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2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2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

この問題は前の回答

① 123 456 (789はのせない

② 147 258 (369はのせない

③ 168 357 (249はのせない

④ 267 348 (159はのせない

でやると一枚除外した9が偽物でないと破綻してこの方法では確に断できない

ちなみにおさらいするとこれは偽物の金貨一枚見つけ不良品交じりの天秤4つそれぞれ一回使い確実に金貨9枚から一枚見つけるやり方でときかたきこうなる

上の数字は9枚の金貨に番号を振っっている

一つ不良品の天秤があっても四つの天秤に両方で調べていないの一枚づつのせて金貨を入れ替えて片方三枚両方で六枚で全部入れ替えて金貨全部確実来を得るため金貨一枚を除外し8枚のうち一枚3回づつか確かめていけば確実に本物の天秤二つで偽物と本物が判定されどれが不良品かわかりこれを全ての天秤で上の数字のように天秤に片方三枚両方で六枚いれてやれば確実に偽物一枚はこの条件で見つけられる

しかしこの問題は9枚のうち2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最小と最大の数もとめよ

当然最小の答えは天秤に1枚づつ計2枚天秤に乗せその2枚が9枚のうちの偽物なら三回はかれば判明して最小無駄なく天秤を使うのは最低四回使うから四回が最小だけど

そして肝心の無駄なく天秤を使い9枚のうち2枚の偽金貨を確実に見つけれる最小と最大数

となれば考えるべきは最悪のケース

無駄なく天秤を使うために複数の金貨を載せた時不良品のランダムが最悪な事になったとき

例えば金貨全てに2から9の番号をつけ①が不良品だとして

① 123 456

② 123 456

③ 123 456

この場合なら1と2と2が偽物として①不良品重いと動けば乗せた金貨全て図り直しになる

9枚のうち1枚偽物なら1枚除外し残り八枚一枚三回秤にかければ二つの正常な天秤で偽物は確実に判明するか除外した一枚が消去法で偽物と判明する

しかしこのやりかただと二枚の偽物と不良品のぬ判定全て同じという最悪のケースになるとそれに含まる金貨全て図り直しという

無駄な天秤の使用するケースが生まれてしまうため無駄のない少ない数天秤を使いどのパターンでも無駄なく天秤を使う最大の回数がわからなくなる

なのでこの答えを出す際は無駄なく天秤を使いつつ運の悪いパターンも視野に入れ天秤を使い不良品一つを見つける必要がある

なので最初はこう

これは最初に偽物二枚が金貨に交じり①が不良品

① 12　34

② 12　34

③ 12　34

これならどの最悪のケースでも不良品がかかりつつ入れ替えてしまえば二分の一で偽物がわかる

何故最初が最悪のケースと考えるのかは最初が上手いいく効率的になり天秤を使う数を無駄なく使い確実に判断できる最小と最大の数にはならない

例えば

① 12　34

② 12　34

③ 12　34

で1と3が偽物でランダム性が釣り合うなら使っていない④の天秤に

④14　32

と2と4を入れ替えれば1と3が偽物と判明する

これは最悪のケースで①②③の判定が同じでも④でその判定が同じ4枚のうち2枚を入れ替えれば二分の一の確率なのであと一回でいい

なのでこの方法なら無駄のない最小は最初の一枚つつづと同じ天秤使用数は4回

このやり方で一つの天秤4枚つづはかればどんな最悪のケースでも無駄なく天秤を使い9枚のうち2枚の偽物を判断でき無駄なく天秤を使う数の答えを出せる

①が不良品で1と6偽物なら不良品の判定がどうあろうと確実に④で①が不良品と判明し④ 1　2　で1が偽物と判明する　

① 12 34

② 12 34

③ 12 34

④ 1 2

もう一度②　56 78

で56のどちらかが偽物で

もう一度③5　6

で偽物は5と判明でこのケースでは6回で9枚のうち2枚が偽物とわかる

①が不良品で1と6偽物でこのパターンなら

①　23 57

② 23 57

③ 23 57

④ 23 57

で不良品が釣り合い①②③の判定が同じでも④ 23 57　で④のみが確実に正常な天秤とわかり23 57　が本物と判明する

なので無駄ない数天秤を使うなら今後は確実に正常な天秤④のみ使い

もう一度④ 14 68

で釣り合い④ 18 64片方の一枚づつを入れ替えれば1と6が偽物と判明し天秤使用数6回

①が不良品で1と9偽物でこのパターンなら

①　23 57

② 23 57

③ 23 57

④ 23 57

で不良品が釣り合い①②③の判定が同じでも④ 23 57　で④のみが確実に正常な天秤とわかり23 57　が本物と判明する

なので無駄ない数天秤を使うなら今後は確実に正常な天秤④のみ使い

もう一度④ 14 68

重いほう再度④ 1　4で1と残った9が偽物と判明し天秤使用数6回

なのでこの問題の答えは最小天秤使用数4回最大天秤使用数6回となる

普通に算数の問題だな

まあいいかまたほんの少し賢くなれたし

流石に資料教本知識なしも教養なしに全て頭で考えるのは中々苦労した

私高学歴でも高い教育欠片も体験していないし

最初は9枚から2枚の偽物をこの天秤使って探し出すというのだったけど無理だったので最小使用数と最大使用数を出す問題にしたら上手くいった

わかりにくかったらごめんね

There are nine gold coins that look the same. Two 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 give a random result from "tilts to the right", "tilts to the left", and "balances".

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

You can use each scale only once, and then use any scale you like as many times as you like. Find the minimum and maximum number of scales that can be used to determine the number of counterfeit coins out of the nine without waste.

