I was wondering if someone would be able to help me create a function that would given a number would return the tribonacci number that corresponds to that number for example
10 would return 44
15 would return 927
Generally, you should show what you have already tried and then let us know what problem you are having, so we can point you to a fix. Asking Stack Overflow to do this for you from scratch is poor manners.
But, I love Clojure and these problems are fun. First of all, there are so many ways to do this, and just googling Clojure fibonacci will get you started in many many different directions for this.
Here is a quick example using reduce:
(defn tri [x]
(last (reduce
(fn [[a b c] n]
(conj [b c] (+ a b c))) [0 0 1] (range (- x 3)))))
You could also use iterate:
(nth (iterate (fn [[a b c]] [b c (+ a b c)]) [0 0 1]) 5)
This returns the 5th from the first set of tribonacci numbers, where [0 0 1] is the initial set.
To get an exact result from X'th index, also wrap it in a function as above, like so:
(defn tri-iterate [x]
(last
(nth (iterate (fn [[a b c]] [b c (+ a b c)]) [0 0 1]) (- x 3))))
You can do this in many other ways as well.
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.
A reduce call has its f argument first. Visually speaking, this is often the biggest part of the form.
e.g.
(reduce
(fn [[longest current] x]
(let [tail (last current)
next-seq (if (or (not tail) (> x tail))
(conj current x)
[x])
new-longest (if (> (count next-seq) (count longest))
next-seq
longest)]
[new-longest next-seq]))
[[][]]
col))
The problem is, the val argument (in this case [[][]]) and col argument come afterward, below, and it's a long way for your eyes to travel to match those with the parameters of f.
It would look more readable to me if it were in this order instead:
(reduceb val col
(fn [x y]
...))
Should I implement this macro, or am I approaching this entirely wrong in the first place?
You certainly shouldn't write that macro, since it is easily written as a function instead. I'm not super keen on writing it as a function, either, though; if you really want to pair the reduce with its last two args, you could write:
(-> (fn [x y]
...)
(reduce init coll))
Personally when I need a large function like this, I find that a comma actually serves as a good visual anchor, and makes it easier to tell that two forms are on that last line:
(reduce (fn [x y]
...)
init, coll)
Better still is usually to not write such a large reduce in the first place. Here you're combining at least two steps into one rather large and difficult step, by trying to find all at once the longest decreasing subsequence. Instead, try splitting the collection up into decreasing subsequences, and then take the largest one.
(defn decreasing-subsequences [xs]
(lazy-seq
(cond (empty? xs) []
(not (next xs)) (list xs)
:else (let [[x & [y :as more]] xs
remainder (decreasing-subsequences more)]
(if (> y x)
(cons [x] remainder)
(cons (cons x (first remainder)) (rest remainder)))))))
Then you can replace your reduce with:
(apply max-key count (decreasing-subsequences xs))
Now, the lazy function is not particularly shorter than your reduce, but it is doing one single thing, which means it can be understood more easily; also, it has a name (giving you a hint as to what it's supposed to do), and it can be reused in contexts where you're looking for some other property based on decreasing subsequences, not just the longest. You can even reuse it more often than that, if you replace the > in (> y x) with a function parameter, allowing you to split up into subsequences based on any predicate. Plus, as mentioned it is lazy, so you can use it in situations where a reduce of any sort would be impossible.
Speaking of ease of understanding, as you can see I misunderstood what your function is supposed to do when reading it. I'll leave as an exercise for you the task of converting this to strictly-increasing subsequences, where it looked to me like you were computing decreasing subsequences.
You don't have to use reduce or recursion to get the descending (or ascending) sequences. Here we are returning all the descending sequences in order from longest to shortest:
(def in [3 2 1 0 -1 2 7 6 7 6 5 4 3 2])
(defn descending-sequences [xs]
(->> xs
(partition 2 1)
(map (juxt (fn [[x y]] (> x y)) identity))
(partition-by first)
(filter ffirst)
(map #(let [xs' (mapcat second %)]
(take-nth 2 (cons (first xs') xs'))))
(sort-by (comp - count))))
(descending-sequences in)
;;=> ((7 6 5 4 3 2) (3 2 1 0 -1) (7 6))
(partition 2 1) gives every possible comparison and partition-by allows you to mark out the runs of continuous decreases. At this point you can already see the answer and the rest of the code is removing the baggage that is no longer needed.
