Reaching consensus under a deadline

Abstract

Group decisions are often complicated by a deadline. For example, in committee hiring decisions the deadline might be the next start of a budget, or the beginning of a semester. It may be that if no candidate is supported by a strong majority, the default is to hire no one - an option that may cost dearly. As a result, committee members might prefer to agree on a reasonable, if not necessarily the best, candidate, to avoid unfilled positions. In this paper we propose a model for the above scenario—Consensus Under a Deadline (CUD)—based on a time-bounded iterative voting process. We provide convergence guarantees and an analysis of the quality of the final decision. An extensive experimental study demonstrates more subtle features of CUDs, e.g., the difference between two simple types of committee member behavior, lazy vs. proactive voters. Finally, a user study examines the differences between the behavior of rational voting bots and real voters, concluding that it may often be best to have bots play on the voters’ behalf.

Appendices

A proofs of theorems, lemmata and formal statements

Theorem 1

In any $CUD(\mathcal {F}^{IMaj}_\sigma , \tau )$ with $\sigma \in (\frac{n}{2},n]$ and consistent utility functions, either the default alternative is set at $t=0$, or a valid alternative becomes the winner at time $t\in [0:\tau ]$.

Proof

Proof of Theorem 1.

First, we prove by contradiction that if there are possible winners at the beginning of the process, then the process converges with some valid alternative as a winner. This implies that at stage $\tau$ the set of possible winners is not empty, nor is it empty at the final stage 0, since there is a winner. Suppose that it does not hold: the process does not converge, even if there were possible winners at the beginning. In other words, at the final step $t=0$ the set of possible winners is empty, although at the initial step $t=\tau$ it is not empty. This can happen if, at some time step between the beginning at $t=\tau$ and the end at $t=0$, the set of possible winners becomes empty (including the time step $t=0$). Consider the time step $\tau '$, such that for all $\tau ''>\tau '$ the set of possible winners is not empty, while for $\tau '$ it is empty.

Consider the preceding iteration of our game protocol, i.e., the time step $\tau '+1$. There are two possible scenarios: (i) no voter changes the vote, and (ii) some voter changes the vote.

First, we assume that at the time step $\tau '+1$ there are no voters who wish to change their votes. Since at $\tau '+1$ the set of possible winners is not empty, for all the voters the utility is not zero under the current strategy. Since their utility at time $\tau '+1$ reflects the outcome of time $\tau '$, the set of possible winners could not become empty at time step $\tau '$.

Second, if the second scenario has occurred, at the time step $\tau '+1$ there are some voters who do change their vote. The only reason for a voter to change her vote is to improve her utility by switching away from an alternative that is either no longer in the set of possible winners, or will be removed from the set at the next time-slice. Such a switch, however, necessitates that the voter’s newly chosen alternative remains in the set of possible winners at the next time-slice. Otherwise, the voter’s utility from this new choice would not be positive, and no decision switch would occur. In particular, this implies that the set of possible winners is not empty. The obtained contradiction proves that if the set of possible winners is not empty at $\tau$, it cannot become empty at 0, therefore there is a valid alternative that is declared as a winner.

What remains is to show consistency of our protocol, i.e., that if at the time step $\tau$ the set of possible winners is empty, it must be empty at the time step 0 as well. Denote $c \in C^+$ an alternative that at time step $\tau$ has the maximum score $s_c$ among all other alternatives. Since the set of possible winners is empty at the time step $\tau$, it implies that $\sigma -s_c \ge \tau +1$. Even if at each step until the deadline there will be a voter that changes her vote for this alternative c, at time 0 this alternative will have obtained only $\tau$ more votes. As a result, the score of the alternative c at time step 0 can be at most $s_c+\tau$, which would imply that $\sigma -s_c-\tau \ge 1$ and $c\not \in \widehat{W}(\mathbf {s},0)$. As the score of c is maximal possible at time $t=\tau$, the same holds for all other alternatives. Hence, no alternative is a member of $\widehat{W}(\mathbf {s},0)$—it is empty.

