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[Ozle]

Yep, done it i think

[Asper]

Yep, done it i think

I'm curious to see your solution (preferably spoilered)

[eHalcyon]

I'm pretty sure I've solved this before.  Maybe it was a variant though.  I will try again.

[Ozle]

Yep, done it i think

I'm eager to see your solution (preferably spoilered)

Will do, pretty awkward to explain without a spreadsheet....

Involves weighing 4 v 4 balls

[Ozle]

Technically i remembered most of it rater than solved it

[Axxle]

Technically i remembered most of it rater than solved it

Are you sure then? I'm trying to work it out but the difficulty seems to be You don't know if the bowl-shaped hat is lighter or heavier.

[eHalcyon]

Technically i remembered most of it rater than solved it

Are you sure then? I'm trying to work it out but the difficulty seems to be You don't know if the bowl-shaped hat is lighter or heavier.

Yeah, I think you're right.  IIRC, it's very doable to find the counterfeit, but sometimes you can't determine whether it is heavier or lighter.

[eHalcyon]

Here is a partial answer for the easier scenario from initial weighing.  I've solved the other scenario in the past (caveat: can't remember if I was able to determine heavier or lighter) but I never remember it and it takes me a while to get it again.


Weigh 4 vs. 4
- if the two sides are the same, go to A
- if one side is heavier, go to B

A) The counterfeit is in the unweighed group of 4.  The other 8 are guaranteed real.  Weigh 3 of the potential counterfeits vs. 3 real.
- if the two sides are the same, go to AA
- if one side is heavier, go to AB

AA) The counterfeit is the one potential that you did not weigh.  Weigh it vs. 1 real to determine whether it is heavier or lighter.

AB) The counterfeit is in the group of 3 you just weighed and you know whether it is heavier or lighter.  Weigh 1 of them vs. another 1 of them.
- if the two sides are the same, the counterfeit is the one you did not weigh.
- if one side is heavier, you know which of the 2 is the counterfeit.

B) The counterfeit is among the 8 you weighed.  The other 4 are guaranteed real.

What now?

[Asper]

Here is a partial answer for the easier scenario from initial weighing.  I've solved the other scenario in the past (caveat: can't remember if I was able to determine heavier or lighter) but I never remember it and it takes me a while to get it again.


Weigh 4 vs. 4
- if the two sides are the same, go to A
- if one side is heavier, go to B

A) The counterfeit is in the unweighed group of 4.  The other 8 are guaranteed real.  Weigh 3 of the potential counterfeits vs. 3 real.
- if the two sides are the same, go to AA
- if one side is heavier, go to AB

AA) The counterfeit is the one potential that you did not weigh.  Weigh it vs. 1 real to determine whether it is heavier or lighter.

AB) The counterfeit is in the group of 3 you just weighed and you know whether it is heavier or lighter.  Weigh 1 of them vs. another 1 of them.
- if the two sides are the same, the counterfeit is the one you did not weigh.
- if one side is heavier, you know which of the 2 is the counterfeit.

B) The counterfeit is among the 8 you weighed.  The other 4 are guaranteed real.

What now?

If you would like to have that information, your solution so far is    right  .

[eHalcyon]

I know it is right so far.

I think I have the rest.  Just finalizing.

[DStu]

What now?

You have 4 potentially heavy (H) and 4 potentially light (L) hats. And four real (R)

You weight HHLL vs RRRL
a) equal: the C must be in the other HHL.  Weight HH, either one is heavier or it's the L.
b) HHLL is heavier.  Then it's HH from the right or L from the left, weight HH and see a)
c) HHLL is lighter.  Then it's LL, weight them.

[eHalcyon]

Here is what I had before:

Weigh 4 vs. 4
- if the two sides are the same, go to A
- if one side is heavier, go to B

A) The counterfeit is in the unweighed group of 4.  The other 8 are guaranteed real. 
Weigh 3 of the potential counterfeits vs. 3 real.
- if the two sides are the same, go to AA
- if one side is heavier, go to AB

AA) The counterfeit is the one potential that you did not weigh. 
Weigh it vs. 1 real to determine whether it is heavier or lighter.

