I am trying to compute certain score, which is the sum of all variables (e.g., $a$, $b$, $c$). The higher the value of a variable, the score will be higher. However, I have one variable, say $a$, which is in the unit of seconds (time), representing the speed. I expect the score to be higher if it is in higher speed, which means I will need to inverse/reverse the value of $a$.

There are two ways which I can think of:

(1) $ \frac{1}{x} $

(2) $ 1 - (\frac{1}{1 + x}) $

Any comments if these are the right ways?


closed as unclear what you're asking by JMoravitz, Namaste, Paul Frost, Xander Henderson, HK Lee Sep 1 '18 at 2:38

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  • 1
    $\begingroup$ The first function is decreasing on $\Bbb R^+$ while the second is increasing. They will give opposite sort orders. $\endgroup$ – dxiv Aug 31 '18 at 19:31
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    $\begingroup$ How about more simply multiplying everything in the list by $-1$... $\endgroup$ – JMoravitz Aug 31 '18 at 19:41
  • $\begingroup$ Just checking: this is not about sorting the list of numbers? $\endgroup$ – Robert Soupe Sep 1 '18 at 0:22
  • $\begingroup$ no, it is not about sorting. $\endgroup$ – Stuart Peterson Sep 1 '18 at 7:29

(2) will not work. It will leave the numbers in the same order as before. (1) works fine. So will $-x$, or if you want positive numbers $a-x$ where $a$ is larger than the largest of your values, or $-x^3$ or many other things. You just need some decreasing function of $x$ and you are there. Without another criterion nobody can say which is the right way.

  • $\begingroup$ another criterion? such as? $\endgroup$ – Stuart Peterson Aug 31 '18 at 21:00
  • $\begingroup$ What else you want. I gave a few examples. All will reverse the ordering of the input. How would you choose between them? $\endgroup$ – Ross Millikan Sep 1 '18 at 0:19

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