Math and Logic Puzzles

Sudo_Nym · Wed Aug 03, 2016 6:29 am

Suppose I've got an urn that contains 150 black marbles, and 75 white marbles, and I have a pile of black marbles sitting off to the side. I randomly draw 2 marbles from the urn, and depending on what color combo they are, I do one of the following three things:

1) If both marbles are black, I throw one away and put the other back.
2) If both marbles are white, I throw both away and put a black marble back in the urn.
3) If I draw one of each, I throw the black marble away, and put the white marble back in the urn.

Because each operation reduces the number of marbles by one, eventually, I'll have only 1 marble left in the urn. What are the odds that the last marble will be black?

Sudo_Nym · Wed Aug 03, 2016 10:55 am

In post 3649, serrapaladin wrote:
In post 3647, Sudo_Nym wrote:Suppose I've got an urn that contains 150 black marbles, and 75 white marbles, and I have a pile of black marbles sitting off to the side. I randomly draw 2 marbles from the urn, and depending on what color combo they are, I do one of the following three things:

1) If both marbles are black, I throw one away and put the other back.
2) If both marbles are white, I throw both away and put a black marble back in the urn.
3) If I draw one of each, I throw the black marble away, and put the white marble back in the urn.

Because each operation reduces the number of marbles by one, eventually, I'll have only 1 marble left in the urn. What are the odds that the last marble will be black?

White marbles can only be removed in pairs, so eventually you'll get down to 1 white marble, which cannot be removed.

Spoiler:

Sudo_Nym · Thu Aug 04, 2016 4:49 am

It's been a bit since I graduated, so forgive me if I slip up in my notation or something.

Let x be a sequence of bounded sequences (i.e. x = {x_1, x_2, ....}, where each x_n is a sequence that is bounded both above and below). Then let y = {y_1, y_2, ...}, where y_n = {x_1_n, x_2_n, ...} (That is, each y_n is a sequence of the nth elements of each x sequence). If A = [min(y_1), min(y_2), ...] and B = [max(x_1), max(x_2), ...], then which is larger: max(A) or min(B)?

Sudo_Nym · Fri Aug 05, 2016 4:59 am

Suppose you've got 5 holes in a straight line in your backyard, and there's a fox living in one of them, but you don't know which one. Each day, you have enough time to check just one of the holes for the fox, and every night, the fox moves to a different hole. The fox always moves to a hole that is adjacent to the previous hole. Which hole should you check each day to catch the fox in the quickest time?

Sudo_Nym · Fri Aug 05, 2016 7:50 am

Spoiler: Milittos and Shadow Dancer's solution

Spoiler: serrapaladin's solution

Sudo_Nym · Mon Aug 08, 2016 3:39 am

There's a square table on a track so that the top can rotate. In each corner of the table is a well, and each well contains a tumbler which is either face up or face down. You can't see into the wells, but with each move, you can reach into two wells, feel the orientation of the tumblers, then flip either or both if you wish. After each move, however, the table rotates and stops on an arbitrary side so that you don't know it's orientation compared to the previous move. What's the smallest number of moves necessary to make all the tumblers face in the same direction?

Sudo_Nym · Mon Aug 08, 2016 10:42 am

Assume that if they're all facing the same direction, you're told that you've won, even if you don't know that you've accomplished the goal.

Sudo_Nym · Mon Aug 08, 2016 12:17 pm

Who's correct. I had a slightly different order, but it had basically the same iidea- take advantage of the fact that you're told if you win to make a pure information move so that you can force the position to UDUD regardless of start, then flip the diagonals.

Suppose you've got a rail 1 mile long (5280 feet). You then push the rails so that the endpoints are 1 foot closer, making the rail bend up in the middle to accommodate this. How far off the ground is the middle of the rail?

Sudo_Nym · Mon Aug 08, 2016 12:31 pm

Sorry, I assume it bends in an arc. So the length of the beam is still 1 mile as measured along the beam, but the ends are 5279 feet apart as measured along the ground.

Sudo_Nym · Wed Aug 10, 2016 8:23 am

I hadn't bothered to account for the earth's curvature, since I figured that over 1 mile, its basically negligible. Maybe I'm more astonished than I should be that pushing the endpoints 1 foot closer together makes the arc bow up ~44 feet in the middle.

Sudo_Nym · Wed Aug 10, 2016 12:07 pm

In a chess game, white's moves go 1. f3, 2. Kf2, 3. Kg3, 4. Kh4. On the fourth move, black checkmates him. What were black's moves?

Sudo_Nym · Wed Aug 10, 2016 12:45 pm

In post 3685, Aronis wrote:
In post 3684, Sudo_Nym wrote:In a chess game, white's moves go 1. f3, 2. Kf2, 3. Kg3, 4. Kh4. On the fourth move, black checkmates him. What were black's moves?
I think this should work?
d5, knightf6, Qd6, Qf4

Not checkmate, because g4 blocks the check.

