PUZZLE TIME

No.

Girlfriend dilemma.

At the north pole, ezpz. I pick ferryboats.

>>7790490
>No.

wrong

>>7790493
I was anticipating this reply, and I hope someone posts the correct solution soon and then I'll call you out on your bullshit.

>>7790492
Apparently I can't read.
You are 1 km north of a cirle of latitude on the south pole with a circumference of 1 km or an integer fraction thereof.

>>7790501
*near the south pole

>>7790498
It's not bullshit. The solution is pretty easy actually

everywhere if we just move the referencepoint north/south pole? Due du symmetry? girlfriend dilemma

>>7790501
This. Next riddle OP.

>>7790501
You mean: 1km/2*pi*k where k is a natural number.

But it's correct. Ferryboats:

Two ferryboats start at the same time on opposite sides of the river, traveling perpendicular to the shore. They both travel at constant speed, but one is faster than the other. They pass at a point 720m from the nearest shore. Each boats rests 10 minutes before returning back. On the return trip they meet 400m from the other shore.
How wide is the river?

Next puzzles:
1. Handshakes and Networks
2. Where are you going?
3. Desert Crossing

>>7790523
1760m

>>7790545
Correct. Your pick?

>>7790546
Let me pick it for you if you don't mind. Handshakes and Networks:

At a physics convention the number of physicists who shook hands an odd number of times, is even. Prove it (without using graph theory). Advanced version: Prove it using induction.

Next problems to pick:
1. The thick coin
2. Find the Oddball
3. Hole in Sphere

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And here is The Thick coin:

How thick should a coin be to have a 1/3 chance of landing on the edge?

Iff you can solve this in less than 20 seconds in your head, you're smarter than his majesty pic related.

>>7790568
height of cylinder must be equal to one side.

2pi r *d = pi r^2
d = 1/2 r

>>7790563
Isn't this missing something like "all physicists shake hands with everyone else once"?

>>7790571
Not even close. Consider all the possibilities how a coin can hit the ground.

>>7790573
No, the formulation is sufficient.

>>7790563
attribute a number to every physicist - how many hands he shaked
The sum of all those numbers is even - since you've counted every handshake twice.
If sum of some numbers is even, there must have been evenly many odd numbers among them.

>>7790589
Correct. Now try the advanced version.

What do you pick? Oddball or Hole in Sphere?

>>7790595
Oddball

>>7790568
Should one assume the coin loses all momentum as soon as it hits the ground and then it basically drops dead to its resting position?
Or some more complex bounce bounce bounce with a constant dampening factor?

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>>7790597
Oddball:

You have a balance scale and 9 balls. One ball is heavier than the other 8. What is the minimum amount of weightings you need to find the odd ball?

>>7790523
are you tricking sci into doing your homework?

>>7790602
No bounce-bounce. Bounce-bounce wouldn't even matter since in the long run you'd get the same amount of states with our without.

>>7790606
Two

>>7790614
Choose:

1. The Broken Bar
2. Probability of Matches
3. Dividing the Cake (easiest)

>>7790568
diameter/sqrt(3)

>>7790563
>At a physics convention the number of physicists who shook hands an odd number of times, is even. Prove it

You mean like out of 4 physicists, 1 of them shook everyone else's hand once?

>>7790606
>One ball is heavier than the other 8.

Individually or combined?

Individually = 2
Combined = 1

>>7790623
That would be way too thick. Think of it as a sphere.

>>7790523
Mind posting the resolution?

>>7790624
>You mean like out of 4 physicists, 1 of them shook everyone else's hand once?

Yes f.e., because then he would have shaken hands 3 times.

>>7790606
No matter how I split this I need three measurements..

>>7790634
see >>7790545

The first time the slower boats meeet the faster one, it has traveled 720m and the boats together have traveled one river length. The second time it has traveled the length of the river + 400m, and the boats together have traveled 3 river lengths. So 3x720 = river + 400, subtract 400 and river is 1760.

>>7790637
Even number of physicists: 4
Each shakes hands an odd number of times:. 3, 1, 1, 1

>>7790648
correct :) that's an example of the theorem

>>7790621
I will pick for you. Probability of Matches:

A deck of 52 cards is laid out in a row, face-up. Beneath it another deck is laid out. What is the probability that there are exactly n number of matches (i.e. queen of hearts beneath queen of hearts).

