Oneiromancer
Level: Legendary Smitemaster

Rank Points: 2936
Registered: 03-29-2003
IP: Logged

Okay, this is the same kind of puzzle as the introduction to Dugan's level 9...only not trivial.

You have a 3 gallon bottle and a 5 gallon bottle. You want to measure out exactly 4 gallons of water. How?

To forestall any questions, there is an unlimited supply of water you can draw from.

I wanted to make an amusing story to go along with this...but I decided to spare you all.

Oh yes, and there are at least 2 solutions to this. For extra credit you can give both of them.

Game on,

(Note: there is no such thing as extra credit in this game.)

[Edited by Oneiromancer on 12-22-2003 at 07:41 PM GMT]

____________________________
"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

levelthirteen
Level: Master Delver
Rank Points: 109
Registered: 02-20-2003
IP: Logged

File: 3134cd203.txt (715 bytes)
Downloaded 180 times.
License: Other
From: Unspecified

Ascii spreadsheet attached.

Oneiromancer
Level: Legendary Smitemaster

Rank Points: 2936
Registered: 03-29-2003
IP: Logged

Looks good to me, your turn.

Game on,

____________________________
"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

levelthirteen
Level: Master Delver
Rank Points: 109
Registered: 02-20-2003
IP: Logged

On a standard minesweeper board (30x16 with 99 mines), what's chance of getting exactly one zero and 8 nonzero safe squares on randomly selected first move?

The_Red_Hawk
Level: Smitemaster
Rank Points: 783
Registered: 09-02-2003
IP: Logged

For Oneiro's puzzle, I would like to point out a way of solving it that is not generally known. I don't know if this is the one levelthirteen did though, because I can't download the spreadsheet.

You simply fill both full, then empty them by half, leaving you with four. You empty them by half by pouring out the water, then stopping when you see the edge on the far side.

____________________________
Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

.....the king of the skies.....

eytanz
Level: Smitemaster

Rank Points: 2708
Registered: 02-05-2003
IP: Logged

I'm not sure what you mean by that last part - and you're assuming that the containers have standard shapes. If one is donut shaped, for instance, and the other is curled in the form of a lower-case e, how would you ever know when you've emptied half? The standard solutions (which levelthirteen provides versions of) don't depend on anything like that, and are bound to be accurate.

Also, you're assuming that the person doing the measuring isn't blind, or in the dark, or filling and emptying the containers by proxy (say, using impercise mechanical arms). None of these problems affect the standard solutions, but your solution won't work with any of them.

____________________________
I got my avatar back! Yay!

Oneiromancer
Level: Legendary Smitemaster

Rank Points: 2936
Registered: 03-29-2003
IP: Logged

Yeah, I kind of forgot to give the standard statement of "you have no way of measuring volume any other way than through the containers that you are given" or something like that. It's a normal logic puzzle, not a "fuzzy" logic puzzle.

Game on,

____________________________
"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

Schik
Level: Legendary Smitemaster

Rank Points: 5383
Registered: 02-04-2003
IP: Logged

levelthirteen wrote:
On a standard minesweeper board (30x16 with 99 mines), what's chance of getting exactly one zero and 8 nonzero safe squares on randomly selected first move?

This one's tough, and I haven't yet come up with a solution. Experimentation is telling me that the chance is a touch over 3% - say, about 3.06575 percent. Getting to that number with just pen and paper - that's another issue.

____________________________
The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi

Scott
Level: Smitemaster
Rank Points: 578
Registered: 02-12-2003
IP: Logged

3% seems a bit high to me. I have played many games of mine sweeper and never had it happen. Unless its been coded in such a way as to make this never happen which I guess is possible.

The_Red_Hawk
Level: Smitemaster
Rank Points: 783
Registered: 09-02-2003
IP: Logged

Speaking of minesweeper, I've noticed that my first click is never a mine. Is that coded, or just luck?

____________________________
Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

.....the king of the skies.....

Scott
Level: Smitemaster
Rank Points: 578
Registered: 02-12-2003
IP: Logged

Its coded that way. Easy way to test that is make a custom game and put in the max mines you can. You can never hit a mine the first click.

eytanz
Level: Smitemaster

Rank Points: 2708
Registered: 02-05-2003
IP: Logged

It's coded. If you click a mine, it's moved somewhere random.

