How lower limit.? Shouldn't t exactly be $2^{2n}$ when $f(a,b)$ is exact?

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5 votes

A computer program computes a function $\: f \: \{0, 1\}^* \times \{0, 1\}^* \rightarrow \{0, 1\}^*$. Suppose $f(a, b)$ ahs length $\mid b \mid ^2$, where $\mid a \mid$ and $\mid b \mid$ are the lengths of $a$ and $b$. Suppose, using this program, the following computation is performed.

x="01" for i=1, ... , n do x=f("01", x)

Suppose at the end, the length of the string $x$ is $t$. Which of the following is TRUE (assume $n \geq 10)?

- $ t \leq 2n$
- $n < t \leq n^2$
- $n^2 < t \leq n^{\log_2 n}$
- $ n^{\log_2 n} < t \leq 2^{(2n)}$
- $2^{(2n)} < t$

0 votes

Initially $x$ is of $2$ bits.

In next iteration of loop $x$ has size $4$

lly, after $n$ iterations(at the end of loop ) the size of $x$ is $2^{(2*2*2*2*.......n \space times)}$

ie, $t=2^{2^n}$

option $A, B, C$ is wrong as they are small compared to $t$.

option $E$ is wrong as it exceeds the value of $t$.

So, answer is option $D$

In next iteration of loop $x$ has size $4$

lly, after $n$ iterations(at the end of loop ) the size of $x$ is $2^{(2*2*2*2*.......n \space times)}$

ie, $t=2^{2^n}$

option $A, B, C$ is wrong as they are small compared to $t$.

option $E$ is wrong as it exceeds the value of $t$.

So, answer is option $D$