The Professor

[Gifthorse]

I still dont like the deductive solution

What about something that sounds like this

Since the prof. got 9 distinct responses, he himself cannot shake eight hands and therefore his wife cannot shake zero.

And since the person who actually shakes eight cannot shake his wifes zero, then we know the prof. and his wife shook at least one.

After this my brain and wording is failing me. . .

[Werebear]

Ahhh, but if the numbers used in the original solution weren't right, then this is the RECALCULATION of the wife's possible handshakes. So Professor=4. How many choices do we have now for the wife? Is there a formula that descibes her possibilities?

Werebear

[cha]

Ideas for Werebear -
Let's say the guest who reported 0 originally left the note, meaning s/he actually shook 1 hand. We know the one who shook eight hands shook hands with everyone but himself and his spouse, so he and 0 are a couple. Whose hand could she have shook, without meaning a second guest's count was also off? Same argument for the other pairings...

However - "we have already said the professor himself shook 4 hands" - No, we really haven't. We worked the original puzzle as third person observers without direct knowledge of how many people the professor shook hands with. The professor may have shook hands with five people, including the mystery guest who then under-reported.

So the mystery guest could have been someone the prof shook hands with who forgot one, or someone the prof did not shake hands with who remembered incorrectly.

I don't see an answer here, other than "somebody".

[Werebear]

...We, the puzzle solvers, 'figured out' the professor shook 4 hands. I would hope the professor himself remembered how many hands he shook.... we're counting the 4 handshakes for the professor a given.

Werebear

[Buffy]

My dad got 9 handshakes.Someone else I asked got 49 handshakes.

[CrystyB]

and i shook about 25 hands myself too.

[Gifthorse]

This is to me one of the best puzzles

It is simple wording and difficult reasoning

I hope it is solved soon

The puzzle is always easy when the solution is there

Also, have any of you posting people tried this with nine other friends?!!

[slice of pi]

Here lies the problem with only 5 couples. The professor got 9 different answers. If you pair the couples with letters A&B C&D E&F G&H I&J. Starting with A&B(assuming A&B are the professor and his wife),
A..C, D, E, F, G, H, I, J....8
B..C, D, E, F, G, H, I, J....8
C..A, B, E, F, G, H, I....7
D..A, B, E, F, G, H....6
E..A, B, C, D, G....5
F..A, B, C, D....4
G..A, B, C....3
H..A, B....2
I..A....1
J..No hands shaken

When you move down the line to J, J shakes 0 hands. Here lies the problem. If you look at the hands shaken by A&B, J has allready shaken 2 hands, and I who supposed to have shaken one hand has shaken 3 hands, and so forth. I have two other possibilities:

The word problem says that the professor and his wife invited 4 other couples. It could be possible that they each invited 4 couples making a total of 9 couples. At the end of the evening it said that the professor asked each other person, not, he asked each person. Starting with his wife and then asking every other person (all the ladies to keep it simple) he would get 9 different answers.

Then there is a second possible way of looking at this. The Professor and his wife (Hosts) invited 4 other couples (Guests) then you have 9 "Others" for a total of 19 people at the party. The problem says that, "During the evening, various guests and hosts greeted some of the others with handshakes. At the end of the evening, the Professor asked each other person at the party how many different people's hands he/she had shaken; he got nine different answers." But I still don't know how many hands the wife shook!



[This message has been edited by slice of pi (edited 06-05-2002 03:11 AM).]

[Werebear]

Please read the previous solutions. You are including the wife in the nine answers... the 'nine' are the professor and his eight guests. the wife is a repeat of one of the nine numbers.

Werebear

[slice of pi]

I have to disagree. There are ten people (five couples) at the party not nine people. The professor did not ask himself how many hands he had shaken. Like I showed in my chart, in order to have nine different answers the professor and his wife had to shake everyones hands. As you go down the line subtracting one person to make 7 then 6 and so forth the number of answers actually start repeating because in my example, J in the chart shakes 0 hands to make nine different answers, yet A&B have allready shaken his hands making J's answer two hands not zero. Same with (letter)I, his hand count in the chart is one. Look higher in the chart and you will see that A,B,and C have shaken his hands so (letter)I actually has three handshakes not one, which is why I think there are more than 5 couples at the party.

