commit allerede lavet opgaver

Diskret Mat/Inclusion-exclusion/main.typ

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+#title[Inclusion/Exclusion]
+Section 8.5+8.6
+
+The hat problem: Suppose everyone one in a classroom has a hat, and they put their hat in a box when they enter, and then grab a random hat when they leave. What is the probability that everybody gets their own hat. Answer is $1/n!$
+
+What about the probability that _someone_ will not get their own hat: $1-1/n!$
+
+What about the probability that someone will get their own hat, or that no one? Need inclusion/exclusion. The probability that no one will get their own hat is $37.8%$ pretty much no matter how many people.
+
+*Remember:*
+
+$|A union B| = |A| + |B| - |A inter B|$ 
+
+and 
+
+$|A union B union C| = |A|+|B|+|C| - |A inter B| - |A inter C| - |B inter C| + |A inter B inter C|$
+
+== Formula
+$ |A_1 union A_2 union dots union A_n| = |A_1| + |A_2| + dots + |A_n| - |A_1 inter A_2| - |A_1 inter A_3| - ... - |A_(n-1) inter A_n|\ 
++ |A_1 inter A_2 inter A_3| + |A_1 inter A_2 inter A_4| + dots + |A_(n-2) inter A_(n-1) inter A_n)|\
+- |A_1 inter A_2 inter A_3 inter A_4| - dots - |A_(n-3) inter A_(n-2) inter A_(n-1) inter A_(n)|\
++ |A_1 inter A_2 inter A_3inter A_4 inter A_5| + dots + |A_(n-4) inter A_(n-3) inter A_(n-2) inter A_(n-1) inter A_(n)|\ 
+dots dots (-1)^(n+1) |A_1 inter A_2 inter dots inter A_n| $
+
+$ =sum_(i=1)^n |A_i|- sum_(i<=i_1<i_2<=n) |A_i_1 inter A_i_2| + dots + (-1)^(k+1) sum_(i_1<i_2<dots<i_k)|A_i_1 inter A_i_2 inter dots inter A_i_k| + dots + (-1)^(n+1)|A_1 inter dots inter A_n| $
+
+Proof: Focus on any $x in A_1 union A_2 union dots union A_n$ and see how many times it is counted. Let $k$ be the number of $i$ such that $x in A_i$
+
+$ -mat(k;0) + mat(k;1)-mat(k;2)+mat(k;3)-mat(k;4)+mat(k;5)-dots + (-1)^(k+1) mat(k;k) + 1 = 1 $
+(The reason we write $+1$ is because $-mat(k;0)=-1$)
+
+Ignoring the $+1$, the rest is a row in Pascals triangle, and we now the alternating sum (+, then -, then +) is 0, so we know the whole thing will be 1.
+
+Consider a set `A` with `N` elements. Properties $P_1, P_2, P_3, dots, P_n$ which means that some elements have the property and some doesn't. $N(P_3 P_5 P_6 P_9) = $ number of elements with each of the properties $P_3 P_5 P_6 P_9$
+
+$N(P_(3) ' P_(5) ' P_6 ' P_9 ') = $ the number of elements with none of the properties $P_3 P_5 P_6 P_9$
+
+*Inclusion/Exclusion:*
+$ N(P_1 'P_2 'dots P_n ') = N - sum^n_(i=1)N(P_i) + sum_(i_1 < i_2)N(P_i_1 P_i_2) dots + (-1)^n N(P_1 P_2 dots P_n) $
+
+*Proof:* Put $A_i=$ the elements with property $P_i$ 
+
+
+`m` balls in `n` boxes. We want to count the number of ways to put `m` balls into `n` boxes such that no box is empty. It follows $m>=n$.
+
+$A = $ all functions of $S->B={1,2,3,dots,n}$
+$A_i = $ all functions such that `i` is not in the image.
+
+$ |A\\ union_(i=1)^n A_i| = |A|-|union_(i=1)^n A_i| = n^m-mat(n;1)(n-1)^m +mat(n;2)(n-2)^m-dots (-1)^n mat(n;n)(n-n)^m $
+
+= Permutations of an n-set {1,2,3,4}
+Normally a permutation of a set is the elements put on a row (like here $2,3,1,4$ and  $3,1,4,2$).
+
+You can think of permutations as a bijection (different elements should be mapped to different elements, and every element needs to be mapped to an element) from ${1,2,3,4}$ to ${1,2,3,4}$:
+$ mat(1,2,3,4;2,3,1,4) $
+Notice the 4, mapped unto itself, it's called a fixed point. The 4's would also be a fixed point here:
+$ mat(1,2,4,3;2,3,4,1) $
+But then its a different set at the top.
+
+== Derangement
+A bijection ($pi$) of a set `S` unto itself with no fixed points.
+
+That is $pi(i) eq.not i$
+
+The hat problem: to find the chances of noone getting their own hat, just  use derangements.
+
+== Counting permutations
+$A$ is the set of all permutations of ${1,2,dots,n}$
+
+$A_i =$ the set of permutations $pi$ such that $pi(i)=i$
+
+$ 
+D_n &=|A\\(union_(i=1)^n) = |A| - sum_(i=1)^n + sum_(i_1<i_2)|A_i_i inter A_i_2| - dots (-1)^n|A_1 inter A_2 inter dots inter A_n|\ 
+= n! - mat(n;1)(n-1)! + mat(n;2)(n-2)! - &mat(n;3)(n-3)!+dots (-1)^k mat(n;k)(n-k)! plus.minus dots |A_1 inter A_2 inter dots inter A_n|\
+&= sum_(k=0)^n (-1)^k mat(n;k)(n-k)!\
+&= sum_(k=0)^n (-1)^k n!/(k!(n-k)!) (n-k)!\
+&=n! sum_(k=0)^n (-1)^k/k!\
+&= k! [1- 1/1! + 1/2! - 1/3! + dots (-1)^n 1/(n!)]
+$
+$D_n$ is the permutations
+
+$ 
+1&=1/2+1/4+1/8+1/16+1/32+dots\
+1&=0.99999999dots\
+exp(x)&=1+x/1!+x^2/2!+x^3/3!+dots\
+exp(1)&=e=1+1/2!+1/3!+dots\
+exp(-1)&=e^-1=1/e=1-1/2!+1/3!-1/4!+dots
+$
+$exp(-1)$ is what we had before, so
+$ D_n &= k! [1- 1/1! + 1/2! - 1/3! + dots (-1)^n 1/(n!)]=n! dot 1/e\
+&= n! 1/(2.7dots)= 0.378 n!
+$
+
+= Counting primes
+The number of primes $<= 100$. The same as saying $100-"not primes"$
+
+Put $ A&={1,2,dots,100}\ A_2 &="The numbers divisible by 2"\ A_3&="The numbers divisible by 3"\ A_5&="The numbers divisible by 5"\ A_7&="The numbers divisible by 7" $
+
+Number of primes 
+$ &=100-|A_2 union A_3 union A_5 union A_7|-1+4 - |A_2|-|A_3|-|A_5|-|A_7|+|A_2 inter A_3|+dots\
+&=103-50-33-20-14+16+14+6+dots=25
+$
+($-1$ because 1 is not a prime, and $+4$ because else we're excluding $2,3,5,7$)