Presumably, that does have some effect on the answer to the question above, which seems like an extremely difficult puzzle to me, unless there's some trick I'm not aware of.

It's probably not too difficult to write a computer program to calculate the answer for you (though I can't see an easier way at the moment than using brute force to generate all the possible boards, and count good starting positions - which will take several weeks on a modern computer - but that doesn't mean there isn't one). But I don't think this thread should be about the sort of puzzles that can't be solved using one's brain (and maybe pen and paper) alone. And if that's possible here, I have no idea how.


[Edited by eytanz on 12-23-2003 at 05:35 AM GMT]

____________________________
I got my avatar back! Yay!

Schik
Level: Legendary Smitemaster

Rank Points: 5383
Registered: 02-04-2003
IP: Logged

Well let's all put our brains together and see what we can come up with.

I'll start with the obvious - the 5x5 area centered around the randomly selected point (S) must look like this:
XXXXX
XOOOX
XOSOX
XOOOX
XXXXX
Where O is definitely NOT a bomb, and X may or may not be. Every O must be connected to at least one bomb.

We also know that if S is in the edge rows/columns, it's automatically out, as it doesn't have 8 neighbors. Likewise if it's in the next-to-edge rows/columns, it's out because at least one of its neighbors can't have any bombs adjacent to it. So right there 168 out of 480 starting points cannot possibly fit the criteria of the puzzle.

Let's say that we're using "standard" Minesweeper rules, so S is by definition NOT a bomb. Thus the 99 bombs are randomly placed in (30*16-1) = 479 spaces. The odds that all the spaces marked O are *not* bombs is 380/479 * 379/478 * 378/477 * 377/476 * 376/475 * 375/474 * 374/473 * 373/472.

Now the tough part - of the remaining 471 tiles on the board, 99 are bombs. What are the chances that each O is adjacent to at least one bomb? I'll sleep on it and pass the torch on to the next person.

Side note - With the above "rules" - removing edge/near edges and the easily calculated inner ring of "O"s, we're down to about 5.4% already. If my 3% guess earlier is correct, then most often than not, if S isn't on/near the edges and there are no bombs adjacent to S, then each O is adjacent to at least one bomb.


____________________________
The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi

levelthirteen
Level: Master Delver
Rank Points: 109
Registered: 02-20-2003
IP: Logged

There are 36856(out of 65536) ways to place mines in the 'X' so that all of the 'O' squares are next to a mine.
If all of the puzzles in this thread need to be solvable by hand, then this is probably invalid.

agaricus5
Level: Smitemaster
Rank Points: 1838
Registered: 02-04-2003
IP: Logged

levelthirteen wrote:
There are 36856(out of 65536) ways to place mines in the 'X' so that all of the 'O' squares are next to a mine.
If all of the puzzles in this thread need to be solvable by hand, then this is probably invalid.

Why?

He/she may not be faster than a computer, but a human could do it eventually as well. You just used a faster and more efficient method.

____________________________
Resident Medic/Mycologist

Schik
Level: Legendary Smitemaster

Rank Points: 5383
Registered: 02-04-2003
IP: Logged

Argh... so we need to sum up the chances of these 36856 possibilities.... I wont' be able to do it in the next couple days, but when I get a chance if nobody's solved it yet, I'll write a little code to figger it out.


____________________________
The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi

DiMono
Level: Smitemaster

Rank Points: 1181
Registered: 09-13-2003
IP: Logged

There must be at least 4 mines on the outside, out of 16

There are 47 ways for 4 of the outside squares to be mines. For most of the other ways to do this, it's always some number * these 47. There are basically 2^12 outside combinations, so we're looking at 81 * 2^12 which is 4096 * 47 = 192512

Yes, there are only 65536 total combinations, but this includes the mirror images. I'm going somewhere with this, don't worry.

There are 128 total ways to have 4 mines in the 16 spaces, so the total number of combinations including mirror images is 524288 (4096 * 128).

So 192512 / 524288 = 0.367
When I run the calculations in a previous post I get ~0.1545
Multiplying the numbers together gives me ~.0566, so it's about 1/17.6

Edit: I just ran 100 tries and didn't get any, so it seems my calculations were on crack, but maybe someone else can figure stuff out based on them.

