i'm trying to find a function that, given S a set of integer and I an integer, return all the subsets of S that sum to I
is there such a function somewhere in clojure-contrib or in another library ?
if no, could anyone please give me some hints to write it the clojure way?
Isn't this the subset sum problem, a classic NP-complete problem?
In which case, I'd just generate every possible distinct subset of S, and see which subsets sums to I.
I think it is the subset sum problem, as #MrBones suggests. Here's a brute force attempt using https://github.com/clojure/math.combinatorics (lein: [org.clojure/math.combinatorics "0.0.7"]):
(require '[clojure.math.combinatorics :as c])
(defn subset-sum [s n]
"Return all the subsets of s that sum to n."
(->> (c/subsets s)
(filter #(pos? (count %))) ; ignore empty set since (+) == 0
(filter #(= n (apply + %)))))
(def s #{1 2 45 -3 0 14 25 3 7 15})
(subset-sum s 13)
; ((1 -3 15) (2 -3 14) (0 1 -3 15) (0 2 -3 14) (1 2 3 7) (0 1 2 3 7))
(subset-sum s 0)
; ((0) (-3 3) (0 -3 3) (1 2 -3) (0 1 2 -3))
These "subsets" are just lists. Could convert back to sets, but I didn't bother.
You can generate the subsets of a set like this:
(defn subsets [s]
(if (seq s)
(let [f (first s), srs (subsets (disj s f))]
(concat srs (map #(conj % f) srs)))
(list #{})))
The idea is to choose an element from the set s: the first, f, will do. Then we recursively find the subsets of everything else, srs. srs comprises all the subsets without f. By adding f to each of them, we get all the subsets with f. And together, that's the lot. Finally, if we can't choose an element because there aren't any, the only subset is the empty one.
All that remains to do is to filter out from all the subsets the ones that sum to n. A function to test this is
(fn [s] (= n (reduce + s)))
It is not worth naming.
Putting this together, the function we want is
(defn subsets-summing-to [s n]
(filter
(fn [xs] (= n (reduce + xs)))
(subsets s)))
Notes
Since the answer is a sequence of sets, we can make it lazier by changing concat into lazy-cat. map is lazy anyway.
We may appear to be generating a lot of sets, but remember that they share storage: the space cost of keeping another set differing by a single element is (almost) constant.
The empty set sums to zero in Clojure arithmetic.
Related
I am trying to write a lazy seq to generate the Collatz sequence for a given input int.
I love this function because is so cleanly maps to the mathematical definition:
(defn collatz
"Returns a lazy seq of the Collatz sequence starting at n and ending at 1 (if
ever)."
[n]
(letfn [(next-term [x]
(if (even? x)
(/ x 2)
(inc (* 3 x))))]
(iterate next-term n)))
The problem is that this produces infinite seqs because of how the Collatz sequence behaves:
(take 10 (collatz 5))
=> (5 16 8 4 2 1 4 2 1 4)
I could easily drop the cycle by adding (take-while #(not= 1 %) ...), but the 1 is part of the sequence. All the other ways I've thought to trim the cycle after the one are ugly and obfuscate the mathematical heart of the Collatz sequence.
(I've considered storing the seen values in an atom and using that in a take-while predicate, or just storing a flag in an atom to similar effect. But I feel like there is some better, more beautiful, less intrusive way to do what I want here.)
So my question: What are clean ways to detect and trim cycles in infinite seqs? Or, could I generate my lazy seq in a way (perhaps using for) that automatically trims when it reaches 1 (inclusive)?
