Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
If $a$ and $b$ are different elegant programs (minimal program for some output), is their joint Kolmogorov complexity the sum of their individual complexities, i.e. $K(a,b) = K(a) + K(b)$?
$\begingroup$@YuvalFilmus A Turing machine $M$ is elegant if $|\langle M\rangle|\leq|\langle M'\rangle|$ for every other Turing machine $M'$ that computes the same function.$\endgroup$
$\begingroup$I'd expect the answer to be "no". Most strings are incompressible, so it should be possible to find two very similar strings for which the shortest program is just print "The string". But, now, the shortest program that prints both strings is essentially print "The string" twice except make these small changes the second time, giving $K(a,b) = K(a) + \text{a bit} \ll K(a)+K(b)\approx 2K(a)$. In general, $K(a,b)=K(a)+K(b|a)+O(\log K(a,b))$.$\endgroup$
print "The string"
. But, now, the shortest program that prints both strings is essentiallyprint "The string" twice except make these small changes the second time
, giving $K(a,b) = K(a) + \text{a bit} \ll K(a)+K(b)\approx 2K(a)$. In general, $K(a,b)=K(a)+K(b|a)+O(\log K(a,b))$. $\endgroup$