sgignoux / sicp-solutions.net

Used to be codology.net

sicp-solutions.net (used to be codology.net)

I have renamed the site to reflect the fact that it seems now only about providing my solutions to the SICP books.

Static site by: https://gohugo.io

Theme:

• https://github.com/nodejh/hugo-theme-cactus-plus
• https://github.com/olOwOlo/hugo-theme-even

sicp-solutions.net's Issues

Fix graphviz plugin

The plugin to get Graphviz is broken after migrating to new themes.

Example:

• https://codology.net/post/sicp-solution-exercise-1-14/

Ex 2.1 Bug

The solution has a bug. For the reason, check here. In short, you need to use abs on the result of gcd.

Ex 1.10

Both g(n) and h(n) should be piecewise functions where

• g(n) = 2 ^ n (when n > 0); 0 (when n = 0);
• h(n) = 2 ^ h(n-1) (when n > 1); 2 (when n = 1); 0 (when n = 0)

Will you have more solutions

Hi, will there be more solutions for the remaining exercises?
These are very nice and useful, and I want to thank you for these nice solutions so far @sgignoux

mistake in your solution of the Exercise 1.13

Dear Sébastien,
I believe I have found a mistake in your solution of the Exercise 1.13
from SICP.

In the proof, you state that 0 < ψ < 1. Since ψ = (1 - sqrt(5)) / 2 ≃
−0.62, this is clearly not the case. You should add an absolute value to
the proof.

Exercise 1.14 Typo

Towards the end of the solution for 1.14 (below the text "by continuing to apply the formulas") it is stated that the summands should be:

T(n-10i,1) for T(n,3)
T(n-25i,1) for T(n,4)
T(n-50i, 1) for T(n,5)

But it seems it should be:

T(n-10i,2) for T(n,3)
T(n-25i,3) for T(n,4)
T(n-50i,4) for T(n,5)

Ex 1.13 - concluding statement.

Your solution to ex 1.13 helped me a lot to understand it; thank you.

In your concluding statement

This show that φ^n/√5 will never be further away from the value of Fib(n) than 1

I would change this to

This shows that φ^n/√5 will never be further away from the value of Fib(n) than 1/2

Since the previous statement bounds the value of φ^n/√5 to Fib(n) +/- 1/2

Exercise 2.10: off by one

Interval boundaries should be checked as well:

#;1> (div-interval (make-interval 1 2) (make-interval 22 0))

Error: (/) division by zero
1.0
0

        Call history:

        <eval>    [div-interval] (lower-bound y)
        <eval>    [lower-bound] (min (car i) (cdr i))
        <eval>    [lower-bound] (car i)
        <eval>    [lower-bound] (cdr i)
        <eval>    [div-interval] (mul-interval x (make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y))))
        <eval>    [div-interval] (make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y)))
        <eval>    [div-interval] (/ 1.0 (upper-bound y))
        <eval>    [div-interval] (upper-bound y)
        <eval>    [upper-bound] (max (car i) (cdr i))
        <eval>    [upper-bound] (car i)
        <eval>    [upper-bound] (cdr i)
        <eval>    [div-interval] (/ 1.0 (lower-bound y))
        <eval>    [div-interval] (lower-bound y)
        <eval>    [lower-bound] (min (car i) (cdr i))
        <eval>    [lower-bound] (car i)
        <eval>    [lower-bound] (cdr i) <--

Exercise 1.6

For some reason, it doesn't go into an infinite loop for me, ( I use the Pretty-big interpreter in DrRacket)
and the result is almost the same except for the natural numbers that have a natural square root, s.t (sqrt 4)
with normal if (sqrt 4) gives 2.0000xxxxxx which x's represent some garbage of numbers
but with the If procedure, it produces the natural number 2 w/o any fractions

Typo Error in Exercise 1.36

I think this is just a minor typo issue, but I noticed you were using (log 100) instead of (log 1000) as stated in the problem which is changing the results.

Link:
https://github.com/albondad/codology.net/blob/master/content/post/SICP-Solution-Exercise-1-36.md

Error in Ex2.6 Solution - Markdown

You gave the following solution :
(define one (lambda (f) (lambda (x) (f (f x)))))

But this is infact 2 in Church numerals.
You can check the correct solution from the cummunity-scheme-wiki : http://community.schemewiki.org/?sicp-ex-2.6

Typo in solution 1.14

Hi, thank you for making your solutions and work available. It is nice having your insight available as I work through the wizard book!

In your solution for 1.14 you have a slight typo in the below section

Let's break it down:

- There are 4 green nodes for `(c x 2)` corresponding to how many time you can subtract a dime from 12, plus one.
- Then for each of the green node, there is the option of using only dime, which is the case that we looked at first.

You say dime, but in the example you are showing nickels.