Weghted Votng Systems Elssa Brown Chrssy Donovan Charles Noneman June 5, 2009 Votng systems can be deceptve. For nstance, a votng system mght consst of four people, n whch three of the people have 2 votes, one has vote and the number of votes needed to pass a proposal s 4. It turns out that the person wth vote wll never have an effect on the outcome, despte the fact that he or she has a vote. In ths paper, we analyze weghted votng systems usng Banzhaf s defnton of power n order to fnd out how power can be dvded among n voters, a queston posed n a paper by John Tolle. Tolle counted the number of ways power can be dstrbuted among groups of three and four voters (though hs condtons are dfferent from ours). At the end of the paper, we provde an upper bound for the number of ways power can be dstrbuted to n voters. Terms A weghted votng system conssts of a group of voters. Each voter s gven a number of votes, whch s known as a weght. A weghted votng system also ncludes the number of votes needed to pass a proposal, or a quota. We call the voters players and denote the players by P,..., P n. We wrte the weghts gven to each player as v,..., v n, where v corresponds to P. Furthermore, v v 2... v n. In other words, P sgnfes the player wth the largest weght, whereas P n stands for the player wth the smallest weght. The quota s represented by q, and must be greater than half of the total number of votes. If the quota were exactly half (or less than half), then many problems arse. All of the nformaton about a weghted votng system can be captured by the notaton [q; v,..., v n ]. We defne a coalton as any subset of the set of players {P,..., P n }. We say a coalton wns (or s wnnng) when the sum of the weghts of each player n the coalton meets or exceeds the quota. We say a coalton loses (or s losng) when the sum of the weghts of each player n the coalton does not meet the quota. An assgnment s a functon that assgns to each coalton the label W or L. A vald assgnment s an assgnment n whch there are no contradctons arsng from the labels gven to each coalton. That s, there exsts some [q; v,..., v n ] such that all coaltons the assgnment labels W are wnnng and all coaltons labeled L are losng.
We say a coalton s wnnable f there exsts a vald assgnment n whch the coalton s labeled W. We day t s non-wnnable f there s no vald assgnment n whch t s labeled W. That s, t s labeled L n all vald assgnments. 2 Two Players Suppose we have two players. We know that there are four coaltons, snce coaltons are smply subsets of the set of two players, {P, P 2 }. Thus, the four coaltons are, {P }, {P 2 } and {P, P 2 }. But we don t know any of the specfcs of the votng system. That s, we don t know the weghts of the players or the quota. So, nstead, we have to fnd out whch of the followng assgnments are possble for a 2-player votng system. Assgnment 2 3 4 5 6 7 8 L L L L L L L L {P } W W W W L L L L {P 2 } W W L L W W L L {P, P 2 } W L W L W L W L It s easy to see that some assgnments are not vald. Assgnments where s wnnng are not ncluded n the table for ths reason. We know these assgnments are nvald because the sum of the weghts of the players n s 0. 0 cannot be greater than or equal to the quota because the quota s postve n all cases. There are more nvald assgnments n the table. For example, assgnments and 2 say that both {P } and {P 2 } n are wnnng. However, ths s mpossble because we know that q > 2 v. If both of these coaltons are wnnng, then each player possesses more than 2 of the total votes. Assgnment 4 s nvald because t says that {P } wns but {P, P 2 } loses. Ths s mpossble snce {P, P 2 } must have at least as many votes as {P }. Assgnment 6 s nvald because t says that {P 2 } wns and {P } loses. Ths s agan mpossble snce v v 2. = Thus, there are only three vald assgnments for a 2-player votng system. assgnments 3, 7 and 8: L, W, L, W, L, L, L, W and L, L, L, L. These are represented by 3 Vald Assgnments When we decde whether an assgnment s vald, we apply one or more of the followng concepts. Theorem 3.. If the sum of the weghts of the players n coalton A s greater than or equal to the sum of the weghts of the players n coalton B and B s a wnnng coalton, then A s a wnnng coalton. Proof. B s a wnnng coalton f and only f the sum of the weghts of the players n B s greater than or equal to the quota. By hypothess, the sum of the weghts of the players n A s greater than or equal to the sum of the weghts of the players n B. Therefore, the sum of the weghts of the players n A s greater than or equal to the quota. Thus, A s a wnnng coalton. 2
