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Nuntar
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Congratulations :)

Well, now that three people have got it I think it's time to reveal the hidden rule -- in a secret tag of course in case anyone who hasn't got it yet still wants to work on it.

Click here to view the secret text


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[Last edited by Nuntar at 12-17-2008 01:01 PM]
12-17-2008 at 02:14 AM
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Rabscuttle
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There are more than one sets with Q in them: E I N Q S U
12-17-2008 at 02:15 AM
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RoboBob3000
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Ah, I was wrong. I thought that each set of letters could produce a word that contained every letter in the set twice. I thought CINOT produced CONSTRICTION.

EINST != tinniness. That doesn't work for your rule.

EINST == intestines, by my rule. It also works for your rule.

Regardless, I think anybody who wants the turn should take it. I never have a puzzle ready to go.

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[Last edited by RoboBob3000 at 12-17-2008 03:32 AM]
12-17-2008 at 03:32 AM
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TripleM
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When I visited San Francisco recently I went to the Exploratorium. Two things caused my attention - one was about a coil of copper wire - you put your hands on it, and immediately pull them away as it feels burning hot. Only to find afterwards the coils alternate warm and cool. My hand still felt funny a minute later, so who knows why the human brain works like that.

Anyway, the second was the following puzzle. A little wooden box, with two marbles inside on something resembling a skateboarding halfpipe, with a small bump in the middle. Your aim is to get the marbles in the slots at the ends (top and bottom of picture); one on each side.

Aftering spending several minutes trying everything I could think of, I gave up. 2 minutes later, I came back and solved it straight away. What do you do?

(Warning - handdrawn picture approaching..)



[Last edited by TripleM at 12-18-2008 11:21 AM]
12-18-2008 at 11:20 AM
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Sillyman
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Soooo... completely sealed in?

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FNORD
12-18-2008 at 05:59 PM
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stigant
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Is the top open? If yes, then I'd reach in, pick up the marbles and put them in the slots. Something tells me that's against the rules, though.

But, if the top is closed, then I'd turn the box upside down, and spin it.

Spinning it right side up might work as well.

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12-18-2008 at 06:23 PM
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TripleM
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Yeah, top was closed.

stigant wrote:
Spinning it right side up might work as well.

Yup. Spinning is the answer. Very cool puzzle, in my opinion. (And I'm out of mod points at the moment).
12-18-2008 at 08:07 PM
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TripleM
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Nuntar wrote:
*gets impatient for stigant to take his turn*

This thread has basically every puzzle ever invented in it, a while ago it reached the point of 'if anyone has a puzzle to post. that's a bonus' ;)'
12-21-2008 at 12:28 AM
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stigant
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Alright:

The bad guy captures 100 good guys and makes this deal with them: He will line them up so that they are all facing the same direction and can see everybody who is in front of them (ie, the guy at the back of the line can see the other 99, the guy right in front of him can see 98 other guys, the guy at the front of the line can't see anybody at all). Each guy will have a either a blue or red hat (chosen by the bad guy), and nobody can see his own hat. Then, starting at the end of the line (ie with the guy who can see everybody else) and working forward, the bad guy will ask each good guy what color hat he is wearing. All the other good guys will hear what each other good guy says. After everybody has identified their hat color, all who were correct will be released, and all who were incorrect will be summarily executed. They may plan their strategy ahead of time, but the bad guy will be listening and can choose hats accordingly. How can the good guys maximize the number of survivors?

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12-21-2008 at 04:01 AM
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Sillyman
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Already done: The first guy says red if there is odd red, blue if there is odd blue. Then everybody else says what they know their hat color to be from all the previous people's statements, if it has been said that there are odd red, but they can see/have heard even red, their hat must be red, otherwise, it's blue.

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FNORD
12-21-2008 at 06:46 AM
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lopsidation
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Aren't you supposed to wait 24 hours if you already know it?

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12-21-2008 at 06:58 PM
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Maurog
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Nah, it was already solved in this thread several times. Haven't you read the previous 75 pages?

