text:
November 3, 2017 October 31, 2017 Kris
**This week's prompt is amnesia. Enjoy!**
"Don't know what happened. Found him like this…"
"The Nyx got him, looks like. It'd been hours…"
Everything hurt. Every muscle and tendon. Hell, every molecule in my body screamed with agony. I was vaguely away of noises–voices–but I couldn't open my eyes. If I opened them, my head would explode. I was sure of it.
"Let me see him."
That rumble of a voice had me opening my eyes. I was in an unfamiliar room–dark but cozy–and three people stood at the end of the bed. The two smaller guys wore their worry plain on their faces, but the big man was staring right at me. He noticed me looking, and quickly rounded the end of the bed.
"Ben, baby."
I blinked–even that hurt–and tried to speak. It came out as a croak, and I cleared my throat and tried again. "Where am I?"
The big guy froze, and concern washed his features. After a second, he approached cautiously. "Ben?"
"How do you know my name?" Panic flooded my veins, but when I tried to push myself out of bed, I fell back against the mattress, too weak to hold myself up. "Who are you? What's going on?"
"Shit," one of the other guys muttered. I didn't turn to look at which one. "The Nyx wiped his memory."
"Leave us." The other two scrambled to do the big man's bidding, and I heard the soft snick as the door latched closed.
Fuck. Now I was trapped in the room, too weak to defend myself, with a big man who looked like he could snap me in half. He was gorgeous, all shoulders and muscles and scowl. But definitely dangerous. I wanted to be elsewhere.
"You're safe here," he rumbled, never taking his dark gaze off me. "I promise. No one is going to hurt you."
And just like that, I believed him. I didn't understand it, since my brain told me I should be terrified. But I wasn't. His words, his voice, just soothed me. I relaxed a fraction, but I still kept myself ready. I needed to be ready, just in case.
He crossed the room, and I tracked him, but he only opened a small dorm sized fridge and pulled out a bottle of water. He made a show of cracking the seal, showing me it hadn't been tampered with. I took it from him when he handed it over, and gratefully swallowed some. It felt so good, I took a larger drink.
"Easy now," he murmured, stepping close to the bed.
I took another smaller sip, and then wiggled a little until I could sit up against the headboard. He slowly lowered himself to the mattress, perching on the edge. I appreciated the courtesy.
"What's the last thing you remember?"
Odd question, but I searched my brain anyway. After a minute, I realized there was a hole in my memory. Definitely something missing. Had I hit my head? I reached up and felt around, but there was no bump. Had I been drugged? Oh god, was that why I was here?
"Ben?"
My name in his voice startled me, but I settled again. Something about this man comforted me, and I didn't think it should be that easy, but there was no doubt about it. I took another tiny drink.
He frowned. "And what happened on Tuesday?"
I couldn't help the small, proud grin. "I aced Professor Millhouse's midterm."
His breath whooshed out and he squeezed his eyes shut. Then whispered, "Damn."
"Four years ago," he muttered, half to himself. "Before the veil ripped and all the unnatural flooded into the natural world."
"What?" I pushed at the blankets, suddenly desperate to get up. He was fucking nuts.
"Ben, stop." His voice was a command and a plea, and I obeyed instinctively, my body going still.
"What in the hell is going on?"
"Two days ago, you went after a Nyx. It shouldn't have been a big deal, but we lost contact. Peter and Remy found you a few hours ago, nearly dead." He leaned in just a little, and I could see the pain in his eyes. "We thought it killed you."
I couldn't deal with the raw emotion on his face, in his voice. "Nyx?"
"A soulless, blood sucking demon. Vicious, nasty things. You've dealt with hundreds, though, so I wasn't worried. But then you didn't come back."
"I…" I had no idea what to say to that. Even though his words didn't make sense, there was something about them, the way they settled into me, that I just knew it to be true. But how? Why? I still didn't know what was going on, and the panic that had been bubbling under the surface spilled over. I started hyperventilating, the edges of my vision dimming, and my heart pounded so hard it hurt.
He touched my wrist, and I jerked my gaze to his. I was lost in his deep, dark eyes. The worry and pain and love and affection there. They were so familiar. I knew those eyes. I didn't know how, or why, but I knew this man who was staring at me like I hung the moon.
Unbidden, his name came to my lips. "Lucas?"
The grin that bloomed on his face transformed his features. And like a flood, memories crashed into me. The past four years. When the veil between the worlds ripped and a host of unholy, terrifying creatures came racing into our world. How nearly half the humans had died in the first onslaught, but Lucas and I survived. Banned together with others, and set about protecting those who were left.
"Oh baby. There you are."
I started crying, because I couldn't do anything else. I'd nearly left him to this world alone, and that was the one thing I vowed never to do. I clung to him, and he held me just as tightly.
We'd be okay. There wasn't any other choice.


text:
Wolfgang Clement
Former German Federal Minister of Economics and Labour
Wolfgang Clement is a renowned economist with profound knowledge of the political establishment, where he remained for nearly four decades. As former Prime Minister of North Rhine-Westphalia he was super-minister for Economy and Labour under Chancellor Gerhard Schröder, before becoming a business manager in the private sector.
After graduation Wolfgang began his journalism career at the Westfälische Rundschau and studied law at the University of Münster. He then went on to work as a law clerk and assistant at the Institute of Procedural Law at the University of Marburg. In 1968 Wolfgang returned to Westfälische Rundschau, first as political editor, then as head of politics and finally as vice editor in chief. In 1986, he took the role as editor in chief at the Hamburger Morgenpost and remained there until 1989.
From 1970 to 2008, Wolfgang was a member of the SPD. Up until 1989 he held various positions at the state and federal level. In 1989 he became the head of the Chancellery Office in Dusseldorf. In 1995 he beccame Minister of Economic Affairs, Technology and Transportation. Finally he succeeded Johannes Rau as Prime Minister of North Rhine-Westphalia in 1998.
In the 2000 election, Wolfgang was confirmed in his position, but changed to Berlin in 2002 to become Minister of Economics and Labour under Chancellor Gerhard Schröder, a position created especially for him. In his new office, he faced the challenge of implementing the proposals of the Hartz Commission and to initiate major reforms of the labor market. Although this lost him a significant number of votes within the SPD for the re-election as deputy party leader, he received praise from many economists for his swift action.
He has received several awards for his political activities, including the European trade prize in 2001.
As a pragmatic realist, Wolfgang is one of the most intriguing figures in the Cabinet, and is referred to several times as the only conceivable successor to Gerhard Schröder. His direct and occasionally brash nature makes him a very popular party leader to the media. After 38 years of membership, Wolfgang left the SPD in 2008.

redpajama_set_name: RedPajamaCommonCrawl


This Living History features an interview with Ambassador Kathleen Stephens who served as the United States Ambassador to the Republic of Korea from 2008 to 2011. Ambassador Stephens was the first female ambassador and one of the few Korean-speaking senior U.S. diplomats to serve in that position since the opening of the U.S. embassy in Seoul in 1949.
In the first of this two-part interview series with Ambassador Stephens, she reflects on her experience serving as a young Peace Corps volunteer from 1975 to 1977 in rural South Chungcheong province and as a junior diplomatic officer in the early 1980s at the U.S. Embassy in Seoul.
During these formative time periods, Ambassador Stephens witnessed South Korea's rapid modernization and economic development firsthand. Not only was Ambassador Stephens a witness to South Korea's incredible growth over the span of just one generation (or "the miracle of Han River") but she was personally involved in making history at the ground level as a diplomat. In this interview, she gives an account of her experience dealing with student protests and sit-ins in Korea as a junior diplomat at the U.S. embassy during a period of tumultuous change in South Korea's democratic history.
Please see more below on this interview with Ambassador Stephens. This is the first part of her story.
Click on the interview questions below to navigate to the relevant answer section in the video.
Ambassador Kathleen Stephens is the president and CEO of the Korea Economic Institute of America. Previously, she served as a career diplomat within the United States Foreign Service from 1978 to 2015. She was the U.S. charge d'affaires in India from 2014 to 2015 and the Under Secretary of State for Public Diplomacy and Public Affairs in 2012. She served as the U.S. ambassador to the Republic of Korea from 2008 to 2011. Before moving to Seoul, Ambassador Stephens served in a number of policy positions in Washington, working as the Principal Deputy Assistant Secretary of State for East Asian and Pacific Affairs from 2005 to 2007, the Deputy Assistant Secretary of State for European and Eurasian Affairs from 2003 to 2005, the director of the Department of State Office of Ecology and Terrestrial Conservation from 2001 to 2003, the deputy chief of mission at the U.S. embassy in Lisbon, Portugal, from 1998 to 2001, and the Principal Officer at the U.S. consulate general in Belfast, Northern Ireland, from 1995-1998. She has also been posted to U.S. missions in Croatia, Serbia, South Korea, Trinidad and Tobago, and China. She holds a B.A. in East Asian Studies from Prescott College in Prescott, Arizona and an M.A. from Harvard University's Kennedy School.
Sang Jun Lee is a research associate in the Korea Chair and helped with the production of this post.


Ricardo Bessa is another of our best Manga style illustrators. Ricardo specialises in keenly detailed caricatures, dynamic perspectives and rich, atmospheric colour schemes.
Ricardo's portfolio is a wonderland of strange beasts, enchanted landscapes and passionately detailed character studies drawn from life and mythology.


Shocking news emerged that the UK's leading independent provider of healthcare waste management had been failing to dispose of 350 tonnes of medical waste, which included human body parts; sparking a major national incident and highlighting the importance of the waste duty of care.
This blog explains what the 'Duty of Care' is, providing you with practical guidance on how to meet your waste duty of care requirements. The 'Duty of Care' applies to anyone who produces, carries, keeps, dispose of, treats, imports or has control of waste. This process begins at the point you intend to discard something and produce waste. So more than likely it will apply to you and your business….


text:
\section{Introduction and Overview}
The theory of computation has been extended to the quantum setting; a notable example being the conception of a quantum Turing machine \cite{Mller2007QuantumKC,doi:10.1137/S0097539796300921}. Similarly, algorithmic information theory has been generalized to the quantum realm \cite{Vitnyi2001QuantumKC, Berthiaume:2001:QKC:2942985.2943376, Mller2007QuantumKC,2014JMP....55h2205B, brudno}. It hence seems natural to extend algorithmic randomness, the study of randomness of infinite bitstrings using notions from computation and information, to the quantum realm \cite{unpublished, qpl, bhojraj2020quantum}.

A coherent, infinite sequence of qubits (called a \emph{state}\cite{unpublished}), is the quantum version of an infinite bitstring. With this notion of a state in hand, one can define randomness for states similarly to how randomness is defined for infinite bitstrings.

Martin-L{\"o}f randomness (which is equivalent to Solovay randomness) and Schnorr randomness for infinite bitstrings, defined using the concept of `effective null sets', have characterizations in terms of initial segment prefix-free Kolmogorov complexity (denoted by $K$) \cite{misc,misc1, DBLP:series/txtcs/Calude02}; the initial segments of random infinite bitstrings are incompressible in the sense of $K$.

Quantum Solovay randomness and quantum Schnorr randomness for states have been defined using the quantum versions of effective null sets \cite{unpublished, bhojraj2020quantum}. Analogously to the classical situation, one may explore the connections between quantum Solovay randomness and quantum Schnorr randomness and the initial segment descriptive complexity of states (a length $n$ initial segment of a state is a density matrix on $n$ qubits).

Motivated by this, we asked whether there is a quantum analogue of $K$ which yields a characterization of quantum Solovay and quantum Schnorr randomness. We define $QK$, a complexity measure for density matrices based on prefix-free, classical Turing machines. The abbreviation $QK$ stands for `quantum-K', reflecting our intention of developing a quantum analogue of $K$, the classical prefix-free Kolmogorov complexity.

To the best of our knowledge, all notions of quantum Kolmogorov complexity developed so far, with one exception\cite{Vitnyi2001QuantumKC}, exclusively use machines which are not prefix-free (plain classical machines or quantum Turing machines) \cite{doi:10.1137/S0097539796300921, Berthiaume:2001:QKC:2942985.2943376,Mller2007QuantumKC}. $K_Q$, a notion developed in \cite{Vitnyi2001QuantumKC} uses a quantum Turing machine, $Q$ together with the classical prefix-free Kolmogorov complexity in its definition.

We give an overview of the main points in the paper.

In Section \ref{sec:2} we introduce quantum-K ($QK$) for density matrices and some of its properties. Theorem \ref{thm:67} (generalized in Lemma \ref{lem:K}) shows that $QK$ agrees with $K$ on the classical qubitstrings. Theorem \ref{thm:8} is a tight upper bound for $QK$ similar to that for $K$. Theorem \ref{thm:count} is a counting condition similar to that for $QC$ \cite{Berthiaume:2001:QKC:2942985.2943376}, $C$ and $K$ \cite{misc, misc1}.

Section \ref{sec:4} reviews quantum algorithmic randomness: a recently developed \cite{bhojraj2020quantum, unpublished} theory of randomness for states (infinite qubitstrings) using quantum versions of the classical `effectively null set'.

Section \ref{sec:3}, the main focus of this paper, connects $QK$ with two quantum algorithmic randomness notions: weak Solovay randomness and quantum Schnorr randomness, defined in Section \ref{sec:4}. Two important characterizations show that the initial segments of Martin-L{\"o}f randoms (equivalently, of Solovay randoms) are asymptotically incompressible in the sense of $K$: the Chaitin characterization (See \cite{Chaitin:1987:ITR:24912.24913} and theorem 3.2.21 in \cite{misc}),
\[X  \text{ is Martin-L{\"o}f random} \iff \text{lim}_{n}  K(X\upharpoonright n)-n= \infty,\]
and the Levin\textendash Schnorr characterization (See theorem 3.2.9 in \cite{misc}),
\[X  \text{ is Martin-L{\"o}f random} \iff \exists c \forall n [ K(X\upharpoonright n)>n-c].\]
(Characterizations having the former form  will be called `Chaitin type' and those having the latter form will be called `Levin\textendash Schnorr type').
We investigate the extent to which these classical characterizations carry over to weak Solovay randoms and $QK$.