AB) The counterfeit is in the group of 3 you just weighed and you now know whether it is heavier or lighter based on result A. 
Weigh 1 of the potentials vs. another 1 of the potentials.
- if the two sides are the same, the counterfeit is the one you did not weigh.
- if one side is heavier, you know which of the 2 is the counterfeit.

And here is the rest of it:

B) The counterfeit is among the 8 you weighed.  4 are "heavy" and 4 are "light" based on the initial result.  The other 4 are guaranteed real.
Weigh 2 "heavy" and 2 "light" (let this be side X) vs. 3 real and 1 "heavy" (let this be side Y).  Note that 2 "light" and 1 "heavy" are left out.
- if the two sides are the same, go to BA.
- if side X is heavier, go to BB.
- if side Y is heavier, go to BC.

BA) None of the ones you just weighed are counterfeit, so the counterfeit is either the single unweighed "heavy" or one of the two unweighed "light".
Weigh the two remaining "light" ones against each other.
- if they weigh the same, the unweighed "heavy" is the counterfeit.
- if one is lighter, that one is the counterfeit.

BB) The counterfeit is one of the two "heavy" you weighed on side X.  Weigh them against each other.  The one that is heavier is the counterfeit.

BC) The counterfeit is either the "heavy" you weighed on side Y or one of the two "light" you weighed on side X.  Weigh the two "light" ones against each other.
- if they weigh the same, the "heavy" is counterfeit.
- if one is lighter, that one is the counterfeit.

I don't think I made any mistakes, but no guarantees.


PPE: did I just get ninja'd in a much pithier solution?

[eHalcyon]

PPE: did I just get ninja'd in a much pithier solution?

Yeah I did.   I used an extra heavy instead of an extra light though!

[DStu]

PPE: did I just get ninja'd in a much pithier solution?

Don't mknow what "pithier" means, but it's the same solution (mod symmetries).

[eHalcyon]

PPE: did I just get ninja'd in a much pithier solution?

Don't mknow what "pithier" means, but it's the same solution (mod symmetries).

It means you were much more concise.

[Asper]

You are both right, of course. I'm curious DStu, did you remember or find out?

[DStu]

You are both right, of course. I'm curious DStu, did you remember or find out?

I dont know if i knew this variant at all or just the one where you knew if its heavier. Just renemvered splitting 4 4, and then just tried until it fits

[Asper]

You are both right, of course. I'm curious DStu, did you remember or find out?

I dont know if i knew this variant at all or just the one where you knew if its heavier. Just renemvered splitting 4 4, and then just tried until it fits

Cool. I recall i needed an entire day back then at school

[Avin]

If any of you lack the capacity to solve problems involving infinitely countable hats (I know I do), here's one of my favourite puzzles (this one is by Raymond Smullyan; I had to retranslate the version I have, so sorry if the English is wonky):

In 1918, the day the First World War armistice was signed, three married couples celebrated the occasion dining together. Each husband is the brother of one wife, and each wife is the sister of one husband; that is, there are three brother-sister pairs in the group. We know the following:

-Helen is exactly twenty-six weeks older than her husband, who was born in august.
-The sister of Mr. White, who is married to the brother-in-law of Helen’s brother, got married to him on her birthday, which is in January.
-Marguerite White isn’t as tall as William Black.
-Arthur’s sister is taller than Beatrice.
-John is fifty years old.

What is Mrs. Brown first name?

I don't know if anyone will try to solve this one or not, but just in case, here's a hint that normally should get you to the answer:
The date is relevant.

Oh interesting! Building from Jack Rudd's result then that Helen's brother is Mr. White:

The second bullet point basically just says "Mr. White's sister, Helen, who is married to her husband, got married on her January birthday."
The first bullet point says that Helen's husband was born in August, and they are 26 weeks apart in age.

Note that August 1st is exactly 26 weeks after January 31st, on a non-leap year. So Helen and her husband who were born on the same year, were not born on a leap year. John however, was born in 1868 according to point 5, which is a leap year, so John isn't Helen's husband.

