Say I have nodes A,B,C and D. If choose to start with A then B I can make a total of 2^(4-2) subsets:

For A then B

{A, B}
{A, B, C}
{A, B, D}
{A, B, C, D}

I can do the same by choosing first A then C:

For A then C

{A, C}
{A, B, C}
{A, C, D}
{A, B, C, D}

But as you can see there is an overlap, {A,B,C} and {A,B,C,D} occur in both lists.

How do I calculate the number of distinct subsets, keeping in mind that I would want to compare any number of these groups of subsets of length 2^(N-2)?

I can do this by remembering all the subsets generated for A then B and then only add distinct ones for A then C but I was hoping there is a shortcut since I will be working with a large number of nodes and I'm only interested in the number not in the actual subsets.

A little more information might help: the nodes are actually in a partially ordered graph, and the above mentioned Y will be equal to the number of edges in this graph.

  • $\begingroup$ I'm not sure I understand. Are you trying to enumerate the number of sets having at least two elements? If so, I don't think you can do better than $2^N - \binom{N}{0} - \binom{N}{1} = 2^N - 1 - N$. $\endgroup$ – Austin Mohr Jan 20 '12 at 12:28
  • $\begingroup$ No I'm looking for the number of sets which have 2 specific elements, for example A and B, and then I want to know the number of sets which have another 2 specific elements, for example B and C. This is obviously the same number, but then I want to know how many distinct subsets these two sets of subsets have. $\endgroup$ – Roy T. Jan 20 '12 at 12:48

EDIT: I've rephrased the answer to keep a more compact answer. I keep the mention to the binomial coefficient for consistency with the comments.

So the question boils down to this: if I have N nodes and Y subsets of size $2^{(N-2)}$ how many of these subsets are actually distinct?

At first, I thought it was possible to use the binomial coefficient $\left(\begin{array}{c}n\\ k\end{array}\right)$, but this is clearly not the case.

As mentioned by the other answers and comments, the inclusion/exclusion principle should be used, and the general formula is: $$ \mathbf{card}(\bigcup^n_{i=1} A_i) = \sum^n_{k=1} (-1)^{k+1} \left(\sum_{1 \leq i_1 < \cdots < i_k \leq n} \mathbf{card}(A_{i_1} \cap \cdots \cap A_{i_k}) \right) $$

So we just need to define the $A_i$. Given a set of nodes $N$, we are interested in the subsets generated by two letters $X, Y \in N$, and we write:

$$ Z_{XY} = \{ Z \subseteq N \mid X \in Z \land Y \in Z\} $$

In this case, the question becomes: given a set of generators $G = \{\{X_1, Y_1\}, \cdots \{X_k, Y_k\}\}$ (note that the number $k$ stands for $Y~$ in the original question), what is the cardinality of $Z_{X_1Y_1} \cup \cdots \cup Z_{X_kY_k}$?

If we write $A_i$ for $Z_{X_iY_i}$, we have $$ \mathbf{card}(A_{i_1} \cap \cdots \cap A_{i_n}) = 2 ^{N - \mathbf{card}(\{X_{i_1}, Y_{i_1}, \cdots, X_{i_n}, Y_{i_n}\})} $$ because the cardinality of the intersection corresponds to the number of subsets possible without the generators. Using this formula and the general formula for inclusion/exclusion, you can calculate the cardinality of the set.

For instance, we have: $$ \begin{align*} \mathbf{card}(Z_{X_1Y_1} \cup Z_{X_2Y_2}) & = \mathbf{card}(Z_{X_1Y_1}) + \mathbf{card}(Z_{X_2Y_2}) - \mathbf{card}(Z_{X_1Y_1} \cap Z_{X_2Y_2}) \\ & = 2^{N-2} + 2^{N-2} - 2^{N - \mathbf{card}(\{X_1, Y_1, X_2, Y_2\})} \end{align*}$$

Similarly: $$ \begin{align*} \mathbf{card}(Z_{X_1Y_1} \cup Z_{X_2Y_2} \cup Z_{X_3Y_3}) & = \mathbf{card}(Z_{X_1Y_1}) + \mathbf{card}(Z_{X_2Y_2}) + \mathbf{card}(Z_{X_3Y_3}) \\ & - ~ \mathbf{card}(Z_{X_1Y_1} \cap Z_{X_2Y_2}) - ~ \mathbf{card}(Z_{X_1Y_1} \cap Z_{X_3Y_3}) - ~ \mathbf{card}(Z_{X_2Y_2} \cap Z_{X_3Y_3}) \\ & + ~ \mathbf{card}(Z_{X_1Y_1} \cap Z_{X_2Y_2} \cap Z_{X_3Y_3}) \\ & = 2^{N-2} + 2^{N-2} + 2^{N-2} \\ & - ~2^{N - \mathbf{card}(\{X_1, Y_1, X_2, Y_2\})} - 2^{N - \mathbf{card}(\{X_1, Y_1, X_3, Y_3\})} - 2^{N - \mathbf{card}(\{X_2, Y_2, X_3, Y_3\})} \\ & + ~2^{N - \mathbf{card}(\{X_1, Y_1, X_2, Y_2, X_3, Y_3\})} \end{align*}$$

