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# Roll Them Out

## Question

In a long narrow chute there are 8 balls: 4 black ones on the left and 4 white ones - slightly larger - on the right.  In the middle of the chute there is a small niche that can hold 1 ball of either color.  The chute's right end has an opening large enough for a black but not a white ball.  The chute is long enough to hold 11 balls.  Explain how you will roll out all of the black balls.  What is the smallest the chute can be?

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I hope someone solves it, I'm at an end, with it.

Lets try it together.

Hidden Content

Again, to obvious to be true.

But what is wrong?

Since the niche needn't be in the center of the chute, 7 + niche + 4 = 12 works.

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Just to clarify, does this ask whether you could extract the black balls if the chute could hold fewer than 11 balls and if so what is the smallest such number?

Trivial for length of 15. Can't yet do it for 13 or 11.

Edited by bonanova
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If I understand the picture properly, the chute looks like this, with the hole on the Right end:

BBBB /niche\ WWWW

and I think balls can the niche freely even when the niche is occupied

BBB /B\ WWWW
BBBWWWW/B\
BBBWWWW/\
BB /B\ WWWW
BBWWWW /B\
BBWWWW /\
B /B\ WWWW
BWWWW /B\
BWWWW /\
/B\WWWW
WWWW/B\
WWWW

So the chute has to be long enough to hold 4 balls to right of Niche and 7 balls to left.  That's enough to hold 11, but the OP says the Niche is to be in "the middle", which implies the same size on either side of the Niche. I guess I need 15, made up of 7 to the left, then the niche, then 7 to the right (of which I only need 4)

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You can also do it by extending the niche to hold three balls. Same as adding 2 the length of the left side, as CaptainEd notes.

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Just to clarify, does this ask whether you could extract the black balls if the chute could hold fewer than 11 balls and if so what is the smallest such number?

Hidden Content

yes, it does.

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Durn! I still can't do it, even with 6 on a side plus a one-ball niche. I can easily get to BWWW /_\ BBBW

but I don't see how to get that rightmost W out of the B's way, without putting 7 on the left side and one in the niche. I'm assuming that inserting/removing a ball into/out of the niche requires that there be no ball at the niche's position in the main chute.

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Durn! I still can't do it, even with 6 on a side plus a one-ball niche. I can easily get to BWWW /_\ BBBW

but I don't see how to get that rightmost W out of the B's way, without putting 7 on the left side and one in the niche. I'm assuming that inserting/removing a ball into/out of the niche requires that there be no ball at the niche's position in the main chute.

What size of chute are you using?

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Sorry, with a chute of length 13: 6 spaces on the left, one space in the center for the 1-ball niche, and 6 spaces on the right. As Bonanova said earlier, it's trivial with 15-length chute. If the niche can slide around, we can do it with fewer spaces. You said that the niche is "in the middle". I have assumed that meant "equidistant from both ends".

BMAD, are we allowed to assume an asymmetrical chute? (ie, with the niche in a position specified by us?)

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I am unsure of what you mean exactly. But the niche must remain stationary and equidistant between the balls, this is not to say that the chute is of equal length on both sides.

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Denote an empty section of tube by "_" underscore.
Denote the niche by a " " blank space if it's empty and by a lower case b (black)or w (white) if it contains a ball.
There must be 7 spaces to its left and 4 spaces to its right.

BBBB___ WWWW

BBB____bWWWW  4 spaces needed to right of niche
BBBWWWWb____  7 spaces needed to left of niche
BBBWWWW ____  B    1 black ball extracted

BB_____bWWWW
BB_WWWWb____
BB_WWWW ____  BB   2 black balls extracted

B______bWWWW
B__WWWWb____
B__WWWW ____  BBB  3 black balls extracted

_______bWWWW
___WWWWb____
___WWWW ____  BBBB 4 black balls extracted

If the niche has the width of one ball, the length of the tube (including niche) must be 12 or greater.

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I am unsure of what you mean exactly. But the niche must remain stationary and equidistant between the balls, this is not to say that the chute is of equal length on both sides.

I think "remain equidistant between the balls" is difficult to understand. Literally it means the distance between the niche and the balls is the same. The same as what? And if distances remain the same, how can anything move? I think what is meant is that initially the niche lies between the two sets of balls. I also think the only way for a black ball to change from being to the left of a white ball to being on its right is to lie in the niche while the white  ball moves over it to the left. I also think the niche must contribute to the length of the tube by one unit (ball width.)  And the tube and niche remain stationary while (only) balls move. Are these assumptions in line with the OP?

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I think what is meant is that initially the niche lies between the two sets of balls.

Yes, only has to be at the start.

I also think the only way for a black ball to change from being to the left of a white ball to being on its right is to lie in the niche while the white  ball moves over it to the left.

Not true, the white ball could also use the niche And the black one go to the right.

I also think the niche must contribute to the length of the tube by one unit (ball width.)

Yes but not left or right.

And the tube and niche remain stationary

Yes.

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BMAD this is a great puzzle.
I hope someone solves it, I'm at an end, with it.

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I hope someone solves it, I'm at an end, with it.

Lets try it together.

Number the balls left to right B4 B3 B2 B1  W4 W3 W2 W1.
- There is no reason to alternate the position of the black balls, the first one that goes out is B1. -> All other Bs are left from B1.
- As long as a white is right to B1, B1 cannot go out.

Just before B1 goes out, we must get one of these situations.:
a) B1 in the niche, all other 7 on the left in any order
b) W1 in the niche, again all other 7 on the left, B1 in the leading position

=> 7 left, niche in the middle, 7 right

I do not claim a) and/or b) can be reached (though they nost probably are). I claim these are necessary conditions so there is no solution for a chute of less than 15.

Again, to obvious to be true.

But what is wrong?

Edited by harey
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I hope someone solves it, I'm at an end, with it.

Lets try it together.

Hidden Content

Again, to obvious to be true.

But what is wrong?

It can be done in a smaller chute.

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I hope someone solves it, I'm at an end, with it.

Lets try it together.

Hidden Content

Again, to obvious to be true.

But what is wrong?

It can be done in a smaller chute.

Sure. Applying quantum physics.

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I hope someone solves it, I'm at an end, with it.

Lets try it together.

Hidden Content

Again, to obvious to be true.

But what is wrong?

It can be done in a smaller chute.

Sure. Applying quantum physics.

I am not sure what you mean, but I am saying in terms of the amount of balls that can be fitted into the chute, this chute can be smaller.

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I googled some definitions of "middle" and "center". If this is THE solution, than the dictionaries are pretty wrong.

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@harey, we have language barriers here, and I think "in the middle" in the OP is safe to take as meaning "between the groups of white and black balls."

@BMAD,  did you mark my previous post as the correct answer?
It only found a solution for length 12, and there was an earlier indication that a shorter chute was possible.

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@harey, we have language barriers here, and I think "in the middle" in the OP is safe to take as meaning "between the groups of white and black balls."

@BMAD,  did you mark my previous post as the correct answer?
It only found a solution for length 12, and there was an earlier indication that a shorter chute was possible.

hmm, I see where I remarked that  a chute shorter than 15 were possible but I am unsure of a solution of a size less than 12.

Edited by BMAD
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I guess I misunderstood then. It sounds like the OP asks how to do it with 11, and then ask what the minimum length would be (even shorter.)
What was confusing was trying to do it with 11.

Good puzzle.

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Oh, i think the issue is the assumption (or a poor translation) behind what i meant when i said it is long enough to hold 11 balls. Clearly a chute that can hold 12 can also hold 11. The task was to find the smallest such chute that can hold 11 balls and allow the black balls to exit.

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