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Discrete Math Exam Review
Terms in this set (57)
A ballot in which the votes are asked to rank the candidates in order of preference.
A ballot in which ties are not allowed.
Election of first place votes.
The candidate with the most first place votes.
If Candidate X is preferred by the voters over each of the other candidates in a head to head comparison, then candidate X should win.
Borda Count Method
Each place on a ballot is assigned points. In an election with N candidates, we give 1 point for last place, 2 points for second from last place, and so on.
Independence of Irrelevant Alternatives Criteria
If a Candidate X is winner of an election in a recount, one of the non-winning candidates is removed from the ballots, then X should still be a winner of the election.
Round #1: Count 1st place votes for each candidate, just as you would in the plurality method. If a candidate has a majority of first-place votes, that candidate is winner. Otherwise, eliminate the candidate (or candidates if there is a tie) with the fewest first place votes.
Round #2: Cross out the name(s) of the candidates eliminated from the preference and recount the first place votes (Remember that when a candidate is eliminated from the preference schedule, in each column the candidates below move up a spot.
If a Candidate X is a winner of an election and, in a reelection, the only changes in the ballots are changes that favor X (and only X), then X should remain a winner of the election.
The voters in a weighted voting system.
The votes each player controls a certain number of votes or the total number of votes in a coalition.
The minimum number of votes needed to pass a motion (yes-no vote).
***Can't be less than or equal to half the total voters. [7 : 5,4,4,2]
***Can't be larger than total votes. [17: 5, 4, 4, 2]
The player's weight is bigger than or equal to the quota.
---> [11: 12,5,4]
A player with no power.
---> [30: 10, 10, 10, 9]
If a motion can not pass unless a player votes in favor of the motion, then that player has veto power.
---> [12: 9, 5, 4, 2]
Any set of players that might join forces and vote the same way. The coalition consisting of all players is called a GRAND COALITION.
Some coalitions have enough to win and some don't. We called the former ... and the latter LOSING COALITIONS.
In a winning coalition, a player is said to be a critical player for the coalition if the coalition must have that player's votes to win.
The player that contributes the votes that run what was a losing coalition into a winning coalition.
(or the booty) the informal name we will give to the items being divided, is denoted by S.
Each player has an internalized value system.
The player's value system conforms to basic laws of arithmetic.
Willing participants accept all outcomes.
Each participant does not know the value system of the other participant.
Each player has a right to their fair share.
Items can be divided. The set S is divisible.
Items can not be divided. The set S is indivisible.
1) First person cuts a slice they valued as a fair share.
2) Second person examines the piece.
3) Each remaining person, in turn, can either pass or trim the piece.
4) After last person has made decision, last person to trim the slice receives it. If no one has modified the slice, then the person who cut it receives it.
5) Whoever receives the piece leaves with their piece and process repeats with the remaining people. Continue until only 2 people remain; they can divide what is left with the divider-chooser method.
Method of Sealed Bids
1) Bidding: Each player makes a bid (in $) for each of the items in the estate, given an honest assessment of the actual value of each item. Submits their own bid in envelope.
2) Allocation: Each item will go to the highest bidder (tie can be broken w/ coin flip)
3) 1st Settlement: Depending on what items (if any) a play gets in step 2, they will owe money to or be owed money by the estate.
4) Division of the Surplus: Surplus is common money that belongs to the estate, and thus to be divided equally among the players.
5) Final Settlement: Obtained by adding the surplus money to the 1st settlement obtained in step 3.
Method of Markers
1) Bidding: Each players independently divides the array into N segments by placing markers along the array.
2) Allocations: Scan array left to right until first marker is located. Player owning that marker goes 1st and gets the 1st segment in his bid. That players markers are removed, and we continue scanning left to right looking for the second marker.
Two critical elements in the definition of the word. We are dividing and assigning things. We are doing this on a proportional basis and in a planned, organized fashion.
The term we will use to describe the players involved in the apportionment. There are N states in any apportionment problem.
This term describes the set of M identical, indivisible objects that are being divided among the N states.
Each state has a population P: which is used as the basis for the apportionment of the seats to the states.
The quota rounded down; denoted by LQ.
The quota rounded up and denoted by UQ.
1) Calculate each state's standard quota.
2) Give each state its lower quota.
3) Give the surplus seats to the state with the largest fractional parts until there are no more surplus seats.
Most serious flaw of Hamilton's method. Occurs when an increase in the total number of seats being apportioned forces a state to lose one of its seats.
When state A loses a seat to state B even thous the population of A grew at a higher rate than the population of B.
The addition of a new state with its fair share of seats can, in and of itself, affect the apportionment of other states.
1) Find a suitable divisor D. (A suitable or MODIFIED DIVISOR is a divisor that produces an apportionment of exactly M seats when the quotas (population divided by D) are rounded down.)
2) Each state is apportioned its lower quota.
1) Find a suitable divisor D. (A suitable or MODIFIED DIVISOR is a divisor that produces an apportionment of exactly M seats when the quotas (population divided by D) are rounded up.)
2) Each state is apportioned its upper quota.
1) Find a "suitable" divisor D. (A suitable or MODIFIED DIVISOR is a divisor that produces an apportionment of exactly M seats when the quotas (population divided by D) are rounded conventionally.)
2) Start with standard divisor to find standard quota.
3) Use traditional rounding.
- If yes, then done.
- If seats apportioned too many, make SD bigger.
- If seats apportioned is too small, make SD smaller.
Two vertices are said to be adjacent if there is an edge joining them.
A path that passes through every edge of a graph once and only once. A path starts and ends at different vertices.
Circuit that passes through every edge of a graph once and only once. A circuit starts and ends at the same vertex.
Euler's Path Theorem
If a graph is connected and has exactly two odd vertices, then it has a Euler Path (at least one, usually more). Any such path must start at one of the odd vertices and end at the other one.
Euler's Sum of Degree's Theorem
The sum of the degrees of all the vertices of a graph equals twice the number of edges (and therefore is an even number). A graph always has an even number of odd vertices.
Visits every vertex on the graph once and only once.
Circuit that visits each vertex of the graph once and only once (at the end, of course, the circuit must return to the starting vertex).
If a connected graph has N vertices (N>2) and all of them have degrees greater than or equal to N/2, then the graph has a Hamilton circuit.
Go Cheap Strategy
Start from the hot city. From there, go to the city that is the cheapest to get to. From each new city, go to the next city that is cheapest to get to. When there are no more new cities to go to, go back home.
1) Make a list of all the possible Hamilton circuits of the graph.
2) For each Hamilton circuit, calculate its total weight (add the weights of all the edges in the circuit.
3) Choose an optimal circuit (there is always more than one optimal circuit to choose from).
1) From the starting vertex, go to its nearest neighbor ( take the edge with the smallest weight to the next vertex)
2) From each vertex go to its nearest neighbor, choosing only among the vertices that haven't yet been visited. (If there are edges with equal weights, choose at random). Keep doing this until all the vertices have been visited.
1) Pick the cheapest edge (with smallest weight) available.
2) Pick next cheapest link available.
3) Continue picking / marking cheapest unmarked link available that doesn't (a) close a circuit or (b) create three edges coming out of a single vertex.
A subgraph of a network that:
- Connects all the vertices of the network
- Has no circuits.
Minimum Spanning Tree
Among all spanning trees of a weighted network, one with the least total weight.
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