Theorem 3.2. If {P, P 2,..., P k } s a wnnng coalton, then any superset of ths coalton s wnnng. Proof. The hypothess states that k j=0 v j q. Suppose a player jons the coalton {P, P 2,..., P k }. When a player jons a coalton, the weght of that player, x, s added to the number of votes already held by the k coalton. Snce the weght of every player s greater than or equal to 0, + x q. Therefore, the coalton remans wnnng. Theorem 3.3. If {P, P 2,..., P k } s a wnnng coalton, then any coalton that s dsjont from ths coalton s a losng coalton. Proof. Suppose {P, P 2,..., P k } s a wnnng coalton. Suppose there s a coalton d that contans none of P, P 2,..., P k, but s also wnnng. Ths mples that the sum of the weghts of the players n both n {P, P 2,..., P k } and the coalton d meet or exceed the quota. Snce q > 2 v, both coaltons have more than half of the total number of votes. Ths means that together these coaltons have more votes than the total number of votes. Ths s clearly mpossble. Therefore, only one of two dsjont coaltons can be a wnnng coalton. If {P, P 2,..., P k } s wnnng, any coalton d that s dsjont s losng. 4 Wnnables To understand wnnable coaltons, t s enough to understand non-wnnable coaltons. As an example of non-wnnable coaltons, consder the followng lemma. j=0 = v j Lemma 4.. All coaltons of sze that do not contan P are losng n all vald assgnments. Proof. If {P }, where, wns, then P wns because v v. However, we know by 3.3 that two dsjont coaltons cannot both wn. Snce assumng {P } where wns produces a contradcton, {P } must be losng n all vald assgnments. Ths means that all coaltons of sze besdes {P } are non-wnnable. Here s a more general descrpton of non-wnnables. Theorem 4.2. If two coaltons A and B are dsjont and the sum of the votes of the players n A must be greater than or equal to the sum of the votes of the players n B, then B s always losng. Therefore, t s a non-wnnable. Proof. There are two cases to consder. Frst, suppose A s a wnnng coalton. Then, by 3.3, B s not a wnnng coalton. Second, suppose A s a losng coalton. If B were to be wnnng, then A would also be a wnnng coalton by 3.. Agan, ths s a contradcton of 3.3. Therefore, B s losng when A s losng. Snce B s losng regardless of whether A s wnnng or losng, t s a non-wnnable. Based on ths theorem, we can determne how many coaltons are wnnable. It wll be convenent for the followng proofs to convert coaltons to player lsts. To form a player lst, smply take all of the players 3
subscrpts and put these numbers n a sequence ordered from smallest to largest. For example, the coalton {P, P 3 } becomes, 3. If a player lst s named A, call the frst number of the player lst A. In general, call the th number n the lst A. Lemma 4.3. In an n-player votng system, consder a coalton c of sze s that produces player lst C. c s wnnable coalton a of sze s wth player lst A such that A < C {,..., s} and A C j, j {,..., s}. Proof. Suppose not. Suppose that c s wnnable and coalton a of sze s such that A < C {,..., s} s s and A C j, j {,..., s}. Snce A < C, then v A v C. Thus, v C. We know that a and c are dsjont snce A C j, j {,..., s}. Thus a has at least as many votes as c and they are dsjont. By 4.2, c never wns. Ths s a contradcton of the statement that c s wnnable, because a wnnable coalton wns n at least one vald assgment. To establsh the number of wnnable coaltons n a n-player votng system, we convert player lsts nto paths through a lattce grd. For player lst C of sze s n an n-player system, the path wll run from (0, 0) to (n s, s) on a s (n s) grd. Thnk of the constructon of the lattce path from C as a seres of n moves. On move, the move s up f C and the move s rght f C. It s easy to see that there s a bjecton from all player lsts of sze s n an n-player system (and