Riddle Needs Solved....Now!

Miscellaneous and Personal Stuff I Must Share
21

Wow, guys, good work! My favorite is “Hat-Trick”… I tell that riddle to people all the time. It is just easy enough to figure out, but just hard enough to force you to think for a moment.

“Tactor Beam” is a real stumper for me. Bryan, could you explain your reasoning on Timing is Everything? That one always bothered me.

Also, lying defeated is not phrased poorly, though it makes the “usual” assumption that a lone dichotomy in answering exists.[spoiler] If purple-solers always lie, asking a person what color soles they have means, no matter what color soles they have, they must answer green (greens will honestly say they have green soles, and purples will lie and say they have green soles). This means that the second man, as translating, must be telling the truth and have green soles (in order to be lying, the native-speaker would have ha d to have said, “I have purple soles”, which is an impossible case). Asking the thrid man what color soles the second man has then gives this away completely.

To Bryan: I am not sure that “yellow” is really within the scope of the riddle.[/spoiler]

nightime, now that you know what [**php] does, maybe you can see what [**spoiler] does. :wink:

Enlightenment is my second favorite riddle ever, though this one actually stumps many people. The solution is very pleasing to me with four switches rather than three, though the methodology isn’t really any different. (four is the limit to the switches, however; given five there can be no guarantee of success, though one can reach a 50/50 proposition with it).

Flight Plight is the sort of riddle that irritates me. It is almost too complicated to just rattle off, but it is trivial with paper and a decent diagram.

I do not get Heads Up… I thought I did, but since it isn’t a prob or stat riddle, then I am truly stumped. The only thing I’ve ever been able to think of is…Don’t stack them heads up, stack them on their sides like a roll of quarters lying longways… all have the same number of heads… zero! But this takes liberty with the word “pile” I think, and feels like cheating. I do not like this riddle. I wish someone could tell me what it was, pretty please? I would be very disappointed if the only answer was “feel the coin for the face”… grrr.
Eat and Run: AFAICT, all but the very first one can be guaranteed to be saved. The strategy is simply to call out the color of the hat in front of you. The first man then has a 50/50 shot at being right, but I can see no way to guarantee his life.

22

nightime, that solution for Heads Up was the only thing I could think of, too, but it just doesn’t seem right. You beat me to posting the answer to Three Dots, but here’s the extra credit:

[Spoiler]Assume there are only:
Two green dots. If this were the case, it would be the same situation as before: the people with the green dots could guess their own color. Since this didn’t happen, there must be at least:

Three green dots. However, three people would see two green dots and one red dot. Since there must be at least three green dots, and one person obviously doesn’t have one of them, any of the three people with green dots could guess their own color. But this didn’t happen, so there must be:

Four green dots. Since no one could immediately guess their color, everyone must have a green dot.

Interesting note: this problem can also be solved even if the natives leave your blindfold on![/spoiler]

For Tractor Beam, this is the only thing I could think of, and I admit it’s pretty slim:

The pier used to be a bridge out to the island. The tractor was simply driven over the bridge. However, an unfortunate tidal wave wiped out most of the bridge soon afterwards, leaving only enough bridge left to convert to a pier.

23

Oops, that’s erislover’s solution to Head’s Up. Nightime still beat me to Three Dots.

24

This wouldn’t guarantee that all but one are saved. Since you can only call out the next person’s color or your own color (assuming you’re not last in line), you can guess your own color and save yourself, but the next guy won’t have any idea what his color is. If every other person guessed the color of the person in front of him, then you could save approximately 50% of those people (if their color is the same as the person’s ahead of them), and you could also save all the other people, since they know their color. So, ~75% could be saved with that method.

Can anyone think of a solution that can save more people?

