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\begin{document}
\title{Invertible Fractional Calculus}% Force line breaks with \
\author{Luke A. Siemens}
\email{luke.siemens@lsiemens.com}
\noaffiliation
\date{\today}
\maketitle

Fractional calculus has two aspects that I find deeply unsatisfying: the multiple incompatible definitions and that composition of operator with itself forms a semigroup not an abelian group. In this paper I demonstrate a series of conditions under which a version of fractional calculus can be uniquely defined and forms an abelian group. First I will explicitly define which operator will be extended to a fractional version. It must reproduce differntiation and integration and also have the desired algebraic properties. Then I will use Ramanujan's Master Theorem (RMT) to define the extension of the operator, and investigate its properties.

Let the proposed fractional calculus operator $J^\alpha$ have the following properties: for $\alpha \in \mathbb{Z}$ it can reproduce repeated integration and differentiation, it is a linear operator, that $J^\alpha$ forms an abelian group when acting on some suitable subset of differentiable functions and it is analytic in the parameter $\alpha$. First let us identify a suitable operator to extend into a fractional calculus with nice algebra, and function space for which this operator is well behaved. Let the operator $J^k$ for $k \in \mathbb{Z}$ be,
\begin{equation}
J^k f(x) := \begin{cases} \frac{1}{(k-1)!}\int_{-\infty}^x (x - t)^{k - 1}f(t)dt & k \geq 1 \\ f(x) & k = 0 \\ \frac{d^{\left|k\right|}}{dx^{\left|k\right|}}f(x) & k \leq -1 \end{cases}
\label{integer_calculus}
\end{equation}
Using this operator a space of functions where the operator is well behaved is any function that is bounded by positive exponentials. Let us call this set $\mathbb{S}$,
\begin{equation}
\mathbb{S} := \left\lbrace f \in C^\omega(\mathbb{C}) \middle| (\forall n \in \mathbb{Z}^+)(\exists a_n \in \mathbb{M}, b_n \in \mathbb{R}), b_n > 0, \frac{d^n}{dx^n}f(x) \in \mathbb{B}(a_n, b_n) \right\rbrace
\label{exponentialy_bounded}
\end{equation}
Where the sets $\mathbb{M}$ and $\mathbb{B}$ are defined as follows,
\[\mathbb{M} := \left\lbrace a \in C(\mathbb{R}) \middle| (\forall x, y \in \mathbb{R}), x > y, a(x) > a(y) \geq 0 \right\rbrace\]
\[\mathbb{B}(a, b) := \left\lbrace f \in C^\omega(\mathbb{C}) \middle| (\forall x, x_0 \in \mathbb{R}), x \leq x_0, |f(x)| \leq a(x_0)e^{bx} \right\rbrace\]
Some properties of this operator and the space on which it acts is investigated in the appendix. Note that set $\mathbb{S}$ is a subset of $C^{\omega}(\mathbb{C})$, it is a vector space and if $f(x) \in \mathbb{S}$ then $J^k f(x) \in \mathbb{S}$. Also if $f(x) \in \mathbb{S}$ then $J^k J^m f(x) = J^{k + m} f(x)$, where $k, m \in \mathbb{Z}$.

\section{Fractional extension using RMT}
Now that we have a suitable linear operator to extend, we need a procedure to actually extend the operator. We will use RMT for this purpose as given by G. H. Hardy \cite[p.~186]{hardy1940ramanujan}. Given a sequence $\phi(k)$, where $k \in \mathbb{Z}^+$, then
\[g(u) = \sum_{k=0}^\infty \frac{\phi(k)(-u)^k}{k!}\]
If the series converges and its Mellin transform exists then the following result holds,
\[\int_0^{\infty} u^{s-1}g(u)du = \Gamma(s)\phi(-s)\]
where $\phi(-s)$ is the analytic interpolation of the sequence $\phi(k)$ subject to some growth constraints \cite[p.~188--189]{hardy1940ramanujan}.
Using RMT fractional calculus can be defined in the following manner. For some function $f(x) \in \mathbb{S}$ consider the action of the operator $J^k$ at a point $x_0$. Every version of fractional calculus consists of producing an interpolation over some or all of this sequence. RMT can be used to find one such interpolation. Let us define $\phi(k)$ in terms of a function $f(x) \in \mathbb{S}$ ,
\[\phi(k) = \left. J^{-k}f(x)\right|_{x = x_0} = \left. \frac{d^k}{dx^k}f(x) \right|_{x = x_0}\]
In this case $g(u)$ is,
\[g(u) = \sum_{k=0}^\infty \frac{(-u)^k}{k!} \left. \frac{d^k}{dx^k}f(x)\right|_{x=x_0}\]
Note that $g(u)$ is the Taylor expansion of $f(x_0 - u)$ in terms of $u$. Now using $f(x_0 - u)$ in the Mellin transform,
\[\int_0^{\infty} u^{\alpha-1}f(x_0 - u)du = \Gamma(\alpha)\phi(-\alpha) = \left. \Gamma(\alpha)J^{\alpha}f(x)\right|_{x=x_0}\]
Using the substitution $t = x_0 - u$, the integral becomes,
\[\int_{x_0}^{-\infty} -(x_0 - t)^{\alpha-1}f(t)dt = \left. \Gamma(\alpha)J^{\alpha}f(x) \right|_{x=x_0}\]
Finally rearranging, and applying for arbitrary $x$,
\[J^{\alpha}f(x) = \frac{1}{\Gamma(\alpha)} \int_{-\infty}^{x} (x - t)^{\alpha-1} f(t)dt\]
The operator $J^{\alpha}$ is only defined on the region $\mathfrak{R}(\alpha) \geq 1$, but since it is constructed by RMT it is analytic on this region. So the complete Operator must be defined in terms of the analytic continuation of the interpolation produced by RMT. In the case of this operator its domain can be extended by using integration by parts given $f(x) \in \mathbb{S}$,
\[J^{\alpha}\frac{d}{dx}f(x) = \frac{1}{\Gamma(\alpha)}\int_{-\infty}^x (x-t)^{\alpha-1} \left( \frac{d}{dt} f(t) \right)dt = \frac{\alpha-1}{\Gamma(\alpha)}\int_{-\infty}^x (x-t)^{\alpha-2}f(t)dt=\frac{1}{\Gamma(\alpha-1)}\int_{-\infty}^x (x-t)^{\alpha-2}f(t)dt = J^{\alpha-1}f(x)\]
Using this repeatedly allows for $J^{\alpha}$ to be extended to the region $\mathfrak{R}(\alpha-n) \geq 1$ for $n \in \mathbb{Z}^+$,
\begin{equation}
J^{\alpha}\frac{d^n}{dx^n}f(x)=J^{\alpha-n}f(x)
\label{analytic_continuation}
\end{equation}
Therefore using RMT to interpolate between the derivatives of $f(x)$ yields an analytic function that interpolates between the derivatives and integrals. However, the operator $J^{\alpha}$ produced from RMT is not unique.
\[\psi(\alpha) \in C^{\omega}(\mathbb{C}), \psi(k) = 0, k \in \mathbb{Z}^-\]
\[\left. J'^{\alpha}f(x)\right|_{x=x_0} = \left. J^{\alpha}f(x)\right|_{x=x_0} + \psi(\alpha)\]
Evaluating the function at the points $k \in \mathbb{Z}^-$
\[\left. J'^{\alpha}f(x)\right|_{x=x_0} = \left. J^{\alpha}f(x)\right|_{x=x_0} + \psi(\alpha) = \left. J^{\alpha}f(x)\right|_{x=x_0}, k \in \mathbb{Z}^-\]
This new function $\left. J'^{\alpha}f(x) \right|_{x=x_0}$ also satisfies the basic properties expected of fractional calculus. This demonstrates that using RMT to define a fractional calculus, produces many of the algebraic properties I am looking for, but it is not unique. In order to define a unique fractional calculus operator using RMT an additional constraint must be added. Let us denote an arbitrary compatible fractional calculus operator as $I^\alpha$. Define a new operator $R^\alpha$ as,
\[R^\alpha f(x) = I^\alpha f(x) - J^\alpha f(x)\]
where $J^\alpha$ is the operator found using RMT. If we can prove that $R^\alpha = 0$ when some additional constraint is applied, then that constraint would force the operator $I^\alpha$ to be uniquely defined. To start apply the generalized Leibniz rule, given in \textit{Fractional Integrals and Derivatives} \cite[p.~280]{samko1993fractional}, to the operator $I^\alpha$,
\[I^\alpha f(x)g(x) = \sum_{n=0}^\infty \binom{-\alpha}{n}\left( I^{\alpha + n}f(x) \right)\left( \frac{d^n}{dx^n} g(x)\right) = \sum_{n=0}^\infty \binom{-\alpha}{n}\left( \left(J^{\alpha + n} + R^{\alpha + n}\right)f(x) \right)\left( \frac{d^n}{dx^n} g(x)\right)\]
Then subtracting $J^\alpha f(x)g(x)$ from both sides,
\begin{equation}
I^\alpha f(x)g(x) - \sum_{n=0}^\infty \binom{-\alpha}{n}\left( J^{\alpha + n}f(x) \right)\left( \frac{d^n}{dx^n} g(x)\right) = R^\alpha f(x)g(x) = \sum_{n=0}^\infty \binom{-\alpha}{n}\left( R^{\alpha + n}f(x) \right)\left( \frac{d^n}{dx^n} g(x)\right)
\label{RemainderOperatorProductRule}
\end{equation}
Now that we are setup, look at the set of ODEs $\frac{d^n}{dx^n}f(x) = f(x)$. From this the following general statement can be made,
\[\exists f(x) \forall n \in \mathbb{Z}, \frac{d^n}{dx^n}f(x) = f(x), f(0) = 1\]
given that negative integers in the index $n$ are interpreted as repeated integrals of the form $\int_{-\infty}^x f(t)dt$. This statement is true and admits only one solution $f(x) = e^x$. Let us generalize this statement to a fractional form,
\begin{equation}
\exists f(x) \forall \alpha \in \mathbb{C}, \frac{d^\alpha}{dx^\alpha}f(x) = f(x), f(0) = 1
\label{ConstaintOnFractionalCalculus}
\end{equation}
From statement \eqref{ConstaintOnFractionalCalculus} we can conclude that if $f(x)$ exists it must be $f(x) = e^x$, since it necessitates that $\frac{d}{dx}f(x) = f(x)$ and $f(0) = 1$. We will now require that statement \eqref{ConstaintOnFractionalCalculus} applies to fractional calculus. Using this condition $R^\alpha$ can be calculated in the case where $f(x) = e^x$. Using this constraint $I^\alpha e^x = e^x$, but since $J^\alpha e^x = e^x$ then $R^\alpha e^x = 0$. Now apply this result to \eqref{RemainderOperatorProductRule} with $f(x) = e^x$ and $g(x) = e^{-x}h(x)$ with $h(x) \in \mathbb{S}$.
\[R^\alpha h(x) = \sum_{n=0}^\infty \binom{-\alpha}{n}\left( R^{\alpha + n}e^x \right)\left( \frac{d^n}{dx^n} h(x)e^{-x}\right) = \sum_{n=0}^\infty \binom{-\alpha}{n} 0 \left( \frac{d^n}{dx^n} h(x)e^{-x}\right) = 0\]
Given statement \eqref{ConstaintOnFractionalCalculus} is true, then $R^\alpha = 0$, so the operator $J^{\alpha}$ is the compatible fractional calculus operator. This operator still is only defined for $\mathfrak{\alpha} \geq 1$. Using the analytic continuation of the functions produced by RMT \eqref{analytic_continuation} a general form of the operator can be defined as,
\begin{equation}
J^{\alpha}f(x) = \frac{1}{\Gamma(\alpha+n)}\int_{-\infty}^x (x-t)^{\alpha+n-1}\left( \frac{d^n}{dt^n} f(t) \right)dt
\label{fractional_calculus}
\end{equation}
where $f \in \mathbb{S}$, $n \in \mathbb{Z}^+$, $\alpha \in \mathbb{C}$ and $\mathfrak{R}(\alpha + n) \geq 1$.

