From: Ramakrishnan Muthukrishnan <vu3rdd@gmail.com>
Date: Fri, 17 Sep 2010 14:13:01 +0000 (+0530)
Subject: solution to 2.72
X-Git-Url: https://git.rkrishnan.org/pf/reliability?a=commitdiff_plain;h=0064ce01edcdd2463864e7b3dfcb905962e41f0f;p=sicp.git

solution to 2.72
---

diff --git a/src/sicp/ex2_72.rkt b/src/sicp/ex2_72.rkt
new file mode 100644
index 0000000..e9c06da
--- /dev/null
+++ b/src/sicp/ex2_72.rkt
@@ -0,0 +1,22 @@
+#lang racket
+
+#|
+Q:
+
+Consider the encoding procedure that you designed in exercise 2.68. What is the order of growth
+in the number of steps needed to encode a symbol? Be sure to include the number of steps needed
+to search the symbol list at each node encountered. To answer this question in general is difficult. 
+
+Consider the special case where the relative frequencies of the n symbols are as described in 
+exercise 2.71, and give the order of growth (as a function of n) of the number of steps needed
+to encode the most frequent and least frequent symbols in the alphabet. 
+|#
+
+#|
+A.
+
+encode-symbol, has to go down to the last level, n is the depth of the tree, where n
+is the number of symbols. On every node, it cons'es a number. Also there is the member? function
+which spends max of n steps to find out if a given symbol is a member of the set. This is done for
+every symbol in the message. So the worst case growth is of the order O(n * n).
+|#