Lemma 1

Let $c \notin \widehat{W}$ at step t, then $c \notin \widehat{W}$ at any step $t'<t$.

Proof

Proof of Lemma 1.

Given that $s_c^t<\sigma -t$ and at each step the candidate c can get at most one vote, then $s_c^{t'}<s_c^t+(t-t')<\sigma -t+(t-t')=\sigma -t'$ for any $t'<t$. Therefore, $c \notin \widehat{W}$ at step $t'$.

Lemma 2

If $s_c^{t+1}<s_c^t$, then $c \in \widehat{W}(s^t,t)$.

Proof

Proof of Lemma 2.

Given that $s_c^{t+1}<s_c^t$, there exists a voter, i, who changed her vote in favor of c at time step t. Let $c'$ denote a candidate that she voted for at the preceding time step $t+1$, and let us assume that the lemma’s conclusion does not hold. That is, let us assume that $c \notin \widehat{W}(s^t,t)$.

Notice that, except c, no candidate increased his score from the time step $t+1$ to the time step t. This is because only one voter was given the chance to change her vote, and thus $s^t=s^{t+1}-c'+c$. As a result, $\widehat{W}(s^{t+1},t)\supseteq \widehat{W}(s^t,t)$. Therefore, either (but not both) of the following holds:

• $a_i(\mathbf {top}_i(\widehat{W}(s^{t+1},t)), \mathbf {top}_i(\widehat{W}(s^t,t)))$

• $\mathbf {top}_i(\widehat{W}(s^{t+1},t))= \mathbf {top}_i(\widehat{W}(s^t,t))$.

A voter changes her vote only if it increases her utility, and conditions above indicate that a lazy voter’s utility (Definition 1) will not change between $s^{t+1}$ and $s^t$. Thus, voter i can not be a lazy voter.

Now, combining the fact that $s^t=s^{t+1}-c'+c$ with our attempt to assume that $c \notin \widehat{W}(s^t,t)$, we conclude that no candidate in the set $\widehat{W}(s^t,t)$ has a score higher that he has in $s^{t+1}$. More formally, $\forall \widehat{c}\in \widehat{W}(s^t,t),\ \ s^t_{\widehat{c}}\le s^{t+1}_{\widehat{c}}$. As a result, switching from $c'$ to c would not have been the preferred move of a proactive voter (Definition 2), as it does not change the utility of the set of possible outcomes.

We conclude that if the set of voters consists of lazy and/or proactive voters, then the assumption $c\notin \widehat{W}(s^t,t)$ leads to a contradiction of no voter having an incentive to change her vote. Thus, $c\in \widehat{W}(s^t,t)$ must hold.

Lemma 3

If a voter j at the time step t votes for candidate $c \in \widehat{W}(s^t,t)$, then $c=\mathbf {top}_i(\widehat{W}(s^t,t))$.

Proof

Proof of Lemma 3.

First notice that all voters initially vote for their top choice, thus the lemma’s conclusion holds for $t=\tau$. However, let us assume, to the contrary, that the lemma does not hold in general. In particular, it would imply that there is a time step t such that for any $t'>t$ (i.e., preceding steps) the statement of the lemma is fulfilled, and at step t there is a voter j such that: (i) she votes for $c\in \widehat{W}(s^t,t)$; (ii) there is a candidate $c' \succ _{j} c$ in the set $\widehat{W}(s^t,t)$.

Lemma 1 implies that $\widehat{W}(s^t,t) \subseteq \widehat{W}(s^{t+1},t+1).$ If voter j did not change her vote at time t, then by maximality of t holds $c=\mathbf {top}_i\in \widehat{W}(s^{t+1},t+1)$ and it implies that $c=\mathbf {top}_i\in \widehat{W}(s^t,t)$, which we assumed not to hold. Thus, j must have changed her vote at time step t. However, since there is $c' \succ _{j} c$ in the set $\widehat{W}(s^t,t)$, both lazy and proactive consistent utility (see Definitions 1, 2) from c is lower than it is from $c'$ at time step t, which contradicts optimality of choice in voting decisions (see Game Protocol 1, line 8).