\section{Properties of $J^{\alpha}$}
Now that we have $J^{\alpha}$ uniquely defined we can check if it maintains the properties that were built into the operator $J^k$. Note that the operator \eqref{fractional_calculus} is a linear operator for all $\alpha \in \mathbb{C}$ and that for $\alpha \in \mathbb{Z}$ it is equivalent to the operator $J^k$. Given $f \in \mathbb{S}$, $\alpha \in \mathbb{C}$, $n \in \mathbb{Z}^+$ and $\mathfrak{R}(\alpha + n) \geq 1$ then using Libnitz's integral rule,
\[\frac{d}{dx} J^{\alpha} f(x) = \frac{d}{dx}\frac{1}{\Gamma(\alpha + n)}\int_{-\infty}^x (x-t)^{\alpha+n-1} \left( \frac{d^n}{dt^n} f(t) \right)dt = \frac{\alpha+n-1}{\Gamma(\alpha+n)}\int_{-\infty}^x (x-t)^{\alpha+n-2} \left( \frac{d^n}{dt^n} f(t) \right)dt = J^{\alpha-1}f(x)\]
and using integration by parts,
\[J^{\alpha} \left( \frac{d}{dx} f(x) \right) = \frac{1}{\Gamma(\alpha+n)}\int_{-\infty}^x (x-t)^{\alpha+n-1} \left( \frac{d^{n+1}}{dt^{n+1}} f(t) \right)dt = -\frac{1-n-\alpha}{\Gamma(\alpha+n)}\int_{-\infty}^x (x-t)^{\alpha+n-2} \left( \frac{d^n}{dt^n} f(t) \right)dt = J^{\alpha-1}f(x)\]
So the operator $J^\alpha$ and derivative commute, and applying this property repeatedly yields $\frac{d^m}{dx^m}J^{\alpha}f(x) = J^{\alpha-m}f(x) = J^{\alpha}\frac{d^m}{dx^m}f(x)$, where $m \in \mathbb{Z}^+$.

Given $f \in \mathbb{S}$, $\alpha, \beta \in \mathbb{C}$, $n,m \in \mathbb{Z}^+$, $\mathfrak{R}(\alpha + n) \geq 1$ and $\mathfrak{R}(\beta + m) \geq 1$ then,
\[J^{\alpha}J^{\beta} f(x) = \frac{1}{\Gamma(\alpha + n)}\int_{-\infty}^x (x-t)^{\alpha+n-1} \left( \frac{d^n}{dt^n} \frac{1}{\Gamma(\beta+m)}\int_{-\infty}^t (t-s)^{\beta+m-1} \left( \frac{d^m}{ds^m} f(s) \right)ds \right)dt\]
Rearranging and using the fact that derivatives commute with $J^{\alpha}$,
\[J^{\alpha}J^{\beta} f(x) = \frac{d^{n+m}}{dx^{n+m}}\frac{1}{\Gamma(\alpha+n)\Gamma(\beta+m)}\int_{-\infty}^x \int_{-\infty}^t (x-t)^{\alpha+n-1}(t-s)^{\beta+m-1} f(s) ds dt\]
Swapping the order of integration,
\[J^{\alpha}J^{\beta} f(x) = \frac{d^{n+m}}{dx^{n+m}}\frac{1}{\Gamma(\alpha+n)\Gamma(\beta+m)} \int_{-\infty}^x \int_{s}^x (x-t)^{\alpha+n-1}(t-s)^{\beta+m-1} f(s) dt ds\]
Using a change of variables with $t=x-(x-s)r$ and rearranging yields,
\[J^{\alpha}J^{\beta} f(x) = \frac{d^{n+m}}{dx^{n+m}}\frac{1}{\Gamma(\alpha+n)\Gamma(\beta+m)} \int_{-\infty}^x (x-s)^{\alpha+\beta+n+m-1} f(s)ds \int_0^1 r^{\alpha+n-1}(1 - r)^{\beta+m-1}dr\]
Identifying the beta integral ,$\int_0^1 r^{\alpha+n-1}(1-r)^{\beta+m-1}dr=\frac{\Gamma(\alpha+n)\Gamma(\beta+m)}{\Gamma(\alpha+\beta+n+m)}$, and then rearranging,
\[J^{\alpha}J^{\beta} f(x) = \frac{d^{n+m}}{dx^{n+m}}\frac{1}{\Gamma(\alpha+\beta+n+m)} \int_{-\infty}^x (x-s)^{\alpha+\beta+n+m-1} f(s)ds = \frac{d^{n+m}}{dx^{n+m}} J^{\alpha+\beta+n+m} f(x) = J^{\alpha + \beta} f(x)\]
Relaxing constraints on $J^\alpha$ to allow Schwartz distributions enables us to reconstruct other standard fractional calculus definitions. Denote the Heaviside function as $H(x)$. The Riemann-Liouville fractional integral ${}_aI_x^\alpha$ can be defined as,
\[{}_aI_x^\alpha f(x) = \frac{1}{\Gamma(\alpha)}\int_a^x (x - t)^{\alpha - 1}f(t)dt = \frac{1}{\Gamma(\alpha)}\int_{-\infty}^x (x - t)^{\alpha - 1}H(t - a)f(t)dt = J^\alpha \left(H(x - a)f(x)\right)\]
The Riemann-Liouville fractional derivative ${}_aD_x^\alpha f(x)$, where $n = \lceil \alpha \rceil$
\[{}_aD_x^\alpha f(x) = \frac{d^n}{dx^n} {}_aI_x^{n - \alpha} f(x) = \frac{d^n}{dx^n} J^{n - \alpha} \left(H(x - a)f(x)\right) = J^{-\alpha} \left(H(x - a)f(x)\right)\]
The Caputo derivative ${}_a^C D_x^\alpha f(x)$ is,
\[{}_a^C D_x^\alpha f(x) = \frac{1}{\Gamma(n - \alpha)} \int_a^x (x - t)^{n - \alpha - 1} \left( \frac{d^n}{dt^n}f(t) \right) dt = J^{n - \alpha} \left(H(x - a)\frac{d^n}{dx^n}f(x)\right) \]
So $J^\alpha$ is a commuting linear operator. With the property $J^{\alpha}J^{\beta} f(x) =J^{\alpha+\beta} f(x) =J^{\alpha}J^{\beta} f(x)$, where $\alpha,\beta \in \mathbb{C}$ and $f \in \mathbb{S}$. Also given Schwartz distributions are allowed, when $a = -\infty$ and $f(x) \in \mathbb{S}$ then, $J^\alpha f(x) = {}_{a}I_x^\alpha f(x) = {}_{a}D_x^{-\alpha} f(x) = {}_{a}^C D_x^{-\alpha} f(x)$.

\section{Conclusion}
We have seen that RMT can be used to extend a calculus operator to a fractional version, and that the fractional calculus operator described is compatible with multiple definitions of fractional calculus operators given the constraints applied. Also this fractional calculus operator uniquely satisfies the conditions we imposed on fractional calculus when statement \eqref{ConstaintOnFractionalCalculus} is added as an additional constraint. And finally the operator can be generalized. If we modified the operator $J^k$ to use a different lower limit of integration and found a corresponding space of functions on which the operator is invertible then it should be possible to apply RMT to define a unique fractional calculus on that set of functions. So if the lower limit of integration was set to $\infty i$ the resulting operator should be a version of fractional calculus defined on a set that includes periodic functions.

\clearpage

\appendix*
\section{The integral as an invertible linear operator}
As a starting point for defining a fractional calculus we will first construct an integral operator having properties compatible with what I am looking for in fractional calculus. This operator that we will call $J^k$ should be a commuting linear operator that can reproduce repeated integration and differentiation. Also we will construct a function space $\mathbb{S}$ where the operator $J^k$ is well-behaved.

\subsection{Definitions}
Let the operator $J^k$ for $k \in \mathbb{Z}$ be,
\[ J^k f(x) := \begin{cases} \frac{1}{(k-1)!}\int_{-\infty}^x (x - t)^{k - 1}f(t)dt & k \geq 1 \\ f(x) & k = 0 \\ \frac{d^{\left|k\right|}}{dx^{\left|k\right|}}f(x) & k \leq -1 \end{cases} \]
Using this operator any function for which its absolute value and the absolute value of its derivatives are bounded by exponentials of the form $ae^{bx}$, with $a > 0$ and $b > 0$, are well-behaved. This set of functions can be expanded to produce a function space. Let us call this space set $\mathbb{S}$,
\[ \mathbb{S} := \left\lbrace f \in C^\omega(\mathbb{C}) \middle| (\forall n \in \mathbb{Z}^+)(\exists a_n \in \mathbb{M}, b_n \in \mathbb{R}), b_n > 0, \frac{d^n}{dx^n}f(x) \in \mathbb{B}(a_n, b_n) \right\rbrace \]
Where $\mathbb{M}$ is a set of positive monotonic increasing functions and $\mathbb{B}$ is a set of analytic functions for which the absolute value of the function along the real line is bounded by the exponential $e^{bx}$ rescaled by the function $a(x_0)$. The sets $\mathbb{M}$ and $\mathbb{B}$ are defined as follows,
\[\mathbb{M} := \left\lbrace a \in C(\mathbb{R}) \middle| (\forall x, y \in \mathbb{R}), x > y, a(x) > a(y) \geq 0 \right\rbrace\]
\[\mathbb{B}(a, b) := \left\lbrace f \in C^\omega(\mathbb{C}) \middle| (\forall x, x_0 \in \mathbb{R}), x \leq x_0, |f(x)| \leq a(x_0)e^{bx} \right\rbrace\]

\subsection{Properties of $\mathbb{S}$}
Looking at the properties of the space $\mathbb{S}$ let us assume the following. Given $f, g \in \mathbb{S}$ by definition there exists functions such that the following is true for $x_0 \geq x$, with $a_n, c_n \in \mathbb{M}$ and $b_n, d_n \in \mathbb{R}$
\[\left| \frac{d^n}{dx^n}f(x) \right| \leq a_n(x_0)e^{b_n x}, \left| \frac{d^n}{dx^n} g(x) \right| \leq c_n(x_0)e^{d_n x}\]
to simplify the arguments we will assume that $b_n \geq d_n$.
\[\left| \frac{d^n}{dx^n} \left( f(x) + g(x) \right) \right| \leq \left| \frac{d^n}{dx^n} f(x) \right| + \left| \frac{d^n}{dx^n} g(x) \right| \leq a_n(x_0)e^{b_n x} + c_n(x_0)e^{d_n x} \leq \left(a_n(x_0)e^{(b_n - d_n) x_0} + c_n(x_0)\right)e^{d_n x}\]
\[\left| \frac{d^n}{dx^n} (\alpha f(x)) \right| \leq \left|\alpha\right| \left| \frac{d^n}{dx^n} f(x) \right| \leq \left| \alpha \right| a_n(x_0)e^{b_n x}\]
So the set $\mathbb{S}$ is a function space. Now investigating the action of derivatives and integrals on this space, let us check if the set $\mathbb{S}$ is closed under repeated differentiation.
\[\left| \frac{d^n}{dx^n} \left(\frac{d^m}{dx^m} f(x)\right) \right| \leq \left| \frac{d^{n+m}}{dx^{n+m}} f(x) \right| \leq a_{n+m}(x_0)e^{b_{n+m} x}\]
In the case of integration we will check for closure in two steps, first if $n = 0$
\[\left| \frac{d^0}{dx^0} \left( \int_{-\infty}^x f(t)dt \right) \right| \leq \int_{-\infty}^x \left| f(t) \right|dt \leq \int_{-\infty}^x a_0(x_0)e^{b_0 t}dt = a_0(x_0)\frac{1}{b_0}e^{b_0 x}\]
and then in the case of $n \geq 1$, using Leibniz’s integral rule,
\[\left| \frac{d^n}{dx^n} \left( \int_{-\infty}^x f(t)dt \right) \right| \leq \left| \frac{d^{n-1}}{dx^{n-1}} f(x) \right| \leq a_{n-1}(x_0)e^{b_{n-1} x}\]
Using the result above repeatedly to extend the result to repeated integration. In the case of $m>n$,
\[\left| \frac{d^n}{dx^n} \left(\frac{1}{(m-1)!} \int_{-\infty}^x (x - t)^{m-1} f(t)dt\right) \right| \leq a_0(x_0)\frac{1}{{b_0}^{m-n}}e^{b_0 x}\]
and in the case of $n \leq m$,
\[\left| \frac{d^n}{dx^n} \left(\frac{1}{(m-1)!} \int_{-\infty}^x (x - t)^{m-1} f(t)dt\right) \right| \leq a_{n-m}(x_0)e^{b_{n-m} x}\]
This shows that repeated integrals and derivatives of functions in $\mathbb{S}$ exist and are in the space $\mathbb{S}$.

