Schik
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agaricus5 wrote:
House numbers? Is this an American thing, by chance? (I guess by the dollar sign)

I have never heard of getting house numbers for money before.

Most houses here have their number marked on them - generally using little metal numbers you can buy at a hardware store. They cost money.

Here is an example picture of one.


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agaricus5
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Schik wrote:

agaricus5 wrote:
House numbers? Is this an American thing, by chance? (I guess by the dollar sign)

I have never heard of getting house numbers for money before.

Most houses here have their number marked on them - generally using little metal numbers you can buy at a hardware store. They cost money.

Here is an example picture of one.

Okay, okay.

I do understand what a house number is, thank you. I do have one in front of my house as well.

Saying that, I understand the logic now.

I thought at first that it was "[number] units of something", not "a/several unit(s) of it that happen(s) to have the value of [number]".

What happens when the store decids to have a half-price sale, though?

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Frodo
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Sure - I am not around for a bit, and they all get the answer.

Great job, Schik and DiMono. Thanks for posting the explanation too, DiMono - it was the exact same way I solved it, years ago.

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Schik
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And since Onei has declared it to be my turn again, here's a new one:

Dan and Steve were talking one day...

"Dan, I've finished tracing my family tree back from the year 500 AD, and I found one interesting guy".
"What's so interesting about him?" asked Dan.
"Well, he was x years old in the year x^2 (x squared) and he had a son who was y years old in the year y^3 (y cubed)".
Dan looked perplexed. "Sorry Steve, but I can't solve for x or y".
"Well, he was your age when his son was born." said Steve.
"You're right" said Dan. "He was an interesting guy! But I still can't solve for x or y".

How old is Dan?

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Watcher
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Dan is 60 years old.

The son can be born in 720, 990, 1320 or 1716, corresponding to y equal to 9, 10, 11, or 12. All other values give dates before 500 or after 2004. (Actually, 8 doesn't, it gives 504, but there wouldn't be room for the father.)

If y = 9, the father was born in 650 or 702, for x equal to 26 or 27.
If y = 10, the father was born in 930, for x equal to 31.
If y = 11, the father was born in 1260, for x equal to 36.
If y = 12, the father was born in 1640, for x equal to 41.

Only if the father's age is 60 when his son is born is there not enough information to determine the birthyears.

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Schik
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Watcher wrote:
Only if the father's age is 60 when his son is born is there not enough information to determine the birthyears.

Very good. You're up.


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Watcher
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I don't think we've had this one yet.

Continue the sequence:

0, 1, 10, 2, 100, 11, 1000, 3, 20, 101, 10000, 12...

I'd like at least four more terms.

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DiMono
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Not to sound pushy, but wasn't there a rule about not having pure math puzzles in this thread?

[Edited by DiMono on 08-06-2004 at 10:05 PM GMT: one of those words wasn't what I meant to type...]

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Oneiromancer
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Yeah, we've gone over this before, but over 38 pages it's okay to forget. Basically, it depends on the level of math skills involved. If it only requires basic high school skills (like up to algebra and geometry) then it's okay. (Rule #4 on the first post has this in it.) It was puzzles like those requiring trigonometry or Fermat's Theorem, etc., or ones that pretty much required writing a program to solve that were over the top.

At first glance, this just seems to be a math logic puzzle and so it is appropriate.

Game on,

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KevG
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Since the 24 hour rule has long played out and this thread has pretty much ground to a halt, I'll go ahead and give the answer. (Actually, I didn't remember it, even though I saw it on rec.puzzles. I looked it up here: http://www.research.att.com/~njas/sequences/ .)

The answer is 100000, 1001, 110, 4.

Why?


Should I wait for confirmation or go ahead and post my puzzle?

Watcher
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Your answer is correct. Go ahead and post your puzzle.

[Edited by Watcher on 08-18-2004 at 11:57 AM GMT]

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KevG
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O.K. here we go:

A prisoner is scheduled to be executed. However, he is given a chance to save himself. He is led to a room. He is handed twelve balls and a hat is placed on his head so he can't see what color it is. He is then shown a table. The table is perfectly square. On it are four identical coffe mugs set in a square. Under each mug is a penny. The prisoner does not know which are facing heads up and which are facing tails up, but is told that not all are facing the same way. He will be allowed 5 moves. On each move he may remove two of the mugs and adjust the coins as he sees fit. He then will hand the mugs to the guards. They will replace the mugs and rotate the table so that the prisoner does not know which of the mugs he previously picked. After that he makes another move. After any move if the coins are all facing the same way the man will win and be freed. If he runs out of turns he will be executed. What should his strategy be? Assuming optimal strategy, what are his chances of survival?

