nyuuzyou/wikis huggingface dataset:
Jordan Curve Characterization of Simply Connected Set

== Theorem ==
Let $D \subseteq \R^2$ be an open path-connected subset of the Euclidean plane.
Then $D$ is simply connected, the following condition holds:
:For all Jordan curves $\gamma : \closedint 0 1 \to \R^2$ with $\Img \gamma \subseteq D$, we have:
:$\Int \gamma \subseteq D$
Here $\Img \gamma$ denotes the image of $\gamma$, and $\Int \gamma$ denotes the interior of $f$.
== Proof ==
=== Sufficient condition ===
Suppose that $D$ is simply connected.
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve with $\Img \gamma \subseteq D$.
By definition of simple connectedness, there exists a constant loop $c : \closedint 0 1 \to D$ such that $\gamma$ and $c$ are path-homotopic in $D$.
Let $H : \closedint 0 1 \times \closedint 0 1 \to D$ be the path homotopy between $\gamma$ and $c$.
From Interior of Jordan Curve is Subset of Image of Null-Homotopy, it follows that:
:$\Int \gamma \subseteq \Img H$
As:
:$\Img H \subseteq D$
it follows that:
:$\Int \gamma \subseteq D$
=== Necessary condition ===
Suppose that for all Jordan curves $\gamma : \closedint 0 1 \to \R^2$ with $\Img \gamma \subseteq D$, we have: :$\Int \gamma \subseteq D$
Let $f : \closedint 0 1 \to D$ be a loop in $D$.
By definition of closed set, it follows that the complement $\relcomp {\R^2} D$ is closed in $\R^2$.
From Closed Real Interval is Compact, it follows that $\closedint 0 1$ is compact.
From Continuous Mappings preserve Compact Subsets, it follows that $\Img f$ is compact.
From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that:
:$\map d {\Img f, \relcomp {\R^2} D} > 0$
where :$\map d {X,Y}$ denotes the Euclidean distance between the sets $X$ and $Y$.
From Jordan Curve Bounding Loop in Euclidean Plane, there exists a Jordan curve $\gamma : \closedint 0 1 \to \R^2$ such that:
:$\Img f \subseteq \Int \gamma$
and for all $t \in \closedint 0 1$:
:$\map d{\map \gamma t, \Img f} < \map d {\Img f, \relcomp {\R^2} D} / 2$
It follows that:
:$\Img \gamma \subseteq D$
From Interior of Jordan Curve is Simply Connected, it follows that $\Img \gamma$ is simply connected.
By definition of simple connectedness, there exists a constant loop $c : \closedint 0 1 \to D$ such that $f$ and $c$ are path-homotopic in $\Int \gamma$.
Let $H : \closedint 0 1 \times \closedint 0 1 \to \Int \gamma$ be the path homotopy between $f$ and $c$ in $\Int \gamma$.
As $\Img \gamma \subseteq D$, it follows that the $H$ is also a path homotopy between $f$ and $c$ in $D$.
By definition of simple connectedness, it follows that $D$ is simply connected.
== Sources ==
* : $10$: Separation Theorems in the Plane: $\S 62$: Invariance of Domain
Category:Simply Connected Spaces
Jordan Curve Image Equals Set Homeomorphic to Circle

== Theorem ==
Let $C \subseteq \R^2$ be a subset of the Euclidean space $\R^2$.
Let $\mathbb S^1$ denote the unit circle whose center is at the origin of the Euclidean space $\R^2$.
Let $\tau_C$ and $\tau_{\mathbb S^1}$ be the subspace topologies on $C$ respectively $\mathbb S^1$ induced by the Euclidean topology on $\R^2$.
Then $\struct{ C, \tau_C }$ is homeomorphic to $\struct { \mathbb S^1 , \tau_{\mathbb S^1} }$, there exists a Jordan curve $\gamma : \closedint 0 1 \to \R^2$ with image equal to $C$.
== Proof ==
By definition of simple loop, a Jordan curve is a simple loop in $\R^2$.
Euclidean Space is Complete Metric Space and Metric Space is Hausdorff shows that $\R^2$ is a Hausdorff space.
The result now follows from Simple Loop Image Equals Set Homeomorphic to Circle.
Category:Circles
Category:Examples of Homeomorphisms
Category:Jordan Curves
Jordan Curve Theorem

