harey

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About harey

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  • Birthday February 26

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  1. Roll Them Out

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

    Sure. Applying quantum physics.
  3. A loaded coin

    One possibility: However, in theory, it can go forever..  
  4. Roll Them Out

    Again, to obvious to be true. But what is wrong?
  5. Magical Tennis

    A typo again...I get used. Thinking it over, it is much more complicated.
  6. Magical Tennis

    So obvious that I fear I am missing something:
  7. Prove that you solved sudoku

    @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.
  8. The Coc'ks Clock

    Please delete. 10-11:
  9. Prove that you solved sudoku

    @phil1882: The solution does not require a computer. I do not see well encrypting decrypting by hand.  All you need: scissors, paper, pencil.
  10. Prove that you solved sudoku

    @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.
  11. Prove that you solved sudoku

    @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?
  12. Prove that you solved sudoku

    Well... suppose I want the proof now and I am not willing to wait unless someone else solves it. Hint: scissors might come handy.
  13. You solved a particularly hard sudoku and you are eager to prove it. Just you do not want to give me any hint, nor, God bless, reveal the solution. How will you proceed?
  14. Random bullets

    Bonanova wrote: @kman, I have argued this analysis in another venue, and my doubter objects to the proof. He accepts that 1/n is the probability that initially the first bullet is fastest, but not subsequently. He asks for proof that the first collision does not alter the statistical properties of the surviving bullets -- he says that 1/(n-2) cannot simply be assumed for the first remaining bullet to be the fastest of the remaining bullets. I pulled my hair out trying to disabuse him of that doubt. If I claim that I still have quite correct reactions after drinking a glass of wine, it is on you to prove me that I am wrong. If I claim that I still have quite correct reactions after drinking two bottles of whisky, it is on me to prove  that I am right. In this case, it is on him to prove that the statistical properties change. Anyway, both methods use the same proceeding: Step 1) build a table of speeds Step 2) split the table into 2 parts: 'to reject' and 'to reexamine' Step 3) reexamine the part 'to reexamine'. Where they differ is the definition of 'to reject'. One method rejects vn< ... <v3<v2<v1. One method rejects v1>vi (i=2,3, ... ,n). Interpretation of partial products: (number of cases we retained/rejected)/(total before retention/rejection). The partial products of the second method describe what happens collision after collision; the first method might be much less clear, depending on proceedings in step 3. Bonanova wrote: For example, in the case of n=4, the probability for at least one collision is not 3/4, it's 23/24. I agree that p(at least one collision) is not 3/4. I agree that p(at least one collision) is 23/24. Just I do not consider ALL cases with at least one collision: I deselect cases v1>vi (i=2,3, ... ,n). For n=4, 3/4 of cases will remain. It is NOT the full set of cases with at least one collision.