I do not exactly understand what skipping the middle pair means, and the program works in situations where a cycle is created when I test it. How do I fix this? Learn more. Asked 2 months ago. Active 2 months ago. Viewed 1k times.
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Anne Welcome to SO! There was a pretty similar question asked yesterday and the solution might be helpful. Active Oldest Votes. Sign up or log in Sign up using Google. Sign up using Facebook.
It only takes a minute to sign up. My program currently passes all the check50 tests up until I have to lock in the pairs to the locked array. You must lock all pairs in order which check50 confirms mine areup until the point that locking in the next pair would cause a loop in the votes system. My logic is this. In order to not have a loop, there must be at least one candidate that is not beaten by any other candidate in a two candidate stand off. Therefore, I must stop locking pairs at the point in which the loser for that pair is the final candidate that hasn't yet lost if indeed this ever happens.
To achieve this, I first create an array of unique losing candidates ,but omit the final candidate to lose if indeed this happens. I then proceed to copy each pair into the array, only after having checked if the loser for that pair is in my list of 'permitted loser'.
There very well could be an issue with the code, and if so then I will just continue trying to correct it myself, but I just wanted to get some other opinions on the logic. Clearly if there is a hole in the logic then checking and re-checking my code is pointless. I don't exactly understand your logic, I found something that feels somewhat simpler in an answer on the new cs50 ed platform.Bsod errors
So my code checks the nodes reachable from B, over any number of steps over already locked edges avoiding some duplicate checks, but as there are no cycles so far, it would at least be finite. If A is in that set reachable from B, I discard the pair.
This logic does not work, because the loops can have branches. So you could have loops with branches that make them look like a 6 or a 9. The simplest way I found to Lock pairs into the candidate graph in order, without creating cycles, was to create a function that detects loops by iterating throughout the whole loop via two nested for loops.
One for loop to check if the original strand of code forms a loop another for loop to see if a branch of that code created a loop.
These for loops reset with goto commands after they found the next iteration of the loop. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Asked 6 months ago. Active 23 days ago. Viewed 3k times. I'm looking for a little help to check my logic on this problem set.Ranked pairs RP or the Tideman method is an electoral system developed in by Nicolaus Tideman that selects a single winner using votes that express preferences.
RP can also be used to create a sorted list of winners. If there is a candidate who is preferred over the other candidates, when compared in turn with each of the others, RP guarantees that candidate will win.Fxs vs fxo
Because of this property, RP is, by definition, a Condorcet method. RP can also be used to create a sorted list of preferred candidates. To create a sorted list, repeatedly use RP to select a winner, remove that winner from the list of candidates, and repeat to find the next runner up, and so forth. To tally the votes, consider each voter's preferences. B, one for A in A vs. C, and one for B in B vs.Ark ragnarok clay location
Voters may also express indifference e. Once tallied the majorities can be determined. The pairs of winners, called the "majorities", are then sorted from the largest majority to the smallest majority. A majority for x over y precedes a majority for z over w if and only if one of the following conditions holds:. Condorcet cycle evaluation can be visualized by drawing an arrow from the pair's winner to the pair's loser in a directed graph.
Using the sorted list above, lock in each pair in turn unless the pair will create a circularity in the graph for example, where A is more than B, B is more than C, but C is more than A.
In the resulting graph for the locked pairs, the source corresponds to the winner. A source is bound to exist because the graph is a directed acyclic graph by construction, and such graphs always have sources.
In the absence of pairwise ties, the source is also unique because whenever two nodes appear as sources, there would be no valid reason not to connect them, leaving only one of them as a source. Imagine that Tennessee is having an election on the location of its capital.
The population of Tennessee is concentrated around its four major cities, which are spread throughout the state. For this example, suppose that the entire electorate lives in these four cities and that everyone wants to live as near to the capital as possible.
Note that absolute counts of votes can be used, or percentages of the total number of votes; it makes no difference since it is the ratio of votes between two candidates that matters.
The votes are then sorted. Knoxville will be added first, followed by Nashville vs. Every "lock in" would add another arrow to the graph showing the relationship between the candidates.
Here is the final graph where arrows point away from the winner. In this example, Nashville is the winner using RP, followed by Chattanooga, Knoxville, and Memphis in second, third, and fourth places respectively. In the example election, the winner is Nashville.Nestle competitors analysis
This would be true for any Condorcet method. Using the First-past-the-post voting and some other systems, Memphis would have won the election by having the most people, even though Nashville won every simulated pairwise election outright. Using Instant-runoff voting in this example would result in Knoxville winning even though more people preferred Nashville over Knoxville. Of the formal voting criteriathe ranked pairs method passes the majority criterionthe monotonicity criterionthe Smith criterion which implies the Condorcet criterionthe Condorcet loser criterionand the independence of clones criterion.
