Long Postings break RHP thread editor

When a posting of many lines is submitted to a thread, an erroneous screen is always reached, although the posting still goes in. For instance see my latest post in
http://www.redhotpawn.com/board/showthread.php?threadid=128880&page=1

This gave the following screen when I submitted it. (note all the instances of the words "Testing"😉, I anticipate this message will give a similar error:

Testing Interesting approach PBE6, I got 66 positions, with the following proof: It simplifies things to work with the position of a hand represented by a number between 0 and 12, where 0 and 12 are midnight. The following equation will produce such a number: Pos_1 = H + X/12 + F/12 where H and X are Integers between 0 and 11 (inclusive), and 0<=F<1 In such a position, H will correspond to hours, and x/12 + f/12 to \"fractions of an hour\", the definition ensures that the fractional part cannot exceed 1. So let us assume Hand_A of the clock is at such a position, and first of all let us assume that it is the Hour Hand. We know that Hour = H, and, working in fractional minutes we can state: Minutes = 5(X + F) (note that Minutes cannot exceed 60) If Hand_A is the Hour Hand, then Hand_B must be the minute hand, and in that case, to show 5(X + F) minutes, it must be at: Pos_2 = X + F But, assuming this is an ambiguous time, Hand B could alternatively be an Hour Hand Noting that X ranges from 0 to 11 and F from O to 1, so from the mapping of position to time, if this is the case the time will be X hours, and F*60 minutes. So if Hand B showed the hour, the Hand_A would be the minute hand, to show F*60 minutes, it would point to: Pos_3 = 12F Because they refer to the same hand, Pos_1 and Pos_3 must be equivalent, so: 12F = H + X/12 + F/12 which rearranges to: F = (H + X/12)*144/143 We can therefore cancel F out of the first equation and rearrange to give: Pos_1 = (12H + X)*144/143 So, every combination of H and X is an ambiguous time, giving 12^2 = 144 combinations. Of these 144, for some Pos_1 and Pos_2 will be identical, meaning that while we don\'t know which hand is the hour and which is the minute we do know the time. Noting that in Pos 1, H is the hour, and in Pos 2 X is the hour, and for the times to be identical the hours must be equivalent, therefore for these times X=H So there are 12 combinations where the two hands are in the same place This leaves 144 - 12 = 132 combinations But, because there are two ways round the hands can go for each position, there are only half this number of unique combinations so: Unique, ambiguous hand combinations = 132/2 = 66
Testing Interesting approach PBE6, I got 66 positions, with the following proof: It simplifies things to work with the position of a hand represented by a number between 0 and 12, where 0 and 12 are midnight. The following equation will produce such a number: Pos_1 = H + X/12 + F/12 where H and X are Integers between 0 and 11 (inclusive), and 0<=F<1 In such a position, H will correspond to hours, and x/12 + f/12 to \"fractions of an hour\", the definition ensures that the fractional part cannot exceed 1. So let us assume Hand_A of the clock is at such a position, and first of all let us assume that it is the Hour Hand. We know that Hour = H, and, working in fractional minutes we can state: Minutes = 5(X + F) (note that Minutes cannot exceed 60) If Hand_A is the Hour Hand, then Hand_B must be the minute hand, and in that case, to show 5(X + F) minutes, it must be at: Pos_2 = X + F But, assuming this is an ambiguous time, Hand B could alternatively be an Hour Hand Noting that X ranges from 0 to 11 and F from O to 1, so from the mapping of position to time, if this is the case the time will be X hours, and F*60 minutes. So if Hand B showed the hour, the Hand_A would be the minute hand, to show F*60 minutes, it would point to: Pos_3 = 12F Because they refer to the same hand, Pos_1 and Pos_3 must be equivalent, so: 12F = H + X/12 + F/12 which rearranges to: F = (H + X/12)*144/143 We can therefore cancel F out of the first equation and rearrange to give: Pos_1 = (12H + X)*144/143 So, every combination of H and X is an ambiguous time, giving 12^2 = 144 combinations. Of these 144, for some Pos_1 and Pos_2 will be identical, meaning that while we don\'t know which hand is the hour and which is the minute we do know the time. Noting that in Pos 1, H is the hour, and in Pos 2 X is the hour, and for the times to be identical the hours must be equivalent, therefore for these times X=H So there are 12 combinations where the two hands are in the same place This leaves 144 - 12 = 132 combinations But, because there are two ways round the hands can go for each position, there are only half this number of unique combinations so: Unique, ambiguous hand combinations = 132/2 = 66
Testing
Testing

Originally posted by iamatiger
When a posting of many lines is submitted to a thread, an erroneous screen is always reached, although the posting still goes in. For instance see my latest post in
http://www.redhotpawn.com/board/showthread.php?threadid=128880&page=1

This gave the following screen when I submitted it. (note all the instances of the words "Testing"😉, I anticipate this ...[text shortened]... ombinations so: Unique, ambiguous hand combinations = 132/2 = 66
Testing
Testing

Originally posted by Phlabibit
I don't know about longer, but like a PM it would help to know 1999/2000 available

That way you can cut and save for the next post like I often do with a PM message.

P-.

Originally posted by iamatiger
It lets long postings go in, in their entirety no matter how long, it just gives an error screen so you think it didn't work (and the preview doesn't work properly either). You can experiment with it by cutting and pasting my message above into a new post! 🙂

I vote for allowing long postings myself, sometimes there is a lot to say.