I have the following bit of code that produces the correct results:
(ns scratch.core
(require [clojure.string :as str :only (split-lines join split)]))
(defn numberify [str]
(vec (map read-string (str/split str #" "))))
(defn process [acc sticks]
(let [smallest (apply min sticks)
cuts (filter #(> % 0) (map #(- % smallest) sticks))]
(if (empty? cuts)
acc
(process (conj acc (count cuts)) cuts))))
(defn print-result [[x & xs]]
(prn x)
(if (seq xs)
(recur xs)))
(let [input "8\n1 2 3 4 3 3 2 1"
lines (str/split-lines input)
length (read-string (first lines))
inputs (first (rest lines))]
(print-result (process [length] (numberify inputs))))
The process function above recursively calls itself until the sequence sticks is empty?.
I am curious to know if I could have used something like take-while or some other technique to make the code more succinct?
If ever I need to do some work on a sequence until it is empty then I use recursion but I can't help thinking there is a better way.
Your core problem can be described as
stop if count of sticks is zero
accumulate count of sticks
subtract the smallest stick from each of sticks
filter positive sticks
go back to 1.
Identify the smallest sub-problem as steps 3 and 4 and put a box around it
(defn cuts [sticks]
(let [smallest (apply min sticks)]
(filter pos? (map #(- % smallest) sticks))))
Notice that sticks don't change between steps 5 and 3, that cuts is a fn sticks->sticks, so use iterate to put a box around that:
(defn process [sticks]
(->> (iterate cuts sticks)
;; ----- 8< -------------------
This gives an infinite seq of sticks, (cuts sticks), (cuts (cuts sticks)) and so on
Incorporate step 1 and 2
(defn process [sticks]
(->> (iterate cuts sticks)
(map count) ;; count each sticks
(take-while pos?))) ;; accumulate while counts are positive
(process [1 2 3 4 3 3 2 1])
;-> (8 6 4 1)
Behind the scene this algorithm hardly differs from the one you posted, since lazy seqs are a delayed implementation of recursion. It is more idiomatic though, more modular, uses take-while for cancellation which adds to its expressiveness. Also it doesn't require one to pass the initial count and does the right thing if sticks is empty. I hope it is what you were looking for.
I think the way your code is written is a very lispy way of doing it. Certainly there are many many examples in The Little Schema that follow this format of reduction/recursion.
To replace recursion, I usually look for a solution that involves using higher order functions, in this case reduce. It replaces the min calls each iteration with a single sort at the start.
(defn process [sticks]
(drop-last (reduce (fn [a i]
(let [n (- (last a) (count i))]
(conj a n)))
[(count sticks)]
(partition-by identity (sort sticks)))))
(process [1 2 3 4 3 3 2 1])
=> (8 6 4 1)
I've changed the algorithm to fit reduce by grouping the same numbers after sorting, and then counting each group and reducing the count size.
Related
I am learning clojure. While solving one of the problem, I had to use first + filter. I noted that the filter is running unnecessarily for all the inputs.
How can I make the filter to run lazily so that it need not apply the predicate for the whole input.
The below is an example showing that it is not lazy,
(defn filter-even
[n]
(println n)
(= (mod n 2) 0))
(first (filter filter-even (range 1 4)))
The above code prints
1
2
3
Whereas it need not go beyond 2. How can we make it lazy?
This happens because range is a chunked sequence:
(chunked-seq? (range 1))
=> true
And it will actually take the first 32 elements if available:
(first (filter filter-even (range 1 100)))
1
2
. . .
30
31
32
=> 2
This overview shows an unchunk function that prevents this from happening. Unfortunately, it isn't standard:
(defn unchunk [s]
(when (seq s)
(lazy-seq
(cons (first s)
(unchunk (next s))))))
(first (filter filter-even (unchunk (range 1 100))))
2
=> 2
Or, you could apply list to it since lists aren't chunked:
(first (filter filter-even (apply list (range 1 100))))
2
=> 2
But then obviously, the entire collection needs to be realized pre-filtering.
This honestly isn't something that I've ever been too concerned about though. The filtering function usually isn't too expensive, and 32 element chunks aren't that big in the grand scheme of things.
I have implemented the Sieve of Eratosthenes using Clojure's standard library.
(defn primes [below]
(remove (set (mapcat #(range (* % %) below %)
(range 3 (Math/sqrt below) 2)))
(cons 2 (range 3 below 2))))
I think this should be amenable to parallelism as there is no recursion and the reducer versions of remove and mapcat can be dropped in. Here is what I came up with:
(defn pprimes [below]
(r/foldcat
(r/remove
(into #{} (r/mapcat #(range (* % %) below %)
(into [] (range 3 (Math/sqrt below) 2))))
(into [] (cons 2 (range 3 below 2))))))
I've poured the initial set and the generated multiples into vectors as I understand that LazySeqs can't be folded. Also, r/foldcat is used to finally realize the collection.