== Theorem ==
Let $\gamma: \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\Img \gamma$ denote the image of $\gamma$.
Then $\R^2 \setminus \Img \gamma$ is a union of two disjoint connected components.
Both components are open in $\R^2$, and both components have $\Img \gamma$ as their boundary.
One component is bounded, and is called the interior of $\gamma$.
The other component is unbounded, and is called the exterior of $\gamma$.
=== General Result ===
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $\S3.1$: Simple Closed Curves
* 
* 
Category:Jordan Curve Theorem
Category:Jordan Curves
Jordan Curve Theorem/General Result

== Theorem ==
Let $M$ be a connected manifold of dimension $n - 1$ without boundary.
Let $M$ be embedded in Euclidean space $\R^n$.
Then $M$ divides $\R^n$ into an inside and an outside.
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:Jordan Curve Theorem
Jordan Curve Theorem/Historical Note

== Historical Note on Jordan Curve Theorem ==
The Jordan Curve Theorem was stated by in $1893$, who provided a purported proof for it.
This, however, was complicated and difficult to follow, and it was considered at the time to be incomplete and invalid.
In $1905$, produced what was then considered to be a rigorous and complete proof, which was subsequently accepted by the mathematical community.
Some recent thought suggests that 's proof has been criticised unfairly, and that it is in fact valid after all.
== Sources ==
* 
* 
* 
Category:Historical Notes
Jordan Curve and Jordan Arc form Two Jordan Curves/Corollary

== Corollary to Jordan Curve and Jordan Arc form Two Jordan Curves ==
Let $\gamma: \closedint a b \to \R^2$ be a Jordan curve, where $\closedint a b$ is the closed real interval between $a, b \in \R$ with $a < b$.
Let the interior of $\gamma$ be denoted $\map {\operatorname {Int} } \gamma$.
Let the image of $\gamma$ be denoted $\Img \gamma$.
Let $\sigma: \closedint c d \to \R^2$ be a Jordan arc such that:
:$\map \sigma c, \map \sigma d \in \Img \gamma$
and:
:$\forall t \in \openint c d: \map \sigma t \in \map {\operatorname {Int} } \gamma$
Let $t_1 = \map {\gamma^{-1} } {\map \sigma c}$.
Let $t_2 = \map {\gamma^{-1} } {\map \sigma d}$.
Let $t_1 < t_2$.
Define:
:$-\sigma: \closedint c d \to \Img \sigma$ by $-\map \sigma t = \map \sigma {c + d - t}$
Let $*$ denote concatenation of paths.
Let $\gamma \restriction_{\closedint a {t_1} }$ denote the restriction of $\gamma$ to $\closedint a {t_1}$.
Define:
:$\gamma_1 = \gamma {\restriction_{\closedint a {t_2} } } * \paren {-\sigma} * \gamma {\restriction_{\closedint {t_1} b} }$
Define:
:$\gamma_2 = \gamma {\restriction_{\closedint {t_2} {t_1} } } * \sigma$
Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that:
:$\Int {\gamma_1} \cup \Int {\gamma_2} \cup \paren { \Img \sigma \setminus \set { \map \sigma c , \map \sigma d } } = \Int \gamma$
== Proof ==
We prove that $\gamma_1, \gamma_2$ are Jordan curves in the same way as in the proof of Jordan Curve and Jordan Arc form Two Jordan Curves.
Then, it follows from the defintions of $\sigma$ that $\Img {\gamma_1}, \Img {\gamma_2}$ are identical in this corollary and in the main theorem.
As the interior and exterior of Jordan curves only depend on their images, the conclusion of this corollary follows from the main theorem.
Category:Jordan Curves
Category:Jordan Arcs
Jordan Decomposition Theorem

