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help understanding simple recursion

PostPosted: Fri May 13, 2016 10:49 am
by philV
I am long time programmer wanting to learn LISP (Clojure), watching the 1986 MIT class on the "Structure and Interpretation of Computer Programs" on youtube. 23 minutes 56 seconds into the lecture 1B ;;

He gives a scheme recursive program to add 2 numbers, any insight would be appreciated!

Scheme version:

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   (IF (= X 0)
     (1+ (+ (-1+ X) Y))))

(I've translated to Clojure and added some print statements).

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(defn z [x y]
  (if (= x 0)
    (do (println "x is zero x: " x " y: " y)
    (do (println "else x: " x " y: " y)
      (inc (z (dec x) y)))))

here is the output:

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user> (z 4 9)
else x:  4  y:  9
else x:  3  y:  9
else x:  2  y:  9
else x:  1  y:  9
x is zero x:  0  y:  9

This I do not understand: the code says (= x 0) then y. When x is 0 y is 9 not 13!

Re: help understanding simple recursion

PostPosted: Sat May 14, 2016 2:13 pm
by David Mullen
Repeatedly printing X and Y doesn't reveal the whole process, whereas explicit nesting might make things clearer. The call

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(z 4 9)

can be re-expressed, via the function definition, as:

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(inc (z (dec 4) 9))

Evaluating the innermost form, this becomes:

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(inc (z 3 9))

Rewriting the call to Z, like we did before, we have:

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(inc (inc (z (dec 3) 9)))

And so on—manually evaluating the innermost form every time:

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   (inc (inc (z 2 9)))
=> (inc (inc (inc (z (dec 2) 9))))
=> (inc (inc (inc (z 1 9))))
=> (inc (inc (inc (inc (z (dec 1) 9)))))
=> (inc (inc (inc (inc (z 0 9)))))
=> (inc (inc (inc (inc 9))))
=> (inc (inc (inc 10)))
=> (inc (inc 11))
=> (inc 12)
=> 13

Re: help understanding simple recursion

PostPosted: Sat May 14, 2016 4:17 pm
by philV
Thanks for replying! I understand your walk through of the recursion, it is obviously correct, but seems quite counter-intuitive.

At some point state is (inc (inc (inc (inc (z 0 9))))) and then the "If" resolves to y, which seemed to me to be the culmination and the answer is 9! Since the other branch of the "if" will never again be reached it appears to me to it would therefore be of no further interest. But in fact the evaluation of that unreachable branch does continue as you show below. Both branches of the "If" are resolved to a primitive even though one is unreachable from the code!