Thus, there are, just as in [14], 2(m − q) + 2n(q + 1) + 4 voters preferring c **to all other candidates, and** 2n(q **+ 1) + 5 candidates preferring** c
′.

For each 1 ≤ i ≤ n and 1 ≤ j ≤ 2(q **+ 1), we add one voter with preferences**

Ti ≻ x 3 i,j,1 **≻ · · · ≻** x 3 i,j,k−1 ≻ c **≻ · · ·** .

For each 1 ≤ i ≤ m and 1 ≤ j ≤ **2, we add one voter with preferences**

$${\mathrm{j,1}}$$

bi ≻ x 4 i,j,1 **≻ · · · ≻** x 4 i,j,k−1 ≻ c
′ **≻ · · ·** .

The above construction is similar to that given in [14]. We note that in this case, we must ensure that we do not delete the buffer candidates in our case for deleting candidates. We thus add the following voters.

For each 1 ≤ i ≤ 2(m − q) + 2n(q **+ 1) + 4, one voter with preferences**

x
1
i,1 **≻ · · · ≻** x
1
i,k ≻ c **≻ · · ·**
d for $1\leq i\leq m$ and $1\leq j$ . 
and for 1 ≤ i ≤ m and 1 ≤ j ≤ **2, one voter with preferences**

x 4

i,j,1 **≻ · · · ≻** x 4 i,j,k ≻ c **≻ · · ·** .

This ensures that c
′cannot gain in relation to c **by deleting any buffer candidates.**
We show that there is an addition of at most q voters to ({c, c′} ∪ B, V ) such that c **is excluded from**
winning if and only if there exists a hitting set of S of at most q **elements. Let** S
′ **be a hitting set of** S
of size q**. In the election (**S
′ ∪ B ∪ {c, c′}, V ), c receives 2(m − k) + 2n(q **+ 1) + 4 approvals,** c
′**receives**
2(m − k) + 2n(q + 1) + 5 approvals, each candidate si ∈ S receives at most 2n(q **+ 1) + 2 approvals, and each**
buffer candidate receives at most 2 approvals. Thus, c
′ wins this election and c **is excluded from winning.**
In contrast, let D be a subset, of size at most q, of S such that c is not a winner of (D ∪ B ∪ {c, c′}, V ).

We first note that if b ∈ S or b is a buffer candidate, that c beats b. Thus, if c **is excluded from winning** this election, then c
′ must beat c. In (D ∪ {c, c′}, V ), c
′receives 2n(q + 1) + 5 + 2(m − ||D||**) approvals**
and c receives 2(m − q) + 2n(q + 1) + 4 + 2(q + 1)m′ approvals, where m′is the number of sets in S **that** are not hit by D. Since c is excluded from winning, 2(m − q) + 2(q + 1)m′ ≤ 2(m − ||D||**), which implies** (q + 1)m′ + ||D|| − q ≤ 0. So m′ = 0 and ||D|| corresponds to a hitting set of S **of size at most** q.

As in [14], this reduction also shows that k**-approval is computationally resistant to destructive control**
by deleting voters. In this case, we start with the election ({c, c′}∪B ∪S, V **). Destructive control by deleting**
at most m − q candidates is possible if and only if there exists a hitting set of S of at most q **elements. We** note here that the additional voters approving the buffer candidates to c **is required in this case to prevent** one from deleting buffer candidates. In this case, deleting buffer candidates would be useless because c **would** gain one approval while c
′ **would gain at most one approval.**
Theorem 21 k**-veto elections are computationally resistant to destructive control by either adding or deleting candidates.**
Proof: Consider an instance of Hitting Set, where we are given a set S = {s1, . . . , sm}, n **subsets of** S, T1, . . . , Tn**, and positive integer 1** ≤ q ≤ m.