• Current Events & Politics
    Welcome Guest
    Please read before posting:
    Forum Guidelines Bluelight Rules
  • Current Events & Politics Moderators: deficiT | tryptakid | Foreigner

Questions With Answers (cf. the "Universe Threads")

Fjones said:
NSFW:
6,210,001,000
Click to expand...


NSFW:
Very good, Sir. Your answer is correct.



Wow, time for a New Puzzler already. Let's see...in honor of Fjones, I post a Marble Puzzle (only one that can be solved by a human being!).


Alan and Bob play a game with a pile n marbles [assume that n - 1 is not a multiple of 4]. They take turns removing either 1, 2, or 3 marbles. The person who takes the last marble loses.


If Alan plays first, then who wins?
 
Binge Artist said:
NSFW:
Very good, Sir. Your answer is correct.



Wow, time for a New Puzzler already. Let's see...in honor of Fjones, I post a Marble Puzzle (only one that can be solved by a human being!).


Alan and Bob play a game with a pile n marbles [assume that n - 1 is not a multiple of 4]. They take turns removing either 1, 2, or 3 marbles. The person who takes the last marble loses.


If Alan plays first, then who wins?
Click to expand...

NSFW:
Alan wins. If 2 marbles, Alan picks 1 and wins. if 3, Alan picks 2 and wins. if 4 marbles, Alan picks 3 and wins.

If 6 marbles, Alan picks 1, leaving 5. Now Bob must leave 2, 3, or 4 marbles for Alan's turn, which has just been shown to be a win for Alan.

If 7 marbles, Alan picks 2, leaving 5.

If 8 marbles, Alan picks 3, leaving 5.

Any of the instances where n-1 is a multiple of 4 result in a loss for whoever's turn it is, the case of 1 marble being obvious, and the case of 5 being shown above. The case for 9, 13, etc is obvious, because if there are 9 marbles (or 13 or 17 etc.), suppose player 1 removes X marbles. Player 2 then removes 4-X marbles, thereby reducing the number of total marbles by 4 for each cycle until eventually player 1 will have to make a move with 5 marbles remaining.

Thus, all player 1 has to do in any game where the number of marbles n-1 is NOT a multiple of 4 is to remove the necessary number of marbles such that the remaining number of marbles IS one more than a multiple of 4.
This was a bit wordy and cumbersome but it works.
 
Last edited:
Marble game (or pennies or whatever. 15 of any object).

Players take turns removing marbles. Whoever takes the last one loses.

The actual game usually does not involve lines paper, that is just for rules clarification.

marbles.jpg


On your turn, you can remove any or all marbles from ONE ROW, and one row only. So you may remove any single marble, or the entire bottom row (or any row) or any part of a row. But you may never remove marbles from more than 1 row in your turn.

Who has the advantage? Player 1 (goes first), or player 2? What should each player's strategy be?
 
Fjones said:
Marble game (or pennies or whatever. 15 of any object).

Players take turns removing marbles. Whoever takes the last one loses.

The actual game usually does not involve lines paper, that is just for rules clarification.

marbles.jpg


On your turn, you can remove any or all marbles from ONE ROW, and one row only. So you may remove any single marble, or the entire bottom row (or any row) or any part of a row. But you may never remove marbles from more than 1 row in your turn.

Who has the advantage? Player 1 (goes first), or player 2? What should each player's strategy be?
Click to expand...
NSFW:
THe second player because it is a non transitive game i think. Can't really be bothered to work it out for certain of give some more formal proof as it's late. Am I right?
 
rpm, I have no idea what "transitivity" means in the context of this game.
 
A transitive game is one in which move a is better than b, b better than c and so on. It seems then that the second player will have the advantage as they will always be able to trump the previous players move. I may be miles of here, sorry.
 
Binge Artist said:
Gut instincts on 1/2 involve analogous situations like this:

Pile A is a fully shuffled stack of 1 million cards, numbered from 1 to 1 million.

Pile B is a single card, numbered zero.


Some goofball is going to pick a card from one of the two Piles, and if you guess the card's number, the goofball gonna give you a million bucks (and the goofball picks the Pile based on
the toss of a coin).


Does any zero-like number pop out at you?

_____________________________________________________________

So, the question is exactly how analagous is this?

Suppose Pile A had two cards, 1 & 2, and suppose Pile B was just the zero card. Is the situation identical to the situation faced by the sleeping woman?
Click to expand...


I am not following you here.

And I am going on record as saying that the sleeping beauty problem is 1/3.
 
