I can't understand why this lazy-seq causes a stackoverflow, and why not when you pass the sequence to dorun:
(defn very-lazy [s]
(lazy-seq
(if (seq s)
[(first s) (very-lazy (rest s))]
[])))
(dorun (very-lazy (range 200000000)))
>nil
(take 2 (very-lazy (range 20000000))
>...(1577 (java.lang.StackOverflowError
If it's lazy then take 2 should cause the lazy seq to iterate only two times, why doesn't happen and why dorun works?
In your example function returns lazyseq (0 (1 (2 (3 (...))))). That's why dorun runs without stackoverflow (there is sequence of two elements 0 and (1 (2 (...))) that dorun doesn't evaluate) and second fails (it returns infinite nested sequences that repl tries evaluate to print out).
I guess you're looking for this solution
(defn very-lazy [s]
(lazy-seq
(if (seq s)
(cons (first s) (very-lazy (rest s)))
[])))
(take 10 (very-lazy (range 200000000)))
-> (0 1 2 3 4 5 6 7 8 9)
Related
I want to write a function in Racket which takes an amount of money and a list of specific bill-values, and then returns a list with the amount of bills used of every type to make the given amount in total. For example (calc 415 (list 100 10 5 2 1)) should return '(4 1 1 0 0).
I tried it this way but this doesn't work :/ I think I haven't fully understood what you can / can't do with set! in Racket, to be honest.
(define (calc n xs)
(cond ((null? xs) (list))
((not (pair? xs))
(define y n)
(begin (set! n (- n (* xs (floor (/ n xs)))))
(list (floor (/ y xs))) ))
(else (append (calc n (car xs))
(calc n (cdr xs))))))
Your procedure does too much and you use mutation which is uneccesary. If you split the problem up.
(define (calc-one-bill n bill)
...)
;; test
(calc-one-bill 450 100) ; ==> 4
(calc-one-bill 450 50) ; ==> 9
Then you can make:
(define (calc-new-n n bill amount)
...)
(calc-new-n 450 100 4) ; ==> 50
(calc-new-n 450 50 9) ; ==> 0
Then you can reduce your original implememntation like this:
(define (calc n bills)
(if (null? bills)
(if (zero? n)
'()
(error "The unit needs to be the last element in the bills list"))
(let* ((bill (car bills))
(amount (calc-one-bill n bill)))
(cons amount
(calc (calc-new-n n bill amount)
(cdr bills))))))
This will always choose the solution with fewest bills, just as your version seems to do. Both versions requires that the last element in the bill passed is the unit 1. For a more complex method, that works with (calc 406 (list 100 10 5 2)) and that potentially can find all combinations of solutions, see Will's answer.
This problem calls for some straightforward recursive non-deterministic programming.
We start with a given amount, and a given list of bill denominations, with unlimited amounts of each bill, apparently (otherwise, it'd be a different problem).
At each point in time, we can either use the biggest bill, or not.
If we use it, the total sum lessens by the bill's value.
If the total is 0, we've got our solution!
If the total is negative, it is invalid, so we should abandon this path.
The code here will follow another answer of mine, which finds out the total amount of solutions (which are more than one, for your example as well). We will just have to mind the solutions themselves as well, whereas the code mentioned above only counted them.
We can code this one as a recursive-backtracking procedure, calling a callback with each successfully found solution from inside the deepest level of recursion (tantamount to the most deeply nested loop in the nested loops structure created with recursion, which is the essence of recursive backtracking):
(define (change sum bills callback)
(let loop ([sum sum] [sol '()] [bills bills]) ; "sol" for "solution"
(cond
((zero? sum) (callback sol)) ; process a solution found
((< sum 0) #f)
((null? bills) #f)
(else
(apply
(lambda (b . bs) ; the "loop":
;; 1. ; either use the first
(loop (- sum b) (cons b sol) bills) ; denomination,
;; 2. ; or,
(loop sum sol bs)) ; after backtracking, don't!
