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Old 2021-04-03, 02:21   #1
Xyzzy
 
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Default April 2021

https://www.research.ibm.com/haifa/p...April2021.html
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Old 2021-04-03, 05:17   #2
Dieter
 
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Originally Posted by Xyzzy View Post
Your goal: Find an n such that there is a set of unwinnable numbers for seven steps (i.e., the set is of size n-7). In your answer, supply the number n and the elements of the unwinnable set.

If my understanding is correct, "seven steps" has to be replaced by "seven spins".
What do you mean?
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Old 2021-04-04, 08:26   #3
tgan
 
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Quote:
Originally Posted by Dieter View Post
Your goal: Find an n such that there is a set of unwinnable numbers for seven steps (i.e., the set is of size n-7). In your answer, supply the number n and the elements of the unwinnable set.

If my understanding is correct, "seven steps" has to be replaced by "seven spins".
What do you mean?
I also think you are correct
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Old 2021-04-04, 12:33   #4
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On each spin, the wheel moves forward q steps in a clockwise direction
. . .
So, if the wheel is right before 1 and performs 3 steps, it ends up on 3.
Wouldn't it end up on 7 ("before the 1" defined as traveling over it during the spin)? If it started betwen the 8 and 1 and the wheel was rotated clockwise three steps, the arrow would end up on the 6.
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Old 2021-04-04, 13:32   #5
Dieter
 
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Originally Posted by EdH View Post
Wouldn't it end up on 7 ("before the 1" defined as traveling over it during the spin)? If it started betwen the 8 and 1 and the wheel was rotated clockwise three steps, the arrow would end up on the 6.
Three remarks:

- See #2: seven spins, not seven steps
- The wheel turns anti-clockwise. Or the arrow turns clockwise.
- What does mean: "there is a set of unwinnable numbers"? "There is at least one set" or "there is exactly one set"?
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Old 2021-04-04, 13:39   #6
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Quote:
Originally Posted by EdH View Post
Wouldn't it end up on 7 ("before the 1" defined as traveling over it during the spin)? If it started betwen the 8 and 1 and the wheel was rotated clockwise three steps, the arrow would end up on the 6.
I noticed that also. As to "steps," the problem says (my emphasis)
Quote:
On each spin, the wheel moves forward q steps in a clockwise direction and the number / prize reached is eliminated from the wheel (i.e., the player does not get it).

Each step moves the wheel a little less than one number forward (so the wheel comes to rest on a number and not between two). So, if the wheel is right before 1 and performs 3 steps, it ends up on 3.
I am unable to reconcile this usage of "steps" with the usage WRT "unwinnable sets."

I also note
Quote:
Each step moves the wheel a little less than one number forward (so the wheel comes to rest on a number and not between two).
The phrase "a little less" is ambiguous. I'm not sure whether the step size remains in constant proportion to the distance from one number to the next. If the choice of q (number of steps in a spin) is not bounded above, it would seem that the only way to insure you never land between numbers is to make the step size an irrational multiple of pi radians...
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Old 2021-04-04, 13:41   #7
Dieter
 
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Quote:
Originally Posted by EdH View Post
Wouldn't it end up on 7 ("before the 1" defined as traveling over it during the spin)? If it started betwen the 8 and 1 and the wheel was rotated clockwise three steps, the arrow would end up on the 6.
If I replace "clockwise" by "anti-clockwise" it is possible to confirm the three unwinnable sets of the example.
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Old 2021-04-04, 15:28   #8
uau
 
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Originally Posted by Dr Sardonicus View Post
I noticed that also.
From the example it seems clear that the intended meaning is that the arrow moves clockwise around the wheel.
Quote:
I am unable to reconcile this usage of "steps" with the usage WRT "unwinnable sets."
Those refer to different things with "step".
Quote:
The phrase "a little less" is ambiguous.
I'm pretty sure the intended meaning is "infinitesimally small". That is, the phrasing is only intended to indicate which of the two sectors is the one removed, not meant to imply that any finite multiple of it would ever reach the size of a full sector.
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Old 2021-04-04, 20:11   #9
Dieter
 
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I'm pretty sure the intended meaning is "infinitesimally small". That is, the phrasing is only intended to indicate which of the two sectors is the one removed, not meant to imply that any finite multiple of it would ever reach the size of a full sector.[/QUOTE]

That's the only interpretation making sense. Otherwise the puzzlemaster had to concretize the "a little less than one number" - 0,9999 or so. q will become very big.
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Old 2021-04-05, 13:22   #10
Dr Sardonicus
 
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Quote:
Originally Posted by uau View Post
From the example it seems clear that the intended meaning is that the arrow moves clockwise around the wheel.
But it says (my emphasis)
Quote:
A player faced with the wheel chooses some number q and starts spinning the wheel k times. On each spin, the wheel moves forward q steps in a clockwise direction
Perhaps they need a proofreader...

Quote:
I'm pretty sure the intended meaning is "infinitesimally small". That is, the phrasing is only intended to indicate which of the two sectors is the one removed, not meant to imply that any finite multiple of it would ever reach the size of a full sector.
I was merely quibbling over the length of a "step." The number of "steps" any actual mechanical "wheel of fortune" could take in one spin would be fairly limited. The condition that this is fixed (q steps per spin) is unusual, but that's what the problem says. This requirement reminds me of the "problem of Josephus."
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Old 2021-04-05, 15:21   #11
Kebbaj
 
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Quote:
Originally Posted by Dr Sardonicus View Post
But it says (my emphasis)Perhaps they need a proofreader...

I was merely quibbling over the length of a "step." The number of "steps" any actual mechanical "wheel of fortune" could take in one spin would be fairly limited. The condition that this is fixed (q steps per spin) is unusual, but that's what the problem says. This requirement reminds me of the "problem of Josephus."
Indeed it is Josephus Problem.
In example 1 of the wheel with q = 5:
1 round removes the 5
2nd round remove the 3
3rd round removes the 8 and remains 1,2,3,6,7
The next round is the 7 which will jump.
....
But what I don't understand is example 2.
"a set of k numbers unwinnable"?

Last fiddled with by Kebbaj on 2021-04-05 at 15:26
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