How do I do mutually recursive definitions in Clojure?
Here is a code in Scala to find prime numbers which uses recursive definitions:
val odds: Stream[Int] = cons(3, odds map { _ + 2 })
val primes: Stream[Int] = cons(2, odds filter isPrime)
def primeDivisors(n: Int) =
primes takeWhile { _ <= Math.ceil(Math.sqrt(n))} filter { n % _ == 0 }
def isPrime(n: Int) = primeDivisors(n) isEmpty
primes take 10
I translated this to Clojure:
(def odds (iterate #(+ % 2) 3))
(def primes (cons 2 (filter is-prime odds)))
(defn prime-divisors [n]
(filter #(zero? (mod n %))
(take-while #(<= % (Math/ceil (Math/sqrt n)))
primes)))
(defn is-prime [n] (empty? (prime-divisors n)))
(take 10 primes)
But writing the definitions in the Clojure REPL one by one gives
java.lang.Exception: Unable to resolve symbol: is-prime in this context (NO_SOURCE_FILE:8)
after I write (def primes (cons 2 (filter is-prime odds))).
Is there a way to do mutually recursive definitions in Clojure?
You need to (declare is-prime) before the first time you reference it.
This is referred to as a "forward declaration".
Greg's answer is correct. However you have to rearrange your code: (def odds ...) (declare primes) (defn prime-divisors ...) (defn prime? ...) (def primes ...). This should do the trick.
The problem is, that the definition of primes is not a function. It is executed immediately and hence tries to dereference the prime? Var which is not bound, yet. Hence the exception. Re-arranging should take care of that.
(Disclaimer: I haven't checked that the code works with the re-arrangement.)
And I think prime? is broken. (prime? 2) should give false, no?
Related
I need help with an assignment that uses Clojure. It is very small but the language is a bit confusing to understand. I need to create a function that behaves like count without actually using the count funtion. I know a loop can be involved with it somehow but I am at a lost because nothing I have tried even gets my code to work. I expect it to output the number of elements in list. For example:
(defn functionname []
...
...)
(println(functionname '(1 4 8)))
Output:3
Here is what I have so far:
(defn functionname [n]
(def n 0)
(def x 0)
(while (< x n)
do
()
)
)
(println(functionname '(1 4 8)))
It's not much but I think it goes something like this.
This implementation takes the first element of the list and runs a sum until it can't anymore and then returns the sum.
(defn recount [list-to-count]
(loop [xs list-to-count sum 0]
(if (first xs)
(recur (rest xs) (inc sum))
sum
)))
user=> (recount '(3 4 5 9))
4
A couple more example implementations:
(defn not-count [coll]
(reduce + (map (constantly 1) coll)))
or:
(defn not-count [coll]
(reduce (fn [a _] (inc a)) 0 coll))
or:
(defn not-count [coll]
(apply + (map (fn [_] 1) coll)))
result:
(not-count '(5 7 8 1))
=> 4
I personally like the first one with reduce and constantly.
I have implemented a lazy prime number generator (nextprime returns the next prime starting from the number passed):
(defn allprimes
([] (allprimes 2))
([x] (lazy-seq (cons (nextprime x) (allprimes (nextprime x))))))
Let's assume nextprime is a costly function, in order not to execute it twice I have tried binding it to a symbol:
(defn allprimes
([] (allprimes 2))
([x] (let [next (nextprime x)]
(cons next (lazy-seq (allprimes (next)))))))
But this does not work (java.lang.Long cannot be cast to clojure.lang.IFn). Why?
Also, is there any difference between (cons n (lazy-seq ...)) and (lazy-seq (cons n ...)) ?
edit: thanks Kyle for pointing the error in the first question. If parentheses are removed from next, it works.
You have extra parenthesis around (next) : next is then called as a function IFn though next is a Long...
For your second question, the las example in clojure doc gives details on the difference between the 2 solutions.
Roughly (cons n (lazy-seq ...)) will always evaluate n, even if it is not consumed, and (lazy-seq (cons n ...)) is fully lazy. It can matter if n is not just a number but some function that may be computation intensive.
Let's say you have a recursive function defined in a let block:
(let [fib (fn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2)))))]
(fib 42))
This can be mechanically transformed to utilize memoize:
Wrap the fn form in a call to memoize.
Move the function name in as the 1st argument.
Pass the function into itself wherever it is called.
Rebind the function symbol to do the same using partial.
Transforming the above code leads to:
(let [fib (memoize
(fn [fib n]
(if (< n 2)
n
(+ (fib fib (- n 1))
(fib fib (- n 2))))))
fib (partial fib fib)]
(fib 42))
This works, but feels overly complicated. The question is: Is there a simpler way?
