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You don't need a full sort. It would work, but it's overkill.
Essentially you have an array (or list, or collection of some sort) of card values, indexed by player.
The standard linear-time algorithm for finding the max or min of an array and remembering where you saw it...
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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Very roughly in pseudo-code for traditional languages (no particular language)
nPlayers = 7;
int cardForPlayer[nPlayers] = (cards for each of the players);
int bestPlayerNo = 1;
int bestScore = cardForPlayer[0];
for (int playerIndex = 1; playerIndex++; playerIndex < nPlayers)
if (cardForPlayer[playerIndex] > bestScore)
{
bestScore = cardForPlayer[playerIndex];
bestPlayerNo = playerIndex + 1;
}
For a simple functional style language (no particular language)
cardForPlayer = List p1card, p2card, ..., p7card
bestScore = max cardForPlayer
bestPlayerNo = (FindIndex cardForPlayer bestScore) + 1
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Give OP's seniority and rep, I didn't condescend to spelling out the algorithm.
I've lost track of the number of times and different languages I've implemented it...
There is a useful variant where you start with a sentinel value ("minus infinity" or whatever) and compare every one including the first.
This handles the case where there is no item that qualifies for selection.
In this example, it would be easy to skip a player if he has no card. What if nobody had any cards?
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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You have a "list" of players.
The list is always processed in order; so the "first" element in the list should be the the player that plays first, etc.
After the last player, you cycle back to the first. A "pass" just bumps the player index.
I wrote a GO game on that basis; for 2 to 7 players; humans and bots. 7 was an arbitrary limit (The Warring States).
The "top" player is the one with the .Max something; unless it's tied.
It was only in wine that he laid down no limit for himself, but he did not allow himself to be confused by it.
― Confucian Analects: Rules of Confucius about his food
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Suppose your operations on arrays were constrained as follows:
- .get(i) accesses the element at index position 𝑖 in read-only fashion.
- .length() returns the length of the array.
- .flip(i) flips the array up to index position 𝑖. Thus, the {1,2,3,4}.flip(2) operation changes the array to {3,2,1,4}.
Only these three operations are available to you. Also assume that the three operations are in 𝑂(1).
Develop an algorithm that sorts a given array in 𝑂(𝑛 × 𝑙𝑜𝑔 𝑛).
Suppose you have an array of length 𝑖 + 1. You want to correctly sort the element at position 𝑖. The array is already sorted up to 𝑖 - 1. The correct position of the element at position 𝑖 is index position 𝑗. Find a sequence of flip operations that inserts the element in the correct location.
Problem: I have no idea how to begin
modified 14-Dec-20 13:06pm.
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If you can figure out the last paragraph, you'll have a way to move an element into the correct position when it's out of place. That element might only be part way down the array, in which case you can move it up by only doing flips as far as that element, leaving the elements after it alone.
Start with an actual example, say 1235674. You want to move 4 up into the position currently occupied by 5. Clearly, 4 must be included in the first flip, else it'll never move. So flip(6) is the first operation, which yields 4765321. Now the ball is in your court.
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They don't say to flip one way or another. Looks more like a rotate. Poor spec.
It was only in wine that he laid down no limit for himself, but he did not allow himself to be confused by it.
― Confucian Analects: Rules of Confucius about his food
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Read on the lines of bubble sort. That should help solving this
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Could help me solve this problem pls?
The job is to calculate the degree of discontent of passengers on a bus.
Passengers get on the bus, all at the same point and say at which point they want to stop (that point is the mileage from the point where they get on the bus to where they want to disembark), but the bus can only make k stops, that is, it is a value limited number of stops. The discontent will be calculated as follows (x-y) * 2 where x is the place that each passenger chose to stay and y where the bus stopped. The K will be defined after calculating the discontent of each passenger
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1. Nobody here is going to do your homework for you.
2. That's a maths problem, not a programming problem.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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If we do not take into account the growth of passenger dissatisfaction due to an excessively large number of stops, then in a first approximation the formula will look like this ((x - y) / k) * 2 But this is just my opinion.
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Hi,
I am building a small web app that allow stamp collectors to swap their stamps. The main problem I am trying to solve is 1-on-1 (direct) swap. Since usually A needs something from B, but B not always need something from A.
But the solution is to find a circular swap, for example:
A needs from B
B needs from C
C needs from A
result: everyone is happy.
I have a list of collectors, and for each a list of stamps they got, and a list of stamp they are looking for.