Lemma 4

If there is a voter that changes her ballot at time t from voting for c to voting for $c'$, then $s_c^{t+1} \le s_{c'}^{t+1}$.

Proof

Proof of Lemma 4.

Let us assume the opposite to the lemma’s conclusion, that is $s_{c}^{t+1}>s_{c'}^{t+1}$ and consequently, $s_c^{t+1} \ge s_{c'}^{t+1}+1=s_{c'}^t$. As a result, according to Lemma 2, $c' \in \widehat{W}(s^t,t)$. However, since $s_{c'}^t$ is sufficient to become a possible winner at time t and $s_c^{t+1}\ge s_{c'}^t$, holds that $c \in \widehat{W} (s^{t+1},t)$. Note that, according to Lemma 1, the above implies that $c,c' \in \widehat{W}(s^{t+1},t+1)$.

Now, Lemma 3 implies that if a voter votes for a candidate in the set $\widehat{W}(s^{t+1},t+1)$, then she prefers this candidate over all other possible winners. Therefore, similarly to the proof of Lemma 2, neither a lazy nor a proactive voter would change their votes. The obtained contradiction proves the lemma.

Corollary 1

Let $\mathbf {a}=(a_1,\ldots ,a_n)$ be the truthful profile, let $\tau$ be the deadline time, and let $\mathbf {b}$ be the ballot profile induced by $\mathbf {a}$, i.e., $b_i=\mathbf {top}_i(C)$. CUD stops with some $w\in C^+$ if and only if there is an alternative $c\in C^+$ so that $\mathbf {sc}_c(\mathbf {b})\ge \sigma -\tau$.

Proof

Proof of corollary 1.

First, let us assume the right-hand side of the “if and only if” statement, and show that CUD stops with a non-default alternative. The condition $\mathbf {sc}_c(\mathbf {b})\ge \sigma -\tau$ implies that c is a possible winner by the definition of $\widehat{W}(\mathbf {b},\tau )$. Thus, at time step $t=\tau$ the set of possible winners contains at least one alternative. As a result, by Theorem 1, the process converges with some (non-default) alternative chosen as the winner.

Now, let us deal with the opposite direction of the Theorem’s implication. Let CUD stop with some $w \in C^+$. Then this candidate w achieved $\sigma$ votes in no more than $\tau$ steps. At each step he could get 1 vote at most, that is, he achieved no more than $\tau$ votes. Thus, for the initial score of w it must hold that $\mathbf {sc}_w(\mathbf {b})\ge \sigma -\tau$.

Theorem 2

Let $\mathbf {a}=(a_1,\ldots ,a_n)$ be the truthful profile, let $\tau$ be the deadline time, and let $\mathbf {b}$ be the ballot profile induced by $\mathbf {a}$, i.e., $b_i=\mathbf {top}_i(C)$). If there is an alternative $c\in C^+$ so that $\mathbf {sc}_c(\mathbf {b})\ge \max \left\{ \left\lfloor \frac{n}{2}\right\rfloor +1,\sigma -\tau \right\}$ then CUD terminates with c as the winner.

Proof

Proof of Theorem 2.

Note that at any step such that all voters whose top-choice is candidate c, vote for c, for any other candidate it is true that $s_{c'}^t \le n- s_c^t \le \left\lfloor \frac{n}{2}\right\rfloor$ and $s_{c'}^t \le s_c^t -1$.

Thus, if CUD terminates with a winner other than c, it implies that c loses some votes of those whose top-choice is c.

Consider t such that for every $t' \ge t$ all voters whose top-choice is c vote for c and at step t one of them changes her vote to $c'$. Thus, given that no one of them has changed their vote before, $s_{c'}^t<s_c^t$ which contradicts Lemma 4. That is, there is no such t.

Therefore, candidate c retains the same number of votes, $\mathbf {sc}_c^\tau$, until step 0, which implies that at that last step he has more votes than any other candidate.

