Most programming languages have functions for getting bits but I need to do it on a calculator so I need to understand how to do it mathematically. Basically I need a formula for getting the $n$th bit from and integer, $i$.
So given: $i = 6$ and $n = 3$, I would get $1$ because binary for $6$ is $110$.
How might I manage this?
Given $i$ you want to throw away the bottom $n-1$ bits, so integer divide by $2^{n-1}$. Then take $\bmod 2$ to get the low order bit.
Another very useful formula for the $k$-th bit of number $b=b_{n}\cdots b_{0}$ is $$b_{k}=\left\lfloor\frac{b}{2^k}\right\rfloor-2\left\lfloor\frac{b}{2^{k+1}}\right\rfloor.$$ This identity can be easily verified by using that, for $b=\sum_{i=0}^{n}b_{i}2^{i}$, we have $\left\lfloor\frac{b}{2^k}\right\rfloor=\sum_{i=k}^{n}b_{i}2^{i-k}$.
b>>k, and $\left\lfloor\frac{b}{2^{k+1}}\right\rfloor$ to b>>k+1, which truncates the last bit after the division by $k$, $2*\left\lfloor\frac{b}{2^{k+1}}\right\rfloor$ is equivalent to b>>k+1<<1, which is b>>k&~1, and (b>>k)-(b>>k&~1)=b>>k&1, the $k$th bit.
– DroneBetter
Jul 14 '23 at 14:48
If you can use mod then its easy: ((value - (value mod (2^n))) / (2^n)) mod 2
Otherwise bit0 is (1+(-1^(value+1)))/2 - applicable for any integer value or any function that returns integer value. Attempts to emulate mod of a function may lead to result which also contain mod operation.
To get n-th bit,
first lets shift value/function: shift right is (valuable - bit 0) / 2 - apply n times.
next get bit0 as described.
Most calculators have a way to truncate fractions. So to get bit #3 here are the steps
Example: Find the bit #5 of $111$.
$$2^5=32$$ So $$ 111/32=3.46875 $$ The truncated value is $3$. Now, $3$ divided by $2$ leaves $1$ as remainder (odd number). So bit #5 is $1$.
Repeat for bit #4. $$ 2^4=16$$. So $$ 111/16 = 6.9375 $$ Truncated answer is $6$ which is even. So bit #4 is zero.