Theorem \ref{thm:7} is a Chaitin type of characterization of
weak Solovay randomness ($\rho$ is weak Solovay random $\iff$ lim$_{n}QK^{\epsilon}(\rho_{n})-n = \infty$).
This shows that the Levin\textendash Schnorr condition ($\forall n[QK^{\epsilon}(\rho_{n})>^{+}n]$) is implied by weak Solovay randomness.

Theorem \ref{thm:39} shows both Chaitin and Levin\textendash Schnorr type characterizations of weak Solovay randomness when restricting attention to a specific class of states. It is worth noting that Theorem \ref{thm:39} uses the proof of our main result (Theorem 2.11) in \cite{bhojraj2020quantum}.

For general states, subsection \ref{subsect:weak} shows that the Levin\textendash Schnorr condition implies something slightly weaker than weak Solovay randomness.

While $K$ plays well with Solovay randomness, $K_C$, a version of $K$ using a computable measure machine, $C$ (a prefix-free Turing machine whose domain has computable Lebesgue measure) gives a Levin\textendash Schnorr  characterization of Schnorr randomness (See theorem 7.1.15 in \cite{misc1}). Motivated by this, we introduce $QK_C$ a version of $QK$ using computable measure machines in subsection \ref{schn}.

It turns out that $QK_C$ yields not just a Levin\textendash Schnorr type (Theorem \ref{thm:schnor}), but also a Chaitin type (Theorem \ref{thm:Chaitin}) characterization of quantum Schnorr randomness.

Theorem \ref{thm:Chaitin} together with Theorem \ref{thm:677} and lemma 3.9 in \cite{bhojraj2020quantum}, implies that Schnorr randoms have a Chaitin type characterization in terms of $K_C$ (Theorem \ref{class}). So, results in the quantum realm imply a new result in the classical setting.

In summary, we introduce $QK$ and show that the initial segments of weak Solovay random and quantum Schnorr random states are incompressible in the sense of $QK$.

\section{The Definition and Properties of QK}
\label{sec:2}
 We assume familiarity with the notions of density matrix (See for example, \cite{Nielsen:2011:QCQ:1972505}), prefix-free Kolmogorov complexity ($K$) and $\mathbb{U}$, the universal prefix-free (or self-delimiting) Turing machine (See \cite{misc,misc1, DBLP:series/txtcs/Calude02}).

 The output of $\mathbb{U}$ can be interpreted as unordered tuples of complex algebraic vectors (equivalently, finite subsets of natural numbers).

The notation $\mathbb{U}(\sigma)\downarrow = F$ means that $ \mathbb{U}(\sigma)$ outputs the index of $F$ with respect to some fixed canonical indexing of finite subsets of the naturals. We will never use an ordering on the elements of $F$ in any of our arguments: $F$ will be used to define an orthogonal projection : $\sum_{v \in F} |v\big>\big<v|$ which clearly does not depend on an ordering on $F$.

As explained in \cite{unpublished}, the quantum analogue of a bitstring of length $n$ is a density matrix on $\mathbb{C}^{2^n}$.

For a density matrix, $\tau$, let $|\tau|$ denote the $s$ such that $\tau$ is a transformation on $\mathbb{C}^{2^s}$. For any $n$, $\mathbb{C}^{2^n}_{alg}$ is the space of elements of $\mathbb{C}^{2^n}$ with complex algebraic entries.

Logarithms will always be base 2. The notation $\leq^+ , \geq^+, =^+$ will be used for `upto additive constant' relations.

For $\epsilon >0 $, $QK^{\epsilon}(\tau)$ is defined to be
\begin{defn}
\label{defn:4}
$QK^{\epsilon}(\tau) := $ inf  $\{|\sigma|+$log$|F|: \mathbb{U}(\sigma)\downarrow = F$, a orthonormal set in $\mathbb{C}^{2^{|\tau|}}_{alg}$ and $\sum_{v \in F} \big<v |\tau |v \big> > \epsilon \}$

\end{defn}
 The term $\sum_{v \in F} \big<v |\tau |v \big>$ is the squared length of the `projection of $\tau$ onto span($F$)' which also equals the probability of getting an outcome of `1' when measuring $\tau$ with the observable given by the Hermitian projection onto span$(F)$\cite{Nielsen:2011:QCQ:1972505}. Although it is useful to intuitively think of $\sum_{v \in F} \big<v |\tau |v \big>$ as the `projection of $\tau$ onto span($F$)', we use quotes as $\tau$ is a convex combination of possibly \emph{multiple} unit vectors, while the notion `projection onto a subspace' refers usually to a single vector.

 Note that for a given $\tau$, $QK^{\epsilon}(\tau)$ is determined by the classical prefix-free complexities and dimensions of those subspaces, span$(F)$, such that the projection of $\tau$ onto span$(F)$ has squared length atleast $\epsilon$. I.e., $QK^{\epsilon}(\tau)$ depends only on the $K$-complexities and ranks of those projective measurements of $\tau$ such that the probability of getting an outcome of `$1$' is atleast $\epsilon$. Roughly speaking, $QK^{\epsilon}(\tau)$ depends on the dimensions and prefix-free complexities of subspaces which are $\epsilon$ `close' to $\tau$.

 This is in contrast to $QC^{\epsilon}(\tau)$ which depends on the quantum complexities of density matrices, not classical prefix-free complexities of subspaces, which are $\epsilon$ close to $\tau$ (Recall that $QC$ is based on quantum Turing machines) \cite{Berthiaume:2001:QKC:2942985.2943376}. Also, while the rank of the approximating projection is taken into consideration in $QK$, the rank of the approximating density matrix is not taken into account in $QC$.

 So, $QC^{\epsilon}(\tau)$ quantifies the quantum complexity of approximating $\tau$ by density matrices upto $\epsilon$ while $QK^{\epsilon}(\tau)$ measures the sum of the prefix-free complexity and the logarithm of the dimension of subspaces $\epsilon$ close to $\tau$.

A test demonstrating the quantum non-randomness of a state, $\rho$ uses computable sequences of projections of `small rank' which are $\epsilon$-close to initial segments (density matrices) of $\rho$ (See Section \ref{sec:4}). It hence seems plausible that a complexity measure for a density matrix, $\tau$ must reflect the complexities and ranks of projections $\epsilon$-close to $\tau$ in order to play well with quantum randomness notions for states.

We mention that our $QK$ is entirely different from the $ \overline{QK_M} $ and $\overline{QK^{\delta}_M}$ notions defined in Definition 3.1.1 in \cite{Mller2007QuantumKC} using quantum Turing machines.

$QK^{\epsilon}$ would not be a `natural' complexity notion for density matrices if the following theorem did not hold:
\begin{thm}
\label{thm:67}
 Fix a rational $\epsilon$. $K(\sigma)= QK^{\epsilon}(|\sigma\big>\big<\sigma|)$ holds for all classical bitstrings $\sigma$, upto an additive constant depending only on $\epsilon$.
\end{thm}
We isolate here a simple but useful property which will be used for proving Theorem \ref{thm:67}.
\begin{lem}
\label{lem:30}
Let $n$ be a natural number, $E=(e_{i})_{i=1}^{2^{n}}$ be any orthonormal basis for $\mathbb{C}^{2^{n}}$ and $F$ be any Hermitian projection matrix acting on $\mathbb{C}^{2^{n}}$. For any $\delta>0$, let  \[S^{\delta}_{E,F}:= \{e_{i} \in E: \big<e_{i}|F|e_{i}\big> > \delta \}.\]
Then, $|S^{\delta}_{E,F}| < \delta^{-1} \text{Tr}(F)$.
\end{lem}
\begin{proof}
Note that since $F$ is a Hermitian projection, $\big<e_{i}|F|e_{i}\big>=\big<Fe_{i}|F e_{i}\big> = |F e_{i}|^{2}\geq 0$. So,
\[\delta|S^{\delta}_{E,F}| < \sum_{e_{i} \in S^{\delta}_{E,F}}\big<e_{i}|F|e_{i}\big> \leq \sum_{i \leq 2^{n}}\big<e_{i}|F|e_{i}\big> = \text{Tr}(F).\]
\end{proof}
\begin{proof}

We now prove Theorem \ref{thm:67}, the idea of which is as follows: Given a classical bitstring and a subspace `close' to it, we find a subspace spanned only by classical bitstrings `close' to this subspace. Then we compress each of the spanning classical strings and show that the string we began with must be one of these.
Fix a rational $\epsilon$. Consider the machine $P$ doing the following:
\begin{enumerate}
    \item On input $\pi$, $P$ searches for $\pi=\sigma \tau$ such that
    $\mathbb{U}(\sigma)\downarrow = F$, an orthonormal set, $F \subseteq \mathbb{C}^{2^{n}}_{alg}$ for some $n$ and $|\tau|= \lceil \text{log}(\epsilon^{-1}|F|)\rceil$.
    \item Letting $O:= \sum_{v\in F} |v\big>\big<v|$ and $E$, the standard basis of $\mathbb{C}^{2^{n}}$, find the set $S^{\epsilon}_{E,O}$ from Lemma \ref{lem:30}.
    \item Take a canonical surjective map $g$ from the set of bitstrings of length $\lceil \text{log}(\epsilon^{-1}|F|)\rceil$ onto $S^{\epsilon}_{E,O}$. ($g$ exists since  $|S^{\epsilon}_{E,O}|<\epsilon^{-1}|F|$ by \ref{lem:30}).  Output $g(\tau)$.
\end{enumerate}
We first show that $P$ is prefix-free. Suppose $\pi$ and $\pi'$ are in the domain of $P$ and $\pi \preceq \pi'$. Then, $\pi=\sigma \tau$ and $\pi'=\sigma'\tau'$ and $\sigma$ and $\sigma'$ are in the domain of $\mathbb{U}$. $\pi \preceq \pi'$ implies that $\sigma \preceq \sigma'$ or $\sigma' \preceq \sigma$. But as $\mathbb{U}$ is prefix-free, $\sigma = \sigma'$ must hold. Since the computations $\mathbb{U}(\pi)$, and $\mathbb{U}(\pi')$ and not stuck forever at (1), it must be that $|\tau|= \lceil \text{log}(\epsilon^{-1}|F|)\rceil$ and $|\tau'|= \lceil \text{log}(\epsilon^{-1}|F'|)\rceil$ where $\mathbb{U}(\sigma)\downarrow=F=F'=\mathbb{U}(\sigma')\downarrow$. So, $\tau$ and $\tau'$ have the same length implying that $\pi=\pi'$.
\\
Now, let $\sigma \in 2^n$ be any classical bitstring. Let $\lambda$ and $F \subseteq \mathbb{C}^{2^{n}}_{alg}$ a orthonormal set such that $|\lambda|+ \text{log}(|F|)=QK^{\epsilon}(|\sigma\big>\big<\sigma|)$,  $\sum_{v \in F} \big<v |\sigma\big>\big<\sigma|v \big> > \epsilon$ and $\mathbb{U}(\lambda)=F$. Let $O:= \sum_{v\in F} |v\big>\big<v|$. Note that since
$\epsilon<\sum_{v \in F} \big<v |\sigma\big>\big<\sigma|v \big> = \big<\sigma|O|\sigma\big>$, $\sigma \in S^{\epsilon}_{E,O}$ where $E$ is the standard basis. Let $\tau$ be a length $ \lceil \text{log}(\epsilon^{-1}|F|)\rceil$ string such that $g(\tau)=\sigma$. Then, we see that $P(\lambda \tau) = \sigma $. \[\text{K}(\sigma)\leq^+   |\lambda|+|\tau|\leq^+  |\lambda|+\text{log}(\epsilon^{-1})+ \text{log}(|F|) \leq^+ QK^{\epsilon}(|\sigma\big>\big<\sigma|)\]
This establishes one direction. Note that the additive constant depends on $\epsilon$. The constant (zero) in the other direction turns out to be independent of $\epsilon$: Given some classical bitstring $\sigma$, let $\mathbb{U}(\pi)=\sigma$ and $|\pi|=\text{K}(\sigma)$. Then, letting $F=\{\sigma\}$ in \ref{defn:4}, $QK^{\epsilon}(\sigma)\leq QK^{1}(\sigma)\leq |\pi|=K(\sigma)$, for any $\epsilon>0$.