The below looks like a more or less literal translation of the definition and gives the result you want:
(defn collatz-iter [x]
(cond (= x 1) nil
(even? x) (/ x 2)
:else (inc (* 3 x))))
(defn collatz [n]
(take-while some? (iterate collatz-iter n)))
(collatz 12) ;; => (12 6 3 10 5 16 8 4 2 1)
Basically, you can use nil as the value to stop the sequence, thus keeping the final 1.
you could also use another approach, which is recursive lazy-seq generation. That is quite common for this class of tasks, doesn't break the lazy sequence abstraction and avoids intermediate sequences' creation:
(defn collatz [n]
(if (== n 1)
(list 1)
(lazy-seq (cons n (collatz (if (even? n)
(/ n 2)
(inc (* 3 n))))))))
user> (collatz 12)
;;=> (12 6 3 10 5 16 8 4 2 1)
If I run this code, I will get an error "ArityException Wrong number of args (0) passed to: core/max"
(apply max (filter #(zero? (mod % 7)) (range 1 3)))
However, if I run this code
(apply max (filter #(zero? (mod % 7)) (range 1 10)))
then I get the result 7.
Is there anyone who can help me to figure out this problem?
(filter #(zero? (mod % 7)) (range 1 3))
this, produces an empty sequence.
However, max must be called with at least one argument. When you apply an empty sequence to it, it's called with zero arguments, and this produces the arity exception.
You could do something like this:
(defn my [a b]
(let [result (filter #(zero? (mod % 7)) (range a b))]
(if (zero? (count result))
nil ; or 0 or.. whatever
(apply max result))))
apply and reduce
Because the question came up, here's a short explanation of the difference between apply and reduce.
They are two totally different concepts, however, in the following case both do the same job when combined with max.
let xs be any collection of numbers.
(apply max xs) equals (reduce max xs)
apply
Usually functions are called with a number of arguments, so one can call max like so: (max 3), or (max 5 9), or (max 4 1 3) ... As noticed before: just (max) would be an arity exception.
Apply however, lets someone call a function passing the arguments in the form of a collection. So in correspondence to the last example, the following is possible: (apply max [3]), or (apply max [5 9]), or (apply max [4 1 3]) ... Just (apply max []) or even (apply max) would lead to the same arity exception as above. This is useful in many cases.
reduce
Reduce in contrast is a bit trickier. Along with map and filter it's absolutely essential for functional programming languages like Clojure.
The main idea of reduce is to walk through a collection, in each step desired information from the current item is processed and added to a memo or accumulator.
Say, one wants to find out the sum of all numbers in a collection.
Let xs be [3 4 5 23 9 4].
(reduce + xs) would do the job.
more explicitly one could write: (reduce (fn [memo value] (+ memo value)) xs)
The function which is passed as the first argument to reduce expects two parameters: The first one is the memo, the second one the value of the current item in the collection. The function is now called for each item in the collection. The return value of the function is saved as the memo.
Note: that the first value of the collection is used as an initial value of the memo, hence the iteration starts with the second value of the collection. Here's what it is doing:
(+ 3 4) ; memo is 7 now
(+ 7 5) ; memo is 12 now
(+ 12 23) ; memo is 35 now
(+ 35 9) ; memo is 44 now
(+ 44 4) ; memo is 48 now
(There's also a way to specify the start value of the memo, see clojuredocs.org for more details)
This works equally with max. In each iteration the value of the current item is compared with the memo. Each time the highest value is saved to the memo: Hence the memo in this case represents the "maximum value until now".
So (reduce max [4 1 3 5 2]) is calculated like this:
(max 4 1) ; memo is 4
(max 4 3) ; memo is 4
(max 4 5) ; memo is 5
(max 5 2) ; memo is 5
so?
Which one to use now? It showed that there's not really a notable difference in the time that (reduce max (range 100000000)) and (apply max (range 100000000)) take. Anyways, the apply solution looks easier to me, but that's just my opinion.
There are no numbers divisible by 7 between 1 and 3, the result of filter in your first example returns an empty sequence, which means that the first example if calling (apply max []) which is the same as calling (max). max requires at least one parameter, hence the ArityException.
A couple of options to fix it:
(last (filter #(zero? (mod % 7)) (range 1 3))
or
(if-let [by-7 (seq (filter #(zero? (mod % 7)) (range 1 3)))]
(apply max by-7)
nil ;; or whatever value in case the collection is empty
)
According to the error message, the number of arguments that are passed to max is 0, and that is wrong. I guess it makes sense because it's impossible to compute the maximum for an empty list.
The reason why max gets no arguments is that there are no numbers divisible by 7 between 1 and 3.
