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Forum Index : Microcontroller and PC projects : Algorithm Request: Comparison against a range that wraps
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vegipete Guru  Joined: 29/01/2013 Location: CanadaPosts: 1132 |
Does anyone have or can figure out an _elegant_ algorithm for determining if two values are within a certain range of each other when the allowable range wraps around? For example, suppose I want to know if two times are within 5 minutes of each other. Say 12:30 and and 12:27. "IF ABS(time1 - time2) < deltatime THEN true" works fine except... it fails with, say, 12:58 and 1:02 (assuming a 12 hour clock.) I have something with 5 IF statements, plus some MODs, which just strikes me as ugly. Is there a better way? Visit Vegipete's *Mite Library for cool programs. |
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Mixtel90 Guru Joined: 05/10/2019 Location: United KingdomPosts: 7937 |
Convert all times to seconds since a fixed point, do the calculation then convert back? Mick Zilog Inside! nascom.info for Nascom & Gemini Preliminary MMBasic docs & my PCB designs |
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jirsoft Guru Joined: 18/09/2020 Location: Czech RepublicPosts: 533 |
Is the difference between those 2 times 0:04 (12:58am, 1:02pm) or 11:56 (12:58pm, 1:02am)? You need to set the border conditions. But I'm not expert for 12 hour clock... Jiri Napoleon Commander and SimplEd for CMM2 (GitHub), CMM2.fun |
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| vegipete Guru Joined: 29/01/2013 Location: CanadaPosts: 1132 |
I only give time as an example. Compass direction in degrees is much the same. 359 degrees is very close to 1 degree. Or screen pixels, where the edges wrap to the other side. I haven't come up with a simple criteria yet for when to 'unwrap' one of the limits. For example, with 359 degrees, you would subtract 360, giving -1, which _is_ close to 1. Visit Vegipete's *Mite Library for cool programs. |
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| Mixtel90 Guru Joined: 05/10/2019 Location: United KingdomPosts: 7937 |
You don't allow it to wrap round. :) a = b*(1*(b<360)) Any use? Have I got the right idea? a is always equal to b until b=360 then a=0 Mick Zilog Inside! nascom.info for Nascom & Gemini Preliminary MMBasic docs & my PCB designs |
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| Volhout Guru Joined: 05/03/2018 Location: NetherlandsPosts: 5089 |
To determine the difference, you subtract a from b To determine the difference in a case a wrap: add full range to b. Then subtraxt Then take modulus range PicomiteVGA PETSCII ROBOTS |
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| vegipete Guru Joined: 29/01/2013 Location: CanadaPosts: 1132 |
Here's an example with degrees: 'target_angle' is where you want to be. 'current_angle' is where you are. 'tolerance' is whether you are close enough. All angles are in degrees MOD 360, ie from 0 to 359.99whatever angle_min = target_angle - tolerance This won't work with wrapping.angle_max = target_angle + tolerance if (angle_min < current_angle) AND (current_angle < target_max) then DoSomething Instead angle_min = target_angle - tolerance works, but seems so convoluted and crazy. But maybe that's what is required.angle_max = target_angle + tolerance if angle_min < 0 then if current_angle > angle_min+360 OR current_angle < angle_max then DoSomething elseif angle_max >= 360 then if current_angle > angle_min OR current_angle < angle_max-360 then DoSomething elseif angle_min < current_angle AND current_angle < angle_max then DoSomething endif Visit Vegipete's *Mite Library for cool programs. |
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phil99 Guru Joined: 11/02/2018 Location: AustraliaPosts: 2640 |
Vegipete's concept is the same as one I used for time, though different code and it works well. For angles another method is to convert numbers over 180 to negative. But only if they are within the "window" distance of 0 / 360. Eg. if (X > 180) AND (Y-window < 0) then X = X - 360 if (Y > 180) AND (X-window < 0) then Y = Y - 360 Then perform the tolerance window test. Phil Edit I think that's how it went, many years ago. Edited 2021-06-25 12:02 by phil99 |