\end{proof}
 \begin{defn}
\label{defn:78}
A `system' $B= ((b^{n}_{0},b^{n}_{1}))_{n\in \mathbb{N}}$ is a sequence of orthonormal bases for $\mathbb{C}^{2}$ such that each $b^{n}_{i}$ is complex algebraic and the sequence $ ((b^{n}_{0},b^{n}_{1}))_{n\in \mathbb{N}}$ is computable.
\end{defn}
\begin{remark}\label{rem:K}
Let $B= ((b^{n}_{0},b^{n}_{1}))_{n\in \mathbb{N}}$ be a  system, as in \ref{defn:78}. Let $A_B$ be the set of all pure states, $\sigma$ such that $\sigma$ is a product tensor of elements from $B$. For example, $b^{1}_{0}\otimes b^{2}_{1}\otimes b^{3}_{1}\otimes b^{4}_{0} \in A_B$. Then, the previous theorem generalizes to the following: Fix a rational $\epsilon$, a $B$ and a $A_B$ as above. $K(\sigma)=^+ QK^{\epsilon}(|\sigma\big>\big<\sigma|)$ holds for all $\sigma \in A_B$, upto an additive constant depending only on $B$ and $\epsilon$. Here, $K(\sigma)$ is defined in the obvious way. For example, $K(b^{1}_{0}\otimes b^{2}_{1}\otimes b^{3}_{1}\otimes b^{4}_{0})=K(0110)$. This is proved by replacing $S^{\epsilon}_{E,O}$ with $S^{\epsilon}_{B,O}$ in the proof of Theorem \ref{thm:67}.
\end{remark}
The following lemma can be proved similarly to Theorem \ref{thm:67}.
\begin{lem}
\label{lem:K}
Fix a rational $\epsilon$ and let $(B_{n})_n$ be a computable sequence such that $B_n$ is a orthonormal basis for $\mathbb{C}^{2^{n}}$ composed of algebraic complex vectors. Then, for all $\sigma \in \bigcup_{n} B_{n}$, $K(\sigma)= QK^{\epsilon}(|\sigma\big>\big<\sigma|)$, upto an additive constant depending only on $\epsilon$ and $(B_{n})_n$.
\end{lem}
Note that $K(\sigma)$ is well-defined as $\sigma$ is complex algebraic.
The following Theorem \ref{thm:8} agrees nicely with the upper bound for $K$ in the classical setting: for all strings $x$, $K(x) \leq |x|+K(|x|)+1 $ (See theorem 2.2.9 in \cite{misc}.).
\begin{thm}
\label{thm:8}
There is a constant $d>0$ such that for any $\epsilon$ and any $\tau$, QK$^{\epsilon}(\tau) \leq |\tau| + $ K$(|\tau|) + d$.
\end{thm}
\begin{proof}
Let $k>1$. Let $P$ be the prefix-free Turing machine which on input $\pi$, such that $\mathbb{U}(\pi)=n$ outputs $E=(e_{i})_{i=1}^{2^{n}}$, the standard computational basis of $\mathbb{C}^{2^{n}}$.
\end{proof}
It may seem that this upper bound, given by the apparently inefficient device of using $2^{|\tau|}$ many orthonormal vectors to approximate $\tau$, can be improved. However, the bound is tight by Theorem \ref{thm:67} together with the classical counting theorem (see \cite{misc1}, theorem 3.7.6.).
\\
As we shall see later, the unique tracial state $\tau = (\tau_{n})_{n\in \mathbb{N}}$ where for all $n$, $\tau_n$ is the $2^n$ by $2^n$ diagonal matrix with $2^{-n}$ along the diagonal is quantum Martin-L{\"o}f random. Theorem \ref{thm:31} shows that its initial segments achieve the upper bound given by Theorem \ref{thm:8}.
\begin{thm}
\label{thm:31}
Let $k$ be any natural number. There is a constant $t$ such that for all $n$, $QK^{2^{-k}}(\tau_{n})\geq n+K(n)-t$.
\end{thm}
\begin{proof}
Fix a $k$ and suppose towards a contradiction that for all $t \in \mathbb{N}$, there is a $n_{t} $ such that $QK^{2^{-k}}(\tau_{n_{t}})< n_{t}+K(n_{t})-t$. So, for all $t$, there are $F_{t} \subseteq \mathbb{C}^{2^{n_{t}}}$ and $\sigma_{t}$ such that $\mathbb{U}(\sigma_{t})=F_{t}$ and

\[2^{-k}<\sum_{v\in F_{t}} \big<v|\tau_{n_{t}}|v\big>= 2^{-n_{t}}|F_{t}|,\]
and
\[|\sigma_{t}|+\text{log}(|F_{t}|) < n_{t}+K(n_{t})-t.\]
\\
Taking log on both sides of the first inequality and inserting in the second gives that for all $t$, $n_t$ and $\sigma_{t}$,
\begin{align}
    \label{eq:43}
    && t-k+|\sigma_{t}| <  K(n_{t}).
\end{align}
Now, define a prefix-free machine $M$ as follows. On input $\pi$, $M$ checks if $\mathbb{U}(\pi)$ halts and outputs a orthonormal set $F\subseteq \mathbb{C}^{2^{n}}$ for some $n$. If so, then $M(\pi)=n$. Let r be the coding constant of $M$. Note that for all $t$, $M(\sigma_{t})=n_t$. So, $K(n_t) \leq |\sigma_{t}|+r.$
Together with \eqref{eq:43}, we have that for all $t$,
$t-k+|\sigma_{t}|< |\sigma_{t}|+r$. So, $t-k<r$ for all $t$, a contradiction.

\end{proof}

In contrast to Lemma \ref{lem:K}, we have,
\begin{lem}
Fix an $\epsilon>0$ and an $n\in \mathbb{N}$. It is not true that for all $\sigma$, complex algebraic pure states in $\mathbb{C}^{2^{n}}$, $QK^{\epsilon}(|\sigma\big>\big<\sigma|)=^{+} K(\sigma)$.
\end{lem}
\begin{proof}
Clearly, for all $\sigma$, complex algebraic pure states, $QK^{\epsilon}(|\sigma\big>\big<\sigma|)\leq^{+} K(\sigma)$ holds.
Suppose that for some $\epsilon$ and $n\in \mathbb{N}$, for all $\sigma \in \mathbb{C}^{2^{n}}_{alg}$, pure states, $QK^{\epsilon}(|\sigma\big>\big<\sigma|)\geq^{+} K(\sigma)$ holds.
By Theorem \ref{thm:8}, for all $\sigma \in \mathbb{C}^{2^{n}}_{alg}$, pure,
$K(n)+n\geq^{+}QK^{\epsilon}(|\sigma\big>\big<\sigma|)\geq^{+} K(\sigma)$. This is a contradiction as there are only finitely many programs of length atmost $n+K(n)$ but there are infinitely many complex algebraic pure states, $\sigma$ of length $n$.
\end{proof}Analogously to $QC$ \cite{Berthiaume:2001:QKC:2942985.2943376,brudno} a `counting condition' also holds for $QK$: the cardinality of a orthonormal set of vectors with bounded complexity has an upper bound depending on the complexity bound. The counting condition for $QK$ is established in a different fashion than that for $QC$ (which uses entropy inequalities like  Holevo's-chi \cite{Berthiaume:2001:QKC:2942985.2943376} and Fanne's inequality \cite{brudno}). This reflects once again that $QC$ invloves approximating a density matrix by another density matrix while $QK$ involves `projecting' a density matrix onto a  subspace.
\begin{thm}
\label{thm:count}
Let $V=(v_{i})_{i=1}^{N} \subset \mathbb{C}^{2^{s}}$ be a collection of orthonormal vectors with $QK^{\epsilon}(|v_{i}\big>\big<v_{i}|)\leq B$ for all $i$. Then, $N\leq \epsilon^{-1}2^{B}$.
\end{thm}
\begin{proof}
For each $v_i$, we have $\sigma_i$ and $F_i$,\[F_{i}= \sum_{t\in A_{i}} |t\big>\big<t|,\]
with $A_{i}\subset \mathbb{C}^{2^{s}}$ orthonormal, such that $\big<v_{i}|F_{i}|v_{i}\big>>\epsilon$, $\mathbb{U}(\sigma_{i})=F_i$ and $|\sigma_{i}|+$log$|A_{i}|\leq B$. Let $D\subseteq \{1,2\cdots N\}$ be maximal such that $F_{i}\neq F_j$ for $i,j$ in $D$. ($D \neq \{1,2\cdots N\}$ may hold as there may be $i,j$ with $F_{i}=F_j$). Let $F$ be the orthogonal projector onto the subspace spanned by $A:=\bigcup_{i \in D}A_{i}$. Then, $A$ has dimension atmost $\sum_{i \in D}|A_{i}|$. By $|A_{i}| \leq 2^{-|\sigma_{i}|}2^{B}$ for all $i$ and noting that $\sigma_{i}\neq \sigma_j$ for $i,j$ in $D$,
\[\text{Tr}(F) \leq \sum_{i \in D}|A_{i}| \leq \sum_{i \in D} 2^{-|\sigma_{i}|}2^{B}\leq 2^{B}\sum_{\sigma \in \text{dom}(\mathbb{U})}2^{-|\sigma|} \leq  2^B.\]

The reason behind summing over $i \in D$, rather than over $i \leq N$ was to get the second to last inequality. By the maximality of $D$, $A =\bigcup_{i \leq N}A_{i}$ and so, $A_i$ is a subspace of $A$ for all $i\leq N$. So, $\big<v_{i}|F|v_{i}\big>\geq \big<v_{i}|F_{i}|v_{i}\big> >\epsilon $ for all $i \leq N$. By orthonormality of $V$,

\[\epsilon N < \sum_{i} \big<v_{i}|F|v_{i}\big> \leq \text{Tr}(F).\]
\end{proof}

\section{Quantum algorithmic randomness}
\label{sec:4}
We briefly review quantum algorithmic randomness. All definitions in this section are from \cite{unpublished} or \cite{bhojraj2020quantum} unless indicated otherwise.

While it is clear what one means by an infinite sequence of bits, it is not immediately obvious how one would formalize the notion of an infinite sequence of qubits. To describe this, many authors have independently come up with the notion of a \emph{state} \cite{unpublished,article,brudno}. We will need the one given by Nies and Scholz \cite{unpublished}.
\begin{defn}
A \emph{state}, $\rho=(\rho_n)_{n\in \mathbb{N}}$ is an
infinite sequence of density matrices such that $\rho_{n} \in \mathbb{C}^{2^{n} \times 2^{n}}$ and $\forall n$,  $PT_{\mathbb{C}^{2}}(\rho_n)=\rho_{n-1}$.
\end{defn}
The idea is that $\rho$ represents an infinite sequence of qubits whose first $n$ qubits are given by $\rho_n$. Here, $PT_{\mathbb{C}^{2}}$ denotes the partial trace which `traces out' the last qubit from $\mathbb{C}^{2^n}$. The definition requires $\rho$ to be coherent in the sense that for all $n$, $\rho_n$, when `restricted' via the partial trace to its first $n-1$ qubits, has the same measurement statistics as the state on $n-1$ qubits given by $\rho_{n-1}$.
The following state will be the quantum analogue of Lebesgue measure.
\begin{defn}
\label{def:tr}
Let $\tau=(\tau_n)_{n\in \mathbb{N}}$ be the state given by setting $\tau_n = \otimes_{i=1}^n I$ where $I$ is the two by two identity matrix.
\end{defn}
\begin{defn}
A \emph{special projection} is a hermitian projection matrix with complex algebraic entries.
\end{defn}
Since the complex algebraic numbers (roots of polynomials with rational coefficients) have a computable presentation, we may identify a special projection with a natural number and hence talk about computable sequences of special projections.
\begin{defn}
\label{defn:sigclass}
A quantum $\Sigma_{1}^0$
set (or q-$\Sigma_{1}^0$
set for short) G is a computable
sequence of special projections $G=(p_{i})_{i\in \mathbb{N}}$ such that $p_i$ is $2^i$ by $2^i$ and range$(p_i \otimes I) \subseteq$ range $(p_{i+1})$ for all $i\in \mathbb{N}$.

\end{defn}
\begin{defn}
If $\rho$ is a state and $G=(p_{n})_{n\in \mathbb{N}}$ a q-$\Sigma_{1}^0$
set as above, then $\rho(G):=\lim_{n}$ Tr$(\rho_{n}p_n)$.
\end{defn}

\begin{defn}
A \emph{quantum Martin-L{\"o}f test} (q-MLT) is a computable sequence, $(S_{m})_{m \in \mathbb{N}}$ of q-$\Sigma_{1}^0$ classes such that $\tau(S_m)$ is less than or equal to $2^{-m}$ for all m, where $\tau$ is as in Definition \ref{def:tr}.
\end{defn}

\begin{defn}
$\rho$ is \emph{q-MLR} if for any q-MLT $(S_{m})_{m \in \mathbb{N}}$, $\inf_{m \in \mathbb{N}}\rho(S_m)=0$.
\end{defn}
Roughly speaking, a state is q-MLR if it cannot be `detected by projective measurements of arbitrarily small rank'.
\begin{defn}
$\rho$ is said to \emph{fail  the q-MLT $(S_{m})_{m \in \mathbb{N}}$, at order $\delta$}, if $\inf_{m \in \mathbb{N}}\rho(S_m)>\delta$. $\rho$ is said to \emph{pass  the q-MLT $(S_{m})_{m \in \mathbb{N}}$ at order $\delta$} if it does not fail it at $\delta$.
\end{defn}
So, $\rho$ is q-MLR if it passes all q-MLTs at all $\delta>0$. Quantum Martin-L{\"o}f randomness is modelled on the classical notion:
An infinite bitstring $X$ is said to \emph{pass} the Martin-L{\"o}f test $(U_{n})_n$ if $X \notin \bigcap_{n}U_n$ and is said to be \emph{Martin-L{\"o}f random (MLR)} if it passes all Martin-L{\"o}f tests (See 3.2.1 in \cite{misc}).

A related notion is Solovay randomness. A computable sequence of $\Sigma^{0}_1$ classes, $(S_{n})_n$ is a \emph{Solovay test} if $\sum_{n} \mu(S_{n})$, the sum of the Lebesgue measures of the $S_n$s is finite. An infinite bitstring $X$ \emph{passes} $(S_{n})_n$ if $X\in S_n$ for infinitely many $n$ (See 3.2.18 in \cite{misc}).

We obtain a notion of a quantum Solovay test by replacing `$\Sigma^{0}_1$ class' and `Lebesgue measure' in the definition of classical Solovay tests with `quantum-$\Sigma^{0}_1$ set' and $\tau$ respectively. The following definitions are from \cite{bhojraj2020quantum} unless indicated otherwise:
\begin{defn}\label{defn:1}
A uniformly computable sequence of quantum-$\Sigma^{0}_1$ sets, $(S^{k})_{k\in\omega}$ is a \emph{ quantum-Solovay test} if  $\sum_{k\in \omega} \tau(S^{k}) <\infty.$\end{defn}\begin{defn}\label{defn:2}
For $0<\delta<1$, a state $\rho$ \emph{fails the Solovay test $(S^k)_{k\in\omega}$ at level $\delta$} if there are infinitely many $k$ such that $\rho(S^k)>\delta$.
\end{defn}
\begin{defn}
A state $\rho$ \emph{passes the Solovay test $(S^k)_{k\in\omega}$} if for all $\delta>0$, $\rho$ does not fail $(S^k)_{k\in\omega}$ at level $\delta$. I.e., lim$_{k}\rho(S^{k})=0$.
\end{defn}
\begin{defn}
A state $\rho$ is \emph{quantum Solovay random} if it passes all quantum Solovay tests.\end{defn}
It is remarkable that $X$ is MLR if and only if it passes all Solovay tests (See 3.2.19 in \cite{misc}). This is also true in the quantum realm\cite{bhojraj2020quantum}.

