@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.
@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.
@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?
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.