Krishh
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If he can actually do anything with the balls:
Take two mugs, put a ball in each, switch the pennies the same side up. One the second turn choose two mugs that don't have balls in them.
Chance of success: 100%

If the balls are something you accidentaly left in from another puzzle and have no use in this one the optimal strategy is to choose two mugs at the opposite corners and turn the pennies the same side up. Repeat each turn. This way there is a 50% chance each turn to get a pair (effectivly making the whole puzzle function the same way as with 2 mugs), and I think that the chance of success is 50% too, but I might be wrong. (in which case another solver just has to calculate the percent correctly using my solution)

And I don't have any new puzzles in mind currently, so feel free to post one.

[Edited by Krishh on 08-18-2004 at 12:57 PM GMT]

KevG
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The balls and hat are intended as humorous references to other well-known puzzles that have appeared previously in the thread. (And pretty much everywhere else.) They have no bearing on the puzzle. Besides, the guards are replacing the mugs and wouldn't allow him to place the balls there anyway.

Always picking two diagonally opposite mugs is not the optimal strategy. While it will give you a much more than 50% chance of survival, there's a strategy that is better.

DiMono
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Take the two mugs nearest to you and lift them up. If they're the same, flip them both. If they're different, flip one of them. Let's say you're making them tails.

On the second move, pick up two diagonal mugs. If they're both tails, then there is precisely one coin still showing heads. If there's one of each, then flip the heads over to tails, and either you are freed or there is NOW precisely one coin still showing heads.

At this point, if not already freed, you have only a 1/8 chance of being executed if you always flip the same two diagonal mugs.

Explanation:

Given that they don't all start the same way, there are 14 possible arrangmenets, 6 of which have two turned each way, 8 with three turned one way.

If two turned each way
There are 1/3 odds you'll see two turned the same way when you see your initial mugs. This means there's a 1/3 chance of immediate freedom given 2 each way. If you don't see two the same initially, you have a 15/16 chance of subsequently seeing and flipping the offender. This means you have a 5/8 + 1/3 chance of freedom if they start with two each way, or 23/24.

If three turned one way
There are 1/2 odds you'll see the one turned the wrong way, and a 1/2 chance you'll flip it, so there's a 1/4 chance you'll gain immediate freedom. If you don't flip it, you must flip a coin on your second move, leaving you with 7/8 * 1/4 chance of freedom.

If you don't see the one turned the wrong way, you have a 15/16 * 1/2 chance of freedom, so that's 15/32 + 1/4 + 6/32 = 29/32 chance of survival.

The sum of odds

(23/24 * 6/14) + (29/32 * 8/14)
23/56 + 29/56
52/56
93% chance of survival

[Edited by DiMono on 08-18-2004 at 03:47 PM GMT]

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wackhead_uk
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Seeing as I cant figure out a puzzle andihaventdonetheaboveone I will post something I heard from a friend. Not a real puzzle, but funny, and it uses a lot of synonyms, so it might not sound as good when you read it, but it is good if you tell it to somebody.

A man is sentenced to death. He is well known for escaping, through varous methods. They therefore decide to put him in a room impossible to escape from. It has only a light set into the roof for light, which he cannot get off or use in any way to escape with. They give him only a table to sleep on, so he cannot use the blankets to escape out of the window (Which, incidentally, is 200ft off the ground). They weld the door shut and give him nothing apart from his clothes.
The next morning, they go in, and he has escaped. How?

Here's the answer, don't blame me for the "oh ha ha" factor of it:



Now if any of you didn't feel like hitting your own head against the wall, tell me now.

KevG
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DiMono wrote:
Take the two mugs nearest to you and lift them up. If they're the same, flip them both. If they're different, flip one of them. Let's say you're making them tails.

On the second move, pick up two diagonal mugs. If they're both tails, then there is precisely one coin still showing heads. If there's one of each, then flip the heads over to tails, and either you are freed or there is NOW precisely one coin still showing heads.

At this point, if not already freed, you have only a 1/8 chance of being executed if you always flip the same two diagonal mugs.

There's a better strategy than this available. Too early to say any more than that

Krishh
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Now if any of you didn't feel like hitting your own head against the wall, tell me now.

Ouch. You should seriously consider sending that to Really Bad Jokes. It would fit right in.

Nillo
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Krishh wrote:
Ouch. You should seriously consider sending that to Really Bad Jokes. It would fit right in.

That site rocks.

I really liked #651.

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Mattcrampy
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Hmmm, I'll have a crack.

First turn: Pick up diagonal ones. Flip them so they're the same way. Say, heads.

Second turn: Flip the two nearest you. Make sure the left one is heads, the right one is tails. Now you either have a diagonal arrangement, an arrangement where one side's heads and one's tails, or you have three heads and one tail.

Third turn: Flip the two nearest you. If they're both heads, you've either got a three-on-one or a two-sided one. Leave them. If one's heads and one's tails, flip the tail, to make a guaranteed three-on-one (assuming you don't walk free at this stage). If both are tails, flip them and you walk free.