[Werebear]

Please listen to me. I know there are ten people at the party. But only nine of these people we know how many hands they shook. When you make your chart, you listed the wife as a known factor - shaking eight hands. You can't just do that. That's why you got incorrect answers. That's like me filling in a crossword clue with an incorrect answer, and then complaining that no other answers fit. Are you with me?

The professor and his wife had four couples over. Eight people as guests, two hosts. Ten people. Is this not correct? The professor got nine unique answers on how many handshakes.

Who were the nine answers from? There are three, and only three choices.

1) Seven guests, the wife, and the professor. Not likely. No puzzle in there at all.
2) The eight guests and the wife. Well, if we know the wife, then why ask how many she shook?
3) The eight guests and the professor. This gives us nine unique answers, the wife is unknown. THIS IS THE PUZZLE THAT WAS SOLVED THE FIRST DAY. See solution already laid out four times in previous posts, please. Complete with graphs, charts, and 27 8x10 color glossy photographs with circles and arrows and a paragraph on the back describing each one.

Werebear

I'd say lock this forum, but then we'd only get 10 more cropping up. *grin*

[BigBird]

A misconception here - The Professor asked the 8 guests and his wife and got 9 answers. He did not ask himself. HE knows how many hands his wife shook. The puzzle is for US to work it out.

But I agree that the answer is 4

To reiterate in an attempt to make this clear for those who need it....

The responses are obviously 0 through 8
Man "8" did not shake his own hand or his wifes hand - BUT SHOOK EVERYONE ELSES
Therefore his wife MUST have been the person who shook No Hands - There is nobody else who it could be.

Apply the same logic to Man 7,6 and 5 with their respective spouses at 1, 2 and 3

That leaves the person who shook 4 hands with no partner.

This persons spouse (by the logic above) MUST have also shaken 4 hands.

Since there were no doubling up of numbers, the professor must be the other person to shake 4 hands, therefore the person who said they shook 4 hands must be his wife. QED

[Cadmium]

I don't feel like typing. Here's your prove



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People who think they know everything really annoy those of us that do.

[Cadmium]

Hmm, this was meant for slice of pi but he removed his post. Well, maybe it's usefull in the future .

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People who think they know everything really annoy those of us that do.

[cha]

Quote: "A misconception here - The Professor asked the 8 guests and his wife and got 9 answers. He did not ask himself. HE knows how many hands his wife shook. The puzzle is for US to work it out.

This was the point I was trying to make in response to Werebear's twist on the puzzle. Suppose the prof actually shook five hands. If someone gave the prof an incorrect response originally, the prof would have gotten 9 different responses, and WE would be operating on only the info given in the original puzzle, and ASSUME the prof shook four hands. So any of the guests (except the one who reported eight), in Werebear's sub-puzzle, could have forgotten shaking the prof's hand and left the note.

[This message has been edited by cha (edited 06-06-2002 06:48 PM).]

[Andylkl]

One handshake.

[kartelite]

isn't it about time we DROPPED the thread, there is nothing left to be said, it has been worked out different ways over and over, still someone thinks they are smart and has another answer. as was noted the first day, the answer is 4. THE END.

[MacadamiamaN]

If the answer is 4, why is the puzzle still in the "unsolved puzzles" category? It may simply be that the site has not been updated since the last news post on May 14th. As i've only been a member for under a month, I'm not familiar with how often the site gets updated. Is the site updated a lot?

[Cadmium]

It depends. Sometimes it gets updated every two weeks, sometimes every month. So to answer your question, it is possible that solved puzzles are still in the unsolved puzzles category.

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People who think they know everything really annoy those of us that do.

[BearTalon]

I wish I had been here sooner. I read all the posts, but skimmed through some and so I don't know if anyone ever proved it using elimination. It's been assumed in most posts, but I don't recall a proof showing WHY the person who shook 8 hands is paired with the 0 shaker; 7,1; 6,2; 5,3; 4,4.

So, here's what I know. The problem is written correctly and there are no semantic twists in it. Everyone is assumed to have answered correctly. The professor DOES NOT need to know how many hands he's shaken and if he didn't, he can figure it out. The answer *IS* 4, as most of us agree.

9 different answers, no person can shake hands with more than 8 people so the answers are 0 through 8.