bills)))))
It is to be called through e.g. one of
;; construct `the-callback` for `solve` and call
;; (solve ...params the-callback)
;; where `the-callback` is an exit continuation
(define (first-solution solve . params)
(call/cc (lambda (return)
(apply solve (append params ; use `return` as
(list return)))))) ; the callback
(define (n-solutions n solve . params) ; n assumed an integer
(let ([res '()]) ; n <= 0 gets ALL solutions
(call/cc (lambda (break)
(apply solve (append params
(list (lambda (sol)
(set! res (cons sol res))
(set! n (- n 1))
(cond ((zero? n) (break)))))))))
(reverse res)))
Testing,
> (first-solution change 406 (list 100 10 5 2))
'(2 2 2 100 100 100 100)
> (n-solutions 7 change 415 (list 100 10 5 2 1))
'((5 10 100 100 100 100)
(1 2 2 10 100 100 100 100)
(1 1 1 2 10 100 100 100 100)
(1 1 1 1 1 10 100 100 100 100)
(5 5 5 100 100 100 100)
(1 2 2 5 5 100 100 100 100)
(1 1 1 2 5 5 100 100 100 100))
Regarding how this code is structured, cf. How to generate all the permutations of elements in a list one at a time in Lisp? It creates nested loops with the solution being accessible in the innermost loop's body.
Regarding how to code up a non-deterministic algorithm (making all possible choices at once) in a proper functional way, see How to do a powerset in DrRacket? and How to find partitions of a list in Scheme.
I solved it this way now :)
(define (calc n xs)
(define (calcAssist n xs usedBills)
(cond ((null? xs) usedBills)
((pair? xs)
(calcAssist (- n (* (car xs) (floor (/ n (car xs)))))
(cdr xs)
(append usedBills
(list (floor (/ n (car xs)))))))
(else
(if ((= (- n (* xs (floor (/ n xs)))) 0))
(append usedBills (list (floor (/ n xs))))
(display "No solution")))))
(calcAssist n xs (list)))
Testing:
> (calc 415 (list 100 10 5 2 1))
'(4 1 1 0 0)
I think this is the first program I wrote when learning FORTRAN! Here is a version which makes no bones about using everything Racket has to offer (or, at least, everything I know about). As such it's probably a terrible homework solution, and it's certainly prettier than the FORTRAN I wrote in 1984.
Note that this version doesn't search, so it will get remainders even when it does not need to. It never gets a remainder if the lowest denomination is 1, of course.
(define/contract (denominations-of amount denominations)
;; split amount into units of denominations, returning the split
;; in descending order of denomination, and any remainder (if there is
;; no 1 denomination there will generally be a remainder).
(-> natural-number/c (listof (integer-in 1 #f))
(values (listof natural-number/c) natural-number/c))
(let handle-one-denomination ([current amount]
[remaining-denominations (sort denominations >)]
[so-far '()])
;; handle a single denomination: current is the balance,
;; remaining-denominations is the denominations left (descending order)
;; so-far is the list of amounts of each denomination we've accumulated
;; so far, which is in ascending order of denomination
(if (null? remaining-denominations)
;; we are done: return the reversed accumulator and anything left over
(values (reverse so-far) current)
(match-let ([(cons first-denomination rest-of-the-denominations)
remaining-denominations])
(if (> first-denomination current)
;; if the first denomination is more than the balance, just
;; accumulate a 0 for it and loop on the rest
(handle-one-denomination current rest-of-the-denominations
(cons 0 so-far))
;; otherwise work out how much of it we need and how much is left
(let-values ([(q r)
(quotient/remainder current first-denomination)])
;; and loop on the remainder accumulating the number of bills
;; we needed
(handle-one-denomination r rest-of-the-denominations
(cons q so-far))))))))
I am working on program related to the different of dealing with even numbers in C and lisp , finished my c program but still having troubles with lisp
isprime function is defined and I need help in:
define function primesinlist that returns unique prime numbers in a lis
here what i got so far ,
any help with that please?
