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 Posted: Tue Aug 20, 2013 4:31 pm    Post subject: Giuseppe Zanotti Vente-jrhtrkj7 Minimum number of weighing to find the imperfect bag In my module they have given some sort shortcut to evaluate the outcome which namely for $n$ bags: In 1st case whether $3^a \lt n \lt 3^a+1$,afterwards the minimum numeral of weighing required is $a+1$. In the second case, we need $2^a \lt n \lt 2^a+1$ which gives the minimum of $a+1$ weighing. However I couldn't emulate up this near. Could anybody explain this approximate to me?Also,in privation of the information nearly whether the imperfect bag is cumbersome or lighter than the other ones,[url=http://ventechaussuresgiuseppezanotti.webs.com/]Giuseppe Zanotti Vente[/url],how accurate could I pinpoint the imperfect sack? For a two pan balance, you can split the bags into three batches, A, B,[url=http://giuseppezanottiventediscount.webs.com/?]Giuseppe?Zanotti?Sneakers[/url], and C. You fathom batch A against B (which had better have the same digit of bags among them). If they balance, the imperfect sack is among batch C. Are told whether the imperfect sack namely heavy alternatively light? If so, you know which of A alternatively B it is surrounded So an weighing decreases the team of feasible bags forward a factor 3. If a spring balance is used you can put half the bags aboard the scale a while back If the result is an exact digit of pounds, the imperfect an namely not aboard the scale. If the result namely never an exact numeral of pounds, the faulty one namely on the scale. The numeral of feasible bags decreases by a factor of 2. Addition: With eight measure A1, A2, A3 vs B1, B2, B3. If they balance,always six are appealing so measure A1 vs C1,afterwards A1 vs C2 and you know which is wrong and how. If they don't balance,mention the A's outweigh the B's. Weigh A1, B1 against A2, B2. If they balance,meter A3 against C1 and you know the answer. If not assume A1, B1 namely heavy Then either A1 is heavy alternatively B2 is light. Weigh A1 against C1. In always cases we find the bad bag and know if it namely heavy alternatively light in three weighings. You can actually do better: discern the 12 coins problem for 12 coins amid three weighings. For the two pan balance,afterwards you have some more work to do. As that namely impartial an more morsel of information, you ought be learned to find a strategy that only adds one weighing. If the digit of bags is a multiple of 3 you can just fathom B vs C the 1st period you obtain contention between A and B. If you have some known agreeable bags from a previous weighing, you can fill out with them. Can you chart out what to do if you have 8 bags,amount three aboard each side and find the left side cumbersome For the spring balance it doesn matter. Ross Millikan Oct 3 '11 by 16:48 The minimum number of weighings you would need to ascertain the imperfect bag among the first case is 2: pick two bags by random and compare them. If you happen to strike"aboard the defective sack they aspiration not balance; afterwards meter the 1st of the two bags against one of the bags that you haven't weighted ahead If they balance,[url=http://giuseppezanottisneakerssite.webs.com/]Chaussure Giuseppe Zanotti[/url], the imperfect bag namely the second sack you picked; if they don't balance,afterwards the first sack is the faulty one. Likewise,[url=http://giuseppezanottibottesnew.webs.com/]Giuseppe Zanotti Sneakers[/url], you would need a minimum of an weighing with a spring fraught scale to make sure a imperfect bag: pick a sack by random, and weigh it. If the weight is never exactly one pound,afterwards that's the imperfect sack and you're done. Of lesson the problem is that these methods are not guaranteed to find the imperfect bag among"small"digit of steps; the second manner would require along best one weighing,at worst $n1$ weighings, and the expected number of weighings namely probably elsewhere approach $\lceil\fracn12\rceil$. Rather, the problem is asking you for a systematic manner which will make sure the defective bag and which namely,among some sense, "optimal" (probably associative to the expected digit of weighings), and asking you what namely the "best case scenario"in those algorithms. But you probably don't have the tools essential to determine what means namely"optimal", which makes the question rather illposed among my opinion. That said, Ross's answer namely probably the intended solution: if $3^a \lt n\leq 3^a+1$,separate the sacks into three piles with $3^a$,[url=http://giuseppezanottisandales2013fall.webs.com/]Giuseppe Zanotti[/url], $3^a$, and $n2\times 3^a$ sacks each and weight the 1st two. In the best case scenario, they balance and you can eliminate always those bags, leaving you with $3^a1 \lt n2\times 3^a \leq 3^a$ bags to consider. Lather, rinse, and