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$. 
Then there exists measures $\mu^+$ and $\mu^-$ on $\struct {X, \Sigma}$ such that: 
:$\mu = \mu^+ - \mu^-$
Further, at least one of $\mu^+$ and $\mu^-$ is finite.
== Proof ==
From the definition of a Hahn decomposition, the set $P$ is $\mu$-positive, the set $N$ is $\mu$-negative and: 
:$X = P \cup N$ 
with $P$ and $N$ disjoint.
From Sigma-Algebra Closed under Countable Intersection, we have: 
:$A \cap P \in \Sigma$
and:
:$A \cap N \in \Sigma$
for each $A \in \Sigma$.
Now, for each $A \in \Sigma$, define: 
:$\map {\mu^+} A = \map \mu {A \cap P}$
and:
:$\map {\mu^-} A = -\map \mu {A \cap N}$
We verify that $\mu^+$ and $\mu^-$ are indeed measures by first showing that they are signed measures.
=== Lemma ===
We show that these are in fact measures by showing that $\mu^+ \ge 0$ and $\mu^- \ge 0$. 
From Intersection is Subset, we have: 
:$A \cap P \subseteq P$
Since $P$ is a $\mu$-positive set, we have:
:$\map \mu {A \cap P} \ge 0$
for each $A \in \Sigma$. 
From Intersection is Subset, we have: 
:$A \cap N \subseteq N$
Since $N$ is a $\mu$-negative set, we have: 
:$\map \mu {A \cap N} \le 0$
for each $A \in \Sigma$.
So:
:$\map {\mu^+} A \ge 0$
and:
:$\map {\mu^-} A \ge 0$
for each $A \in \Sigma$.
So, from Non-Negative Signed Measure is Measure, we have: 
:$\mu^+$ and $\mu^-$ are measures.
We now verify that at least one of the measures are finite. 
From Intersection with Subset is Subset, we have: 
:$\map {\mu^+} X = \map \mu {X \cap P} = \map \mu P$
and:
:$\map {\mu^-} X = -\map \mu {X \cap N} = -\map \mu N$
From the definition of a finite measure, we show that either: 
:$\map \mu P < \infty$
or:
:$-\map \mu N < \infty$
Since $\mu$ is countably additive, we have that: 
:$\map \mu X = \map \mu P + \map \mu N$
From Set is Subset of Itself, we have: 
:$P \subseteq P$
so, since $P$ is $\mu$-positive, we have: 
:$\map \mu P \ge 0$
Applying Set is Subset of Itself again we have:
:$N \subseteq N$
so, since $N$ is $\mu$-negative, we have: 
:$\map \mu N \le 0$
By the definition of a signed measure, we cannot have both $\map \mu P = +\infty$ and $\map \mu N = -\infty$, so at least one of $\map \mu P$ and $\map \mu N$ is finite.
So, either: 
:$\map {\mu^+} X = \map \mu P < \infty$
or:
:$\map {\mu^-} X = -\map \mu N < \infty$
That is, at least one of $\mu^+$ and $\mu^-$ are finite.
Finally, we verify that: 
:$\mu = \mu^+ - \mu^-$
Note that $A \cap P$ and $A \cap N$ are disjoint with: 
:$A = \paren {A \cap P} \cup \paren {A \cap N}$
so:
So:
:$\mu = \mu^+ - \mu^-$
as required.
== Sources ==
* : $4.1$
Category:Jordan Decomposition Theorem
Category:Signed Measures
Jordan Decomposition Theorem/Lemma