An \emph{interval Solovay test} is a Solovay test, $(S_{n})_n$ such that each $S_n$ is generated by a finite collection of strings (See 3.2.22 in \cite{misc}). Its quantum version is:

\begin{defn}\cite{unpublished}
A \emph{strong Solovay test} is a computable sequence of special projections $(S^{m})_{m}$ such that $\sum_{m}\tau(S^{m}) < \infty$. A state $\rho$ fails $(S^{m})_{m}$ at $\epsilon$ if for infinitely many $m$, $\rho(S^{m})>\epsilon$.
\end{defn}
\begin{defn}\cite{unpublished}
A state $\rho$ is \emph{weak Solovay random} if it passes all strong quantum Solovay tests.\end{defn}
It is open whether weak Solovay randomness is equivalent to q-MLR. We need the notion of a computable real number to talk about Schnorr randomness: For the purposes of this paper, a function, $f$ from the natural numbers to the rationals is said to be \emph{computable} if there is a Turing machine, $\phi$ such that on input $n$, $\phi$ halts and outputs $f(n)$ (See Theorem 5.1.2 in \cite{misc1}). Note here that we interpret the output of a Turing machine as a rational number.
\begin{defn}
A sequence $(a_n)_{n\in \mathbb{N}}$ is said to be \emph{computable} if there is a computable function $f$, such that $f(n)=a_n$.

\end{defn}

\begin{defn}
A real number $r$ is said to be \emph{computable} if there is a computable function $f$ such that for all $n$, $|f(n)-r|<2^{-n}$.
\end{defn}

By 7.2.21 and 7.2.22 in \cite{misc1}, a Schnorr test may be defined as:
\begin{defn}
A \emph{Schnorr test} is an interval Solovay test, $(S^{m})_{m}$ such that $\sum_{m}\mu(S^{m}) $ is a computable real number.
\end{defn}
An infinite bitstring passes a Schnorr test if it does not fail it (using the same notion of failing as in the Solovay test). We mimic this notion in the quantum setting.
\begin{defn}
\label{defn:Schnorr}
A \emph{quantum Schnorr test} is a strong Solovay test, $(S^{m})_{m}$ such that $\sum_{m}\tau(S^{m}) $ is a computable real number. A state is quantum Schnorr random if it passes all Schnorr tests.
\end{defn}

\section{Relating QK to randomness }
\label{sec:3}
\subsection{A Chaitin type result}

Theorem \ref{thm:7} is a Chaitin type characterization of the weak Solovay random states in terms of $QK$. (Chaitin's result in the classical setting says that an infinite bitstring $X$ is Solovay random if and only lim$_{n}K(X\upharpoonright n) - n = \infty$).
\begin{thm}
\label{thm:7}
A state $\rho = (\rho_{n})_{n}$ is weak Solovay random if and only if \[\forall \epsilon>0 \forall c >0 \forall^{\infty} n  QK^{\epsilon}(\rho_{n}) \geq n+c.\]
\end{thm}

\begin{proof}
($\Longleftarrow)$: Suppose for a contradiction that $\rho$ fails a strong Solovay test $(S_m)_{m}$ at $\epsilon>0$. The idea will be to use the subspaces given by the $ S_m $s, to approximate $\rho$. More, precisely, the $F$ appearing in the definition of $QK^{\epsilon}$ will be the orthonormal vectors given by the projection $S_m$ for an appropriate $m$. The details are as follows.  Let $M$ be the prefix-free machine doing the following. On input $\sigma$, if $\mathbb{U}(\sigma)=m$, then output $(v_i)_{i}$ where \[S_{m}= \sum_{i} |v_{i}\big>\big<v_{i}|.\] Let $c_{M}$ be it's coding constant. Take an $m$ such that Tr$(\rho_{n_{m}}S_{m})>\epsilon$ (Notation: $n_{m}$ is the natural number $n$ such that $S_m$ is a projection on $n$ qubits.). By the choice of $m$, \[\text{QK}^{\epsilon}(\rho_{n_{m}}) \leq \text{K}(m)+ c_{M}+ \text{log}(2^{n_{m}}\tau(S_{m}))= n_{m}+ \text{K}(m) - f(m) + c_{M},\]
where $f$ is the function: $f(m)= -\text{log}(\tau(S_m))$. As $f$ is computable and as $\sum_{m} 2^{-f(m)} < \infty $ by the definition of a strong Solovay test, Lemma 3.12.2 in \cite{misc1} implies that for all $m$, $\text{K}(m) - f(m) \leq q$ for some constant $q$. Noting that we may assume the sequence $n_{m}$ to be strictly increasing in $m$ and  letting $c:= q+c_{M}+1$, we see that $\exists^{\infty} n$  such that $ \text{QK}^{\epsilon}(\rho_{n})< n+c.$
\\
($\Longrightarrow$):
Suppose toward a contradiction that there is a $\epsilon>0$ and a constant $c>0$ such that there are infinitely many $n$ with $ \text{QK}^{\epsilon}(\rho_{n})< n+c.$  Define a strong Solovay test $S$ as follows. Let $T$ be the set of all $\sigma$ such that $\mathbb{U}(\sigma)$ halts and outputs an orthonormal set $F_{\sigma}\subseteq  \mathbb{C}^{2^{n_{\sigma}}}$  such that $|\sigma|+$ log $|F_{\sigma}|< n_{\sigma} + c$. For all $\sigma \in T$, let \[P_{\sigma}:= \sum_{v\in F_{\sigma}} |v \big>\big<v |\] and let $S:= (P_{\sigma})_{\sigma \in T}$. For all $\sigma \in T$, $2^{|\sigma|}|F_{\sigma}|< 2^{n_{\sigma} + c}$. So, $ \tau(P_{\sigma})= 2^{-n_{\sigma}}|F_{\sigma}|< 2^{c-|\sigma| }$.  So,
\[\sum_{\sigma \in T}\tau(P_{\sigma}) < 2^{c}\sum_{\sigma \in T}2^{-|\sigma|} < 2^{c}\sum_{\sigma: \mathbb{U}(\sigma)\downarrow} 2^{-|\sigma|} < \infty, \]
since $\mathbb{U}$ is prefix-free. This shows that $S$ is a strong Solovay test. For any $n$ such that $ \text{QK}^{\epsilon}(\rho_{n})< n+c$, there is a $\sigma \in T$ such that Tr$(P_{\sigma} \rho_{n})>\epsilon$. So, $\rho$ fails $S$ at $\epsilon$.
\end{proof}
The following corollary shows the equivalence of weak Solovay and q-ML randomness for a specific type of states.
Let $B= ((b^{n}_{0},b^{n}_{1}))_{n\in \mathbb{N}}$ be a system ( Definition \ref{defn:78}). Let $A^{\infty}_B$ be the set of all states which are limits of elements from $A_B$ as in
\ref{rem:K}. For example, $b^{1}_{0}\otimes b^{2}_{1}\otimes b^{3}_{0}\otimes b^{4}_{1}\cdots =: \rho \in A^{\infty}_B$.

\begin{cor}
\label{cor}
For any $B$, weak Solovay randomness is equivalent to q-MLR on $A^{\infty}_B$.
\end{cor}
\begin{proof}
Fix a system $B= ((b^{n}_{0},b^{n}_{1}))_{n\in \mathbb{N}}$ and let $\rho\in A^{\infty}_B$ be weak Solovay random. Let $\rho'$ be the bitstring induced by $\rho$. I.e., for example if $\rho=b^{1}_{0}\otimes b^{2}_{1}\otimes b^{3}_{0}\otimes b^{4}_{1} \cdots$, then $\rho':= 0101\cdots$.  By Theorem \ref{thm:7}, for $\epsilon=0.5$ \[  \forall c >0 \forall^{\infty} n  QK^{0.5}(\rho_{n}) \geq n+c.\] By Remark \ref{rem:K}, $K(\rho'\upharpoonright n)= MK^{0.5}(\rho_{n})$ upto a constant depending only on $B$. So,
\[  \forall c >0 \forall^{\infty} n  K(\rho'\upharpoonright n) \geq n+c.\]
By Chaitin's result, \cite{Chaitin:1987:ITR:24912.24913} $\rho'$ is MLR. Now, by an easy modification of 3.13 from \cite{unpublished}, $\rho$ is q-MLR. We already know that q-MLR implies weak Solovay randomness for any state from before.
\end{proof}

\subsection{Chaitin and Levin\textendash Schnorr type results}
It turns out that weak Solovay randomness is equivalent to q-MLR and has both Chaitin ((3) in Theorem \ref{thm:39} ) and Levin\textendash Schnorr ((4) in Theorem \ref{thm:39}) type characterizations in terms of $QK$ when the states are restricted to a certain class, $\mathcal{L}$ defined below. To define this class we need to consider the halting set over the halting set : $\emptyset''=(\emptyset')'$ (See\cite{misc}). Let $\mathcal{L}$ denote the union of the two classes of states.
\begin{enumerate}
    \item States in $A^{\infty}_B$ for some $B$, as in Corollary \ref{cor}
    \item States which do not Turing compute $\emptyset''$.
    \end{enumerate}
 Nies and Barmpalias (in personal communication) have shown that q-MLR is equivalent to weak quantum Solovay randomness for states which do not compute $\emptyset''$. The same equivalence also holds on $A^{\infty}_B$ by Corollary \ref{cor}. This similarity motivates our study of $\mathcal{L}$.
\begin{thm}
\label{thm:39}
If $\rho=(\rho_{n})_{n} \in  \mathcal{L}$, then the following are equivalent
\begin{enumerate}
\item $\rho$ is q-MLR.
    \item $\rho$ is weak Solovay random.
    \item $ \forall \epsilon>0 \forall c >0 \forall^{\infty} n,  QK^{\epsilon}(\rho_{n}) \geq n+c.$
    \item $\forall \epsilon>0\exists c \forall n, QK^{\epsilon}(\rho_{n})>n-c .$
\end{enumerate}
\end{thm}
\begin{proof}
(1)$\iff$(2) follows from the previous remarks.\\
(4)$\Longrightarrow$(1):\begin{proof} First, let $\rho \in A^{\infty}_B$ for some $B$ and let (4) hold. By the same argument as in Corollary \ref{cor}, we get that
\[  \exists c >0 \forall  n  ,K(\rho'\upharpoonright n) \geq n-c .\]
The classical Levin\textendash Schnorr result \cite{misc1} implies that $\rho'$ is MLR. Using once more 3.13 in \cite{unpublished} as in \ref{cor}, we see that $\rho$ is q-MLR. Now suppose $\rho$ does not Turing compute $\emptyset''$. We will show that (4) implies (2). Suppose for a contradiction that $(S^{m})_{m}$ is a strong Solovay test which $\rho$ fails at $\epsilon'>0$. By Theorem 2.11 in \cite{bhojraj2020quantum}, we can effectively compute a q-MLT $(G^{m})_{m}$ which $\rho$ fails at some rational $\epsilon>0$. Let $g(m):=$ the least $s$ such that Tr$(\rho_sG^{2m}_{s})>\epsilon$. As $\rho$ computes $g$, by Martin's high domination theorem (see \cite{misc1} for a proof), there is a total computable function $f$ such that $\exists^{\infty} g(n)<f(n)$. We may assume that $f(t)>3t$ for all $t$ by taking the max of 2 computable functions. Fix this $f$ (non-uniformly) and consider the following machine, $M$:\\ On input $0^{m}1$, $M$ outputs $F^{m}$ where $F^{m}$ is such that
\[G^{2m}_{f(m)}= \sum_{v \in F^{m}}|v\big>\big<v|.\]
Clearly $M$ is prefix free. Let $l-1$ be it's coding constant. Let $t$ be so that $f(t)>g(t)$. Let $F^t$ be defined similarly to $F^m$ above. Then, by definition of $g$, we have that \[\epsilon< \sum_{v \in F^{t}}\big<v|\rho_{f(t)}|v\big>.\] $M(0^{t}1)=F^{t}$ and so, there is a $\pi$ such that $|\pi|\leq t+l $ and $\mathbb{U}(\pi)=F^{t}$. Also note that $|F^{t}|\leq 2^{f(t)-2t}$ by the definition of a q-MLT. So,
$
\text{QK}^{\epsilon}(\rho_{f(t)}) \leq t+l +f(t)-2t=f(t)-t+l.$

Recall that $t$ was an arbitrary element of the infinite set $\{s: f(s)>g(s)\}$. So, for infinitely many $t$s, there is an $n=f(t)$ such that $\text{QK}^{\epsilon}(\rho_{n}) \leq n-t+l,$  contradicting $(4)$.
\end{proof}
$(3)\Longrightarrow(4)$ is obvious and $(2)\Longrightarrow(3)$ was done in Theorem \ref{thm:7}.
\end{proof}
We apply the preceding theorem to get the following quantum analog of a classical result, Proposition 3.2.14 in \cite{misc}.
\begin{thm}
\label{thm:compset}
Let $C$ be an infinite computable set, $\rho \in \mathcal{L}$ and $\epsilon>0$. If there is a $d$ such that for all $m\in C$, $QK^{\epsilon}(\rho_m)>m-d$, then $\rho$ is weak Solovay random.
\end{thm}
\begin{proof}
Let $M$ be the machine doing the following: On input $\sigma$, check if $\mathbb{U}(\sigma)=F$, an orthonormal set $F\subseteq \mathbb{C}^{2^{n}}$. If such a $F$ and $n$ exist, compute $s$ such that $n+s$ is the least element of $C$ greater than $n$ and output the set:
\[T:= \{v \otimes \pi: v \in F, \pi \in 2^{s}\}.\]
Note that $|T|=2^{s}|F|$. It is easy to see that $M$ is prefix-free. Let $l$ be it's coding constant. Suppose for a contradiction that $\rho$ is not weak Solovay random. \ref{thm:39} implies that $\forall c$, $\exists n_{c}$ such that $QK^{\epsilon}(\rho_{n_{c}}) \leq n_{c}-c.$ Let $c$ be arbitrary and take such an $n:=n_c$. There is a $\sigma$ and $F$ such that $\mathbb{U}(\sigma)=F \in \mathbb{C}^{2^{n}}$, $|\sigma|+$log$(|F|)\leq n-c$ and \[\sum_{v\in F} \big<v|\rho_{n}|v\big> >\epsilon.\] Let $t=n+s$ be the least element of $C$ greater than $n$. On input $\sigma$, $M$ outputs $T$ as above.
Note that \[Q:=\sum_{w\in T}|w\big>\big<w|= \sum_{v\in F, \pi\in 2^{s}}|v\big>\big<v|\otimes |\pi\big>\big<\pi|=\big(\sum_{v\in F }|v\big>\big<v|\big)\otimes \big( \sum_{\pi\in 2^{s}}|\pi\big>\big<\pi|\big)=W\otimes I,
\]
where $W:=\sum_{v\in F}|v\big>\big<v|$ and $I$ be the identity on $\mathbb{C}^{2^{s}}$.
Then, by the coherence property of states, \[\sum_{w\in T} \big<w|\rho_{t}|w\big>=\text{Tr}(\rho_{t}Q)=\text{Tr}(\rho_{t}[W\otimes I])=\text{Tr}(\rho_{n}W)>\epsilon.\]
Consequently, \[QK^{\epsilon}(\rho_{t})\leq |\sigma|+  \text{log}(|T|)+ l = |\sigma|+  \text{log}(|F|)+s+ l\leq n-c+s+l=t-c+l.\]
Since $d$ and $l$ were constants and $c$ was arbitrary, this contradicts the assumption.
\end{proof}

