FRaznki
A minor query.
You say you begin by touching two of the corners. You later say, "Either way -- after the first spin -- you will have three glasses facing the same direction".
Now here is the query:
Let us suppose that the four at the beginning are three one way up and one the other. Let us then suppose that you select the diagonal that contains one of each. Your instructions then were that you turn one of these over. Suppose that we call the three the same 'A' and the solitaire 'B'. Now, if you turn over the 'A' , making it a 'B' - and there is a 50/50 chance that you will - you will be left with two 'A's and two 'B's. Surely this is hardly consistent with, "Either way -- after the first spin -- you will have three glasses facing the same direction"?
P.S.
Franki (Sorry about the typo in the previous post)
You also wrote, "If they are different -- leave them be -- and spin again".
But again, surely this is exactly where you and Adrian faulted me - [to misquote Adrian] "You could keep getting that 'different' all day". (It is, after all, another 50/50 chance, just as mine was) I would suggest that this is the same 'lack of guarantee' that I was - rightly - corrected on.
Sorry if this sounds like being picky, but...
.
P.P.S.
And - as is only fair - here is mine to be perused and queried.
There are only three possible arrangements of these glasses
1)
A B
B A
2)
A A
B B
3)
A A
A B
As the solution allows either all upside down or all the right way up, it doesn't matter which is which as they are symmetrical (i.e. a reversal of all four makes no real difference) and as the lazy susan rotates there can be no preference as to which way around it is. So 1) 2) and 3) above cover all possible arrangements other than 'all four the same'.
Attempt one: a pair of opposites (i.e. diagonals or corners) are selected and, if the same, both are reversed; if different any one of the two is reversed.
Arrangement 1) becomes ?'all four the same'.
Arrangement 2) becomes an arrangement 3).
Arrangement 3) becomes - ?'all four the same'. - Or becomes an arrangement 1). - Or remains an arrangement 3).
However, we know in the case of the arrangement 3)'s what the majority is, it being the same as that which we increased by the change.
Attempt two: again we check a diagonal and reverse both if the same. If different we reverse that one known to be in the minority.
This makes arrangement 1) all four the same -
and makes arrangement 3) all four the same if the pair are different.
If they are the same it remains arrangement 3) with the minority known. Any arrangement 2) becomes arrangement 3) with the minority known. Thus we now only have Arrangement 3) with the minority known.
Attempt three: again we check a pair of opposites (i.e. diagonals) If different, we reverse the one known to be the minority, making all four the same. If the same, we reverse any one of them. This leaves only arrangement 2)
Attempt four: we check an adjacent pair (i.e. an edge or side) if they are the same we reverse both, making all four the same. If they are different we reverse both, leaving only arrangement 1)
Attempt five: We check an opposite pair (i.e. a diagonal) and reverse both, making all four the same.
Mungo
My solution may be incorrect.
I really cannot tell right now -- but your question has me stumped for the moment (I'm working in my head -- and I gotta get out the glasses.
I got a lot of stuff to do. I'll get back to you tomorrow.
That will give me a chance to look at your solution.
I'm now pretty much convinced that my original solution is wrong. I've revised it and will publish it after I study Mungo's.
(And I may steal a couple of ideas from him if I find that after more testing my new solution does not work. :wink: But I think it does!)
Frank
My second query was a nonsense. It would have been a valid comment had you intended to keep trying the same approach until you did get the pair you wanted, but on re-reading, that is not what you were saying at all. First query stands, second query retracted - and I promise to read more carefully in the future.