\subsection{Properties of $J^k$}
Considering the operator $J^k$ operating on functions in the space $\mathbb{S}$. Note that the operator is a linear operator for all $k \in \mathbb{Z}$ and that for $\left| k \right| > 1$ the operator $J^k$ can be constructed by repeated application of either $J^1$ or $J^{-1}$. Also note that the set $\mathbb{S}$ is closed under the action of $J^k$. Given that the function $f(x) \in \mathbb{S}$, and using the fundamental theorem of calculus and Libnitz’s integral rule we can show that,
\[J^1J^{-1} f(x) = \int_{-\infty}^x \frac{d}{dt} f(t)dt = f(x) - 0 = J^0 f(x)\]
\[J^{-1}J^1 f(x) = \frac{d}{dx} \int_{-\infty}^x f(t)dt = f(x)\frac{d}{dx}x - 0 = J^0 f(x)\]
So the following identity is true $J^1J^{-1} f(x) = J^0 f(x) = J^{-1}J^1 f(x)$. Applying this identity repeatedly and using the fact that $J^k$ can be constructed by repeated application of $J^1$ and $J^{-1}$ leads to the result,
\begin{equation}
J^kJ^m f(x) = J^{k+m} f(x)
\label{additiveProperty}
\end{equation}

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\usepackage{color}
\newcommand{\red}[1]{{\color{red} #1}}
\usepackage{booktabs}
\title{Red Scare!}
\begin{document}
\maketitle
\begin{abstract}
Oh no! All I wanted was to write a straightforward reachability exercise.
But some of the vertices have turned red, giving me cruel ideas for much harder questions!
\end{abstract}
\section{Problems}
In every problem of this exercise, we consider a graph $G$ with vertex set $V(G)$ and edge set $E(G)$.
The graph can be directed or undirected.
Every graph in this exercise is simple (no multiple edges between any pair of vertices) and unweighted.
We fix the notation $n=|V(G)|$ and $m=|E(G)|$.

\begin{marginfigure}
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 \caption{Example graph $G_{\text{ex}}$ corresponding to the file {\tt G-ex}.}
\end{marginfigure}

Every graph comes with two specified vertices $s,t\in V(G)$ called the \emph{start} and \emph{end} vertices, and a subset $R\subseteq V(G)$ of \emph{red} vertices.
In particular, $R$ can inlude $s$ and $t$.
We fix the notation $r= |R|$.
In the example graph $G_{\text{ex}}$, we have $s=0$, $t=3$, and $R=\{4,5,7\}$.

An \emph{$s,t$-path} is a sequence of distinct vertices $v_1,\ldots, v_l$ such that $v_1=s$, $v_l=t$, and $v_iv_{i+1}\in E(G)$ for each $i\in\{1,\ldots,l-1\}$.
The \emph{length} of such a path is $l-1$, the number of edges.
Note that this definition requires the vertices on a path to be distinct, this is sometimes called a \emph{simple} path.

The problems we want solved for each graph are the following:
\begin{description}
  \item[None] Return the length of a shortest $s,t$-path internally avoiding $R$.
    To be precise, let $P$ be the set of $s,t$-paths using no vertices from $R$ except maybe $s$ and $t$ themselves. Let $l(p)$ denote the length of a path $p$.
    Return $\min\{\,l(p)\colon p\in P\,\}$.
    If no such path exists, return `-1'.
    Note that if the edge $st$ exists then the answer is 1, no matter the colour of $s$ or $t$.
    In $G_{\text{ex}}$, the answer is 3 (because of the path 0, 1, 2, 3.)
  \item[Some] Return `true' if there is a path from $s$ to $t$ that includes at least one vertex from $R$.
    Otherwise, return `false.'
    In $G_{\text{ex}}$, the answer is `yes' (in fact, two such paths exist: the path 0, 4, 3 and the path 0, 5, 6, 7, 3.)
  \item [Many] Return the maximum number of red vertices on any path from $s$ to $t$.
    To be precise, let $P$ be the set of $s,t$-paths and let $r(p)$ denote the number of red vertices  on a path $p$.
    Return $\max\{\,r(p)\colon p\in P\,\}$.
    If no path from $s$ to $t$ exists, return `-1'.
    In $G_{\text{ex}}$, the answer is `2' (because of the path 0, 5, 6, 7, 3.)
  \item [Few] Return the minimum number of red vertices on any path from $s$ to $t$.
    To be precise, let $P$ be the set of $s,t$-paths and let $r(p)$ denote the number of red vertices  on a path $p$.
    Return $\min\{\,r(p)\colon p\in P\,\}$.
    If no path from $s$ to $t$ exists, return `-1'.
    In $G_{\text{ex}}$, the answer is 0 (because of the path 0, 1, 2, 3.)
  \item [Alternate] Return `true' if there is a path from $s$ to $t$ that alternates between red and non-red vertices.
    To be precise, a path $v_1,\ldots, v_l$ is \emph{alternating} if for each $i\in\{1,\ldots,l-1\}$, exactly one endpoint of the edge $v_iv_{i+1}$ is red.
    Otherwise, return `false.'
    In $G_{\text{ex}}$, the answer is `yes' (because of the path 0, 5, 6, 7, 3.)
\end{description}

\subsection{Requirements}
\paragraph{Solved instances.}
For three of the problems (I’m not telling which), you need to be able to handle \emph{all} instances.
For the remaining two problems, you will not be able to solve all instances, but should be able to solve roughly half of them.
Your solutions must run in polynomial time.
If you do have a polynomial-time implementation that takes more than 1 hour on some instance, just abort it and report this in your report.\sidenote{However, this should not happen. As a guideline, I have a non-optimised Python implementation that solves all the instances in a single run in half an hour on a low-powered 2012 laptop.}

For two of the problems (I’m not telling which), you will not be able to write an algorithm that works for all graphs, because the problems are hard in general.
For one of those problems, you should be able to argue for computational hardness with a simple reduction.
The other problem will probably mystify you.
A sophisticated hardness argument exists in the research literature (but not in the course text book)---you are welcome to try to find it and include a reference in your report.
But you are not required to find an explanation, and absolutely not required to come up with the reduction yourself.

\paragraph{Universality.}
Your algorithms must run in polynomial time on a well-defined class of graphs.
“Well-defined class” means something like “all graphs,”\sidenote{“All graphs” means all simple, loop-less, unweighted graphs.} “directed graphs,” “undirected graphs,” “bipartite graphs,” “acyclic graphs,” “graphs of bounded treewidth,” “planar graphs,” “expanders” or even a combination of these.

In particular, you are allowed to do something like this:
\begin{quotation}
  \vspace*{-3ex} \begin{tabbing}
  if \= (isBipartite($G$)) then\\
  \> $\cdots\qquad\qquad$\=\# run the Strumpf--Chosa algorithm \\
  else \\
  \> print(``?!'') \>\# problem is NP-hard for non-bipartite graphs, so give up
\end{tabbing}
\end{quotation}

On the other hand, you are not allowed to base your algorithm on specific knowledge of which graphs are in the {\tt data} directory,
For an extreme example, the following would not be allowed:
\begin{quotation}
  \vspace*{-3ex}
  \begin{tabbing}
  if \= (filename == "rusty-1-17")) then\\
  \> print("14") \=\# solved by hand\\
  else \\
  \> print(``?'') \>\# no idea what to do
\end{tabbing}
\end{quotation}

\paragraph{Libraries.}
This exercise focusses on choosing \emph{between} algorithms, not implementing them.
Thus, you are \emph{not} required to write these algorithms from scratch.
For instance, if you need a minimum spanning tree, and you already have a working implementation of Prim’s algorithm, you are free to reuse that.
In particular, you are free to use a library implementation of these algorithms.
You are also free to use implementations from books or from other people, provided that you are not violating any intellectual property rights.
(It goes without saying that you properly cite the source of these implementations in your report.)

You are highly encouraged to use your own implementation of standard graph algorithms that you may have made for some other exercise.
If you do this, then separate that implementation in your source code, maybe by leaving it in its own file.
Attribute all the original authors in the source code.

\paragraph{Deliverables.}
Hand in
\begin{enumerate}
  \item a report; follow the skeleton in {\tt doc/report.pdf}.
  \item a text file {\tt results.txt} with all the results, as specified in the report.
  \item the programs you have written to answer the questions, including any scripts you need to run these programs on the instances, and a README file that explains how to recreate {\tt results.txt} by running your programs.
\end{enumerate}

\newpage
\section{Appendix: Gallery of Graphs}

This gallery consists of descriptions and drawings of many of the graphs in the {\tt data} directory.
Ideally, these descriptions are useful for finding mistakes in your code.
In particular, for many of these graphs it is obvious what the correct answers are.
Some of the graphs are \emph{random} graph---they have no structure and are pretty boring.
Others, such as the Word graphs, have a lot of structure.

\subsection{Individual graphs}
\begin{marginfigure}
 \begin{tikzpicture}[scale=.4]
   \node  at (1,-2) {\tt P3-0};
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=below:$0$, label = above:$s$] at (0,0) {};
   \node (1) [label=below:$1$, fill=red, draw, inner sep =1.5pt] at (1,0) {};
   \node (2) [label=below:$2$, label = above:$t$] at (2,0) {};
   \draw (0) -- (1) -- (2);
 \end{scope}
 \end{tikzpicture}
\quad
 \begin{tikzpicture}[scale=.4]
   \node  at (1,-2) {\tt P3-1};
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=below:$0$, label = above:$s$] at (0,0) {};
   \node (1) [label=below:$1$, label = above:$t$] at (1,0) {};
   \node (2) [label=below:$2$, fill=red, draw, inner sep =1.5pt] at (2,0) {};
   \draw (0) -- (1) -- (2);
 \end{scope}
 \end{tikzpicture}

 \bigskip
 \begin{tikzpicture}[scale=.4]
   \node  at (0,-2) {\tt K3-0};
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=135:$0$, label = 45:$s$] at (90:1cm) {};
   \node (1) [label=210:$1$, fill=red, draw, inner sep =1.5pt] at (210:1cm) {};
   \node (2) [label=285:$2$, label = 15:$t$] at (330:1cm) {};
   \draw (0) -- (1) -- (2) -- (0);
 \end{scope}
 \end{tikzpicture}
\quad
 \begin{tikzpicture}[scale=.4]
   \node  at (0,-2) {\tt K3-1};
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=135:$0$, label = 45:$s$] at (90:1cm) {};
   \node (1) [label=210:$1$, fill=red, draw, inner sep =1.5pt] at (210:1cm) {};
   \node (2) [label=285:$2$, label = 15:$t$] at (330:1cm) {};
   \draw [->] (0) -- (1);
   \draw [->] (1) -- (2);
   \draw [->] (0) -- (2);
 \end{scope}
 \end{tikzpicture}
\quad
 \begin{tikzpicture}[scale=.4]
   \node  at (0,-2) {\tt K3-2};
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=135:$0$, label = 45:$s$] at (90:1cm) {};
   \node (1) [label=210:$1$, fill=red, draw, inner sep =1.5pt] at (210:1cm) {};
   \node (2) [label=285:$2$, label = 15:$t$] at (330:1cm) {};
   \draw [->] (1) -- (0);
   \draw [->] (1) -- (2);
   \draw [->] (0) -- (2);
 \end{scope}
 \end{tikzpicture}
 \caption{Paths and triangles with various choices of orientation and redness.}
 \end{marginfigure}

 The {\tt data} directory contains a small number of individual graphs, typicall of very small size.
 This includes $G_{\text{ex}}$, a number of small graphs of 3 vertices shown to the right, and an all-red dodecahedron.
 It is a good idea to use these graphs initially to ensure that your parser works, your graph data structure makes sense, etc.
These graphs also serve as an invitation to create more toy examples by hand while you test your code, so ensure that everything works on very small graphs.
Keep some good, clear drawings of `your' graphs around, they help immensely when finding mistakes.

