yashsingh1/xdaforums huggingface dataset:
[Off-topic] stephj:
Please accept my apologies for the belated reply to this, just back from two weeks in Brittany, in NW France.

Nos vacances etaient magnifique! La Bretagne est tres belle!

The phone and SIM is an office phone and the company contract prohibits roaming.

Back to the problem…………

[MENTION=744528]dbls[/MENTION]fstr.

Well done! Both answers are correct.

The wagon switching problem is a bit of a mind bender to solve. Your eleven images of the moves actually translate into 18 individual moves. A great effort! But it is possible to do it in two moves less if you pull ‘A’ and park it in the bottom spur first, then carry out a similar series of shifts, with ‘A’ in the bottom to start with.

But that’s just being picky. A tremendous effort just to solve it at all! 18 moves will do just fine!

Regarding the bolt problem, correct again. After one turn the bolts are in the original position, as the motion of the screws cancels the action of the other.

For any number of full turns clockwise or anticlockwise the same thing happens. Even in part turns the horizontal distance between the bolt heads remains the same.
[Off-topic] Seraz007:
no one? i hadz sadz :(:rolleyes::crying::rolleyes::o
[Off-topic] Dblfstr:
I see, if this were my job (moving the train cars) I would eventually find the most efficient way to relocate them. Keeping track of all movements and locations in so many steps was nearly impossible without drawing it out. I did it in my head (finally) but thought, that can't be right, I must have missed something. Once I drew out the movements, I was relieved that I was correct. Still, very difficult.

The screw problem was similar because you must think in a 3D space, with three simultaneous rotations, not easy for some. However, thinking about the rotation in relative terms makes it quite simple.

Hope you had fun in France.

---------- Post added at 08:14 AM ---------- Previous post was at 08:11 AM ----------

They are all easy, once you know the answer. I stand by my resolution. In a world that defies logic, its nor too far fetched to believe the girl could carry the tree, in her pocket even.
[Off-topic] Seraz007:
well the answer is 2 :rolleyes::rolleyes:

half of 2 is 1 and she takes back 1 so it remains 2 everytime :rolleyes::p :silly::silly::D:D
[Off-topic] Dblfstr:
That is the obvious answer. However, your wording was incorrect in the problem:

"...

(if she chooses 10 she gives the guard 5)

~~the girl also takes ONLY ONE APPLE FROM THE AMOUNT SHE GAVE THE GUARD~~ (so she has 4 remaining>>5-1=4 duuhh )

..."

Should have said (so she has 6 remaining (5+1 = 6 duuhh)

Your description in the problem is

Girl: 10

Girl: 5 Guard: 5

Girl 4 Guard: 5

When in actuality it is

Girl: 10

Girl: 5 Guard: 5

Girl: 6 Guard:: 4

That one word she... :D
[Off-topic] Seraz007:
^^^DayuM!^^^ :rolleyes: :D:o:o
[Off-topic] saket_oo:
zero
[Off-topic] stephj:
Peter and Paul race each other over 100 yards.

Paul beats Peter by 10 yards.

They decide to re-run the race, but to even things out a bit, Paul starts from 10 yards behind the start line.

Assuming they run at the same speed as before, who wins?
[Off-topic] Seraz007:
Tie? :confused::cool::eek:
[Off-topic] stephj:
That's a guess. You're not thinking hard enough.

The winner is either Peter or Paul. But which one, and why?
[Off-topic] Dblfstr:
First we calculate the relative velocity of Peter and Paul. The unit is arbitrary, we will just use a ratio.

Peter's velocity is 9/10 of Paul's velocity

If Paul traveled 100 yards in time (t): V1 = 100/t; t = 100/V1

Peter would travel 90 yards in time (t): V2 = 90/t; V2 = 90/ (100/V1) = 90V1 / 100 = (9/10) *V1

t being equal, V2 = (9/10)V1

So we have that. Now we need to know how far Peter will travel in the time it takes Paul to travel 110 yards.