Sudo_Nym · Wed Aug 10, 2016 1:23 pm

"g4" is notation for "move the pawn at g2 to g4". In standard notation, pawn moves don't get a piece notation, they just get their destination square denoted unless there's ambiguity over which pawn can move into that square.

Sudo_Nym · Thu Aug 11, 2016 3:11 am

I think that's why I like this as a chess puzzle. White's play is so suicidal that it seems like it should be easy to figure out how to checkmate in 4, but the play to find it is so counter-intuitive that most people wouldn't think of it.

Sudo_Nym · Thu Aug 11, 2016 3:47 am

Spoiler:

Sudo_Nym · Fri Aug 12, 2016 7:43 am

Chess puzzles are nice, I guess.

White to move, mate in 2.

Sudo_Nym · Fri Aug 12, 2016 8:45 am

Indeed. Think I started off a bit easy for my technically second chess problem.

White to move, mate in 2. Let me know if you guys actually like chess puzzles, so I don't flood if people want to do other stuff instead.

Sudo_Nym · Fri Aug 12, 2016 9:03 am

Correct. And very well then:

White to move, mate in 2.

Sudo_Nym · Fri Aug 12, 2016 9:11 am

In post 3700, Who wrote:
Spoiler:
Qd4.
If black blocks with pawn, Qd5#. Else f7#.

Spoiler:

Sudo_Nym · Fri Aug 12, 2016 9:40 am

inspiratieloos has it.

Mate in 2.

Sudo_Nym · Fri Aug 12, 2016 10:51 am

Implosion gets that one. I'd keep score if I wasn't so lazy. Instead, have another:

Mate in 2.

Sudo_Nym · Fri Aug 12, 2016 11:52 am

Spoiler:

Mate in 2.

Sudo_Nym · Sat Aug 13, 2016 5:04 am

Yeah, serra has it. 1. Ra1 Kxg6 (forced) 2. Qb1#

Let's shake it up. Try a mate in 3.

Sudo_Nym · Sat Aug 13, 2016 6:57 am

Well done, Faalcon.

Another mate in 3.

Sudo_Nym · Sat Aug 13, 2016 1:34 pm

Very well done, Faalcon.

Here's a simpler one for the chess novices who still want to participate:

Mate in 6

Sudo_Nym · Sat Aug 13, 2016 2:02 pm

StrangerCoug gets on the board! A more difficult one:

Mate in 3

Sudo_Nym · Sat Aug 13, 2016 4:38 pm

In post 3736, itlepip wrote:
Bc6 Kd4 Nc5 if e5 Rd6 if anything else Rf4#

Spoiler:

Sudo_Nym · Mon Aug 15, 2016 3:13 am

A hint, for people that want one:

Spoiler:

Sudo_Nym · Mon Aug 15, 2016 3:41 am

In post 3741, itlepip wrote:Not d4 d3? Every line I have seen fails due to those squares.

It turns out that marching down the board in a lot of lines. Either way, you have to stop black from doing that.

Sudo_Nym · Wed Aug 17, 2016 12:39 pm

Too hard? Should I give more hints or just give the solution?

Sudo_Nym · Wed Aug 17, 2016 1:31 pm

Alright, here's a clue:

black escapes most variations by marching down the board, either to e4 to d4, and then escaping the net. White needs to find a way to block black's king from getting down the board, without moving any of the pieces that will be necessary for the mating move, and has two moves to get into position. Maybe there are some other concerns white has in this position that can be disregarded if they don't actually stop white from getting into position.

Sudo_Nym · Thu Aug 18, 2016 2:56 am

Spoiler:

Sudo_Nym · Wed Aug 24, 2016 3:13 am

Sorry to interrupt, but:

Mate in 2.

Sudo_Nym · Wed Aug 24, 2016 4:41 am

Congrats to Inspirateloos.

Mate in 2.

Sudo_Nym · Wed Aug 24, 2016 5:39 am

Indeed. Here's another:

Mate in 2.

Sudo_Nym · Wed Aug 24, 2016 6:44 am

Indeed; Qa1, and then there's nothing black can do to escape. I feel like Sam Loyd has the idea to pick ridiculous looking moves, and then figure out a situation where that's the solution to a puzzle.

Mate in 2.

Sudo_Nym · Wed Aug 24, 2016 7:38 am

Mate in 2.

Sudo_Nym · Wed Aug 24, 2016 7:57 am

Correct, Rhinox. Getting the Rook out of the way while covering the promotion square leaves black with no outs. Perhaps my predilection for puzzles where the solution is the most ludicrous looking move is making things too easy. Let's shake this up:

It turns out there's a missing pawn in the problem! White to add a pawn to the board somewhere, then mate in 2.