== Lemma ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$. 
For each $A \in \Sigma$, define: 
:$\map {\mu^+} A = \map \mu {P \cap A}$
and:
:$\map {\mu^-} A = -\map \mu {N \cap A}$
Then:
:$\mu^+$ and $\mu^-$ are signed measures.
== Proof ==
We verify both of the conditions given in the definition of a signed measure.
=== Proof of $(1)$ ===
We have, from Intersection with Empty Set: 
:$\map {\mu^+} \O = \map \mu \O = 0$
verifying $(1)$ for $\mu^+$.
We also have:
:$\map {\mu^-} \O = -\map \mu \O = 0$
verifying $(2)$ for $\mu^-$.
=== Proof of $(2)$ ===
Let $\sequence {D_n}_{n \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets. 
We have: 
Since:
:$D_i \cap D_j = \O$ whenever $i \ne j$
we have: 
:$\paren {P \cap D_i} \cap \paren {P \cap D_j} = \O$
from Intersection with Empty Set.
So, since $\mu$ is countably additive, we have: 
That is: 
:$\ds \map {\mu^+} {\bigcup_{n \mathop = 1}^\infty D_n} = \sum_{n \mathop = 1}^\infty \map {\mu^+} {D_n}$
for any sequence $\sequence {D_n}_{n \mathop \in \N}$ of pairwise disjoint $\Sigma$-measurable sets, so $(2)$ is satisfied for $\mu^+$.
We also have: 
As before, the sets $N \cap D_n$ are pairwise disjoint. 
Since $\mu$ is countably additive, we have: 
That is: 
:$\ds \map {\mu^-} {\bigcup_{n \mathop = 1}^\infty D_n} = \sum_{n \mathop = 1}^\infty \map {\mu^-} {D_n}$
for any sequence $\sequence {D_n}_{n \mathop \in \N}$ of pairwise disjoint $\Sigma$-measurable sets, so $(2)$ is satisfied for $\mu^-$.
So $\mu^+$ and $\mu^-$ satisfy both conditions $(1)$ and $(2)$, and so are both signed measures.
Category:Jordan Decomposition Theorem
Jordan Decomposition of Finite Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be measurable.
Let $\mu$ be a finite signed measure on $\struct {X, \Sigma}$. 
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are finite measures.
== Proof ==
From the definition of Jordan decomposition, we have: 
:$\mu = \mu^+ - \mu^-$
with at least one of $\mu^+$ and $\mu^-$ finite.
From the definition of a finite signed measure, we have: 
:$\cmod {\map \mu X} < \infty$
We show that: 
:if exactly one of $\mu^+$ and $\mu^-$ is finite, then $\mu$ is not a finite signed measure
that is: 
:if exactly one of $\mu^+$ and $\mu^-$ is finite, then $\map \mu X = \infty$ or $\map \mu X = -\infty$.
At which point we will have a contradiction, giving:
:$\mu^+$ and $\mu^-$ are both finite.
 suppose exactly one of $\mu^+$ and $\mu^-$ is finite.
Consider first the case of $\mu^+$ non-finite.
Then $\mu^-$ is finite, so:
:$\map {\mu^-} X < \infty$
Since $\mu^+$ is not finite, we have: 
:$\map {\mu^+} X = \infty$
From the definition of extended real subtraction, we then have: 
:$\map \mu X = \map {\mu^+} X - \map {\mu^-} X = \infty$
which contradicts: 
:$\cmod {\map \mu X} < \infty$
So we have a contradiction in the case of $\mu^+$ non-finite.
Now suppose that $\mu^-$ is non-finite.
Then $\mu^+$ is finite, so:
:$\map {\mu^+} X < \infty$
Since $\mu^-$ is not finite, we have: 
:$\map {\mu^-} X = \infty$
From the definition of extended real subtraction, we then have: 
:$\map \mu X = \map {\mu^+} X - \map {\mu^-} X = -\infty$
which contradicts: 
:$\cmod {\map \mu X} < \infty$
So we have a contradiction in the case of $\mu^-$ non-finite.
So we must have that both $\mu^+$ and $\mu^-$ are finite, completing the proof.
Category:Finite Signed Measures
Jordan Matrix/Examples/Arbitrary Example 1

== Example of Jordan Matrix ==
This is an arbitrary example of an order $4$ Jordan matrix:
:$\quad \begin {pmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end {pmatrix}$
where it is understood that $\lambda \ne 0$.
== Sources ==
* 
* 
Category:Examples of Jordan Matrices
Jordan Polygon Interior and Exterior Criterion

