Suggested Answer
February 3, 2010
1. Consider a general room assignment problem with
N
students and
N
rooms. Answer the following
questions.
(a) Suppose that the initial assignment of rooms is already eﬃcient. Show that TTC algorithm leads to
the same allocation of rooms, i.e. every student ends up with his or her initially assigned room.
Deﬁnition:
An allocation
x
∈
X
is
eﬃcient
if there is no allocation
y
∈
X
such that
(1)
y
j
%
j
x
j
for all
j
∈
N
, and
(2)
y
j
±
j
x
j
for some
j
.
Answer: Shorter Proof
Assume that the TTC algorithm leads to another allocation
x
0
∈
X
(then show contradiction later)
where
x
0
6
=
x
. Since the TTC algorithm leads to a unique core allocation which is individually
rational (see Lecture Note of Room Assignment p.2226) we have
x
0
i
²
i
x
i
for all
i
∈
N
(deﬁnition of
individually rationality). However, since we assumed
x
0
6
=
x
, we must have
x
0
i
±
i
x
i
for some
i
∈
N
.
This contradicts the eﬃciency of allocation
x
.
Answer: Longer Proof
Assume that the TTC algorithm leads to another allocation
x
0
∈
X
. (then show contradiction later)
Also assume that the initial allocation is (
h
1
,h
2
,...,h
N

1
,h
N
)
We know that in each step of TTC algorithm, it creates
(1) trading cycle(s),
(2) selftrading cycle(s), or
(3) both of (1) and (2)
Since
x
0
6
=
x
, in some step(s) of TTC algorithm, there must be at least one trading cycle (otherwise
x
0
=
x
). In such a trading cycle, a group of members (
k
1
,k
2
,...,k
J
) creates a trading cycle
h
k
1
→
by
k
1
h
k
2
→
by
k
2
··· →
by
k
J

1
h
k
J
→
by
k
J
h
k
1
.
(trading cycle)
Thus, compared to their initial rooms, each of group members (
k
1
,k
2
,...,k
J
) get a strictly better
room (i.e.
x
0
i
±
i
x
i
for trading cycle students). Also, students who create selftrading cycles get their
initial assignments (i.e.
x
0
i
∼
i
x
i
for selftrading cycle students). This contradicts the eﬃciency of
allocation
x
.
1