thus all coaltons of sze s) to all lattce paths of length n wth s up moves. Lemma 4.4. A coalton s non-wnnable f and only f the lattce path whch represents that coalton has no ponts that are above the y = x lne. Proof. Consder a non-wnnable coalton c wth player lst C. Snce c s non-wnnable, by 4.3, there s a coalton a wth player lst A such that A < C {,..., s} and A C j, j {,..., s}. In the lattce path that represents c, there s at least one rght move before every up move. Ths s because for each up move expressng the fact that C C, there s a unque rght move expressng the fact that A C. The rght move happens before the up move because A < C. If there s at least one rght move before every up move n the lattce path representng c, then ths lattce path cannot cross the lne y = x. The lattce path at most touches the lne y = x. Consder a lattce path that does not go above the y = x lne. Let C be the player lst that corresponds to ths lattce path. Let c be the coalton that corresponds to player lst C. Construct player lst A such that t contans the numbers that correspond to the frst c rght moves of the lattce path. Because the lattce path does not goes above the lne y = x at any pont, there are at least as many rght moves as up moves. Furthermore, there s a rght move before every up move. So for each, C > A because the th rght move happens before the th up move. Thus, A s a player lst such that A < C {,..., c } and A C j, j {,..., c }. Ths establshes by 4.3 that c s non-wnnable. Wth these proofs, we can at last fnd the number of wnnable coaltons. = v A = Theorem 4.5. In an n-player system, there are ( ) n s wnnable coaltons of sze s, where s n 2. Proof. We know that there s a bjecton from all wnnable coaltons of sze s to all lattce paths the run from (0, 0) to (n s, s) through an (s) (n s) grd and have at least one pont above the lne y = x. Therefore, a count of these paths provdes a count of the wnnable coaltons. In order to do ths, we convert 4
these paths nto other paths that are easer to count. Here s how the converson works. Consder such a lattce path. Fnd the frst pont that s above y = x. Change every edge after ths pont from an up move to a rght move or from a rght move to an up move. Ths acton wll result n a new path that goes rght s tmes and up n s+ tmes. Ths s because the path must have moved up tmes and to the rght tmes to reach the lne y = x, for some nteger. Once t reached the lne y = x, t moved up. Therefore, the unaltered part of the path travels up + tmes and to the rght tmes. It must go another (n s) moves to the rght and s ( + ) moves up to reach the pont (n s, s). However, these moves have been swtched. The new path goes up ( + ) + (n s ) = n s + tmes and to the rght ()+(s ) = s tmes. Ths means that we now have a path on an (n s+) (s ) grd. Ths process establshes a bjecton between paths that have at least one pont over y = x on the (s) (n s) grd and paths that have at least one pont over y = x on the (n s + ) (s ) grd. So, we count the paths through the new grd. In fact, all paths through the new grd must go above y = x because we know that s n 2, so the grd s taller than t s long. There are ( n s ) paths n ths new grd. Every path must go to the rght (s ) tmes and the path has a total n moves. Thus, constructng a path n ths grd s equvalent to choosng whch (s ) postons of a sequence wth n postons wll be up moves. Therefore, there are ( n s ) wnnable coaltons of sze s. Theorem 4.6. In an n-player votng system, there are ( ) n s wnnable coaltons of sze s, where s > n Proof. Snce s > n s, the (s) (n s) grd s taller than t s long. Therefore, all paths that go from (0, 0) to (n s, s) go above the y = x lne. Ths s because (n s, s) s above the lne y = x. Thus, all coaltons of sze s, where s > n 2, are wnnable. By the same combnatoral argument as above, there are ( n s) such paths. Snce these paths correspond to coaltons of sze s, there ( n s) wnnable coaltons of sze s. These proofs tell us how many wnnable coaltons of a gven sze there are. To determne how many wnnable coaltons there are n total, we sum over s. 2. Theorem 4.7. The number of wnnable coaltons n an n-player system s 2 n ( ) n. n 2 n ( ) n Proof. We know that there are = 2 n coaltons n an n-player system. When s = n 2, the expresson =0 for the number of wnnables of sze s s ( ) n. When s = n +, the expresson for the number of wnnables of sze s s ( n s ) s (. Thus, the mssng term s n total number of wnnable coaltons s 2 n ( n n 2 ). 2 n 2 ). Ths term, then, counts the non-wnnables. Therefore, the Although the followng proof does not relate to our later work, we nclude ths addtonal result, whch makes a connecton wth the Catalan numbers. Theorem 4.8. For k < n 2, ( n k ) = k =0 ( ) n 2 C, where C s the th Catalan number. k 5