Fjones said:
Who has the advantage? Player 1 (goes first), or player 2? What should each player's strategy be?
Click to expand...


NSFW:
I believe the first player can force a win with any of these three first moves: 1. Remove the single coin from the first row, 2. Remove one coin from the third row, or 3. remove one coin from the fifth row. The plan for victory, I believe, is very easy to "intuitively" grasp, but somewhat cumbersome to rigorously outline, but it involves writing the number of pennies in each column in binary form.

I studied a very similar game in a combinatorics class. The only difference was that the winner took the last piece. However, the same strategy seems to apply, only with minor modification
 
Last edited:
Binge Artist said:
NSFW:
I believe the first player can force a win with any of these three first moves: 1. Remove the single coin from the first row, 2. Remove one coin from the third row, or 3. remove one coin from the fifth row. The plan for victory, I believe, is very easy to "intuitively" grasp, but somewhat cumbersome to rigorously outline, but it involves writing the number of pennies in each column in binary form.

I studied a very similar game in a combinatorics class. The only difference was that the winner took the last piece. However, the same strategy seems to apply, only with minor modification
Click to expand...

Yes. Very impressive. You are exactly right. Any one of those 3 moves and player 1 can force a win. Any other move, and player 2 can force a win.
 
Ah...very nice RedLeader.


On a friendly puzzle thread for DRUG ADDICTS, you go and post a puzzle whose solution is evidently journal material 8)


On a related note, IIRC, that coin puzzle Fjones posted may have been a research problem not toooo many years ago. In fact, I think John Conway has written papers on it.
 
You asked for something hard. I gave you something hard. Something hard, which kept you entertained for a long while. Come now, Bingey...don't tell me you're general equation is like Ax + By + C = 0...

Dude, every puzzle I post in here has journal articles written about them. I just reword it so you cannot google it ;)

I'll post another soon. Tell me how hard you want it to be.
 
New Puzzler, in honor of the recent Cannabis Holliday, 4/20.


Let S be the set of all "prime factorials". That is S = {2F, 3F, 5F, 7F, 11F...} Note: normally I would write "!" instead of "F", but Bluelight doesn't seem to like "!" for some reason...

Anyway, the puzzle is this: Using only multiplication and division and members of S, can one create an expression equal to 420/2010?

Examples:

1/2 = 2F/(2F*2F)

1/3 = 2F/3F

1/10 = (2F*3F)/5F
 
Last edited:
Time to bump this thread for Summer '10 fun.

Problem ("Sums and Squares"): Assume that a_1, a_2, ..., a_16 are positive real numbers whose sum is 100 and the sum of whose squares is 1000. Let X be the set of all positive real numbers X such that a_i < X for all X in X. Find min(X).

Difficulty Level: :X:X:X
 
Problem ("Coin-Tossing"): Assume three men, Mr. X, Mr. Y and Mr Z, arrive with amounts of coins, x, y and z, to play a game. The rules of the game are as follows:

(1) In each round, each player tosses one coin. In the case of a tie (HHH, TTT) coins are retossed. Repeat until a toss yields not a tie (HHT, HTH, HTT, THH, THT, TTH).

(2) The odd player out (e.g. the "T" in HHT) keeps all 3 coins to himself from that round.

Given values x, y and z, what is the average/expected number of tosses before the first player goes broke?

Note that we ask for "tosses" and not "rounds," as multiple tosses can occur within multiple rounds to break ties.

Difficulty Level: :X:X:X:X:X

(I can give a hint on this one, but the hint might just drive one even more mad.)
 
RedLeader said:
Time to bump this thread for Summer '10 fun.

Problem ("Sums and Squares"): Assume that a_1, a_2, ..., a_16 are positive real numbers whose sum is 100 and the sum of whose squares is 1000. Let X be the set of all positive real numbers X such that a_i < X for all X in X. Find min(X).

Difficulty Level: :X:X:X
Click to expand...


Either I'm missreading this, or the problem is trivial.

NSFW:
Answer: 25.

Basically, what we want to do is maximize a_16.

Let m be the average of a_1, a_2, a_3...a_15.

Then,

15m + a_16 = 100, and

15m^2 + (a_16)^2 < or = (a_1)^2 + (a_2)^ + ... + (a_16)^2 = 1,000.

Solving this quadratic inequality, we have that m must be in [5, 7.5], with a smaller m implying a bigger a_16. And indeed, we get a_16 = 25 and a_1 = a_2 = a_3 = ... = a_15 = 5.
 
You must log in or register to reply here.
Top