\subsection{Word graphs}
In the word graphs, each vertex represents a five-letter word of English.
For $k\in\{1,2\}$, an edge joins $u$ and $v$ if the corresponding words are anagrams, or if they differ in exactly $k$ positions.
For instance ``begin'' and ``binge'' are neighbours, and so are ``turns'' and ``terns'' for $k=1$.

The word graphs  come in two flavours.
The \emph{rusty word} graphs are guaranteed to include ``begin,'' ``ender,'' and ``rusty.''
The vertex corresponding to ``rusty'' is coloured red, no other vertices are red.

The filenames are {\tt rusty-$k$-$n$}.
\begin{figure}
\[
\begin{tikzpicture}[
    xscale = 1.5,
    every node/.style={ draw,rectangle,
      rounded corners, inner sep = 1.5pt, font =\sf\small }
    ]
  \node (rungs) at (3,2) {RUNGS};
  \node (sings) at (1,2) {SINGS};
  \node (begin) [label = below:$s$] at (0,0) {BEGIN};
  \node (rents) at (4,1) {RENTS};
  \node (ender) [label = below:$t$] at (4,0) {ENDER};
  \node (rests) at (5,1) {RESTS};
  \node (binge) at (1,0) {BINGE};
  \node (singe) at (1,1) {SINGE};
  \node (rings) at (2,2) {RINGS};
  \node (rusty) [fill = red] at (5,3) {RUSTY};
  \node (rente) at (3,1) {RENTE};
  \node (runts) at (4,2) {RUNTS};
  \node (enter) at (3,0) {ENTER};
  \node (rusts) at (5,2) {RUSTS};
  \node (tinge) at (2,0) {TINGE};
  \node (tings) at (2,1) {TINGS};
  \node (runty) at (4,3) {RUNTY};
\draw (rungs) -- (runts) ;
\draw (sings) -- (tings) ;
\draw (sings) -- (rings) ;
\draw (sings) -- (singe) ;
\draw (begin) -- (binge) ;
\draw (rents) -- (rests) ;
\draw (rents) -- (rente) ;
\draw (rents) -- (runts) ;
\draw (ender) -- (enter) ;
\draw (rests) -- (rusts) ;
\draw (binge) -- (tinge) ;
\draw (binge) -- (singe) ;
\draw (singe) -- (tinge) ;
\draw (rings) -- (tings) ;
\draw (rusty) -- (rusts) ;
\draw (rusty) -- (runty) ;
\draw (rente) -- (enter) ;
\draw (runts) -- (rusts) ;
\draw (runts) -- (runty) ;
\draw (tinge) -- (tings) ;
\draw (rungs) -- (rings) ;
  \end{tikzpicture}
\]
\caption{{\tt rusty-1-17}}
\end{figure}

The \emph{common word} use the same adjacency structure, and always include `start' and `ender.'
A word is red if it is uncommon (like `ender'), just under half the words are uncommon.
The filenames for these graphs are {\tt common-$n$}.

\subsection{Grids}

\begin{marginfigure}
\begin{tikzpicture}[scale = .5, every node/.style={circle, fill, inner sep =1.5pt}]
  \foreach \x in {0,1,2,3,4} {
    \foreach \y in {0,1,2,3,4} {
      \node (\x_\y) at (\x,\y) {};
    }
  }
  \node [label = left:$s$] at (0,0) {};
  \node [label = right:$t$] at (4,4) {};
  \draw (1_4) -- (2_4);
  \draw (1_4) -- (0_4);
  \draw (1_4) -- (1_3);
  \draw (1_4) -- (0_3);
  \draw (1_3) -- (2_3);
  \draw (1_3) -- (2_4);
  \draw (1_3) -- (1_2);
  \draw (1_3) -- (0_2);
  \draw (1_3) -- (0_3);
  \draw (1_2) -- (2_2);
  \draw (1_2) -- (2_3);
  \draw (1_2) -- (1_1);
  \draw (1_2) -- (0_2);
  \draw (1_2) -- (0_1);
  \draw (1_1) -- (2_1);
  \draw (1_1) -- (2_2);
  \draw (1_1) -- (1_0);
  \draw (1_1) -- (0_0);
  \draw (1_1) -- (0_1);
  \draw (1_0) -- (2_0);
  \draw (1_0) -- (2_1);
  \draw (1_0) -- (0_0);
  \draw (0_4) -- (0_3);
  \draw (0_2) -- (0_3);
  \draw (0_2) -- (0_1);
  \draw (0_0) -- (0_1);
  \draw (3_1) -- (3_0);
  \draw (3_1) -- (3_2);
  \draw (3_1) -- (2_0);
  \draw (3_1) -- (2_1);
  \draw (3_1) -- (4_2);
  \draw (3_1) -- (4_1);
  \draw (3_0) -- (2_0);
  \draw (3_0) -- (4_0);
  \draw (3_0) -- (4_1);
  \draw (3_3) -- (3_2);
  \draw (3_3) -- (3_4);
  \draw (3_3) -- (2_2);
  \draw (3_3) -- (2_3);
  \draw (3_3) -- (4_3);
  \draw (3_3) -- (4_4);
  \draw (3_2) -- (2_1);
  \draw (3_2) -- (2_2);
  \draw (3_2) -- (4_2);
  \draw (3_2) -- (4_3);
  \draw (3_4) -- (2_3);
  \draw (3_4) -- (2_4);
  \draw (3_4) -- (4_4);
  \draw (2_0) -- (2_1);
  \draw (2_1) -- (2_2);
  \draw (2_2) -- (2_3);
  \draw (2_3) -- (2_4);
  \draw (4_2) -- (4_1);
  \draw (4_2) -- (4_3);
  \draw (4_3) -- (4_4);
  \draw (4_0) -- (4_1);
  \node [fill=red,draw,inner sep = 1.5pt] at (1,0) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (1,1) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (1,2) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (1,3) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (3,1) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (3,2) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (3,3) {};
  \node [fill=red,draw,inner sep = 1.5pt] at (3,4) {};
\end{tikzpicture}
\caption{The grid for $N=5$, represented by {\tt grid-5-0}.}
\end{marginfigure}

The Grid graphs consist of $N^2$ vertices that represent integer coordinates $(x,y)$ for $x,y\in\{0,\ldots,N-1\}$.
Each vertex $(x,y)$ is connected to $(x-1,y)$, $(x,y-1)$, and $(x-1,y-1)$, provided that these vertices exist.
The red vertices form a maze-like structure in the graph:
Every second row is red, except for the top- or bottommost vertex, alternatingly.
There is a unique $s,t$-path avoiding all red vertices, and a shortest alternating path following the diagonal.

Grid graphs of various sizes are represented by {\tt grid-$N$-0}.
Each of these graphs comes with two variants.
In {\tt grid-$N$-1}, some random red vertices have turned non-red (so there are `holes' in the hedges).
In {\tt grid-$N$-2}, some random non-red vertices have turned red (so some passages are blocked).

\subsection{Walls}

\begin{marginfigure}
 \begin{tikzpicture}[scale=.4]
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=below:$0$, label=left:$s$] at (0,0) {};
   \node (1) [label=below:$1$] at (1,0) {};
   \node (2) [label=below:$2$] at (2,0) {};
   \node (3) [label=below:$3$,fill=red,draw,inner sep = 1.5pt] at (3,0) {};
   \node (4) [label=above:$4$] at (3,1) {};
   \node (5) [label=above:$5$] at (2,1) {};
   \node (6) [label=above:$6$] at (1,1) {};
   \node (7) [label=above:$7$, label=left:$t$] at (0,1) {};
   \draw (0) -- (1) -- (2) -- (3) -- (4) -- (5) -- (6) -- (7) -- (0);
 \end{scope}
 \end{tikzpicture}
 \caption{The single-brick wall, {\tt wall-p-1}.}
 \end{marginfigure}

Bricks are arranged like a wall of height $2$.
Here are three bricks with overlap $1$:
\begin{marginfigure}
 \begin{tikzpicture}[scale=.4]
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=below:$0$] at (0,0) {};
   \node (1) [label=below:$1$] at (1,0) {};
   \node (2) [label=below:$$] at (2,0) {};
   \node (3) [label=below:$$] at (3,0) {};
   \node (4) [label=above:$4$] at (3,1) {};
   \node (5) [label=above:$5$] at (2,1) {};
   \node (6) [label=above:$6$] at (1,1) {};
   \node (7) [label=above:$7$] at (0,1) {};
   \node (8) [label=below:$$] at (4,0) {};
   \node (9) [label=below:$$] at (5,0) {};
   \node (10) [label=below:\small$10$]  at (5,-1) {};
   \node (11) [label=below:\small$11$]  at (4,-1) {};
   \node (12) [label=below:\small$12$]  at (3,-1) {};
   \node (13) [label=below:\small$13$]  at (2,-1) {};
   \node (14) [label=below:\small$14$]  at (6,0) {};
   \node (15) [label=below:\small$15$,fill=red,draw,inner sep = 1.5pt]  at (7,0) {};
   \node (16) [label=above:\small$16$]  at (7,1) {};
   \node (17) [label=above:\small$17$]  at (6,1) {};
   \node (18) [label=above:\small$18$]  at (5,1) {};
   \node (19) [label=above:\small$19$]  at (4,1) {};
   \draw (0) -- (1) -- (2) -- (3) -- (4) -- (5) -- (6) -- (7) -- (0);
   \draw (3)--(8)--(9)--(10)--(11)--(12)--(13)--(2);
   \draw (9)--(14)--(15)--(16)--(17)--(18)--(19)--(8);
 \end{scope}
 \end{tikzpicture}
 \caption{Three bricks with overlap 1, {\tt wall-p-3}.}
 \end{marginfigure}

 \begin{marginfigure}
 \begin{tikzpicture}[scale=.4]
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=below:$0$, label=left:$s$] at (0,0) {};
   \node (1) [label=below:$1$] at (1,0) {};
   \node (2) [label=below:$2$] at (2,0) {};
   \node (3) [label=below:$$] at (3,0) {};
   \node (4) [label=above:$4$] at (3,1) {};
   \node (5) [label=above:$5$] at (2,1) {};
   \node (6) [label=above:$6$] at (1,1) {};
   \node (7) [label=above:$7$, label=left:$t$] at (0,1) {};
   \node (8) [label=above:$8$] at (4,0) {};
   \node (9) [label=above:$9$] at (5,0) {};
   \node (10) [label=below:\small$$]  at (6,0) {};
   \node (11) [label=below:\small$11$]  at (6,-1) {};
   \node (12) [label=below:\small$12$]  at (5,-1) {};
   \node (13) [label=below:\small$13$]  at (4,-1) {};
   \node (14) [label=below:\small$14$]  at (3,-1) {};
   \node (15) [label=below:\small$15$]  at (7,0) {};
   \node (16) [label=below:\small$16$]  at (8,0) {};
   \node (17) [label=below:\small$17$,fill=red,draw,inner sep = 1.5pt]  at (9,0) {};
   \node (18) [label=above:\small$18$]  at (9,1) {};
   \node (19) [label=above:\small$19$]  at (8,1) {};
   \node (20) [label=above:\small$20$]  at (7,1) {};
   \node (21) [label=above:\small$21$]  at (6,1) {};
   \draw (0) -- (1) -- (2) -- (3) -- (4) -- (5) -- (6) -- (7) -- (0);
   \draw (3)--(8)--(9)--(10)--(11)--(12)--(13)--(14)--(3);
   \draw (10)--(15)--(16)--(17)--(18)--(19)--(20)--(21)--(10);
 \end{scope}
 \end{tikzpicture}
 \caption{Three bricks with overlap 0, {\tt wall-z-3}.}
 \end{marginfigure}

