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Kulan: Knightfall's Crisis in Bluffside Game [OOC]
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<blockquote data-quote="Neurotic" data-source="post: 7816328" data-attributes="member: 24380"><p>[USER=6749263]@Envisioner[/USER] : lets try this. You need to pick the letter A from the jumble of 9 letters (simplification)</p><p>JustinCae(sorry [USER=6776182]@JustinCase[/USER], suitably only one A <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /> )</p><p></p><p>You could go like this</p><p>Is the first letter an 'a'? J => NO</p><p>Is the second letter an 'a'? u => NO </p><p></p><p>This depends on the combination (i.e. reAson will be quicker than JustinCae, but slower than Announcement) - but on average, you need N/2 tries (in your case N = 9 letters, in game, we have 15 phrases)</p><p></p><p>Or you could go "binary tree"</p><p>Split into two groups of roughly half</p><p><strong>Justi </strong>vs <strong>nCae</strong></p><p>You ask</p><p>Is there an A in the first group</p><p>if YES, you take that group and repeat</p><p>if NO, you take the other one and repeat => The advantage here is that there are no "empty" castings, you always GAIN information</p><p></p><p>Continuing the logic:</p><p><strong>nC</strong> vs <strong>ae</strong></p><p>And finally, you split ae and ask is this an 'a' </p><p></p><p>So, for the length of 9 you need only 3 castings and the algorithm is logarithmic so for 15, you need 4, for 32 5, 64 6, 1024 10, 1 000 000 only 20.</p><p></p><p>Compare to </p><p>15 needs 7 (on average)</p><p>32 needs 16</p><p>1024 needs 512</p><p>1 000 000 needs 500 000</p><p></p><p>As I said, you may not know it, but Relgar could <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /> The advantages of characters with skills different than our own...</p></blockquote><p></p>
[QUOTE="Neurotic, post: 7816328, member: 24380"] [USER=6749263]@Envisioner[/USER] : lets try this. You need to pick the letter A from the jumble of 9 letters (simplification) JustinCae(sorry [USER=6776182]@JustinCase[/USER], suitably only one A :) ) You could go like this Is the first letter an 'a'? J => NO Is the second letter an 'a'? u => NO This depends on the combination (i.e. reAson will be quicker than JustinCae, but slower than Announcement) - but on average, you need N/2 tries (in your case N = 9 letters, in game, we have 15 phrases) Or you could go "binary tree" Split into two groups of roughly half [B]Justi [/B]vs [B]nCae[/B] You ask Is there an A in the first group if YES, you take that group and repeat if NO, you take the other one and repeat => The advantage here is that there are no "empty" castings, you always GAIN information Continuing the logic: [B]nC[/B] vs [B]ae[/B] And finally, you split ae and ask is this an 'a' So, for the length of 9 you need only 3 castings and the algorithm is logarithmic so for 15, you need 4, for 32 5, 64 6, 1024 10, 1 000 000 only 20. Compare to 15 needs 7 (on average) 32 needs 16 1024 needs 512 1 000 000 needs 500 000 As I said, you may not know it, but Relgar could :) The advantages of characters with skills different than our own... [/QUOTE]
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