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I'm an undercover :) computer science teacher


Sep
16
answered Dichotomy Paradox for the Running Man.
Sep
14
answered Probability that 20 sided die beats 12 sided die with reroll
Aug
25
comment TOTAL is not Recursively Enumerable
@user137481 $n$ is the input of $R$, and if M halts, then it halts for some $n$, then $R$ will not be total.
Aug
25
answered TOTAL is not Recursively Enumerable
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
added 2 characters in body
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
added 1135 characters in body
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
added 123 characters in body
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 $1-\lfloor \frac{b_i}{2^{|a_i|-1}}\rfloor$.