This should keep you busy.

>>7790646
Thanks

>>7790614
>>7790621
how do you find the ball with two weightings?

>>7790563
Let n be the number of physicists who shook hands an odd number of times.
- A handshake between 2 people both with an odd number of previous handshakes will subtract 2 from n.
- A handshake between 2 people both with an even number of handshakes will add 2 to n.
- A handshake between 2 people with an even and an odd number of handshakes respectively will not change n.
Therefore no handshake will change the parity of n. At the start noone has shaken hands so n is 0 (i.e. even).

This counts as proof by induction, right?

>>7790679
Weigh 2 groups of 3 balls against each other, you now know in which of the 3 groups the heavy ball is. Then weigh 2 of the balls from that group against each other.

>>7790679
Their are 9 different possible states:
>Ball 1 is heavier
>Ball 2 is heavier
>...
>Ball 9 is heavier


A weighing can give 3 different results
>Left side is heavier
>Right side is heavier
>The sides are in balance

For each weighing you have to distribute the remaining states evenly to the results. This way you divide the number of states by 3 every weighing.

>>7790693
And if the two balls are heavier than the one? How do you which one it is without a third weighing?

>>7790723
Only 1 ball is heavier than all the others, that's stated in the problem.

>>7790693
3 groups of 3 balls (9 total)
Weigh only 2 groups against each other.

If those 2 groups are evenly weighted then the non-weighted group contains the heavier ball.

The rest is as you said, just adding the rest of the info left out. This of course assumes that all other balls are evenly weighted and not all random weights.

>>7790568
thickness = r * 3^0.5

>>7790726
So one side of the scales has one normal ball plus the heavier ball, the other side has one normal ball. You now know the heavier ball is one of two balls, how do you know which one?

>>7790563
2 * coin radius * tan (pi/3)

>>7790733
Nonono, the point is you weigh TWO balls from the three against each other, and leave one on the side. One ball on one side of the scale, one ball on the other side of the scale, and one not on the scale.

>>7790745
Oh, I misread "weigh two against each other" as "weigh two against the other"

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>>7790568
should be like this.

>>7790729
Radius/[Thickness/2] = sin(pi/4)

>>7790757
That's exactly what I did >>7790623 but it's not correct. I would like some explanation on why the reduction to a randomly rotated rectangle doesn't work.

>>7790768
https://www.seas.harvard.edu/softmat/downloads/2011-10.pdf

>>7790479
Just a bit above the south pole. There's infinitely many points.

>>7790568
What, did he solve in in 20 seconds in his head or something? He's a genius, but no way.

>>7790779
i think it is more intuitive to consider rotation around only one axis. This is how a coin toss is done, and it is more in line with angular momentum conservation.
I dont like his answer.

>>7790815
I'm having a bit of trouble imagining a uniformly random orientation of a sphere, but indeed I think the coin would have to spin unreasonably fast around its axis for that.

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>>7790688
Correct.

>>7790768
Because it's in 3 dimensions. Answer is 1/(2*sqrt(2)), see >>7790779

That von Neumann solved it in less than 20 seconds blows my mind. Very strong understanding of mathematics.

Another questions for the physicists (in case you aren't already familiar), 2D Random Walk:

Starting from origin a particle has a 1/4 chance to move either east, west, south, or north. What is the probability that the particle will return to origin after number of steps n.

>>7790844
0 if n is odd

>>7790479
define south on the north pole you fag

>>7790875
Anywhere.

>>7790875
Anywhere along a great circle.

>>7790875
towards the south pole through the earths core

>>7790807
You don't understand von Neumann's sheer mental ability.

>Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Upon reaching the other train, the fly turns around and continues towards the first train. How many kilometers does the fly travel before getting squashed in the collision of the two trains?

>When posed with a variant of this question involving a fly and two bicycles, John von Neumann is reputed to have immediately answered with the correct result. When subsequently asked if he had heard the short-cut solution, he answered no, that his immediate answer had been a result of explicitly summing the series

More:

>As far as I could tell, von Neumann was able on once reading a book or article to quote it back verbatim; moreover he could do it years later without hesitation. He could also translate it at no diminution in speed from its original language into English. On one occasion I tested his ability by asking him to tell me how the 'Tale of Two Cities' started. Whereupon, without pause, he immediately began to recite the first chapter and continued until asked to stop after about ten or fifteen minutes.

