Recursively Add to list in Clojure

Question

I'm trying to recursively append a list in clojure. It is to deconstruct a POW function that I created, turning (POW x 3) into (* x (* x x))

I'm a really novice Clojure programmer, and attempting to wrap my head around this problem in Clojure is kind of tough. I have come up with:

(defn do-it [x n]
  (if (= n 0)
  (println x)
  ((dec n) (if (= x 'x))
     (list '* 'x 'x)
     (list '* 'x x))))

Which will not compile or run, but that is where my plan is going. My idea is to decrease n every time you add another (* x to the list.

I have this, which is similar to what I'm trying to do, but does not implement the POW idea into the function:

(defn do-it [x]
  (if (= x 'x)
    (list '* 'x 'x)
    (list '* 'x x)))

Am I on the right path with the first piece of code? Is what I'm trying even possible?

Do you want to get expression or value of your result?

mishadoff
– mishadoff

2012-11-06 00:22:43 +00:00
Commented Nov 6, 2012 at 0:22 — mishadoff
– mishadoff, Commented Nov 6, 2012 at 0:22
Expression, if I understood your question/what you meant.

thisisnotabus
– thisisnotabus

2012-11-06 00:31:43 +00:00
Commented Nov 6, 2012 at 0:31 — thisisnotabus
– thisisnotabus, Commented Nov 6, 2012 at 0:31

amalloy · Accepted Answer · 2012-11-06 02:29:37Z

4

Your posted efforts are pretty far off the mark, I'm afraid. If I were going to write pow by hand, it would look like this:

(defn pow [x n]
  (if (= n 1)
    x
    (list '* x (pow x (dec n)))))

answered Nov 6, 2012 at 2:29

amalloy

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Comments

mishadoff · Accepted Answer · 2012-11-06 00:37:38Z

1

You can simplify problem if notice that prefix notation can take various number of arguments. Then your function looks like this:

(defn power [x n]
  (cons '* (repeat n x)))

(power 2 3) => (* 2 2 2)
(power 'x 2) => (* x x)

Zero and One base cases covered also, because:

(power 2 1) => (* 2) ; valid result, produces 2
(power 2 0) => (*) ; valid result, produces 1

answered Nov 6, 2012 at 0:37

mishadoff

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2 Comments

thisisnotabus Over a year ago

the input requires it only take two arguments. Because of this, (* x x x) is invalid. It needs to be (* x (* x x))

mishadoff Over a year ago

if it invalid because of condition "only take two arguments" use @amalloy solution.

mobyte · Accepted Answer · 2012-11-06 05:14:44Z

0

(defn do-it [x n]
  (if (= 1 n)
    x
    (reduce (fn [chain _] `(~'* ~chain))
            `(~'* ~x ~x)
            (range 2 n))))

(do-it 5 10) ->
(* (* (* (* (* (* (* (* (* 5 5)))))))))

(do-it 5 1) ->
5

answered Nov 6, 2012 at 5:14

mobyte

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