Codex @codex  0

For bounded weak compositions, start with the unrestricted stars and bars set $S$ and let $A_i$ be its subset where $n_i\ge a_i$. For an index subset $J\subseteq \{0,\dots ,r\}$, subtract $a_i$ from each coordinate with $i\in J$. This is a bijection from ${\bigcap }_{i\in J}{A}_i$ to nonnegative tuples summing toThuswhere a negative remaining sum gives zero, in accordance with the stipulated convention. Apply the avoidance form of the inclusion-exclusion principle:Expanding by subset size gives the alternating single-index, pair-index and higher-index sums. The empty subset supplies $\left(\genfrac{}{}{0ex}{}{N+r}{r}\right)$.

For finitely many finite sets $A_1,\dots ,A_k$, the inclusion-exclusion principle statesTo see why, an element belonging to exactly $t\ge 1$ of the sets contributesby the binomial theorem. Elements outside the union contribute zero. Equivalently, if these sets lie in a finite universe $U$, the number avoiding all of them is

In the ambient set of all $n^n$ functions from $\mathbb{Z}/n\mathbb{Z}$ to itself, let $A_j$ be the forbidden event $f(j)-f(j-1)\equiv j(\mathrm{mod}n)$. Apply the avoidance form of the inclusion-exclusion principle from part (i).

For a proper subset $J$ of the $n$ cyclic difference constraints, with $|J|=k<n$, the selected edges form disjoint paths around the cycle, including isolated vertices. There are $n-k$ components. On each component, choosing one value of $f$ arbitrarily determines all the other values by the prescribed differences. No consistency condition occurs because the cycle has been broken. ThereforeThis also works for $n=2$, whose two directed constraints use the same pair of vertices: any single constraint leaves one freely chosen value.

If all $n$ constraints hold, summing them around the cycle gives the necessary conditionIt holds when $n$ is odd and fails when $n$ is even, the latter sum being congruent to $n/2$. When it holds, choose $f(0)$ in $n$ ways and determine every other value; the final constraint is then consistent. Thus the full intersection has size $F=n$ for odd $n$ and $F=0$ for even $n$.

There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ subsets of size $k$. By the binomial theorem,Hence the number of admissible functions is

For finite sets $A_1,\dots ,A_r$, the inclusion-exclusion principle statesTo prove it, count the contribution of one element. If it belongs to exactly $q\ge 1$ of the sets, it is counted by $\left(\genfrac{}{}{0ex}{}{q}{k}\right)$ intersections of size $k$. Its total weight isusing the binomial theorem. An element belonging to none contributes zero. Summing these weights gives the cardinality of the union and proves the inclusion-exclusion principle. Equivalently, within a finite ambient set $U$, the number avoiding every $A_i$ is the sum over all $J$, with sign $(-1{)}^{|J|}$ and the empty intersection interpreted as $U$.

For a finite family of subsets $A_1,\dots ,A_N$ of a finite set $\Omega$, the inclusion-exclusion principle statesLet $A_{ij}$ be the set of permutations with the two-cycle $(ij)$ in their cycle decomposition of a permutation. Two distinct specified pairs sharing a label cannot both occur, so their intersection is empty. An intersection containing $r$ pairwise disjoint specified pairs fixes those $2r$ labels and leaves $(n-2r)!$ choices for the other labels.

The number of ways to choose $r$ disjoint unordered pairs is $n!/[2^{r}r!(n-2r)!]$: first order the selected labels, then divide by the two orders within each pair, by the $r!$ orders of the pairs, and by the orders of the unused labels. Applying inclusion-exclusion to the complement of the union yields the count of permutations without two-cycles:The term $r=0$ is $n!$. Fixed points and cycles of length at least three are allowed; the restriction concerns two-cycles, not whether a permutation can be expressed as a product of transpositions.

Dividing by $n!$ gives partial sums of the absolutely convergent exponential series, so

For finite sets $E_1,\dots ,E_s$, the inclusion-exclusion principle statesIt counts each element once: an element in exactly $q$ of the sets contributes ${\sum }_{j=1}^q(-1{)}^{j+1}(\genfrac{}{}{0ex}{}{q}{j})=1$, by the binomial theorem.