Elimination Proof:

Let's say the people who shook *8* hands and *1* hand are a couple, A & B respectively. That means that A shook everyone else's hand, and B duplicated one of those. So, counting from here, all 10 people would have at least 1 handshake (8, 2 and the rest 1's), so there is no person giving ZERO shakes. So A's spouse must have given less than one shake (anything else greater than 1 shake would have the same duplicating result). So, A's spouse B gave zero shakes.

Now that A & B have been decided, similar logic is applied to spouses C & D: if C shakes 7 and D shakes 2, and no more are given to A & B, there is now no-one with only one handshake. So C & D give 7 and 1. Etc.

You wind up with the couples handshakes paired like this - 8:0, 7:1, 6:2, 5:3, 4:4. There are exactly 9 distinct answers. The professor asked everyone else ("each other person" - another way of saying "everyone else"), which includes his wife. I don't know where people got the idea that he didn't ask her or he would already know (did they stick together all night?).

Because the professor got all 9 distinct answers, by elimination his own handshakes must be the other possible 4. Based on the proof, shown in so many ways by other posts, that pairs him with the other 4, which his wife received.

(To kill it totally - if the professor gave any other number than 4 shakes, he would have taken one of the unique answers and the answers to his question would have included two 4's, violating the rules of the puzzle.

Now - Let's wait for a new puzle.

[Cadmium]

And that makes 100 replies on a puzzle that has been solved on the first page .

btw, Werebear, is that you .

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People who think they know everything really annoy those of us that do.

[Werebear]

*bonk Cadmium on the head with a Timex-Sinclair 1000*

Werebear

[Cadmium]

I hope you broke it, Werebear! Those computers are a pain in the *ss.

And I'm fine, thanx for asking.

*wishes he hadn't lied about being OK, it's more KO*

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People who think they know everything really annoy those of us that do.

[DinnerJr]

I also ended up with the Professor's Wife shaking four hands.

If you make a 10x10 grid of possible handshakes and start checking off who shook whose hand, you'll see a couple of immediate necessities:

The nine answers that the professor got were: 0,1,2,3,4,5,6,7,8. We know that you can't possibly shake more than 10 people's hands, since only 10 people were at the party. To shake 10 hands, you'd have to shake your own hand. That's not allowed. To shake 9 hands, you'd have to shake your partner's hand. Also not allowed. So maximum handshaking is 8.

Start with couple A (let's call them Bill and Hillary). Bill was very friendly and shook everyone's hand. He's the 8. Following that, Hillary is the only person at the party who could have shook 0 hands (everyone else at the party shook at least Bill's 1 hand).

Following that logic (not sure if it's ok to give all of it away), you can deduce that all couples, when taken together, shook a total of 8 hands, and the Professor and his wife each shook 4 hands.

[Werebear]

*Bear starts beating his head against the IBM buildlings in Endicott* (hey, they're emptying out fast anyways... )

Werebear

[SaberKitty]

just wondering, why are we still debating the answer to this puzzle? were the bazillian times it was repeated throughout this thread and the 8 or so locked ones not sufficient? are we just bored?
and why don't any of the professor threads have "not adaquite for real life situations" on them? maybe it's because i'm still new-ish, but whenever I lurked(which mostly was here) all the threads with the answers had that attached... just wondering...because if it's not obvious, i'm pretty bored too- i just don't want to argue about an easy puzzle.
-SK

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"You mean you have other words?" cried the bird happily. "Well, by all means, use them. You're certainly not doing very well with the ones you have now."

[Beartalon]

Wow - I was the 100th person to post a reply to an already solved problem.

I was a bit overzealous (finally a puzzle for which I knew the answer) after reading all the different proofs and wanted to add my own.

Now here's the real reason for this post - a variation for which I don't yet have a solution (perhaps this should be a different thread, moderators?):

What happens if instead of couples, we were talking about a triplets convention?
Five sets of triplets (unrelated to each other set) are having a party in one of their rooms after hours.

After greetings are exchanged, it's noted that no two related triplets shook hands.
One individual of one set of triplets asks everyone else (all 14), "How many hands did you shake?"

Assuming everyone else answers truthfully
A. Is it possible to get 14 unique answers?
B1. If so, how many hands did the asker shake?
B2. If not, how many duplicates are there?
C. Is there a pattern to the number of shakes assigned in each group of triplets?

I don't yet know the answer - this is totally off the top of my head.