If you want the ascending sequences instead then you only need to change the < to a >:
;;=> ((-1 2 7) (6 7))
If, as in the question, you only want the longest sequence then put a first as the last function call in the thread last macro. Alternatively replace the sort-by with:
(apply max-key count)
For maximum readability you can name the operations:
(defn greatest-continuous [op xs]
(let [op-pair? (fn [[x y]] (op x y))
take-every-second #(take-nth 2 (cons (first %) %))
make-canonical #(take-every-second (apply concat %))]
(->> xs
(partition 2 1)
(partition-by op-pair?)
(filter (comp op-pair? first))
(map make-canonical)
(apply max-key count))))
I feel your pain...they can be hard to read.
I see 2 possible improvements. The simplest is to write a wrapper similar to the Plumatic Plumbing defnk style:
(fnk-reduce { :fn (fn [state val] ... <new state value>)
:init []
:coll some-collection } )
so the function call has a single map arg, where each of the 3 pieces is labelled & can come in any order in the map literal.
Another possibility is to just extract the reducing fn and give it a name. This can be either internal or external to the code expression containing the reduce:
(let [glommer (fn [state value] (into state value)) ]
(reduce glommer #{} some-coll))
or possibly
(defn glommer [state value] (into state value))
(reduce glommer #{} some-coll))
As always, anything that increases clarity is preferred. If you haven't noticed already, I'm a big fan of Martin Fowler's idea of Introduce Explaining Variable refactoring. :)
I will apologize in advance for posting a longer solution to something where you wanted more brevity/clarity.
We are in the new age of clojure transducers and it appears a bit that your solution was passing the "longest" and "current" forward for record-keeping. Rather than passing that state forward, a stateful transducer would do the trick.
(def longest-decreasing
(fn [rf]
(let [longest (volatile! [])
current (volatile! [])
tail (volatile! nil)]
(fn
([] (rf))
([result] (transduce identity rf result))
([result x] (do (if (or (nil? #tail) (< x #tail))
(if (> (count (vswap! current conj (vreset! tail x)))
(count #longest))
(vreset! longest #current))
(vreset! current [(vreset! tail x)]))
#longest)))))))
Before you dismiss this approach, realize that it just gives you the right answer and you can do some different things with it:
(def coll [2 1 10 9 8 40])
(transduce longest-decreasing conj coll) ;; => [10 9 8]
(transduce longest-decreasing + coll) ;; => 27
(reductions (longest-decreasing conj) [] coll) ;; => ([] [2] [2 1] [2 1] [2 1] [10 9 8] [10 9 8])
Again, I know that this may appear longer but the potential to compose this with other transducers might be worth the effort (not sure if my airity 1 breaks that??)
I believe that iterate can be a more readable substitute for reduce. For example here is the iteratee function that iterate will use to solve this problem:
(defn step-state-hof [op]
(fn [{:keys [unprocessed current answer]}]
(let [[x y & more] unprocessed]
(let [next-current (if (op x y)
(conj current y)
[y])
next-answer (if (> (count next-current) (count answer))
next-current
answer)]
{:unprocessed (cons y more)
:current next-current
:answer next-answer}))))
current is built up until it becomes longer than answer, in which case a new answer is created. Whenever the condition op is not satisfied we start again building up a new current.
iterate itself returns an infinite sequence, so needs to be stopped when the iteratee has been called the right number of times:
(def in [3 2 1 0 -1 2 7 6 7 6 5 4 3 2])
(->> (iterate (step-state-hof >) {:unprocessed (rest in)
:current (vec (take 1 in))})
(drop (- (count in) 2))
first
:answer)
;;=> [7 6 5 4 3 2]
Often you would use a drop-while or take-while to short circuit just when the answer has been obtained. We could so that here however there is no short circuiting required as we know in advance that the inner function of step-state-hof needs to be called (- (count in) 1) times. That is one less than the count because it is processing two elements at a time. Note that first is forcing the final call.
