Question

#lang eopl

(define (vectorSum V b e)    ; b is starting index, and e is ending index
  (cond
    [ (eqv? b e) vector-ref V b]
    [ (> b e)
      (eopl:error 'vectorSum "starting index must be smaller than or equal to the end index")]
    [ else (+ (vector-ref V b) (vectorSum V (+ b 1) e))]))


(define A #(1 1 1 1 1))

When I try this, I am getting the wrong result. What’s the problem here?

> (vectorSum A 0 4)
8
> (vectorSum A 0 1)
2
> (vectorSum A 0 3)
6
> (vectorSum A 1 3)
5
> (vectorSum A 1 2)
3

> (vectorSum A 0 1)
2
> (vectorSum A 1 2)
3

Take (vectorSum A 0 3), when I expanded the recursion, I thought it was supposed to be

+ 1 + VectorSum (1 3)
    + 1 + VectorSum (2, 3)
        + 1 + VectorSum (3, 3)
            + 1   (I hit the first case, there is no more recursion)
= 4

Instead, I get 6. Why?

Thanks.

---

Look at 0,1 and 1,2 the answers aren’t equal.

Answer 1

Your answer should look like this:

(define (vectorSum V b e)
  (cond ((eqv? b e)
         (vector-ref V b))
        ((> b e)
         (eopl:error 'partialVectorSum "starting index must be smaller than or equal to the end index"))
        (else (+ (vector-ref V b) (vectorSum V (+ b 1) e)))))

It was a simple mistake – you forgot a couple of parenthesis in this line:

[ (eqv? b e) vector-ref V b]

It should have been:

[ (eqv? b e) (vector-ref V b) ]

Without those parenthesis, you’re not actually calling the vector-ref procedure, instead you’re listing some symbols and returning the final one, b in this case. Remember to always surround a procedure call and its arguments between parenthesis, just as you did in the else part.