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 Post subject: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 12:49 am 
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Hey everyone! I would like to pose a fun question.

If the original "The Orb" puzzle has 4 rows of beads, with the rows containing 8, 20, 20, and 8 beads,:

How many beads would be in a version of The Orb with one more row of beads, but still having 8 beads in the polar rows?

How many beads would be in a version of The Orb with two more rows of beads, but still having 8 beads in the polar rows?

I already know the answers, but it was fun for me to figure out, so I wanted to share the enjoyment here. :)

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 Post subject: Re: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 3:09 am 
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I did something like this, for fun though.

I think the beads for one more layer would be 22 around the equater and 19 for each row around that layer, 8-19-22-19-8: 76 beads

An Orb with 2 extra layers would be 8-18-21-21-18-8: 94 beads

I can't actually remember the numbers but I can hope these are right.

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 Post subject: Re: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 11:03 am 
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The number of beads in each row does need to be even in order to turn the halves.

Here's a hint:

If the top row has 8 beads, then that determines the space between all rows. So additional rows increase the size of the sphere. By how much?

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 Post subject: Re: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 1:54 pm 
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8-20-20-8

8-20-26-20-8

8-22-30-30-22-8

Carl

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 Post subject: Re: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 2:25 pm 
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wwwmwww wrote:
8-20-20-8

8-20-26-20-8

8-22-30-30-22-8

Carl


Yes!
Fun, right? :)

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 Post subject: Re: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 4:35 pm 
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JasonSmith wrote:
Yes!
Fun, right? :)
Yes... though I never changed the size of my sphere... just the size of my beads.

Assume a sphere or radius = 1.

Then the width of a bead becomes:

(2*pi*sin(90/n))/8

Where n is the total number of rows.

I then calculate the circumference of each row and divide by this width to get the number of beads. However I think I would say all these puzzles are fudged to a degree as this math doesn't produce whole numbers so you need to round the numbers to the nearest even whole number. Note if you round up on a given row that row should then be used to determine the bead width and you should re-check the others. I.e. this row is now the most tightly packed.

I think you could end up in a state where a row which isn't the most tightly pack is rounded down by more then 1 whole number and that amount of slop I think could cause problems. With the ones I just calculated that didn't happen. And I've now seen your Silver Orb thread so I certainly see where the question comes from. Very nice looking puzzle.

I'm guessing you are about to post the 6 row version too?

Carl

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 Post subject: Re: A puzzle design puzzle about The Orb
PostPosted: Fri Jul 26, 2013 5:58 pm 
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Yes, if the slop adds to one, then you can pull a bead across the dividing line and cheat.

On the silver orb, I actually fudged to:
8-22-26-22-8
to make the second rows tighter. According to my calculations they were going to be too loose for comfort.

For my fudge, those rows drop down the sphere by a smidge to make them longer, and come back up to the right place again when they reach the dividing line. Just enough for two more beads, but a surprisingly small cheat.

The slop on the next orb is very small in comparison, far from adding up to one. So no fudging will be needed I think.

Another interesting consideration is that the more rows you have, the more rows are adding slop, so the tighter things need to be.


Yes, I'm doing one more of these in between Radiolarians. :)

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