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.
Related
This is a simple attempt to reproduce some code that Ross Ihaka gave as an example of poor R performance. I was curious as to whether Clojure's persistent data structures would offer any improvement.
(https://www.stat.auckland.ac.nz/~ihaka/downloads/JSM-2010.pdf)
However , I'm not even getting to first base, with a Stack Overflow reported, and not much else to go by. Any ideas? Apologies in advance if the question has an obvious answer I've missed...
; Testing Ross Ihaka's example of poor R performance
; against Clojure, to see if persisntent data structures help
(def dd (repeat 60000 '(0 0 0 0)))
(defn repl-row [d i new-r]
(concat (take (dec i) d) (list new-r) (drop i d)))
(defn changerows [d new-r]
(loop [i 10000
data d]
(if (zero? i)
data
(let [j (rand-int 60000)
newdata (repl-row data j new-r)]
(recur (dec i) newdata)))))
user=> (changerows dd '(1 2 3 4))
StackOverflowError clojure.lang.Numbers.isPos (Numbers.java:96)
Further, if anyone has any ideas how persistent functional data structures can be used to best advantage in the example above, I'd be very keen to hear. The speedup reported not using immutable structures (link above) was about 500%!
Looking at the stack trace for the StackOverflowError, this seems to be an "exploding thunk" (lazy/suspended calculation) problem that isn't obviously related to the recursion in your example:
java.lang.StackOverflowError
at clojure.lang.RT.seq(RT.java:528)
at clojure.core$seq__5124.invokeStatic(core.clj:137)
at clojure.core$concat$cat__5217$fn__5218.invoke(core.clj:726)
at clojure.lang.LazySeq.sval(LazySeq.java:40)
at clojure.lang.LazySeq.seq(LazySeq.java:49)
at clojure.lang.RT.seq(RT.java:528)
at clojure.core$seq__5124.invokeStatic(core.clj:137)
at clojure.core$take$fn__5630.invoke(core.clj:2876)
Changing this line to realize newdata into a vector resolves the issue:
(recur (dec i) (vec newdata))
This workaround is to address the use of concat in repl-row, by forcing concat's lazy sequence to be realized in each step. concat returns lazy sequences, and in your loop/recur you're passing in the lazy/unevaluated results of previous concat calls as input to subsequent concat calls which returns more lazy sequences based on previous, unrealized lazy sequences. The final concat-produced lazy sequence isn't realized until the loop finishes, which results in a stack overflow due its dependence on thousands of previous concat-produced lazy sequences.
Further, if anyone has any ideas how persistent functional data structures can be used to best advantage in the example above, I'd be very keen to hear.
Since it seems the usage of concat here is to simply replace an element in a collection, we can get the same effect by using a vector and assoc-ing the new item into the correct position of the vector:
(def dd (vec (repeat 60000 '(0 0 0 0))))
(defn changerows [d new-r]
(loop [i 10000
data d]
(if (zero? i)
data
(let [j (rand-int 60000)
newdata (assoc data j new-r)]
(recur (dec i) newdata)))))
Notice there's no more repl-row function, we just assoc into data using the index and the new value. After some rudimentary benchmarking with time, this approach seems to be many times faster:
"Elapsed time: 75836.412166 msecs" ;; original sample w/fixed overflow
"Elapsed time: 2.984481 msecs" ;; using vector+assoc instead of concat
And here's another way to solve it by viewing the series of replacements as an infinite sequence of replacement steps, then sampling from that sequence:
(defn random-replace [replacement coll]
(assoc coll (rand-int (count coll)) replacement))
(->> (iterate (partial random-replace '(1 2 3 4)) dd)
(drop 10000) ;; could also use `nth` function
(first))
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)
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.
I am learning Clojure and trying to solve Project's Euler (http://projecteuler.net/) problems using this language.
Second problem asks to find the sum of the even-valued terms in Fibonacci sequence whose values do not exceed four million.
I've tried several approaches and would find next one most accurate if I could find where it's broken. Now it returns 0. I am pretty sure there is a problem with take-while condition but can't figure it out.
(reduce +
(take-while (and even? (partial < 4000000))
(map first (iterate (fn [[a b]] [b (+ a b)]) [0 1]))))
To compose multiple predicates in this way, you can use every-pred:
(every-pred even? (partial > 4000000))
The return value of this expression is a function that takes an argument and returns true if it is both even and greater than 4000000, false otherwise.
user> ((partial < 4000000) 1)
false
Partial puts the static arguments first and the free ones at the end, so it's building the opposite of what you want. It is essentially producing #(< 4000000 %) instead of #(< % 4000000) as you intended, So just change the > to <:
user> (reduce +
(take-while (and even? (partial > 4000000))
(map first (iterate (fn [[a b]] [b (+ a b)]) [0 1]))))
9227464
or perhaps it would be more clear to use the anonymous function form directly:
user> (reduce +
(take-while (and even? #(< % 4000000))
(map first (iterate (fn [[a b]] [b (+ a b)]) [0 1]))))
9227464
Now that we have covered a bit about partial, let's break down a working solution. I'll use the thread-last macro ->> to show each step separately.
user> (->> (iterate (fn [[a b]] [b (+ a b)]) [0 1]) ;; start with the fibs
(map first) ;; keep only the answer
(take-while #(< % 4000000)) ;; stop when they get too big
(filter even?) ;; take only the even? ones
(reduce +)) ;; sum it all together.
4613732
From this we can see that we don't actually want to compose the predicates evan? and less-than-4000000 on a take-while because this would stop as soon as either condition was true leaving only the number zero. Rather we want to use one of the predicates as a limit and the other as a filter.
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) "")))