Proof. As was shown prevously, there are ( n s ) lattces paths n the (s) (n s) grd that go from (0, 0) to (n s, s) and go above the lne y = x when s < n 2. We can count these paths n another way. Agan we count the paths by fndng the frst pont on each path whch s over y = x. All of these ponts le on y = x +. There are two ways that a path can nclude (, + ): the path ncludes (, ) or t ncludes (, ). Snce we only want to count paths where (, + ) s the frst pont above y = x, we only count the paths that nclude (, ). Addtonally, we only count paths contanng (, ) f they have been below y = x pror to ths pont. There are C such paths. Snce all relevent paths to (, + ) came drectly from (, ), there are C relevent paths to (, + ). Now we must fnd the number of paths from (, + ) to (s, n s). Ths s smply ( ) n (+2) s (+) snce the path to (, + ) has length + 2( and s + from the bottom. So the number of paths where (, + ) s the frst pont above y = x s: C n 2 ) k. A sum over all results n the followng formula. s ( ) n 2 C s The two methods of countng these paths are equal. Therefore, or =0 ( ) n s ( ) n 2 = C s s ( ) n = k =0 k =0 ( ) n 2 C k 5 Banzhaf Power Index We have explaned a lot about vald assgnments, but we have not addressed the queston of how much power does a player n a gven weghted votng system have. To begn, consder the followng example. Suppose we have a commttee consstng of a presdent, a vce presdent, a treasurer and a peon. Durng meetngs, the commttee needs to make decsons. In order to reflect the mportance of each member, they make a weghted votng system. The presdent s assgned 2 votes. The vce presdent and treasurer are also each assgned 2 votes, whle the peon gets. They decde that a smple majorty of 4 votes s needed to pass a moton. The presdent and the vce presdent realze that they would lke more bacon snacks at commttee meetngs. As they have 4 votes between them, ther votes meets the quota, so the opnons of the other members do not matter. If, however, the presdent or vce presdent has a change of heart, then the moton for more bacon snacks wll not pass. Therefore ther opnons do matter. Ths makes us notce somethng: all t takes to make a decson s two people who are not the peon votng n agreement. In fact, the peon s vote never affects the outcome of a vote. When we gave the peon vote and the rest of the commttee 2 votes, we thought we were gvng them twce as much power as the peon. Instead, we have created a system where three people have an equal 6
amount of power and one person has none. Clearly, the number of votes that a player has s not an accurate ndcator of power. So, what s a good way to defne power? In order to fgure out how much power a player n a votng system has, we use the Banzhaf Power Index. Smply put, ths s a measure of how many tmes a player s opnon s essental, or crtcal, to the success of a coalton that he or she s n. We saw that n the coalton consstng of both the presdent and vce presdent both players are essental to the success of the coalton. Should we remove ether one of them, there would no longer be enough votes to meet the quota. Ths stuaton s called a crtcal nstance. A crtcal nstance for a player occurs when removng that player from a coalton causes that coalton to go from wnnng to losng. In the Banzhaf Power Index, a player s power s defned as the rato of the player s crtcal nstances to the total number of crtcal nstances. A power dstrbuton s a vector of these ratos. 