 \begin{marginfigure}
 \begin{tikzpicture}[scale=.4]
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (0) [label=below:$0$] at (0,0) {};
   \node (1) [label=below:$1$] at (1,0) {};
   \node (2) [label=below:$2$] at (2,0) {};
   \node (3) [label=below:$3$] at (3,0) {};
   \node (4) [label=above:$4$] at (3,1) {};
   \node (5) [label=above:$5$] at (2,1) {};
   \node (6) [label=above:$6$] at (1,1) {};
   \node (7) [label=above:$7$] at (0,1) {};
   \node (8) [label=below:$8$] at (4,0) {};
   \node (9) [label=below:$9$] at (5,0) {};
   \node (10) [label=below:$10$]  at (6,0) {};
   \node (11) [label=below:$11$,fill=red,draw,inner sep = 1.5pt]  at (7,0) {};
   \node (12) [label=above:$12$]  at (7,1) {};
   \node (13) [label=above:$13$]  at (6,1) {};
   \node (14) [label=above:$14$]  at (5,1) {};
   \node (15) [label=above:$15$]  at (4,1) {};
   \draw (0) -- (1) -- (2) -- (3) -- (4) -- (5) -- (6) -- (7) -- (0);
   \draw (3)--(8)--(9)--(10)--(11)--(12)--(13)--(14)--(15)--(8);
 \end{scope}
 \end{tikzpicture}
 \caption{Two bricks with negative overlap, {\tt wall-n-2}.}
 \end{marginfigure}

The Wall graphs are a family consisting of $N$ overlapping 8-cycles called \emph{bricks}.
The bricks are laid in a wall of height 2, with various intervals of overlap.
Each wall has a single red vertex $w$, the rightmost vertex at the same level as vertex $0$.
These graphs are interesting instances for finding paths from $s$ to $t$ through the red vertex.
The should help you avoid some obvious pitfalls when developing an algorithm for the problem \emph{Some}.

The Walls with overlap 1, called {\tt brick-1-$N$}, allow an $s,t$-path through $w$.

The Walls with overlap 0, called {\tt brick-0-$N$}, allow a walk from $s$ to $t$ through $w$, but this walk will use $N-2$ vertices twice.
In particular, such a walk it not a path, and your algorithm for Problem \emph{Some} should not be fooled by it.

The walls with negative overlap, called {\tt brick-n-$N$} also allow a walk from $s$ to $t$ through $w$, but this walk would use $N-2$ edges twice.
Again, such walk is not a path.

\section{Ski}
SkiFree\footnote{Read more at Wikipedia’s SkiFree page, the author’s site {\tt ski.ihoc.net}, or (amazingly) SkiFree Fan Fiction at {\tt www.fanfiction.net/game/SkiFree/}.} is an ancient computer game by Chris Pirih, part of the Microsoft Entertainment Pack in 1991, and going back to VT100 VAX/VMS terminals, ultimately inspired by Activision’s \emph{Skiing} for the Atari 2600 console.

Time to show you can handle yourself off-pist. Get from the start to the goal, avoiding the trees, dogs rocks etc.
Beware of the dreaded Yeti who is told to lurk at the red nodes of the mountain and will certainly chase you down and eat you if you pass him.

In each level, the player moves down, and either one step left or right.
(Some moves are blocked by obstacles.)

\begin{figure*}[h]
\includegraphics[width=7cm]{ski.png}
\quad
\begin{tikzpicture}[scale = .5, every node/.style={circle, fill, inner sep =1.5pt}]
  \node  (0) [label = above:$0\,\,s$] at (0,0) {};
  \node  (1) [label = above:$1$] at (-1,-1) {};
  \node  (2) [label = above:$2$] at (1,-1) {};
  \node  (3) [label = above:$3$] at (-2,-2) {};
  \node  (4) at (0,-2) {};
  \node  (5) [label = above:$5$] at (2,-2) {};
  \node  (6) [label = above:$6$] at (-3,-3) {};
  \node  (7) at (-1,-3) {};
  \node  (8) at (1,-3) {};
  \node  (9) [label = above:$9$] at (3,-3) {};
  \node (10) at (-4,-4) {};
  \node (11) at (-2,-4) {};
  \node (12) [label = left:$12$, fill = red, draw] at (0,-4) {};
  \node (13) at (2,-4) {};
  \node (14) at (4,-4) {};
  \node (15) at (-5,-5) {};
  \node (16) at (-3,-5) {};
  \node (17) at (-1,-5) {};
  \node (18) at (1,-5) {};
  \node (19) at (3,-5) {};
  \node (20) at (5,-5) {};
  \node (21) at (-6,-6) {};
  \node (22) at (-4,-6) {};
  \node (23) at (-2,-6) {};
  \node (24) at (0,-6) {};
  \node (25) at (2,-6) {};
  \node (26) at (4,-6) {};
  \node (27) at (6,-6) {};
  \node (29) at (-5,-7) {};
  \node (30) at (-3,-7) {};
  \node (31) at (-1,-7) {};
  \node (32) at (1,-7) {};
  \node (33) at (3,-7) {};
  \node (34) at (5,-7) {};
  \node (35) at (7,-7) {};
  \node (36) [label= below:$36\,\,t$] at (0,-8) {};
  \draw [->] (0) -- (1);
  \draw [->] (0) -- (2);
  \draw [->] (1) -- (3);
  \draw [->] (2) -- (4);
  \draw [->] (2) -- (5);
  \draw [->] (3) -- (6);
  \draw [->] (3) -- (7);
  \draw [->] (4) -- (7);
  \draw [->] (4) -- (8);
  \draw [->] (5) -- (8);
  \draw [->] (6) -- (11);
  \draw [->] (7) -- (11);
  \draw [->] (8) -- (12);
  \draw [->] (9) -- (13);
  \draw [->] (9) -- (14);
  \draw [->] (10) -- (15);
  \draw [->] (11) -- (16);
  \draw [->] (12) -- (18);
  \draw [->] (13) -- (18);
  \draw [->] (14) -- (19);
  \draw [->] (14) -- (20);
  \draw [->] (15) -- (22);
  \draw [->] (16) -- (22);
  \draw [->] (16) -- (23);
  \draw [->] (17) -- (23);
  \draw [->] (17) -- (24);
  \draw [->] (18) -- (24);
  \draw [->] (18) -- (25);
  \draw [->] (19) -- (25);
  \draw [->] (19) -- (26);
  \draw [->] (20) -- (26);
  \draw [->] (20) -- (27);
  \draw [->] (21) -- (29);
  \draw [->] (22) -- (29);
  \draw [->] (23) -- (30);
  \draw [->] (23) -- (31);
  \draw [->] (24) -- (31);
  \draw [->] (24) -- (32);
  \draw [->] (25) -- (33);
  \draw [->] (26) -- (33);
  \draw [->] (26) -- (34);
  \draw [->] (27) -- (35);
  \draw [->] (29) -- (36);
  \draw [->] (30) -- (36);
  \draw [->] (31) -- (36);
  \draw [->] (32) -- (36);
  \draw [->] (33) -- (36);
  \draw [->] (34) -- (36);
  \draw [->] (35) -- (36);
\end{tikzpicture}
\caption{A Ski level and the corresponding graph.
  This graph is described in {\tt ski-illustration}.
Yeti is lurking at node 12.
Graphics from {\tt ski.ihoc.net}.}
\end{figure*}

\subsection{Increasing numbers}
Each \emph{Increase} graph is generated from a sequence $a_1,\ldots, a_n$ of unique integers with $0\leq\alpha_i\leq 2n$.
(The random process is this: Pick a subset of size $n$ from $\{0,\ldots, 2n\}$ and arrange the elements in random order.)
We set $s=\alpha_1$ and $t=\alpha_n$.
Odd numbers are red.
There is an edge from  $\alpha_i$ to $\alpha_j$ if $i<j$ and $\alpha_i<\alpha_j$.

\begin{figure}
 \begin{tikzpicture}[]
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (2)  [label=below:$2$, label = left:$s$] at (0,0) {};
   \node (13) [label=below:$13$, fill=red, draw] at (1,0) {};
   \node (1)  [label=below:$1$,  fill=red, draw] at (2,0) {};
   \node (11) [label=below:$11$, fill=red, draw] at (3,0) {};
   \node (15) [label=below:$15$, fill=red, draw] at (4,0) {};
   \node (3)  [label=below:$3$,  fill=red, draw] at (5,0) {};
   \node (4)  [label=below:$4$] at (6,0) {};
   \node (5)  [label=below:$5$, label = right:$t$, fill=red, draw] at (7,0) {};
   \draw [->] (2) edge[] (13);
   \draw [->] (2) edge [bend left] (11);
   \draw [->] (2) edge [bend left] (15);
   \draw [->] (2) edge [bend left] (3);
   \draw [->] (2) edge [bend left] (4);
   \draw [->] (2) edge [bend left] (5);
   \draw [->] (13) edge [bend left] (15);
   \draw [->] (1) edge [] (11);
   \draw [->] (1) edge [bend left] (15);
   \draw [->] (1) edge [bend left] (3);
   \draw [->] (1) edge [bend left] (4);
   \draw [->] (1) edge [bend left] (5);
   \draw [->] (11) edge []  (15);
   \draw [->] (3) edge [] (4);
   \draw [->] (3) edge [bend left] (5);
   \draw [->] (4) edge [] (5);
 \end{scope}
 \end{tikzpicture}
 \caption{\tt increase-n8-1.0}
  \end{figure}

\begin{figure}
 \begin{tikzpicture}[]
 \begin{scope}[every node/.style={circle, fill, inner sep =1.5pt}]
   \node (11)  [label=below:$11$, label = left:$s$] at (0,0) {};
   \node (15) [label=below:$15$, fill=red, draw] at (1,0) {};
   \node (1)  [label=below:$1$,  fill=red, draw] at (2,0) {};
   \node (8) [label=below:$8$] at (3,0) {};
   \node (13) [label=below:$13$, fill=red, draw] at (4,0) {};
   \node (4)  [label=below:$4$] at (5,0) {};
   \node (6)  [label=below:$6$] at (6,0) {};
   \node (12)  [label=below:$12$, label = right:$t$] at (7,0) {};
   \draw [->] (11) edge[] (15);
   \draw [->] (11) edge [bend left] (13);
   \draw [->] (11) edge [bend left] (12);
   \draw [->] (1) edge [] (8);
   \draw [->] (1) edge [bend left] (13);
   \draw [->] (1) edge [bend left] (3);
   \draw [->] (1) edge [bend left] (4);
   \draw [->] (1) edge [bend left] (6);
   \draw [->] (1) edge [bend left] (12);
   \draw [->] (8) edge [] (13);
   \draw [->] (8) edge [bend left] (12);
   \draw [->] (4) edge [] (6);
   \draw [->] (4) edge [bend left] (12);
 \end{scope}
 \end{tikzpicture}
 \caption{\tt increase-n8-2.0}
  \end{figure}