(defun comprimento (lista)
(if (null lista)
0
(1+ (comprimento (rest lista)))))
(defun primesinlist (number-list)
(let ((result ()))
(dolist (number number-list)
(when (isprime number)
( number result)))
(nreverse result)))
You need to either flatten the argument before processing:
(defun primesinlist (number-list)
(let ((result ()))
(dolist (number (flatten number-list))
(when (isprime number)
(push number result)))
(delete-duplicates (nreverse result))))
or, if you want to avoid consing up a fresh list, flatten it as you go:
(defun primesinlist (number-list)
(let ((result ()))
(labels ((f (l)
(dolist (x l)
(etypecase x
(integer (when (isprime x)
(push x result)))
(list (f x))))))
(f number-list))
(delete-duplicates (nreverse result))))
To count distinct primes, take the length of the list returned by primesinlist.
Alternatively, you can use count-if:
(count-if #'isprime (delete-duplicates (flatten number-list)))
It sounds like you've already got a primality test implemented, but for sake of completeness, lets add a very simple one that just tries to divide a number by the numbers less than it up to its square root:
(defun primep (x)
"Very simple implementation of a primality test. Checks
for each n above 1 and below (sqrt x) whether n divides x.
Example:
(mapcar 'primep '(2 3 4 5 6 7 8 9 10 11 12 13))
;=> (T T NIL T NIL T NIL NIL NIL T NIL T)
"
(do ((sqrt-x (sqrt x))
(i 2 (1+ i)))
((> i sqrt-x) t)
(when (zerop (mod x i))
(return nil))))
Now, you need a way to flatten a potentially nested list of lists into a single list. When approaching this problem, I usually find it a bit easier to think in terms of trees built of cons-cells. Here's an efficient flattening function that returns a completely new list. That is, it doesn't share any structure with the original tree. That can be useful, especially if we want to modify the resulting structure later, without modifying the original input.
(defun flatten-tree (x &optional (tail '()))
"Efficiently flatten a tree of cons cells into
a list of all the non-NIL leafs of the tree. A completely
fresh list is returned.
Examples:
(flatten-tree nil) ;=> ()
(flatten-tree 1) ;=> (1)
(flatten-tree '(1 (2 (3)) (4) 5)) ;=> (1 2 3 4 5)
(flatten-tree '(1 () () 5)) ;=> (1 5)
"
(cond
((null x) tail)
((atom x) (list* x tail))
((consp x) (flatten-tree (car x)
(flatten-tree (cdr x) tail)))))
Now it's just a matter of flatting a list, removing the number that are not prime, and removing duplicates from that list. Common Lisp includes functions for doing these things, namely remove-if-not and remove-duplicates. Those are the "safe" versions that don't modify their input arguments. Since we know that the flattened list is freshly generated, we can use their (potentially) destructive counterparts, delete-if-not and delete-duplicates.
There's a caveat when you're removing duplicate elements, though. If you have a list like (1 3 5 3), there are two possible results that could be returned (assuming you keep all the other elements in order): (1 3 5) and (1 5 3). That is, you can either remove the the later duplicate or the earlier duplicate. In general, you have the question of "which one should be left behind?" Common Lisp, by default, removes the earlier duplicate and leaves the last occurrence. That behavior can be customized by the :from-end keyword argument. It can be nice to duplicate that behavior in your own API.
So, here's a function that puts all those considerations together.
(defun primes-in-tree (tree &key from-end)
"Flatten the tree, remove elements which are not prime numbers,
using FROM-END to determine whether earlier or later occurrences
are kept in the list.
Examples:
(primes-in-list '(2 (7 4) ((3 3) 5) 6 7))
;;=> (2 3 5 7)
(primes-in-list '(2 (7 4) ((3 3) 5) 6 7) :from-end t)
;;=> (2 7 3 5)"
;; Because FLATTEN-TREE returns a fresh list, it's OK
;; to use the destructive functions DELETE-IF-NOT and
;; DELETE-DUPLICATES.