As a child he also used to multiply 8+ digit numbers in his head for fun.

>>7790844
0 if n is odd.

if n is even ...
we'll use the multinomial coefficient to calculate the nb of ways to place n distinct objects (step1, step2, ..., step n) into 4 boxes (north, south, east, west).
M(n,s,e,w)

This will give you the number of ways to go back to origin after n steps
Sum i = 0 to n/2-1 of M(n/2-i,n/2-i, i,i)
If you divide by 4^n you get the probability.

I'm too lazy to try to simplify this...

>>7790991
This is truly phenomenal!
I'm so glad for humanity that such people can exist.
... but it does make me feel stupid and insignificant ...

>>7791320
>Sum i = 0 to n/2-1 of M(n/2-i,n/2-i, i,i)
I meant sum i = 0 to n/2 and not sum i = 0 to n/2-1

>>7790991
During his graduate studies, and after working with von Neumann, Gillies became a fan of the book "One-upmanship" by Stephen Potter. John von Neumann was also a fan of this work, and was extremely successful at impressing others with his intelligence. An apocryphal math problem asks about a bumble bee flying back and forth between two approaching trains, and how far did it fly before colliding? When von Neumann gave the correct answer, the questioner asked if he used a standard time/rate-of-travel trick, and he replied, 'no, I summed the infinite series in my head' to impress the questioner. This method of impressing and astonishing others appealed to both Gillies and von Neumann.

>>7790670
Should we assume the decks are shuffled or is this supposed to be a trick question?

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>>7792112
Yes, the decks are shuffled.

Here's an easy one. The Flight around the World:

There is a group of planes on a small island. The tank of each is just large enough to take it halfway around the globe. Any amount of fuel can be transferred in flight. The island is the only fuel loss, and there is no time lost in refueling either on land or in the air.
What is the minimum number of plans to make a trip around the globe in a great circle, assuming the have the same ground speed and fuel consumption is constant?

>>7792252
>The island is the only fuel loss

meant to say fuel source. sorry, it's early

>>7792252
Three. Send two flying in one direction, exchange fuel at one half pi, let one crash and the other continue. when that plane gets to pi send a third plane in the opposite direction and exchange fuel at 3/4 pi.

>>7792578
now without crashing planes

>>7790670
((52 Choose n) / 52! )^2

>>7792578
kek'd heartily

>>7790991
I came

>>7792596
provably wrong because the prob of having exactly 51 matching is 0.
it s not possible to have 51 without the 52nd also matching...

>>7794126
There, I gave it more thought...
There are a few steps involved
Nb of permutations for card 1 to match is 51!
Nb of permutations for card 1 and card 2 is 50!
Nb of permutations for card 1, 2 and 3 is 49!
...
Once you figure that out, you try to calculate the nb of permutations to have at least one card with a match (let's call that Z).
Have in mind that P(A u B) = P(A)+P(B)-P(A and B) ... use the generalized version with 52 events. and make sure you consider all intersections.
Z=sum (j=1 to 52) (-1)^(j-1) (j in 52) (52-j)!

52! - Z is the number of permutations with no match.
We also need to develop a similar expression for nb of permutations with no match for any nb of cards.

Prob (n matches) = (n in 52) * (nb of permutations with no match in 52-n cards) / 52!

>>7790991
The solution of this is 75 km yes?

>>7795161
Half that, since both trains are moving 50kmh, they meet after 30 minutes.

>>7795178
Never mind, I'm retarded.

>>7790991
>Supposedly a genius
>Can't see that the trains are going to colide after half an hour
>Explicitly sums a series instead

Clearly he was autistic af.

Will continue to posts some riddles and games, if you are interested.

A man decides to shop on the town with his wife one evening.
They visit twelve stores.

At each store,
the man spends half of the money he had before he walked in to the store,
plus, 15 dollars. At the end of the trip,
the man has only 3 dollars left,
so he decides to return home with his wife.

At the start of the trip, exactly how much money did the man originally have?

(These are not mine by the way)
You're at a popular Italian Restaurant with your friends after a sports game.
A waitress takes your ticket to the chef. You've just ordered seven pizzas,
for the guys and gals.