== Theorem ==
Let $P$ be a polygon embedded in $\R^2$.
Let $q \in \R^2 \setminus \partial P$, where $\partial P$ denotes the boundary of $P$.
Let $\mathbf v \in \R^2 \setminus \set \bszero$ be a non-zero vector.
Let $\LL = \set {q + s \mathbf v: s \in \R_{\ge 0} }$ be a ray with start point $q$.
Let $\map N q \in \N$ be the number of crossings between $\LL$ and $\partial P$.
Then:
:$(1): \quad$ $q \in \Int P$, $\map N q \equiv 1 \pmod 2$
:$(2): \quad$ $q \in \Ext P$, $\map N q \equiv 0 \pmod 2$
Here, $\Int P$ and $\Ext P$ denote the interior and exterior of $\partial P$, when $\partial P$ is considered as a Jordan curve.
== Proof ==
From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve.
From the Jordan Polygon Theorem, it follows that $\Int P$ and $\Ext P$ are path-connected.
Then, Jordan Polygon Parity Lemma shows that $\map N q = \map {\operatorname{par} } q$, where $\map {\operatorname{par} } q$ denotes the parity of $q$.
From Jordan Polygon Theorem, it follows that $\Ext P$ is unbounded, while $\Int P$ is bounded.
As $\partial P$ is the image of a Jordan curve, it follows from Continuous Image of Compact Space is Compact/Corollary 2 that $\partial P$ is also bounded.
Let $\map {B_R} \bszero$ be the open ball with radius $R$ and center $\bszero$.
Then, there exists $R \in \R_{>0}$ such that $\Int P \cup \partial P \subseteq \map {B_R} \bszero$.
If $q_0 \in \R^2 \setminus \map {B_R} \bszero$, then $q_0 \in \Ext P$.
Then, the ray $\set {q_0 + s q_0: s \in \R_{\ge 0} } \subseteq \R^2 \setminus \map {B_R} \bszero$, so there are zero crossings between the ray and $\partial P$.
From Jordan Polygon Parity Lemma, it follows that $\map {\operatorname{par} } {q_0} = 0$.
As $\Ext P$ is path-connected, it follows from the lemma that for all $q \in \Ext P$, we have $\map {\operatorname{par} } q = 0$.
If $q_1 \in \R^2 \setminus \partial P$ with $\map {\operatorname{par} } q = 1$, it follows that $q_1 \notin \Ext P$.
As $\R^2 \setminus \partial P = \Int P \cup \Ext P$, it follows that $q_1 \in \Int P$.
From Jordan Polygon Parity Lemma:
:$\forall q \in \Int P: \map {\operatorname{par} } q = 1$
So if instead $q_0 \in \R^2 \setminus \partial P$ with $\map {\operatorname{par} } q = 0$, the only possibility is that $q_0 \in \Ext P$.
== Motivation ==
This criterion is used by computer graphics programs to determine whether a given point lies inside a polygonal region.
== Sources ==
Jeff Ericsson: Computational Topology (dead link?)
 
