ふうよしまた少し賢くなったぞ Ok, I'm a little smarter now

問題4 Problem 4

海外の皆様日本語の文章の後に英語訳があります

For those of you overseas, there is an English translation after the Japanese text.

The problem is a little different, so give it a try.

問題が少し違うよ

ぜひ挑戦してね

またほんの少し賢く慣れたぞ

見た目が同じ9枚の金貨がある。その中の4枚は他よりわずかに軽い偽金貨である。

天秤を使って偽金貨を特定したい。

ここには天秤が4つある。しかし、この中の1つの天秤は不良品である。

不良品は“右に傾く”、“左に傾く”、“釣り合う”の中からランダムな結果を出してしまう。

4つの天秤のうちどれが不良品かは分からない。

天秤をそれぞれ1回づつからなず使い後は好きな天秤を好きな回数使い9枚のうち4枚の偽金貨を天秤を使う数を無駄なく使い確実に判断できる天秤使用数最小と最大の数もとめよ

ではさっそく最小の無駄のない天秤使用数は

1234が偽物で①が不良品として①②③の判定が同じなら④が正常とわかり今後これのみ使うまでは同じ

① 12 34

② 12 34

③ 12 34

④ 12 34

なので12か34をと思うかもしれないがここは一気に1234をまとめて④に残りの5678をのせ

④ 1234 5678で偽物判定の1234のうち34と本の判定の78を入れかえ④にのせ④ 1278 5634で釣り合い1234と判明する

なので計6回か最小となる

次は1枚偽物が混じったとき

1234が偽物で①が不良品として①②③の判定が同じなら④が正常とわかり今後これのみ使うまでは同じ

① 15 67

② 15 67

③ 15 67

④ 15 67

となり偽物判定の15を一枚づつ正常な④にのせ④ 1 5 で1が偽物と判明する

次は残りの2348④に乗せ④23 48となり38を入れ替え④28 43で天秤に変化があるなら変えていない24が偽物と確定する

次にどちらか確実に偽物の38を④に一枚づつのせ④3 8で3が偽物と確定し1234の4枚の偽物が判明する

計8回天秤を使用

次は2枚偽物が混じったとき

1234が偽物で①が不良品として①②③の判定が同じなら④が正常とわかり今後これのみ使うまでは同じ

① 12 56

② 12 56

③ 12 56

④ 12 56

となり偽物判定の12を一枚づつ④に乗せ④ 1 2 で釣りあい偽物2枚が確定する

次は残りの3478を④に乗せ④34 78の偽物判定の34を一枚づつ④に乗せ④ 3 4で釣りあい偽物2枚が確定する

これで偽物1234が判明し

天秤使用回数計7回

最後は3枚偽物が混じったとき

1234が偽物で①が不良品として①②③の判定が同じなら④が正常とわかり今後これのみ使うまでは同じ

① 12 35

② 12 35

③ 12 35

④ 12 35

となり偽物判定の12を一枚づつ④に乗せ④ 1 2 で釣りあい偽物2枚が確定する

これまた前回と同じく偽物3が本物認定されかねないが最後まで聞いてね

残りは4678となる

残りの4678を④に乗せ④ 46 78となり偽物反対の46を一枚づつ④に乗せ④ 4 6となり4が偽のと判明残るは9と偽物の混じる可能性のある35なので9は除外して35一枚づつ④に乗せ④ 3 5で偽物が3と確定する仮に釣り合うなら9が偽物と確定する

これで天秤使用数計8回

この問題は無駄なく天秤を使い天秤使用数最小と最大もとめる問題なので答えは最小6回最大8回となる

2枚偽物が混じったとき

何故天秤使用回数計7回になったのか最初から解いたらそうなっただけで専門知識皆無なので正直わかりません

専門家以下学者よりもさらに以下の私に聞かれてもわからないので各自研究している先生に聞いてください

2024 10 29現在作家志望の私の仕事でも領域でありませんので

I'm a little smarter now

There are nine gold coins that look the same. Four of them are counterfeit and slightly lighter than the others.

You want to use a scale to identify the counterfeit coins.

There are four scales here. However, one of them is defective.

The defective scale will produce a random result from among "tilts to the right", "tilts to the left", and "balances".

You don't know which of the four scales is defective.

Use each scale once, but not more than once. After that, use any scale you like as many times as you like, and use the 4 fake gold coins out of the 9 without waste. Find the minimum and maximum number of scales you can use to make a reliable judgment.

Now, let's find the minimum number of scales you can use without waste.

If 1234 is fake and ① is defective, and the judgments of ①, ②, and ③ are the same, then ④ is normal and you will only use this one from now on.

① 12 34

② 12 34

③ 12 34

④ 12 34

So you might think 12 or 34, but here you can put 1234 together and put the remaining 5678 on ④.

④ 1234 5678, swap the 34 of the 1234 judged as fake with the 78 judged as genuine, and put it on ④.④ 1278 The balance is 5634 and it turns out to be 1234

So the total is 6 times, or the minimum

Next, when one fake is mixed in

If 1234 is fake and ① is defective, and if the judgments of ①, ②, and ③ are the same, it is understood that ④ is normal and it will remain the same until it is used only from now on

① 15 67

② 15 67

③ 15 67

④ 15 67

Then, place one of the fake 15s on the normal ④, and ④ 1 5 reveals that 1 is fake

Next, place the remaining 2348 on ④, which becomes ④ 23 48, and swap 38. If there is a change in the balance at ④ 28 43, it is confirmed that the unchanged 24 is fake

Next, place one of the fake 38s on ④, and ④ 3 8 confirms that 3 is fake, and 1234 are found to be fake

Use the balance 8 times

Next, when 2 fakes are mixed in

1234 is fake and ① is defective, and if ①, ②, and ③ are judged the same, ④ is found to be normal, and it will remain the same until it is used alone from now on

① 12 56

② 12 56

③ 12 56

④ 12 56

Then, place one of the fake 12s on ④, and balance with ④ 1 2, confirming that there are 2 fakes

Next, place the remaining 3478 on ④, and place one of the fake 34s from ④ 34 78 on ④, and ④ 3 4 balances and confirms that there are two fakes

This confirms that 1234 are fakes

7 times in total when the balance is used

Finally, when 3 fakes are mixed in

1234 is fake and ① is defective, and if the judgments of ①, ②, and ③ are the same, ④ is found to be normal, and it will remain the same until it is used only from now on

① 12 35

② 12 35

③ 12 35

④ 12 35

Then, put one of the fake 12s on ④, and ④ 1 2 balances and confirms that there are two fakes

This also means that fake 3 may be recognized as the real one, just like last time, but please listen to the end

The remaining number is 4678

The remaining 4678 is put on ④, and ④ 46 78, and then put one of the fake 46s on ④, and ④ 4 6, and 4 is found to be fake

The remaining number is 9 and 35, which may be mixed with fakes, so exclude 9 and put one of the 35s on ④, and ④ 3 5 confirms that the fake is 3. If it balances, 9 is confirmed as a fake.

This means that the balance is used a total of 8 times.

This problem is about using the balance efficiently and finding the minimum and maximum number of balances used, so the answer is a minimum of 6 and a maximum of 8.

When 2 fakes are mixed in

I honestly don't know why the total number of balances used is 7 because I solved it from the beginning and it turned out that way. I have no specialized knowledge, so I don't know what to do if you ask me, even less than an expert or a scholar, so please ask a professor who is researching it.

2024 10 29 Currently, I am an aspiring writer, and it is not in my field of work, so