero? step)
(let [first-t (-> start ((if (neg? step) math/floor math/ceil))
(- start) (/ step))
t-step (/ 1 (math/abs step))]
(take-while #((if (neg? step) > <) (+ start (* step %)) stop)
(iterate (partial + t-step) first-t)))))
(defn crossed-tiles [[x y :as p] p2 m]
(let [[dx dy :as diff-vec] (map - p2 p)
ipf (fn [getter]
(integer-points (getter p) (getter p2) (getter diff-vec)))
x-int-ps (ipf first)
y-int-ps (ipf second)
get-tiles (fn [[x-indent y-indent] t]
(->> [(+ x-indent x (* t dx)) (+ y-indent y (* t dy))]
(get-tile m)))]
(concat (map (partial get-tiles [0.5 0]) x-int-ps)
(map (partial get-tiles [0 0.5]) y-int-ps))))
(defn clear-line?
"Returns true if the line between p and p2 passes over only
walkable? tiles in m, otherwise false."
[p p2 m]
(every? walkable? (crossed-tiles p p2 m)))
(defn clear-path?
"Returns true if a circular object with radius r can move
between p and p2, passing over only walkable? tiles in m,
otherwise false.
Note: Does not currently work for objects with a radius >= 0.5."
[p p2 r m]
(let [diff-vec (map (partial * r) (gmath/normalize (map - p2 p)))
ortho1 ((fn [[x y]] (list (- y) x)) diff-vec)
ortho2 ((fn [[x y]] (list y (- x))) diff-vec)]
(and (clear-line? (map + ortho1 p) (map + ortho1 p2) m)
(clear-line? (map + ortho2 p) (map + ortho2 p2) m))))
(defn straighten-path
"Given a path in the map m, remove unnecessary nodes of
the path. A node is removed if one can pass freely
between the previous and the next node."
([m path]
(if (> (count path) 2) (straighten-path m path nil) path))
([m [from mid to & tail] acc]
(if to
(if (clear-path? from to 0.49 m)
(recur m (list* from to tail) acc)
(recur m (list* mid to tail) (conj acc from)))
(reverse (conj acc from mid)))))
(defn to-mid-points [path]
(map (partial map (partial + 0.5)) path))
(defn to-tiles [path]
(map (partial map int) path))
(defn a*
"A* search for a grid of squares, mat. Tries to find a
path from start to goal using only walkable? tiles.
start and goal are vectors of indices into the grid,
not points in R^2."
[mat start goal factor]
(let [width (count mat)
height (count (first mat))]
(letfn [(h [{pos :pos}] (* factor (gmath/distance pos goal)))
(g [{:keys [pos parent]}]
(if parent
(+ (:g parent) (gmath/distance pos (parent :pos)))
0))
(make-node [parent pos]
(let [node {:pos pos :parent parent}
g (g node) h (h node)
f (+ g h)]
(assoc node :f f :g g :h h)))
(get-path
([node] (get-path node ()))
([{:keys [pos parent]} path]
(if parent
(recur parent (conj path pos))
(conj path pos))))
(free-tile? [tile]
(let [type (get-in mat (vec tile))]
(and type (walkable? type))))
(expand [closed pos]
(let [adj [[1 0] [0 1] [-1 0] [0 -1]]
add-pos (partial mapv + pos)]
(->> (take 4 (partition 2 1 (cycle adj)))
(map (fn [[t t2]]
(list* (map + t t2) (map add-pos [t t2]))))
(map (fn [[d t t2]]
(if (every? free-tile? [t t2]) d nil)))
(remove nil?)
(concat adj)
(map add-pos)
(remove (fn [[x y :as tile]]
(or (closed tile) (neg? x) (neg? y)
(>= x width) (>= y height)
(not (walkable? (get-in mat tile)))))))))
(add-to-open [open tile->node [{:keys [pos f] :as node} & more]]
(if node
(if (or (not (contains? open pos))
(< f (open pos)))
(recur (assoc open pos f)
(assoc tile->node pos node)
more)
(recur open tile->node more))
{:open open :tile->node tile->node}))]
(let [start-node (make-node nil start)]
(loop [closed #{}
open (pm/priority-map start (:f start-node))
tile->node {start start-node}]
(let [[curr _] (peek open) curr-node (tile->node curr)]
(when curr
(if (= curr goal)
(get-path curr-node)
(let [exp-tiles (expand closed curr)
exp-nodes (map (partial make-node curr-node) exp-tiles)
{:keys [open tile->node]}
(add-to-open (pop open) tile->node exp-nodes)]
(recur (conj closed curr) open tile->node))))))))))
(defn find-path [mat start goal]
(let [start-tile (point->tile start)
goal-tile (point->tile goal)
path (a* mat start-tile goal-tile)
point-path (to-mid-points path)
full-path (concat [start] point-path [goal])
final-path (rest (straighten-path mat full-path))]
final-path))
I recommend the Clojure High Performance Programming book for addressing questions like yours.
There are functions to unbox primitives (byte, short, int, long, float, double).
Warn-on-reflection does not apply to numeric type reflection / failure to optimize numeric code. There is a lib to force warnings for numeric reflection - primitive-math.
You can declare the types of function arguments and function return values (defn ^Integer foo [^Integer x ^Integer y] (+ x y)).
Avoid apply if you want performance.
Avoid varargs (a common reason to need apply) if you want performance. Varargs functions create garbage on every invocation (in order to construct the args map, which usually is not used outside the function body). partial always constructs a varargs function. Consider replacing the varargs (partial * x) with #(* x %), the latter can be optimized much more aggressively.