My problem is that this is a little slower than the first version.
(time (first (primes 1000000))) ;=> approx 26000 seconds
(time (first (pprimes 1000000))) ;=> approx 28500 seconds
Is there too much overhead from the coordinating processes or am I using reducers wrong?
Thanks to leetwinski this seems to work:
(defn pprimes2 [below]
(r/foldcat
(r/remove
(into #{} (r/foldcat (r/map #(range (* % %) below %)
(into [] (range 3 (Math/sqrt below) 2)))))
(into [] (cons 2 (range 3 below 2))))))
Apparently I needed to add another fold operation in order to map #(range (* % %) below %) in parallel.
(time (first (pprimes 1000000))) ;=> approx 28500 seconds
(time (first (pprimes2 1000000))) ;=> approx 7500 seconds
Edit: The above code doesn't work. r/foldcat isn't concatenating the composite numbers it is just returning a vector of the multiples for each prime number. The final result is a vector of 2 and all the odd numbers. Replacing r/map with r/mapcat gives the correct answer but it is again slower than the original primes.
as far as remember, the r/mapcat and r/remove are not parallel themselves, thy are just producing foldable collections, which are in turn can be subject to parallelization by r/fold. in your case the only parallel operation is r/foldcat, which is according to documentation "Equivalent to (fold cat append! coll)", meaning that you just potentially do append! in parallel, which isn't what you want at all.
To make it parallel you should probably use r/fold with remove as a reduce function and concat as a combine function, but it won't really make your code faster i guess, due to the nature of your algorithm (i mean you will try to remove a big set of items from every chunk of a collection)
For planning a dinner menu a user wants to have a rotation 3 days meat and then one vegetarian entree. The meat rotation is beef, chicken, fish, pork. The veg rotation is stirfry then tofu. Is there a better (more elegant, concise, or readable) way to generate a meal sequence than what I have below?
(defn menu [n] (let [meat (cycle '[:beef :chicken :fish :pork])
veg (cycle '[:stirfry :tofu])]
(take n (mapcat seq (interleave (partition 3 meat) (partition 1 veg))))))
this is slightly shorter and easier to read i guess (and probably faster):
(defn menu [n]
(let [meat (cycle '[:beef :chicken :fish :pork])
veg (cycle '[:stirfry :tofu])]
(take n (rest (mapcat conj (partition 3 meat) veg)))))
in repl:
user> (menu 10)
(:beef :chicken :fish :tofu :pork :beef :chicken :stirfry :fish :pork)
as mapcat can take a number of collection, it will do
(conj (:beef :chicken :fish) :stirfry)
(conj (:pork :beef :chicken) :tofu) and so on
you just need to drop the first value, as conj with lists adds an element to the beginning.
if it is possible to start with veg meal, then you can also lose rest, making it shorter.
(take n (mapcat conj (partition 3 meat) veg))
I'd be tempted to generalise the problem:
Accept any number of kinds of entree interleaved in any proportion.
Generate the whole lazy sequence, leaving the client to clip off as
many as required.
You can supply the data as alternate number and sequence of dishes. For your example, we have
(def dishes [3 [:beef :chicken :fish :pork]
1 [:stirfry :tofu]])
We can adapt your solution to generate the endless lazy sequence of dishes:
(defn rotate [nlist-pairs]
(let [pairs (partition 2 nlist-pairs)
sources (map (fn [[n xs]] (partition n (cycle xs)))
pairs)]
(apply concat (apply interleave sources))))
To get the first ten ...
(take 10 (rotate dishes))
;(:beef :chicken :fish :stirfry :pork :beef :chicken :tofu :fish :pork)
Below is my answer for 4clojure Problem 108
I'm able to pass the first three tests but the last test times out. The code runs really, really slowly on this last test. What exactly is causing this?
((fn [& coll] (loop [coll coll m {}]
(do
(let [ct (count coll)
ns (mapv first coll)
m' (reduce #(update-in %1 [%2] (fnil inc 0)) m ns)]
(println m')
(if (some #(<= ct %) (mapv m' ns))
(apply min (map first (filter #(>= (val %) ct) m')))
(recur (mapv rest coll) m'))))))
(map #(* % % %) (range)) ;; perfect cubes
(filter #(zero? (bit-and % (dec %))) (range)) ;; powers of 2
(iterate inc 20))
You are gathering the next value from every input on every iteration (recur (mapv rest coll) m')
One of your inputs generates values extremely slowly, and accellarates to very high values very quickly: (filter #(zero? (bit-and % (dec %))) (range)).
Your code is spending most of its time discovering powers of two by incrementing by one and testing the bits.
You don't need a map of all inputs with counts of occurrences. You don't need to find the next value for items that are not the lowest found so far. I won't post a solution since it is an exercise, but eliminating the lowest non matched value on each iteration should be a start.
In addition to the other good answers here, you're doing a bunch of math, but all numbers are boxed as objects rather than being used as primitives. Many tips for doing this better here.