Another edit: Clinical trials gave me 20 successes in 1488 attempts, which is close enough that I'll say 1/75

Incidentally, I encountered exactly one 7 in all those trials, and one 6 as well, so they're pretty rare

[Edited by DiMono on 12-25-2003 at 03:06 AM GMT]

____________________________
Deploy the... I think it's a yellow button... it's usually flashing... it makes the engines go... WHOOSH!

levelthirteen
Level: Master Delver
Rank Points: 109
Registered: 02-20-2003
IP: Logged

Have a hint. It's a 28k listing of all 36856 choices.

Schik
Level: Legendary Smitemaster

Rank Points: 5383
Registered: 02-04-2003
IP: Logged

I'm really not understanding your calculations. At all.

DiMono wrote:
There are 47 ways for 4 of the outside squares to be mines.

I'm with you here.

For most of the other ways to do this, it's always some number * these 47. There are basically 2^12 outside combinations, so we're looking at 81 * 2^12 which is 4096 * 47 = 192512

But.... the total number of possible combinations of those 16 squares is 2^16 = 65536. You're saying there's 3x as many than that that actually work.

Yes, there are only 65536 total combinations, but this includes the mirror images. I'm going somewhere with this, don't worry.

The 65536 already counts mirror images, as it's counting EVERY possible combination. I'm really not following this.

Bah, let me just take a stab at it.

Number of bombs : number of combinations that work
0 : 0
1 : 0
2 : 0
3 : 0
4 : 47
5 : 472
6 : 1994
7 : 4836
8 : 7664
9 : 8496
10 : 6834
11 : 4052
12 : 1768
13 : 556
14 : 120
15 : 16
16 : 1

Number of bombs : chance of getting that number of bombs:
4 : (99/471 * 98/470 * 97/469 * 96/468) * (372/467 * 371/466 * 370/465 * 369/464 * 368/463 * 367/462 * 366/461 * 365/460 * 364/459 * 363/458 * 362/457 * 361/456)

That's long. Let's try to generalize.
x : (99!/(99-x)!) / (471!/(471-x)!) * (372!/(372-(16-x))!) / ((471-x)!/455!)

Hmm, I'm not sure if that's any easier.

denomiator will allways be 4533693136063779600000000000000000000000000, so I'll just list the numerators:
4 : 530347823907473880000000000000000000000
5 : 139565216817756310000000000000000000000
6 : 36240691659859375000000000000000000000
7 : 9284805301286286200000000000000000000
8 : 2346709032193236900000000000000000000
9 : 585069923094752290000000000000000000
10: 143869653220021040000000000000000000
11: 34889370944364775000000000000000000
12: 8343110443217665300000000000000000
13: 1967074819945628100000000000000000
14: 457211985176551460000000000000000
15: 104752072075490230000000000000000
16: 23653693694465535000000000000000

Now multiply each of those numerators by the # of combinations with that # of bombs:
4 : 24926347723651272000000000000000000000000
5 : 65874782337980979000000000000000000000000
6 : 72263939169759592000000000000000000000000
7 : 44901318437020480000000000000000000000000
8 : 17985178022728968000000000000000000000000
9 : 4970754066613015000000000000000000000000
10: 983205210105623840000000000000000000000
11: 141371731066566050000000000000000000000
12: 14750619263608832000000000000000000000
13: 1093693599889769300000000000000000000
14: 54865438221186171000000000000000000
15: 1676033153207843800000000000000000
16: 23653693694465535000000000000000

Add those all up, and put the denominator back in, and we have 232062797576915062579780335000000000000000/4533693136063779600000000000000000000000000

I'll reduce that to 5156951057264779168439563/100748736356972880000000000 before quitting.

Multiply that by the stuff in my first post - 312/480 * 380/479 * 379/478 * 378/477 * 377/476 * 376/475 * 375/474 * 374/473 * 373/472 = 125953458562065200000000/
1254310534691298400000000 = 314883646405163/3135776336728246.

So my answer is 5156951057264779168439563/100748736356972880000000000 * 314883646405163/3135776336728246.



____________________________
The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi

levelthirteen
Level: Master Delver
Rank Points: 109
Registered: 02-20-2003
IP: Logged

Schik wrote:
I'm really not understanding your calculations. At all.