25

Action Auction

The deal is that both the highest and the second highest pay their bids, but only the highest bidder gets the $100. So, the bidding will quickly proceed to $100. At that point, everyone will drop out except the highest bidder ($100) and second highest bidder (say, $99) – everyone else drops out because now they’re losing money, even if they win the highest bid. So the second highest bidder owes $99, and the highest bidder breaks even. The second highest bidder, not wanting to pay $99, bids $101, becomes the highest bidder and nets a $1 loss ($101 bid - $100 prize). However, the original highest bidder, now the second highest bidder, owes $100. So he bids $102, becomes the highest bidder and nets a $2 loss. The other person then ups his bid to $103 to net a $3 loss, and so on and so forth. The bidding will continue until one of the two bidders gets tired and stops bidding. Prof. Chaos has to pay $100, but will be paid at least $99 (if the bidding stops at $100 and the second highest bidder just accepts his loss). If the bidding continues, Prof. Chaos stands to make a lot of money.

26

Action Auction

The deal is that both the highest and the second highest pay their bids, but only the highest bidder gets the $100. So, the bidding will quickly proceed to $100. At that point, everyone will drop out except the highest bidder ($100) and second highest bidder (say, $99) – everyone else drops out because now they’re losing money, even if they win the highest bid. So the second highest bidder owes $99, and the highest bidder breaks even. The second highest bidder, not wanting to pay $99, bids $101, becomes the highest bidder and nets a $1 loss ($101 bid - $100 prize). However, the original highest bidder, now the second highest bidder, owes $100. So he bids $102, becomes the highest bidder and nets a $2 loss. The other person then ups his bid to $103 to net a $3 loss, and so on and so forth. The bidding will continue until one of the two bidders gets tired and stops bidding. Prof. Chaos has to pay $100, but will be paid at least $99 (if the bidding stops at $100 and the second highest bidder just accepts his loss). If the bidding continues, Prof. Chaos stands to make a lot of money.

27

For Heads Up

Pick 20 coins. Flip them all and put them in a pile. Leave the rest of the coins unflipped in the other pile. If you got all tails, then they’re now all heads and there are still 20 heads in the other pile. If you got n heads, there are now 20-n heads in each pile.

For Three’s a Crowd

They will never come to an agreement. As Bryan Ekers posted, A will get $99.99, B $0.01, and C $0.00. But then C would say to B “I’ll take a penny and give you $99.99” (or I’ll take n and give you 100-n) and B & C would both be happier, so they’ll switch. But then A would say to one of them (whichever stood to make less) “I’ll give you k and take 100-k.” And it could go on forever, because no distribution is stable. For any distribution, the 2 people with the least amount of money (even if some amounts are equal) could agree to give each other more and cut out the person with the most money. So they’d make counteroffers forever.

For More Hats

I can get up to a 7/8 chance of winning. If 3 is red, 1 & 2 pass & 3 says red. If it’s blue and 2 is red, 1 passes & 2 guesses red. If 2 & 3 are blue then 1 guesses. Only 1/4 of the time 2 & 3 both have blue & 1 is forced to guess, and for half of those the guess is wrong.

For Eat and Run, I couldn’t do better than jmizzou’s answer.

Does anyone have anything for Tractor Beam or Ties that Bind? I’m guessing jmizzou’s Tractor Beam answer doesn’t work, since the next riddle starts with the phrase “After using the same method that the tractor took to get to the Island.”

28

Ah, jeez, that Eat and Run solution is incorrect! Hmm, indeed. More hypothesis:

[spoiler]If one is limited to speaking only one word, then all but the first may be saved by saying the decimal-equivalent of a binary number that represents the 9 hats he sees (with, for instance, black = 1 and white = 0). This will kill one man, but save all the others.

Contrariwise, if one is only allowed to speak colors, then the agreement could be as such: if you have the same color hat as me, I will say my color in a deep voice; if you have the opposite color as me, I will say it in a high voice. This way they still say their own color, but also pass along information as to what is in front of them. The first man still only has a 50/50 shot at living as his job is to only speak the color in front of him.[/spoiler]Think I got it there. :slight_smile:

29

Oh wow, that Heads Up solution is fantastic! So simple!!! :smack: But, knock knock, the riddles aren’t really linked in any way. If they were, this would be the lone example.

…
God, that solution to Heads Up really is great, though.

30

My More Hats answer is wrong - I missed the word “simultaneously.”

