Comments on: SICP section 1.2.2

By: eliben

eliben — Tue, 13 May 2008 16:27:02 +0000

Anh,

My solution gives correct values for all the elements inside the triangle. As far as my understanding goes, there is no required value for elements outside the triangle, and the exercise in SICP doesn’t enforce anything specific.

Your solution is shorter and gives 0 for outside the triangle.

By: Anh

Anh — Wed, 23 Apr 2008 03:23:40 +0000

sorry somehow it got cut off, and my x = your col and vice versa
(define (pascal x y) (cond ((= 1 x) 1) ((> x y) 0) (else (+ (pascal (- x 1) (- y 1)) (pascal x (- y 1))))))

By: Anh

Anh — Wed, 23 Apr 2008 03:16:08 +0000

Hi,

I just started reading SICP as well. Looks like your version of the pascal function gives some wrong answers, e.g. (pascal 2 4) gives 2 instead of 0.

Here’s mine in Scheme

(define (pascal x y)
(cond ((= 1 x) 1)

By: eliben

eliben — Fri, 06 Jul 2007 09:25:38 +0000

It is necessary for the case of a wrong column in row 1 (say, column 2). Without the test, it goes into an infinite loop (instead of just giving a wrong answer)

By: Jaime

Jaime — Fri, 06 Jul 2007 08:57:42 +0000

Hi Eli,

i’ve been doing these exercises on paper since i don’t have a scheme interpreter right here. For the Pascal’s Triangle i came up with a solution quite similar to yours (besides that i start counting from zero). The only big difference is the initial test, is it necessary to test that you’re in the first row? I mean, the first level matches the other 2 conditions (first col and row = col).