I take risks in answering since I am not a scholar but I don't think so. You pretty much did the standard thing which in fine is a partial application of memoization through a fixed point combinator.
You could try to fiddle with macros though (for simple cases it could be easy, syntax-quote would do name resolution for you and you could operate on that). I'll try once I get home.
edit: went back home and tried out stuff, this seems to be ok-ish for simple cases
(defmacro memoize-rec [form]
(let [[fn* fname params & body] form
params-with-fname (vec (cons fname params))]
`(let [f# (memoize (fn ~params-with-fname
(let [~fname (partial ~fname ~fname)] ~#body)))]
(partial f# f#))))
;; (clojure.pprint/pprint (macroexpand '(memoize-rec (fn f [x] (str (f x))))))
((memoize-rec (fn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2)))))) 75) ;; instant
((fn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2))))) 75) ;; slooooooow
simpler than what i thought!
I'm not sure this is "simpler" per se, but I thought I'd share an approach I took to re-implement letfn for a CPS transformer I wrote.
The key is to introduce the variables, but delay assigning them values until they are all in scope. Basically, what you would like to write is:
(let [f nil]
(set! f (memoize (fn []
<body-of-f>)))
(f))
Of course this doesn't work as is, because let bindings are immutable in Clojure. We can get around that, though, by using a reference type — for example, a promise:
(let [f (promise)]
(deliver! f (memoize (fn []
<body-of-f>)))
(#f))
But this still falls short, because we must replace every instance of f in <body-of-f> with (deref f). But we can solve this by introducing another function that invokes the function stored in the promise. So the entire solution might look like this:
(let [f* (promise)]
(letfn [(f []
(#f*))]
(deliver f* (memoize (fn []
<body-of-f>)))
(f)))
If you have a set of mutually-recursive functions:
(let [f* (promise)
g* (promise)]
(letfn [(f []
(#f*))
(g []
(#g*))]
(deliver f* (memoize (fn []
(g))))
(deliver g* (memoize (fn []
(f))))
(f)))
Obviously that's a lot of boiler-plate. But it's pretty trivial to construct a macro that gives you letfn-style syntax.
Yes, there is a simpler way.
It is not a functional transformation, but builds on the impurity allowed in clojure.
(defn fib [n]
(if (< n 2)
n
(+ (#'fib (- n 1))
(#'fib (- n 2)))))
(def fib (memoize fib))
First step defines fib in almost the normal way, but recursive calls are made using whatever the var fib contains. Then fib is redefined, becoming the memoized version of its old self.
I would suppose that clojure idiomatic way will be using recur
(def factorial
(fn [n]
(loop [cnt n acc 1]
(if (zero? cnt)
acc
(recur (dec cnt) (* acc cnt))
;; Memoized recursive function, a mash-up of memoize and fn
(defmacro mrfn
"Returns an anonymous function like `fn` but recursive calls to the given `name` within
`body` use a memoized version of the function, potentially improving performance (see
`memoize`). Only simple argument symbols are supported, not varargs or destructing or
multiple arities. Memoized recursion requires explicit calls to `name` so the `body`
should not use recur to the top level."
[name args & body]
{:pre [(simple-symbol? name) (vector? args) (seq args) (every? simple-symbol? args)]}
(let [akey (if (= (count args) 1) (first args) args)]
;; name becomes extra arg to support recursive memoized calls
`(let [f# (fn [~name ~#args] ~#body)
mem# (atom {})]
(fn mr# [~#args]
(if-let [e# (find #mem# ~akey)]
(val e#)
(let [ret# (f# mr# ~#args)]
(swap! mem# assoc ~akey ret#)
ret#))))))
;; only change is fn to mrfn
(let [fib (mrfn fib [n]
(if (< n 2)
n
(+ (fib (- n 1))
(fib (- n 2)))))]
(fib 42))
Timings on my oldish Mac:
original, Elapsed time: 14089.417441 msecs
mrfn version, Elapsed time: 0.220748 msecs
I am going through the Clojure in Action book and code similar to that below is given for a function that returns all pairs of numbers below m whose sum is a prime (assume prime? is given):
(defn pairs-for-primes [m]
(let [z (range 0 m)]
(for [a z b z :when (prime? (+ a b))]
(list a b))))
How would one generalize that to return the n-tuples of all numbers below m whose sum is a prime?
