In the setting of error-correcting codes with feedback, Alice wishes to communicate a $k$-bit message $x$ to Bob by sending a sequence of bits over a channel while noiselessly receiving feedback from Bob. It has been long known (Berlekamp, 1964) that in this model, Bob can still correctly determine $x$ even if $\approx \frac13$ of Alice's bits are flipped adversarially. This improves upon the classical setting without feedback, where recovery is not possible for error fractions exceeding $\frac14$. The original feedback setting assumes that after transmitting each bit, Alice knows (via feedback) what bit Bob received. In this work, our focus in on the limited feedback model, where Bob is only allowed to send a few bits at a small number of pre-designated points in the protocol. For any desired $\epsilon > 0$, we construct a coding scheme that tolerates a fraction $ 1/3-\epsilon$ of bit flips relying only on $O_\epsilon(\log k)$ bits of feedback from Bob sent in a fixed $O_\epsilon(1)$ number of rounds. We complement this with a matching lower bound showing that $\Omega(\log k)$ bits of feedback are necessary to recover from an error fraction exceeding $1/4$ (the threshold without any feedback), and for schemes resilient to a fraction $1/3-\epsilon$ of bit flips, the number of rounds must grow as $\epsilon \to 0$. We also study (and resolve) the question for the simpler model of erasures. We show that $O_\epsilon(\log k)$ bits of feedback spread over $O_\epsilon(1)$ rounds suffice to tolerate a fraction $(1-\epsilon)$ of erasures. Likewise, our $\Omega(\log k)$ lower bound applies for erasure fractions exceeding $1/2$, and an increasing number of rounds are required as the erasure fraction approaches $1$.

翻译：在设置有反馈的错误校正代码时,爱丽丝希望通过在频道上发送一系列比特(kkk$-bit)信息($xx$)向鲍勃传递一个美元比特的信息,同时在一条频道上发送一系列比特,不受噪音影响地接收鲍勃的反馈。人们早就知道(Berlekamp, 1964),在这个模型中,鲍勃仍然可以正确地确定美元x美元,即使爱丽丝的位数是翻转的对称值。如果任何想要的 $\acrox1\forc13$的比特级设置没有反馈,那么对于错误分数不能超过$1/3-plc14$。最初的反馈假设假设,爱丽丝(通过反馈)在每分送一个比特的比特点上,我们只允许鲍勃在协议中少量的预指定点上发送几位数。对于任何想要的 $=lon > 美元比特的比值, 我们的比值(1/3-plon$) 的比值只能依靠$的比特的反馈(我们的比值) ===xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx