Blondbeard
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Yes. That is the obvious That isn't the answer I was looking for. I'll add some hints later on.

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[Last edited by Blondbeard at 10-28-2006 09:29 AM]

TripleM
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Hmm.. I was about to (in fact, I did, then I deleted it) guess that it was a proof of Pythagoras' Theorem, but that is actually a slightly different picture.

krammer
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File: puzzleblondbeard.PNG (2.4 KB)
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To save those on slow connections, I managed to reduce the size of your file to less than 0.5% of the one you posted. Please don't use bmp unless absolutely necessary...

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Maurog
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Hmm, is it maybe the fact that if you "unflip" the triangles, you get a square again?


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Blondbeard
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Well... I don't know how to treat this. The correct answer has been posted, but in a slightly odd form. In fact it is a proof of Pythagoras Theorem. There are multiple pictures wich proves the theorem, but I like this one. Name the triangles sides a, b and c (which is also the side of the large square). It should be easy to see that a squared plus b squared is indeed equal to c squared. Since you actually did post this, Tripple M, I say you gave the correct answer.

zex20913
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There are literally an infinite number of proofs for the Pythagorean Theorem. At least, according to one book that listed thousands of them.

I see this proof now (had to think about it for a bit before I discovered (a-b)^2), but the proof I learned first (and best) comes from Maurog's picture, where we have (a+b)^2. I just see that more clearly.

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Citrus
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If that was in 3D, it would make quite a nice vase.

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TripleM
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Yeah, I guess it does work with this picture as well, though the proof using the other picture (which is almost identical to yours except the triangles are flipped the other way, so its easy to mix them up by accident, which is why I still posted that guess ) seems to be a lot more clear. Oh well, now I'll have to think up a puzzle..

TripleM
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OK, I haven't come up with anything yet, but I'll continue to think; I've heard so many good ones but none come to mind right now. If anyone else knows one, feel free to take over.

jamesdenem
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Well, I guess since nobody else has any ideas, I might as well post one.

What is the next number(Shown as '?') of the following sequence, and why?

4, 3, 3, 5, 4, 4, ?

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jbluestein
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jamesdenem wrote:
Well, I guess since nobody else has any ideas, I might as well post one.

What is the next number(Shown as '?') of the following sequence, and why?

4, 3, 3, 5, 4, 4, ?

3.

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[Last edited by jbluestein at 11-02-2006 07:19 PM]

Kevin_P86
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Yes, duh. I was thinking:

Instead of


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+++++[>+++++<-]>[>+++>++++>+++++<<<-]>.>+.>-------.<++++.+++++.

jamesdenem
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Pretty quick response, jbluestein got the number, and Kevin_P86 got the reason. You two can decide amond yourselves who wants to post the next puzzle, and if neither of you do, I have a much more difficult one in mind...

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Kevin_P86
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jbluestein got it first, so I think it should be his turn. But I can come up with something if he prefers...

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jbluestein
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Kevin_P86 wrote:
jbluestein got it first, so I think it should be his turn. But I can come up with something if he prefers...

Go for it. Sick baby means little sleep. Means brain no work too good.


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Kevin_P86
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Ok. This is likely one that everyone has seen before at some point, but it's still a good one. Your task is to draw a continuous curving line (in two dimensions) that passes through each line segment of this diagram exactly once (i.e. it must start and end at the same point). Is it possible? What if the line need not start and end at the same point? Is it possible then?



To avoid confusion, I have labeled the 16 line segments that need to be crossed. The order in which the segments are crossed is unimportant.

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[Last edited by Kevin_P86 at 11-02-2006 10:38 PM]

coppro
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Using geometry, I don't even need to try to tell you that it isn't possible, because the two bottom sections each have 5 surrounding segments, and there are 9 on the outside. Since 5 and 9 are odd numbers, 3 odd numbers makes it unsolvable, whether or not it is continuous.

Kevin_P86
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And the middle box on the top has an odd number (5) as well

Your turn, coppro.

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agaricus5
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coppro wrote:
Using geometry, I don't even need to try to tell you that it isn't possible, because the two bottom sections each have 5 surrounding segments, and there are 9 on the outside. Since 5 and 9 are odd numbers, 3 odd numbers makes it unsolvable, whether or not it is continuous.

I think you can also convert the diagram to a graph, with the spaces inside the rectangle as nodes, and edges as arcs, and use this to solve the puzzle. There are 9 arcs to the outside, and three nodes with 5 arcs inside the rectangle. So, the puzzle is unsolvable, because you will end up with at least two "unpaired" arcs, where a "pair" of arcs is a pair that allows you to enter and then leave a particular node. The unpaired arcs allow you to enter a node, but there won't be any arcs left leaving from that node (it's an odd, single arc), so you can't leave.

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Kevin_P86
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agaricus5 wrote:

coppro wrote:
Using geometry, I don't even need to try to tell you that it isn't possible, because the two bottom sections each have 5 surrounding segments, and there are 9 on the outside. Since 5 and 9 are odd numbers, 3 odd numbers makes it unsolvable, whether or not it is continuous.

I think you can also convert the diagram to a graph, with the spaces inside the rectangle as nodes, and edges as arcs, and use this to solve the puzzle. There are 9 arcs to the outside, and three nodes with 5 arcs inside the rectangle. So, the puzzle is unsolvable, because you will end up with at least two "unpaired" arcs, where a "pair" of arcs is a pair that allows you to enter and then leave a particular node. The unpaired arcs allow you to enter a node, but there won't be any arcs left leaving from that node (it's an odd, single arc), so you can't leave.

Yes, exactly. After converting it to a graph the object is to find an Euler Circuit in the first case, and an Euler Path in the second case. We have theorems which state an Euler Circuit exists iff all vertices have even degree (and the graph is connected, of course), and we have an Euler Path iff 0 or 2 vertices have an odd degree (and the graph is connected).

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Syntax
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Kevin_P86 wrote:
Yes, duh. I was thinking:

Click here to view the secret text

×"One", "Two", "Three", ...

Instead of

Click here to view the secret text

×"Zero", "One", "Two", "Three", ...

There are -1 types of people in the world... those who initialise arrays with 1 vs those who initialise them with 0

Kevin_P86
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Syntax wrote:

Kevin_P86 wrote:
Yes, duh. I was thinking:

Click here to view the secret text

×"One", "Two", "Three", ...

Instead of

Click here to view the secret text

×"Zero", "One", "Two", "Three", ...

There are -1 types of people in the world... those who initialise arrays with 1 vs those who initialise them with 0

I don't think I've EVER initialized an array with 1's... and I've initialized A LOT of arrays in my day! I assumed it would start with 1 because that's the only way I've seen this puzzle before...

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MartianInvader
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File: puzzle.jpg (67.1 KB)
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Kevin_P86 wrote:

To avoid confusion, I have labeled the 16 line segments that need to be crossed. The order in which the segments are crossed is unimportant.

Hehe... you managed to cause confusion with the labeling. Thought I'd post this since it took me a little while to make. It's what happens when you don't read the puzzle closely enough.



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[Last edited by MartianInvader at 11-04-2006 12:36 AM]

Maurog
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Actually, there is a solution, and all the people using Euler are wrong. The answer lies in Kevin_86's ingenious use of the word "segment" as opposed to say "edge". These things are segments, they have width and height, and there are two ways to cross any of them - horizontally and vertically. Here is an example of a solution crossing all the segments and ending in the same spot:

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Mattcrampy
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I don't care if this is the intended answer or not, that's damn clever.

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agaricus5
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Maurog wrote:
Actually, there is a solution, and all the people using Euler are wrong. The answer lies in Kevin_86's ingenious use of the word "segment" as opposed to say "edge". These things are segments, they have width and height, and there are two ways to cross any of them - horizontally and vertically.

That is an ingenious interpretation of the word "segment", I agree, but I don't agree with you about your definition of the word "segment". A segment is a portion or piece of something, but the word "segment" doesn't necessarily define the dimensions of an object. If I told you I am thinking of a "line segment", would you say the "line" I'm thinking of has one or two dimensions?

Therefore, you can't say that either answer is wrong; you can only say that each answer is correct, given a certain assumption about the puzzle.

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[Last edited by agaricus5 at 11-04-2006 02:40 PM]

Tahnan
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(poke poke)

RoboBob3000
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The answer is Eppoke, Oklahoma.

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Mattcrampy
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RoboBob3000 wrote:
The answer is Eppoke, Oklahoma.

But what is the question? "What is a surprisingly good holiday destination?" "What town will give birth to the Antichrist?" "What town (including state) is an anagram of 'A Meek Polka Hoop'?"

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agaricus5
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RoboBob3000 wrote:
The answer is Eppoke, Oklahoma.

Actually, the answer is this.

Edit: Is there something wrong with this joke here? No-one posted a puzzle, so I can't see the harm in replying to Rabscuttle.

Of course, it might be a random negative modder, but the question is then: "why"?

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[Last edited by agaricus5 at 11-12-2006 04:40 PM]