07 Apr '10 19:54

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"ðŸ˜‰, 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

Testing

Testing

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"ðŸ˜‰, 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 = 66Testing

Testing