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 ID:86368   Nov 26 2009, 7:01 pm Keywords: dd, humour, ninjas A well-known result in the field of ninja mathematics is the Conservation Ninjutsu Law, which holds that the total amount of ninjutsu present in a group of ninja is independant of the number of ninja present. This post presents a means of calculuating the relative amount of ninjutsu ninja of various 'levels' possess, using the well-known D&D model (Version 3.5, as version 4.0 has not yet received significant peer review). The D&D model holds that a creature of difficulty 20 is equivalent to 2 creatures of difficulty 18, 4 creatures of difficulty 16, and 8 of difficulty 14, and so on [1]. Assuming that all these groups have equivalent ninjutsu (From the conservation of ninjutsu law) allows us to calculate the ninjutsu of each individual ninja in each group relative to each other group, assuming ninjutsu is distributed evenly amongst group members. From the table on [1], we can determine the following equivalency: 1 level 20 = 2 * level 18 = 4 * level 16 = 8 * level 14 = 16 * level 12 = 32 * level 10 = 64 * level 8 = 128 * level 6 = 256 * level 4 = 512 * level 2 = 1024 * level 1. In general, 1 level 20 ninja is worth 2**(20-(k+1)) level k ninjas. If we assign a level 1 ninja an (arbitrary) ninjutsu of one nin, then the ninjutsu of a level k ninja will, therefore, be 2**(k/2) - note that this function does not work for a level 1 ninja - a discontinuity in ninja mathematics? More research is required. [1] Dungeon Master's Guide, page 49, table 3-1
 #1 Nov 26 2009, 8:55 pm Bahaha.
 #2 Nov 27 2009, 6:44 pm a pirate must have nayed this ...although this would prove useful for a pirate to know. typical of pirates, biting the hand that feeds them
 #3 Nov 27 2009, 8:07 pm Zaole wrote: a pirate must have nayed this ...although this would prove useful for a pirate to know. typical of pirates, biting the hand that feeds them While I am a pirate, I nayed this because I can't trust D&D; I lost my first mate last time I did that.