the input configuration is A.B.~C (in this case, AB = -2). All other sums of excitation are similarly defined. The fibers from input A may be separated into four parts: 1. Ac is the sum of excitations from input A from fibers NOT inhibited by either inputs B or C. 2. A is the sum of excitations from input A from fibers NOT inhibited by input B, but which may be inhibited by input C. 3. A is defined similarly to Ag 4. A is simply the total sum of excitations from input A. To make clear the reason for the definitions, we consider a particular 2-Venn diagram which we label as follows: Venn A Imagine that a fiber from input A must break up into four parts and pass through this 2-Venn before ending on a neuron, as in figure 9. After each fiber leaves the Venn diagram, it may break up into many parts and become either excitatory or inhibitory on the neuron. The two boundaries of the areas of the Venn diagram represent inhibitions of the fibers through them from inputs B and C. In a space of the 2-Venn we write a number which equals the numerical value of excitation of the neuron from the fiber passing through that space. As an example, we write for the 2-Venn of figure 9 representing fibers from input A on the neuron: Venn A and our definitions become: 43, 4=3+1=4, A=2+3=5, A=2+1+3-1=5. All 3 inputs to a neuron may be represented in this manner. Figure 9. Inhibitions from B and C may be viewed as forming a Venn diagram. Returning to our problem of constructing a neuron for figure 10; we write, for example, AB-A_+B since only those fibers from A not inhibited by B, and those BA' fibers from B not inhibited by A, will excite the neuron when ABC is the input configuration. Clearly, the ~A~B.~C space of the Venn diagram must contain a zero, since for this input configuration, the neuron receives no excitation. The equations of figure 10 indicate that five variables may be chosen arbitrarily. For example, let: Figure 10. An example of a Venn diagram for which a neuron must be constructed. For each input to the neuron there exists a 2-Venn that describes the number of excitations and inhibitions on a neuron and how these fibers are inhibited. Thus, if there exist three 2-Venns that satisfy the above equations, then the neuron may certainly be constructed. Suppose the 2-Venns are filled by first satisfying equation 7, then equation 6, and so on up to equation 1. This might be done as follows: It is readily seen that the number in each space of every 2-Venn is determined by the value assigned to each of the 12 variables of equations 1-7. Moreover, there is always a sufficient number of spaces in the 2-Venns to satisfy the equations since the number of variables is precisely equal to the number of spaces in the three 2-Venns. In our example, the 2-Venns are: 网 B Venn C The neuron which may be constructed directly from these Venns is: ( 9 ) + 2 ( 2 ) + ♪(§). A (8-1)-Venn has 28-1 In general, the number of variables such as A, A, CAB, etc. will be + spaces; therefore, the total (2-1). In a note on page 17 we such as those of equations 1-7, may always be chosen at will to construct the required (S-1)-Venns, and a neuron may always be constructed for any such choice. As a result, we have the Theorem: Excitatory and inhibitory fibers to a neuron plus inhibitory interaction among them is sufficient for the construction of a neuron which computes the logical functions, for changing thresholds, described by any Venn diagram which requires spontaneous firing of the neuron for 0≤0. An interesting problem is the construction of neurons which are cheapest in the sense that they receive a minimum number of fibers from the input. Suppose, for example, that excitatory and inhibitory fibers and inhibitions of input fibers cost 14 apiece. In the special case of = 3, the three 2-Venns are: F = | ABC + 2 | AC - ABC + 24 - ABC + 3 | 4 - Ag - AC + ABC | + BAC + 2 | BC - BAC|+2|AB - A 3 | B + Ag – AB – Bc + BAC | + | ABC - BAC + ABC which may be solved to obtain the cheapest neuron. We have no explicit method for solving this equation for all 8. |