Let 8个金币 = G1,G2,G3,------G8 1) weigh G1 , G2; there are two results: equal weight or not equal weight if equal weight, go to step 2; if not equal (Let G1 1a) weigh G1G3, G2G4 ; if equal, go to 1b If not equal (G1G3 1b) weigh G1,G3; if =, G2 is the heavier one, and G4 is the lighter one -----Answer 1c) if not equal, G3 is重假币, and G1 is 轻假币-----Answer 2) weigh G3, G4; there are two results: equal weight or not equal weight if equal weight go to step 3; if not equal (Let G3 2a) weigh G3G5, G4G6; if equal go to 2b, if not equal (G1G3 2b) weigh G3,G5; if =, G4 is 重假币, and G6 is 轻假币-----Answer 2c) if not equal G5 is 重假币, and G3 is 轻假币-----Answer 3) weigh G5, G6, if equal weight go to step 4; if not equal (Let G5 3a) weigh G5G7, G6G8 if equal go to 3b, if not equal (G5G7 3b) weigh G5, G7 if =, G6 is 重假币, and G8 is轻假币-----Answer 3c) if not equal G7 is 重假币, and G5 is 轻假币-----Answer 4) weigh G7, G8, (G7, G8 must be two 假币) ------ Answer
Let 9个金币 = G1,G2,G3,------G9 1) weigh G1G3G5G7, G2G4G6G8; there are two results: equal weight or not equal weight if equal weight, A9 is a 真金币, go to step 2; if not equal, A9 is a 假金币, (Let G1G3G5G7), go to 1a 1a) weigh G1G3G9, G2G4G6 ; if equal (G2,G4,G6 are 真金币, and G9 is a 重假币), go to 1b If not equal, G1G3G9轻假币, go to 1c 1b) weigh G1,G3; the lighter one is a 轻假币 ( we knew G9 is a 重假币 ----- Answer 1c) if not equal G9 is a 轻假币, and G1,G3 are 真金币, Weigh G2,G4 if =, G6 is a 重假币, (we knew G9 is a 轻假币) ----- Answer 2) weigh G1G2G9,G3G4G6; there are two results: equal weight or not equal weight if equal, 2 假币 must in G4,G6 or G5,G7; go to step 3; if not equal, we know G1,G2,G9 are 真金币,(Let G1G2G9), go to 2a 2a) weigh G3,G4; if =, G6 is a重假币, if not =, the heavier one is a 重假币 ---Answer 2b) weigh G5,G7; if =, G8 is a 轻假币; if not =, the lighter one is a 轻假币 ---Answer 3) weigh G4, G6, if equal weight go to step 4; if not equal; the heavier one is a 重假币 and the lighter one is a 轻假币 ----Answer 4) weigh G5,G7; the heavier one is a 重假币 and the lighter one is a 轻假币 ----Answer
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first part, 1) weigh G1 , G2; there are two results: equal weight or not equal weight if equal weight, go to step 2; if not equal (Let G1 1a) weigh G1G3, G2G4 ; if equal, go to 1b If not equal (G1G3 1b) weigh G1,G3; if =, G2 is the heavier one, and G4 is the lighter one -----Answer 1c) if not equal, G3 is重假币, and G1 is 轻假币-----Answer
It looks strange to me that if G1 So in particular, after 1a, you get G1G3 |
1c does not seem right. |
I did not figure out the 9 coin case yet, but there might be errors in this 8 coin case as well, please check: Let us simply denote the coins by {1,2,3,4,5,6,7,8}. Furthermore, if we identify coin i to be the heavier fake, we put iH; or to be the lighter fake, then iL; or to be normal, then iN. There are only one H, one L and 6Ns. There is one fact we will use again and again: if a weigh of several coins, for example {i,j,k} vs. {p,q,r} yields a result like {i,j,k}>{p,q,r}, then none of i,j,k can be the lighter one and none of p,q,r can be the heavier one. (*) Another fact will be used repeatedly is that if a weigh containing more than 1 coin on each side yields equality, then either one side contains all normal coins or both fake coins. (**) And of course the trivial fact that if {i}={j} then both i, j are normal. (***) Step 1. Weigh {1,2,3} vs. {4,5,6} (Weigh #1) if the result is equal, go to Step 2a. Otherwise, go to Step 2b. Step 2a. we will take the next TWO weighs as: Weigh #2: {1} vs. {2} Weigh #3: {4} vs. {5} If the results of both #2 and #3 are equal, that means 1,2,4,5 are all normal. Therefore, 3N, 6N. (otherwise we won't have equality of weigh #1). Therefore we only need to take Weigh #4:{7} vs. {8}, which will tell us the final answer. If {1}>{2} and {4}={5}, this implies 4N, 5N. And therefore 6N (otherwise #1 can't be equal, using fact **). And therefore both fakes are in {1,2,3}. The last Weigh #4: {3} vs. {4} will tell us the final