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SU-CS254 MAR032025

Last edited: August 8, 2025

Today: let’s smash and together. Recall:

Recall:

\(x \in L \implies \exists y, V\qty(x,y) = 1\), \(x \not \in L \implies \forall y V\qty(x,y) = 0\)

\(x \in L \implies \text{Pr}\qty [V\qty(x,r) = 1] \geq \frac{2}{3}, x \not\in L \implies \text{Pr}\qty [V\qty(x,r) = 0] \geq \frac{2}{3}\)

Consider a new quantifier:

SU-CS254B APR072025

Last edited: August 8, 2025

Recall the relationship between time and space; in particular:

\begin{equation} \text{TIME}\qty(t\qty(n)) \subseteq \text{SPACE}\qty(t\qty(n)) \end{equation}

“space is more valuable than time.”

On time vs. space

\begin{equation} \text{TIME}\qty(\qty(n)) \subseteq \text{SPACE} \qty( \frac{t\qty(n)}{\log \qty(t\qty(n))}) \end{equation}

\begin{equation} \text{TIME}\qty(\qty(n)) \subseteq \text{SPACE}\qty(t\qty(n)) \end{equation}

that is: space is strictly more valuable than time. Oh, and also:

\begin{equation} \text{TIME}\qty(t\qty(n)) \subseteq \text{SPACE}\qty(\sqrt{t\qty(n) \log t\qty(n)}) \end{equation}

Key idea: let’s transform the previous thing into a graph problem!


Suppose \(M\) is the time \(T\) truing machine, and consider its configure at time \(t < T\). Key idea: we can tradeoff time for space! notably, we will solve the same problem using a different algorithm, not modify the original machine.

SU-CS254B APR092025

Last edited: August 8, 2025