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I'm an undercover :) computer science teacher


Aug
2
comment Find the greatest number such that when 76151 and 226 are divided by it, the remainders are all alike?
Or you both make a mistake by making a wrong hypothesis...
Aug
2
comment Find the greatest number such that when 76151 and 226 are divided by it, the remainders are all alike?
@SamiBenRomdhane If it's not explicitly written, you can't make such a strong hypothesis ! If the OP forgot about it, it's his responsibility.
Aug
2
answered Find the greatest number such that when 76151 and 226 are divided by it, the remainders are all alike?
Aug
1
comment Estimating the mean and variance of numbers assigned to each person in population of one billion
Estimators works on binary distribution : each people has a card with Yes/No or 0/1. But you can't use an estimator to guess what will happen for a card with "a number" (imagine a population with all cards with $0$ except one with $10^{10^{10}}$. Most estimators will be completely wrong.)
Jul
29
comment Problems On Many-one Reducible
@NicholasKorman Yes, don't prove it false if it's right !
Jul
29
answered Problems On Many-one Reducible
Jul
28
revised Breaking chocolate bars game
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Jul
26
revised Breaking chocolate bars game
added 1 character in body
Jul
26
comment Breaking chocolate bars game
@mathlove I edited my answer to answer your last comment.
Jul
26
revised Breaking chocolate bars game
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Jul
24
comment Breaking chocolate bars game
@mathlove and by "remove" I mean split of course :) or cut...
Jul
24
comment Breaking chocolate bars game
@mathlove : any good move for A is a move that only drop the value by 1 (because B will have to (at least) add 1 on his turn. This is always possible when the value is positive. From that and the table, you can deduce you always can do that by choosing any piece $(a,b)$ with $a\ge 2^{|b|}$ and remove a (almost) square $(2^{|b|-1},b)$. You're sure to win easily.
Jul
24
comment Breaking chocolate bars game
@mathlove. By induction. You just need to see that $v(a,b)=v(2a,2b)=v(2a,2b+1)=v(2a+1,2b)=v(2a+1,2b+1)$. And $v(n,1)=-v(1,n)=n-1$. The difficult step is to realise that cutting along a border is always bad. But you need to be familiar with the Conway theory (read "On numbers and games" by J.H.Conway).
Jul
23
revised Breaking chocolate bars game
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Jul
23
revised Breaking chocolate bars game
add values
Jul
23
comment Breaking chocolate bars game
@mathlove Yes, if there are several pieces, the sum (of the values of each piece) must be positive. This is not exactly as your conjecture. And remember that if $a_i<b_i$, the value is $1-\lfloor \frac{b_i}{2^{|a_i|-1}}\rfloor$.
Jul
22
revised Breaking chocolate bars game
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Jul
22
answered Breaking chocolate bars game
Jul
21
awarded  Yearling
Jul
9
answered An analytical proof that the sequence from the chord's iterational method is monotonic