Theorem 3

Let $\mathbf {a}$ be the truthful profile of voters participating in a CUD. Assuming that it is well-defined for the CUD instance, the following bounds can be placed on the additive Price of Anarchy, $PoA^+$, depending on the ratio of the deadline timeout $\tau$ and the number of voters n:

• 1. If $\tau \le \sigma - \left\lfloor \frac{n}{2}\right\rfloor$, then

  

• 2. If $\sigma -\left\lfloor \frac{n}{2}\right\rfloor<\tau <\sigma$, then

  

• 3. If $\tau \ge \sigma$, then

  

Proof

Proof of Theorem 3.

Case when $\tau \le \sigma - \big \lfloor \frac{n}{2}\big \rfloor$.

Note that, even if at each step every voter would change her vote in favour of the same alternative c, this alternative c would get no more than $\tau$ points.

$$\begin{aligned} s_c^{\tau }+\tau \ge \sigma \Leftrightarrow s_c^{\tau } \ge \sigma - \tau \ge \big \lfloor \frac{n}{2}\big \rfloor \end{aligned}$$

Note that, there can be at most two alternatives with score $\big \lfloor \frac{n}{2}\big \rfloor$. If c is the only one, then he is the winner, hence $PoA^+(\mathbf {a})=0$. Suppose there are two such alternatives: w and c. If they have equal scores at $\tau$: $s_w^\tau =s_c^\tau =\big \lfloor \frac{n}{2}\big \rfloor$, then, whoever wins, $PoA^+(\mathbf {a})=0$. Another possibility is that there are two such alternatives, w and c, such that w has more points, i.e., $s_w^\tau =s_c^\tau +1$, and all other alternatives have 0 points. Alternative c would win only if every supporter of w would change to c, which would take all $\tau$ stages. But, from those who initially voted for w, no voter would change her vote to c, since they are better off by keeping their votes for w. Hence, w will win, and consequently, $PoA^+(\mathbf {a})=0$.

Case when $\sigma -\left\lfloor \frac{n}{2}\right\rfloor<\tau <\sigma$

Let $\omega$ denote the plurality winner at the time step $t=\tau$ and c denote the winner at the time step $t=0$, and let us assume the contrary to the Theorem, i.e., $PoA^+(\mathbf {a})>\left\lfloor \frac{n}{2}\right\rfloor +\tau -\sigma$. In particular, it would mean that $\omega \ne c$ and

$$\begin{aligned} s_{\omega }^\tau -s_c^\tau >\left\lfloor \frac{n}{2}\right\rfloor +\tau - \sigma \end{aligned}$$

(4)

In $\tau$ steps candidate c obtains at most $\tau$ votes and becomes a winner, that is $s_c^\tau +\tau \ge \sigma$. Therefore, $s_c^\tau \ge \sigma -\tau$. Combining this with Eq. 4, we obtain:

$$\begin{aligned} s_{\omega }^\tau >\left\lfloor \frac{n}{2}\right\rfloor +\tau - \sigma +s_c^\tau \ge \left\lfloor \frac{n}{2}\right\rfloor +\tau -\sigma +\sigma -\tau =\left\lfloor \frac{n}{2}\right\rfloor \end{aligned}$$

Thus, $s_{\omega }^\tau >\left\lfloor \frac{n}{2}\right\rfloor$ and, consequently, $s_{\omega }^\tau \ge \left\lfloor \frac{n}{2}\right\rfloor +1$. However, according to Theorem 2, this means that $\omega$ is the declared winner of the CUD at time step $t=0$. Which contradicts key part of our assumption: $c\ne \omega$. We thus must conclude that $PoA^+(\mathbf {a}) \le \left\lfloor \frac{n}{2}\right\rfloor +\tau -\sigma$, as is per the Theorem.

Case $\tau \ge \sigma$.

Notice again that if $PoA^+ \ne 0$ then the winner at the time step $t=\tau$ (denoted by $\omega$) and the winner at the time step $t=0$ (denoted by c) must be different. Now, Theorem 2 implies that at the time step $t=\tau$ the truthful profile winner, $\omega$, can have at most $\left\lfloor \frac{n}{2}\right\rfloor$ votes, otherwise it must be the winner at the time step $t=0$ as well.