This problem is the previous answer.

① 123 456 (789 cannot be placed on the scale.

② 147 258 (369 cannot be placed on the scale.

③ 168 357 (249 cannot be placed on the scale.

④ 267 348 (159 cannot be placed

If you do this, the 9 you removed must be a fake, otherwise it will fail and you cannot be sure with this method

To recap, this is a method to find one fake gold coin and use each of the four scales with defective coins once to find one out of the 9 gold coins, and it will look like this

The numbers above are numbered for the 9 gold coins

Even if there is one defective scale, you do not check it on both scales, but place one on each of the four scales and swap the gold coins, one on each side for three coins, and swap all six coins on both sides to get a sure result for all the gold coins. If you remove one gold coin and check each of the 8 coins three times, you will definitely be able to determine which is fake and which is genuine with two real scales, and you will know which is defective, and you can do this with all the scales, with three on each side as shown in the numbers above. If you put six coins on each side, you can definitely find one fake coin under these conditions.

However, for this problem, you need to find the minimum and maximum number of coins that can be used to determine the number of counterfeit coins out of nine without waste.

Obviously, the minimum answer is to put two coins on the balance, one on each side, and if those two coins are counterfeit out of the nine, you can find them by weighing them three times, so it's the minimum number to use the balance without waste, so four times is the minimum.

And the important thing is the minimum and maximum number of coins that can be used to find two counterfeit coins out of nine without waste.

So, you should consider the worst case.

When multiple coins are placed on the balance to use the balance without waste, the random number of defective coins becomes the worst.

For example, if all the coins are numbered from 2 to 9 and ① is the defective coin,

① 123 456

② 123 456

③ 123 456

In this case, 1, 2, and 2 are fakes, and if ① the defective item is heavy and you move it, all the gold coins you put on it will have to be re-weighed.

If one of the nine coins is fake, remove it and put the remaining eight coins on the scale three times, and the fake will definitely be identified with two normal scales, or the one you removed will be identified as a fake by a process of elimination.

However, if you do it this way, in the worst case scenario where the two fakes and the defective item are all judged to be the same, all the gold coins included in it will have to be re-weighed.

Since this method creates a case where a wasteful scale is used, you will need to use a few scales without waste, and you will not know the maximum number of times to use the scale without waste in any pattern.

So when coming up with this answer, you need to use the scale without waste, while also considering unlucky patterns, and use the scale to find one defective item.

So at first, it's like this.

This is where two fakes are mixed in with the gold coins at first, and ① is the defective item.

① 12　34

② 12　34

③ 12　34

In this case, in any worst case scenario, if you replace the defective product, you can easily find out that it is a fake.

The reason why the first case is considered the worst case is that the first case is good and efficient, and the number of numbers used on the scale is not wasted, so it is not the minimum and maximum numbers that can be reliably determined.

For example,

① 12　34

② 12　34

③ 12　34

If 1 and 3 are fake and the randomness is balanced, then swapping 2 and 4 with ④14　32

into the unused scale ④ will reveal that 1 and 3 are fake.

This is the worst case scenario, and even if the judgments of ①, ②, and ③ are the same, if you swap 2 of the 4 cards with the same judgment in ④, the probability is 50%, so you only need to do it one more time.

So, with this method, the minimum number of scales to use without waste is 4 times, the same as the first one in each case.

If you weigh 4 cards in a row in this way, you can use the scale without waste in the worst case scenario, and you can determine that 2 of the 9 cards are fake, and you can get the answer to the number of scales to use without waste.

If ① is a defective product and 1 and 6 are fake, then no matter how the defective product is judged, ① will definitely be found to be defective in ④, and 1 will be found to be fake in ④ 1　2.

① 12 34

② 12 34

③ 12 34

④ 1 2

Again ②　56 78

Then either 56 is fake

Then again ③ 5 6

The fake is found to be 5, so in this case, after 6 tries, 2 out of 9 are found to be fake

If ① is a defective product and 1 and 6 are fakes, and this pattern is used

①　23 57

② 23 57

③ 23 57

④ 23 57

The defective product is balanced, and even if the judgments of ①, ②, and ③ are the same, ④ 23 57 is found to be the only correct balance, and 23 57 is found to be the real one

So, if you want to use a number balance that doesn't waste, from now on, only use the correct balance ④

Once again ④ 14 68

Balance it, and ④ 18 64 If you swap one of each, 1 and 6 are found to be fake, and the number of balances used is 6

If ① is a defective product and 1 and 9 are fakes, and this pattern is used

①　23 57

② 23 57

③ 23 57

④ 23 57

The defective item is balanced, and even if the judgments of ①, ②, and ③ are the same, ④ 23 57 proves that only ④ is a correct balance and 23 57 is the real one.

So if you use a balance that doesn't waste anything, from now on you should only use the correct balance ④.

Again, ④ 14 68

The heavier one again, ④ 1 4 proves that 1 and the remaining 9 are fake, and the number of balance uses is 6.

So the answer to this problem is the minimum number of balance uses is 4 and the maximum number of balance uses is 6.

It's just a normal arithmetic problem.

Well, it's okay, I've become a little smarter again.

As expected, it was quite difficult to think everything in my head without any reference materials, textbooks, knowledge, or education.

I'm highly educated, but I haven't experienced any high education.

At first, I was asked to use this balance to find 2 fakes from 9 pieces, but it was impossible, so I made it a problem of finding the minimum and maximum number of uses and it worked.

Sorry if it's hard to understand