5. Fndng Crtcal Instances In Secton 2, we saw that there are only three vald assgnments for a two-player system:. {P } and {P, P 2 } wn. 2. Only {P, P 2 } wns. 3. No coalton wns. In the frst assgnment, P s crtcal n both coaltons. Furthermore, P 2 s not crtcal n the coalton {P, P 2 }. Therefore, P has power of 2 2 whle P 2 has power of 0 2. Thus, the power dstrbuton s, 0. In the second assgnment, both players are crtcal to the wnnng coalton. So, of the two crtcal nstances, each player clams one. Ths produces a power dstrbuton of 2, 2. In the last assgnment no one wns. Therefore, there are no crtcal nstances, yeldng a power dstrbuton of 0, 0. 6 Examples of Power Dstrbutons We see that a power dstrbuton depends entrely on the assgnment of a gven votng system. To fnd all of the power dstrbutons for an n-player system, we need to fnd all of the vald assgnments. To do ths, we analyze the coaltons of an n-player system wth the followng method.. Label all non-wnnables L. 2. Assgn c, an unlabeled (and therefore wnnable) coalton, to wnnng. 3. Remember the partal assgnment as ths pont. 4. If t s possble to label other coaltons wnnng or losng based on the theorems of Secton 3, then do t. 7
5. If there are any remanng unlabeled coaltons, recursvely run ths process by returnng to step 2. 6. Ths wll produce all vald assgnments that contan c as wnnng. 7. Return to the partal assgnment n step 3 and set c as losng. 8. If there are any unlabeled coaltons go back to step 2. Wth ths method, we are able to fnd all of the vald assgnments for the three-player and four-player systems. From here, we can calculate the power dstrbutons. These results follow. Showng the detals of the vald assgnment method s dffcult n ths format and of lttle value. However, n order to provde some nsght nto the producton of these results, the neccessary elements of the assgnments are ncluded n the charts. That s to say, f all the coaltons n the Coaltons Set to Wnnng cell are set to wnnng and all of the coaltons n the Coaltons Set to Losng cell are set to losng, then by applyng the rules n secton 3, the entre vald assgnment s constructed. Also, these lstngs have the potental to llumnate the general n-player system. 6. Three Players There are sx vald assgnments and fve unque power dstrbutons. Table : Vald Assgnments n a 3-Player System Assgnment Coaltons Set to Wnnng Coaltons Set to Losng Resultng Wnnng Coaltons {P } None {P }, {P, P 2 }, {P, P 3 }, {P, P 2, P 3 } 2 {P 2, P 3 } {P } {P, P 2 }, {P, P 3 }, {P 2, P 3 }, {P, P 2, P 3 } 3 {P, P 3 } {P }, {P 2, P 3 } {P, P 2 }, {P, P 3 }, {P, P 2, P 3 } 4 {P, P 2 } {P, P 3 } {P, P 2 }, {P, P 2, P 3 } 5 {P, P 2, P 3 } {P, P 2 } {P, P 2, P 3 } 6 None {P, P 2, P 3 } None Table 2: Power Dstrbutons n a 3-Player System Assgnment Power Dstrbuton, 0, 0 2, 3, 3 3 3 5, 5, 5 4 2, 2, 0 5 3, 3, 3 6 0, 0, 0 6.2 Four Players There are 5 vald assgnments and 3 unque power dstrbutons. 8
Table 3: Vald Assgnments n a 4-Player System Assgnment Coaltons Set to Wnnng Coaltons Set to Losng Resultng Wnnng Coaltons {P } None {P }, {P, P 2 }, {P, P 3 }, {P, P 4 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P, P 2, P 3, P 4 } 2 {P 2, P 3 } {P } {P, P 2 }, {P, P 3 }, {P 2, P 3 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P 2, P 3, P 4 }, {P, P 2, P 3, P 4 } 3 {P, P 4 }, {P 2, P 3, P 4 } {P } {P, P 2 }, {P, P 3 }, {P, P 4 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P 2, P 3, P 4 }, {P, P 2, P 3, P 4 } 4 {P, P 4 } {P }, {P 2, P 3, P 4 } {P, P 2 }, {P, P 3 }, {P, P 4 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P, P 2, P 3, P 4 } 5 {P, P 3 }, {P 2, P 3, P 4 } {P, P 4 }, {P 2, P 3 } {P, P 2 }, {P, P 3 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P 2, P 3, P 4 }, {P, P 2, P 3, P 4 } 6 {P, P 3 } {P, P 4 }, {P 2, P 3 }, {P 2, P 3, P 4 } {P, P 2 }, {P, P 3 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P, P 2, P 3, P 4 } 7 {P, P 2 }, {P 2, P 3, P 4 } {P, P 3 } {P, P 2 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P 2, P 3, P 4 }, {P, P 2, P 3, P 4 } 8 {P, P 2 }, {P, P 3, P 4 } {P, P 3 }, {P 2, P 3, P 4 } {P, P 2 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P, P 2, P 3, P 4 } 9 {P, P 2 } {P, P 3 }, {P 2, P 3, P 4 }, {P, P 2 }, {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 } {P, P 2, P 3, P 4 } 0 {P 2, P 3, P 4 } {P, P 2 } {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P 2, P 3, P 4 }, {P, P 2, P 3, P 4 } {P, P 3, P 4 } {P, P 2 }, {P 2, P 3, P 4 } {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 3, P 4 }, {P, P 2, P 3, P 4 } 2 {P, P 2, P 4 } {P, P 2 }, {P, P 3, P 4 } {P, P 2, P 3 }, {P, P 2, P 4 }, {P, P 2, P 3, P 4 } 3 {P, P 2, P 3 } {P, P 2, P 4 } {P, P 2, P 3 }, {P, P 2, P 3, P 4 } 4 {P, P 2, P 3, P 4 } {P, P 2, P 3 } {P, P 2, P 3, P 4 } 5 None {P, P 2, P 3, P 4 } None Table 4: Power Dstrbutons n a 4-Player System Assgnment Power Dstrbuton, 0, 0, 0 9