For No Vacancy and No Vacancies,

I think there’s not enough information. For the first one, suppose there are an infinite number of rooms, with 20 unoccupied and the rest occupied. (This is the same setup as the Heads Up problem, so I should be able to say this). Then there is a room available for me. But suppose there are an infinite number of occupied rooms, and no unoccupied rooms. Then I don’t fit. (That’s like saying "suppose there are an infinite # of coins with heads up. How many of those coins have tails up? Zero.) For the No Vacancies riddle, you obviously can’t fit an infinite number of boarders if you can’t fit one. But suppose you have an infinite number of occupied rooms, and the occupants all leave and split into 2 equal groups (e.g. those in odd-numbered rooms and those in even-numbered rooms). Now one group of the forest people arrive and go back to their old rooms. Now the other group arrives. Of course they can still go back to their old rooms. (They could even have kept their room keys.) So everyone fits, or no one fits, depending on information that is not given in the problem. So the answer is: not enough info.

Some of the other riddles are linked, erislover. Trouble in River City starts this way:“After asking which door he would send his Mother-In-Law through and choosing the other one (Open Door Policy), you are allowed to walk through the Door of Freedom.”

31

Ties that Bind (careful! reading this will cause you to smack your forehead):

String1: light both ends, String 2: light one end. When string one burns out (30 min), light the other end of String 2 (which had 30 left, but by lighting the other end, you count 15). Incidentally, this doesn’t even need to be a “theoretical” riddle as by looping the first strig you can light all three ends simultaneously. Unfortunately not my solution. This riddle somehow came up in GQ and someone figured it out there.

32

Oh, and erislover, thanks.

The reasoning behind the Timing is Everything answer:

Look at it from the car’s point of view. It gets home 20 minutes early. It didn’t have to drive the distance between where it picked Joe up & the train station, and it didn’t have to drive back that distance. So driving that distance twice takes 20 minutes. So driving that distance takes 10 minutes. So it was a 10 minute drive from the train station. Since it usually arrives at the train station at 5, it was 4:50. So Joe had been walking for 50 minutes.

33

Wow, that Heads Up soultion is nice. Erislover’s second solution for Eat and Run is probably the answer.

Light Bridge:

[spoiler]#2 and #1 go across the bridge (2 minutes). #2 returns alone (4 minutes). #5 and #10 go across the bridge (14 minutes). #1 returns (15 minutes). #2 and #1 go across the bridge (17 minutes). Graphically:

          +----------+  1,2,5,10 (0 minutes)

1,2 ±---------+ 5,10 (2 minutes)
1 ±---------+ 2,5,10 (4 minutes)
1,5,10 ±---------+ 2 (14 minutes)
5,10 ±---------+ 1,2 (15 minutes)
1,2,5,10 ±---------+ (17 minutes)
[/spoiler]

34

No vacancy/ies

The first one- just bump everyone up one room. Then give room 1 to the new guest. When another infinite number of people come, just put all the current guests in the room numbered double to the one they are in (1>2, 2>4, 3>6, etc.) and the new guests have an infinte number of odd rooms to use.

35

What about the answer to the riddle of what happened to the “to be” in the title of this thread?

36

Great Dave, alternatively, for those of us particularly nerdy (and still saving room for a third and fourth batch of infinite guests) Place first infinite group in rooms that are a power of 2. Place the next batch in rooms that are multiples of 3. Pleace the third group in prime numbered rooms that aren’t room 2 or 3. A fourth group can fit in the rest; I have no idea what numbers are left so easily described, but they are infinite, too.

37

Oh, knock knock, most of them are linked in that they are told in a story-like format, but solutions to one aren’t buried in the others. If it were, the answer would have been given in Flight plight, except that it would completely contradict Tractor Islad’s assumptions. See what I mean?