So for Paul's 110 = V1*t; t = 110/V1

For Peter's distance (x) in time (t) at V2

x = V2 * t substitute for t

x = V2 * (110 / V1)

x = (9/10)V1 * (110/V1) substitute for V2

x = (9V1 / 10) * (110 / V1)

x = 990 / 10) V1 cancels out

x = 99 yards

Peter would travel 99 yards in the time it takes Paul to travel 110.

Paul would still win!

If Paul backed up 12 yards instead of 10, Peter would win

(9*112) / 10 = 100.8 yards
[Off-topic] stephj:
Can't argue with that, but you appear to have got the names crossed over in the middle.

Paul is the faster runner, he runs 100 yards in the time taken for Peter to do 90 yards.

With Paul starting 10 yards back, after the time taken to run the original race, they are dead level 10 yards from the finish line.

As Paul is the faster runner, he just pips Peter to the line, to win by a yard.
[Off-topic] Dblfstr:
I think you typed this as I was editing. I admit I did get the names backwards a few times. But I fixed it :D

Either way, the calculations are sound.

But really if you think about it. It is simple. Its just a ratio of Paul's Velocity to Peter's velocity. Paul/Peter, 100/90 = 110/x = 110/99

---------- Post added at 01:30 PM ---------- Previous post was at 01:24 PM ----------

Just as a traffic light turns green, a waiting car starts off with a constant acceleration of 6.00m/s^2.

At the instant the car begins to accelerate, a truck with a constant velocity of 21.0m/s passes in the next lane.

How far will the car travel before it overtakes the truck?

How fast will the car be traveling when it overtakes the truck?
[Off-topic] veeman:
147 meters and 42 m/s?
[Off-topic] Ichigo:
This is probably wrong, but this is what I got:

Time for car to pass truck = 7 seconds

Using that...

The car is going 42m/s when it overtakes the truck

The car traveled 147m before overtaking the truck
[Off-topic] Dblfstr:
Calculation or it didn't happen :P
[Off-topic] Ichigo:
d = displacement

Car - d = ½at² (a = 6.00)

Truck - d = 21t

½at² = 21t

½at= 21

½(6)t = 2

3t = 21

t =7

Car's speed - 6.00 * 7 = 42m/s

Car's distance - 21.00 * 7 = 147m

Yes, I know I didn't show the units in my equations. :p

---------- Post added at 03:29 PM ---------- Previous post was at 03:22 PM ----------

Dang. You got it before me.
[Off-topic] veeman:
D=vt

D=0.5at^2

vt= 0.5at^2

(21)(t)= 0.5(6)t^2

21t - 3t^2 = 0

-3t(t-7)=0

t= 0 or 7 but since it can't be 0, it must be 7

Plug back into 1st equation.

D=vt

D=(21)(7)

D=147
[Off-topic] Dblfstr:
Fantastic! Solve for t, given a(car) = 6ms^2 and v(truck) = 21 m/2

1/2at^2 = vt

t = 7 seconds

car accelerating at 6m/s^2 for 7 seconds reaches a velocity of 42m/s (would give average velocity of 21m/s, equal to that of the truck :cool: )

the truck and/or car traveling at 21m/s for 7 seconds wold travel 147 meters.

---------- Post added at 02:51 PM ---------- Previous post was at 02:50 PM ----------

First lol, Great Job!

See post above for other kudos!
[Off-topic] veeman:
I love physics problems. :thumbup:
[Off-topic] Dblfstr:
This thread needs more of them.
[Off-topic] Dirk:
Well played, and correct! :)

The next one will be more challenging :D
[Off-topic] Jays2Kings:
Lol I was going to say his house was 5 miles long.
[Off-topic] Dirk:
I might have accepted it as an answer! It's certainly shows lateral thinking.

Ok, gotta roll out the big guns...

This is my favourite. Solve it and you may find it saving your life one day!

You are shipwrecked and find yourself washed up on a desert island. You crawl up out of the sea on the south end of the island and take stock of the land before you. The island is covered in vegetation, with no beaches to speak of and about a twenty minute walk in diameter.