repeat. The "bad" case occurs if the bags don't balance,for afterwards you will need accompanying weighings to determine the defective sack but once you determine if it is cumbersome alternatively light, you ambition be capable to make sure which third of the sacks has the faulty bag surrounded each subsequent step In the bestcasethroughout scenario, the bags balance amid every tread so you always eliminate twothirds of the bags amid each tread Since $3^a\lt n\leq 3^a+1$, it aspiration take you $a+1$ weighings to find the imperfect bag. How do we make sure if the defective sack is light or ponderous Suppose that somewhere along the line but never by the 1st weighing), the bunch among pan A does not balance with the cluster amid pan B. Because every bag you've set alongside so distant namely known to be good you have enough knowntobegood bags to contrast against the cluster among pan A. Compare them. If they balance,afterwards the imperfect sack is among the cluster that was amid pan B, and it is ponderous if pan B was heavier than pan A, and lighter if pan A was heavier than pan B. Now proceed with the bunch among pan B as before barely swiftly you know if the imperfect bag namely heavy alternatively light, so by every subsequent weighing you know precise which third contains the imperfect sack If the cluster amid A does not balance with the knowntobegood bags,afterwards you know A contains the defective sack and you know that it is light if cluster A namely lighter,ponderous whether bunch A is cumbersome. This will work always, unless you are at your 1st weighing, and the cluster of "setaside" bags namely not enough to compare with always of cluster A alternatively cluster B. In that case, take half alternatively as approach to an half as you can) the sacks among A, and move them "out"; take one half of the sacks within B, and move them to pan A; and afterwards take enough knowntobegood sacks from bunch C, and area them among B. Three things can happen: (i) the pans are balanced now; then the imperfect sack is within the cluster of sacks from A you took out,[url=http://giuseppezanottitalonsenvente.webs.com/]Giuseppe Zanotti Sneakers[/url], and you know whether it namely ponderous alternatively light (depending aboard if A was heavier than B alternatively lighter than B, respectively); continue as forward (ii) The pans are unbalanced as they were before; then the faulty bag namely in the cluster of sacks from A and B that you did never move; you still don't know whether the imperfect sack namely cumbersome or light (you'll must graph it out behind but you know which third of the sacks to continue with for the afterward tread Or (iii) The pans are unbalanced,merely amid the inverse way than they were before; this can only happen if the imperfect bag namely between the cluster of sacks you moved from B to A; and this aspiration differentiate you if it namely heavy or light. Once you know whether the imperfect bag namely cumbersome alternatively light,by each stage you only absence an weighing to know which third contains the imperfect bag. In the second case, with $2^a \lt n \leq 2^a+1$, you detach the bags into a cluster with $2^a$ bags and an with $n2^a\leq 2^a$ bags. Weight the first: whether the outcome namely $2^a$ units, these are always agreeable bags", and the bad sack namely within the second heap If the sequel is never $2^a$ units,then the bad sack is among this bunch Eliminate the bags that we know are good lather, rinse, and repeat. It ambition take you $a+1$ steps to find the imperfect bag since every step eliminates almost half the remaining bags.相关的主题文章: [url=http://munilaw.com/2008/12/blog-round-two.html#comments]http://munilaw.com/2008/12/blog-round-two.html#comments[/url] [url=http://www.61jingling.com/bbs/forum.php?mod=viewthread&tid=64334&pid=83356&page=11&extra=page=1#pid83356]http://www.61jingling.com/bbs/forum.php?mod=viewthread&tid=64334&pid=83356&page=11&extra=page=1#pid83356[/url] [url=http://www.safezs.com/plugin.php?id=eis_machine:code&a=task&referer=http://www.safezs.com/forum.php?mod=post&%3Baction=reply&%3Bfid=52&%3Btid=16471&%3Bextra=page%253D1&%3Breplysubmit=yes&%3Binfloat=yes&%3Bhandlekey=fastpost]http://www.safezs.com/plugin.php?id=eis_machine:code&a=task&referer=http://www.safezs.com/forum.php?mod=post&%3Baction=reply&%3Bfid=52&%3Btid=16471&%3Bextra=page%253D1&%3Breplysubmit=yes&%3Binfloat=yes&%3Bhandlekey=fastpost[/url] [url=http://www.vatneart.net/index.php?album=prints&image=Imagination_MG_0806.jpg]http://www.vatneart.net/index.php?album=prints&image=Imagination_MG_0806.jpg[/url] [url=http://www.txquyi.com/forum.php?mod=viewthread&tid=26442&pid=40052&page=1&extra=page=1#pid40052]http://www.txquyi.com/forum.php?mod=viewthread&tid=26442&pid=40052&page=1&extra=page=1#pid40052[/url]
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