There is a tradeoff with using primitive jvm single-type arrays (they are mutible and fixed in length, which can lead to more complex and brittle code), but they will perform better than the standard clojure sequential types, and are available if all else fails to get the performance you need.
Also, use criterium to compare various implementations of your code, it has a bunch of tricks to help rule out the random things that affect execution time so you can see what really performs best in a tight loop.
Also, regarding your representation of a point as [x y] - you can reduce the space and lookup overhead of the collection holding them with (defrecord point [x y]) (as long as you know they will remain two elements only, and you don't mind changing your code to ask for (:x point) or (:y point)). You could further optimize by making or using a simple two-number java class (with the tradeoff of losing immutibility).
I'm trying to write a procedure that makes use of the distributive property of an algebraic expression to simplify it:
(dist '(+ x y (exp x) (* x 5) y (* y 6)))
=> (+ (* x (+ 1 5))
(* y (+ 1 1 6))
(exp x))
(dist '(+ (* x y) x y))
=> (+ (* x (+ y 1))
y)
; or
=> (+ (* y (+ x 1))
x)
As the second example shows, there can be more than one possible outcome, I don't need to enumerate them all, just a valid one. I'm wondering if someone could provide me with at least a qualitative description of how they would start attacking this problem? Thanks :)
Oleg Kiselyov's pmatch macro makes distributing a factor across terms pretty easy:
(define dist
(λ (expr)
(pmatch expr
[(* ,factor (+ . ,addends))
`(+ ,#(map (λ (addend)
(list factor addend))
addends))]
[else
expr])))
(dist '(* 5 (+ x y))) => (+ (5 x) (5 y))
The main trick is to match a pattern and extract elements from the expression from the corresponding slots in the pattern. This requires a cond and let with tricky expressions to cdr to the right place in the list and car out the right element. pmatch writes that cond and let for you.
Factoring out common terms is harder because you have to look at all the subexpressions to find the common factors and then pull them out:
(define factor-out-common-factors
(λ (expr)
(pmatch expr
[(+ . ,terms) (guard (for-all (λ (t) (eq? '* (car t)))
terms))
(let ([commons (common-factors terms)])
`(* ,#commons (+ ,#(remove-all commons (map cdr terms)))))]
[else
expr])))
(define common-factors
(λ (exprs)
(let ([exprs (map cdr exprs)]) ; remove * at start of each expr
(fold-right (λ (factor acc)
(if (for-all (λ (e) (member factor e))
exprs)
(cons factor acc)
acc))
'()
(uniq (apply append exprs))))))
(define uniq
(λ (ls)
(fold-right (λ (x acc)
(if (member x acc)
acc
(cons x acc)))
'()
ls)))
(factor-out-common-factors '(+ (* 2 x) (* 2 y)))
=> (* 2 (+ (x) (y)))
The output could be cleaned up some more, this doesn't cover factoring out a 1, and remove-all is missing, but I'll leave all that to you.
A very general approach:
(dist expr var-list)
=> expr factored using terms in var-list
dist would have to know about "distributable" functions like +,-,*,/,etc and how each of them behave. If, say, it only knew about the first four, then :
(dist expr var-list
(if (empty? var-list) expr
(let* ([new-expr (factor expr (first var-list))])
(return "(* var " (dist new-expr (rest var-list)))))
That "return "(* var " " is not correct syntax, but you probably already knew that. I'm not a racket or lisp expert by any means, but basically this comes down to string processing? In any case, factor needs to be fleshed out so that it removes a single var from * functions and all of the var from + functions (replacing them with 1). It also needs to be smart enough to only do it when there are at least two replacements (otherwise we haven't actually done anything).
How can I create a method which takes two numbers and prepare a list from first number to second number. The first number is always positive and less than second number? I tried the following but the I am not sure how to have a global variable in Scheme to hold previous values.
(define preplist
(let ((temp '()))
(lambda (x y)
(cond ((= x y) (append temp (list x)))
(else (append temp (list x))
(display x)
(preplist (+ x 1) y))))))
Expected result is: (preplist 3 7) => (3 4 5 6 7)
Can some one please help to resolve this problem?
The solution for (x, y) can be computed as: put x on the front of (x+1, y). It is thus clearly recursive. Like this:
(define (preplist x y)
(if (= x y)
(list y)
(cons x (preplist (+ x 1) y))))
See, it works:
> (preplist 1 4)
(1 2 3 4)
> (preplist 5 7)
(5 6 7)
There are several mistakes in your code, for starters you don't need a global variable defined in a let for storing the result, it's enough to build the answer as you advance in the recursion. And don't use append in this case, if the solution template is followed closely, a cons will suffice for building the output list.
You should stick to the recipe for building a new list recursively; this is how the problem should be solved using that recipe, it's perhaps a bit more idiomatic like this:
(define preplist
(lambda (x y)
(cond ((> x y) ; if the exit condition is met
empty) ; then return the empty list
(else ; otherwise
(cons x ; cons the current element
(preplist (add1 x) y)))))) ; and advance the recursion
An altogether different approach would be to write a tail-recursive solution. This is more efficient because a constant amount of stack is used. It doesn't follow the design recipe as outlined above, but is somewhat more similar to the solution you had in mind - but bear in mind that this doesn't use global variables (only a named let for the iteration) and the solution is accumulated and passed around as a parameter:
(define (preplist x y)
(let loop ((i y) ; named let for iteration
(acc empty)) ; define and initialize parameters
(if (> x i) ; if exit condition is met
acc ; return accumulated value
(loop (sub1 i) ; otherwise advance recursion
(cons i acc))))) ; and add to the accumulator
Of course, as pointed by #dyoo in the comments, in a practical setting you'd use the built-in range procedure which does basically the same as the preplist procedure.