At the same time, it must be that $s_c^\tau \ge 1$. Otherwise, no voter would be able to switch to c at any time, as there will be at least one other candidate with a higher score than c (the current winner) and, thus, Lemma 4 would prevent the switch. Since possible winners never lose votes, but only gain them, which means that the score of c will never drop to zero either. Thus, $PoA^+(a) \le \left\lfloor \frac{n}{2}\right\rfloor -1$, as required.

Lemma 5

The last two bounds in Theorem 3are tight. For all $\tau$ and n that satisfy the conditions of Eqs. 2and 3, there exists a truthful profile $\mathbf {a}$ such that the corresponding bound holds as an equality.

Proof

Proof of Lemma 5.

Table 8 provides an example of a voting profile that proves that the bounds in Case 2 ($\sigma -\left\lfloor \frac{n}{2}\right\rfloor<\tau <\sigma$) of Theorem 3 are tight. All voters are grouped into 3 Blocks. Voters in Block-1 prefer the candidate c over $c_1$ and over all other candidates; voters in Block-2 prefer candidate $\omega$ over c and over all other candidates; finally, each voter in Block-3 prefers some distinct candidate (but not $\omega$) over c and over all other candidates. We assume that there are $\sigma -\tau$ voters in Block-1, $\left\lfloor \frac{n}{2}\right\rfloor$ voters in Block-2, and the rest of the voters are in Block-3. It is assumed that $\sigma -\tau \ge 2$, so, there are at least 2 voters in Block-1.

Table 8 Proof of Lemma 5: voters’ preferences

Notice that we can indeed construct such a preference profile given that $\sigma - \left\lfloor \frac{n}{2}\right\rfloor <\tau$, that is, $\sigma - \tau < \left\lfloor \frac{n}{2}\right\rfloor$. In particular, the size of Block-3 is $k=n-\left\lfloor \frac{n}{2}\right\rfloor -\sigma + \tau$, and, therefore, every candidate from $\{c_1,c_2,...c_k\}$ appears as a top-choice only once.

We assume that $\sigma - \tau \ge 2$ and n is odd. Then, $\widehat{W}(s^{\tau },\tau )=\{c,w\}$. Thus, at any time step prior to $\sigma - \left\lfloor \frac{n}{2}\right\rfloor$ voters from the Block-3 will want to change their vote to c, since they currently vote for a candidate outside the set of possible winner $\widehat{W}$ and for all of these voters $c\succ w$. No candidate from Block-1 or Block-2 will want to change their votes during that time.

As a result, for $t=\left\lfloor \frac{n}{2}\right\rfloor -\sigma + \tau$ holds $s_c^{t}=s_w^{t}=\left\lfloor \frac{n}{2}\right\rfloor$. After this step all voters from Block 1 and all except one from Block 3 vote for c, and all the voters from Block 2 vote for w. At the next step, $t-1$, all the voters will want to change their vote since the set of possible winners $\widehat{W}(s^{t-1},t-1)$ can only contain candidates with $\left\lfloor \frac{n}{2}\right\rfloor +1$ votes.

Ties among voters who wish to change their vote are broken randomly, and with probability $\frac{\left\lfloor \frac{n}{2}\right\rfloor +1}{n}$ a voter will be chosen who will change her vote to c. This will make c the only possible winner, which, more formally, means that $\widehat{W}(s^{\left\lfloor \frac{n}{2}\right\rfloor -\sigma +\tau -1},$

$\left\lfloor \frac{n}{2}\right\rfloor -\sigma +\tau -1)=\{c\}$. Hence, c will be the winner of the entire election process, the final winner. Given that c and w are the only possible winners, and w is a Plurality winner at time $\tau$, $PoA^+=\left\lfloor \frac{n}{2}\right\rfloor -\sigma +\tau$.

Table 9 provides an example of a voting profile that proves that the bounds in the Case 3 ($\tau \ge \sigma$) of Theorem 3 are tight.

