@ -50,7 +50,7 @@ Or by the composition of all gate applications up to this point: $(f_l \circ f_{
\section{A Bit of Randomness}
\subsection{Single Bits in Superposition}
Many real world problems are believed (if $\mathbf{P}\neq\mathbf{NP}$) to don't be efficiently solvable on fully deterministic computers like the model described above. Fortunately, it turns out that if we allow for some randomness in our algorithms, we're often able to efficiently find solutions for such hard problems with sufficiently large success probabilities. Often times, the error probabilities can even be made exponentially small. For this reason, we also want to introduce randomness into our model. Algorithms or computational models harnessing the power of randomness are usually called \emph{probabilistic}.
Many real world problems are believed to not be efficiently solvable on fully deterministic computers like the model described above (if $\mathbf{P}\neq\mathbf{NP}$). Fortunately, it turns out that if we allow for some randomness in our algorithms, we're often able to efficiently find solutions for such hard problems with sufficiently large success probabilities. Often times, the error probabilities can even be made exponentially small. For this reason, we also want to introduce randomness into our model. Algorithms or computational models harnessing the power of randomness are usually called \emph{probabilistic}.
Again, we start with simple one bit systems. Later, we'll see how to expand the following methods to full bit vectors/registers. In the deterministic single bit model above, the state transition of a bit $b$ in step $t$ is defined by $f_t(b)\in\{0,1\}$. Now, the transition function (or gate) is simply allowed to flip an unfair coin and either output 0 or 1 for heads or tails respectively. Of course, the state of $b$ prior to the transition should have an effect on the computation. That is, why we allow different (unfair) coins for either $b =0$ or $b =1$. To distinguish between deterministic and probabilistic transition functions, we will denote the latter by $\ptrans(b)\in\{0,1\}$. Or to reformulate this idea: Depending on the value of $b$, the output of $\ptrans(b)$ follows one of two Bernoulli trials. There are 4 possible transitions with probabilities $p_{00}$, $p_{01}$, $p_{10}$ and $p_{11}$, where $p_{ij}$ is the probability of $b$ transitioning form $i$ to $j$. Obviously, $\sum_j p_{ij}=1$ always needs to be satisfied.
Tom Krüger