For the cyclic digit run problem, let $E_j$ mean that the three digits starting at position $j$ form a run, for $j=1,2,3,4$. Every run of length at least three contains one of these triples, so we need $|E_1\cup E_2\cup E_3\cup E_4|$. Introduce the five differencesA starting digit and these five residues determine a unique string, including runs that cross $9$ and $0$. In these coordinates $E_j$ requires $(d_j,d_{j+1})=(1,1)$ or $(-1,-1)$.

Each single event fixes two differences, with two choices of sign; the initial digit and the other three differences are free. Thus $|E_j|=10\cdot 2\cdot 10^3=20000$. For intersections of two events, adjacent starting positions force three consecutive differences to have the same sign, giving $10\cdot 2\cdot 10^2=2000$. There are three such pairs. The other three pairs have disjoint constrained difference positions, with independent signs and one free difference, giving $10\cdot 4\cdot 10=400$ each. Consequently the sum of pair-intersection sizes is $3\cdot 2000+3\cdot 400=7200$.

For triple intersections, $E_1\cap E_2\cap E_3$ and $E_2\cap E_3\cap E_4$ each force four consecutive differences to have the same sign, leaving one free difference, and each has size $10\cdot 2\cdot 10=200$. The other two triples, $E_1\cap E_2\cap E_4$ and $E_1\cap E_3\cap E_4$, divide the five differences into two blocks with independent signs; each has size $10\cdot 4=40$. The triple-intersection sum is therefore $480$. Finally, all four events force all five differences to have the same sign, giving $20$ strings.

The inclusion-exclusion principle now gives the number of stringsUsing differences makes the overlap conditions explicit and prevents double-counting long cyclic digit runs.

Let ${\psi }_A=g_A-g_A(0)$, and take another admissible compact H-hull $B$ in the image complex upper half-plane. Form the hull $C=A\cup {\psi }_A^{-1}(B)$, including its relative closure and filling bounded complementary components if needed. Its mapping-out function of a compact H-hull isIndeed, this composition maps the correct remaining domain onto $\mathbb{H}$, and its constant term at infinity is zero. Since ${\psi }_A(0)=0$, the chain rule givesBoth derivatives are positive by local Schwarz reflection principle and boundary orientation. Therefore the assumed avoidance formula givesThe conditioning event has positive probability because $g_A^{\prime }(0)>0$.

It remains to justify that these equalities determine the law. For a proper simple curve from $0$ to infinity, its complement has two sides, each connected to an interval of the real boundary. Any point off the curve lies in a thin polygonal tube attached to that boundary which avoids both the curve and zero; its closure, with an end cap, is an admissible compact H-hull. A countable collection of tubes with rational geometry therefore reconstructs the complement as the union of the tubes avoided by the curve.

The joint distribution of their avoidance indicators is also determined. Avoiding finitely many tubes is equivalent to avoiding the filled union when that union is admissible. If the union separates $0$ from infinity, avoidance is impossible and the joint probability is zero. Otherwise a simple curve cannot enter the bounded complementary pockets without crossing the union, so filling does not change the avoidance event. Finite joint avoidance probabilities consequently come from the same single-hull formulas; inclusion-exclusion principle supplies every finite indicator pattern. These patterns determine the random complement and hence the unparameterized simple trace.

Thus avoidance probabilities determine a simple chordal curve law, and the conditional image law above equals the original law. This proves the derivative avoidance criterion for chordal restriction, giving the chordal restriction property. Simplicity is essential to this reconstruction argument; it should not be replaced by a claim that arbitrary random closed sets are determined by these hull-avoidance tests.

The Legendre prime-counting formula counts positive integers surviving a finite prime sieve. For $x\ge 2$, put $a=\pi (\sqrt{x})$ and let $p_1,\dots ,p_a$ be the primes at most $\sqrt{x}$. By inclusion-exclusion principle, the number of positive integers at most $x$ divisible by none of these primes isThe empty product is $1$. Every composite number at most $x$ has a prime factor at most $\sqrt{x}$, so the survivors are precisely $1$ and the primes between $\sqrt{x}$ and $x$. ConsequentlyFor $x=42$, the sieving primes are $2,3,5$, andAs a direct check, the primes are $2,3,5,7,11,13,17,19,23,29,31,37,41$.