CMPSCI 250: Introduction to Computation
David Mix Barrington
Questions and Answers on Homework Assignment #5
HW#4 due on paper in class, Monday 7 March 2011
Question text is in black, answers in blue.
- Question 5.1, posted 7 March: I have a difficulty with
the definition in Problem 4.6.3. If I look at C(6,6,5), it should be false
because it has the same value as C(1,6,5), which is false because 1 != 6. But
there is a natural r such that a = rm + b, namely r = 0. Have I got something
No, I (in the book) have something wrong. Please add the additional
condition that C(a,b,m) is FALSE whenever b ≥ m.
- Question 5.2, posted 18 March: I read the correction
above, but I'm still confused. If C(a,b,m) is always false when b
≥ m, how can it be true if and only if the natural r exists? For
example, if m = 5, a = 11, and b = 6, the natural r satisfies a = rm
+ b but the new condition makes C(11,6,5) false.
You are right -- the desired condition, that C(a,b,m) is true if and only if
that natural r exists such that a = rm + b, is now only true for b
< m and you only need to prove it in this case.
- Question 5.3, posted 22 March: In 4.7.3, do I
have to use the
No, you may use any combination of
- Question 5.4, posted 22 March: For 4.11.5, I don't see
any pattern in the lengths for n=0, n=1, and n=2.
Look further, at n=3 and n=4. (I know I didn't
draw the n=4 case in the book but you should be able to draw it.)
In order to make a recursive proof you are going to have to figure
out what the n+1 case's path has to do with the n case's path. The
new path is roughly twice as long as the old one -- can you
Last modified 22 March 2011