Sharks drawn by your presence are circling the waters, making any chance of swimming off the island impossible. A sudden storm blows in and an unfortunate lightning strike starts a fire on the far North side of the island. Even more unfortunate is that the wind is blowing the fire right at you!

Before long the fire covers the entire width of the island and is edging towards you. There are no tools and no time time to dig a shelter.

How will i survive the fire?, you ask yourself.

How indeed...?

A little visual help:
[Off-topic] deliriousDroid:
When you say no chance of swimming off the island. Does that mean absolutely no swimming at all?

Sent from my SilverBullet 'HTCDesireZ running CM7'
[Off-topic] Dirk:
If you so much as go near the water you will be killed until you are dead!

Good to get that one out of the way :D

Edit: Going to bed now so i'll see how everyone is doing with it tomorrow!
[Off-topic] deliriousDroid:
Am I also to assume that your not a pole vaulting champion? Jump over the fire

Sent from my SilverBullet 'HTCDesireZ running CM7'
[Off-topic] LordManhattan:
If 1 is 2 what is 2? (A lot of people fails on this one)

Edit: No takers? :P Don't be afraid! :D
[Off-topic] deveron3:
Um, 1?
[Off-topic] deveron3:
Edit: deleted
[Off-topic] deveron3:
Stop drop and roll. ;)
[Off-topic] LordManhattan:
Yup! You would be amazed how many says 2, 3 and 4 :P
[Off-topic] Maxamillion:
put the fire out with water from the sea?
[Off-topic] :
Grab a shark, rip off it's head and use it's blood to put out the fire..

Sent from my Frosty phone using the xda app
[Off-topic] deveron3:
No mate, you learn to fly. ;)
[Off-topic] Jays2Kings:
Use your flame-retardant suit if you have one. If not, do a rain dance :D
[Off-topic] xaccers:
Start a small fire on the South coast, when it has cleared a patch of vegetation, jump over the small fire and let it burn away from you towards the large fire?

If April showers bring Mayflowers, what do Mayflowers bring?
[Off-topic] jorim.tielemans:
Don't forget the wind
[Off-topic] Cheer98:
Here's another quick question:

If you would die tomorrow, why not today? :p
[Off-topic] mercianary:
You survive the fire because your not there, i am.
[Off-topic] Dirk:
Close enough! Take a long stick to the fire on the north side and set fire to it. Run back the way you came and start your own fire. Now you have the original fire coming up behind you and your new fire burning away from you. Obviously vegetation can't burn twice so the fire moving away from you is clearing an area for you that's safe to stand on!

Easy! ;)

I like that one because it's pure logic, no trickery. And thinking like that could save your life one day!

Well played xaccers.
[Off-topic] veeman:
A bear walks south for one kilometer, then it walks west for one kilometer, then it walks north for one kilometer and ends up at the same point from which it started. What color was the bear?
[Off-topic] Ichigo:
Blonde!
[Off-topic] veeman:
Wat... Lol, no. There's an actual logical answer to this. :D
[Off-topic] Seraz007:
White?
[Off-topic] Ichigo:
Lol I was confused. I though maybe it was blonde since it went south, then west, then north and somehow ended up in the same spot that it started.
[Off-topic] veeman:
Here's another one. A pirate sailing in a ship throws a cannonball into the ocean. Does the water level...

A. Rise

B. Sink

C. Stay the same

Provide your reasoning!

---------- Post added at 11:37 PM ---------- Previous post was at 11:35 PM ----------

Yes, the bear is white because it's a polar bear.

You can only go south, west, then north and end up in the same place at the north pole.
[Off-topic] Ichigo:
Well, one would obviously have to say rise due to the water displacement. The water would rise by the volume of the cannonball...

Unless a whale swallowed the cannonball or something, which would he very unlikely... I feel I'm missing something though...
[Off-topic] veeman:
You're missing something. Keep thinking.
[Off-topic] Ichigo:
I was too quick to answering... I should've thought about that one more thoroughly.

---------- Post added at 12:45 AM ---------- Previous post was at 12:42 AM ----------

Well, my first thought was that more water was evaporating at that moment than the volume of the cannon ball...

