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harey

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Posts posted by harey

  1. When midnight strikes

    in New Logic/Math Puzzles

    Posted

    6 hours ago, bonanova said:

    The key question is this: can a coin that is kept at a certain event ever be discarded at a later event?

    If a coin can be discarded, it does not mean it will be discarded. I would reformulate it: The key question is this: will all coins that are kept at a certain event ever be discarded at a later event?

    BTW, we can establish a bijection between Al's and Bert's coins. The coins bear green numbers. After each step, Al renumbers them and assigns them blue numbers 1, 3, 5, ... His blue numbers will match the (green) numbers in Bert's box. For any number of steps. I think it is legal to assume it is true even for N-> inf.

    Another way to prove that Bert's box will not be empty: graphical presentation. The number of coins in his box is a straight line (at 45 degrees). How is that it suddenly drops to 0?

    And maybe a corollary: Bert never discards more coins that he receives. How is that when he has, let's say 8 coins, he can have less in a later stage?

    If we reason with {coins} and {events}, don't you see a 1-1 relation?

  3. When midnight strikes

    in New Logic/Math Puzzles

    Posted · Edited by harey
    corrected a typo

    12 hours ago, bonanova said:

     

      Hide contents

    Al:
    On step 1 Al discards coin 1. On step 2 A discards coin 2. ... On step N Al discards coin N. As N -> inf, every coin is discarded.
    If there were a coin in Al's box at midnight it would have a number. What number would that be?

    Lets start with this one.

    If every coin is discarded, it comes to claim that inf - inf =0

    Remember the hotel with an infinite number of rooms all occupied? An infinity of guests arrive, you move the person from room 1 to room 2, from room 2 to room 4, from room 3 to room 6... getting an infinity of empty rooms.

    Now, an infinity of guests leave. Is the hotel empty? And why?

     

  5. When midnight strikes

    in New Logic/Math Puzzles

    Posted · Edited by harey
    For the pleasure to put it in bold.

    After N steps, they will have received 2*N coins and withdrawn N coins. At that moment, there will be 2*N-N coins in the box.

    If the box is empty at midnight, this implies:

    limit(2*N-N)(for N->inf) = 0

    At least a little bit surprising.

    @ThunderCloud

    I have some troubles to refute your argument. If you remove an infinity of finite numbers from infinity of finite numbers, it does not imply no finite number remain. (Not sure I am convincing and clear enough.)

    Counterargument: Al removed all coins 1 - N, coins > N remain. If N -> inf, numbering looses it's sense, but he did not remove all coins.

    As for Charlie, I am ruminating, too. The first idea: every number will remain with p=1/2. Wrong, 1 will be more likely removed than 99.

    2nd idea:
    1st step, 2 coins: p(removing 1)=1/2
    2nd step, 3 coins: p(removing 1)=p(1 was not removed in the first step) * 1/3 = 1/2 * 1/3 = 1/6, p(1 remaining after 2nd step)=1 - 1/2 - 1/6 = 1/3
    3rd step, 4 coins:  p(removing 1)=p(1 not yet removed) * 1/4 = 1/3 * 1/4 = 1/12, p(1 remaining after 3rd step)=1 - 1/2 - 1/6 - 1/12 =

    I will not venture further, but this will not converge to 0. (Compare to 1 - 1/2 - 1/4 - 1/8...)

     

  7. Soldiers in a field

    in New Logic/Math Puzzles

    Posted

    On 1/5/2018 at 11:50 PM, CaptainEd said:

    Maybe this is clearer and more accurate than my previous try.

    Point one: 

     

      Hide contents

     

    The shortest distance is between one pair, A and B. They are a mutual pair, as the shortest distance from each goes to the other.

     

    Point two:

     

      Hide contents

     

    Consider the node whose smallest distance is greater than anyone else’s. Call that soldier A. A’s smallest distance is to a node, call it B.

    Which soldier is pointing at A? Call it Z. Because AB is A’s least distance, AB < Ax for all x in S - {A,B}

    Because ZA is Z’s least distance, ZA < Zx for all x in S - {A,Z}

    Because  A’s smallest distance is greater than anyone else’s, ZA < AB. But that contradicts the fact that AB < AZ. 

    However, it is possible that Z = B, as AB <= AZ and ZA <= AB. 

    Result: the node with the highest lowest distance is in a mutual pair OR A is not watched.

    What about the next highest least distance? Call it node C, its successor D and predecessor Y.

    CD < Cx for all x in S - {C,D}

    YC < Yx for all x in S - {C,Y}

    Because C’s smallest distance is greater than anyone other than A or B, YC < CD, but that contradicts that CD < YC.

    This can only happen if Y = D. So CD are a mutual pair OR C is not watched.

    The same argument proceeds one pair at a time, until an unwatched soldier is found, OR one soldier remains. That soldier is not watched.

     

     

     

    Cannot you shorten it?

    - if no one else watches A nor B, remove A and B and start over (this happens if A and B are very near and all other are far away enough)
    - if someone else watches A and/or B, at least one is not being watched (evident; if proof needed, start with 3 and continue with recursion)

  11. In Soviet Russia

    in Jokes

    Posted

    Two judges meet in Moscow, one is laughing, laughing, laughing... The second asks why.
    1st judge: I heard an awfully funny joke.
    2nd judge: Tell me!
    1st judge: I cannot. I sentenced the man who told it to 5 years.
     

  17. Roll Them Out

    in New Logic/Math Puzzles

    Posted · Edited by harey

    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?

  21. Prove that you solved sudoku

    in New Logic/Math Puzzles

    Posted · Edited by harey

    @Jasen I think you got it, but SO confusing.

    1) Insert the solution on small pieces of paper into the original grid, numbers down. (You better use a non-transparent paper.)
    2) For each row/column/square I ask, collect your papers and show them to me in ascending order: (With the original numbers, 1-9 will be used exactly once.)
    3) Put your pieces of paper back.

  24. Prove that you solved sudoku

    in New Logic/Math Puzzles

    Posted · Edited by harey

    @CaptainEd  Good work.... But why so complicated? No need for a third person.

    <spoiler>

    1) The solver secretly creates a matrix with the complete solution. Known numbers are preceded by a star.

    2) The challenger writes a program that takes as input this matrix. The program displays numbers preceded by a star and blanks for numbers not preceded by a star and makes the necessary checks. (If the solver fears the program would display everything, it can be tested on another grid.)

    3) The solver wipes the harddisk (optional).

    </spoiler>

    Almost there. Just the computerized solution does not have the beauty of the manual solution - as I said, it is an intermediate step. How can it be done without a computer? All you need: scissors, paper, pencil.

  25. Prove that you solved sudoku

    in New Logic/Math Puzzles

    Posted · Edited by harey

    @Jason

    Not bad, we might come to the solution this way in 2-3 steps. (Just YOU solved the sudoku, so YOU enter the answer).

    Hint: Be a little more specific about the program. How should I write the program that you cannot fool it by entering  shifted 1 2 3 4 5 6 7 8 9 for every line/column?

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