(delete-duplicates
(delete-if-not 'primep (flatten-tree list))
:from-end from-end))
I have a small exercise in Lisp:
Write a function test-delta with parameters delta and lst, which will
check if the difference between successive elements in lst is smaller than
delta. Write the function in two ways:
recursively
using a mapping function
I have no problem writing that function recursively, but I don't know which mapping function I should use. All the standard mapping functions work with only one element of the list at a time. reduce cannot be used either, because I do not have some operation to use between successive elements. What function could I use here?
All standard functions are working only with one element at time.
Reduce function cannot be use either
because i do not have some operation to use between to elements.
There's already an answer by uselpa showing that you can do this with reduce, but it feels a bit awkward to me to bend reduce to this case.
It's much more natural, in my opinion, to recognize that the standard mapping functions actually let you work with multiple lists. I'll show mapcar and loop first, and then every, which I think is the real winner here. Finally, just for completeness, I've also included maplist.
mapcar
The standard mapcar can take more than one list, which means that you can take elements from two different lists at once. Of particular note, it could take a list and (rest list). E.g.,
(let ((list '(1 2 3 4 5 6)))
(mapcar 'cons
list
(rest list)))
;=> ((1 . 2) (2 . 3) (3 . 4) (4 . 5) (5 . 6))
loop
You can use loop to do the same sort of thing:
(loop
with l = '(1 2 3 4 5 6)
for a in l
for b in (rest l)
collect (cons a b))
;=> ((1 . 2) (2 . 3) (3 . 4) (4 . 5) (5 . 6))
There are some other variations on loop that you can use, but some of them have less conventient results. E.g., you could loop for (a b) on list, but then you get a (perhaps) unexpected final binding of your variables:
(loop for (a b) on '(1 2 3 4 5 6)
collect (list a b))
;=> ((1 2) (2 3) (3 4) (4 5) (5 6) (6 NIL))
This is similar to what maplist will give you.
every
I think the real winners here, though, are going to the be every, some, notevery, and notany functions. These, like mapcar can take more than one list as an argument. This means that your problem can simply be:
(let ((delta 4)
(lst '(1 2 4 7 9)))
(every (lambda (x y)
(< (abs (- x y)) delta))
lst
(rest lst)))
;=> T
(let ((delta 2)
(lst '(1 2 4 7 9)))
(every (lambda (x y)
(< (abs (- x y)) delta))
lst
(rest lst)))
;=> NIL
maplist
You could also do this with maplist, which works on successive tails of the list, which means you'd have access to each element and the one following. This has the same 6 NIL at the end that the second loop solution did, though. E.g.:
(maplist (lambda (tail)
(list (first tail)
(second tail)))
'(1 2 3 4 5 6))
;=> ((1 2) (2 3) (3 4) (4 5) (5 6) (6 NIL))
reduce can be used:
(defun testdelta (delta lst)
(reduce
(lambda (r e)
(if (< (abs (- r e)) delta)
e
(return-from testdelta nil)))
lst)
t)
or, without return-from (but possibly slower):
(defun testdelta (delta lst)
(and
(reduce
(lambda (r e)
(and r (if (< (abs (- r e)) delta) e nil)))
lst)
t))
The ClojureDocs page for lazy-seq gives an example of generating a lazy-seq of all positive numbers:
(defn positive-numbers
([] (positive-numbers 1))
([n] (cons n (lazy-seq (positive-numbers (inc n))))))
This lazy-seq can be evaluated for pretty large indexes without throwing a StackOverflowError (unlike the sieve example on the same page):
user=> (nth (positive-numbers) 99999999)
100000000
If only recur can be used to avoid consuming stack frames in a recursive function, how is it possible this lazy-seq example can seemingly call itself without overflowing the stack?