Sudo_Nym · Wed Aug 24, 2016 8:52 am

Well, it has to involve some element of the unexpected, or it's not much of a puzzle, is it?

Sudo_Nym · Wed Aug 24, 2016 9:29 am

Elekitu has it.

White is kicking himself- after completing his last move, he notices that he could have checkmated black instead! What move did white just play, and what move should he played to checkmate instead?

Sudo_Nym · Thu Aug 25, 2016 11:49 am

SS has it. I normally enjoy watching the though process, but I won't hold it against him.

A bit of a simpler one along the same lines:

Assuming it's white to move, what must black's last move have been?

Sudo_Nym · Thu Aug 25, 2016 1:17 pm

Faalcon is correct! Let's have a little fun:

Mate in 0.

Sudo_Nym · Fri Aug 26, 2016 5:09 am

Mate in 2.

Sudo_Nym · Fri Aug 26, 2016 5:39 am

Sorry, I accidentally swapped the color of one of the pawns that invalidated the solution. Should be better now.

Sudo_Nym · Fri Aug 26, 2016 6:07 am

@Rhinox: Correct!

@S_S: Stop breaking kayfabe!

Another mate in 2

Sudo_Nym · Fri Aug 26, 2016 6:17 am

Is that a serious guess, or are you just looking for a ludicrous move?

Sudo_Nym · Fri Aug 26, 2016 6:27 am

Nope, Qg2 is correct. I'm just wondering if people are actually working on the puzzles or just guessing silly moves at this point. Call me paranoid. That said, if there's somebody out there who has a fun puzzle that uses completely expected moves, feel free.

Mate in 2.

Sudo_Nym · Fri Aug 26, 2016 7:13 am

Sukima is correct.

Mate in 2.

Sudo_Nym · Fri Aug 26, 2016 7:40 am

Elekitu has this one. I'm going to take a break on these for a little bit, I think.

Sudo_Nym · Fri Aug 26, 2016 9:14 am

Spoiler:

Sudo_Nym · Fri Aug 26, 2016 2:31 pm

Are you sure it's 8 moves?

Spoiler:

Sudo_Nym · Sat Aug 27, 2016 5:17 am

In post 3843, Elekitu wrote:
Spoiler:
1. Qg3 [whatever]
2. Qb8#

1. Ra5+ Bxa5
2. b5#

1. d6+ Ka7
2. Qa2#

Spoiler:

Sudo_Nym · Sat Aug 27, 2016 5:37 am

Ya got me, ya tricksy devil

Sudo_Nym · Sun Aug 28, 2016 4:25 am

Well, I suppose that's long enough to wait.

Mate in 3.

Sudo_Nym · Sun Aug 28, 2016 5:40 am

Faalcon is correct.

Mate in 2.

Sudo_Nym · Sun Aug 28, 2016 9:05 am

In post 3851, itlepip wrote:
Qxd6 nx rx else rb1

If Qxd6+, then a5 and there's no mate.

Sudo_Nym · Sun Aug 28, 2016 9:09 am

In post 3853, Something_Smart wrote:
Spoiler:
1. Qe6
1. ...Nxe6 2.Na7#
1. ...a5 2. Na7#
1. ...Nb6 2. Rxc5#
1. ... Kb6 2. Rb1#
1. ...c4 2. Rb1#
1. ...Ne8 2. Na7#

1. Qe6 cxd4 and there's no mate.

Sudo_Nym · Sun Aug 28, 2016 9:27 am

In post 3855, Something_Smart wrote:
In post 3854, Sudo_Nym wrote:
In post 3853, Something_Smart wrote:
Spoiler:
1. Qe6
1. ...Nxe6 2.Na7#
1. ...a5 2. Na7#
1. ...Nb6 2. Rxc5#
1. ... Kb6 2. Rb1#
1. ...c4 2. Rb1#
1. ...Ne8 2. Na7#

1. Qe6 cxd4 and there's no mate.

Spoiler:
Qg6 then

most of the lines are the same, and if cxd4 Qb1 is mate

That's the solution.

Basically, covering the knight gives you mate in most lines, you just need specifically Qg6 for the cxd4 line.

More mate in 2 action.

Sudo_Nym · Sun Aug 28, 2016 1:05 pm

S_S is correct.

What was the last move by each player?

Sudo_Nym · Sun Aug 28, 2016 2:41 pm

In post 3860, Faalcon wrote:
Spoiler:
black played Ba7 and white played Nxa7

Assuming you meant to flip the board, that's correct.

Mate in 1, but for who? Who's turn is it, and what is the mate?

