SICP Exercise 2.92: Quick Update

If you remember, back in what I called the Naïve Approach, I gave an example subtraction of two polynomials:

((5y2 + 2y - 1)x2 + (2y2 + y + 2)x - 3) - ((5x2 + 2x)y2 + (2x2 + x)y - (x2 - 2x + 5))

...and showed how the naïve approach (i.e. expressing one polynomial in the same indeterminate as the other by simply creating a new polynomial of the required indeterminate with only a zero-order term with the coefficient being the first polynomial) failed to correctly simplify the results, giving:

((5y2 + 2y - 1)x2 + (2y2 + y + 2)x - ((5x2 + 2x)y2 + (2x2 + x)y - (x2 - 2x - 2))

...instead of:

2

Well, here's the output from my Scheme interpreter this morning:

> (define poly-1
    (make-polynomial-from-coeffs
     'x
     (list (make-polynomial-from-coeffs
            'y
            (list (make-integer 5) (make-integer 2) (make-integer -1)))
           (make-polynomial-from-coeffs
            'y
            (list (make-integer 2) (make-integer 1) (make-integer 2)))
           (make-integer -3))))
> (define poly-2
    (make-polynomial-from-coeffs
     'y
     (list (make-polynomial-from-coeffs
            'x
            (list (make-integer 5) (make-integer 2) zero))
           (make-polynomial-from-coeffs
            'x
            (list (make-integer 2) (make-integer 1) zero))
           (make-polynomial-from-coeffs
            'x
            (list (make-integer -1) (make-integer 2) (make-integer -5))))))
> (sub poly-1 poly-2)
(integer . 2)

Write-up to follow. Unfortunately I'm away this weekend, so it may not be posted until sometime next week. However, I'm pretty chuffed I've finally found the time to finish this exercise off!