Other than that, I'm not sure.
[Off-topic] roofrider:
Stays the same as the ball is thrown from the ship.
[Off-topic] Ichigo:
But the objects on the ship do not displace the water. You being in a boat and you being in the water is two totally different things...
[Off-topic] roofrider:
I don't know, my brain is rusty but that was the first thing that came to my mind. :p

I think they will displace water but the amount displaced might not be the same, not sure though. I will displace some amount of water even if i'm on a boat because i weigh you know.
[Off-topic] Ichigo:
Lol yeah, I'm assuming there would be some displacement, but they wouldn't be equal...
[Off-topic] roofrider:
I'm sticking with option C though. :cowboy:
[Off-topic] veeman:
That's a key part of the problem.(but not quite right) Keep at it.

No one has given the right answer along with the right reasoning.
[Off-topic] ArmorD:
This has something to do with iron(assuming the cannonball is iron?) being more dense than water... So the ball displaces around 8th of the weight of the water(of the same volume as the ball*) and so on...

I'm gonna say the water level goes down. I'm not very good at explaining physics stuff in English :D I am studying physics in university but this one is really hard :eek:

Sent from my panda
[Off-topic] veeman:
The two assumptions for the cannonball is that it sinks in water and is smaller than the ship. Those are the only things you need to know about the ball.
[Off-topic] Dblfstr:
The volume of water displaced is equal to the volume of the object that is submerged.

The cannonball, being denser than water, would displace more water by its mass than by its volume.

The water level would sink.

An object floating displaces water equal to its mass.

An object submerged displaces water equal to its volume.
[Off-topic] veeman:
Yes, you're correct. The cannonball does displace more water with its mass than with its volume. When the cannonball is thrown off the ship, the ship rises and the water level decreases. The water level increased by the volume of the cannonball is less than the water level decreased by it being dropped off the ship and the ship rising.
[Off-topic] Dblfstr:
This one is hard:

A rope rests on two platforms which are both inclined at an angle theta (which you are

free to pick), as shown. The rope has uniform mass density, and its coefficient of

friction with the platforms is 1. The system has left-right symmetry. What is the

largest possible fraction of the rope that does not touch the platforms? What angle theta

allows this maximum value?
[Off-topic] veeman:
Is there a diagram? I need to visualize that.

Edit: Oops, didn't see the picture. *facepalm*

---------- Post added at 09:56 AM ---------- Previous post was at 09:48 AM ----------

The mass and/or length of the rope isn't necessary to solve this problem?
[Off-topic] Dblfstr:
No, it is arbitrary. mass = m, length = l

Good Luck

HINT: Find the angle first.

You do not need length. Just a fraction of it, f.

length/length = 1

f is the fraction not touching the platform, (1 - f) is the fraction that is touching the platform.
[Off-topic] Dblfstr:
Oh, good. It is sill stumping folks.

Here is another hint.

Let the total mass of the rope be m, and let a fraction f of it hang in the air. Consider the right half of this section. Its weight, (f/2)mg, must be balanced by the vertical component, Tsin(theta), of the tension at the point where it joins the part of the rope touching the right platform. The tension at this point therefore equals T= (f/2)mg/sin(theta)
[Off-topic] Ichigo:
I know I had no chance so I cheated. From what I saw in the answer, I didn't know quite enough physics(formulas) to answer the question. I was happy to know I did understand all the trig and geometry going on. :)
[Off-topic] veeman:
Do I need set Ff (force of friction) equal to Tension? So it's just barely stationary.
[Off-topic] Dblfstr:
The force of friction has to not only balance the tension, but also the gravitational force along the plane.

---------- Post added at 01:45 PM ---------- Previous post was at 01:42 PM ----------

It is hard, but once you understand what is going on, it is actually very simple.

I always have to draw a free body diagram with any problem like this. Visualizing helps me think and understand what is happening.