Fourth turn: Take a diagonal. If one's heads and the other's tails, flip the tail to make a three-on-one. Now you have a three-on-one for sure, if you're unlucky enough to still be here.

Fifth turn: You have a 1 in 2 chance of being freed. Pick a diagonal and pray.

Hang on, this is worse than DiMono's...

Matt

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KevG
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Mattcrampy wrote:
Hang on, this is worse than DiMono's...

Yeah, it is In particular, you took 4 moves to find the same information DiMono found in 2 moves.

[Edited by KevG on 08-19-2004 at 09:08 AM GMT]

KevG
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KevG wrote:
There's a better strategy than this available. Too early to say any more than that

I don't think it's too early anymore. Since the optimal strategy gives a better than 93% chance of success, I don't think it's much of a spoiler to say that it's possible to guarantee that the coins are turned all the same way.

Also, in my reply to Matt I strongly implied that knowing the exact state of the coins is important. It definitely is.

I'll post a secreted spoiler in a few days for closures sake if no one is interested in attempting this. Then either someone can post a new puzzle or we can pull the life-support on this thread.

DiMono
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Alright, picking up from my previous method, with 3 heads and 1 tails after two flips:

Flip 3: turn over two adjacent cups. If they're different, you know what the odd one out is, so flip it and go free. If they're not different, then flip one; it doesn't matter which. There are now two of each, either diagonal from each other, or next to each other. Remember which coin you flipped to tails: let's say the left one.

Flip 4: turn over two adjacent cups. If they're the same, flip them both, you're free. If there's one of each, but the tails is on the right, do nothing, because you're free on flip 5. If there's one of each with tails on the left, flip them both.

Flip 5: turn over two diagonal cups. If they're the same, flip them. If not, then you were looking at the same two coins on flip 4 as on flip 3, and there's nothing you can do about it.

Freedom: .5 + 1/3 + 1/8 = 96% given three coins after 2 turns, which ends up at about 98% overall. Not quite there yet, but closer...

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eytanz
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As far as I can see, as long as you're just selecting mugs with no way of marking them you can never know the exact state of the coins, because there's always a possibility that there's a coin you never see.

Therefore, the only way it would be possible to guaruntee a 100% survival chance is if you had some way of influencing mug selection to make sure you see all the coins. The question, then, is what exactly allowed when you are "adjusting the coins as you see fit". Are you allowed to move coins in addition to flipping them?

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DiMono
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1) Pick two diagonals and make sure they're both heads.

2) Pick up two adjacent and see what they are. Don't flip either of them, but note what they are. If they're both heads, you have 3H and 1T. If they're different, you either have 2H2H diagonally or 3H1T.

3) Pick up two diagonals and see what they are. If they're different, flip the tail over and go free. If they're both tails, flip them and go free. If they're both heads, it depends what happened on flip 2.

3.1) If you flipped up two heads in step 2, then flip one of the heads to tails. You are now looking at some orientation of

HH
TT

From there you pick up two adjacent and flip them both, then pick up two diagonal and flip them both, and go free.

3.2) If you flipped up one of each in step 2, then flip both of them to tails. You either go free, or have 1H3T. You then proceed as directed by 3.1, swapping heads for tails.

100% free after 6, 96% free after 5... getting there.

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DiMono
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I got it!

1) Pick up two diagonals and make sure they're both heads. Either you flipped one or you didn't, it doesn't matter:

H  -> H   OR  H  -> H   OR  T  -> H
 T     H       H     H       T     H

2) If you're not free, pick up two diagonals and see what they are. If they're different or both tails, make them both heads and go free. If they're both heads, then flip them both to tails. If you're not free, then you are necessarily looking at some orientation of:

TT
HT

3) Pick up two diagonals and see what they are. If they're different, flip the heads over and go free. If they're the same, flip one over. You're now looking at some orientation of:

TT
HH

4) Pick up two adjacents and flip them both. If they were the same, you're free. If they weren't the same, you now have some orientation of:

HT
TH

5) Pick up two diagonals and flip them both, and go free.

[Edited by DiMono on 08-22-2004 at 02:58 PM GMT]

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eytanz
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Oh! What I didn't see was that you will be freed after any turn, not just after the last turn. D'oh. Well done, DiMono.

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DiMono
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eytanz wrote:
Oh! What I didn't see was that you will be freed after any turn, not just after the last turn. D'oh. Well done, DiMono.

Thanks. Now I need to do some stuff and come up with a puzzle. See you all soon!

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KevG
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Bingo

DiMono
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Alright, a simple word puzzle to keep you all occupied. Call it a break from the math-oriented puzzles we've been seeing recently.

What animal comes next in the sequence?

eel, cat, lion, ibex, panda, skunk, ?

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