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TassyJim Guru Joined: 07/08/2011 Location: AustraliaPosts: 6283 |
The simplest is diff = (new_angle-old_angle + 360) mod 360 You will have to decide if you want negative results or all positive. It depends on the usage. MOD gives integer results so either multiply by 10 then MOD then divide by 10 etc or use another way to do MOD dim float old_angle, new_angle, diff do input "both angles:", old_angle,new_angle diff = (new_angle-old_angle + 360) mod 360 ' MOD gives integer results print diff diff = new_angle-old_angle + 360 if diff >= 360 then diff = diff - 360 print diff loop edit: diff = (((new_angle-old_angle)*10 + 3600+1800) mod 3600)/10-180 ' 1 decimal place and negatives Jim Edited 2021-06-25 15:56 by TassyJim VK7JH MMedit |
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| phil99 Guru Joined: 11/02/2018 Location: AustraliaPosts: 2640 |
Found my old code fragment for compass. my previous post is messed up. 'window comparator window=10 : x=356 : y=3 if (y>=360-window) and (x<window) then : y=y-360 :endif if (x>=360-window) and (y<window) then : x=x-360 :endif ? abs(x-y) if (abs(x-y) =< window) then : ?"within range" :else :? "No match" :endif |
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| jirsoft Guru Joined: 18/09/2020 Location: Czech RepublicPosts: 533 |
I will prepare both values with MOD to proper range 0-MAX, so you will have for example 2 values 0-359 for angle or 0:00-11:59 for 12-hour clock time. Then if smallest difference min(max(a,b)-min(a,b), WRAP+min(a,b)-max(a,b)) where WRAP is 360 for angle and 12:00 for time...Jiri Napoleon Commander and SimplEd for CMM2 (GitHub), CMM2.fun |
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CaptainBoing Guru Joined: 07/09/2016 Location: United KingdomPosts: 2170 |
Two ways I use. As already alluded-to, convert the times to unix/epoch time then compare (later MMBasics have Epoch time built in) but here is a Function I use all the time on smaller Mites If Abs(UnixTime(d1$)-UnixTime(d2$))<=300 Then... ... or (as already alluded-to) use DateDiff (which uses UnixTime at its core) and the time period of your choice to do the same comparison If Abs(DateDiff("m",d1$,d2$))<=5 Then ... Neither of the above methods are fazed by the wrap around from 23:59 to 00:00 Edited 2021-06-25 19:17 by CaptainBoing |
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| Volhout Guru Joined: 05/03/2018 Location: NetherlandsPosts: 5089 |
Couldn't put it any more compact... PicomiteVGA PETSCII ROBOTS |
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| vegipete Guru Joined: 29/01/2013 Location: CanadaPosts: 1132 |
There's some awesome ideas! Jiri's version seems to be the hardest to break - I haven't been able to. I split it into 2 lines to simplify it slightly. Phill's only works within the window. (Which suits my request.) His can't return an arbitrary difference. TassyJim's only works 'forwards'. (359,1 is OK, 1,359 is not) The neat thing is that all three have an underlying similarity. Plus they are more elegant than my brute force IF EVERYTHING method. Thanks! dim float old_angle, new_angle, diff dim window = 10 do input "both angles:", old_angle,new_angle print "Phill99" if (new_angle >= 360-window) AND (old_angle < window) then new_angle = new_angle - 360 if (old_angle >= 360-window) AND (new_angle < window) then old_angle = old_angle - 360 diff = abs(new_angle - old_angle) print diff print "TassyJim" diff = new_angle-old_angle + 360 if diff >= 360 then diff = diff - 360 print diff print "Jiri" diff = max(old_angle,new_angle)-min(old_angle,new_angle) diff = min(diff, 360-diff) print diff loop Edit: might be able to simplify Jiri's some more: diff = abs(old_angle - new_angle) diff = min(diff, 360-diff) Edited 2021-06-26 04:01 by vegipete Visit Vegipete's *Mite Library for cool programs. |
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