Sudo_Nym · Sun Aug 28, 2016 4:11 pm

S_S is correct

Mate in 2.

Sudo_Nym · Tue Aug 30, 2016 7:28 am

Sorry I'm late, but Sukima is right.

Sudo_Nym · Thu Sep 01, 2016 5:08 am

We are airy little creatures,
All of different voices and features:
One of us in glass is set,
One of us you'll find in jet,
Another you may see in tin,
And the fourth a box within;
If the fifth you should pursue,
It can never fly from you.
What are we?

Sudo_Nym · Tue Sep 06, 2016 3:46 am

Put point A at 1/3 the distance, and point B at 2/3 the distance.

Pick up 1000 bananas, walk to point A. The camel eats 333 bananas on the way, so leave the remaining 667 bananas at A. Go back to the start, pick up 1000 bananas, go to point B, picking up 333 bananas at A on the way. The camal eats 333 bananas from A to B, so now there's 1000 bananas at the start, 334 bananas at A, and 667 bananas at B. Go back to start, pick up 1000 bananas. Walk to the end, picking up 333 bananas at A, and 333 Bananas at B. Now you have 1 banana at A, 334 at B, and 667 bananas at the finish. So it's possible to get 667 bananas to the end point this way.

Put A at 1/4, B at 1/2, C at 3/4:

Code: Select all

Start
3000  2000  1000      0     0

A
0      750    500   250     0

B
0        0    750   500     0

C
0        0      0   750     0

Finish 
0        0      0     0   750

So you can get more bananas with 3 stops than with 4. I don't care to work out right now what the optimal number of stops is, but there's at least those two worked out.

Sudo_Nym · Fri Sep 23, 2016 10:45 am

Suppose 2n points are arranged on a plane so that no three are collinear, and then half are colored red and half are blue. Will it always be possible to connect a red dot to a blue dot, in pairs, so that none of the connecting lines intersect?

Sudo_Nym · Sun Sep 25, 2016 7:18 am

Spoiler:

Sudo_Nym · Fri Sep 30, 2016 7:56 am

Suppose we randomly select x and y on the interval [0, 1], uniformly distributed. What is the probability that mean(x,y) <= .1?

Sudo_Nym · Tue Jan 03, 2017 10:08 am

Suppose we're going to play a game. The rules of the game are such: using an ordinary deck of cards whose cards have been shuffled, we each call out a sequence of colors (say, Red/Black/Red), then deal out the cards one at a time. Any time three cards in a row are dealt in a way that matches one of our called sequences, we take all three cards as a trick, and then continue dealing until all the cards are dealt, and then the player with the most tricks win.

Assuming I call RBR, what sequence should you call to maximize your chance at winning?

There is a general rule that allows P2 to maximize winnings against any sequence, if somebody feels like working that out as well.

Sudo_Nym · Fri Jan 06, 2017 4:47 am

Yeah, that's the basic logic. BRR beats RBR 922 times out of 1000 in the simulation, if you're curious, and you're right about why it works. Cards instead of coin flips, if you're curious, party because you can "steal" cards, but not really recorded flips, and also because the finite number of cards in the deck raises P2's odds somewhat.

Sudo_Nym · Fri Jan 06, 2017 2:35 pm

Two men jointly own x cows. They sell the cows for x dollars a piece, and use the money to buy sheep at $12 each. They don't have exactly enough money from the sale of the cows to buy an integer number of sheep, so they buy a lamb with the remainder, then divide the flock so that each man has an equal number of animals. They both decide to sell their flocks at market value; the man who agreed to take the lamb is a little shortchanged, though, so the other man gives him his harmonica to make the deal even. How much was the harmonica worth?

Sudo_Nym · Fri Jan 06, 2017 5:16 pm

Implosion is correct.

Sudo_Nym · Fri Jan 06, 2017 6:19 pm

Suppose you're a King, and one day, messengers from two different lands come bearing cash to bribe you for your support. The first messenger has an infinite number of envelopes; the first envelope contains $1, the second contains $2, the nth envelope contains $n, and so on. The second messenger also has an infinite number of envelopes, such that the first contains $2, the second contains $4, and the nth contains $2n.

The second messenger claims that you should thus aid him, because for each envelope the first gave you, his envelope contains twice as much cash, and therefore, his bribe is greater. The first counters that his envelopes contain every integer amount of dollars, while the second messenger only brought half the integers, and therefore his own bribe is greater. Which one is correct?

Yes, I know it's silly.

Sudo_Nym · Mon Jun 19, 2017 9:33 am

If you count the number 1, the first 9 primes are 1, 2, 3, 5, 7, 11, 13, 17, and 19. Is it possible to construct a 3x3 magic square from these nine digits?

Sudo_Nym · Mon Jun 19, 2017 10:54 am

Spoiler: Felissan