2 3, 3, 3, 0 3 2, 6, 6, 6 4 7 0, 0, 0, 0 5 5 2, 4, 4, 2 6 3 5, 5, 5, 0 7 3, 3, 6, 6 8 2, 3 0, 0, 0 9 2, 2, 0, 0 0 4, 4, 4, 4 2 5, 5, 5, 5 2 3 8, 3 8, 8, 8 3 3, 3, 3, 0 4 4, 4, 4, 4 5 0, 0, 0, 0 We made an attempt at fndng the vald assgnments for the fve player system. However, due to the dffculty of ths task, we abandoned t and turned to a more general queston. 7 Upper Bound As we have sad, t s relatvely easy to determne the vald assgnments for a votng system wth four or fewer players. Wth fve players, t becomes long and tresome; the sx-player system s even worse. Therefore, we changed our goal from fndng the vald assgnments to fndng a reasonable upper bound on the number of vald assgments. The smplest upper bound s the sze of the power set of the set of the coaltons of an n-player system, 2 2n. Each set of coaltons corresponds to the set of wnnng coaltons for some assgnment. Thus, ths expresson counts every assgnment. However, 2 2n s massve for n 3 and many of these assgnments are nvald. Tne next step s to remove some of these nvald assgnments. We know, for nstance, that should be consdered losng n a vald assgnment. In other words, any assgnment that labels as wnnng s nvald. Therefore, we can take out of consderaton. Ths makes the upper bound 2 2n. We extend ths dea to all non-wnnable coaltons. In 2 2n, we allow all nonempty coaltons to be consdered wnnng. However, we know that some nonempty coaltons are losng n every vald assgnment. A better upper bound only counts assgnments n whch these coaltons are losng. Ths means that only wnnable coaltons are allowed to be wnnng, whch makes our second upper bound 2 2n ( n ) 2. Let s compare these two upper bounds. 0
n 2 2n 2 2n ( n ) 2 Actual 3 28 32 6 4 32,768,024 5 5 2,47,483,648 4,94,304 63 6 9,223,372,036,854,775,808 7,592,86,044,46?? 7. Incorporatng Basc Concepts nto the Upper Bound Unfortunately, the second upper bound stll greatly overcounts the number of vald assgnments. One bg problem s that t does not take nto account the basc concepts that were dscussed earler n the paper. By takng these deas nto consderaton, we can construct a better upper bound. Thanks to the basc concepts, f we know the sze of c, then we know the status of coaltons that are ether supersets of c or dsjont from c. We also know how many coaltons of these types there are. Moreover, we know how many coaltons there are n total. Therefore, we know precsely how many other coaltons could be ether wnnng or losng. Compared wth the earler upper bound, there are fewer coaltons that we are unsure of. These are the coaltons that are nether completely dsjont from c nor exactly a superset of c. Of course, f we do know whch players are part of c, then we mght be able determne the status of some of these coaltons. But, from the standpont of the algorthm, whch only knows the sze of c, all of these coaltons are uncertan. To further reduce the number of uncertan coaltons, we can say that all coaltons that are smaller than c are losng. Ths s permssble because we count from smaller to larger coaltons. In other words, we start wth coaltons of sze, the smallest nonempty coaltons. We consder each coalton of sze n the manner descrbed above. Then, we move to coaltons of sze 2. Snce we have counted all of the vald assgnments (and some nvald assgnments) that have coaltons of sze labeled as wnnng, we can say coaltons of ths sze are losng wthout runnng the rsk of mssng a vald assgnment. So, n order to count the number of uncertan coaltons when we consder a coalton c of sze s to be wnnng, we consder three quanttes. Frst, the number of coaltons of greater or equal sze to c. n ( ) n C(s) = Second, the number of coaltons of greater or equal sze that are dsjont from c. n s ( ) n s D(s) = Thrd, the number of supersets of c. =s =s n s ( ) n s S(s) = =
