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(5.46)

EXAMPLE 26-1

(continued)

z 3 2 = 00100101

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Figure 21.9 (a) Learning XOR using the 2-input MLP denoted in the figure as MLP(2), 4-input MLP, denoted as MLP(4), quaternion-valued geometric nets GMLP0 2 0 and GMLP2 0 0 and P-QMLP; (b) comparison of the geometric neural network GMLP0 2 0 , the 2-input MLP, denoted MLP(2), and the 4-input MLP, denoted MLP(4).





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The example responds to all errors that might arise when loading the CSS le from the internet, in onLoadError(). If there s an error parsing the CSS for any reason, you catch the errors in the catch block of onLoadSuccess(). A timer is also used to stop waiting for a long load (perhaps a bit contrived, as a very long delay will eventually trigger a load error on its own), calling onTimeout() if the program is tired of waiting. All of these log a message with trace(), and all of them fall back on a recovery strategy using a built-in stylesheet, set in setDefaultStyle(). Another approach to this problem is to add the text into the text eld at the beginning and redraw the text with the appropriate style after the stylesheet loads successfully. This way, the user can see the content without delay. On the other hand, it might produce a visible icker as the text is redrawn. These kinds of considerations are at the core of an error-handling strategy.

00010101 = 00110000,

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- 2 2 = 1 - 0 and zh - z3 = 0 - 1. From Vx, the XOR difference is easily computed. In addition, Problem 21 shows that Vx determines the modular difference. Consequently, all of the relevant difference information is contained in Vz. However, for convenience we retain the modular difference. We refer to Wang s precise differential as the signed differential. This differential provides more information than the modular and XOR differential combined, and, therefore, it allows us to have correspondingly greater control over the results at each step. However, it is important to realize that there is still a great deal of freedom in choosing values that satisfy a given signed differential. Consider, for example, 8-bit bytes y and y. Then the modular difference is y - y (mod a8). Suppose Vy is specified. Then for each + in Vy, the corresponding bit of y must be 1, and the corresponding bit of y rnust be 0. Similarly, for each - in Vy, the corresponding bit of y must be 0, and the corresponding bit of y must be 1. But for any bit position that is not a 1 in y @ y the bits of y and y must agree, and the value, of the bits in each such position is arbitrary. That is, we have one bit of freedom for each . i n Vy, and no freedom whatsoever for each + and - . To make this more concrete, suppose we require that Vz be . .+-. . . . . Then z and z in (5.46) satisfy this requirement, as do each of t he pairs

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