Nice Misery avatar btw.

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12-22-2008 at 05:47 AM
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Nuntar
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Is it okay to post a chess puzzle? I would assume most people here know the rules of chess, and that's all a puzzle like this needs -- you don't have to be good at the strategies or anything like that.


It is White's turn. Can he castle?

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12-22-2008 at 10:44 AM
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Maurog
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Unknown.
Click here to view the secret text

Click here to view the secret text


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[Last edited by Maurog at 12-22-2008 02:15 PM : Ah, nevermind, I see why this isn't possible.]
12-22-2008 at 01:34 PM
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Nuntar
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You've got the right beginning, but you can take the analysis further.

Hint:
Click here to view the secret text


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12-22-2008 at 01:44 PM
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stigant
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Ok, suppose, as Maurog did, that a white pawn made its way over to the left side of the board and was promoted to a rook. It can't have been either the A or B pawns: The B pawn is still at home and the A pawn is on A3. There's no way any other pawn can end up there (other than the B pawn, which we can see is still on B2). So any pawn which promoted at A8 must have captured at least 2 black pieces. Now, there is also a white pawn at B4. Together, these two pawns (the promoted pawn and the B4 pawn) must have captured at least 4 black pieces. Either the promoted pawn was the original C pawn, and captured 2 to get over to the A file, and the B4 pawn was the original D pawn and captured 2 to get to the B file, or the promoted pawn was the original D pawn and captured 3 while the B4 pawn was the original C pawn and captured 1 piece. There are 12 black pieces on the board, so only 4 are missing. This means that all 4 must have been captured by white pawns. The black, light squared bishop could not possibly have moved since the B7 and D7 pawns are still at home, so black must have promoted his missing pawn to a bishop at some point. But more to the point, the original light squared bishop must have been captured on C8 by a white pawn. But there is no way that a white pawn could have occupied either B7 or D7 because the black pawns are still on those squares.

So, no, white cannot castle because there is no way that he could have gotten his H rook out without moving his king at some point, nor could he have promoted a pawn into a rook.

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12-22-2008 at 05:24 PM
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stigant
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Even more to the point, why would he want to?
1. Qe2+ Kd4
2. Rd1+
leads to mate after a couple of useless blocking moves by black.

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12-22-2008 at 05:29 PM
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Sillyman
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Hang on... that means the position is impossible if we are looking from the standard angle. As Maurog said, the board must be reversed. Or it could be a nonstandard imbalanced placement in which case the king could probably castle.

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FNORD
12-22-2008 at 05:34 PM
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stigant
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I don't have a complete move order that achieves this position, but black could have promoted his A pawn to a bishop on D1 after capturing 3 of white's pieces (white is missing 4 pieces, so that's possible), so unless you have some other objection...

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12-22-2008 at 06:06 PM
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Nuntar
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Stigant is now very close to the solution, but in a problem like this, getting the correct answer (White cannot castle) doesn't count unless you can produce a watertight proof.

Click here to view the secret text


The position is definitely possible, and is the correct way up.

As for why White would want to castle, it's true that Qe2+ is mate in four, but castling is mate in only two. There is, however, another mate in two in the position.

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12-22-2008 at 06:51 PM
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stigant
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Ah, the location of the promotion of the white pawn doesn't matter...

It's the promotion of the black pawn that's impossible. Black must have promoted his A pawn to a light squared bishop (since his original LSB couldn't have moved because the B7 and D7 pawns are still at home). Now, this promotion could have occurred at 3 different squares: B1, D1, and F1 (H1 is impossible because of white's unmoved pawns on the right side). Further, it could not have promoted FROM D2 since it would have been attacking the white king from that square (since the king presumably hasn't moved). In this eventuality, white would have had to move his king or capture the pawn, preventing its promotion. So the black pawn had to promote from either C2 or E2, and in doing so must have captured a piece on the promotion square.