Therefore, the number coaltons that could be ether wnnng or losng n a vald assgnment where c wns s: F (s) = n =s ( ) n n s ( ) n s n s ( ) n s = =s In ths equaton, represents the coalton c, whch s known to be wnnng. However, we can thnk of n s ( ) n s n s ( ) n s c as beng part of the term countng the supersets. In ths way, becomes. Thus, we can reduce these two terms to 2 n s. So, F (s) becomes: F (s) = n =s ( ) n = n s ( ) n s 2 n s =s =0 So, there are at most 2 F (s) vald assgnments n whch coalton c of sze s wns. Of course, ths s not yet the formula that gves an upper bound. formula, we need to present the followng lemmas. In order to explan the complete Lemma 7.. There s only one vald assgnment n whch the coalton consstng solely of P wns. Proof. Every coalton ether contans P (and s therefore a superset) or does not contan P (and s therefore dsjont). Thus, by the basc concepts, f we say {P } s wnnng, we know whether any other coalton s wnnng or losng. Ths completely determnes the assgnment n whch {P } s wnnng. Lemma 7.2. There s only one vald assgment n whch the coalton contanng all of the players loses. Proof. If the coalton contanng all of the players loses, then the sum of the weghts of all players does not meet the quota. Every other coalton contans fewer players than ths coalton. Therefore, every other coalton wll have fewer (or equal) aggregate votes compared to ths sum. Thus, no other coalton can meet the quota. So, the assgnment n whch the coalton of all players loses conssts of all coaltons labeled L. Ths follwong s the formula that gves an upper bound on the number of vald assgnments. 3 + n/2 s=2 ( n ) s 2 F (s) + n s= n/2 + ( ) n 2 F (s) s The prevous two lemmas account for 2 of the 3 assgnments at the begnnng of the formula. Ths 3 also counts the case n whch the coalton of every player s the only coalton that wns. After these extreme cases are taken care of, we focus on coaltons that are n between sze and sze n. To reterate, the formula 2
counts the number of wnnable coaltons of sze s (whch can be found n the secton about the wnnables) and multples ths quantty by the maxmum number of vald assgnments that have a coalton of sze s as wnnng (whch was presented a few paragraphs earler). Let s compare the upper bounds agan. n 2 2n ( n ) 2 New Bound Actual 3 32 5 6 4,024 29 5 5 4,94,304 22,963 63 6 7,592,86,044,46 35,295,402,79?? 7.2 Improvng the Upper Bound There are two man ways to mprove the upper bound. Frst, we can change the way n whch we count the number of ndetermnate coaltons. Instead of consderng all coaltons of sze s or greater, we only consder wnnable coaltons of sze s or greater. Second, we can change the way n whch we sum over the coaltons n order to consder a coalton to be losng mmedately after t s consdered wnnng, rather than watng untl the sze of the coaltons changes. The frst task s to change C(s) so that only wnnable coaltons are counted n the frst place. n 2 ( ) n C(s) = + =s n =Max( n 2 +,s) ( ) n As you can see, f s s less than the floor of n 2, then the countng begns wth coaltons of sze s. If s happens to be greater than the floor of n 2, the expresson Max( n 2 +, s) ensures that we also start countng at s, not at the floor of n 2. We do not need to change the term that counts the supersets of c because c s wnnable and therefore every superset of c s wnnable. Thus, only wnnable coaltons are beng counted by S(s). We do, however, need to change the term that counts the dsjonts of c. Although c s wnnable, a coalton that s dsjont to c may or may not be wnnable. If we do not change D(s) then we wll have the problem of subtractng coaltons that were never added n the frst place. We do not know whch coaltons that are dsjont to c are wnnable. But do know that all coaltons that are bgger than the floor of n 2 are wnnable. Therefore, we count only the dsjont coaltons that are bgger than the floor of n 2. Ths looks lke: D(s) = n s =Max( n 2 +,s) ( ) n s 3
