ai dịch hộ cái Algebra Trignometry Caculus Differential Equation Linear Algebra Time can bend your knees; time can break your heart
Đại số Lượng giác Không biết ( môn này chắc là tương đương với Giải tích ở VN ) Phương trình vi phân Đại số tuyến tính.
không tớ hỏi dịch sang tiếng Pháp cơ bao giờ về VN chơi hả Hoàng, tao 15.12 này về rồi Time can bend your knees; time can break your heart
Không nói ai mà biết được Algèbre Trigonométrie Calcul Equation différentielle Algèbre linéaire Tao chắc hè năm sau mới về được, Noel này ở trường không có ma nào chắc buồn vãi
Vui quá đi mất!!Bác username của chúng ta, sau 1 thời gian vắng bóng(Chắc là đang cưa cây) ở đâu, đã <<Tái xuất giang hồ>>Vui quá đi mất. Bác phải hay lên mạng đi nhé.Có 1 bài thơ bên trang thơ toán, em dịch mãi, mà thấy nó sao sao ấy.thơ chẳng ra thơ, văn xuôi chẳng ra văn xuôi. Bác dịch hộ cái nhé!!! ------------------------------------ Có khi mưa ngoài trời là giọt nước mắt em.
Scooping the Loop Snooper an elementary proof of the undecidability of the halting problem No program can say what another will do. Now, I won't just assert that, I'll prove it to you: I will prove that although you might work til you drop, you can't predict whether a program will stop. Imagine we have a procedure called P that will snoop in the source code of programs to see there aren't infinite loops that go round and around; and P prints the word "Fine!" if no looping is found. You feed in your code, and the input it needs, and then P takes them both and it studies and reads and computes whether things will all end as the should (as opposed to going loopy the way that they could). Well, the truth is that P cannot possibly be, because if you wrote it and gave it to me, I could use it to set up a logical bind that would shatter your reason and scramble your mind. Here's the trick I would use - and it's simple to do. I'd define a procedure - we'll name the thing Q - that would take and program and call P (of course!) to tell if it looped, by reading the source; And if so, Q would simply print "Loop!" and then stop; but if no, Q would go right back to the top, and start off again, looping endlessly back, til the universe dies and is frozen and black. And this program called Q wouldn't stay on the shelf; I would run it, and (fiendishly) feed it itself. What behaviour results when I do this with Q? When it reads its own source, just what will it do? If P warns of loops, Q will print "Loop!" and quit; yet P is supposed to speak truly of it. So if Q's going to quit, then P should say, "Fine!" - which will make Q go back to its very first line! No matter what P would have done, Q will scoop it: Q uses P's output to make P look stupid. If P gets things right then it lies in its tooth; and if it speaks falsely, it's telling the truth! I've created a paradox, neat as can be - and simply by using your putative P. When you assumed P you stepped into a snare; Your assumptions have led you right into my lair. So, how to escape from this logical mess? I don't have to tell you; I'm sure you can guess. By reductio, there cannot possibly be a procedure that acts like the mythical P. You can never discover mechanical means for predicting the acts of computing machines. It's something that cannot be done. So we users must find our own bugs; our computers are losers! by Geoffrey K. Pullum Bác username ơi , dịch hộ bọn em với!!! ------------------------------------ Có khi mưa ngoài trời là giọt nước mắt em.
Các bác dịch sang tiếng Việt hộ em cái nè : Mathematical Excalibur. Thanks trước nhé ! Hoàng Kiên Trung
Mathematical Excalibur là cái báo toán của bọn Hồng Kông. Excalibur hình như không có nghĩa, đó chỉ là tên thanh kiếm của vua Athur. Dịch ra tiếng Việt có thể là "thần kiếm Toán học" chăng ?