Diskret Mat/Inclusion-exclusion/opgaver.typ

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+Section 8.5: 1, *5*, 9, *11*, *15*, *23*, 27
+
+Section 8.6: 3, *5*, *7*, 13, *21*, *25*
+
+= 8.5-5
+Find the number of elements in $A_1 union A_2 union A_3$ if there
+are 100 elements in each set and if
+1. The sets are pairwise disjoint.
+$100+100+100=300$
+2. There are 50 common elements in each pair of sets and no elements in all three sets.
+$100+100+100-50-50-50=150$
+3. There are 50 common elements in each pair of sets and 25 elements in all three sets.
+$100+100+100-50-50-50+25=175$
+4. The sets are equal.
+$100+100+100-100-100-100+100=100$
+
+= 8.5-11
+Find the number of positive integers not exceeding 1000 that are not divisible by $3,17, "or" 35$
+
+$1000-333-58-28+19+9+1-17=610$
+
+= 8.5-15
+How many bit strings of length eight do not contain six consecutive 0s?
+
+$2^8-12-4-1+7+2=248$
+
+= 8.5-23
+Write out the explicit formula given by the principle of inclusion–exclusion for the number of elements in the union of six sets when it is known that no three of these sets have a common intersection.
+$ |A_1union A_2 union A_3 union A_4 union A_5 union A_6| =\ 
+|A_1|+|A_2|+|A_3|+|A_4|+|A_5|+|A_6|-|A_1 inter A_2| - |A_1 inter A_3| - |A_1 inter A_4| - inter |A_1 inter a_5| - |A_1 inter A_6|\
+- |A_2 inter A_3| - |A_2 inter A_4| - |A_2 inter A_5| - |A_2 inter A_6|\
+- |A_3 inter A_4| - |A_3 inter A_5| - |A_3 inter A_6| - |A_4 inter A_5| - |A_4 inter A_6| - |A_5 inter A_6|
+$
+
+= 8.6-5
+Find the number of primes less than 200 using the principle of inclusion–exclusion.
+
+$A_n =$ Numbers divisible by `n`
+$
+    "Number of primes" &= 200 - abs(A_2 union A_3 union A_5 union A_7 union A_11 union A_13) - 1 + 6 \
+                       &= 205 - |A_2| - |A_3| - |A_5| - |A_7| - |A_11| - |A_13| \
+                       &quad + |A_2 inter A_3| + |A_2 inter A_5| + ... + |A_11 inter A_13| \
+                       &quad - |A_2 inter A_3 inter A_5| - ... \
+                       &quad + |A_2 inter A_3 inter A_5 inter A_7| + ... \
+                       &quad - |A_2 inter A_3 inter A_5 inter A_7 inter A_11| - ... \
+                       &quad + |A_2 inter A_3 inter A_5 inter A_7 inter A_11 inter A_13| \
+                       &= 205 - 100 - 66 - 40 - 28 - 18 - 15 \
+                       &quad + 33 + 20 + 14 + 9 + 6 + 4 + 2 + 1 + 1 + 1 + 1 + 1 + 1 \
+                       &quad - 6 - 4 - 2 - 2 - 1 - 1 - 1 - 1 - 1 - 1 \
+                       &quad + 1 + 1 + 1 + 1 + 1 \
+                       &quad - 0 - 0 - 0 - 0 \
+                       &quad + 0 \
+                       &= 46
+$
+= 8.6-7
+ How many positive integers less than 10,000 are not the second or higher power of an integer?
+= 8.6-21