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.viminfo
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# This viminfo file was generated by Vim 7.3.
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# Value of 'encoding' when this file was written
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# hlsearch on (H) or off (h):
~h
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?/invoke
?/instance-class
?/oo-apply-instance
?/create-class
?/define (get
?/make-instance
?/define new
?/define (new
?/new
?/object
?/root
?/root-
?/root-object
?/define (make
?/make-oo-eval
?/problem4
?/instance?
?/application?
?/oo-apply-inst
?/oo-apply-in
?/define (procedure-body
?/compound-p
?/comp
?/displayln
?/eval-make-class
?/begin
?/find
?/(define (create
?/get-methods
?/problem3
?/test
?/CONSTRUCT
?/nothing-yet
?/(define (new
?/PROBLEM 2
?/problem 2
?/PROBLEM 3
?/define (comp
?/not a com
?/wrong
?/method-alist
?/lambda?
?/method1
?/test-problem2-simple
?/define (oo-eval
?/eval-makjjk
?/make-class?
?/eval
?//instance?
?/class?
?/test-problem
?/make
?/method-call
?/makejk
?/problem 1
?/brains
?/console
????
?/statickjjk
?/google.maps
?/zoom
?/diss
?/dropdown
?/heatmap
?/style
?/{
?/index
?/jQuery
?/wild
?/statusline
?/cc
?/pdf
?/army of two
?/per f
?/colorscheme
?/make-frame
?/frame
?/pick-random
?/george
?/zombie-mixin
?/zombie-
?/autonomous
?/create-zombie
?/zombie
?/root-class
?/named-obj
?/get-class
?/create-class?
?/instance-add-mixin!
?/error
?/define (create
?/write-slot
?//write-sl
?/default-me
?/disl
?/eval-make
?/require
?/avatar
?/setup
?/define (me
?/me
?/weapon
?/chainsaw
?/defa
?/shadow
?/make-from
?/(define (make-frame
?/instance-st
?/apply-me
?/define (oo-appl
?/make-primitive-procedure
?/define (make-fram
?/ma
?/default-metaclass
?/make-procedure
# Expression History (newest to oldest):
# Input Line History (newest to oldest):
# Input Line History (newest to oldest):
# Registers:
"0 CHAR 0
for i in chars:
print hex(i)
"1 LINE 0
"2 LINE 0
The two cyphertexts are:
"3 LINE 0
\end{enumerate}
"4 LINE 0
Here we are asked to provide the information
"5 CHAR 0
\answer{1-3 - Detecting Pad Reuse}
\begin{enumerate}[\bf (a)]
\item Left out
\item Those run lengths are generally lower than $\log_2 n + \log_2 \ln n$, but this is irrelevant,
because assuming the text is encoded in \textsc{US-ASCII}, each character is going to be represented
by 8 bits. Therefore longest runs of corresponding bits in binary message are going to be much
higher than random string of bits statistical upper bound $\log_2 n + \log_2 \ln n$ for sufficiently high $n$. Few example values:
\begin{center}
\begin{tabular}{cc}
$\log_2 n + \log_2 \ln n$ & \textbf{longest run (in bits)} \\
\hline
7&8\\
14&16\\
19&40\\
25&64\\
\end{tabular}
\end{center}
Since the statistical properties of English text are inconsistent with hypothesis for random bits, it
can be exploited to solve next exercise.
\item We are given $k = poly(n)$ ciphertexts $C_1, ..., C_k$ each of length $n$ bits. Let's make two observations:
\begin{enumerate}[1.]
\item Form \textbf{(a)} we know that if we have two cyphertexts $C_i$ and $C_j$ such that they were both
encrypted using different OTP, their longest run of the same bits on the same positions will have
length less than $\log_2 n + \log_2 \ln n$ with high probability.
\item From \textbf{(b)} we know that if we have two cyphertext $C_i$ and $C_j$, which were both encrypted
with the same key then the messages, then the corresponding messages $M_i$ and $M_j$ will have
longest run of the same bits on the same positions much higher than $\log_2 n + \log_2 \ln n$ with
high probability. However assuming character on position $u$ is the same for $M_i$ and $M_j$, then
we know that $M_i(u) = M_j(U) \Rightarrow M_i(u) \oplus K = M_j(u) \oplus K \Rightarrow C_i(u) = C_j(u)$,
so encrypting with the same OTP, preserves the length of the longest run. Therefore longest run
for $C_i$ and $C_j$ will be much longer than $\log_2 n + \log_2 \ln n$ for large enough n with high probability.
\end{enumerate}
Combining above observations we deduce that there will exist a pair of ciphertexts, with much longer
longest run of the same bits on the same positions and this pair will be the one that was encoded with the same
OTP. The task is reduced to given $k$ strings finding the two that have the longest run of the same characters
on the same positions. This can be done with the following algorithm:
\paragraph*{Algorithm 1 - Finding two strings with the longest run}
\begin{enumerate}[1.]
\item Given $C_1, ..., C_k$:
\item Extend the alphabet $\Omega$ to $\Omega' = \Omega \cup \{a_1, .. a_{k}\} $ such that $a_1, ..., a_{k}$
are not in $\Omega$.
\item Construct $S = C_1 | a_1 | C_2 | a_2 | ...|C_{k-1} |a_{k-1}| C_k | a_k$, where $|$ represents string
concatenation.
\item Construct suffix tree ($T_S$) for $S$
\item Find deepest non-tree node in $T_S$ and return corresponding positions $p_1, ... p_l$.
\item Return all the substrings $C_i$ such that one of the $p_1, ..., p_l$ corresponds to the
position in $S$ where $C_i$ was concatenated.
\end{enumerate}
"6 CHAR 0
\answer{1-1 - Security Policy for Github}
\paragraph*{Security objective} - protect privacy of users, private repos, always allow users
with correct access rights to access (availability) and prevent unuthorized users from writing to repos.
- allow for auditing
\paragraph*{Principals by role}
Spell out each of these
- admins, repo owner, repo contributor, random user, etc.
\paragraph*{actions}
creating private/public repo, forking, merging, creating user account, accesing repo contents,
sending email to users, requesting a webpage of github, pushing to repo, submiting an issue,
editing issue (only submitter, to prevent impersonation), etc.
Account Creation And Authentication
\paragraph*{Repositores}
private/public repo, forked repo, commit, github datacenter machines, etc.
\paragraph*{CIA}
Make sure we covered all of confidentiality, integrity, availability(DDOS prevention - limit number of accesses to a resource)
\\ What are we preventing versus what are we only detecting
"7 LINE 0