With the simple example given in the original question:

$N = \{A, B, C, D\}$

$Y = Z_{AB} \cup Z_{AC} \cup Z_{BC}$

For AB, we have the subsets $\{\{A,B\},\{A,B,C\},\{A,B,D\},\{A,B,C,D\}\}$

For AC, we have the subsets (I do not include those already in AB): $\{\{A, C\},\{A,C,D\}\}$

For BC, we have the subsets (I do not include those already in AB): $\{\{B, C\},\{B,C,D\}\}$.

Using the formula, we have indeed $\mathbf{card}(Y) = 2^2 + 2^2 + 2^2 - 2 - 2 -2 + 2 =8$.

If we consider BD instead of BC, then we just add the subsets $\{\{B, C\}\}$, which gives $7$, as expected.

  • $\begingroup$ If you quote it like that it would seem the binomial would work, so I updated the question a bit to be clearer. There are Y groups which each contain 2^(N-2) subsets that all start with A-B or A-C etc.. since they are different lengths I don't think I can use the binomial. $\endgroup$ – Roy T. Jan 20 '12 at 12:10
  • $\begingroup$ How do your $n$ and $k$ relate to your $N$ and $Y$? $\endgroup$ – Gerry Myerson Jan 20 '12 at 12:11
  • $\begingroup$ @GerryMyerson: I just wanted to say that if I have N nodes, there are exactly (N K) distincts subsets with K elements, which seemed to be the question at first, since it was mentioned that only the number mattered. But I realize now that I have misunderstood the question. $\endgroup$ – Charles Jan 20 '12 at 12:41
  • $\begingroup$ @RoyT. I hope the reformulation of the question is correct, if it is, I guess it should be possible to find the general formula. $\endgroup$ – Charles Jan 20 '12 at 13:58
  • $\begingroup$ Wow thanks for expanding your answer so much, I'll look at it tomorrow to see if this rhymes with my data :). $\endgroup$ – Roy T. Jan 20 '12 at 16:27

If there are $N$ nodes then there are $2^N$ possible sets, including the empty set and the universal set. Taken together, these sets make up the power set.

You are interested in subsets of the power set which have a give two nodes as elements of each of their elements. There are ${N \choose 2}=\frac{N(N=1)}{2}$ such subsets, each of which has $2^{N-2}$ elements.

If you choose two of these subsets of the power set and look at the the numer of distinct elements of their union, this is the sum of their sizes less the size of their overlap or intersection (to avoid double counting). This depends on whether the two nodes which define the subsets of the power set overlap or not.

  • If they do overlap (your example with $A,B$ and $A,C$) then the value is $2\times 2^{N-2}-2^{N-3} = 3\times 2^{N-3}$. With $N=4$ this gives $6$.

  • If they do not overlap (e.g. $A,B$ and $C,D$) then the value is $2\times 2^{N-2}-2^{N-4} = 7\times 2^{N-4}$. With $N=4$ this gives $7$.


  • It gets more complicated with more cases using inclusion-exclusion. For example with $A,B$ and $A,C$ and $D,E$ the value is $3\times 2^{N-2} -2^{N-3} -2 \times 2^{N-4} +2^{N-5}= 25 \times 2^{N-5}$
  • $\begingroup$ What if some overlap and some others do not? $\endgroup$ – Roy T. Jan 20 '12 at 12:49
  • 1
    $\begingroup$ @Roy: then it get more complicated. In any case you get inclusion-exclusion if you have more than two. I will add something more to my answer $\endgroup$ – Henry Jan 20 '12 at 12:59

Use Inclusion-Exclusion. Number of subsets containing $A$ and ($B$ or $C$) equals number containing $A$ and $B$, plus number containing $A$ and $C$, minus number containing $A$ and $B$ and $C$.

  • $\begingroup$ This is a great idea but I have some trouble when comparing more subsets, for example I seem to be missing one in this specific case: pastebin.com/A3WYsH9U $\endgroup$ – Roy T. Jan 20 '12 at 12:39
  • $\begingroup$ All the posted solutions rely on Inclusion-Exclusion, so my advice is to find a treatment of that topic and study it. It's in pretty nearly every introductory combinatorics and/or discrete math textbook, and undoubtedly there are some good treatments on the web, as well. Happy searching! $\endgroup$ – Gerry Myerson Jan 20 '12 at 23:40
  • $\begingroup$ Thanks Gerry, I'm only finishing up my Bsc. Computer Science so I've only had one discrete mathematics course which didn't include much combinatorics but I'm sure to look into it :) $\endgroup$ – Roy T. Jan 21 '12 at 9:17

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