Also, did not realize it was floating around the internet so freely. I found it embedded in a word doc. :(
[Off-topic] veeman:
Well, I didn't get it but I have it to my friend who is really good at physics and he got an answer today. :(

He got pi/8

He did it without the hints too BTW. I was pretty impressed.
[Off-topic] Dblfstr:
That, my friend, is only half of the answer.
[Off-topic] veeman:
:/ Well, that was the only answer he gave me.

---------- Post added at 09:20 PM ---------- Previous post was at 09:19 PM ----------

Consider a standard chessboard with 64 squares. The Devil is in the room with you. He places one coin on each of the 64 squares, randomly facing heads or tails up. He arbitrarily selects a square on the board, which he calls the Magic Square. Then you have to flip a coin of your choosing, from heads to tails or vice versa. Now, a friend of yours enters the room. Just by looking at the coins, he must tell the Devil the location of the Magic Square. You may discuss any strategy/algorithm with your friend beforehand. What strategy do you use to do this?
[Off-topic] veeman:
This thread died or what?
[Off-topic] 063_XOBX:
What is, trick question for ib4l Alex.
[Off-topic] Dblfstr:
Just to get this straight in my head.

All squares have a coin on them. (64)

The coins are randomly heads and tails.

The devils picks a random square as a magic square.

You randomly flip one coin (not necessarily the magic square).

Your friend must pick the magic square.

I was on vacation the past week.
[Off-topic] veeman:
Yup. That's correct.

It's a good programming puzzle. If you know python or something, I recommend using it.
[Off-topic] Archer:
I take it you know what the magic square is - the devil tells you?
[Off-topic] veeman:
That's correct. The devil tells you which is the magic square.

---------- Post added at 02:04 PM ---------- Previous post was at 02:02 PM ----------

And just to clarify. Your friend enters the room after you have flipped the coin.
[Off-topic] Gogeta:
No more riddles? :(

Sent from my E15i using fapatalk 2
[Off-topic] ironman38102:
This aint 1980s anymore. Now its just trolling people and hacking Android.
[Off-topic] veeman:
Well, no one's answered my problem yet. :P
[Off-topic] stephj:
If all the coins are aligned, then flip the coin on the magic square over left to right across its vertical axis. If you were playing this game in Europe, then you would have to flip the coin top to bottom across its horizontal axis. US coins have a "coin-turn". Sterling or Euro coins don't.

The magic square then appears upside down to all the others, but I assume it's not that simple and the coins are also at random orientations.

In which case it is back to the drawing board..........................
[Off-topic] stephj:
There are probably several ways of doing this. Here's one.

Both you and your friend must become very adept at binary mental arithmetic, which means being able to carry out EOR/XOR (Exclusive OR) operations on binary numbers up to six bits in length. In an XOR operation if the two bits match the result bit is set to 0, if they are different it is set to 1.

Input1: 0011

Input2: 0101

Output: 0110

You look at the board and, numbering the squares from 0 to 63, find the first head, and remember this number. Find the second head and XOR the number of this square with the first. Remember this new value. Repeat the process for all heads on the board, each time XORing the position with the previous carried over value.

This is now the board's hash value.

When the Devil tells you the Magic square, XOR this value with the board's hash value.

This new value is the the number of the coin you must flip. Flip it and leave.

You friend enters, carries out the same process above, but this time his hash value is that of the Magic Square.

Here's the C# code bashed together to test it out. The board is populated with head and tails at random, and the Magic Square is also chosen at random.

using System;
namespace ChessBoard
{
class Program
{
static void Main(string[] args)
{
int i;
int[] Board = new int[64];
int Hash,Magic,Flip;

Random Rand = new Random();