A lazy sequence has the rest of the sequence generating calculation in a thunk. It is not immediately called. As each element (or chunk of elements as the case may be) is requested, a call to the next thunk is made to retrieve the value(s). That thunk may create another thunk to represent the tail of the sequence if it continues. The magic is that (1) these special thunks implement the sequence interface and can transparently be used as such and (2) each thunk is only called once -- its value is cached -- so the realized portion is a sequence of values.
Here it is the general idea without the magic, just good ol' functions:
(defn my-thunk-seq
([] (my-thunk-seq 1))
([n] (list n #(my-thunk-seq (inc n)))))
(defn my-next [s] ((second s)))
(defn my-realize [s n]
(loop [a [], s s, n n]
(if (pos? n)
(recur (conj a (first s)) (my-next s) (dec n))
a)))
user=> (-> (my-thunk-seq) first)
1
user=> (-> (my-thunk-seq) my-next first)
2
user=> (my-realize (my-thunk-seq) 10)
[1 2 3 4 5 6 7 8 9 10]
user=> (count (my-realize (my-thunk-seq) 100000))
100000 ; Level stack consumption
The magic bits happen inside of clojure.lang.LazySeq defined in Java, but we can actually do the magic directly in Clojure (implementation that follows for example purposes), by implementing the interfaces on a type and using an atom to cache.
(deftype MyLazySeq [thunk-mem]
clojure.lang.Seqable
(seq [_]
(if (fn? #thunk-mem)
(swap! thunk-mem (fn [f] (seq (f)))))
#thunk-mem)
;Implementing ISeq is necessary because cons calls seq
;on anyone who does not, which would force realization.
clojure.lang.ISeq
(first [this] (first (seq this)))
(next [this] (next (seq this)))
(more [this] (rest (seq this)))
(cons [this x] (cons x (seq this))))
(defmacro my-lazy-seq [& body]
`(MyLazySeq. (atom (fn [] ~#body))))
Now this already works with take, etc., but as take calls lazy-seq we'll make a my-take that uses my-lazy-seq instead to eliminate any confusion.
(defn my-take
[n coll]
(my-lazy-seq
(when (pos? n)
(when-let [s (seq coll)]
(cons (first s) (my-take (dec n) (rest s)))))))
Now let's make a slow infinite sequence to test the caching behavior.
(defn slow-inc [n] (Thread/sleep 1000) (inc n))
(defn slow-pos-nums
([] (slow-pos-nums 1))
([n] (cons n (my-lazy-seq (slow-pos-nums (slow-inc n))))))
And the REPL test
user=> (def nums (slow-pos-nums))
#'user/nums
user=> (time (doall (my-take 10 nums)))
"Elapsed time: 9000.384616 msecs"
(1 2 3 4 5 6 7 8 9 10)
user=> (time (doall (my-take 10 nums)))
"Elapsed time: 0.043146 msecs"
(1 2 3 4 5 6 7 8 9 10)
Keep in mind that lazy-seq is a macro, and therefore does not evaluate its body when your positive-numbers function is called. In that sense, positive-numbers isn't truly recursive. It returns immediately, and the inner "recursive" call to positive-numbers doesn't happen until the seq is consumed.
user=> (source lazy-seq)
(defmacro lazy-seq
"Takes a body of expressions that returns an ISeq or nil, and yields
a Seqable object that will invoke the body only the first time seq
is called, and will cache the result and return it on all subsequent
seq calls. See also - realized?"
{:added "1.0"}
[& body]
(list 'new 'clojure.lang.LazySeq (list* '^{:once true} fn* [] body)))
I think the trick is that the producer function (positive-numbers) isn't getting called recursively, it doesn't accumulate stack frames as if it was called with basic recursion Little-Schemer style, because LazySeq is invoking it as needed for the individual entries in the sequence. Once a closure gets evaluated for an entry then it can be discarded. So stack frames from previous invocations of the function can get garbage-collected as the code churns through the sequence.
How should I monitor the progress of a mapped function in clojure?
When processing records in an imperative language I often print a message every so often to indicate how far things have gone, e.g. reporting every 1000 records. Essentially this is counting loop repetitions.