answer, because 4 is normal. The other cases of one equality and one inequality in weigh #2 and #3 are similar to the above case. Furthermore, it is IMPOSSIBLE to get two inequalities in #2 and #3. Because if so, we will definitely have one fake in {1,2} and another fake in {4,5}, hence making the first weight equality impossible. This concludes all cases of weigh #1 being equal. Step 2b. WLOG, we can assume the result of #1 is: {1,2,3} > {4,5,6} This implies none of 1,2,3 can be the lighter fake and none of 4,5,6 can be the heavier fake. Furthermore, it also implies that one and only one of 7 and 8 is fake. We will take the next weigh as: Weigh #2: {1,7} vs. {4, 8} If {1,7}={4,8}, since we already know that one of 7, 8 is fake, this equality implies that the other fake must also be in this weigh. Therefore, the rest (2,3,5,6) are all normal. Then we will simply take the other TWO weighs are Weigh #3: {1} vs. {2} and Weigh #4: {4} vs. {2}. One and only one of the results will be inequality and it will tell us the final answer. If {1,7}>{4,8}, this implies none of 1,2,3,7 are light, and none of 4,5,6,8 are heavy. Let us take Weigh #3: {7,8} vs. {1,4}. The result will never be equal (because there is one and only one fake in {7,8}). (#3) If {7,8}>{1,4}, this implies 7H, 8N (the reason is that one of 7 and 8 must be heavy since their side is heavier in this weigh, remember that they canNOT be both normal. Further we knew 8 cannot be heavy.). Therefore all of 1,2,3 are normal. The last weigh #4 will be {4} vs. {5}, if equal, then 6L. Otherwise, the lighter one will show itself. (#3) If {7,8}<{1,4}, this implies 8L, 7N. Therefore 4,5,6 are all normal. Take weigh #4 as {1} vs. {2}. If equal, then 3H. Otherwise the heavier one will show. This concludes the second case of weigh #2. If {1,7}<{4,8}, this implies 1, 7 cannot be heavy, and 4,8 cannot be light. But we knew already from #1, 1,2,3 cannot be light, and 4,5,6 cannot be heavy. Therefore, 1 and 4 are normal. Therefore it is either 7L, 8N or 7N, 8H. Take weigh #3: {7, 8} vs. {1,4}. Again, the result won't be equal. (#3) if {7,8} >{1,4}, then 7N, 8H, therefore 1,2,3 are all normal. Take weigh #4: {4} vs. {5}. (#3) if {7,8} <{1,4}, then 7L, 8N, therefore 4,5,6 are all normal. Take weigh #4: {1} vs. {2}. This exhausted all possbilities. |
Please check for errors again. :) I will use the same notation as above. And again the facts *, **, *** in above post are still true and important in this post. Of course, we now have an additional coin 9. Step 1. Weigh #1. {1,2,3} vs. {4,5,6} If they are equal, go to Step2a, otherwise go to Step 2b. Step2a. When they are equal, it actually tells us that the HL pair must be in one of the 3 groups: {1,2,3} or {4,5,6} or {7,8,9}. Let us take the second Weigh #2: {1,4} vs. {2,5} (#2)If they are equal, then 1,2,4,5 are all normal, since they cannot pair up any more. Therefore, 3N, 6N. Therefore the HL pair is in {7,8,9}. You have two more weighs to decide which is which exactly. (#2)If {1,4}>{2,5}, then we know that one of the fake is here. Therefore neither one is in {7,8,9}. Thus 7N, 8N, 9N. Let us now take Weigh #3: {3,6} vs. {7,8}. The result cannot be equality since 7,8 are normal and 3,6 can't pair. So if {3,6} > {7,8}, then one of 3,6 must be heavy. Therefore 1,4 are normal. And one of 2,5 must be light. The last weigh is simple: {3} vs. {6}. This will decide which one is heavy and thus the light one as well. (#2)If {1,4}<{2,5}, it is symmetric to above case. This concludes the case when weigh #1 yields equality. Step 2b. WLOG, let us assume {1,2,3}>{4,5,6}. Then we know that the 3 groups {1,2,3}, {4,5,6} and {7,8,9} will all have different weights. The group with the H will be the heaviest and the one with the L will be the lightest. Further we know that none of 1,2,3 can be light. And none of 4,5,6 can be heavy. Let us now take weigh #2: {1,4} vs. {7,8} (#2)If {1,4}={7,8}, this implies 7N, 8N. (because 7,8 cannot be HL pair, otherwise