// Populate the board with random Head/Tails = 1/0
for (i = 0; i < 64; i++)
Board[i] = Rand.Next(2);
// Draw the board and generate the hash
Hash = 0;
for (i = 0; i < 64; i++)
{
Console.Write(Board[i]);
if(Board[i]==1)
Hash^=i;
if(i%8==7)
Console.WriteLine();
}
//Print the hash
Console.Write("\nHash = " + Hash.ToString());
// Randomly select the Devil's Magic Square
Magic=Rand.Next(64);
Console.Write(" Magic = " + Magic.ToString());
// Find the coin to flip
Flip = Hash ^ Magic;
Console.WriteLine(" Flip = " + Flip.ToString());
Console.WriteLine();
// Now flip the coin over on the board
Board[Flip] = (Board[Flip] == 0) ? 1 : 0;
// Redraw the board and regenerate the hash
Hash = 0;
for (i = 0; i < 64; i++)
{
Console.Write(Board[i]);
if (Board[i] == 1)
Hash ^=i;
if (i % 8 == 7)
Console.WriteLine();
}
Console.WriteLine();
// The new hash value is the magic square.
Console.WriteLine("\nNew Hash = " + Hash.ToString());
}
}
}

Hash^=i is the C,C++,C# shorthand version of Let Hash = Hash XOR i

And here is the output from a couple of runs.

10010001
10010011
00001010
01010000
00011001
01010001
10010000
01011100

Hash = 11 Magic = 24 Flip = 19

10010001
10010011
00011010
01010000
00011001
01010001
10010000
01011100

New Hash = 24

01111101
10111100
01011000
10110000
10110110
10101110
10101001
11001011

Hash = 51 Magic = 54 Flip = 5

01111001
10111100
01011000
10110000
10110110
10101110
10101001
11001011

New Hash = 54

Incidentally, the method still works if you use the position of the tails instead, as long as your friend uses the same system.

It also works with any size board.

Try it out with a 2x2 board. Be a smartass, and do it your head!
[Off-topic] veeman:
I completely forgot about this thread. Now you get to come up with a puzzle for us to solve. :D
[Off-topic] stephj:
In the days when postage stamps were this cheap, I got a crossed line and heard bits of this conversation.

A: I have just spent exactly £2/$2/2€ on a mixture of 12, 14 and 17 penny/cent stamps.

B: How many of each did you buy?

A: I’ll not tell you that but altogether I bought …. (line fades out)…. stamps.

B: OK. I’ll figure it out……… Wait a minute, I need some more information. Did you buy only one of one kind of stamp?

A: … (line fades out again)..

B: In that case you bought …… (Line goes dead)……

Just how many stamps of each value were bought?
[Off-topic] veeman:
So basically the only relevant information provided is 2 = 0.12x + 0.14y + 0.17z
[Off-topic] gmaster1:
pretty much as theyre the amount of them is all unknown.

im not sure but id just say you can do it with trial and error.

but the Z has to be an even number other wise it wont be exact to £/€/$2

That's right, I'm the premium potaytoman and i seek the TROOOOOF! Problem? :eek:

---------- Post added at 02:08 AM ---------- Previous post was at 02:06 AM ----------

Edit: failed calculations.

That's right, I'm the premium potaytoman and i seek the TROOOOOF! Problem? :eek:
[Off-topic] ssojyeti2:
Does the currency matter?
[Off-topic] gmaster1:
not really. as long as the 2 is something like a pound/dollar and the 12, 14 and 17 are pennys/cents.

That's right, I'm the premium potaytoman and i seek the TROOOOOF! Problem? :eek:
[Off-topic] stephj:
Or 200 = 12x + 14y +17z

The currency is of no importance.

There is more than one solution to the above, but only one of them matches the conversation.
[Off-topic] ssojyeti2:
There needs to be at least 2 of the 17 cent stamps to get an even cost (such as $2), which means that at least one multiple of 34 should be divisible by both 14 and 12, which is not the case
[Off-topic] stephj:
Correct.

Incorrect.

200 minus at least one multiple of 34 should be the sum of multiples of 14 and 12, and the multiple could be one.
[Off-topic] ssojyeti2:
Oops my bad lol...so 3 12-cent, 2 14-cent, and 8 17-cent.
[Off-topic] jegata:
really do not know
[Off-topic] Dirk:
Did you ever hear that monologue from War Of The Worlds about "minds immeasurably superior to ours..."?

I think stephj is trying to ascertain our intelligence level in preparation for taking over the world! :eek: