Basic form
\[ \max(S) = \sum_{T\subseteq S, T \neq \varnothing} (-1)^{|T|-1}\min(T) \]
prove
It does not provide a mathematical proof.
Briefly talk about abstract understanding of perjury:
Consider ranked in descending \ (i \) number, this number will be used as the set \ (T \) when the minimum value is present, \ (T \) all the remaining values are greater than its numbers from selected. Then the solution is to select \ (\ I-Binom. 1 {{} | T | -1} \) .
If \ (i = 1 \) , which is \ (a_i = \ max (S) \) , then it will only be adding \ (1 \) times.
If \ (I>. 1 \) , then it will be a total is calculated \ (\ sum \ limits_ {2 \ not \ mid j, j = 1} ^ {i-1} \ binom {i-1} {j-1 } - \ SUM \ limits_ {2 \ J MID, J = 2. 1-I} ^ {} \ {Binom-I-J. 1. 1} {} \) . The number of combinations of common sense, this thing \ (i-1> 0 \ ) when the answer is \ (0 \) .
In summary, \ (\ max (S) = \ SUM \ limits_ {T \ subseteq S, T \ NEQ \ varnothing} (-1) ^ {| T | -1} \ min (T) \) was established. QED.
Extended 1
\[ \min(S) = \sum_{T\subseteq S, T \neq \varnothing} (-1)^{|T|-1}\max(T) \]
Obviously.
Spread
Order \ (\ max_k (S) \ ) represents \ (S \) of the (\ K) \ large value.
则
\[ \max\ _{k}(S) = \sum_{T\subseteq S, |T| \geq k} (-1)^{|T|-k}\binom {|T|-1}{k-1} \min(T) \]
prove
Similar to the basic form.
Since \ (| T | \ GEQ K \) , so \ (\ min (T) \ Leq \ the MAX_K (S) \) .
For the first descending order of \ (I \) number, the same basic form, this number will be set as \ (T \) when the minimum value is present, \ (T \) all remaining values are from it is greater than the selected number. Then the solution is to select \ (\ I-Binom. 1 {{} | T | -1} \) .
If \ (K I = \) , then it will only be calculated \ (1 \) times, i.e., (T = \ {x \ mid x \ geq a_i \} \) \ time, while \ (\ binom {| T | -1} = {}. 1. 1-K \) .
If \ (I> K \) , then as described above, as the size will \ (| \ | T) the number of times of occurrence of the set of \ (\ binom {i-1 } {| T | -1} \ ) . Each occurrence is calculated \ (\ binom {| T | -1} {k-1} \) times. Therefore, as a size of \ (| T | \) the total contribution appears as a set of \ (\ binom {i-1 } {| T | -1} \ binom {| T | -1} {k-1} = \ {I-Binom. 1. 1-K} {} \ {Binom IK} {| T | -k} \) . So \ (I \) the total contribution to the \ (\ sum \ limits_ {2 \ not \ mid j, j = k} ^ {i-1} \ binom {i-1} {k-1} \ binom {ik } {jk} - \ sum \ limits_ {2 \ mid j, j = 2} ^ {i-1} \ binom {i-1} {k-1} \ binom {ik} {jk} = \ binom {i -1} {k-1} ( \ sum \ limits_ {2 \ not \ mid j, j = k} ^ {i-1} \ binom {ik} {jk} - \ sum \ limits_ {2 \ mid j, ^ {2} = J-I. 1} \ Binom JK} {} {IK) \) . Also, according to the knowledge of the number of combinations, \ (\ SUM \ limits_ {2 \ Not \ MID J, J = K} ^ {I-. 1} \ Binom {IK} {JK} - \ SUM \ limits_ {2 \ MID j, j = 2} ^ { i-1} \ binom {ik} {jk} \)The only thing in \ (ik = 0 \) only for the \ (1 \) , otherwise \ (0 \) .
Application 1
Common applications such as: There \ (n \) variables, the probability of occurrence of each variable is \ (the p-\) . Q. expected time every variable that appears.
Each time may wish to set a variable that appears to \ (t_i \) , then the probability of occurrence of all can be expressed as \ (t \) maximum. One is the emergence of at least \ (t \) minimum.
Then according to the inclusion and exclusion MinMax general form:
\ [\ max (S) = \ sum_ {T \ subseteq S, T \ NEQ \ varnothing} (-1) ^ {| T | -1} \ min (T) \ ]
Meanwhile, according to the desired linear nature, we have:
\ [E (\ max (S)) = \ sum_ {T \ subseteq S, T \ NEQ \ varnothing} (-1) ^ {| T | E} -1 (\ min (T)) \
] and \ (E (\ min (T )) \) obviously well solved. (Apparently \ (\ frac 1p \) right
So the problem is solved.