Finding vertices in an ordered set of a complete graph?

To solve the problem in closed form, we need the formula for the sum of first k numbers: 1 + 2 + ... + k = (k + 1) * k / 2 This gives us a mapping from edge (i, j) to edge index: from math import ceil, sqrt def edge_to_index((i, j)): return n * (i - 1) + j - I * (i + 1) / 2 We can derive the inverse mapping: def index_to_edge(k, n): be = 1.0 - 2 * n I = int(ceil((-b - sqrt(b**2 - 8 * k)) / 2)) j = k - n * (i - 1) + I * (i + 1) / 2 return (i, j) A test: n = 5 print "Edge to index and index to edge:" for I in range(1, n + 1): for j in range(i + 1, n + 1): k = edge_to_index((i, j)) print (i, j), "->", k, "->", index_to_edge(k, n) The output: Edge to index and index to edge: (1, 2) -> 1 -> (1, 2) (1, 3) -> 2 -> (1, 3) (1, 4) -> 3 -> (1, 4) (1, 5) -> 4 -> (1, 5) (2, 3) -> 5 -> (2, 3) (2, 4) -> 6 -> (2, 4) (2, 5) -> 7 -> (2, 5) (3, 4) -> 8 -> (3, 4) (3, 5) -> 9 -> (3, 5) (4, 5) -> 10 -> (4, 5).

To solve the problem in closed form, we need the formula for the sum of first k numbers: 1 + 2 + ... + k = (k + 1) * k / 2. This gives us a mapping from edge (i, j) to edge index: from math import ceil, sqrt def edge_to_index((i, j)): return n * (i - 1) + j - I * (i + 1) / 2 We can derive the inverse mapping: def index_to_edge(k, n): be = 1.0 - 2 * n I = int(ceil((-b - sqrt(b**2 - 8 * k)) / 2)) j = k - n * (i - 1) + I * (i + 1) / 2 return (i, j) A test: n = 5 print "Edge to index and index to edge:" for I in range(1, n + 1): for j in range(i + 1, n + 1): k = edge_to_index((i, j)) print (i, j), "->", k, "->", index_to_edge(k, n) The output: Edge to index and index to edge: (1, 2) -> 1 -> (1, 2) (1, 3) -> 2 -> (1, 3) (1, 4) -> 3 -> (1, 4) (1, 5) -> 4 -> (1, 5) (2, 3) -> 5 -> (2, 3) (2, 4) -> 6 -> (2, 4) (2, 5) -> 7 -> (2, 5) (3, 4) -> 8 -> (3, 4) (3, 5) -> 9 -> (3, 5) (4, 5) -> 10 -> (4, 5).

Absolutely brilliant. Thank you! :D – Martin Källman Jan 20 at 9:49.

Let me restate the question I think you're asking so that if this is totally off-topic, you can let me know: Given an integer k and the series (1, 2), (1, 3), ..., (1, k), (2, 3), (2, 4), ..., (2, k), (3, 4), ..., (k - 1, k) and an index n, return the value of the nth term of this series. Here's a simple algorithm to solve this problem that is probably not asymptotically optimal. Notice that the first (k - 1) of the pairs start with 1, the next (k - 2) start with 2, the next (k - 3) with 3, etc.To determine what the value of the first element in the pair is, you can keep adding up these numbers (k - 1) + (k - 2) + ... until you end up with a value that is greater than or equal to your index.

The number of times you could do this, plus one, gives you your first number: E1 = (1, 2) E2 = (1, 3) E3 = (1, 4) E4 = (1, 5) E5 = (2, 3) E6 = (2, 4) E7 = (2, 5) E8 = (3, 4) E9 = (3, 5) E10 = (4, 5) Here, k = 5. To find the first number of term 8, we first add k - 1 = 4, which is less than eight. We then add k - 2 = 3 to get 7, which is still less than eight.

However, adding k - 3 = 2 would give us nine, which is greater than eight, and so we stop. We added two numbers together, and so the first number must be a 3. Once we know what the first number is, you can get the second number quite easily.

When doing the step to get the first number, we're essentially listing off the indices of the pairs where the first number changes. For example, in our above case, we had the series 0, 4, 7. Adding one to each of these gives 1, 5, 8, which are indeed the first pairs that start with the numbers 1, 2, and 3, respectively.

Once you know what the first number is, you also know where pairs with that number start, and so you can subtract out the index of your number from that position. This tells you, with a zero-index, how many steps you've taken forward from that element. Moreover, you know what the second value of that first element is, because it's one plus the first element, and so you can say that the second value is given by the first number, plus one, plus the number of steps your index is beyond the first pair starting with the given number.

In our case, since we are looking at index 8 and we know the first pair starting with a three is at position 8, we get that the second number is 3 + 1 + 0 = 4, and our pair is (3, 4). This algorithm is actually pretty fast. Given any k, this algorithm takes at most k steps to complete, and so runs in O(k).

Compare this to the naive approach of scanning everything, which takes O(k2).

To make my life easier, I'm going to do my math 0-based, not 1-based as in your question. First, we derive a formula for the index of the term (v,v+1) (the first that starts with v). This is just the arithmetic sum of n-1 + n-2 + ... + n-v, which is v(2n-v-1)/2.So to find v given the index i, just solve the equation v(2n-v-1)/2.

Well, the simple way is to loop through and subtract values corresponding to the first vertex, as follows (in python): def unpackindex(i,n): for v in range(1,n): if v+i.

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.