MATH 2056 Discrete Mathematics II, Fall 2015
October 8, 2015
ASSIGNMENT #4 (Solution)
Problem #8, p. 302: Find the first four terms of the recursively defined sequence
vk = vk−1 + vk−2 + 1, for all integers k ≥ 3
v1 = 1, v2 = 3
Solution:
v3 = v2 + v1 + 1 = 3 + 1 + 1 = 5,
v4 = v3 + v2 + 1 = 5 + 3 + 1 = 9.
The first four terms of the sequence are 1, 3, 5, 9.
Problem #12, p. 302: Let s0 , s1 , s2 , . . . be defined by the formula sn =
Show that the sequence satisfies the recurrence relation
sk =
(−1)n
for all integers n ≥ 0.
n!
−sk−1
.
k
Solution:
−sk−1
=−
k
(−1)k−1
(k−1)!
k
(−1)k−1
=−
k · (k − 1)!
(−1)k−1
=−
k!
(−1)k−1
= (−1) ·
k!
(−1)k
=
k!
= sk
by the definition of sk−1
rewriting the fraction in equivalent form
by the properties of the factorial
factoring out − 1
by the properties of the exponents
by the definition of sk
Problem #16, p. 302: Use the recurrence relation and values for the Tower of Hanoi sequence m1 , m2 , m3 , . . .
discussed in Example 5.6.5 to compute m7 and m8 .
Solution:
m1 = 1,
m2 = 3,
m3 = 7,
m4 = 2m3 + 1 = 2 · 7 + 1 = 15,
m5 = 2m4 + 1 = 2 · 15 + 1 = 31,
m6 = 2m5 + 1 = 2 · 31 + 1 = 63,
m7 = 2m6 + 1 = 2 · 63 + 1 = 127,
m8 = 2m7 + 1 = 2 · 127 + 1 = 255.
MATH 2056 Discrete Mathematics II, Fall 2015
October 8, 2015
Problem #21, p. 303: Double Tower of Hanoi: In this variation of the Tower of Hanoi there are three
poles in a row and 2n disks, two of each of n different sizes, where n is any positive integer. Initially one of
the poles contains all the disks placed on top of each other in pairs of decreasing size. Disks are transferred
one by one from one pole to another, but at no time may a larger disk be placed on top of a smaller disk.
However, a disk can be placed on top of one of the same size. Let tn be the minimum number of moves
needed to transfer a tower of 2n disks from one pole to another.
a. Find t1 and t2 .
b. Find t3 .
c. Find a recurrence for t1 , t2 , t3 , . . ..
Solution:
a. There are 2 disks only, of the same size, when n = 1. Therefore, we can move both directly to the
desired pole. Hence, t1 = 2. (Note that we cannot accomplish this in less than two moves, since there
are 2 disks to be moved, and each move involves one disk only.)
When n = 2, we have 4 disks, 2 larger and 2 smaller. Say, the disks are on pole A and we want to
move them to pole C. The following sequence of moves, using the notation used in class, accomplishes
the task with 6 moves:
A→B
A→B
A→C
A→C
B→C
B→C
To argue that 6 is the minimal number of moves, we have to consider (the fact that) the movement of
the larger disks is not possible before we remove the smaller disks from the initial pole. This requires
at least 2 moves. Further, moving the larger disks to the desired pole requires (at least) 2 moves.
Then, stacking the smaller disks on top of the larger ones requires additional (at least) 2 moves. To
summarize, we have proven that t2 = 6.
b. t3 = 14. To establish this, reason as above. The two largest disks cannot be moved before 4 disks on
top of them are removed from the pole. This requires at least 4 moves. However, with 4 moves we are
going to occupy both other poles, and we still cannot move the largest disks from their initial position.
Thus, we not only have to remove the 4 top disks from the initial pole, but to have them on one of
the other poles. This, as proven in a. requires at least 6 moves. Then, we move the 2 largest disks
to the desired pole in 2 moves (minimal, cannot be done in 1 move), and then we have to move the
4 smaller disks on top of the two largest, which requires no less that 6 moves, as already argued. We
have a total of 14 moves.
c. It is clear from the reasoning in b. that the recurrence is tn = 2tn−1 + 2.