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# [Java] A really hard algorithm implementation

So I wanted to solve this problem:

After a tennis tournament each player was asked how many matches he had.
An athlete can't play more than one match with another athlete.
As an input the only thing you have is the number of athletes and the matches each athlete had. As an output you will have 1 if the tournament was possible to be done according to the athletes answers or 0 if not. For example:

Input: 4 3 3 3 3 Output: 1
Input: 6 2 4 5 5 2 1 Output: 0
Input: 2 1 1 Output: 1
Input: 1 0 Output: 0
Input: 3 1 1 1 Output: 0
Input: 3 2 2 0 Output: 0
Input: 3 4 3 2 Output: 0

the first number of the input is not part of the athletes answer it's the number of athletes that took part in the tournament for example in 6 2 4 5 5 2 1 we have 6 athletes that took part and their answers were 2 4 5 5 2 1

This is what I have wrote so far and it is not working completely:
Code:
``````import java.util.Scanner;
import java.util.Arrays;

public class Tennis {

public static void main(String[] args) {
Scanner input = new Scanner(System.in);

String N;
int count;
int sum = 0;
int max;
int activeAthletes;
int flag;

System.out.printf("Give: ");
N = input.nextLine();

String[] arr = N.split(" ");
int[] array = new int[arr.length];

for (count = 0; count < arr.length; count++) {
array[count] = Integer.parseInt(arr[count]);
//System.out.print(arr[count] + " ");
}

for (count = 1; count < arr.length; count++) {
sum += array[count];
}
//System.out.println("\n" + sum);

activeAthletes = array[0];

for (count = 1; count < array.length; count++) {
if (array[count] == 0) {
activeAthletes--;
}
}

max = array[1];
for (count = 2; count < array.length; count++) {
if (array[count] > max) {
max = array[count];
}
}
// System.out.println(max);

if ((sum % 2 == 0) && (max < activeAthletes)) {
flag = 1;
} else{
flag = 0;
}

System.out.println(flag);
}
}```
```
Okay, so aren't there 3 cases, there must be an even number of matches, no one is allowed to have more than n-1 matches where n is the total number of players and no one is allowed to have greater than half the number of total matches right?

Did I miss a case?

Edit:
I'm in class right now, and quickly looking your code, I don't think you accounted that each athelete can only play n-1 matches (they can't play themself)

Edit2:
Nvm, I just saw your max < athletes in the true case. What cases is it failing on?

Edit3: according to your answers, the case of 1 0, where there is only one athlete, who answers 0, that can't be a tournament
Edited by AchuSaysBlessYou - 4/25/12 at 2:16pm
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Correct me if I'm wrong, but my first instinct is that this is a pretty simple graph/combinatoric problem. Do the numbers violate the handshaking lemma? Given N players, does any player state that they played more than N-1 matches? Does the total number of matches exceed N choose 2? If all three answers are no, the tournament is possible.
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That's exactly it I think, I'll test it and let you know
Quote:
Originally Posted by blangblangÂ

Correct me if I'm wrong, but my first instinct is that this is a pretty simple graph/combinatoric problem. Do the numbers violate the handshaking lemma? Given N players, does any player state that they played more than N-1 matches? Does the total number of matches exceed N choose 2? If all three answers are no, the tournament is possible.

I think the total matches have to be 2 * n choose 2 since each game will be reported twice
 My First Child (15 items)
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Quote:
Originally Posted by AchuSaysBlessYouÂ

I think the total matches have to be 2 * n choose 2 since each game will be reported twice

Good catch, that's exactly right.
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I think the max matches could be n * (n-1) , that your input should not exceed for the tournament to be possible.

For 4 players:-

B: BA,BC,BD :3
C: CA,CB,CD :3
D: DA,CB,CD :3

= (n-1) * n = (4-1) * 4 = 12

For 5 players:-

B: BA,BC,BD,BE :4
C: CA,CB,CD,CE :4
D: DA,CB,CD,CE :4
E: EA,EB,EC,ED :4

= (n-1) * n = (5-1) * 5 = 20

But then as each player gets a max of n-1 matches, you will have to check on false inputs like your last case.
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I think this is actually a lot harder than we are all giving it credit for. The minimum number of matches any one (or more) players played is also a factor.

8 2 1 1 0 0 0 0 0 = 1
8 2 2 0 0 0 0 0 0 = 0

the total number of matches must be less than 2n-1 (.. and %2==0 for input validation), but there is no lower bound. that case there to me looks like a graph problem. hrmmm let me get my notebook and draw this up a bit
Edited by lloyd mcclendon - 4/26/12 at 11:43pm
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As has already been pointed out, the maximum number of games is n(n-1). The more difficult thing is when you have fewer than the maximum. I just ran through plotting all the minimum combinations of n=2 through n=10 in Excel to confirm my suspicion. The minimum number of games is 2(n-1). Here are the results for the some valid combinations of the minimum number of games for 2 through 10 (code used for monospaced font):
Code:
``````2  1 1
3  2 1 1
4  2 2 1 1
5  2 2 2 1 1
6  3 2 2 1 1 1
7  3 3 2 1 1 1 1
7  3 2 2 2 1 1 1
8  3 3 2 2 1 1 1 1
9  3 3 2 2 2 1 1 1 1
9  4 2 2 2 2 1 1 1 1
10 4 3 2 2 2 1 1 1 1 1```
```

There is a clear pattern here for the minimum number of games. The tricky thing is that once you get to a higher value n (this seems to start at 7), the combinations can start shifting. As far as I can tell, this requires a log(2) algorithm that will divide n/2 repeatedly. Any time that the dividend is odd, the player with the most games may steal a game from middle player. However, games played can also be shifted in other ways, as demonstrated by the second example of 9. I haven't thought about this much but it seems like the only time when there is only one combination of minimum games is when n is a power of 2.

Hope that made sense and there's no glaring error there. It's 2 AM and I don't know why I'm up doing this when I have a paper to write.
Edited by ChaoticKinesis - 4/26/12 at 11:10pm
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it's definitely a graph matching problem. It's beyond number crunching as different cases will have different results based on how one thing relates to the other things. You can check up front that n%2==0 && Em < 2n-1, but if that's met, you have to build the graph

lay out the nodes that have a nonzero edge count, draw the edges but leave them unconnected.
for each edge, try to connect it to another node's available edge and mark both of them off
is there _any way to connect the graph such that all edges are connected and no two nodes have more than one connection to each other
Code:
``````2 2 1 1

A B C D
A X 1
B 1 X
C      X
D        X```
```

now .. i'm not even sure if that's the right and correct approach yet, but thinking through it a bit i think it would at least work. as far as coding it, i haven't got much besides a brute forceish type approach with 2 or 3 for loops or recursion .. there's probably a clean way. it reminds me a little bit of the stable marriage problem, but it's not a bipartite graph...
Edited by lloyd mcclendon - 4/26/12 at 11:37pm
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