Date: Apr 12, 2013 2:15 AM
Author: David Park
Subject: Re: "Programming With Mathematica" Exercise help

Hold[(z = 11; a = 9; z + 3)] /. z -> a
ReleaseHold[%]

Hold[a = 11; a = 9; a + 3]
12


David Park
djmpark@comcast.net
http://home.comcast.net/~djmpark/index.html



From: plank.in.sequim@gmail.com [mailto:plank.in.sequim@gmail.com]


I'm loving Paul Wellin's book "Programming with Mathematica: An
Introduction" and am trying to diligently do all the exercises. Most of
them have answers in the back but I'm stuck on Section 4.2, Exercise 2 and
there's no answer given. It gives the following Mathematica code:

z = 11;
a = 9;
z + 3 /. z -> a

14

So "z+3" is being evaluated to 14 and then the substitution has no effect.
He asks how to "use the Hold function in the compound expression to obtain a
value of 12". I don't seem to be able to get this to work. My original
thought was to hold z+3, but then the z in the replacement part gets
evaluated so the replacement is actually 11->3 which doesn't match in the
held z+3 expression. In fact, if you replace "z+3" with Hold[11+3] then
you'll end up with Hold[9+3]. Curiously, this works differently if you use
replace.

In: Replace[Hold[11+3],11->9]
Out: Replace[Hold[11+3],11->9]

In: Hold[11+3]/.11->9
out: Hold[9+3]

I thought these two were supposed to be equivalent so I'm a bit confused
here.

In any event, I've tried all the commonsense ideas I've had and then spent
some time flailing about randomly with Hold's but nothing seems to work
correctly. Can anybody help me understand this? Thanks!

Darrell