Once again, all voters are grouped into three blocks. Voters of Block-1 prefer candidate $\omega$ over c and over all other candidates; Block-2 voters prefer candidate c over $c_1$ and over all other candidates; each voter in Block-3 prefers some distinct candidate (but not $\omega$) over candidate c and other candidates. Let there be $\left\lfloor \frac{n}{2}\right\rfloor$ voters in Block-1, a single voter in Block-2, and the rest of the voters grouped into Block-3. We will assume that n is odd, so that the number of voters in Block-3 is $k=\left\lfloor \frac{n}{2}\right\rfloor$.

Table 9 Proof of Lemma 5: voters’ preferences – point of ref

Now, notice that at the time step $t=\sigma -1$ voters from Block-2 and Block-3 will want to change their vote. In particular, voters from Block-3 will want to change their votes to c, and, because ties among willing voters are broken uniformly at random, with probability $\frac{\left\lfloor \frac{n}{2}\right\rfloor }{\left\lfloor \frac{n}{2}\right\rfloor +1}$ a voter from Block-3 will be granted the opportunity and change her vote in favour of c. Thus, $\widehat{W}(s^{\tau -1},\tau -1)=\{w,c\}$. Analogously to the previously investigated profile, both w and c can be the final winner. Hence, $PoA^+\ge \left\lfloor \frac{n}{2}\right\rfloor -1$. Furthermore, combining this conclusion with an application of Theorem 3 to the constructed profile, we obtain that $PoA^+= \left\lfloor \frac{n}{2}\right\rfloor -1$ for the constructed profile. This yields the tightness of Theorem 3 as required. $\square$

Remarks on condorcet winners

In addition to the proofs of our main results, we would like to venture some additional thoughts on refinements of our protocol. Specifically, we would like to make a few notes about its relationship to Condorcet winners.

Lemma 6

Protocol1 is not Condorcet-consistent.

Proof

Consider a preference profile, where an alternative $c\in C$ is the second choice for all voters, and the top choices divide equally among ${a_1,a_2,a_3}\subseteq C\setminus \{c\}$. Furthermore, assume that c is not the default alternative.

It is easy to see that c is a Condorcet winner. However, since the initial score of c is zero and, as a consequence of Lemma 4, no voter will change her ballot in favour of c, and it will persistently remain outside of the set of possible winners. Hence, Protocol 1 will either result in a default or any other alternative in $C\setminus \{c\}$. That is, Protocol 1 is not Condorcet-consistent.

However, it is marginally easier for a Condorcet winner to survive our Protocol and become its winner, as the following augmentation of Theorem 2 states.

Conjecture 1

Let $\mathbf {a}=(a_1,\ldots ,a_n)$ be the truthful profile, let $\tau$ be the deadline time, and let $\mathbf {b}$ be the ballot profile induced by $\mathbf {a}$, i.e., $b_i=\mathbf {top}_i(C)$). Let $c\in C^+$ be a Condorcet winner under the profile $\mathbf {a}$. If in addition holds that $\mathbf {sc}_c(\mathbf {b})\ge \max \left\{ \left\lceil \frac{n}{2}\right\rceil -1,\sigma -\tau \right\}$ then CUD terminates with c as the winner.

Full simulation results

See Tables 10, 11, 12 and 13.

Table 10 Number of vote changes: the average and standard deviation of the number of vote changes required to reach a consensus

Table 11 The average and standard deviations of the upper bounds for the additive price of anarchy for 10 voters for different initial $\tau$ (time until the deadline). The $PoA^+$ is identical for both lazy and proactive voters

Table 12 The average and standard deviations of the upper bounds for the additive price of anarchy for 20 voters for different initial $\tau$ (time until the deadline). The $PoA^+$ is identical for both lazy and proactive voters

Table 13 The average and standard deviations of the upper bounds for the additive price of anarchy for 30 voters for different initial $\tau$ (time until the deadline). The $PoA^+$ is identical for both lazy and proactive voters