I was wondering what approaches I could take to this in clojure where I am mapping a function over my sequence of records. In this case printing the message (and even keeping count of the progress) seem to be essentially side-effects.
What I have come up with so far looks like:
(defn report
[report-every val cnt]
(if (= 0 (mod cnt report-every))
(println "Done" cnt))
val)
(defn report-progress
[report-every aseq]
(map (fn [val cnt]
(report report-every val cnt))
aseq
(iterate inc 1)))
For example:
user> (doall (report-progress 2 (range 10)))
Done 2
Done 4
Done 6
Done 8
Done 10
(0 1 2 3 4 5 6 7 8 9)
Are there other (better) ways of achieving this effect?
Are there any pitfalls in what I am doing? (I think I am preserving laziness and not holding the head for example.)
The great thing about clojure is you can attach the reporting to the data itself instead of the code that does the computing. This allows you to separate these logically distinct parts. Here is a chunk from my misc.clj that I find I use in just about every project:
(defn seq-counter
"calls callback after every n'th entry in sequence is evaluated.
Optionally takes another callback to call once the seq is fully evaluated."
([sequence n callback]
(map #(do (if (= (rem %1 n) 0) (callback)) %2) (iterate inc 1) sequence))
([sequence n callback finished-callback]
(drop-last (lazy-cat (seq-counter sequence n callback)
(lazy-seq (cons (finished-callback) ()))))))
then wrap the reporter around your data and then pass the result to the processing function.
(map process-data (seq-counter inc-progress input))
I would probably perform the reporting in an agent. Something like this:
(defn report [a]
(println "Done " s)
(+ 1 s))
(let [reports (agent 0)]
(map #(do (send reports report)
(process-data %))
data-to-process)
I don't know of any existing ways of doing that, maybe it would be a good idea to browse clojure.contrib documentation to look if there's already something. In the meantime, I've looked at your example and cleared it up a little bit.
(defn report [cnt]
(when (even? cnt)
(println "Done" cnt)))
(defn report-progress []
(let [aseq (range 10)]
(doall (map report (take (count aseq) (iterate inc 1))))
aseq))
You're heading in the right direction, even though this example is too simple. This gave me an idea about a more generalized version of your report-progress function. This function would take a map-like function, the function to be mapped, a report function and a set of collections (or a seed value and a collection for testing reduce).
(defn report-progress [m f r & colls]
(let [result (apply m
(fn [& args]
(let [v (apply f args)]
(apply r v args) v))
colls)]
(if (seq? result)
(doall result)
result)))
The seq? part is there only for use with reduce which doesn't
necessarily returns a sequence. With this function, we can rewrite your
example like this:
user>
(report-progress
map
(fn [_ v] v)
(fn [result cnt _]
(when (even? cnt)
(println "Done" cnt)))
(iterate inc 1)
(range 10))
Done 2
Done 4
Done 6
Done 8
Done 10
(0 1 2 3 4 5 6 7 8 9)
Test the filter function:
user>
(report-progress
filter
odd?
(fn [result cnt]
(when (even? cnt)
(println "Done" cnt)))
(range 10))
Done 0
Done 2
Done 4
Done 6
Done 8
(1 3 5 7 9)
And even the reduce function:
user>
(report-progress
reduce
+
(fn [result s v]
(when (even? s)
(println "Done" s)))
2
(repeat 10 1))
Done 2
Done 4
Done 6
Done 8
Done 10
12
I have had this problem with some slow-running apps (e.g. database ETL, etc). I solved it by adding the function (tupelo.misc/dot ...) to the tupelo library. Sample:
(ns xxx.core
(:require [tupelo.misc :as tm]))
(tm/dots-config! {:decimation 10} )
(tm/with-dots
(doseq [ii (range 2345)]
(tm/dot)
(Thread/sleep 5)))
Output:
0 ....................................................................................................
1000 ....................................................................................................
2000 ...................................
2345 total
API docs for the tupelo.misc namespace can be found here.