This is a really inefficient way of counting powers of 2:
(filter #(zero? (bit-and % (dec %))) (range))
This is essentially counting from 0 to infinity, testing each number along the way to see if it's a power of two. The further you get into the sequence, the more expensive each call to rest is.
Given that it's the test input, and you can't change it, I think you need to re-think your approach. Rather than calling (mapv rest coll), you probably only want to call rest on the sequence with the smallest first value.
I wrote a piece of code to count the leading hash(#) character of a line, which is much like a heading line in Markdown
### Line one -> return 3
######## Line two -> return 6 (Only care about the first 6 characters.
Version 1
(defn
count-leading-hash
[line]
(let [cnt (count (take-while #(= % \#) line))]
(if (> cnt 6) 6 cnt)))
Version 2
(defn
count-leading-hash
[line]
(loop [cnt 0]
(if (and (= (.charAt line cnt) \#) (< cnt 6))
(recur (inc cnt))
cnt)))
I used time to measure both tow implementations, found that the first version based on take-while is 2x faster than version 2. Taken "###### Line one" as input, version 1 took 0.09 msecs, version 2 took about 0.19 msecs.
Question 1. Is it recur that slows down the second implementation?
Question 2. Version 1 is closer to functional programming paradigm , is it?
Question 3. Which one do you prefer? Why? (You're welcome to write your own implementation.)
--Update--
After reading the doc of cloujure, I came up with a new version of this function, and I think it's much clear.
(defn
count-leading-hash
[line]
(->> line (take 6) (take-while #(= \# %)) count))
IMO it isn't useful to take time measurements for small pieces of code
Yes, version 1 is more functional
I prefer version 1 because it is easier to spot errors
I prefer version 1 because it is less code, thus less cost to maintain.
I would write the function like this:
(defn count-leading-hash [line]
(count (take-while #{\#} (take 6 line))))
No, it's the reflection used to invoke .charAt. Call (set! *warn-on-reflection* true) before creating the function, and you'll see the warning.
Insofar as it uses HOFs, sure.
The first, though (if (> cnt 6) 6 cnt) is better written as (min 6 cnt).
1: No. recur is pretty fast. For every function you call, there is a bit of overhead and "noise" from the VM: the REPL needs to parse and evaluate your call for example, or some garbage collection might happen. That's why benchmarks on such tiny bits of code don't mean anything.
Compare with:
(defn
count-leading-hash
[line]
(let [cnt (count (take-while #(= % \#) line))]
(if (> cnt 6) 6 cnt)))
(defn
count-leading-hash2
[line]
(loop [cnt 0]
(if (and (= (.charAt line cnt) \#) (< cnt 6))
(recur (inc cnt))
cnt)))
(def lines ["### Line one" "######## Line two"])
(time (dorun (repeatedly 10000 #(dorun (map count-leading-hash lines)))))
;; "Elapsed time: 620.628 msecs"
;; => nil
(time (dorun (repeatedly 10000 #(dorun (map count-leading-hash2 lines)))))
;; "Elapsed time: 592.721 msecs"
;; => nil
No significant difference.
2: Using loop/recur is not idiomatic in this instance; it's best to use it only when you really need it and use other available functions when you can. There are many useful functions that operate on collections/sequences; check ClojureDocs for a reference and examples. In my experience, people with imperative programming skills who are new to functional programming use loop/recur a lot more than those who have a lot of Clojure experience; loop/recur can be a code smell.
3: I like the first version better. There are lots of different approaches:
;; more expensive, because it iterates n times, where n is the number of #'s
(defn count-leading-hash [line]
(min 6 (count (take-while #(= \# %) line))))
;; takes only at most 6 characters from line, so less expensive
(defn count-leading-hash [line]
(count (take-while #(= \# %) (take 6 line))))
;; instead of an anonymous function, you can use `partial`
(defn count-leading-hash [line]
(count (take-while (partial = \#) (take 6 line))))
edit:
How to decide when to use partial vs an anonymous function?
In terms of performance it doesn't matter, because (partial = \#) evaluates to (fn [& args] (apply = \# args)). #(= \# %) translates to (fn [arg] (= \# arg)). Both are very similar, but partial gives you a function that accepts an arbitrary number of arguments, so in situations where you need it, that's the way to go. partial is the λ (lambda) in lambda calculus. I'd say, use what's easier to read, or partial if you need a function with an arbitrary number of arguments.
Micro-benchmarks on the JVM are almost always misleading, unless you really know what you're doing. So, I wouldn't put too much weight on the relative performance of your two solutions.
The first solution is more idiomatic. You only really see explicit loop/recur in Clojure code when it's the only reasonable alternative. In this case, clearly, there is a reasonable alternative.
Another option, if you're comfortable with regular expressions:
(defn count-leading-hash [line]
(count (or (re-find #"^#{1,6}" line) "")))