Ranked pairs fails the consistency criterion and the participation criterion. While ranked pairs is not fully independent of irrelevant alternativesit still satisfies local independence of irrelevant alternatives.
Ranked pairs fails independence of irrelevant alternatives.Lotus esprit v8 for sale
It only takes a minute to sign up. Can anyone help check what is wrong with my logic? My logic is that the program will check whether a winner in any pairs has been a loser or not. If count has been changed, then pairs would not be locked to avoid the cycle. Consider voting of three candidates A, B, C where sorted pairs as follows: 1.
Here, first pair meets your condition for incrementing the value of count but there looks no harm in assigning true to this combination. Hint: Consider an example of candidates, shape a cyclic graph and write down your pairs matrix.
You may be required to navigate recursively through the locked matrix to avoid cyclic graph. Sign up to join this community.
The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Ask Question. Asked 2 months ago. Active 2 months ago. Viewed times. Active Oldest Votes. I hope it gives you some idea to proceed with the issue.
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With some help from TomKarzes, I managed to refine my code slightly, but for some reason I am now unable to fulfil one of the question's requirements i. Even when I use my original code without Tom's inputit still does not work anymore, so I am really confused like When I run it through Check50, a built-in checker to see if my code has fulfilled the question's requirements, this is what I get:. If anyone can take the time to explain to me, in terms of efficiency and perhaps "design of code", when recursion is preferred to iteration and vice versa, that would be greatly appreciated as well!
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Feedback post: New moderator reinstatement and appeal process revisions. The new moderator agreement is now live for moderators to accept across the…. Allow bountied questions to be closed by regular users. Hot Network Questions.You already know about plurality elections, which follow a very simple algorithm for determining the winner of an election: every voter gets one vote, and the candidate with the most votes wins. But the plurality vote does have some disadvantages.
What happens, for instance, in an election with three candidates, and the ballots below are cast? A plurality vote would here declare a tie between Alice and Bob, since each has two votes. But is that the right outcome? In a ranked-choice system, voters can vote for more than one candidate. Instead of just voting for their top choice, they can rank the candidates in order of preference. The resulting ballots might therefore look like the below.
Here, each voter, in addition to specifying their first preference candidate, has also indicated their second and third choices. And now, what was previously a tied election could now have a winner.
The race was originally tied between Alice and Bob. But the voter who chose Charlie preferred Alice over Bob, so Alice could here be declared the winner. Ranked choice voting can also solve yet another potential drawback of plurality voting. Take a look at the following ballots. Who should win this election? In a plurality vote where each voter chooses their first preference only, Charlie wins this election with four votes compared to only three for Bob and two for Alice. Alice, however, might reasonably make the argument that she should be the winner of the election instead of Charlie: after all, of the nine voters, a majority five of them preferred Alice over Charlie, so most people would be happier with Alice as the winner instead of Charlie.
The graph for the above election, then, would look like the below. Likewise, the other arrows mean that more voters prefer Alice to Charlie, and more voters prefer Charlie to Bob. In this case, the source is Alice — Alice is the only one who has no arrow pointing at her, which means nobody is preferred head-to-head over Alice.
Alice is thus declared the winner of the election. Consider the below ballots. Between Alice and Bob, Alice is preferred over Bob by a margin. Between Bob and Charlie, Bob is preferred over Charlie by a margin. But between Charlie and Alice, Charlie is preferred over Alice by a margin.
It only takes a minute to sign up. I've been going through the CS50x course. It was doable so far but now I'm just stuck on the Tideman problem from p-set 3.
Tideman is a voting method that picks the Condorcet winner of an election, i. Here's a link to the problem description page. Yet, I am unable to understand what exactly 'middle pair' means here. I've documented the code as best as I can. However, I'm a total beginner here. Feel free to correct me wherever required.
Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Locking pairs in the Tideman problem Ask Question. Asked 4 months ago. Active 4 months ago. Viewed 1k times. Advitiay Anand Advitiay Anand 1 1 1 bronze badge. Hi, I am having the same issue as you did. Did you figure out in the end? I used a recursive function in the end.
Here's my 2 cents. Draw a complicated test case on a piece of paper, and use it as a reference to write a recursive function. Also, you could ask the experts at CS50's Discussion Forum us. Middle pairs can create cycles. Also, its really just a trio of for loops and a few if statements that you need.
With cycles, all that matters is that charlie beat alice, and alice beat bob, and charlie beat alice by a higher number than bob beat charlie. Active Oldest Votes. Featured on Meta. The new moderator agreement is now live for moderators to accept across the…. Feedback post: New moderator reinstatement and appeal process revisions.
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