Translating formulas into clausal form (1) forall x (reads(x) -> literate(x)) !reads(x) or literate(x) (2) forall y (dolphin(y) -> !literate(y)) !dolphin(y) or !literate(y) (3) exists z (dolphin(z) and intelligent(z)) dolphin(a) and intelligent(a) (3.1) dolphin(a) (3.2) intelligent(a) Conclusion: (4) exists w (intelligent(w) and !reads(w)) negated conclusion: !(exists w (intelligent(w) and !reads(w))) forall w (!intelligent(w) or reads(w)) !intelligent(w) or reads(w) Proof by refutation using resolution: (3.2) intelligent(a) (4) !intelligent(w) or reads(w) ----------------------------------------------------------- (5) reads(a) (substituting w by a) (2) !dolphin(y) or !literate(y) (3.1) dolphin(a) ----------------------------------------------------------- (6) !literate(a) (substituting y by a) (1) !reads(x) or literate(x) (5) reads(a) ----------------------------------------------------------- (7) literate(a) (substituting x by a) (6) !literate(a) (7) literate(a) ----------------------------------------------------------- (8) empty clause
Operator Patrick / Thomas | Sally \ / | \ Overtime Output = low Output = hight No / Yes \ 3, 6, 7 5 / \ Output = high Output = low 1, 4 2, 8 (1) For the first attribute Operator: 4/8*(-2/4*log(2/4)-2/4*log(2/4))+3/8*(-3/8*log(1))+1/8*(-1/1*log(1)) = 0.5 Machine: 2/8*(-1/2*log(1/2)-1/2*log(1/2))+3/8*(-2/3*log(2/3)-1/3*log(1/3))+ +3/8*(-2/3log(2/3)-1/3log(1/3)) = 0.94 Overtime: 3/8*(-3/3*log(1))+5/8*(-3/5*log(3/5)-2/5*log(2/5)) = 0.61 So operator is selected. (2) For Operator = Patrick Machine: 2/4*(-1/2*log(1/2)-1/2*log(1/2))+2/4*(-1/2*log(1/2)-1/2*log(1/2)) = 1 Overtime: 2/4*(-2/2*log(2/2))+2/4*(-2/2*log(2/2)) = 0 So Overtime is selected.
from x(AI0) = a * x(AI1) + b * x(AI2) and x(BI0) = a * x(BI1) + b * x(BI2) we get 0.8 = a * 0 + b * 1 and 1.4 = a * 1 + b * 1 hence, a = 0.6, b = 0.8. (we are using a for alpha and b for beta here) so x(CI0) = a * x(CI1) + b * x(CI2) = 0.6 * 1 + 0.8 * (-2) = -1 x(DI0) = a * x(DI1) + b * x(DI2) = 0.6 * 0 + 0.8 * (-2) = -1.6 Because the expected value of the x coordinate of point D of the unknown object (-1.6) does not match the actual value of that coordinate (-1.8), the unknown object does not fit the templates.
+-----------+-----------+---------+---------+---------+---------+----------+ | Variables | Initial W | ex. 1 | ex. 2 | ex. 3 | ex. 4 | Final W | +-----------+-----------+---------+---------+---------+---------+----------+ | A | | 0.000 | 0.000 | 1.000 | 1.000 | | | B | | 0.000 | 1.000 | 0.000 | 1.000 | | | Des. Out | | 0.000 | 1.000 | 1.000 | 0.000 | | +-----------+-----------+---------+---------+---------+---------+----------+ | W a->c | 0.100 | 0.100 | 0.100 | 0.100 | 0.100 | 0.100 | | W a->d | 0.200 | 0.200 | 0.200 | 0.200 | 0.200 | 0.199 | | W b->c | -0.100 | -0.100 | -0.100 | -0.100 | -0.100 | -0.100 | | W b->d | 0.050 | 0.050 | 0.050 | 0.050 | 0.050 | 0.049 | | W c->e | -0.200 | -0.200 | -0.200 | -0.200 | -0.200 | -0.204 | | W d->e | 0.300 | 0.300 | 0.300 | 0.300 | 0.300 | 0.295 | +-----------+-----------+---------+---------+---------+---------+----------+ | C Output | | 0.3775 | 0.3543 | 0.4013 | 0.3775 | | | D Output | | 0.3775 | 0.3894 | 0.4256 | 0.4378 | | | E Output | | 0.5094 | 0.5115 | 0.5118 | 0.5140 | | +-----------+-----------+---------+---------+---------+---------+----------+ | Beta C | | 0.0255 | -0.0244 | -0.0244 | 0.0257 | | | Beta D | | -0.0382 | 0.0366 | 0.0366 | -0.0385 | | | Beta E | | -0.5094 | 0.4885 | 0.4882 | -0.5140 | | +-----------+-----------+---------+---------+---------+---------+----------+ | D W a->c | | 0.0000 | -0.0000 | -0.0059 | 0.0060 | | | D W a->d | | -0.0000 | 0.0000 | 0.0089 | -0.0095 | | | D W b->c | | 0.0000 | -0.0056 | -0.0000 | 0.0060 | | | D W b->d | | -0.0000 | 0.0087 | 0.0000 | -0.0095 | | | D W c->e | | -0.0481 | 0.0433 | 0.0489 | -0.0485 | | | D W d->e | | -0.0481 | 0.0475 | 0.0519 | -0.0562 | | +-----------+-----------+---------+---------+---------+---------+----------+ Sample Computations for example 1, i.e. A = 0, B = 0: (here e^y means e to the power y) Oc = 1/[1 + e^{y}] where y = A*Wac + B*Wbc + (-1)*1/2 = 0.377541 = 0*0.1 + 0*(-0.1) - 0.5 = -0.5 Od = 1/[1 + e^{y}] where y = A*Wad + B*Wbd + (-1)*1/2 = 0.377541 = 0*0.1 + 0*(0.05) - 0.5 = -0.5 Oe = 1/[1 + e^{y}] where y = Oc*Wcd + Od*Wde = 0.509437 = 0.377541*(-0.2) + 0.377541*(0.3) New Wde (after processing all 4 examples): New Wde := 0.300 -0.0481 + 0.0475 + 0.0519 -0.0562 = 0.295
Part 1 Tent type Quality Standard Fitness T1 5 5/30 = 0.167 T2 10 10/30 = 0.333 T3 15 15/30 = 0.5 sum(qi) = 5 + 10 + 15 = 30 Part 2 Tent type Quality Standard Fitness T1 41 41/150 = 0.273 T2 50 50/150 = 0.333 T3 59 59/150 = 0.393 sum(qi) = 41 + 50 + 59 = 150 Part 3 From Parts 1 and 2 above it's clear that, when the stardard selection method is used, the probability of selecting an individual changes depending on the units in which the fitness is measured (C or F). This is a drawback of the stardard method. Celsius 0.5/0.167 = 2.99 Fahrenheit 0.393/0.273 = 1.44 Note that, with the stardard method, the probability of selecting the fittest individual (T3) is almost 3 times the probability of selecting the least fit individual (T1) when Celsius degrees are used, but only 1 and a half times when Fahrenheit degrees are used.