kube%cogsci@UCBVAX.BERKELEY.EDU (Paul Kube) (05/09/86)
From AIList Digest V4 #119: >Date: Wed, 7 May 86 10:37:36 EDT >From: Bruce Nevin <bnevin@cch.bbn.com> >Subject: net intelligence >A recent letter in Nature (318.14:178-180, 14 November 1985) illustrates >nicely how behavior of a whole may not be predictable from behavior of >its parts. ... >The reductionist engineering prediction would be that the fish could >respond no more quickly than its I/O devices allow, 2*10E-6 seconds. >>From the reductionist point of view, it is inexplicable how the fish >in fact responds in 4*10E-7 seconds. Somewhat reminiscent of the old >saw about it being aerodynamically impossible for the bumblebee to fly! ... > Bruce E. Nevin bnevin@bbncch.arpa Of course, if you have a bad enough theory, it can get pretty hard to figure out bumblebees, fish, or anything else. In this case, however, predicting the behavior of the whole from the behavior of the parts requires nothing more than the most elementary signal detection theory. First, note that the fish does not respond in 4*10^-7 seconds: the latency of the jamming avoidance response (JAR) is only 3*10^-1 seconds. What the fish is able to do is reliably detect temporal disparities in signal arrival on the order of 4*10^-7 seconds, and to do this with arrival-time detectors having standard deviation of error no better than 1*10^-5 seconds. The standard, `reductionist engineering' explication of this goes as follows: The fish has 3*10^-1 seconds to initiate JAR. In this time, it can observe 100 electric organ discharges (EOD's) from the offending fish; its job is to reliably and accurately (>90% confidence within 4*10^-7 seconds) figure out disparities in arrival times of (some component of) the discharges between different regions of its body surface. This will be done by taking the difference in firing time of discharge-arrival detectors which have standard deviation of error of 1*10^-5 seconds. It is well known that the average of N observations of a normally distributed random variable with standard deviation sigma is sigma * sqrt(N) / N; so here the average of the 100 observations of arrival time of a single detector will have standard deviation sqrt(100)/100 * 1*10^-5 = 1*10^-6 seconds (and so 95% confidence intervals of two standard deviations = 2*10^-6 seconds, as reported by Rose and Heiligenberg). Since the standard deviation of the difference of two identically normally distributed random variables is twice the standard deviation of either variable, the temporal disparity measurement has 95% confidence interval of 4*10^-6 seconds. But that's only one pair of detectors, and the fish is paved with detectors. If you want to reduce the 95% confidence interval by another order of magnitude, you just need to average over 100 suitably located detector pairs. (Mechanisms exploiting this fact are also almost certainly responsible for some binaural stereo perception in humans, where the jitter in individual phase-sensitive neurons is much worse than what's required to reliably judge which ear is getting the wavefront first.) --Paul Kube Berkeley EECS kube@berkeley.edu