\subsection{A weak Levin\textendash Schnorr type result}
\label{subsect:weak}
Theorem \ref{thm:7} implies that if $\rho$ is weak-Solovay random then, $ \forall \epsilon>0\exists c \forall n, QK^{\epsilon}(\rho_{n})>n-c$. I.e., being strong-Solovay random implies the Levin\textendash Schnorr  condition. Does this reverse? We give two partial results in this direction: the Levin\textendash Schnorr condition implies that $\rho$ passes all strong-Solovay tests of a certain type.
\begin{defn}
For a rational $s\in(0,1)$, a $s$-strong Solovay test is a strong Solovay test $(S^{r})_{r}$ such that
$\sum_{r}\tau(S^{r})^{s} < \infty $ and $\sum_{r}\tau(S^{r}) $ is a computable real number.
\end{defn}

\begin{thm}
\label{thm:mkssr}
If $ \forall \epsilon>0\exists c \forall n, QK^{\epsilon}(\rho_{n})>n-c$, then $\rho$ passes all $s$-strong Solovay tests for all rational $s\in (0,1)$.
\end{thm}

\begin{proof}
Suppose for a contradiction that $(S^{m})_{m}$ is a $s$-strong Solovay test which $\rho$ fails at $\epsilon>0$ and $\sum_{i} \tau(S^{i})=Q $, computable. For all $m$, let $S^m$ be $2^{n_{m}}$ by $2^{n_{m}}$ and we may let the $n_m$s be distinct. Let $f(m):=-$ log$(\tau(S^m))$ and $g(m):=\lceil s f(m) \rceil$.
Partition $\omega$ into the fibers induced by $g$. ($P$ is a fiber of $g$ if $P=g^{-1}(\{x\})=\{y: g(y)=x\},$ for some $x$.). Note that $g(r)\geq  -s \text{log}\tau(S^{r})$,
and hence
$2^{-g(r)}\leq \tau(S^{r})^{s} .$ In particular, this implies that $P$ is finite. So, there are  countably infinitely many fibers,  $\{P_{1},P_{2}\dots\}$ and $\omega = \bigcup_{m} P_{m},$
where for all $m$, there is an $x_m$ such that $P_{m}=g^{-1}(\{x_{m}\})$ and $m \mapsto x_{m}$ is injective.

The fiber $P$ of $x=g(z)$ can be computed from $x$ as follows. Note that $g(c)=x$ iff, $f(c) \in [s^{-1}(x-1), s^{-1}x]$. As $Q$ is computable, compute an interval $J$ such that, $|J|<2^{-s^{-1}x}$,  $Q\in J$ and $\sum_{r\leq q} 2^{-f(r)}\in J $ for some $q$. So, $f(c)>s^{-1}x$ if $c>q$. $P$ can be computed by evaluating $g$ on $[0,q]$.

The idea is to describe $S^r$ by computing the fiber $P$ containing $r$ and then specifying the location of $r$ in the lexicographical ordering on $P$. As $(S^r)_r$ is a $s$-strong Solovay test, this description of $S^r$ is short enough to derive a contradiction. Consider the  machine, $M$ doing the following:  On input $\lambda$, check if there is a decomposition $\lambda=\pi\sigma$  such that,
\begin{itemize}
    \item There is an $m$ such that $\mathbb{U}(\pi)=0^{g(m)}$.
    \item $|\sigma|=t=\lceil\text{log}(|P|)\rceil$ where  $P$ is the fiber of $g(m)$. (Recall that $P$ can be computed from  $g(m)$.)

\end{itemize}

 If these hold, then order $P$ lexicographically using the ordering on $2^t$ and let $r$ be the $\sigma^{th}$ element in this ordering. Output $F^{r}$ where $F^{r}$ is such that $S^{r}= \sum_{v \in F^{r}}|v\big>\big<v|.$

Note that $M$ is prefix free: Suppose $\lambda,\lambda' \in $ dom$(M)$ as witnessed by $\lambda=\pi\sigma$ and $\lambda=\pi'\sigma'$. So, $M$ finds $m$ and $m'$ such that $\mathbb{U}(\pi)=0^{g(m)}$ and $\mathbb{U}(\pi')=0^{g(m')}$. Let $\pi\sigma \preceq \pi'\sigma'$. Then, it must be that $\pi\preceq\pi'$ or $\pi'\preceq\pi$. Since $\mathbb{U}$ is prefix free, it follows that $\pi=\pi'$. So, $g(m)=g(m')$. Hence, $m$ and $m'$ are in the same fiber, $P$. Letting $t= \lceil \text{log}(|P|)\rceil$, $\lambda,\lambda' \in $ dom$(M)$ implies that $|\sigma|=|\sigma'|=t$.

For each $m\in \omega$, let $r_m$ be any element from $P_m$. Then,
\begin{align}
\label{eq:mkssr3}
    \sum_{m}|P_{m}|2^{-g(r_{m})} = \sum_{m}\sum_{r\in P_{m}} 2^{-g(r)} \leq \sum_{m}\sum_{r\in P_{m}} \tau(S^{r})^{s}= \sum_{i} \tau(S^{i})^{s}< \infty.
\end{align}

Let $h$ be the function defined by, $h(m):=$ log$(|P_{m}|)-g(r_{m})$ where $r_m$ is any representative from $P_m$. By \eqref{eq:mkssr3},
$|P_{m}|2^{-g(r_{m})} \rightarrow 0$ as $m\rightarrow \infty$. So, $h(m) \rightarrow -\infty$ as $m\rightarrow \infty$.
Each fiber is finite and $\rho$ fails the test at $\epsilon$. So, there is an infinite set $I= \{P_{j_{1}}, P_{j_{2}},\dots\}$ such that for all $i$, there is a $t_{i}\in P_{j_{i}}$ with
Tr$(\rho_{n_{t_{i}}}S^{t_{i}})>\epsilon$.  $j_{i} \rightarrow  \infty$ as $i\rightarrow \infty$ and so, $h(j_{i})= $ log$(|P_{j_{i}}|)-g(r_{t_{i}}) \rightarrow -\infty$ as $i\rightarrow \infty$. This asymptotic behavior will be used below to derive a contradiction.

Fix an arbitrary $i$ and a $t=t_{i}\in P_{j_{i}}$ as above.
So, $
\epsilon< \sum_{v \in F^{t}}\big<v|\rho_{n_{t}}|v\big>.$

Let $t$ be the $\sigma^{th}$ element of $P_{j_{i}}$ in the lexicographic ordering used by $M$ and let $\mathbb{U}(\pi)=0^{g(t)}$ and $K(0^{g(t)})=|\pi|$. Then,
$M(\pi\sigma)=F^{t}$ and so, there is a bitstring $\iota$ such that $|\iota|\leq^{+} K(0^{g(t)})+ \lceil \text{log}(|P_{j_{i}}|)\rceil $ and $\mathbb{U}(\iota)=F^{t}$. Note that log$|F^{t}|=$ log(Tr($S^t))=$ log$(\tau(S^{t})) + n_{t}= -f(t) + n_{t}$. Let $d:=(1-s)s^{-1} >0$. So, $\{(\lceil nd \rceil, 0^{n}): n \in \omega\},$
is a bounded request set (see \cite{misc1} for a definition) and hence $K(0^{n})\leq^{+}\lceil nd \rceil$. Using all this, we get that:
\begin{align*}
\text{QK}^{\epsilon}(\rho_{n_{t}})
&\leq^{+} K(0^{g(t)})+\lceil \text{log}(|P_{j_{i}}|)\rceil  -f(t) +  n_{t}\\
&\leq^{+} \lceil d g(t)\rceil+\lceil \text{log}(|P_{j_{i}}|)\rceil -f(t) +  n_{t}\\
&\leq^{+} dsf(t)+\lceil \text{log}(|P_{j_{i}}|)\rceil -f(t) +  n_{t}\\
&=f(t)(ds-1)+\lceil \text{log}(|P_{j_{i}}|)\rceil  +  n_{t}\\
&=-sf(t) +\lceil \text{log}(|P_{j_{i}}|)\rceil  +  n_{t}\\
&\leq^{+}-g(t)+\lceil \text{log}(|P_{j_{i}}|)\rceil  +  n_{t}\\
&=h(j_{i})+n_{t}.
\end{align*}
The last equality follows as $t=t_i$ is in  $P_{j_{i}}$. This means that there is an infinite sequence $(n_{t_{i}})_{i}$ such that
\[\text{QK}^{\epsilon}(\rho_{n_{t_{i}}}) <^{+} n_{t_{i}}+h(j_{i}) .\]
Finally, recall that $h(j_{i})\rightarrow -\infty$ as $i\rightarrow \infty$ and we have a contradiction.
\end{proof}
Theorem \ref{thm:mkssr} can be strengthened by weakening the defining criteria for a $s$-strong Solovay test.
\begin{defn}
\label{def:phi}
Let $s\in(0,1)$ be a rational. Let $\phi$ be any computable, non-decreasing, non-negative function on the reals such that $\phi(sr)\geq^{\times} s\phi(r)$ (I.e., there is a $C>0$ \emph{independent of s}, such that for all $r$,  $C\phi(sr)\geq  s\phi(r)$) and $\sum_{n}2^{-n}\phi(n)<\infty$ (So, $\phi$ does not tend to infinity too fast). A $(\phi,s)$-strong Solovay test is a strong Solovay test $(S^{r})_{r}$ such that
\begin{align}
\label{eq:mkssr6}
    \sum_{r}\dfrac{\tau(S^{r})^{s}}{\phi(-\text{log}(\tau(S^{r})))} < \infty ,
\end{align}

and
\[\sum_{r}\tau(S^{r}) = Q,\] where $Q$ is a computable real number.
\end{defn}
The term in the denominator in \eqref{eq:mkssr6} tends to infinity with $r$ and hence it is easier for a strong Solovay test to be a $(\phi,s)$-strong Solovay test than to be a $s$-strong Solovay test. So, passing all $(\phi,s)$-strong Solovay tests is a more restrictive notion of randomness than passing all $s$-strong Solovay tests. So, the following theorem is an improvement of, and implies Theorem \ref{thm:mkssr}.
\begin{thm}
If $ \forall \epsilon>0\exists c \forall n, QK^{\epsilon}(\rho_{n})>n-c$, then $\rho$ passes all $(\phi,s)$-strong Solovay tests for all rational $s\in (0,1)$ and all $\phi$ as in Definition \ref{def:phi}.
\end{thm}

\begin{proof}
Suppose for a contradiction that  $(S^{m})_{m}$ is a $(\phi,s)$-strong Solovay test which $\rho$ fails at $\epsilon>0$ and $\sum_{i} \tau(S^{i})=Q $, computable. For all $m$, let $S^m$ be $2^{n_{m}}$ by $2^{n_{m}}$ and we may let the $n_m$s be distinct.
For ease of presentation, we do the proof in 2 cases. First, let $s\leq 0.5$.
Let $f(m):=-$ log$(\tau(S^m))$ and let $g(m):=\lceil s f(m) \rceil$.
Partition $\omega$ into the fibers induced by $g$.
Fix some fiber $P$ of some $x=g(r)$. I.e., $r$ is a representative from $P$. Then, $g(r)=\lceil s(- \text{log}\tau(S^{r})) \rceil$. So, $g(r)\geq  -s \text{log}\tau(S^{r}),$
and hence
$2^{-g(r)}\leq \tau(S^{r})^{s} .$
In particular, this implies that each fiber is finite. So, there are  countably infinitely many fibers,  $\{P_{1},P_{2}\dots\}$. So, $\omega = \bigcup_{m} P_{m},$
where for all $m$, there is an $x_m$ such that $P_{m}=g^{-1}(\{x_{m}\})$ and $m \mapsto x_{m}$ is injective.
For each $m\in \omega$, let $r_m$ be any representative from $P_m$.
 For all $r$, $sf(r)\leq g(r)$ and $\phi$ is non-decreasing. So,
\begin{align}
\label{eq:mkssr13}\sum_{m} |P_{m}|\dfrac{2^{-g(r_{m})}}{\phi(g(r_{m}))} = \sum_{m}\sum_{r\in P_{m}} \dfrac{2^{-g(r)}}{\phi(g(r))}\leq  \sum_{m}\sum_{r\in P_{m}} \dfrac{\tau(S^{r})^{s}}{\phi(sf(r))} \leq^{\times} \sum_{r} \dfrac{\tau(S^{r})^{s}}{s\phi(f(r))}< \infty.
\end{align}

The fiber $P$ of $x=g(z)$ can be computed from $x$ for the same reason as in the previous proof. Its idea of `compressing' $S^r$ is also used here.
\\
Consider the  machine, $M$ doing the following:  On input $\lambda$, search for a decomposition $\lambda=\pi\sigma$, such that
\begin{itemize}
    \item $\mathbb{U}(\pi)=0^{g(m)}$ for some $m$.
    \item
    $|\sigma| = t$ where
    $P$ is the fiber containing $m$, (which can be computed from $g(m)$) and $t= \lceil\text{log}(|P|)\rceil$.
\end{itemize}
If found, order $P$ lexicographically using the ordering on $2^t$ and let $r$ be the $\sigma^{th}$ element in this ordering. Output $F^{r}$ where $F^{r}$ is such that
$S^{r}= \sum_{v \in F^{r}}|v\big>\big<v|.$
Note that $M$ is prefix free for the same reason as in the previous proof. Let $l$ be $M$'s coding constant.
Let $h$ be the function defined by, $h(m):=$ log$(|P_{m}|)-g(r_{m})-$log $\phi g(r_{m})$, where $r_m$ is any representative from $P_m$. By \eqref{eq:mkssr13},
\[\dfrac{|P_{m}|2^{-g(r_{m})}}{\phi g(r_{m})} \rightarrow 0\] as $m\rightarrow \infty$. So, $h(m) \rightarrow -\infty$ as $m\rightarrow \infty$.
Each fiber is finite and $\rho$ fails the test at $\epsilon$. So, there is an infinite set $I= \{P_{j_{1}}, P_{j_{2}},\dots\}$ such that for all $i$, there is a $t_{i}\in P_{j_{i}}$ with
Tr$(\rho_{n_{t_{i}}}S^{t_{i}})>\epsilon$.  $j_{i} \rightarrow  \infty$ as $i\rightarrow \infty$ and so, $h(j_{i}) \rightarrow -\infty$ as $i\rightarrow \infty$. This asymptotic behavior will be used below to derive a contradiction.

Fix an arbitrary $i$ and a $t=t_{i}\in P_{j_{i}}$ as above.
So,
\begin{align}
\label{eqn:mkssr1}
\epsilon< \sum_{v \in F^{t}}\big<v|\rho_{n_{t}}|v\big>.
\end{align}

Let $t$ be the $\sigma^{th}$ element of $P_{j_{i}}$ in the lexicographic ordering used by $M$. Let $\pi$ be such that $K(0^{g(t)})=|\pi|$ and $\mathbb{U}(\pi)=0^{g(t)}$. Then,
$M(\pi\sigma)=F^{t}$ and so, there is a bitstring $\kappa$ such that $|\kappa|\leq K(0^{g(t)}) + \lceil \text{log}(|P_{j_{i}}|)\rceil + l$ and $\mathbb{U}(\kappa)=F^{t}$. Note that log$|F^{t}|=$ log(Tr($S^t))=$ log$(\tau(S^{t})) + n_{t}= -f(t) + n_{t}$. So,
\[
\text{QK}^{\epsilon}(\rho_{n_{t}}) \leq K(0^{g(t)})+\lceil \text{log}(|P_{j_{i}}|)\rceil +l -f(t) +  n_{t}.
\]

Note that $\{(n-\lceil\text{log}(\phi(n))\rceil,0^{n}): n\in \omega\}$
is a bounded request set by definition of a $(\phi,s)$ test and so
$K(0^{g(t)})\leq^{+} g(t)-$log$(\phi g(t))$. So,
\[
\text{QK}^{\epsilon}(\rho_{n_{t}}) \leq^{+} g(t)-\text{log}(\phi g(t))+ \lceil \text{log}(|P_{j_{i}}|)\rceil  -f(t) +  n_{t}.
\]
Since
$g(t)-1 < sf(t) $, we have
\begin{align}
    \label{eq:mkssr12}
    -g(t)+1>-sf(t).
\end{align}

Since $s\leq 0.5$ we have that $1-s\geq s$.\\ So,
$g(t)-f(t) \leq sf(t)-f(t)+1 = -(1-s)f(t)+1\leq -sf(t)+1.$
Using \eqref{eq:mkssr12},
\[g(t)-f(t)<-g(t)+2.\]
So,
\[
\text{QK}^{\epsilon}(\rho_{n_{t}}) <^{+} -g(t)+\lceil \text{log}(|P_{j_{i}}|)\rceil -\text{log}(\phi  g(t)) +  n_{t} = h(j_{i}) +  n_{t}.
\]
The equality follows as $t=t_i$ is in  $P_{j_{i}}$. This means that there is an infinite sequence $(n_{t_{i}})_{i}$ such that
$\text{QK}^{\epsilon}(\rho_{n_{t_{i}}}) <^{+} n_{t_{i}}+h(j_{i}).$
Finally, recall that $h(j_{i})\rightarrow -\infty$ as $i\rightarrow \infty$ and we have a contradiction.

Now let $s>0.5$ and let $f,g$ be as in the previous case. Let $b(m):= \lceil (1-s)f(m)\rceil$ and let $C:= \lceil s/(1-s)\rceil + 1$.

Consider the machine $M$ doing the following: on
input $\pi1^{y}0\sigma$, check if the following conditions hold.
\begin{itemize}
    \item
    There is $m$ such that $\mathbb{U}(\pi)=0^{b(m)}$.
    \item If $x=b(m)$,
     $J=(s(1-s)^{-1}(x-1),s(1-s)^{-1}x + 1] \cap \omega$ and $w$ is the $y^{th}$ element of $J$, then there is a $z$ such
     that $g(z)=w$.

     \item
     If $P$ is the fiber of $g$ containing $z$ (P is computable from $w=g(z)$ just as in the previous case) and $t= \lceil\text{log}(|P|)
     \rceil$, then $|\sigma| = t$
     \end{itemize}

     If all the above are met, then order $P$ lexicographically using the ordering on $2^t$ and let $r$ be the $\sigma^{th}$ element in this ordering. Output $F^r$.

Roughly, the idea is as follows: Just as in the previous case, we want to compress $F^r$ where $r$ is the $\sigma^{th}$ number in the fiber of $w=g(m)$. The first step to achieve this is to describe $g(m)$. While in the previous case we used an $\iota$ such that $\mathbb{U}(\iota)=0^{g(m)}$ and $|\iota|=K(0^{g(m)})$), we use here the shorter string $\pi$ where  $\mathbb{U}(\pi)=0^{b(m)}$ and $|\pi|=K(0^{b(m)})$  together with $1^y$ for describing $g(m)$. From $\pi$, we get $x=b(m)$ which in turn gives $J$ which contains $g(m)$. So, $\pi$ along with $y$, the location of $g(m)$ in $J$, describes $g(m)$. After $g(m)$ is found, $F^r$ can be described just as in the previous case. The details are: As $x=b(m)$, $(1-s)f(m)\in (x-1,x]$ and hence $sf(m)$ lies in $(s(1-s)^{-1}(x-1),s(1-s)^{-1}x]$.
So, $\lceil sf(m) \rceil =  g(m)\in J=  (s(1-s)^{-1}(x-1),s(1-s)^{-1}x + 1]  \cap \omega$. Since $|J|\leq C$, $g(m)\in J$ can be determined by specifying $y \leq C$, it's location in $J$. So, $M$ can recover $g(m)$. From this point on, the remaining procedure is the same as in the previous case.
\\
We see that $M$ is prefix-free: Let $\pi1^{y}0\sigma$ and $\pi'1^{y'}0\sigma'$ be in the domain of $M$ and let $\pi1^{y}0\sigma\preceq \pi'1^{y'}0\sigma'$. By the same argument as in case1, $\pi=\pi'$ and  $M$ finds some $m,m'$ with $x=b(m)=b(m')$. It follows that $y=y'$. So, if $w$ is the $y^{th}$ (and  $y'^{th}$)  element of $J$ (as above), then $M$ finds $z,z'$ such that $g(z)=w=g(z')$. So, $z$ and $z'$ are in the same fiber $P$ and it hence follows as in the previous case that $|\sigma|=|\sigma'|$. Define $I$ and $h$ exactly as in the previous case. Fix some $i$ and let $t=t_i$ be an element of $P_{j_{i}}$ such that \eqref{eqn:mkssr1} holds. Let $t$ be the $\sigma^{th}$ element of $P_{j_{i}}$. Let $x=b(t)=\lceil (1-s)f(t)  \rceil$. So, $(1-s)f(t) \in (x-1,x]$ and  $sf(t) \in (s(1-s)^{-1}(x-1),s(1-s)^{-1}x]$. Hence, $g(t) \in (s(1-s)^{-1}(x-1),s(1-s)^{-1}x+1] \cap \omega=J$ and let $g(t)$ be the $y^{th}$ element of $J$.
Let $\pi$ be such that   $\mathbb{U}(\pi)=0^{b(t)}$ and $|\pi|=K(0^{b(t)})$. Then, on input $\pi1^{y}0\sigma$, M finds some $z$ (it could be that $z=t$, but not necessarily) such that $b(t)=b(z)=x$ and then finds that the $y^{th}$ element of $J$ is $g(z')$ for some $z'$ (again, although  $g(z')=g(t)$, it could be that $t=z'$ but not necessarily). Since $z'$ and $t$ are both in $P_{j_{i}}$, $M$ outputs $F^{t}$ after reading $\sigma$. So, there is a $\pi$
 such that $\mathbb{U}(\pi)=F^{t}$ and
 $|\pi|\leq^{+} K(0^{b(t)})+C+\lceil\text{log}(|P_{j_{i}}|)\rceil $.\\
 So, $ \text{QK}^{\epsilon}(\rho_{n_{t}})$
\begin{align*}
&<^{+}K(0^{b(t)})+ \lceil\text
 {log}(|P_{j_{i}}|)\rceil + \text{log}(|F^{t}|)\\
 &\leq^{+}b(t)-\lceil\text{log}(\phi(g(t)))\rceil+\lceil\text{log}(|P_{j_{i}}|)\rceil + n_{t}-f(t)\\
 &\leq^{+}-g(t)-\lceil\text{log}(\phi(g(t)))\rceil+\lceil\text{log}(|P_{j_{i}}|)\rceil + n_{t}
\end{align*}

The last inequality is since, by \eqref{eq:mkssr12} (which holds for any $s$), $b(t)-f(t) \leq (1-s)f(t)-f(t)+1=-sf(t)+1<-g(t)+2$. Since $t=t_{i} \in P_{j_{i}}$, we see that  $ \text{QK}^{\epsilon}(\rho_{n_{t}}) <^{+}h(j_{i})-n_{t_{i}}.$ for all $i$. This gives a contradiction for the same reason as in the previous case.

\end{proof}

\subsection{QK and computable measure machines}
\label{schn}
Schnorr randomness is an important randomness notion in the classical realm\cite{misc,misc1}. While $K$ plays well with Solovay randomness, $K_C$, a version of $K$ using a computable measure machine, $C$ (a prefix-free Turing machine whose domain has computable Lebesgue measure) gives a Levin-Schnorr characterization of Schnorr randomness  (See theorem 7.1.15 in \cite{misc1}).

So, with the intention of connecting it to quantum Schnorr randomness, we define $QK_C$ a version of $QK$ using a computable measure machine, $C$.

Theorem \ref{thm:677} shows that $QK_C$ agrees with $K_C$ on the classical bitstrings. Analogously to the classical case, Theorem \ref{thm:schnor} is a Levin\textendash Schnorr type of characterizations of quantum Schnorr randomness using $QK_C$. Theorem \ref{thm:Chaitin}, a Chaitin type characterization of quantum Schnorr randomness using $QK_C$ implies Theorem \ref{class}, a Chaitin type characterization of \emph{classical} Schnorr randomness in terms of $K_C$.

For $C$ a computable measure machine and $\sigma$ a string, $K_C$ is defined analogously to $K$; $K_C(\sigma):=$inf$\{|\tau|: C(\tau)=\sigma\}.$ The quantum version is:
for $C$, a computable measure machine and a $\epsilon>0$, define
$QK_{C}^{\epsilon}(\tau)$ to be:
\begin{defn}
\label{defn:4}
$QK_{C}^{\epsilon}(\tau) := $ inf  $\{|\sigma|+$log$|F|: C(\sigma)\downarrow = F$, a orthonormal set in $\mathbb{C}^{2^{|\tau|}}_{alg}$ and $\sum_{v \in F} \big<v |\tau |v \big> > \epsilon \}$

\end{defn}
The infimum of the empty set is taken to be $\infty$.
Notation: In this section, $\mu$ denotes Lebesgue measure and $C_t$ denotes $C$ run upto the $t$ steps. We may assume that dom$(C_t)\subseteq 2^t$. By a \emph{sequence}, we mean a countable collection whose elements may possibly be repeated. If $S$ is a sequence, the sum $\sum_{s\in S}$ will be over all elements of $S$, with repetition.