Using that information, if Helen is Mrs. Brown, then we have Helen Brown née White, which means that the other two women must be Marguerite White née Black, and Beatrice Black née White. Since John isn't married to Helen, he must be Mr. White, leaving Arthur as Mr. Black. But this means Arthur's sister is Beatrice, which can't be the case since Arthur's sister must be taller than Beatrice.

So Beatrice must be Mrs. Brown.

[pacovf]

Oh interesting! Building from Jack Rudd's result then that Helen's brother is Mr. White:

The second bullet point basically just says "Mr. White's sister, Helen, who is married to her husband, got married on her January birthday."
The first bullet point says that Helen's husband was born in August, and they are 26 weeks apart in age.

Note that August 1st is exactly 26 weeks after January 31st, on a non-leap year. So Helen and her husband who were born on the same year, were not born on a leap year. John however, was born in 1868 according to point 5, which is a leap year, so John isn't Helen's husband.

Using that information, if Helen is Mrs. Brown, then we have Helen Brown née White, which means that the other two women must be Marguerite White née Black, and Beatrice Black née White. Since John isn't married to Helen, he must be Mr. White, leaving Arthur as Mr. Black. But this means Arthur's sister is Beatrice, which can't be the case since Arthur's sister must be taller than Beatrice.

So Beatrice must be Mrs. Brown.

Yes, that's the answer.

Props to you, good sir/lady! I wasn't able to solve it by myself when I first read it.

EDIT: Actually, John was born somewhere between 12th November 1967 and 11th November 1968, but it doesn't change your conclusion, since if he was Helen's husband, then he would be born during August 1968.

[jotheonah]

Oh interesting! Building from Jack Rudd's result then that Helen's brother is Mr. White:

Sorry, how did we get that?

[jotheonah]

that's actually a very cool puzzle. it depends on deduction hidden in seemingly irrelevant details.

[pacovf]

Oh interesting! Building from Jack Rudd's result then that Helen's brother is Mr. White:

Sorry, how did we get that?

He demonstrated that back in page 7.

[jotheonah]

In 1918, the day the First World War armistice was signed, three married couples celebrated the occasion dining together. Each husband is the brother of one wife, and each wife is the sister of one husband; that is, there are three brother-sister pairs in the group. We know the following:

-Helen is exactly twenty-six weeks older than her husband, who was born in august.
-The sister of Mr. White, who is married to the brother-in-law of Helen’s brother, got married to him on her birthday, which is in January.
-Marguerite White isn’t as tall as William Black.
-Arthur’s sister is taller than Beatrice.
-John is fifty years old.

What is Mrs. Brown first name?

I've got some of the way, but I can't seem to get further off the top of my head.


OK, assuming that Arthur, John and William are the men, and Beatrice, Helen and Marguerite the women, what do we have?

Suppose the sister of Mr White is Beatrice. Then Mr White cannot be Arthur (because Arthur's sister isn't Beatrice), nor can he be William (because William is William Black), so he is John. Thus the men are Arthur Brown, William Black and John White. Also, Beatrice is not married to Helen's brother (because Mr White's sister is married to the brother-in-law of Helen's brother), so must be married to Marguerite's brother. Thus Marguerite is married to Helen's brother and Helen to Beatrice's brother - but this would make Helen Mrs White, when we already know Marguerite is Mrs White.

Therefore the sister of Mr White is Helen. (This fits in with her date of birth - 26 weeks before a date in August is nearly always a date in January.) Now who is Mr White? Suppose he's John. In that case, John is the brother of Helen, Arthur the brother of Marguerite and William the brother of Beatrice. This would in turn imply that John is married to Marguerite, Arthur to Beatrice and William to Helen. In which case the women are Marguerite White, Helen Black and therefore Beatrice Brown.

Alternatively, Mr White could be Arthur. In which case our men are Arthur White, John Brown and William Black. Arthur married Marguerite and his sister is Helen, so the man who married Beatrice has Marguerite as a sister and the man who married Helen has Beatrice as a sister.

Quote tag is our friend here.