\end{document}


\documentclass{article}
\usepackage{tocloft}
\include{common_symbols_and_format}
\renewcommand{\cfttoctitlefont}{\Large\bfseries}
\begin{document}
\logo
\rulename{Two Moving Averages}
\tblofcontents
\ruledescription{This price-only rule takes positions based on two moving averages, and a minimum threshold K. It allocates a positive position when the shorter moving average is above K fraction higher than the other average and vice versa.}
\ruleparameters
{Minimum ratio}{0.01}{Threshold difference in averages.}{$\minimumratio$}
{Short price average length}{2}{Number of days to include in the fast moving average.}{$\averagelengthshort$}
{Long price average length}{2}{Number of extra days to include in slow moving average.}{$\averagelengthlong$}
{Maximum allocation}{1.0}{Maximum size of position to take.}{$Z_{max}$}
\stoptable

\section{Equation}

\begin{equation}
\bigcontribution(\currenttime, \averagelengthshort, \price) = \frac{1}{\averagelengthshort} \sum_{\dummyiterator=0}^{\averagelengthshort-1} \price(\currenttime - \dummyiterator)\\
\end{equation}

\begin{equation}
\bigcontribution(\currenttime, \averagelengthlong, \price) = \frac{1}{\averagelengthlong} \sum_{\dummyiterator=0}^{\averagelengthlong-1} \price(\currenttime - \dummyiterator)\\
\end{equation}

\begin{equation}
\bigcontributiontime = \frac{\bigcontributionshort}{\bigcontributionlong} \\
\end{equation}

\begin{equation}
\position_\currenttime =
\begin{cases}
\mbox{$Z_{max}$,} & \mbox{if } \bigcontributiontime > 1 + 2\minimumratio \\
\mbox{$\frac{\bigcontributiontime - 1 - \minimumratio}{\minimumratio}$,} & \mbox{if } 1 + \minimumratio < \bigcontributiontime < 1 + 2\minimumratio \\
\mbox{0.0,} & \mbox{if } 1/(1 + \minimumratio) < \bigcontributiontime < 1 + \minimumratio \\
\mbox{$-\frac{1/\bigcontributiontime - 1 - \minimumratio}{\minimumratio}$,} & \mbox{if } 1/(1 + 2\minimumratio) < \bigcontributiontime < 1/(1 + \minimumratio) \\
\mbox{$-Z_{max}$,} & \mbox{if } \bigcontributiontime < 1/(1 + 2\minimumratio) \\
\end{cases}
\end{equation}

\hspace{200mm}
\\
\noindent where $\position_\currenttime$ is the portfolio allocation at time $\currenttime$, and $\price = \price(\currenttime)$ is the value of the price series.

\hspace{200mm}
\hspace{200mm}

\keyterms
\furtherlinks

\end{document}


In this section, we deal with only with the FirstOrderType2R case.

  \begin{equation}
    \begin{array}{l}
      \label{eq:full-toto1-ter}
      M x_{k+1} = M x_{k} +h \theta f(x_{k+1},t_{k+1}) +h(1-\theta)f(x_{k},t_{k}) + h r_{k+\gamma} \\[2mm]
      y_{k+\gamma} =  h(t_{k+\gamma},x_{k+\gamma},\lambda _{k+\gamma}) \\[2mm]
      r_{k+\gamma} = g(t_{k+\gamma},\lambda_{k+\gamma})\\[2mm]
    \end{array}
\end{equation}

 \paragraph{Newton's linearization of the first line of~(\ref{eq:full-toto1-ter})} The first line of the  problem~(\ref{eq:full-toto1-ter}) can be written under the form of a residue $\mathcal R$ depending only on $x_{k+1}$ and $r_{k+\gamma}$ such that
\begin{equation}
  \label{eq:full-NL3}
  \mathcal R (x_{k+1},r _{k+\gamma}) =0
\end{equation}
with $$\mathcal R(x,r) = M(x - x_{k}) -h\theta f( x , t_{k+1}) - h(1-\theta)f(x_k,t_k) - h r. $$
The solution of this system of nonlinear equations is sought as a limit of the sequence $\{ x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma} \}_{\alpha \in \NN}$ such that
 \begin{equation}
   \label{eq:full-NL7}
   \begin{cases}
     x^{0}_{k+1} = x_k \\ \\
     r^{0}_{k+\gamma} = (1-\gamma ) r_{k} + \gamma r^0_{k+1}  = r_k \\ \\
     \mathcal R_L( x^{\alpha+1}_{k+1},r^{\alpha+1}_{k+\gamma}) = \mathcal
     R(x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma})  + \left[ \nabla_{x} \mathcal
     R(x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma})\right] (x^{\alpha+1}_{k+1}-x^{\alpha}_{k+1} ) + \\[2mm]
     \qquad\qquad\qquad\qquad\qquad\qquad\left[ \nabla_{r} \mathcal R(x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma})\right] (r^{\alpha+1}_{k+\gamma} - r^{\alpha}_{k+\gamma} ) =0
 \end{cases}
\end{equation}
\begin{ndrva}
  What about $r^0_{k+\gamma}$ ?
\end{ndrva}

The residu free is also defined (useful for implementation only):
\[\mathcal R _{\free}(x) \stackrel{\Delta}{=}  M(x - x_{k}) -h\theta f( x , t_{k+1}) - h(1-\theta)f(x_k,t_k).\]
We get
\begin{equation}
  \mathcal R (x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma}) = \fbox{$\mathcal R^{\alpha}_{k+1} \stackrel{\Delta}{=}  \mathcal R_{\free}(x^{\alpha}_{k+1} )  - h r^{\alpha}_{k+\gamma}$}\label{eq:full-rfree-1}
\end{equation}

\[  \mathcal R
_{\free}(x^{\alpha}_{k+1} )=\fbox{$ \mathcal R _{\free, k+1} ^{\alpha} \stackrel{\Delta}{=}  M(x^{\alpha}_{k+1} - x_{k}) -h\theta f( x^{\alpha}_{k+1} , t_{k+1}) - h(1-\theta)f(x_k,t_k)$}\]

The computation of the Jacobian of $\mathcal R$ with respect to $x$, denoted by $   W^{\alpha}_{k+1}$ leads to
\begin{equation}
   \label{eq:full-NL9}
   \begin{array}{l}
    W^{\alpha}_{k+1} \stackrel{\Delta}{=} \nabla_{x} \mathcal R (x^{\alpha}_{k+1})= M - h  \theta \nabla_{x} f(  x^{\alpha}_{k+1}, t_{k+1} ).\\
 \end{array}
\end{equation}
At each time--step, we have to solve the following linearized problem,
\begin{equation}
   \label{eq:full-NL10}
    \mathcal R^{\alpha}_{k+1} + W^{\alpha}_{k+1} (x^{\alpha+1}_{k+1} -
    x^{\alpha}_{k+1}) - h  (r^{\alpha+1}_{k+\gamma} - r^{\alpha}_{k+\gamma} )  =0 ,
\end{equation}
By using (\ref{eq:full-rfree-1}), we get
\begin{equation}
  \label{eq:full-rfree-2}
  \mathcal R _{\free}(x^{\alpha}_{k+1})  - h  r^{\alpha+1}_{k+\gamma}   + W^{\alpha}_{k+1} (x^{\alpha+1}_{k+1} -
    x^{\alpha}_{k+1})  =0
\end{equation}

%\fbox
{
  \begin{equation}
    \boxed{ x^{\alpha+1}_{k+1} = h(W^{\alpha}_{k+1})^{-1}r^{\alpha+1}_{\gamma+1} +x^\alpha_{\free}}
  \end{equation}
}
with :
\begin{equation}
  \boxed{x^\alpha_{\free}\stackrel{\Delta}{=}x^{\alpha}_{k+1}-(W^{\alpha}_{k+1})^{-1}\mathcal R_{\free,k+1}^{\alpha} \label{eq:full-rfree-12}}
\end{equation}

The matrix $W$ is clearly non singular for small $h$.

Note that the linearization is equivalent to the case (\ref{eq:rfree-2}) and (\ref{eq:rfree-12}) with $\gamma=1$ and replacing $r_{k+1}$ by $r_{k+\gamma}$.

 \paragraph{Newton's linearization of the second  line of~(\ref{eq:full-toto1-ter})}
The same operation is performed with the second equation of (\ref{eq:full-toto1-ter})
\begin{equation}
  \begin{array}{l}
    \mathcal R_y(x,y,\lambda)=y-h(t_{k+\gamma},\gamma x + (1-\gamma) x_k ,\lambda) =0\\ \\
  \end{array}
\end{equation}
which is linearized as
\begin{equation}
  \label{eq:full-NL9}
  \begin{array}{l}
    \mathcal R_{Ly}(x^{\alpha+1}_{k+1},y^{\alpha+1}_{k+\gamma},\lambda^{\alpha+1}_{k+\gamma}) = \mathcal
    R_{y}(x^{\alpha}_{k+1},y^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) +
    (y^{\alpha+1}_{k+\gamma}-y^{\alpha}_{k+\gamma})- \\[2mm] \qquad  \qquad \qquad \qquad  \qquad \qquad
    \gamma C^{\alpha}_{k+1}(x^{\alpha+1}_{k+1}-x^{\alpha}_{k+1}) - D^{\alpha}_{k+\gamma}(\lambda^{\alpha+1}_{k+\gamma}-\lambda^{\alpha}_{k+\gamma})=0
  \end{array}
\end{equation}

This leads to the following linear equation
\begin{equation}
  \boxed{y^{\alpha+1}_{k+\gamma} =  y^{\alpha}_{k+\gamma}
  -\mathcal R^{\alpha}_{y,k+1}+ \\
  \gamma C^{\alpha}_{k+1}(x^{\alpha+1}_{k+1}-x^{\alpha}_{k+1}) +
  D^{\alpha}_{k+\gamma}(\lambda^{\alpha+1}_{k+\gamma}-\lambda^{\alpha}_{k+\gamma})}. \label{eq:full-NL11y}
\end{equation}
with,
\begin{equation}
     \begin{array}{l}
  C^{\alpha}_{k+\gamma} = \nabla_xh(t_{k+1}, x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma} ) \\ \\
  D^{\alpha}_{k+\gamma} = \nabla_{\lambda}h(t_{k+1}, x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma})
 \end{array}
\end{equation}
and
\begin{equation}\fbox{$
\mathcal R^{\alpha}_{yk+1} \stackrel{\Delta}{=} y^{\alpha}_{k+\gamma} - h(x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma})$}
 \end{equation}

Note that the linearization is equivalent to the case (\ref{eq:NL11y}) by replacing $\lambda_{k+1}$ by $\lambda_{k+\gamma}$ and $x_{k+1}$ by $x_{k+\gamma}$.

 \paragraph{Newton's linearization of the third  line of~(\ref{eq:full-toto1-ter})}
The same operation is performed with the third equation of (\ref{eq:full-toto1-ter})
\begin{equation}
  \begin{array}{l}
    \mathcal R_r(r,\lambda)=r-g(\lambda,t_{k+1}) =0\\ \\  \end{array}
\end{equation}
which is linearized as
\begin{equation}
  \label{eq:full-NL9}
  \begin{array}{l}
      \mathcal R_{L\lambda}(r^{\alpha+1}_{k+\gamma},\lambda^{\alpha+1}_{k+\gamma}) = \mathcal
      R_{r,k+\gamma}^{\alpha} + (r^{\alpha+1}_{k+\gamma} - r^{\alpha}_{k+\gamma}) - B^{\alpha}_{k+\gamma}(\lambda^{\alpha+1}_{k+\gamma} -
      \lambda^{\alpha}_{k+\gamma})=0
    \end{array}
  \end{equation}
\begin{equation}
  \label{eq:full-rrL}
  \begin{array}{l}
    \boxed{r^{\alpha+1}_{k+\gamma} = g(\lambda ^{\alpha}_{k+\gamma},t_{k+\gamma}) -B^{\alpha}_{k+\gamma}
      \lambda^{\alpha}_{k+\gamma} + B^{\alpha}_{k+\gamma} \lambda^{\alpha+1}_{k+\gamma}}
  \end{array}
\end{equation}
with,
\begin{equation}
     \begin{array}{l}
  B^{\alpha}_{k+\gamma} = \nabla_{\lambda}g(\lambda ^{\alpha}_{k+\gamma},t_{k+\gamma})
 \end{array}
\end{equation}
and the  residue for $r$:
\begin{equation}
\boxed{\mathcal
      R_{rk+\gamma}^{\alpha} = r^{\alpha}_{k+\gamma} - g(\lambda ^{\alpha}_{k+\gamma},t_{k+\gamma})}
  \end{equation}
Note that the linearization is equivalent to the case (\ref{eq:rrL}) by replacing $\lambda_{k+1}$ by $\lambda_{k+\gamma}$ and $x_{k+1}$ by $x_{k+\gamma}$.

\paragraph{Reduction to a linear relation between  $x^{\alpha+1}_{k+1}$ and
$\lambda^{\alpha+1}_{k+\gamma}$}

Inserting (\ref{eq:full-rrL}) into~(\ref{eq:full-rfree-12}), we get the following linear relation between $x^{\alpha+1}_{k+1}$ and
$\lambda^{\alpha+1}_{k+1}$,

\begin{equation}
   \begin{array}{l}
     x^{\alpha+1}_{k+1} = h(W^{\alpha}_{k+1} )^{-1}\left[g(\lambda^{\alpha}_{k+\gamma},t_{k+\gamma}) +
    B^{\alpha}_{k+\gamma} (\lambda^{\alpha+1}_{k+\gamma} - \lambda^{\alpha}_{k+\gamma}) \right ] +x^\alpha_{free}
\end{array}
\end{equation}
that is
\begin{equation}
  \begin{array}{l}
\boxed{x^{\alpha+1}_{k+1}=x_p + h (W^{\alpha}_{k+1})^{-1}    B^{\alpha}_{k+\gamma} \lambda^{\alpha+1}_{k+\gamma}}
   \end{array}
  \label{eq:full-rfree-13}
\end{equation}
with
\begin{equation}
  \boxed{x_p \stackrel{\Delta}{=}  h(W^{\alpha}_{k+1} )^{-1}\left[g(\lambda^{\alpha}_{k+\gamma},t_{k+\gamma}) -B^{\alpha}_{k+\gamma} (\lambda^{\alpha}_{k+\gamma}) \right ] +x^\alpha_{free}}
\end{equation}

\paragraph{Reduction to a linear relation between  $y^{\alpha+1}_{k+\gamma}$ and
$\lambda^{\alpha+1}_{k+\gamma}$}

Inserting (\ref{eq:full-rfree-13}) into (\ref{eq:full-NL11y}), we get the following linear relation between $y^{\alpha+1}_{k+1}$ and $\lambda^{\alpha+1}_{k+1}$,
\begin{equation}
   \begin{array}{l}
 y^{\alpha+1}_{k+1} = y_p + \left[ h \gamma C^{\alpha}_{k+\gamma} ( W^{\alpha}_{k+1})^{-1}  B^{\alpha}_{k+1} + D^{\alpha}_{k+1} \right]\lambda^{\alpha+1}_{k+1}
   \end{array}
\end{equation}
with
\begin{equation}
y_p = y^{\alpha}_{k+1} -\mathcal R^{\alpha}_{yk+1} + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1}
\end{equation}
that is
\begin{equation}\boxed{
y_p =  h(x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1} }
\end{equation}
\textcolor{red}{
  \begin{equation}
   \boxed{ x_q=(x_p -x^{\alpha}_{k+1})\label{eq:full-xqq}}
  \end{equation}
}

\paragraph{The linear case}
\begin{equation}
  \begin{array}{lcl}
    y_p &=&  h(x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1}\\
        &=&  C^{\alpha}_{k+1} x^{\alpha}_{k+\gamma} + D^{\alpha}_{k+1}\lambda^{\alpha}_{k+\gamma}  + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1} \\
        &=& C^{\alpha}_{k+1}  (x^{\alpha}_{k+\gamma} + \gamma x_p - \gamma x^{\alpha}_{k+1} ) \\
        &=& C^{\alpha}_{k+1}  ((1-\gamma) x_{k} + \gamma x_{free} ) \text {since } x_p =x_{free}
\end{array}
\end{equation}

\paragraph{Implementation details}

For the moment (Feb. 2011), we set $x_q=(1-\gamma) x_{k} + \gamma x_{free} $ in the linear case.
The nonlinear case is not yet implemented since we need to
change the management of \texttt{ H_alpha} Relation to be able to compute the mid--point values.
% things that remain to  do
%
% \begin{itemize}
% \item implement the function \texttt{BlockVector  computeg(t,lambda)} and \texttt{SimpleVector computeh(t,x,lambda)} which takes into account the values of the argument and return and vector
% \item remove temporary computation in Relation of {\verb Xq, \verb g_alpha and \verb H_alpha }. This should be stored somewhere else. (in the  node of the graph)
% \end{itemize}

\clearpage

%%% Local Variables:
%%% mode: latex
%%% TeX-master: "DevNotes"
%%% End:


\section{Results}
\label{Spline_1}

\subsection{Results}

 In our case we simulated the results and found following outputs.
Plotting the second order curve for given data of IMU with just least square method. We get

\begin{figure}
\includegraphics[width=\textwidth]{./figures/lsm.jpg}
\caption{In this case Blue is actual curve traced by IMU this are points with noise.And green is plotted curve. With least square method.}
\end{figure}

Then according to the paper we fused data of camera and IMU so we get following graph.

\begin{figure}
\includegraphics[width=\textwidth]{./figures/AllOutput.jpg}
\caption{Multiple result Graph}
\end{figure}

Here we have 3 curved fitted as shown as red which gives better results and it also uses data of camera shown with black dots. So here according to the above formula scale factor of different section is different it is according to the quality of reading by camera.
For example, in our case
\begin{equation}
\lambda_1=0.5 ,\lambda_2=0.5 ,\lambda_3=0.1
\end{equation}

In 3rd section camera readings were not accurate so taken into smaller scalar value. This camera data further improved the result. And we get minimum error and good fit. With the given spline fitting method.

So if we look into different of error with traditional Least square method with Jung and Taylor method. It gives good results. Almost we get the same curve.

\begin{figure}
\includegraphics[width=\textwidth]{./figures/ErrorC.jpg}
\caption{A. Red curve for method with one curve and least square method.
B. Green curve is using the given method by Jung and Taylor.}
\end{figure}

This graph has error comparison with and without Jung and Taylor method of split curve fitting and scaling camera input.
So it is better to follow the given first method in the paper. It results into considerable reduction in error. which is clearly seen in graph.


\subsubsection{Vertical and Whole-day-ahead Forecasts without Retraining}
\label{vert}

The upper-right in Figure \ref{f:inputs} shows an alternative way to
    generate forecasts for a test day before it has started:
First, a seasonally-adjusted time series $a_t$ is obtained from a vertical
    time series by STL decomposition.
Then, the actual forecasting model, trained on $a_t$, makes an $H$-step-ahead
    prediction.
Lastly, we add the $H$ seasonal na\"{i}ve forecasts for the seasonal component
    $s_t$ to them to obtain the actual predictions for the test day.
Thus, only one training is required per model type, and no real-time data is
    used.
By decomposing the raw time series, all long-term patterns are assumed to be
    in the seasonal component $s_t$, and $a_t$ only contains the level with
    a potential trend and auto-correlations.
The models in this family are:
\begin{enumerate}
\item \textit{fnaive},
      \textit{pnaive}:
          Sum of STL's trend and seasonal components' na\"{i}ve forecasts
\item \textit{vholt},
      \textit{vses}, and
      \textit{vtheta}:
          Exponential smoothing without calibration and seasonal
                       fit
\item \textit{vets}:
          ETS calibrated as described by \cite{hyndman2008b}
\item \textit{varima}:
          ARIMA calibrated as described by \cite{hyndman2008a}
\end{enumerate}
As mentioned in Sub-section \ref{unified_cv}, we include the sum of the
    (seasonal) na\"{i}ve forecasts of the STL's trend and seasonal components
    as forecasts on their own:
For \textit{fnaive}, we tune the "flexible" $ns$ parameter, and for
    \textit{pnaive}, we set it to a "periodic" value.
Thus, we implicitly assume that there is no signal in the remainder $r_t$, and
    predict $0$ for it.
\textit{fnaive} and \textit{pnaive} are two more simple benchmarks.


\graphicspath{{Chapter2/Figs/}}

\chapter{Multi-Omics Factor Analysis (MOFA), a Bayesian model for integration of multi-omics data}

The work described in this Chapter results from a collaboration with Wolfgang Huber's group at the EMBL (Heidelberg, Germany). It has been peer-reviewed and published in \cite{Argelaguet2018}. The method was conceived by Florian Buettner, Oliver Stegle and me. I performed most of the mathematical derivations and implementation, but with significant contributions from Damien Arnol and Britta Velten. The CLL data application was led by Britta Velten whereas the single-cell application was led by me, but with joint contributions in either cases. Florian Buettner, Wolfgang Huber and Oliver Stegle supervised the project.\\
The article was jointly written by Britta Velten and me, with contributions from all authors.

\section{Theoretical foundations}

\subsection*{Mathematical notation} \label{section:mathematical_notation}

\begin{itemize}[noitemsep]
	\item[--] Matrices are denoted with bold capital letters: $\bfW$
	\item[--] Vectors are denoted with bold non-capital letters: $\bfw$. If the vector comes from a matrix, we will use a single index to indicate the row that it comes from. If two indices are used, the first one corresponds to the row and the second one to the column. The symbol '$:$' denotes the entire row/column. For instance, $\bfw_{i}$ refers to the $i$th row from the $\bfW$ matrix, whereas $\bfw_{:,j}$ refers to the $j$th column.
	\item[--] Scalars are denoted with non-bold and non-capital letters: $w$. If the scalar comes from a 1-dimensional array (a vector), a single subscript will indicate its position in the vector. If the scalar comes from a 2-dimensional array, two indices will be shown at the bottom: the first one corresponding to the row and the second one to the column. For instance, $w_{i,j}$ refers to the value from the $i$th row and the $j$th column of the matrix $\bfW$, and $w_i$ to the $i$th value of the vector $\bfw$.
	\item[--] $\boldzero_k$ is a zero vector of length $k$.
	\item[--] $\I_k$ is the identity matrix with rank $k$.
	\item[--] $\E_q[x]$ denotes the expectation of $x$ under the distribution $q$. When the expectations are taken with respect to the same distribution many times, we will avoid cluttered notation and we will instead use $\la x \ra$.
	\item[--] $\Ndist{x}{\mu,\sigma^2}$: $x$ follows a univariate normal distribution with mean $\mu$ and variance $\sigma^2$.
	\item[--] $\Gdist{x}{a,b}$: $x$ follows a gamma distribution with shape and rate parameters $a$ and $b$.
	\item[--] $\Bdist{x}{a, b}$: $x$ follows a beta distribution with shape and rate parameters $a$ and $b$.
	\item[--] $\text{Ber}(x|\theta)$: $x$ follows a Bernoulli distribution with parameter $\theta$.
	\item[--] $\mathds{1}_0$: Dirac delta function centered at 0.
	\item[--] $Tr(\bfX)$: Trace of the matrix \bfX
\end{itemize}

\subsection*{Graphical notation for probabilistic models}

Probabilistic models can be represented in a diagrammatic format (i.e. a graph or a network) that offers a compact visual representation of complicated systems of probability distributions \cite{Bishop2006}. In a graphical model the relationship between the nodes becomes more explicit, namely their conditional independence properties which allow the joint distribution over all variables to be factorised into a series of simpler products involving subsets of variables \cite{Bishop2006}. The basic unit of a network is the node, which represents the different types of variables, including observed variables, unobserved probabilistic variables and unobserved parameters. The nodes are connected by unidirectional edges (arrows) which capture the conditional independence relationship between the variables.