I am trying to find a way, to create a circle, by matching the "have" and "want". The problem is that this may take a very long time with lots of collectors and lots of stamps  .
What would be the best approach for this?
The straight forward algorithm I think (which is obviously not optimized) is to:
1) look what "B" want.
2) find who got what B want (stamp after stamp)
3) for each one, see if he (C) needs something from A
4) if yes, YAY
5) if not, see what C want (stamp after stamp)
6) for each one, see if (D) has something from A
... and so on ...
long procedure...
Any suggestions will be very welcome. Thank you
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Just advertise stamps (for sale) and what the seller wants. Period.
It was only in wine that he laid down no limit for himself, but he did not allow himself to be confused by it.
― Confucian Analects: Rules of Confucius about his food
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In mathematical terms your problem can be stated as finding the cycles of a directed graph. Collectors A, B, C... are nodes in graph and what they want are graph arcs.
Now, if you Google for "finding graph cycles" one of the first results should be Tarjan's strongly connected components algorithm - Wikipedia[^]
You might have already found a solution but anyways...
Mircea
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Thanks!
Yes, you right, I already find this path of solution and working on it.
Thank you for taking the time to answer 
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This is my first question in this forum, so suggestions for improvement are welcome.
I have a data set of arrays which contain Floats between -1 and 1 and a new array of the same format.
The arrays have a length of up to 300.
I have to find the array from the data set that has the smallest distance to the new array.
IMPORTANT: The values of the arrays must remain on their position, while it's alowed to sort the arrays in the data set.
How can I achieve this while keeping an acceptable trade-off between accuracy and scalability?
Of course I could just go through every array but that wouldn't be scalable I guess.
PS: Usually adjacent values have a similar value (so [-0.8, 0.3, -0.7, 0.9] would be very unlikely).
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CLOSED
modified 16-Nov-20 11:38am.
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By identical, I assume you mean isomorphic. See here[^], which speculates that the problem may be NP-intermediate (between P and NP-complete).
Let's call the two graphs G1 and G2. On the surface, the problem seems NP-complete because you have to compare G1 and G2 after renaming the n vertices in G2 using names from G1, so there are n! combinations to try. However, some combinations can be filtered out quickly. For example, a vertex in G2 has to have the same degree (number of edges) as the one it is being compared to in G1. But if this check passes, there's still more checking to do, as in the case of the graph with 6 vertices shown in your link.
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This "same name" talk is confusing. The general problem is that the two graphs have different names for their vertices, so you have to try all possible ways of mapping one graph onto the other. So even if there's there's a vertex A in G1 and a vertex A in G2, that doesn't mean that you don't have to compare A in G1 with B in G2, and so on. You'd only need to compare A to A if you're deliberately being told to test a specific mapping, ignoring all others, or when you're solving the general problem and evaluating each possible mapping in succession.
The neighbours must be the same, not just the degrees. Your example, where all the vertices have the same degree, shows why the problem is difficult. If it's an undirected graph and all the degrees are n-1, then it's easy because both must be complete graphs. If not, then let's say we had a graph with 6 vertices, all of degree 2. How can you tell whether these are two disjoint graphs (two C3's) or a connected bipartite graph? You have to go beyond the vertices' degrees to do that.
On the surface, it looks NP-complete because there are n! possible mappings, and n! is approximated by an exponential function.
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CLOSED
modified 16-Nov-20 11:39am.
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Your first paragraph is correct. So is your second paragraph, but of course all combinations have to be tried before concluding that the graphs are not isomorphic.
Your example shows why the problem is difficult. You have to try all combinations, though optimizations certainly exist. Trivially, the number of vertices in both graphs has to be the same. And if you sort their vertices' degrees, the two sequences have to be the same. But like the 6-vertex graph demonstrated, that isn't enough. There are undoubtedly more optimizations, which is why the article that I linked to speculated that the problem is NP-incomplete (easier than NP-complete). But you'd have to search the net to find the details of those algorithms.
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CLOSED
modified 16-Nov-20 11:39am.
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It appears that I'm more or less answering a homework question, which isn't a good idea for you.
Yes, you have to satisfy the neighbour condition (which will automatically satisfy the degree condition). n is the number of vertices in G2, but G1 must also have n vertices, otherwise the graphs couldn't possibly be isomorphic. There are n! possibilities because you need to try all possible permutations when mapping G2's vertices to G1's. Maybe you get lucky and discover that they're isomorphic when you check the first permutation, or maybe you get unlucky and have to try them all.
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