(41) Round To The Nearest 1/n
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04-12-2018, 01:10 PM
Post: #1
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(41) Round To The Nearest 1/n
Original Blog Entry: https://edspi31415.blogspot.com/2018/04/...arest.html
The program RNEAR rounds a number x to the nearest 1/n. For example, to round x to the nearest 10th, n = 10. To round to the nearest 16th, n = 16. Keystrokes: x [ENTER] y [XEQ] [ALPHA] RNEAR [ALPHA] Code:
Examples: x = π n = 10, result: 3.10 (nearest 10th) n = 1000, result: 3.142 (nearest 1000th) n = 16, result: 3.125 (nearest 16th) |
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04-12-2018, 06:12 PM
(This post was last modified: 04-12-2018 07:04 PM by hth.)
Post: #2
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RE: (41) Round To The Nearest 1/n
What is the magic instruction RFC? I am not familiar with it.
Håkan |
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04-12-2018, 06:57 PM
(This post was last modified: 04-12-2018 07:12 PM by Dieter.)
Post: #3
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RE: (41) Round To The Nearest 1/n
(04-12-2018 06:12 PM)hth Wrote: What is the magic instruction RFC? I do not familiar with it. I think that's supposed to be a "FRC". But, Eddie: why do you split x into integer and fractional part? Why should this be required? Hint re. your blog post: int(x+0,5) rounds to the nearest integer. This even works on a TI-59. ;-) But there is also a way to round a number to display precision: EE INV EE does it. Dieter |
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04-13-2018, 06:27 PM
Post: #4
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RE: (41) Round To The Nearest 1/n
(04-12-2018 08:24 PM)Mike (Stgt) Wrote: I get the same this way: Wow, that's a quite ...creative solution. I'm not sure if I understand completely how this works. For a more straightforward approach see below. (04-12-2018 08:24 PM)Mike (Stgt) Wrote: Note: Do not round "twice", rounding a result to a new n-th. Sure. You once round \(\pi\) and once \(\frac{101}{32}\). So the results are different. (04-12-2018 08:24 PM)Mike (Stgt) Wrote: Dieter's hint int(x+0,5) would take two bytes less. Right – 9 bytes without LBL and END. But here x and n must have identical signs: Code: 01 LBL "RNR" But with 2 bytes more there is a way to handle also different signs: Code: 01 LBL "RNR" Finally, here is a TI-59 version. Data entry is done "the HP way": x [x<>t] n. ;-) Code: 000 LBL Or with the already mentioned EE INV EE rounding method: Code: 000 LBL Of course real TI-59 programmers would avoid "=" and use parentheses instead. ;-) Dieter |
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04-14-2018, 06:31 PM
(This post was last modified: 04-14-2018 06:34 PM by Dieter.)
Post: #5
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RE: (41) Round To The Nearest 1/n
(04-14-2018 08:18 AM)Mike (Stgt) Wrote: About the sign, I am not sure if this is correct (BTW, I discarded the X<>Y previously in line 2): I'd say this is exactly the way it's supposed to be. If you enter \(-\pi\) the output of course also has to be negative. The approximation of \(-\pi\) is –3,125 and not +3,125. So the signs of input and output should match. Dieter |
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04-14-2018, 09:03 PM
Post: #6
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RE: (41) Round To The Nearest 1/n
(04-14-2018 08:09 PM)Mike (Stgt) Wrote: \(\pi\), 16, XEQ "RNR" -> 3,125 -- agreed IMHO that's the correct answer because, of all the infinitely many multiples of 1/(-16), the one nearest to pi is in fact 3.125 ... and rounding x to the nearest multiple of 1/n is the program's stated purpose. <0|ɸ|0> -Joe- |
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04-15-2018, 04:13 PM
(This post was last modified: 04-15-2018 04:15 PM by Dieter.)
Post: #7
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RE: (41) Round To The Nearest 1/n
(04-14-2018 08:09 PM)Mike (Stgt) Wrote: For positive n do doubt. Same for negative n? Sure? How did you prove that? Simple. We are looking for a value of z for which z/n is as close to x as possible. The program then returns z/n. If n<0 then z has to have the opposite sign of x. In simple words: z has to be an integer, but it doesn't have to be a natural number. In our example \(\pi\) rounded to the nearest –1/16 is 3,125, which is –49/–16. Or z=–49. The essential point here: \(z\in\mathbb{Z}\) and not only \(z\in\mathbb{N}\). On the other hand –3,125 cannot be the correct answer as it isn't the closest –1/16 at all. Even 0 is closer to \(\pi\) than –3,125. ;-) Dieter |
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