Similarly to Theorem \ref{thm:67}, we show that $QK_C$ `agrees with' $K_C$ on the classical qubitstrings. In Theorem \ref{thm:677} and its proof, $P$ and $C$ will stand for computable measure machines.
\begin{thm}
\label{thm:677}
 For all rational $\epsilon>0$ and all $C$, there exists a $P$ such that $K_P(\sigma) \leq  QK^{\epsilon}_C(|\sigma\big>\big<\sigma|)+1$ for all classical bitstrings $\sigma$.
\end{thm}

\begin{proof}
The proof is almost identical to that of Theorem \ref{thm:67}. Fix a rational $\epsilon>0$ and a $C$. Consider the machine $P$ from the proof of Theorem \ref{thm:67} but with $\mathbb{U}$ replaced by $C$. We now show that $\mu($dom$(P))$ is computable. Let $\delta>0$ be arbitrary. Since $\mu($dom$(C))$ is computable, find a stage $t$ so that $\mu(\text{dom}(C))-\mu \text{(dom}( C_t )) < \delta.$
(The $t$ can be found as follows: Compute a $q'$ such that $|q-q'|< \delta/2$. So, $q'-\delta/2<q<q'+ \delta/2$. Since $\mu \text{(dom}( C_t ))\nearrow q$, as $t\rightarrow \infty$, we can compute a $t$ such that, $q'-\delta/2<\mu \text{(dom}( C_t ))<q'+ \delta/2$.). We may compute $S$, the set of those strings $\pi=\sigma\tau \in$ dom$(P)$ and $\sigma \in $ dom$(C_t)$. So, dom$(P)\backslash S$ consists of strings $\pi=\sigma\tau$ such that $\sigma\in$ dom$(C)\backslash$dom$(C_t)$. So, it is easy to see that $\mu($dom$(P))$-$\mu(S) < \mu($dom$(C))$-$\mu($dom$(C_t)) <\delta$. As $\delta>0$ was arbitrary, this shows that $\mu($dom$(P))$ is computable.
Now, let $\sigma \in 2^n$ be any classical bitstring such that $QK_C^{\epsilon}(|\sigma\big>\big<\sigma|)<\infty$. Let $\lambda$ and $F \subseteq \mathbb{C}^{2^{n}}_{alg}$ orthonormal such that $|\lambda|+ \text{log}(|F|)=QK_C^{\epsilon}(|\sigma\big>\big<\sigma|)$,  $\sum_{v \in F} \big<v |\sigma\big>\big<\sigma|v \big> > \epsilon$ and $C(\lambda)=F$. Let $O:= \sum_{v\in F} |v\big>\big<v|$. Note that since
$\epsilon<\sum_{v \in F} \big<v |\sigma\big>\big<\sigma|v \big> = \big<\sigma|O|\sigma\big>$, $\sigma \in S^{\epsilon}_{E,O}$ where $E$ is the standard basis. Let $\tau$ be a length $ \lceil \text{log}(\epsilon^{-1}|F|)\rceil$ string such that $g(\tau)=\sigma$. Then, we see that $P(\lambda \tau) = \sigma $. So, \[K_P(\sigma)\leq  |\lambda|+|\tau|\leq |\lambda|+\text{log}(\epsilon^{-1})+1+ \text{log}(|F|) = QK^{\epsilon}_C(|\sigma\big>\big<\sigma|)+1.\]

\end{proof}
\begin{remark}
\label{rem:qkc}
Theorem \ref{thm:677} establishes one direction of the coincidence of $K_C$ and $QK_C$ for classical qubitstrings. In the other direction, take some classical bitstring $\sigma$ with $K_C(\sigma)<\infty$ for some $C$. Let $C(\pi)=\sigma$ and $|\pi|=\text K_C(\sigma)$. Then, letting $F=\{\sigma\}$ in \ref{defn:4}, $QK_C^{\epsilon}(|\sigma\big>\big<\sigma|)\leq QK_C^{1}(|\sigma\big>\big<\sigma|)\leq |\pi|=K_C(\sigma)$, for any $\epsilon>0$.
\end{remark}
\subsection{Quantum Schnorr randomness and $QK_C$}
Theorem \ref{thm:schnor} is a quantum analogue of the classical characterization of Schnorr randomness: $X$ is Schnorr random if and only if for any computable measure machine, $C$, there is a constant $d$ such that for all $n$, $K_{C}(X\upharpoonright n)>n-d$.
\begin{thm}
\label{thm:schnor}
A state $\rho$ is quantum Schnorr random if and only if for any computable measure machine, $C$ and any $\epsilon > 0$, there is a constant $d>0$ such that for all $n$, $QK^{\epsilon}_{C}(\rho_{n})>n-d$.
\end{thm}
\begin{proof}
($\Rightarrow$) We prove it by contraposition. I.e., show that $\rho$ is not quantum Schnorr random if there is a $C$ and an $\epsilon>0$ such that for all $d$, there is an $n=n_d$ such that $QK^{\epsilon}_{C}(\rho_{n}) \leq n-d$. Let $T_{s}$ be the set of all $\sigma$ such that $C_{s}(\sigma)\downarrow = F_{\sigma}$, an orthonormal set such that $|\sigma|+$ log $|F_{\sigma}|< n_{\sigma} $ and $F_{\sigma} \subseteq  \mathbb{C}^{2^{n_{\sigma}}}$ for some $n_{\sigma}$. Let $T=\bigcup_{s}T_{s}.$ For all strings $\sigma$, let $P_{\sigma}:= \sum_{v\in F_{\sigma}} |v \big>\big<v |.$
Let $Q_s$ be the sequence of those $P_{\sigma}$ for $\sigma \in T_s$, $Q$ the sequence of those $P_{\sigma}$ for $\sigma \in T$ and $D_s$ the sequence of those $P_{\sigma}$ for $\sigma \in T\backslash T_s$.
Next, we show that $\alpha:=\sum_{P\in Q} \tau(P)$ is computable by showing how to approximate it within $2^{-k}$ for an arbitrary $k$:  Computably find a $t$ (using the same method as in Theorem \ref{thm:677}) such that $
    \mu(\text{dom}(C )) -\mu \text{(dom}(C_t ))  < 2^{-k}.$
 We show that  $\sum_{P_{\sigma}\in Q_{t}}\tau(P_{\sigma})$ is within $2^{-k}$ of $\alpha$. Note that for all $\sigma \in T$, $2^{|\sigma|}|F_{\sigma}|< 2^{n_{\sigma}}$. So, $ \tau(P_{\sigma})= 2^{-n_{\sigma}}|F_{\sigma}|< 2^{-|\sigma| }$.
\[\alpha -\sum_{P_{\sigma}\in Q_{t}} \tau(P_{\sigma})  = \sum_{P_{\sigma}\in D_t} \tau(P_{\sigma}) =
     \sum_{\sigma \in T/T_{t}}|F_{\sigma}|2^{-n_{\sigma}}
     \leq\sum_{\sigma\in T/T_{t}}2^{-|\sigma|}\]\[
     \leq \sum_{\sigma\in\text{dom}( C)/\text{dom}( C_t ) }2^{-|\sigma|}
     \leq  (\mu(\text{dom}( C ))-\mu (\text{dom}( C_t )))< 2^{-k}\]

Note that $\sum_{P_{\sigma}\in Q_{t}}\tau(P_{\sigma})$ is a rational, uniformly computable in $t$ since dom$(C_t)\subseteq 2^t$ is uniformly computable in $t$. This shows that $Q$ is a quantum Schnorr test. By the assumption, we see that is a infinite sequence $d_1 < d_2 < \cdots$ and a list of distinct natural numbers $n_{d_{1}}, n_{d_{2}} \cdots$ so that for all $i$, there is a $P_i$ in $Q$ such that Tr$(P_i\rho_{n_{d_{i}}})>\epsilon$. So, $\rho$ fails $Q$ at $\epsilon$.

($\Leftarrow$)
We prove it by contraposition. Suppose that $\rho$ fails a quantum-Schnorr test, $(S^{r})_r$ at $\epsilon$. For all $j$, let $s_j$ be the least $t$ such that \[\sum_{i=0}^{t} \tau (S^{i}) > \alpha -2^{-j}.\]
We show how the sequence $(s_j)_j$ can be computed. First, let $\sum_{r}\tau(S^{r})=\alpha$, be a computable real which is not a dyadic rational. $s_j$ may be computed as follows: Note that as $\alpha$ is not a dyadic rational but $\tau(S^i)$ is a dyadic rational for all $i$, we have that
\[\sum_{i=0}^{s_{j}-1} \tau (S^{i}) < \alpha-2^{-j} < \sum_{i=0}^{s_{j}} \tau (S^{i}).\]
By Proposition 5.1.1 in \cite{misc1}, the left cut, $L(\alpha-2^{-j})$ of $\alpha-2^{-j}$ is computable. So, we may search for rationals $q \in L(\alpha-2^{-j}), q' \notin L(\alpha-2^{-j}) $ and for a $t$ such that,
\[\sum_{i=0}^{t-1} \tau (S^{i}) \leq q<q'\leq \sum_{i=0}^{t } \tau (S^{i}).\]
This $t$ is the needed $s_j$.
Now, let $\alpha$ be a dyadic rational. Then, $\alpha - 2^j$ has a finite binary representation and $s_j$ can be directly computed. So, in summary, the $(s_j)_j$ is a computable sequence, after (non-uniformly) knowing whether $\alpha$ is a dyadic rational or not.
For all $r\geq 0$, define special projections
\[ G_{r}:= \sum_{i=s_{r}+1}^{s_{r+1}}S^{i}.\]
So,
\[ \tau(G_{r}) \leq \sum_{i=s_{r}+1}^{\infty}\tau(S^{i}) = \alpha -\sum_{i=0 }^{s_{r}}\tau(S^{i}) < 2^{-r} .\]

Notation: Let each $S^i$ be an operator on $ \mathbb{C}^{2^{n_{i}}}$. Let $n_{r} =$ max$\{n_{i}:s_{r}+1 \leq i \leq s_{r+1}\}$. By tensoring with the identity, we may assume that all
$S^i$, for $s_{r}+1 \leq i \leq s_{r+1}$, are operators on $\mathbb{C}^{2^{n_{r}}}$. Let $F_r \subseteq \mathbb{C}^{2^{n_{r}}}$
be an orthonormal set of complex algebraic vectors spanning the range of $G_r$.
Define a computable measure machine, $C$ as follows.
On input $0^{r}10$, $C$ outputs $F_{2r}$ and on input $0^{r}11$, $C$ outputs $F_{2r+1}$. $C$ is clearly prefix-free and the measure of its domain is $\sum_{r}2^{-r+2}$, which is computable. Since each $G_r$ is a finite sum of the $S^i$s and as $\rho$ fails $(S^i)_i$ at $\epsilon$, there exist infinitely many $r$ such that Tr$(\rho_{n_{r}}G_{r})>\epsilon$. Since  we may let $n_r$ be strictly increasing in $r$, there are infinitely many such $n_r$. Fix such an $n_r$ and let $x= \lfloor r/2 \rfloor$ (I.e., $r=2x$ or $r=2x+1$). Then, QK$^{\epsilon}_{C}(\rho_{n_{r}}) \leq x+2+n_{r}+\text{log}\tau(G_{r}) \leq x+2+n_{r}- 2x.$ So, $\lfloor r/2 \rfloor-2 \leq n_{r}- $QK$^{\epsilon}_{C}(\rho_{n_{r}}) $. Letting $r$ go to infinity completes the proof.

\end{proof}

Theorem \ref{thm:Chaitin} is a Chaitin-type characterization of quantum-Schnorr randomness using QK$^{\epsilon}_{C}$. Together with Theorem \ref{thm:677} and lemma 3.9 in \cite{bhojraj2020quantum}, it implies that Schnorr randoms have a Chaitin type characterization in terms of $K_C$ (Theorem \ref{class}). To the best of our knowledge, this is the first, albeit simple, instance where results in quantum algorithmic randomness are used to prove a new result in the classical theory.
\begin{thm}
\label{thm:Chaitin}
$\rho$ is quantum Schnorr random if and only if for all computable measure machines $C$ and all $\epsilon$, $\forall d \forall^{\infty} n$ QK$^{\epsilon}_{C}(\rho_{n})>n+d$.
\end{thm}
\begin{proof}
($\Rightarrow$)
Suppose toward a contradiction that there is a $C$, an $\epsilon>0$ and $c>0$ such that there are infinitely many $n$ with $ \text{QK}^{\epsilon}_{C}(\rho_{n})\leq n+c.$ Define a quantum Schnorr test $Q$ as follows. Let $T_{s}$ be the set of all $\sigma$ such that $C_{s}(\sigma)\downarrow = F_{\sigma}$, an orthonormal set such that $|\sigma|+$ log $|F_{\sigma}|< n_{\sigma} + c$ and $F_{\sigma} \subseteq  \mathbb{C}^{2^{n_{\sigma}}}$ for some $n_{\sigma}$. Let $T=\bigcup_{s}T_{s}.$ For all strings $\sigma$, let $P_{\sigma}:= \sum_{v\in F_{\sigma}} |v \big>\big<v |.$
Let $Q_s$ be the sequence of those $P_{\sigma}$ for $\sigma \in T_s$ and $Q$ the sequence of those $P_{\sigma}$ for $\sigma \in T$. That $Q$ is a quantum Schnorr test is shown by replacing $2^{-k}$ by $2^{-k-c}$ in the $\Longrightarrow$ direction of the proof of Theorem \ref{thm:schnor}. For any $n$ such that $ \text{QK}^{\epsilon}_{C}(\rho_{n})< n+c$, there is a $\sigma \in T$ such that Tr$(P_{\sigma} \rho_{n})>\epsilon$. So, $\rho$ fails $Q$ at $\epsilon$.

($\Leftarrow$) If $\rho$ is not quantum Schnorr random then by Theorem \ref{thm:schnor}, there is a $C$ and an $\epsilon$ such that $\forall d \exists n$ such that QK$^{\epsilon}_{C}(\rho_{n})\leq n-d$.
\end{proof}
We now show the classical version of Theorem \ref{thm:Chaitin}.
\begin{thm}
\label{class}
An infinite bitstring $X$ is quantum Schnorr random if and only if for all computable measure machines $C$, $\forall d \forall^{\infty} n$ K$_{C}(X\upharpoonright n)>n+d$.
\end{thm}
\begin{proof}
$(\Longrightarrow):$ Suppose first that $X$ is Schnorr random. Then, $\rho:=\rho_X$, the state induced by $X$ is quantum Schnorr random by lemma 3.9 in \cite{bhojraj2020quantum}. Suppose for a contradiction that there is a $C$ and a $d$ such that $\exists^{\infty} n$ such that $K_C(X\upharpoonright n) \leq n+d$. By Remark \ref{rem:qkc}, $\exists^{\infty} n$ such that $QK^{0.5}_C(\rho_n) \leq n+d$, contradicting Theorem \ref{thm:Chaitin}.
($\Longleftarrow): $ Suppose that $X$ is not Schnorr random. Once again, by lemma 3.9 in \cite{bhojraj2020quantum}, we have that $\rho:=\rho_X$ is not quantum Schnorr random. By Theorem \ref{thm:Chaitin}, there is a $C$, an $\epsilon$ and a $d$ such that $\exists^{\infty} n$ such that  QK$^{\epsilon}_{C}(\rho_{n}) \leq n+d$. By Theorem \ref{thm:677}, there is a $P$ such that $\exists^{\infty} n$ such that  $K_{P}(\rho_{n}) \leq n+d+1$, a contradiction.
\end{proof}
\section{Conclusion}
With the intent of developing a quantum version of $K$, we introduced $QK$, a notion of descriptive complexity for density matrices using classical prefix-free Turing machines. Many connections between $K$ and Solovay and Schnorr randomness in the classical theory turned out to have analogous connections connections between $QK$ and weak Solovay and quantum Schnorr randomness.

To the best of our knowledge, the current paper is the only one to study the incompressibility of initial segments (in the sense of prefix-free classical Turing machines) of weak Solovay and quantum Schnorr random states. Nies and Scholz have explored connections between quantum Martin-L{\"o}f randomness and a version of $QC$ using unitary (quantum) machines\cite{unpublished}.