The restaurant offers two choices of pizza, one choice is considered regular, but the other choice is premium, using a different recipe, which makes the pizza weigh slightly more.

All of the pizzas appear the exact same, and you and four of your friends order the usual pizza.
But two of your friends however order premium pizzas of a different recipe.

The chef has baked five pizzas of the same weight, following the same recipe.
The chef has also baked two pizzas of the heavier weight,
which weigh the same to each other,
following the different recipe.

Again, all of the pizzas appear the same.
Consequently,
after they come out of the oven,
he obviously can't tell which are which.

The chef decides to weigh each pizza.
In order to weigh the pizzas,
he needs to place them on a two-pan balance,
and weigh them against each other on both pans.
At most, he can fit three pizzas on each pan.

(You can weigh each at a time,
two at a time on each pan,
or three at a time on each pan,
or any combo of that.)


How many weighing actions are necessary
to tell which are the heavier pizzas?
This problem has more than one solution.

>>7795283
>Visiting stores

Answer is $135138

You can recursively add and multiply, but I wanna model is as so:

3+sum[for x 1 to y](33*(2^x))

For y number of stores, cuz it looks cooler that way...but wait! You can optimize it for use in computing:

I+(I+M*A)*(2^(S)-1)

Where I is initial value
M is multiplicative constant
A is additive constant
S is number of stores to visit

>>7795400
Correct :)

>>7795299
There are a number of ways to do this, as you said. Here's one:

Set one pizza aside and call it A.

Weigh the rest with one half the pizzas on each side. If they are balanced, we know A cannot be a premium, and that each side of the scales has exactly one premium pizza.

With each of the sets of 3 on the scales, set one pizza aside, and see if the other 2 balance. If they do, the one set aside for that group is premium, otherwise the heavier side is premium. This requires one weighing for each of the sets, resulting in a grand total of 3 weighings.

If the scales are UNBALANCED at the first weighing, we must first determine whether A is premium. Suppose side B is heavier than side C. Side C must not contain any premiums, so if you weigh one pizza from side C with pizza A, you can determine whether A is premium.

Then you weigh the pizzas from side B the same way that you split each side in the earlier scenario, e.g. set one aside, weight the other 2. If A is premium, you deduce the other premium the same way as before. If A is NOT premium, then:

If the scales are balanced, both are premium. If they are unbalanced, the lighter one is not premium, and the other 2 are.

This scenario also results in 3 weighings.

So...3 is my guess.

>>7795410
Three is correct :)

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Theres a power tower next to a river, and a house on the other side of said river. (pic related)

We need to connect the house to the power spending the less amount of money.

The costs for the wire are :
$1 a meter for wire over the shore or the bridge.
$2 a meter for wire over the water.

Whats the cheapest way to do it, and if over the water, whats the angle that the wire makes with the shore?

>>7795574
Also how is the angle affected if we change the distances? Does it change?

>>7795574
Consider a triangle
Hyp is the length of the wire if it goes above water.
A is length of base and B the height.
If we can find a triangle for which
2*hyp < A + B then it is cost effective to go above water
with A = h sinx and B = h cosx
we cover all possible triangles
Is it possible to find an angle x that satisfies
2h < h(sinx + cosx)
2 < sinx + cosx
Unfortunately, it is not possible because the max of sinx + cosx is 2^0.5=1.41... so it is never cost-effective to go above water.
Cheapest way to do it is 75$.

>>7796692
ignore that, forgot to consider reduction on the wire that does not go over water...

>>7795574
Is the following accepted as proof?
-Let a,b be sides of a right angled triangle and c be the hypotenuse.

(a-b)^2 + 2c^2 > 0

3c^2 - 2ab > 0 (pythagoras)

3c^2 > 2ab

4c^2 > a^2 + 2ab + b^2 (pythagoras)

2c > a + b

c > (a + b)/2

A problem:
Prove that a sphere with 5 points on its surface must have a closed hemisphere with 4 points in it.

It's a pretty well-known question, so if you know the solution, don't spoil it for others.

>>7796943
draw a circumference around the sphere through any two of the points. This divides the sphere into two hemispheres.

Since there are three points left and only two hemispheres to place them in a basic application of the pidgeonhole principle tells you one must have at least two points, combined with the two on the circumference makes four.