For this thesis we adapted the graphical notations from~\cite{Dietz2010-technical-report-graphs}:

\begin{center}
  \begin{tabular}{m{8cm} m{2cm}}
    Observed variables & \tikz{\node[obs](){$Y$}} \\
    Unobserved probabilistic variables & \tikz{\node[latent](){$\theta$}} \\
    Unobserved parameters & \tikz{\node[latent,double, double distance=1pt](){$\theta$}} \\
    Repetition of node $\theta_n$ for $n\in\llbracket 1;N \rrbracket$ & \tikz{\node[latent](theta){$\theta_n$}; \plate[] {plateN} {(theta)} {$N$};} \\
    Conditional dependency between nodes: $p(Y,\theta) = p(Y|\theta)p(\theta)$ & \tikz{%
            \node[latent]   (theta) {$\theta$};
            \node[obs, xshift=1.5cm] (Y) {$Y$};
            \edge{theta}{Y}}
  \end{tabular}
\end{center}
% For simplicity, fixed hyperparameters are not represented on the graphical model. Unobserved parameters are only represented when optimised together with the unobserved probabilistic variables.

\input{Chapter2/bayes}

\input{Chapter2/factor_analysis}


\documentclass[../main.tex]{subfiles}
\begin{document}

\subsection{Circuit characterization of a matroid}
In the paper of Oxley \cite{ox_paper} it is said that "A set in a matroid that is not independent is called dependent.The hereditary property,(I2), means that a matroid is uniquely determined by its collection of maximal independent sets, which are called bases, or by its collection of minimal dependent sets, which are called circuits." \\
\indent The circuits of a graph are a rather intuitive concept, and so it is natural to define the cycle matroid $M(G)$ of a graph in terms of its circuits. These circuits are the "edge sets of cycles in a graph $G".$ All dependent sets in a matroid contain a circuit.\\
\indent A circuit of a matroid is defined as a minimally dependent set due to the fact that a circuit is a dependent set with the minimum number of edges to be dependent. i.e if you remove one edge we have an independent set.

\begin{defn} By using (I1)–(I3) , it is not difficult to show that the collection $\mathcal{C}$ of circuits of a matroid M has the following three properties:\\
(C1) The empty set is not in $\mathcal{C}$\\
(C2) No member of $\mathcal{C}$ is a proper subset of another member of $\mathcal{C}$\\
(C3) if $ C_1 $ and $ C_2 $ are distinct members of $ C $ and
$ e \in C_1 \cap C_2 $, then $ (C_1 \cup C_2 ) \setminus \{e\} $ contains a member of $\mathcal{C}$
 \end{defn}

\noindent The following two proofs appear in Oxley's text and shows how the circuits define a matroid and how circuits appear in graphs.\cite{ox_book}

 \begin{thm}
 Let M be a matroid and $\mathcal{C}$ be its collection of circuits. Then $\mathcal{C}$ satisfies (C1) - (C3)
  \end{thm}
\begin{proof}
\noindent $(C1)$ is obvious as by $(I1)$ the empty set must always be an independent set.

\noindent $(C2)$ is also straightforward because any $ C \in \mathcal{C} $ is a minimally independent set by definition. Therefore , if there exists a $ C_1 \in \mathcal{C} $ such that $ C_1 \subset C $ then $ C_1 \in \mathcal{C} $ and $ C $ is not a minimally dependent subset of E.

 \vspace{2mm}

 \noindent $(C3)$ Let $ A, B \in \mathcal{C} $ and suppose that (seeking a contradiction) $ (A \cup B) \setminus \{e\} $ where $ e $ is $ \in (A \cap B) $ does not contain a circuit.\\
 \noindent Then $ (A \cup B) \setminus \{e\} $ is independent and therefore in $\mathcal{I}.$

 \vspace{2mm}

\noindent The set $ A \setminus B $ is non-empty.\\
 Let $ s \in A \setminus B    \implies s \in A$\\
 \noindent as $A$ is in $\mathcal{C}$ it is minimally dependent.
 \noindent $\implies A \setminus \{s\} \in \mathcal{I}$ i.e is independent.

 \vspace{2mm}

\noindent Let $J$ be a maximal independent set of $(A \cup B)$ with the following properties: $A \setminus \{s\} \subset J $ and therefore $ \{s\} \notin J $ but as B is a circuit there must be some element $t \in B$ that is not in $J$; $s$ and $t$ are distinct.

 \vspace{2mm}

\noindent $\implies |J| $ must be at most equal to $|(A \cup B) \setminus \{s,t\}| $\\
 $\implies |J| \leq |(A \cup B) \setminus \{s,t\}| = |(A \cup B)| - 2 < |(A \cup B) \setminus \{e\}|$

 \vspace{2mm}

 \noindent Now by (I3) we can substitute elements from $(A \cup B) \setminus \{e\}$ into $(A \cup B) \setminus \{s,t\}$ that are not in $(A \cup B) \setminus \{s,t\}$ but the only elements that fits this condition are $\{s\}$ and $\{t\}$ and introducing either of these elements breaks the independence of $J$.

\noindent Therefore, $(A \cup B) \setminus \{e\}$ must contain a circuit\\
 \end{proof}

\end{document}


\documentclass[12pt]{article} \input{physics1}
\begin{document}

\noindent
Name: \rule[-1ex]{0.55\textwidth}{0.1pt}
NetID: \rule[-1ex]{0.2\textwidth}{0.1pt}

\section*{NYU Physics I---Final Exam}

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
Assume that Professor Hogg has a mass of exactly $70\,\kg$. Now
consider a very realistic, stone statue of Professor Hogg that is
twice his size in every direction (twice as tall, and realistic). What
would be the mass of this statue, approximately?
(From Lecture, 2018-09-06.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
Roughly what is terminal velocity for a 1-liter soda bottle, filled
with soda (or pop or juice or whatever), falling from a tall building?
(From Problem Set 1.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
We spent time talking about two vectors, $\vec{v}_3$ and $\vec{v}_4$,
which were the velocities of the rock on a no-air-resistance trajectory.
What was wrong with this picture, that we drew?
\marginpar{\includegraphics[width=1in]{../jpg/wrong_vectors.png}}
(From Term Exam 1.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
Consider a $1\,\kg$ stone thrown upwards at an initial upward velocity of
magnitude $10\,\mps$. Plot the vertical velocity (with upwards
positive) as a function of time for the first 2 seconds of its
trajectory. Use a gravitational acceleration of $10\,\mpss$. Label
your axes quantitatively. Ignore air resistance.
(From Problem Set 2.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
What is the magnitude $|\vec{N}|$ of the normal force acting on a package of
mass $M$ sitting on the floor of an elevator that is accelerating
upwards with acceleration magnitude $a$. The elevator is an ordinary
elevator in an ordinary building in New York City.
(From Problem Set 3.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A car is going around a banked turn faster than the speed for which
the bank is designed. Therefore the car is relying on lateral friction
to make the turn!  Draw a free-body diagram for the car, showing at
least the gravitational force, the normal force, and the frictional
force.
(From Problem Set 4.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
You have a block of mass $m$ on an inclined plane, inclined at an
angle $\theta=15\,\deg$ to the horizontal. The coefficient of friction
is $\mu=0.9$. What is the magnitude of the frictional force on the
block? The acceleration due to gravity is $g$.
You must leave your answer in terms of $\mu$, $m$, $g$, $\theta$, or
whatever you need to deliver a correct symbolic answer.
Once again, state any assumptions you need to make.
(From Term Exam 2.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A roller coaster zooms over the top of a hill of radius $4\,\m$ (such that the
center-of-mass of the car and people are moving on a circular trajectory of
radius $4\,\m$). At what speed $v$ should the roller coaster go if the people in
the car are going to nearly feel weightless? Use $g=10\,\mpss$, and give
a numerical answer in $\mps$.
(From Lecture 2018-09-27.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A package of mass $m$ is moving at a horizontal velocity of magnitude $v$ at $t=0$
relative to the floor.  It scrapes to a halt on the floor. How much
heat $Q$ is generated by the friction, from $t=0$ until when the
package comes to rest?
(From Problem Set 5.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A $0.16\,\,kg$ ball is dropped from a height of $1\,\m$ onto a concrete
floor. If the bounce takes only $10^{-4}\,\s$, what was the (approximate)
magnitude of the force on the ball from the floor? Give a numerical
answer in $\N$.
(From Lecture 2018-10-11.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
Without using a calculator, estimate the sine of the angle $1.2~\deg$.
Use the small-angle approximation! (\emph{Hint:} Degrees aren't radians.)
(From Term Exam 3.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
If I stretch a spring, I am experiencing Hooke's law. What is the
stress and what is the strain in this case? Answer in words!
(From Lecture 2018-10-18.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A typical adult is holding her or his left arm at a right angle, so
the upper arm is pointing straight down, and the forearm is pointing
horizontally forwards. The hand is oriented palm-up. The arm is
holding a 6 kg grocery bag by its handle in the hand. Draw a diagram
showing the forces acting on the forearm+hand system, at roughly the locations
where they are acting. You can think of it as being the forces on the forearm+hand
bones if that makes more sense.
(From Problem Set 7.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
If you have a potential of the form
$$
U(x) = A\,x^3 - B\,x + C
$$ where $A$ and $B$ and $C$ are positive constants, find a location ($x$ position) at which
the force is zero.
(From Term Exam 4.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
If I tell you that the position $x$ of a particle is moving with
time according to the equation:
\begin{equation}
x(t) = B\,\e^{-\frac{\Gamma}{2}\,t}\,\cos (2\pi\,f\,t + \theta)
\quad ,
\end{equation}
then what are the units of $B$, $\Gamma$, $f$ and $\theta$?
(From Problem Set 8.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A bungy cord has a natural (unstretched) length of $5\,\m$ and it
stretches by $1\,\m$ for every $400\,\N$ of applied force. How much
energy is the bungy cord storing when it is stretched to
a total length of $7\,\m$? Give a numerical answer in $\J$.
(From Problem Set 9.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
How is it that the astronauts in the Space Station are weightless?
Write a grammatically correct answer in one sentence, in fewer than 17
words.  Box your sentence!
(From Term Exam 5.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A car of mass $M$ is accelerating at acceleration $a$ in the $x$
direction. Its center of mass is a height $h$ above the ground.  What
is the net torque on the car, if any, when you consider the reference point to be
a point that is stationary and \emph{on the ground}?
(From Problem Set 10.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
What is escape velocity from the surface of the Earth? That is, how
fast do you have to move to get out of Earth's gravity?
If you don't remember $G$ or $M$ for the Earth (and you shouldn't remember
those!) then remember that $g = G\,M/R^2$ and the radius $R$ of the Earth is
about $6300\,\km$.
Give a numerical answer in $\mps$.
(From Problem Set 11.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
If the Earth were four times less massive than it is, but were still
on a circular orbit at its current orbital radius (semi-major axis),
how much longer or shorter would the year be?  (From Worksheet on
orbits.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
Sketch an orbit of roughly eccentricity 0.9. Most importantly: Show the
point about which the object is orbiting, and make sure your pericenter
and apocenter distances make sense. Don't worry about getting it all right,
just roughly!
(From Term Exam 6.)

\vfill
~
\clearpage

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
What, roughly, is $\gamma$ for the astronauts in the Space Station,
according to a scientist stationary in the Earth rest frame?
(From Problem Set 13.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
A particle of mass $M$ is at rest. It decays into two identical
photons traveling in the positive and negative $x$ directions.
What are the 4-momenta of these two photons?
(From Problem Set 14.)

\vfill

\paragraph{\problemname~\theproblem:}\refstepcounter{problem}%
Two balls of putty of mass $m$ are fired towards each other in outer space.
They collide and stick perfectly, without losing any putty or
radiating any sound or heat or anything else. Is the combined blob
of putty created by this collision slightly more than, slightly less
than, or exactly equal to $2\,m$ in mass? And say why, in fewer than
17 words. Make your answer correct to at least first order in
$\beta^2$. That is, the weakly relativistic limit. But I only need
a qualitative answer, don't calculate anything! (From Lecture
2018-12-13.)

\vfill
~
\end{document}