And no, this is not WereBear, although our birthdates are separated by only one month, one day and one year (+/-) and we have the same occupation, at least one similar interest and come from state (he) and Canadian province (me) both beginning with the word "New" and having coastline on the Atlantic Ocean. Weird.

BearTalon

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The Bear is Grumpy. Feed Him. Feed Him Now.

[This message has been edited by Beartalon (edited 06-11-2002 01:39 AM).]

[Hitchhiker]

*formally* Welcome, BearTalon. I apologize for thinking you might be an alterego of Werebear.

Maybe you and Werebear (and Carebear, if she's still around) should start a Bears' Club, similar to the DogPack. Strength in numbers, and all that.

[Bare]

Would I be welcome, or would this club discriminate against homophones?

[Werebear]

...actually, I found your solution well done, and you even had the answer most consider correct (Except for those who like finding loopholes ) Welcome to the Grey Labyrinth! BTW, I knew a BearTalon on IRC (Undernet) quite a few years ago... you're not him, are you?

Oh, and if you're making me president and dictator-for-life of the Bear Pack, then it is mandated as the first rule that anyone can join. (Except for Orbiting. *grin* Just kidding.) If you're not, then it isn't, and they can't.

Werebear

[Cadmium]

Werebear, why start a Bearpack when you're a fleapack on your own ?

*gets ready to be beaten, kicked and more*

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People who think they know everything really annoy those of us that do.

[cha]

To respond to Beartalon - A is No because if no one can shake hands with on of his or her siblings, that leaves only 12 others available. If one shakes no hands, 13 is the highest number of unique resonses possible.
So, B1 is moot.
B2 would be the person asking could hear a minimum of one number repeated.
C - no real pattern that I could detect. The totals among each set of triplets don't have to add up to the same number...

[Beartalon]

Cha : you're right. I realized no duplicates was impossible (My own fault for writing far too late at night when my math skills diminish). I haven't tried to solve it properly yet to see if I discern any other pattern.

WereBear : I use that nic on IRC although I haven't been on in a long time (a year or so). There was also another user with that nick but he was from the US and I live in Canada. We both used to haunt the same channel - bearcave.

Thanks for everyone's welcome. Now tell me - is being mistaken for WereBear a blessing?

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The Bear is Grumpy. Feed Him. Feed Him Now.

[Werebear]

...of COURSE it is! It's like being mistaken for the President of the Un-- hm. I mean, it's like being mistaken for someone famous, like Michael Jac-- Yecch. I mean, it's like being mistaken for someone really really... oh never mind.

Werebear

[Loreweaver]

Of the 9 poeple (other than the professor) at the party, each person shook a different numher of hands. No person could shake more than 8 hands (9 other people, minus their partner). It does not matter how many hands the Professor shook, bcause he did not answer the poll.

We will give each of the 9 people polled a letter from A to I. Since each of the 9 people polled by the professor replied with a different answer, someone (we will call him/her person I) had to have shaken 8 hands, and one person (person A) had to have shaken no hands. 0, 1, 2, 3, 4, 5, 6, 7, 8 were the nine answers given by the persons answering the professor's poll.

The first round of handshakes comes from person I, who shakes 8 hands (everyone but their partner). Person I is partnered with the person who shook no hands (person A). Otherwise, person I would only be able to shake 7 hands (9 other people, minus their partner and the one who did not shake).

We know off the bat that the professor's wife did not shake 8 hands. She could only shake at most 7 hands (9 people, minus the professor and the person who shook 0 hands). We also know she is not the person who shook 0 hands, otherwise person I could not shake more than 7 hands (9 people, minus the professor's wife and their own partner). So, mark the numbers 8 and 0 off your list.

At this point, everyone but person A has shook at least one hand. The 2nd round of shaking comes from a second guest, person H. After this round of shaking, person H has now shaken 7 hands. Person H has shaken with everyone except person A (who shook no hands) and their own partner. The professor's wife cannot be person H, because then everyone but person A would have shaken at least twice (once from person I, and now from the wife), yet we know someone only shook once. So, mark the number 7 from your list.

Everyone except person A and the partner of person H (which we now call person B) have at least shaken twice (once with person I, and now once with person H). Since person A did not shake any hands, that leaves person B as the only one who has shaken hands exactly once. The wife is not person B, because person B is the partner of person H, and we now know person H is a guest. So, mark the number 1 off your list.