An important open question is whether weak Solovay random states have a Levin\textendash Schnorr characterization in terms of $QK$. Techniques similar to those used in subsection \ref{subsect:weak} may prove to be useful in answering this.

It still remains to find a complexity based characterization of q-MLR states\cite{unpublished}. An important question is whether weak Solovay randomness is equivalent to q-MLR, a positive answer to which will yield a $QK$ based characterizations (namely, those in Theorems \ref{thm:7} and \ref{thm:39}) of q-MLR.

Abbott, Calude and Svozil use value indefiniteness to show that infinite bitstrings resulting from measuring a finite dimensional quantum system satisfy some notions of randomness \cite{DBLP:conf/birthday/AbbottCS15,DBLP:journals/mscs/AbbottCS14,qrng2020 ,abbott2014value}.

Using states instead of a finite dimensional system, and techniques entirely different from theirs, we showed that it is possible to generate a strong form of classical randomness from a quantum source which is not quantum random. More precisely, we construct a computable, non q-MLR state which yields an arithmetically random bitstring with probability one when `measured'\cite{qpl}. Arithmetic randomness is a strong form of classical randomness, strictly stronger than
Martin-L{\"o}f randomness (See 6.8.4 in \cite{misc1}). Roughly speaking, states which yield a Martin-L{\"o}f random  bitstring with probability one when `measured using a computable basis' are defined to be \emph{measurement random} (See \cite{qpl} for the precise definition). It would be interesting to explore the relationship between the initial segment $QK$ complexity and measurement randomness of states.

\section{Acknowledgements}
I am indebted to my PhD thesis advisor, Joseph S. Miller for his encouragement, advice and support. I thank Andr{\'e} Nies for encouraging my study of quantum algorithmic randomness.
\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Film, TV & Theatre
Read Michelle Yeoh's Moving Golden Globes Speech
Michelle Yeoh Honored Her Decades-Long Career in Her Golden Globes Speech
The actress became the second Asian woman to win Best Actress in a Motion Picture – Comedy or Musical.
By Quinci LeGardye Published: Jan 11, 2023
Rich Polk/NBC//Getty Images
Michelle Yeoh took her moment in the spotlight after winning her first Golden Globe.
The beloved actress won tonight for playing Evelyn Wong in the mind-bending multiverse film Everything Everywhere All At Once, receiving long-deserved accolades and recognitions at this point in her decades-long career. During her speech—which she refused to finish early, even speaking back when they began to play her off—the international star reflected on her career and thanked her collaborators on the film, including its directing duo The Daniels and her co-stars Stephanie Hsu, Key Huy Quan, and Jamie Lee Curtis.
"Whew! OK. I'm just gonna stand here and take this all in. Forty years, not letting go of this," she began, before thanking the Hollywood Foreign Press.
"It's been an amazing journey and incredible fight to be here today, but I think it's worth it. I remember when I first came to Hollywood, it was a dream come true until I got here," she said, adding, "I came here and was told, 'You're a minority,' and I was like, 'No that's not possible.'"
"And then, someone said to me: 'You speak English?'" she said, getting real about the discrimination she faced as a young actress. "I mean, forget about them not knowing Korea, Japan, Malaysia, Asia, India. And then I said, 'Yeah, the flight here was about 13 hours long so I learned.'"
"As time went by — I turned 60 last year — and I think all of you women understand this as the days, years, numbers get bigger, the opportunities get smaller, as well," she continued. "I probably was at a time where I thought, 'Come on, girl. You had a really really good run. You worked with some of the best people… It's all good.' Then along came the best gift: Everything, Everywhere All at Once.'"
As the piano music began to play her off, Yeoh told it to "Shut up, please. I can beat you up, and that is serious."
She continued, "I was given this gift of playing this woman who resonated so deeply with me and with so many people. At the end of the day, no matter what universe she was at, she was just fighting for love, for her family."
The Crouching Tiger, Hidden Dragon actress arrived at the Beverly Hilton Hotel in Los Angeles earlier tonight, walking the red carpet in a gorgeous metallic number. The strapless midnight-blue gown was covered with sparkling crystals and featured a double-tiered peplum-style adornment over the floor-length column skirt. She accessorized the gown with a diamond leaf-pattern choker and a chunky diamond bracelet.
Amy Sussman//Getty Images
Quinci LeGardye
Quinci LeGardye is an LA-based freelance writer who covers culture, politics, and mental health through a Black feminist lens. When she isn't writing or checking Twitter, she's probably watching the latest K-drama or giving a concert performance in her car.
Goodbye, Red Carpet Camp. Hello, Glamour
Zendaya Addresses Golden Globes Win After Skipping
Twitter Hilariously Reacts to the Golden Globes
Twitter Has Some Thoughts on Austin's Accent
The Best Beauty Moments from the Golden Globes
The 14 Best-Dressed Stars at the Golden Globes
Did Zendaya Attend the Golden Globes?
Did Lady Gaga Attend the 2023 Golden Globes?
Did Taylor Swift Skip the 2023 Golden Globes?
Rihanna Quietly Arrived at the Golden Globes
Jessica Chastain Wears a Sheer Spiderweb Gown


Braintree Town are expected to have a new striker on board for their curtain-raiser against Farnborough on Saturday.
The Iron are closing in on a deal for a forward from a rival Blue Square South club, ahead of the new season.
They have submitted a seven-day approach for the player and expect the transfer to be wrapped up in time for this weekend's big kick-off.
Braintree have been handed a further boost with the news that striker Jamie Guy should be fit enough to take his place on the bench on Saturday.
It had been feared that the former Colchester United man would be ruled out for several weeks with a foot injury sustained in pre-season.
However, his recovery has been quicker than expected and he could play a part against Farnborough.
Braintree drew with Redbridge in their final pre-season friendly on Monday night.
For all of the latest Braintree Town news, including a full season preview, see this week's Braintree and Witham Times.


I'm always trying to do too much. I enjoy lettering, illustrating, mural painting, graphic design, and learning from other artists.
Don't forget to vote! — November 6th is coming up!
Alphabet styles — Trying to find new styles to letter in!
Calm me down with some green tea. — What's your favorite kind of tea?
Calm me down with some green tea.
Liven up your room with succulents.


text:
Star Wars Roundup
Posted on December 17, 2018 by Mike Glyer
Compiled by Carl Slaughter:
Bootleg Star Wars comic book in China takes liberties
Baylo: the officer who stood up to Darth Vader and Emperor Palpatine
More items follow the jump.
Star Wars animated shorts
"Disney introduces Star Wars to kids with a series of animated shorts" at Engadget.
If you've wanted to bring your kids into the world of Star Wars but weren't entirely sure if they were ready for the movies, Disney's upcoming animated shorts might be your answer. On November 30th, the company will debut six animated shorts on StarWarsKids.com as well as a new Star Wars Kids YouTube channel, with the goal being to introduce younger audiences to Star Wars stories. More shorts are set to follow in December and throughout 2019. Collectively called Star Wars Galaxy of Adventures, the shorts will pair new animation and scenes with dialogue, music and sound effects from the original films.
Mandalorian live action Star Wars series
Males in Last Jedi
Pop Culture Detective theorizes that fan anger at Last Jedi is rooted in dashed expectations of male heroes who are challenged and changed by female heroes.
Anakin Skywalker, The Jedi, and Stoic masculinity.
The Galactic Empire versus freedom
For the vast majority of residents of the Star Wars galaxy, daily life is unaffected by the Sith, the Jedi, and the Rebellion/Resistance. But what about life under the rule of the Empire? What exactly makes the Empire evil? What exactly are Emperor Palpatine's policies and practices that Princess Leia finds so odious, besides destroying her home planet? Disbanding the Senate? Executing child Jedi? Martial law? Excessive commercial regulation that results in the black markets and smuggling and gangs that receive so much attention in the series? Commandeering natural resource to supply such a massive military force? Drafting children into military service to fight and oppress and die for the empire? Princess Padme sums it up: "So this is how liberty dies, with thunderous applause." The Empire is a terrible place to live because its citizens' individual freedom has been suppressed in the name of maintaining galactic order.
Understanding the Star Wars backlash
Wise Crack compares the Star Wars backlash to the Protestant Reformation and Disney to the Catholic Church.
Mark Hamill's famous pitch
"Mark Hamill Confirms Crazy Pitch He Once Made For Infamous 'Star Wars' Character" at Yahoo! News.
Actor Mark Hamill, aka Luke Skywalker, confirmed an off-the-wall idea he had for 1983's "Return of the Jedi" regarding the identity of infamous "Star Wars" bounty hunter Boba Fett: He thought Fett should be Luke's mother.
Hamill replied to a fan's theory Friday, confirming that he had actually pitched the idea to "Star Wars" creator George Lucas. This was before he knew that Leia, played by Carrie Fisher, was Luke's sister.
"I once suggested it to George as the only way we could top Vader being my father. I envisioned her as a double-agent working clandestinely for the Rebels," Hamill said. "SPOILER ALERT: He didn't like my idea."
Mark Hamill reveals what killed Luke
Mark Hamill versus William Shatner
Mark Hamill: Best Star Wars fans are UPFs
This entry was posted in Like Show Business and tagged Carl Slaughter, Mark Hamill, Star Wars by Mike Glyer. Bookmark the permalink.


Ahmadiyya is offensive. They claim that Jesus died in India and thus purposely disrespect Christianity and Islam. Muslims and Christians both believe that Esa (as) was lifted physically towards God in terms of the events of the crucifixtion. Per both religions, Esa (as) is not buried anywhere on planet Earth. Mirza Ghulam Ahmad clashed with Christianity and Hinduism in the 1880's. In 1891, he began clashing with Islam and Sikhism, he criticized every single religion in India.


text:
Philip of Dreux (Philippe de Dreux; 1158–1217) was a French nobleman, Bishop of Beauvais, and figure of the Third Crusade.

He was an active soldier, an ally in the field of Philip Augustus, the French king and his cousin, making him an opponent in campaigns in France and elsewhere of Richard I of England. He was in also in demand as a priest, to make and break marriages. He presided over that of Conrad of Montferrat at the siege of Acre, marrying him to Isabella I of Jerusalem, daughter of Amalric I, whose marriage he annulled. He was also party to the annulment of the marriage between Philip Augustus and Ingeborg of Denmark.

Life
Philip was son of Robert I of Dreux and Agnes of Baudemont, and the brother of Robert II of Dreux.

He first campaigned in Palestine in 1180, in an expedition headed by Henry II of Champagne and Peter I of Courtenay. This attack on Saladin's holdings was ineffectual.

Robert II and Philip of Dreux arrived with forces in Palestine in 1189.

Richard Lionheart bore him a consistent enmity after the Crusade; Philip of Dreux had been one of those relaying the rumour that Richard was responsible for the killing of Conrad of Montferrat. Subsequently Philip had gone to Germany, when Richard was imprisoned, to advocate against setting him free. There, Philip encouraged Richard's captors to treat him poorly, earning the lifelong hatred of Richard, who considered him "a robber and an incendiary".

He was captured by Angevin forces under the mercenary leader Mercadier and Prince John in a Normandy campaign, in 1197. Richard was still refusing to release him a year later, and again early in 1199. When Peter of Capua (who was trying to enlist Richard for the Fourth Crusade) insisted that Richard release Philip, Richard exploded and threatened to castrate Peter, so intense was his hatred of his prisoner Philip. Pope Celestine III was unsympathetic to Philip, confined at Rouen and then, after an escape attempt, at Chinon. He was freed only after Richard's death in 1199, with Richard's successor, John agreeing to exchange him for the captured bishop-elect of Cambrai in 1200.

In 1210 he was in action against the Cathars in southern France, with Renaud de Mouçon, bishop of Chartres, in support of Simon de Montfort.

He drew support from Philip Augustus in his conflict against Renaud de Dammartin, leading to Renaud's 1212 alliance with John. Philip was later a combatant on the victorious French side in 1214 at the Battle of Bouvines. He took a mace to William Longsword, Earl of Salisbury, at an important moment in the battle, leading to the Earl's capture.

In his last year as bishop he founded the Pentemont Abbey, a Cistercian convent whose later buildings in Paris remain to the present day.

References

Sources

External links

1158 births
1217 deaths
Bishops of Beauvais
House of Dreux
Christians of the Third Crusade
12th-century French Roman Catholic bishops
13th-century French Roman Catholic bishops
13th-century peers of France

redpajama_set_name: RedPajamaWikipedia


text:
Lerici and around
The good position of the hotel allows you to quickly visit renowned places.
Town of Roman origin, it is considered the pearl of the Gulf of Poets. It has a medieval castle, now home to a splendid paleontological museum. Famous among english romantic poets of the nineteenth century, there are a lot of mansions, gardens, beaches and an historical centre with many original elements. From here you can take a boat to reach the Cinque Terre.
The Cinque Terre...
Almost at the eastern end of Liguria there are the Cinque Terre, towns hidden in small bays or attached at rocks overlooking the sea, in a landscape unique in the world. Riomaggiore, Manarola, Corniglia, Vernazza and Monterosso are the center of a national park declared World Heritage Site by UNESCO.
...and its town
Manarola, medieval town perched on a rock outcrop, is connectd to Riomaggiore by the splendid Via dell'Amore, a romantic pathway cut out of the sheer rock just a stone throw away from the sea. Vernazza, with its tower-like building, has narrow alleyways leading down to the anchorage which is set a beautiful bay. Monterosso, with its ancient town and beach, it's a thriving cultural centre. Corniglia is a small and pretty fraction of Vernazza.
This striking, sea village is unique of its kind in Liguria: the Doria quay is hemmed in by a number of tower-like houses which effectively turn the harbor in a fortified citadel. The islands of Palmaria, Tino and Tinetto, are located near the town and have many caves. The churches of Saint Pietro and Saint Lorenzo and the Doria castle are worth a visit.
At about 22 miles you can find the town of Colonnata, a fraction of Carrara, located on the slopes ot the Apuan Alps: it is famous around the world for the lardo. On the way you can admire the famous cave of the very sought-after Carrara Marble, also used by the sculptor Michelangelo Buonarroti.
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