... @polygenelubricants : Thanx a lot.. It seems, finally the code is working fine.. Is it safe to usen toHexString function now ? Or, there might be some loopholes with the approach ? – Vivek May 13 '10 at 11:28 ...

... I found that this solution also solved the "Cannot call methods on dialog prior to initialization; attempted to call method 'open'" error that occurs when a dialog is successfully opened, closed, and then the users tries to open the dialog a second time. Thanks @ZOD ...

...he computations of the subproblems overlap. Dynamic programming is typically implemented using tabulation, but can also be implemented using memoization. So as you can see, neither one is a "subset" of the other. A reasonable follow-up question is: What is the difference between tabulation (th...

I have implemented an Ajax request on my website, and I am calling the endpoint from a webpage. It always returns 200 OK , but jQuery executes the error event. I tried a lot of things, but I could not figure out the problem. I am adding my code below: ...

I am trying to compare strings in bash. I already found an answer on how to do it on stackoverflow . In script I am trying, I am using the code submitted by Adam in the mentioned question: ...

I've got a few methods that should call System.exit() on certain inputs. Unfortunately, testing these cases causes JUnit to terminate! Putting the method calls in a new Thread doesn't seem to help, since System.exit() terminates the JVM, not just the current thread. Are there any common patterns...

... @Grinn : user113716 did not list all. He forgot -0 . I do not mean a unary minus with a zero, but the result of that which is a seperate value. You can create it, for instance, by assigning -0 to a variable. – Marco de Wit ...

... If your list contains all non-negative ints, you should take a look at numpy.bincounts: http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html and then probably use np.argmax: a = np.array([1,2,3,1,2,1,1,1,3,2,2,1]) counts = np.bi...

...0 00000000 00000000 00000000 00000000 00000000 00000000 × 2^0 = 1 The smallest representable number greater than 1: 1.0000 00000000 00000000 00000000 00000000 00000000 00000001 × 2^0 = 1 + 2^-52 Therefore: epsilon = (1 + 2^-52) - 1 = 2^-52 Are there any numbers between 0 and epsilon? Plent...

... Technically, that's a bug in the services, isn't it? – holdenweb Aug 19 '15 at 9:52 6 ...