To promote at any of these squares, the A pawn would have to capture at least 3 of white's pieces. White is missing both bishops, a pawn and (presumably) his King side rook. The rook couldn't have been captured by black's A pawn because it can't have moved past the white king. So the A pawn must have captured a pawn and both bishops. White's LSB must be have been captured on the promotion square as it's the only one of the three pieces which could occupy any of the promotion squares. So, which pawn did the black A pawn capture? It can't have been white's A or B pawns (as shown above, they are on A3 and B2) or the F, G or H pawns.

If it was white's C pawn, then white's D pawn is on B4, and must have captured black pieces on C3 and D4, both dark squares. Black is missing a knight, a queen and both of his original bishops. However, both bishops had to be captured at home (C8 and F8) since they can't have moved past the pawns on the 7th rank, and they can't have been captured by a white pawn for the same reason. So the black queen and knight were captured by the white D pawn, the white c pawn was captured by the black A pawn, and the black bishops were captured by non-pawns and the white e pawn must have been promoted to a white rook... where? Either via A7, F7 or H7, all of which would have required the E pawn to capture more material. But then black would have less pieces.

Ok, so the white C pawn must be at B4, and it must have been the white D pawn that got captured by the black A pawn. But how? The A pawn can get to the B file by capturing white's DSB, but it has to capture white's D pawn on the C file, so the D pawn must capture something on the C file as well. It can't be either of the bishops. So it was either a knight or the queen. Meanwhile, the white C pawn captured the other black piece (knight or queen) on either B3 or B4. And that leaves the white E pawn to promote.. but again how? It has to move off it's file by capturing another black piece, but there aren't any left.

EDIT: Oh, black's A pawn can't have captured white's E pawn for similar reasons

So, white can't castle because it's impossible that both sides have promoted a pawn.

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[Last edited by stigant at 12-22-2008 08:12 PM]
12-22-2008 at 08:05 PM
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Nuntar
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:thumbsup Excellent answer!

(Just in case anyone wants to know: this was originally composed as a "dual" problem; you have to find the mate in two and prove that castling doesn't work as an alternative solution. The other mate in two is: )
Click here to view the secret text


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[Last edited by Nuntar at 12-22-2008 08:32 PM]
12-22-2008 at 08:31 PM
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Ok, here's one I came up with myself, though it's certainly possible that it's been thought of before.

My sister-in-law is having a child. We don't know the gender or name yet, so we refer to it as "Gilly" (short for "One with gills"). I've just found out that Merriam Webster defines the term "Nibling" to refer to a niece or nephew in a gender-neutral fashion.

Anyway, I want to make a jigsaw puzzle for my future nibling. I want it to be a pretty simple puzzle. Each piece should be square except for the traditional nubs (pieces which border the edge of the puzzle have no exterior nubs). Nubs are either male (out) or female (in), and male nubs fit into female nubs in the tradtional jigsaw fashion. In a typical jigsaw, most male nubs won't fit into most female nubs. Each male nub has a slightly different shape that will match only a few complementary nubs, or sometimes only 1 female nub. This, along with the picture on the puzzle, typically ensures that there is a unique (up to rotation, although the picture of course usually imposes a symantically unique rotational solution as well) solution to the puzzle. Also, there is a clear back and clear front to each piece (although, I've done jigsaws with pictures on both sides and they are very difficult).

Now, my puzzle pieces will have a distinct front/back orientation. Unfortunatly, I don't have the wherewithall to ensure that all my nubs are different shapes. In fact, I have to make all the nubs semi-circular, and they all have to be the same size (see picture of sample corner puzzle piece). This fact, combined with the nature of the image on the front of the puzzle makes it impossible to tell if two pieces (one with a male nub, one with a female nub) actually fit together without considering all the rest of the pieces in the puzzle. Now, each piece IS distinct, and it's possible to distinguish between particular orientations for each piece, though it is NOT possible to tell which orientation is actually the intended orientation. All of this makes me nervous that my nibling will assemble the puzzle but find the wrong message.