Thus, the new way of countng ndetermnate coaltons s: n 2 ( ) n G(s) = + =s n =Max( n 2 +,s) ( ) n 2 n s n s =Max( n 2 +,s) ( ) n s Here s a comparson between F (s) and G(s). n Bound Usng F(s) Bound Usng G(s) Actual 3 5 5 6 4 29 63 5 5 22,963 6,523 63 6 35,295,402,79 8,459,649,29?? The second task s to consder a coalton c to be losng mmedately after t has been consdered wnnng. Ths s accomplshed by turnng the factor that represents the number of wnnables of a certan sze nto a summaton. n 2 3 + s=2 ( n s ) =0 2 G(s) + n s= n 2 + ( n s) =0 2 G(s) Here s a fnal comparson of upper bounds. 8 References n Prevous Bound New Bound Actual 3 5 0 6 4 63 78 5 5 6,523 6,54 63 6 8,459,649,29,206,353,970?? Mathematcs Magazne: Volume 76, Number, Pages: 33-39 2003 Power Dstrbutons n Four-Player Weghted Votng Systems John Tolle 4
Wkpeda (2009). Catalan Number. Retreved Jun 5, 2009 from: http://en.wkpeda.org/wk/catalan numbers#second proof A Graphcal Analyss We also used graphs to analyze weghted votng systems. In fact, we can completely llustrate the cases of two and three players. To do ths, we thnk of the weghts and the quota as fractons of the total number of votes, nstead of thnkng of these numbers n absolute terms. A. Inequaltes Suppose there are three players. Let the x-axs represent the fracton of the total number of votes that belongs to P 3. The value of the x-coordnate s therefore between 0 and 3. Let the y-axs represent the fracton of the total number of votes that belongs to P 2. The value of y-coordnate s therefore between x and x 2. In other words, P 2 has at least as many votes as P 3 (because v 2 v 3 ) and at most as many votes as P, whch s equal to half of the remanng fracton of votes. Once the x-coordnate and y-coordnate are determned, the fracton of the votes that belongs to P s known, because ths number s x y. Thus, the nequaltes 0 x x 3 and x y 2 defne the area that represents all possble three-player weghted votng systems. 5
We can make a graph of the vald assgnments n a weghted votng system by fndng the condtons under whch each of the wnnable coaltons s wnnng. In the three-player system, the condtons are the followng nequaltes. Suppose {P } wns. Suppose {P, P 2 } wns. Suppose {P, P 3 } wns. Suppose {P 2, P 3 } wns. Suppose {P, P 2, P 3 } wns. q x y y x + ( q) q x y + y q x x q q x y + x q y y q q x + y x + q y y x + q q x y + y + x q How do these nequaltes appear on a graph? Suppose q = 5 6. 6
In ths graph, the green area represents the vald assgnment n whch {P, P 2 } s the smallest coalton that wns. The purple area represents the the vald assgnment n whch {P, P 2, P 3 } s the smallest (and only) coalton that wns. The yellow area represents the vald assgnment n whch {P, P 3 } s the smallest coalton that wns. Fnally, the blue area represents the vald assgnment n whch {P } s the smallest coalton that wns. Two assgnments are mssng: the one n whch {P 2, P 3 } wns and the one n whch no coalton wns. Ths s because these two assgnments do not occur when q s between 2 3 and. In case you cannot see the colors, here are the boundares for each vald assgnment. {P, P 2 } s the smallest coalton that wns. The power dstrbuton s 2, 2, 0. Ths assgnment s bound on the bottom by y = q = 6. It s bound on the rght by x = q = 6. {P, P 2, P 3 } s the smallest (and only) coalton that wns. The power dstrbuton s 3, 3, 3. Ths assgnment s bound on the left by x = q = 6. {P, P 3 } s the smallest coalton that wns. The power dstrbuton s 3 5, 5, 5. Ths assgnment s bound on the top y = q = 6. It s bound on the left by y = x + ( q) = x + 6. {P } wns. The power dstrbuton s, 0, 0. Ths assgnment s bound on the top-rght by y = x + ( q) = x + 6. 7
A.2 Three Dmensonal Graphs In ths secton, the quota wll be gven a dmenson, namely the z-axs, because the quota has a drastc effect on the two-dmensonal graph. The graph below llustrates all possble weghted votng systems wth 3-players. Agan, the colored regons correspond to dfferent vald assgnments. To reterate, the blue area corresponds to the assgnment wth power dstrbuton, 0, 0, red to the one n whch every par of players wn, purple to the one n whch only the coalton of all players wns, green to the one wth power dstrbuton 2, 2, 0 and yellow to the one wth power dstrbuton 3 5, 5, 5. 8