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Caesar 
發表於： Aug 23 2015, 11:45


Loop 發表數： 7,495 所屬群組： 軍團長 註冊日期： 12182004 活躍：29 聲望：2214 
As a side note , I guess we are not sharing the same idea with my solution. Actually the unequal in first trial hints too. But since I've checked with the model answer , and I'm wrong anyway, so I guess no point in explaining my original solution.  

Caesar 
發表於： Aug 23 2015, 11:48


Loop 發表數： 7,495 所屬群組： 軍團長 註冊日期： 12182004 活躍：29 聲望：2214 
Woot !! I get it finally !!  

XxEDxX 
發表於： Aug 23 2015, 11:54


三品官 發表數： 1,467 所屬群組： 太守 註冊日期： 8302011 活躍：11 聲望：507 
"Actually the unequal in first trial hints too." That's why I ask you to think about the case when you get three times (left:heavier). It seems that your method still cannot tell which marble is different and whether it is lighter or heavier than others. I am afraid that the hint in the first trial is not enough. *** "But since I've checked with the model answer , and I'm wrong anyway, so I guess no point in explaining my original solution." I think the model answer may not be unique (that's why I ask), and cannot agree that if your answer is different from the socalled model answer (that you referred to), yours cannot be right. Whatever, thank you very much for your reply to my answer Caesar. I appreciate them very much. 

Caesar 
發表於： Aug 23 2015, 11:54


Loop 發表數： 7,495 所屬群組： 軍團長 註冊日期： 12182004 活躍：29 聲望：2214 
Woot !! I get it finally !!  

XxEDxX 
發表於： Aug 23 2015, 11:54


三品官 發表數： 1,467 所屬群組： 太守 註冊日期： 8302011 活躍：11 聲望：507 
Good to hear that bro. 

Caesar 
發表於： Aug 23 2015, 12:14

Loop 發表數： 7,495 所屬群組： 軍團長 註冊日期： 12182004 活躍：29 聲望：2214 
Yes , I've said urs could be an alternative solution like 2 hours ago.
And yes , I've said I'm wrong like 2 hours ago. So yes n yes.  
Caesar 
發表於： Aug 23 2015, 12:54


Loop 發表數： 7,495 所屬群組： 軍團長 註冊日期： 12182004 活躍：29 聲望：2214 
Altho u just said this isn't ur solution , but I just realised this can actually hit the result .  

Pearltea 
發表於： Aug 23 2015, 12:56

四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
Looks like ED has an alternative solution.
There are some various solutions in the 2nd and 3rd trials, but for the 1st one it has to be 1234 vs 5678, or ABCD vs EFGH. (Whichever is easier). 
XxEDxX 
發表於： Aug 23 2015, 13:04


三品官 發表數： 1,467 所屬群組： 太守 註冊日期： 8302011 活躍：11 聲望：507 
Yes, as one lift is enough to pick the one from the 3 marbles. I am just afraid that "pick 3 from the remaining group" makes some confusion. 

Pearltea 
發表於： Aug 23 2015, 13:08


四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
My understanding was that you meant picking three of the marbles from the 9, 10, 11 & 12 group, so I wasn't confused. 本篇文章已被 Pearltea 於 Aug 23 2015, 13:11 編輯過 

XxEDxX 
發表於： Aug 23 2015, 13:15


三品官 發表數： 1,467 所屬群組： 太守 註冊日期： 8302011 活躍：11 聲望：507 
Thank you very much for your reply. I hope my answer is clear enough. A good game Can you share the other possible answers? 

Pearltea  
四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
Here's mine.
Set aside 9, 10, 11 & 12. 1st weighing: 1,2,3,4 vs 5,6,7,8. (refer 1,2,3,4 as the left side) If Equal 2nd weighing: 1, 9 vs 10, 11. (which is the same method as yours) If Equal 3rd weighing: 1 vs 12 If Unequal 3rd weighing: 10 vs 11 [Before I learned that the more popular question requires finding out whether the unusual one is lighter or heavier, my solution were second weighing 9 vs 10, third weighing 10 vs 11. However that won't solve the problem when 12 is lighter or heavier] If Unequal Set aside 7 & 8 Swap 2 & 4 to the right side, and swap 5 to the left side. 2nd weighing: 1,3,5 vs 2,4,6 If Equal 3rd weighing: 7 vs 8 (we know whether these two are heavier or lighter from the 1st weighing) If Unequal If the Balance shifts This means the swapped marbles contain the unusual one. 3rd weighing: 2 vs 4 (we know whether these two are heavier or lighter from the 1st weighing) If the Balance remains the same This means the nonswapped marbles contain the unusual one. 3rd weighing: 1 vs 3 (we know whether these two are heavier or lighter from the 1st weighing) 本篇文章已被 Pearltea 於 Aug 23 2015, 14:00 編輯過 
Pearltea 
發表於： Aug 23 2015, 13:36

四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
Another solution for the 9, 10, 11 & 12 group:
2nd weighing: 1,2,3 vs 9, 10, 11 3rd weighing: 9 vs 10 (if 2nd weighing is unequal), OR 1 vs 12 (if 2nd weighing is equal) 
XxEDxX 
發表於： Aug 23 2015, 13:49


三品官 發表數： 1,467 所屬群組： 太守 註冊日期： 8302011 活躍：11 聲望：507 
Thank you very much for your sharing, Pearltea. And I agree that the first weighting has to be 1234,5678. Thank you. 

Pearltea  
四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
You're welcome ED.
我在網上找不到其他人用我的solution (For 18). 問了Mr.程序員但他不肯看, 還譏諷的說: "你不是懂VB的嗎? 你自己寫code去verify不就成了?" 我試過, so far每個scenario還可以. 他還說懂寫程式的人不等於懂得玩推理遊戲的人, 方法也可以用程式來找出. 這是甚麼道理 . 如沒有這些logic和scenario如何寫出來? 
Caesar 
發表於： Aug 23 2015, 14:28


Loop 發表數： 7,495 所屬群組： 軍團長 註冊日期： 12182004 活躍：29 聲望：2214 
這畫面很有既視感。  

Pearltea 
發表於： Aug 23 2015, 14:48


四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
怎麼？ Caesar 也這樣跟女友大人對話的嗎？ 

Pearltea 
發表於： Jan 21 2016, 07:31

四品官 發表數： 1,289 所屬群組： 太守 註冊日期： 9222003 活躍：6 聲望：614 
Apologies for resurrecting an old thread  saw an alternative solution to the same problem. The answer was brilliantly thought out:
1, 2, 7, 10 against 3, 4, 6, 9 1, 3, 8, 11 against 2, 5, 6, 7 2, 3, 9, 12 against 1, 4, 5, 8 
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