So the puzzle is this: Is there an arrangement of male/female nubs such that the puzzle has a unique (up to rotational orientation) solution? There are to be 9 pieces which should be assembled into a 3x3 array to complete the puzzle.

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12-23-2008 at 07:49 PM
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Nuntar
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Whew! This is much, much harder than it seems at first sight.

I've got as far as:
Click here to view the secret text


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12-23-2008 at 11:31 PM
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lopsidation
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Also,
Click here to view the secret text


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12-24-2008 at 12:27 AM
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TripleM
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Click here to view the secret text


[Last edited by TripleM at 12-24-2008 01:30 AM]
12-24-2008 at 01:28 AM
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Nuntar
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OK. My proof rests on the two lemmas already given, so I'll formally prove those first.

Lemma 1. There are 70 ways to choose four of the eight possible edge pieces; however, only 18 of those might be a solution.

Proof: as already stated, the four corner pieces must all be different. Hence the edge pieces must, if placed in a row with the straight edge upward, have between them two male and two female nubs on the left, and two of each on the right. This eliminates all but 18 subsets.

Lemma 2. There is no solution unless the four edge pieces are all different as regards the two nubs facing the corner pieces.

Proof: This property holds for 16 of the 18 subsets. The other two are [FFF, FMF, MMM, MFM] and its complement. (The notation is straightforward: FMF for instance means the edge piece with a female nub on the left side, male on the bottom and female on the right, again with the straight edge upward.) In both cases the centre piece will have two nubs of each type, so there are four cases to consider (since for each set of edge pieces, the centre piece might have like nubs adjacent or opposite). After making this post I'll draw some diagrams showing that none of the four cases provides a solution.

So the four edge pieces are all different as regards their corner-facing nubs. Let's place the piece with two female nubs at the top. There are exactly four possible arrangements for the nubs between the edges and corners that make all four edges different and all four corners different; after this post I'll diagram those too.

Given a set of pieces with all corners different and all edges different, the only information we can have as to which of the four possible formations is correct comes from the centre piece and from which of the edge pieces have male and female centre-facing nubs.

Case 1: all four centre-facing nubs are alike. In this case we have no information whatever, so all four formations are possible.

Case 2: three of the four centre-facing nubs are alike. This tells that a particular edge piece must connect to a particular side of the centre piece. But the jigsaw can always be rotated to bring the edge piece with two female nubs to the top, so again all four formations are possible.

Case 3: the centre piece has two nubs of each type. In this case we either know which edge pieces are opposite each other, or that a particular pair are not opposite. But, as the diagrams of the four formations will show, there are two possibilities for which piece is opposite the edge piece with two female nubs, and two possible formations for each. Therefore it is never possible to narrow the jigsaw down to one formation.

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12-24-2008 at 01:49 AM
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And now, those damnable diagrams. Sorry they're so crude.

The first eight, taken in pairs, deal with the four cases mentioned in Lemma 2 and show that each of them provides two possible jigsaws, so is not a solution.

The last four diagrams are the four formations discussed at the end of my proof.

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12-24-2008 at 02:06 AM
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Well, I've come up with a method of my own, though I won't go into massive details.

For ease of writing, I'll replace an 'in' bump by 0 and 'out' bump by 1.

There are two options for the center piece to not have rotational symmetry - either (reading clockwise) 0011 or 0001. (Or 1110, but that's the same).

For the 0011 case, if the edge pieces are called A B C D (in the same clockwise order), replacing them by B A D C gives a new solution.

For the 0001 case, if the edges are A B C D, replacing them by C B A D gives a new solution.

This works since, if you write down how the edges match up, you end up taking the four corners (00 01 10 and 11 in some order), splitting them into pairs, and swapping the second bit in each pair, which still leaves you with 00 01 10 and 11 in some order.

[Last edited by TripleM at 12-24-2008 03:42 AM]
12-24-2008 at 03:41 AM
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Yep, I think both of you have it. TripleM's solution was pretty close to mine, but Nuntar got there first.

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