The third round of handshaking comes from person G. After the round, person G has shaken 6 hands. They have shaken the hand of person I, person H, and 4 others. They cannot shake with person A (who does not shake any hands), person B (who is done shaking hands), nor their own partner. That leaves only persons C, D, E, F, and the professor. The wife is not person G, otherwise persons C, D, E, and F would have shaken hands at least 3 times (from person I, person H, and the wife), yet we know one of them only shook hands twice. Mark the number 6 off your list.

So far, everyone except persons A, B, and the partner of person G (which we will call person C) has shook at least 3 hands. Person A shook no hands, and B only shook 1 hand, leaving C as the only one who shook exactly 2 hands. Since we know person C is the partner of person G, and that person G is a guest, person C is not the professor's wife. Mark the number 2 off your list.

The next guest, person F, shakes hands. Afterward, person F has shaken 5 hands. These are persons I, H, G, and 2 others. They cannot shake with person A (who shook no hands), person B (who is done shaking), person C (who is also done shaking), nor their own partner. This leaves only persons D, E, and the professor. The wife is not person F, otherwise persons D and E would have shaken hands at least 4 times (with persons I, H, G, and the wife), yet we know one of these people only shook hands 3 times. Mark the number 5 off your list.

At this point, everyone except persons A, B, C, and the partner of person F (which we now call person D) have shaken at least 4 times. Person A shook no hands, person B shook 1 hand, and person C only shook 2 hands. This leaves person D as the only one who has shaken exactly 3 hands so far. The wife cannot be person D, because person D is the partner of person F, and we now know person F is a guest. Mark the number 3 from your list.

At this point, person E has shaken hands with exactly 4 people, persons I, H, G, and F. Person E is the only person who has shook just 4 hands. Since all 8 guests are now accounted for (persons A, B, C, D, F, G, H, and I), the professor's wife must be person E, the only letter not yet assigned. Since person E shook 4 hands, that means the professor's wife shook 4 hands.

P.S. If you map this out on paper, drawing lines between the handshakers, you will notice 2 more things. The professor shook exactly 4 hands, and shook the same 4 hands that his wife shook!

[Werebear]

Bravo! Bravo! You got the same solution as the rest of us! Bravo!

NEXT!!!

Werebear

[Beartalon]

<action>
BearTalon hands WereBear a cloth to clean the irony dripping off his keyboard into his lap...
</action>

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The Bear is Grumpy. Feed Him. Feed Him Now.

[Infinite|Insanity]

Ok, sorry ladies and gents. I hate to post another reply to a thread that has already been beaten to death over and over again with a very large stick on earth, then purgatory for some, then heaven(or hell, whichever you prefer), then some other place that may exist after that. I was very annoyed with the answer until I took out my trusty sword and shield (pencil and pad) and beat my shield with the sword. I set out to prove that any combination between couples is possible other than the 8-0, 7-1, 6-2.........

What I did was arrange everyone in a circle: P, W, AH, AW, BH, BW.........

Then I proceeded to number everyone in any random fashion from 0-8. Then I tried to link everyone up accordingly, starting with the Professor's wife and going clockwise (staying within the rules of the puzzle as stated). Obviously, my random numbering did not work, until I numbered everyone according to the commandments of the majority, which Cadmium so aesthetically provided for us.

Now, I hate to admit that I am now a part of the majority....YUCK!BLEH!PITOOY! Everyone who has doubts still, can try this experiment. If anyone can come up with some other answer to disprove that the couples must be 8-0, 7-1, 6-2 etc. and allow for the Professor and Mary Ann, to be any combination other than the 4-4, you will officially become my deity and I will worship your name.

*disclaimer - 1) you must follow the rules stated in the puzzle. 2) I am not going to worship you.

[Cadmium]

Originally posted by Infinite|Insanity:

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aesthetically

---

I love that word .

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People who think they know everything really annoy those of us that do.

[CrystyB]

This just isn't working... EVERYONE that wants to post ANYTHING about this puzzle (the professor) cannot add anything of value. So i suggest whenever anyone feels like it, they check out other crazy puzzles, like SB or MH or some other i just can't remember.

I'm tired of this puzzle, and since noone is going to give us any new ones, let's ressurect the hottest ones we've already been over!

[This message has been edited by CrystyB (edited 06-18-2002 10:30 AM).]