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    <p>If your numbers are discrete, and properly spaced, and drawing from a limited subset, there are other possibilities.</p> <p>If you're guaranteed that a, b, and c are integers for instance, then you can multiply the vector by itself to get an odd power and then dot it with &lt;1, 1, 1>. If we multiply it by itself 4 times, for instance, it will give you &lt; a^5, b^5, c^5 >. If |a| is the largest and |a|=2, then we know that b and c will be 1 or 0, so the value of a^3 will dominate and the dot product will have its sign. For instance, if X= &lt; a=-2, b=1, c=0 > , then X^5 = &lt;-32, 1, 0>. When you dot this with &lt;1, 1, 1> you get -31, whose sign reflects that of the largest absolute value. As the absolute value of the largest number increases, the disparity between it and the other terms will tend to converge - for instance, if we have &lt;-8, 7, 7>, then the algorithm above gives X^5=&lt;-32768, 16807, 16807>, you dot that with &lt;1, 1, 1> and get 846, so the algorithm fails with exponent 5. If we bump the exponent up to 7, we get &lt;-2097152, 823543, 823543>, dotted with &lt;1, 1, 1> gives us -450066, which is the correct answer. Eventually round-off errors will also break this method. But I'm hoping it might give some insights into other alternatives, if you know the limits on your dataset.</p> <p>As a footnote, remember that X^5 = (X*X) * (X*X) * X, so you do one multiply to get X^2, multiply that by itself to get X^4, and then multiply by X - three multiplies total. You need an odd exponent to preserve sign.</p>
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    1. USBenny Smith
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