Category:Topology
Jordan Polygon Theorem

== Theorem ==
Let $P$ be a polygon embedded in $\R^2$.
Denote the boundary of $P$ as $\partial P$.
Then, $\R^2 \setminus \partial P$ is a union of two connected components.
Both components are open in $\R^2$.
One component is bounded, and is called the interior of $P$.
The other component is unbounded, and is called the exterior of $P$.
== Proof ==
=== Lemma ===
We show that $\R^2 \setminus \partial P$ is not path-connected.
Find any $q_1 \in R^2 \setminus \partial P$ and $\theta \in \R$ such that the ray $\LL_\theta = \set {q_1 + s \map g \theta: s \in \R_{\ge 0} }$ has exactly one crossing of $\partial P$.
Find any $q_2 \in \LL_\theta$ that lies on the ray after the crossing, so the ray $\set {q_2 + s \map g \theta: s \in \R_{\ge 0} }$ does not intersect $\partial P$.
Then $\map {\mathrm {par} } {q_1} = 1 \ne 0 = \map {\mathrm {par} } {q_2}$.
From Jordan Polygon Parity Lemma, it follows that $q_1$ and $q_2$ cannot be connected by a path.
As $R^2 \setminus \partial P$ is not path-connected, it follows from the Lemma that $R^2 \setminus \partial P$ is a union of exactly two disjoint path-connected sets, which we denote as $U_1$ and $U_2$.
Let $q \in \R^2 \setminus \partial P$, and let $\map d {q, \partial P}$ be the Euclidean distance between $q$ and $\partial P$.
From Distance between Closed Sets in Euclidean Space, it follows that $\map d {q, \partial P} > 0$.
When we put $\epsilon = \map d {q, \partial P} / 2$, we have $\map {B_\epsilon} q \subseteq \R^2 \setminus \partial P$.
As Open Ball is Convex Set, it follows that $\map {B_\epsilon} q$ is path-connected, so $\map {B_\epsilon} q$ is a subset of either $U_1$ or $U_2$.
Then, both $U_1$ and $U_2$ are open.
From Path-Connected Space is Connected, it follows that $U_1$ and $U_2$ are connected.
Then, $\R^2 \setminus \partial P$ is a union of two components.
From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve $\gamma: \closedint 0 1 \to \R^2$.
From Continuous Image of Compact Space is Compact/Corollary 2, it follows that $\partial P$ is bounded.
That is, there exist $a \in \R^2$ and $R \in \R_{>0}$ such that $\partial P \subseteq \map {B_R} a$.
If $x_1, x_2 \in \R^2 \setminus \map {B_R} a$, $x_1$ to $x_2$ can be joined by a path in $\R^2 \setminus \map {B_R} a$ following:
:the circumference of the two circles with center $a$ and radii $\map d {a, x_1}$ and $\map d {a, x_2}$
:a line segment joining the two circumferences.
Then $\R^2 \setminus \map {B_R} a$ is path-connected, so $\R^2 \setminus \map {B_R} a$ is a subset of one of the components of $\R^2 \setminus \partial P$, say $U_1$.
As $\R^2 \setminus \map {B_R} a \subseteq U_1$, it follows that $U_1$ is unbounded, so $U_1$ is the exterior of $\gamma$.
Then $U_2 \subseteq \map {B_R} a$, so $U_2$ is bounded, which implies that $U_2$ is the interior of $\gamma$.
== Notes ==
This theorem is a special case of the Jordan Curve Theorem.
== Sources ==
Jeff Erickson: Computational Topology 
Category:Topology
Josephus Problem

