COMP360 Problem Set 4 Solution
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Backtracking: Vertex Coloring(10pts)
The following is the search process:
1. A: assign color 1.
2. B: B is connected with A, so assign color 2.
3. C: C is connected with B, so assign color 1.
4. D: D is connected with C, so assign color 2.
5. E: E is connected with C,D, so assign color 3.
6. F, F is connected with A,E, so assign color 2.
7. G, G is connected with A,B,F, so assign color 3.
8. H, H is connected with C,G, so assign color 2.
So it is three colorable.
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2.1
Local Search: Load Balancing(20pts)
(a)
Consider a progress measure Φ defined as the sum of the squares of the loads
on all machines. We claim that after an improving swap move, this quantity
must strictly decrease. Indeed, suppose we execute an improving swap move
on machines Mi and Mj ; suppose the loads on Mi and Mj before the swap
are Ti and Tj respectively, and the loads after the swap are Ti ′ and Tj ′ . Then
the change of Φ is −Ti2 − Tj2 + (Ti ′ )2 + (Ti ′ )2 ; since Ti + Tj = Ti ′ + Tj ′ , and
max(Ti ′ , Tj ′ ) < max(Ti , Tj ), the change is always negative: in other words, Φ
strictly decreases. Since there are only a finite number of ways to partition jobs
across machines, Φ can only decrease a finite number of times. That implies
that the algorithm terminates.
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2.2
(b)
Consider the machine Mi with the maximum load at the end of the algorithm.
If Mi contains a single job, then since this job has to go somewhere in any
solution, the makespan of our solution is the same as the optimal makespan.
Otherwise, Mi has at least two jobs on it. We now claim that the load on Mi
is at most twice the load on any other machine Mj . Since this applies to the
machine Mj of minimum load, it implies that Mi is at most twice the average.
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t . But the optimal makespan is at least the average, therefore this gives
m ∑r r
a factor 2 approximation. Now let’s prove the claim. Suppose Ti > 2Tj . Since
Mi has at least two jobs, the lightest job r on Mi has size tr ≤ 21 Ti . So consider
the swap move in which job r simply moves to Mi . If Ti ′ and Tj ′ are the loads
after this move, we have Ti ′ < Ti and Tj ′ = Tj + tr < 21 Ti + 12 Ti = Ti . Thus
max(Ti ′ , Tj ′ ) < max(Ti , Tj ), so this is an improving swap move, contradicting
the termination of the algorithm.
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Circle radius
Assume y is furthest point of x ∈ P . Since the distance between any other
point to x is no greater than r, our solution is feasible. On the other hand since
r is the distance between x and y, any circle that contains both x and y should
have diameter at least r, which means the radius is at least r/2. Thus, this is a
2-approximation algorithm.
4
Knapsack(15pts)
If we can fit all items into the knapsack, we clearly have obtained an optimal
solution. So let’s assume we can fit in only k < n items. If there is no space left
in the knapsack, since we have been packing only the items with the highest
value per unit space, this is again an optimal solution. WLOG, let’s assume
that we have some leftover space W − V in the end. If our algorithm can take a
−V
fraction of an item, then by adding W
v
to our knapsack value, we would
wk+1 k+1
either match or exceed OPT(unable to take fractional items). Therefore, either
k−1
−V
v
≥ 21 OPT. So this is a 2-approximation
∑i=1 vi ≥ 21 OPT or vk+1 ≥ W
wk+1 k+1
algorithm.
5
TSP(20pts)
Let ∆-TSP-R be the metric TSP problem such that the tour can visit a city
for more than one time. In class, we have argued that the optimal solution for
∆-TSP-R is the same as ∆-TSP. So if you have a c-approximation algorithm for
TSP-R, it is also a c-approximation algorithm for ∆-TSP-R. Remember that
we show in class that one can take any feasible solution of ∆-TSP-R and turn
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it into a feasible solution of ∆-TSP without increasing the cost of the tour by
shortcutting. So this is also a c-approximation algorithm for ∆-TSP.
For the other direction, let’s consider the following algorithm: on input an
instance (X, d) of TSP-R, we compute the new distance function d′ (x, y) as the
length of a shortest path from x to y in a weighted graph that has vertex set X
and weights d(x, y). Note that d′ (x, y) is a distance function that satisfies the
triangle inequality. We also compute the shortest path between any pair x and
y.
We then pass the input (X, d′ ) to our c-approximation algorithm for ∆-TSP,
and find a cycle C ′ such that
costd′ (C ′ ) ≤ c ⋅ OPT∆−T SP (X, d′ )
Note that, for every pair of points (x, y), we have d′ (x, y) ≤ d(x, y) and so
OPT∆−T SP (X, d′ ) = OPT∆−T SP −R (X, d′ ) ≤ OPTT SP −R (X, d).
Finally, we construct a cycle C by replacing each transition x → y in C ′ by
the shortest path from x to y according to d(x,y). Because of the definition of
d′ (x,y), we have
costd (C) = costd′ (C ′ )
and, combining the inequalities we have proved so far,
costd (C) ≤ c ⋅ OPTT SP −R (X, d)
Therefore, we have argued that given a c-approximation algorithm for ∆-TSP,
we can construct a c-approximation algorithm for TSP-R.
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