So my answer is 5156951057264779168439563/100748736356972880000000000 * 314883646405163/3135776336728246.

You win; that totals about 514 / 100000 and I got about 513 / 100000. Source.
That hint was drawn after finding a solution and is mostly useess.

DiMono
Level: Smitemaster

Rank Points: 1181
Registered: 09-13-2003
IP: Logged

That was a sick puzzle, dude. I salute you!



____________________________
Deploy the... I think it's a yellow button... it's usually flashing... it makes the engines go... WHOOSH!

Schik
Level: Legendary Smitemaster

Rank Points: 5383
Registered: 02-04-2003
IP: Logged

Okay, next up:

When five Daisy Scouts received their Scout pins, the troop celebrated with an afternoon tea. Each of the five Daisies had a role in the ceremony, and each of their mothers baked a different type of cookie for the tea. From the following clues, can you determine each Scout's full name, her role in the ceremony, her mother's name, and what type of cookie her mother baked?

1. Carol and the Miller child were both Color Guards in the ceremony.
2. Anne baked lemon cookies and Mrs. Johnson brought chocolate.
3. Melissa is Pam's daughter
4. Abby Brooks was not flag bearer
5. Gwen's mother baked oatmeal cookies
6. Tracy, who is not Lynn's daughter, gave the Farewell speech at the end of the tea
7. Mrs. Miller was not the woman who baked sugar cookies
8. Mary's daughter and the Harris child both gave speeches
9. Kathy Anderson brought both tea and cookies
10. The spice cookies were brought by the mother of the Daisy who gave the Welcome speech

____________________________
The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi

The_Red_Hawk
Level: Smitemaster
Rank Points: 783
Registered: 09-02-2003
IP: Logged

You are supposed to tell us what all the first/last names are, what the duties are, etc. and we have to match them up.

I can only find three last names - Miller, Brooks, and Harris.

____________________________
Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

.....the king of the skies.....

agaricus5
Level: Smitemaster
Rank Points: 1838
Registered: 02-04-2003
IP: Logged

The_Red_Hawk wrote:
You are supposed to tell us what all the first/last names are, what the duties are, etc. and we have to match them up.

I can only find three last names - Miller, Brooks, and Harris.

Maybe there are people with the same surname in the troop.

____________________________
Resident Medic/Mycologist

Oneiromancer
Level: Legendary Smitemaster

Rank Points: 2936
Registered: 03-29-2003
IP: Logged

There are 5 surnames listed...Miller, Johnson, Brooks, Harris, Anderson. Schik's puzzle is perfectly valid at first glance.

Game on,

____________________________
"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

agaricus5
Level: Smitemaster
Rank Points: 1838
Registered: 02-04-2003
IP: Logged

You can miss so much by not looking closely.

I see them now.

____________________________
Resident Medic/Mycologist

Oneiromancer
Level: Legendary Smitemaster

Rank Points: 2936
Registered: 03-29-2003
IP: Logged

Here's my solution:

Carol Anderson was in the Color Guard, and her mom Kathy baked sugar cookies.
Melissa Johnson was a Flag Bearer, and her mom Pam made chocolate.
Abby Brooks gave the Welcome Speech, and her mom Mary baked spice cookies.
Gwen Miller was in the Color Guard, and her mom Lynn baked oatmeal cookies.
Tracy Harris gave the Farewell Speech, and her mom Anne baked lemon cookies.

I always make stupid mistakes when I do these logic puzzles, so I will await confirmation from Schik before I post one. Oh, and I also have to make a 3 hour drive too so it might be a bit.

Game on,

____________________________
"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

Mattcrampy
Level: Smitemaster

Rank Points: 2388
Registered: 05-29-2003
IP: Logged

Seems right, from the information given.

That is, all the clues match. But I'm no Schik, no sir!

Matt

____________________________
What do you call an elephant at the North Pole?

Schik
Level: Legendary Smitemaster

Rank Points: 5383
Registered: 02-04-2003
IP: Logged

You are indeed correct. You're up!

____________________________
The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi

zex20913
Level: Smitemaster

Rank Points: 1723
Registered: 02-04-2003
IP: Logged

Just for future reference, if you submit the puzzle to any other sites or something, you need to make the statement that each set of mother and daughter shares the same last name. It could probably be misconstrued otherwise.

____________________________