(defn all-ntuples-below [n m]
...
for can be used for a sort of "special case" of cartesian product, where you know the sets in advance at compile time. Since you don't actually know the sets you want the product of, you need to use a real cartesian-product function. For example, with clojure.math.combinatorics, you could write
(defn pairs-for-primes [m n]
(let [z (range 0 m)
tuples (apply cartesian-product (repeat n z))]
(filter #(prime? (apply + %)) tuples)))
But perhaps your question is about how to implement a cartesian product? It's not that hard, although the version below is not terribly performant:
(defn cartesian-product [sets]
(cond (empty? sets) (list (list))
(not (next sets)) (map list (first sets))
:else (for [x (first sets)
tuple (cartesian-product (rest sets))]
(cons x tuple))))
You can use take to do that (as pairs-for-primes returns a sequence take will only cause it to calculate the number of items required)
(defn all-ntuples-below [n m]
(take n (pairs-for-primes m)))
Coming from imperative programming languages, I am trying to wrap my head around Clojure in hopes of using it for its multi-threading capability.
One of the problems from 4Clojure is to write a function that generates a list of Fibonacci numbers of length N, for N > 1. I wrote a function, but given my limited background, I would like some input on whether or not this is the best Clojure way of doing things. The code is as follows:
(fn fib [x] (cond
(= x 2) '(1 1)
:else (reverse (conj (reverse (fib (dec x))) (+ (last (fib (dec x))) (-> (fib (dec x)) reverse rest first))))
))
The most idiomatic "functional" way would probably be to create an infinite lazy sequence of fibonacci numbers and then extract the first n values, i.e.:
(take n some-infinite-fibonacci-sequence)
The following link has some very interesting ways of generating fibonnaci sequences along those lines:
http://en.wikibooks.org/wiki/Clojure_Programming/Examples/Lazy_Fibonacci
Finally here is another fun implementation to consider:
(defn fib [n]
(let [next-fib-pair (fn [[a b]] [b (+ a b)])
fib-pairs (iterate next-fib-pair [1 1])
all-fibs (map first fib-pairs)]
(take n all-fibs)))
(fib 6)
=> (1 1 2 3 5 8)
It's not as concise as it could be, but demonstrates quite nicely the use of Clojure's destructuring, lazy sequences and higher order functions to solve the problem.
Here is a version of Fibonacci that I like very much (I took the implementation from the clojure wikibook: http://en.wikibooks.org/wiki/Clojure_Programming)
(def fib-seq (lazy-cat [0 1] (map + (rest fib-seq) fib-seq)))
It works like this: Imagine you already have the infinite sequence of Fibonacci numbers. If you take the tail of the sequence and add it element-wise to the original sequence you get the (tail of the tail of the) Fibonacci sequence
0 1 1 2 3 5 8 ...
1 1 2 3 5 8 ...
-----------------
1 2 3 5 8 13 ...
thus you can use this to calculate the sequence. You need two initial elements [0 1] (or [1 1] depending on where you start the sequence) and then you just map over the two sequences adding the elements. Note that you need lazy sequences here.
I think this is the most elegant and (at least for me) mind stretching implementation.
Edit: The fib function is
(defn fib [n] (nth fib-seq n))
Here's one way of doing it that gives you a bit of exposure to lazy sequences, although it's certainly not really an optimal way of computing the Fibonacci sequence.
Given the definition of the Fibonacci sequence, we can see that it's built up by repeatedly applying the same rule to the base case of '(1 1). The Clojure function iterate sounds like it would be good for this:
user> (doc iterate)
-------------------------
clojure.core/iterate
([f x])
Returns a lazy sequence of x, (f x), (f (f x)) etc. f must be free of side-effects
So for our function we'd want something that takes the values we've computed so far, sums the two most recent, and returns a list of the new value and all the old values.
(fn [[x y & _ :as all]] (cons (+ x y) all))
The argument list here just means that x and y will be bound to the first two values from the list passed as the function's argument, a list containing all arguments after the first two will be bound to _, and the original list passed as an argument to the function can be referred to via all.
Now, iterate will return an infinite sequence of intermediate values, so for our case we'll want to wrap it in something that'll just return the value we're interested in; lazy evaluation will stop the entire infinite sequence being evaluated.
(defn fib [n]
(nth (iterate (fn [[x y & _ :as all]] (cons (+ x y) all)) '(1 1)) (- n 2)))
Note also that this returns the result in the opposite order to your implementation; it's a simple matter to fix this with reverse of course.
Edit: or indeed, as amalloy says, by using vectors:
(defn fib [n]
(nth (iterate (fn [all]
(conj all (->> all (take-last 2) (apply +)))) [1 1])
(- n 2)))
See Christophe Grand's Fibonacci solution in Programming Clojure by Stu Halloway. It is the most elegant solution I have seen.
(defn fibo [] (map first (iterate (fn [[a b]] [b (+ a b)]) [0 1])))
(take 10 (fibo))
Also see
How can I generate the Fibonacci sequence using Clojure?