I have:
(defn keep?
(def sum [])
(loop [i 0]
(when (< i 10)
(conj sum 10)
(recur (inc i))))
sum
)
This just gives me and empty vector even though I am conj-ing 10 onto sum. Is this because it is not in-scope within the Loop? How would I achieve the same outcome. (btw, this example is deliberately simplified)
Thanks
conj does not modify its argument. In fact, without resorting to evil reflection tricks, nothing will modify a vector, it's an immutable data structure. This is one of the fundamental principles of Clojure.
In functional programming, instead of modifying data, we replace it with another immutable value. So you need to use the return value of conj, or it is effectively a noop.
(defn keep?
[]
(loop [i 0 sum []]
(if (< i 10)
(recur (inc i) (conj sum 10))
sum)))
Also, the second arg to defn must always be a vector.
conj is not destructive, it does not alter that collection, returns a new collection with the designated state (reference).
To achieve the desired result, you may:
Define sum in a loop-form, like i is defined, instead of using def
recur (inc i) (conj sum 10) to rebind sum to a new one on every iteration, so that state is built up to what you expect
Once the condition in when is not met, just return sum from your loop and it will bubble up to become the return value of this function. Uh hang on, when does not have an "else-branch", a possible alternative is if
Like so:
(defn keep? []
(loop [i 0
sum []]
(if (< i 10)
(recur (inc i)
(conj sum 10))
sum)))
Just to supplement the other answers, I almost never use the loop function. Here are a few ways to do it using the for function:
; return a lazy sequence
(for [i (range 10) ]
i)
;=> (0 1 2 3 4 5 6 7 8 9)
; return a concrete vector
(vec
(for [i (range 10) ]
i))
;=> [0 1 2 3 4 5 6 7 8 9]
; 'into' is very nice for converting one collection into another
(into #{}
(for [i (range 10) ]
i))
;=> #{0 7 1 4 6 3 2 9 5 8} ; hash-set is unique but unordered
The comprehension:
(for [i (range 5])] i)
... yields: (0 1 2 3 4)
Is there an idiomatic way to get (0 0 1 1 2 4 3 9 4 16) (i.e. the numbers and their squares) using mostly the for comprehension?
The only way I've found so far is doing a:
(apply concat (for [i (range 5)] (list i (* i i))))
Actually, using only for is pretty simple if you consider applying each function (identity and square) for each value.
(for [i (range 5), ; for every value
f [identity #(* % %)]] ; for every function
(f i)) ; apply the function to the value
; => (0 0 1 1 2 4 3 9 4 16)
Since for loops x times, it will return a collection of x values. Multiple nested loops (unless limited by while or when) will give x * y * z * ... results. That is why external concatenation will always be necessary.
A similar correlation between input and output exists with map. However, if multiple collections are given in map, the number of values in the returned collection is the size of the smallest collection parameter.
=> (map (juxt identity #(* % %)) (range 5))
([0 0] [1 1] [2 4] [3 9] [4 16])
Concatenating the results of map is so common mapcat was created. Because of that, one might argue mapcat is a more idiomatic way over for loops.
=> (mapcat (juxt identity #(* % %)) (range 5))
(0 0 1 1 2 4 3 9 4 16)
Although this is just shorthand for apply concat (map, and a forcat function or macro could be created just as easily.
However, if an accumulation over a collection is needed, reduce is usually considered the most idiomatic.
=> (reduce (fn [acc i] (conj acc i (* i i))) [] (range 5))
[0 0 1 1 2 4 3 9 4 16]
Both the for and map options would mean traversing a collection twice, once for the range, and once for concatenating the resulting collection. The reduce option only traverses the range.
Care to share why "using mostly the for comprehension" is a requirement ?
I think you are doing it right.
A slightly compressed way maybe achieved using flatten
(flatten (for [i (range 5)] [ i (* i i) ] ))
But I would get rid of the for comprehension and just use interleave
(let [x (range 5)
y (map #(* % %) x)]
(interleave x y))
Disclaimer: I am just an amateur clojurist ;)
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?