== Classic Problem ==
This is the original Josephus problem, of which all others are variants.
, along with $40$ comrades in arms, have opted to commit suicide rather than fall into the hands of the Romans.
Because he really wants to stay alive, he suggests that they all stand in a circle, and count around it.
Every $3$rd person counted was to drop out of the circle and die.
Where did place himself, and his companion who also did not want to die, so that the two of them would remain the last two standing?
=== Turks and Christians Variant ===
=== Japanese Inheritance Variant ===
== Solution ==
They placed themselves at positions $16$ and $31$ in the circle of $41$.
== Proof ==
On the first time round the circle, we lose the people at positions:
:$3$, $6$, $9$, $12$, $15$, $18$, $21$, $24$, $27$, $30$, $33$, $36$ and $39$.
We are left with:
:$1$, $2$, $4$, $5$, $7$, $8$, $10$, $11$, $13$, $14$, $16$, $17$, $19$, $20$, $22$, $23$, $25$, $26$, $28$, $29$, $31$, $32$, $34$, $35$, $37$, $38$, $40$, $41$
Counting $3$ round from the now eliminated $39$ leads us to $1$, and so the following are eliminated from the above:
:$1$, $5$, $10$, $14$, $19$, $23$, $28$, $32$, $37$, $41$
Now we are left with:
:$2$, $4$, $7$, $8$, $11$, $13$, $16$, $17$, $20$, $22$, $25$, $26$, $29$, $31$, $34$, $35$, $38$, $40$
Counting $3$ round from the now eliminated $41$ leads us to $7$, and so the following are eliminated from the above:
:$7$, $13$, $20$, $26$, $34$, $40$
Now we are left with:
:$2$, $4$, $8$, $11$, $16$, $17$, $22$, $25$, $29$, $31$, $35$, $38$
Counting $3$ round from the now eliminated $40$ leads us to $8$, and so the following are eliminated from the above:
:$8$, $17$, $29$, $38$
Now we are left with:
:$2$, $4$, $11$, $16$, $22$, $25$, $31$, $35$
Counting $3$ round from the now eliminated $38$ leads us to $11$, and so the following are eliminated from the above:
:$11$, $25$
Now we are left with:
:$2$, $4$, $16$, $22$, $31$, $35$
Counting $3$ round from the now eliminated $25$ leads us to $2$, and so the following are eliminated from the above:
:$2$, $22$
Now we are left with:
:$4$, $16$, $31$, $35$
Counting $3$ round from the now eliminated $22$ leads us to $4$, and so the following are eliminated from the above:
:$4$, $35$
Now we are left with:
:$16$, $31$
which is what we wanted to know.
== Historical Note ==
== Sources ==
* : The Josephus Problem: $98$
Category:Josephus Problem
Josephus Problem/Turks and Christians Variant

== Classic Problem ==
On board a ship, tossed in storms and in danger of shipwreck, are $15$ Christians and $15$ Turks.
To lighten the load and save the ship, half are to be thrown overboard.
One of the Christians suggests that all should stand in a circle and every $9$th person counting around the circle should be chosen.
How should the Turks arrange themselves in the circle to ensure that only the Christians die?
== Solution ==
The Turks and Christians should be placed in the following circular order, in which the first person follows the last, and the counting starts with the first person:
:$\text {TTTTCCCCCTTCTTTCTCCTTCCCTCCTTC}$
== Proof ==
On the first time round the circle, we lose the people at positions:
:$9$, $18$, $27$
We are left with:
:$1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $10$, $11$, $12$, $13$, $14$, $15$, $16$, $17$, $19$, $20$, $21$, $22$, $23$, $24$, $25$, $26$, $28$, $29$, $30$
Counting $9$ round from the now eliminated $27$ leads us to $6$, and so the following are eliminated from the above:
:$6$, $16$, $26$
Now we are left with:
:$1$, $2$, $3$, $4$, $5$, $7$, $8$, $10$, $11$, $12$, $13$, $14$, $15$, $17$, $19$, $20$, $21$, $22$, $23$, $24$, $25$, $28$, $29$, $30$
Counting $9$ round from the now eliminated $26$ leads us to $7$, and so the following are eliminated from the above:
:$7$, $19$, $30$
Now we are left with:
:$1$, $2$, $3$, $4$, $5$, $8$, $10$, $11$, $12$, $13$, $14$, $15$, $17$, $20$, $21$, $22$, $23$, $24$, $25$, $28$, $29$
Counting $9$ round from the now eliminated $30$ leads us to $12$, and so the following are eliminated from the above:
:$12$, $24$
Now we are left with:
:$1$, $2$, $3$, $4$, $5$, $8$, $10$, $11$, $13$, $14$, $15$, $17$, $20$, $21$, $22$, $23$, $25$, $28$, $29$
Counting $9$ round from the now eliminated $24$ leads us to $8$, and so the following are eliminated from the above:
:$8$, $22$
Now we are left with:
:$1$, $2$, $3$, $4$, $5$, $10$, $11$, $13$, $14$, $15$, $17$, $20$, $21$, $23$, $25$, $28$, $29$
and these are the positions where the surviving $15$ are to stand.
== Sources ==
* : The Josephus Problem: $99$
Category:Josephus Problem