I wanted this order for the form:
reduce
val, col
f
I was able to figure out that this technically satisfies my requirements:
> (apply reduce
(->>
[0 [1 2 3 4]]
(cons
(fn [acc x]
(+ acc x)))))
10
But it's not the easiest thing to read.
This looks much simpler:
> (defn reduce< [val col f]
(reduce f val col))
nil
> (reduce< 0 [1 2 3 4]
(fn [acc x]
(+ acc x)))
10
(< is shorthand for "parameters are rotated left"). Using reduce<, I can see what's being passed to f by the time my eyes get to the f argument, so I can just focus on reading the f implementation (which may get pretty long). Additionally, if f does get long, I no longer have to visually check the indentation of the val and col arguments to determine that they belong to the reduce symbol way farther up. I personally think this is more readable than binding f to a symbol before calling reduce, especially since fn can still accept a name for clarity.
This is a general solution, but the other answers here provide many good alternative ways to solve the specific problem I gave as an example.
I am trying to obtain the first 10 Fibonacci numbers as follows:
(take 10 (fn fibgen [a b] (cons a (fibgen b (+ a b))) 0 1))
The idea is that fibgen creates a lazy sequence (an infinite stream). Unfortunately this gives the following error:
IllegalArgumentException Don't know how to create ISeq from:user$eval10144$fibgen__10145 clojure.lang.RT.seqFrom (RT.java:528)
How can this be fixed?
I think you just made a typo. Here's what your fibgen function looks like reformatted:
(fn fibgen [a b]
(cons a (fibgen b (+ a b)))
0
1)
This function realizes the entire Fibonacci sequence starting with a and b, then returns 1. What you meant to do was define a function that returns the sequence, call it with 0 and 1, and then take the first ten items from that sequence:
(take 10 ((fn fibgen [a b] (cons a (fibgen b (+ a b)))) 0 1))
If you run this, you'll get an ArithmeticException for integer overflow, because numbers in the Fibonacci sequence quickly leave the range of a 64-bit integer. You can fix this using +':
(take 10 ((fn fibgen [a b] (cons a (fibgen b (+' a b)))) 0 1))
Since Clojure isn't lazy, this will try to realize the entire Fibonacci sequence, which will cause a StackOverflowError. Even though Clojure itself isn't lazy, though, you can create a lazy sequence, which will have basically the same effect in this case:
(take 10 ((fn fibgen [a b] (lazy-seq (cons a (fibgen b (+' a b))))) 0 1))
;;=> (0 1 1 2 3 5 8 13 21 34)
Consider a query function q that returns, with a delay, some (let say ten) results.
Delay function:
(defn dlay [x]
(do
(Thread/sleep 1500)
x))
Query function:
(defn q [pg]
(lazy-seq
(let [a [0 1 2 3 4 5 6 7 8 9 ]]
(println "q")
(map #(+ (* pg 10) %) (dlay a)))))
Wanted behaviour:
I would like to produce an infinite lazy sequence such that when I take a value only needed computations are evaluated
Wrong but explicative example:
(drop 29 (take 30 (mapcat q (range))))
If I'm not wrong, it needs to evaluate every sequence because it really doesn't now how long the sequences will be.
How would you obtain the correct behaviour?
My attempt to correct this behaviour:
(defn getq [coll n]
(nth
(nth coll (quot n 10))
(mod n 10)))
(defn results-seq []
(let [a (map q (range))]
(map (partial getq a)
(iterate inc 0)))) ; using iterate instead of range, this way i don't have a chunked sequence
But
(drop 43 (take 44 (results-seq)))
still realizes the "unneeded" q sequences.
Now, I verified that a is lazy, iterate and map should produce lazy sequences, so the problem must be with getq. But I can't understand really how it breaks my laziness...perhaps does nth realize things while walking through a sequence? If this would be true, is there a viable alternative in this case or my solution suffers from bad design?
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?