So /sci/ any good logic puzzles? hard ones please iv'e been

july 16

>>7919737
That one usually sparks huge debate.But you are right

You're given the string of symbols ZE, from which you can apply the following transformation rules:

1) xE→xEU, meaning that if a string ends with E, you can add a U to the end of it.
2) Zx→Zxx, meaning that you can double the string that comes after Z.
3) xEEEy→xUy, meaning that if you have a row of 3 E's in your string, you can replace them with a U.
4) xUUy→xy, meaning that you can remove any row of 2 U's in your string.

For an example, you can do the following changes: ZE→ZEE (rule 2)→ZEEEE (rule 2 again)→ZUE (rule 3).

The problem is: How would you reach ZU, starting from ZE?

you don't

>>7919838
It's impossible. The amount of E's can never become a multiple of 3.

>>7919729
Cheryl is a fucking slut and Albert and Bernard should be ashamed

>>7919754
Wait, how? The first thing suggests that it isn't May 19 or June 18 because those are the only dates with unique days and it can't be June 17 otherwise Albert would know (Since he knows the month and that's the only one left for June). Since Bernard knows after this it seems like it must be August 17 since this is the only date with a unique day after those three are eliminated.

>>7920482
0.625?

Plz be gentle.

>>7919737
I actually opened the image, read the premise, then just guessed July 16. Good to know all I need is luck to solve puzzles.

>>7921312
You're probably bound to get it eventually.

>>7921313
It wasn't a guess exactly, I just thought that the person would probably try to "hide" the answer by placing it centrally. July 16th's spot feels better than June 18th's too for some reason, the former is lower and thus further along and more "hidden", but it was one of those 2.

People feel uncomfortable putting the right answer near the start or the end and like to hide it. Which is why on average Scantron answers are a little biased towards C as an answer.

Of course it was still luck anyways.

>>7921034
You're right, they're all memeing

>>7921034
You were on the right track, but remember Albert says that he KNOWS that Bernard doesn't know, which means May and June are entirely eliminated from the options. Now that Bernard knows this, he knows from his number which month is correct (he can rule out the 14th because it's the only date that appears in both July and August). Albert knowing this can rule out August because it would've left Bernard with two remaining possibilities.

>>7920482
I'm not a big math nerd but is there a logical reason for it to be above .5?

>>7920482
Optimal cutoff is 0, always reroll.

Here's one that a CS friend told me and i thought it was very clever- relevant coursework makes this kind of thinking trivial but for everyone else it's difficult to grasp.

You are a sort of strategist or a mediator to 100 volunteers playing a game against a sole organizer. The organizer says the rules of the game are simple:

> Each of the 100 people gets a number (1-100 inclusive) and writes it on a dollar bill
>In the next room, there are 100 boxes each labelled with a number (1-100)
>Inside each box is a random dollar that was collected earlier
>The goal is simple: find your dollar
>You have 50 tries to open a box to find your dollar
>Once you enter the room with the boxes, you cannot go back, and only one person in the box room at a time
>If at the end of the game, everyone finds their dollar- you win and everyone gets $100
>If even one person fails to find their dollar, you lose

Your job is to devise a strategy for the people to employ that gives them a chance of winning this game that is greater than the payout (1:100)
Random chance is obviously not the answer- the probability of that working is the same as flipping a coin heads 100 times in a row (1/2^100)

File: Unbalanced_scales.svg.png (12KB, 400x354px) Image search: [Google]

12KB, 400x354px

There are 12 coins, 11 of which are identical and 1 of which is different. You do not know if the different one is heavier or lighter than the others.

You have a balance with you.

Your goal is to find the coin that is different and determine its weight relative to the other coins.

You can use the balance three times.

(The solution I've seen to this on wikipedia is pretty kewl, let's see if /sci/ can come up with another one, try not googling it though).

>>7921405
>Albert knowing this can rule out August because it would've left Bernard with two remaining possibilities.
Huh?

But if B had a 15 he would know it is august and a 17 he would know it is august.

How can Albert know the girl's birthday if all he is choosing between is July 16, August 15 and August 17.

>>7921796
Nvm, got it, I'm stupid.

>>7921805
Btw, is it true this as made for 3rd graders or is this from an olympiad I've heard both.

>>7921034
https://www.youtube.com/watch?v=w3Nzkae_TRU

>>7920482
A random number to how many decimal places?

>>7920482
Let [math] x_1 [/math] be the cutoff of player 1 and [math] x_2 [/math] the cutoff of player 2.
Let [math] p(x_1,x_2) [/math] the probability that player 1 wins if they choose the cutoffs [math] x_1 [/math] and [math] x_2 [/math].
The optimal cutoffs (if they exist, which is not at all obvious) will be the pair [math] (\tilde x_1, \tilde x_2) [/math] such that
[eqn]p(\tilde x_1, \tilde x_2) = \min_{x_2} p(\tilde x_1, x_2) = \max_{x_1} p(x_1 , \tilde x_2) = \frac{1}{2}[/eqn]

>>7921901

What an autistic way of saying P(x>0.5) = P(x<0.5)

>>7921583
I think you need to find the number of permutations of [|1,100|] that can be decomposed in cycles of length inferior to 50

If you do get one such permutation p , you need to open the first chest which contains the number x, than open the chest numbered p(x), etcaetera etcaetera

>>7921583

How do you do this?

The first guy gets lucky and leaves the lids open or something?

>>7921308
why?

>>7921308
0.625 is the maximum chance of winning given your opponent plays the most favorable cutoff for you. It's not the optimal cutoff.

>>7920482
Probability of winning given your cutoff = x and your opponent's cutoff = y is

(xy+y(1-x)^2+x(1-y^2)+(1-x+max(y-x,0))(1-max(y,x)))/2

This has a saddle point at 1/phi, so 1/phi is the optimal cutoff.

>>7922562
>given your opponent plays the most favorable cutoff for you

Surely the maximum chance of winning in this case would be 1?