38

For Sailor’s Delight

Format: when I list #s in order, they go from the head pirate (#1) to the bottom pirate, e.g. 1,0,99 is 1 for the boss pirate, 0 for #2, and 99 #3, and no other pirates live. If it gets down to 2 pirates, #2 will take everything & #1 cannot veto it, so the outcome is 0,100. If it gets down to #3 and he offers #1 more than 0, #1 will take it. So if it gets to #3, #3 offers 1,0,99, and it stops there. #2 knows this will happen, as does #4, so when it gets down to #4 he offers 0,1,0,98, gets yay votes from himself & #2, and it stops there. They all know this will happen, so #5 offers 1,0,1,0,98 and it will stop there if it gets to him. Similarly, #6 offers 0,1,0,1,0,98, #7 offers 1,0,1,0,1,0,97, #8 offers 0,1,0,1,0,1,0,97, #9 offers 1,0,1,0,1,0,1,0,96, and #10 offers 0,1,0,1,0,1,0,1,0,96 and gets yay votes from #2,#4,#6,#8, and himself. So zero pirates die. And the boss was stupid to agree to this means of dividing the money - he gets nothing. It’s more interesting if the pirates are bloodthirsty, and will kill the proposer if the vote is 50-50 (solve for extra credit).

39

Actually, knock knock, the No Vacancy problem suffers from poor wording, but is still solvable. Let’s just say that it is implied in both cases that the hotel is entirely full (i.e. the neon NO is flickering more on than off).

[spoiler] Let all the rooms be numbered 1, 2, 3, … The solution is for each person to give their room to the person in the next lower number (So the person in 2 moves into 3, the one in 59 into 60, etc.) This leaves room 1 unoccupied for the new arrival.

The second problem can be solved in many ways; the most intuitively appealing follows your reasoning a little : All the current occupants move into even-numbered rooms (1 into 2, 2 into 4, 3 into 6, etc.) and the new arrivals go into odd-numbered rooms.

Both of these are concerned with the topic of cardinality of sets, which is pretty much defined in this manner (i.e. a one-to-one relationship can be found between the two sets).
[/spoiler]

I like erislover’s first one for “Eat and Run”. Here’s another one if you’re only allowed to speak the color.

[spoiler]The first person counts the number of black hats. If it’s odd, he says “black”. If even, “white”. Each person after that can figure out their own color based on this and those who’ve gone before.

For example, using 1 for mark^H^H^H^Hblack and 0 for space^H^H^H^H^Hwhite,
the lineup (left is back, right is front) might be :

x01011011 (x is the first guy, about whom we don’t care).
So he sees 5 black hats, says “black” and gets eaten (he had a white hat, actually).

#2 sees an odd number of black hats, so knows his is white, and says so.
#3 sees an even number of black hats, and knows hers must be black, etc.
[/spoiler]

And there’s a similar solution for three colors (not original with me, even though I was able to figure out the method I just described.) Here’s the solution to that:

You add up the total value V of all the hats seen, assigning a color to each number, then speaking V mod 3. Again, each person depends on those who’ve already gone. This is of course exactly the same as the first one (V mod 2) but ‘even’ and ‘odd’ is easier to understand.

My best guess for “Tractor Beam” :

The tractor is made of some low density material or is a reject from “Junkyard Wars”. Either way, it floats.

I got “Heads Up” almost immediately but still can’t get the 4-switch solution.

40

knock knock, I don’t quite agree.

[spoiler] given 1, 2; 2 gets 100 I agree.

Given 123; 1, 0, 99; I agree.

2, being supremely greedy, will never vote for anything other than his own plan, and no player would vote for 0,100,0,0,0,…,0; so there is no way to bribe 2.

Thus, given 1234; 2, 0, 0, 98. Otherwise it is conceivable that 1 won’t agree, since he gains nothing for voting for 4’s plan over 3’s, and by hypothesis 3 loves his life so the only way to guarantee success is to offer 1 two gold.

Given 12345; 3, 0, 0, 98, 0; “what?!?!” you say… well, given the previous resoning, and having 4 know he stands to make 98 if he votes it down, and 3 stands to gain 99 if he votes the others down, 5 is stuck with no gold just to stay alive! Tragedy.

Given 123456; 3,0,0,0,1,96; should be clear enough… 5 gets the shaft in gold if he disagrees, so being greedy he has to answer “aye” to this plan. Vote ties, counting as success.

Given 1234567; 3,0,0,0,1,96,0.

etc etc…[/spoiler] That’s how I’ve always figured it…