weigh #1 would be equal.). Then there are only two possibilities for 1, 4: they must either be 1H,4L or 1N, 4N. (since 1 cannot be light.) Now let us take: weigh #3: {2} vs. {3} and weigh #4: {5} vs. {6} If both results are equal, then 2,3,5,6 are all normal. Therefore 1H, 4L. If {2}>{3} and {5}={6}, then 3N, 2H (since 3 cannot be L), 5N, 6N. Therefore 1N, hence 4N. Hence 9L. If {2}<{3} and {5}={6}, similar to above. If {2}={3} and {5}>{6}, then 2N, 3N, 5N, 6L (since 5 cannot be H), therefore 4N, hence 1N, hence 9L. If {2}={3} and {5}<{6}, again similar. If {2}>{3} and {5}>{6}, then both fakes are here, hence 1N, 4N. Therefore 2H, 6L. And all the other double inequality cases are similar to the last case above. This concludes the cases when weigh #2 yield equality. (#2)If {1,4}>{7,8}, then neither of 7,8 can be H, and neither of 1,4 can be L. Therefore 4N. Let us take weigh #3: {7} vs. {8}. (#3)If {7}={8}, then 7N, 8N. Thus 1H. And L is among 5,6,9. The last weigh will decide it trivially. (#3)If {7}>{8}, then 7N, 8L. Thus H is among 1,2,3. And the last weigh will decide it. If {8}<{7}, similar. (#2)If {1,4}<{7,8}, then neither 7 or 8 can be L, neither 1 or 4 can be H. Thus 1N. Let us again take weigh #3: {7} vs. {8}. (#3)If {7}={8}, then 7N, 8N. Thus 4L. And H is among 2,3,9. The last weigh can decide it. (#3)If {7}<{8}, then 7N, 8H. Thus L is among 4,5,6, and the last weigh can decide it. If {7}>{8}, similar. This exhausts all cases. |
 First let me correct: 1b) weigh G1,G3; --- should be weigh G1G3,G2G4 I’m such an idiot, when I wrote G1 < G2, I only took into account 2 possibilities, that is, either G1 is 轻假币 or G2 is 重假币, and overlooked the possibility that both could be fake. I’m going to play with this puzzle more, as soon as I find more time. P.S. If you call yourself “idiot94”, hate to think what I should call myself. |
Oh, please, do not worry about it :) People make mistakes now and then and I can assure you that I personally make a lot more mistakes than you do ;) There is nothing to feel bad about trying :) As for my name, it is just designed to give people a good laugh, hope you will like the humor :) Wait to see your great solution!  |
Let 8个金币 = G1,G2,G3,------G8 1) weigh G1 , G2; if equal weight, go to step 2; if not equal (Let G1 1a) weigh G1G2, G3G4 ; if equal, G2 is the重假币, and G1 is the轻假币 -----Answer If not equal, G1G2 < G3G4; then, G1 is the 轻假币, and G2 is a 真金币 Weigh G2G3G4,G5G6G7; if =, G8 is the 重假币------- Answer if not equal, G2G3G4 > G5G6G7, weigh G3,G4; the heavier one is the重假币----- Answer if G2G3G4 < G5G6G7; weigh G5,G6; if =, G7 is the 重假币---- Answer if not =, the heavier one is the 重假币----- Answer
2) weigh G1G2G3,G4G5G6; if equal weight, we know G1,G2,G3 are real ones; then, go to step 3; if not equal, (Let G1G2G3 < G4G5G6), then, G3 is the轻假币 weigh G5G6, G7G8; Let G5G6 < G7G8; (if G5G6 > G7G8; weigh G5,G6) weigh G7,G8; the heavier one is the 重假币----- Answer
3) weigh G4G5G7, G5G6G8; if G4G5G7 < G5G6G8; Weigh G1,G5; if =, G6 is the 重假币 and G4 is the 轻假币---Answer if not =, let G1 < G5; G5 is the 重假币 and G4 is the 轻假币---Answer
Let 9个金币 = G1,G2,G3,------G9 1) Weigh G1G3G5G7, G2G4G6G8; there are two results: equal weight or not equal weight if equal weight, A9 is a 真金币, go to step 2; if not equal, A9 is a 假金币, (Let G1G3G5G7), go to 1a 1a) Weigh G1G2G9, G3G4G6 ; if equal, G3,G4,G6,G8 are 真金币, if not =, go to 1ab 1aa) Weigh G1,G3; if G1 < G3, G3 is the 重假币 and G9 is the 轻假币---- Answer if G1 > G3, then, G3 is the 轻假币 and G9 is the 重假币---- Answer 1ab) if G1G2G9 < G3G4G6, G1,G2 are 真金币 and G9 is the 轻假币 ---- Answer weigh G3G4G7, G4G6G8; if G3G4G7 < G4G6G8, weigh G6,G8; the heavier one is the重假币----- Answer if G3G4G7 > G4G6G8, weigh G3,G7; the heavier one is the重假币----- Answer if G1G2G9 > G3G4G6, G1,G2 are 真金币 and G9 is the 重假币 ---- Answer weigh G3G4G7, G4G6G8; if G3G4G7 < G4G6G8, weigh G3,G7; the lighter one is the轻假币----- Answer if G3G4G7 > G4G6G8, weigh G6,G8; the lighter one is the轻假币----- Answer
2) Weigh G1G2G3G9, G5G4G6G8; if =, G1,G2,G3,G5,G7,G9 are 真金币; go to 2a; if not =, go to 2b 2a) weigh G1,G4; if =, weigh G6,G8; the heavier one is the重假币 and the lighter one is the轻假币----- Answer if not =, G1< G4, G4 is the 重假币 weigh G6,G8; the lighter one is the 轻假币----- Answer G1> G4, G4 is the 轻假币 weigh G6,G8; the heavier one is the 重假币----- Answer 2b) if G1G2G3G9 < G5G4G6G8; G3 is the轻假币 Weigh G1G4G5, G6G7G8; if =, weigh G6,G8; the heavier one is the 重假币----- Answer if not =, G1G4G5 < G6G7G8; weigh G4,G5; the heavier one is the重假币 and the lighter one is the轻假币----- Answer if G1G4G5 > G6G7G8; Weigh G6,G7; if =, G8 is the重假币----- Answer if not =, the heavier one is the重假币----- Answer |
I have posted my new solutions on the original question link. Hope that my logic is sound this time. After I finished the problems, I was too exhausted to check them over. I would appreciate your comments. Thank you in advance. |
It is indeed a long and tiring problem. Hope you feel better now :) I just reviewed your new solution. I did not carefully check every single part. But here are some ideas I'd like to share with you: For the 8 coin case, your step 3, there you only considered one possibility "if G4G5G7 < G5G6G8;". You did not discuss other possibilities. But that might be a slippery miss. For the 9 coin case, your step 2, claiming " Weigh G1G2G3G9, G5G4G6G8; if =, G1,G2,G3,G5,G7,G9 are 真金币", I do not see why. If 1 is heavy, 3 is light, it will also give you both equalities in the first two weighs. Again, this might be just a slippery fall. However, there is one important general idea about this kind of problems. Usually, you extract most amount of information by grouping the objects as EVENLY as possible. This way, no matter what the test result is, you will be able to yield roughly the same amount of information. This is similar to many extreme value problems in math: the extreme value is achieved often with some kind of sysmmetry. Why? Because symmetry is special and so is extrema. Therefore they often coincide with each other. When you do not group the objects in the most even way, then it is very likely that in some of the results, you gain too much information while you won't need all the attempts to find the final answer. Yet, in other cases, you gain too little, which would fail the ovreall attempts. Now let us look back, for example, your idea for 8 coins. Your first grouping takes only two coins. In the case when they are equal, you will confirm two good coins. Therefore the fakes are among the remaining 6 and equally likely among them. For 6 coins to have two distinguishable fakes, there are total P(6,2)=6x5=30 possible permutations to put them. Each weigh with our balance can yield 3 kind of results: =, <, or >. Basically hence you get 3 pieces of info. You only have 3 weighs left, you can at most identify 3^3=27 possible combinations with these 3 weighs. That is not enough to tackle 30 possibilities. Although you have two good reference coins now, but I do not think it is enough to compensate for 3 extra possibilities. It is highly unlikely to succeed with this initial grouping. Your idea for the 9 coin case is similar. When you take 4 vs. 4 as the first trial, there are cases you will lose more than you gain by this first weigh. I also suspect it won't work this way. |
Thank you for your valuable suggestions. I have indeed benefitted from them.
Sometimes I tend to omit several steps (or possibilities), though these are probably apparent.
Although I realize that my logic is lacking in precision, I am still attracted by these sorts of problems. I’m sure that I will continue to enjoy working with puzzles like this.
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It is all my pleasure, buddy. :) Glad you like this kind of problems. I will post some "new" ones for you :) They are actually all from my great master constant :). Hope you will like those. |
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