CTF | Collection of CTF solutions | Hacking library

Hi please use RecursiveTask,ForkJoinPool for parallelism. Below code is not final modify on your own way.

The first equal sign in phi(p * q) = phi(p) * phi(q) = (p - 1) * (q - 1) assumes that p and q are coprime (see [1]), while the second equal sign assumes that p and q are prime (see [2], k = 1). p = q violates the first condition, which is why this relation is not valid for p = q.

On the other hand, for k = 2, it follows from [2] phi(p * p) = p * (p - 1), i.e. the relation used in the CTF solution for p = q.

For RSA in practice, however, p != q is a prerequisite, see [3] and [4] (otherwise p and q could be determined quickly: p = q = sqrt(N)).

first of all, if you want to get data by id, use: Flag::find($id);

And in your code, you check not user data, you check data from the database with this: OBR{1FA528F41E8945C}

If you want to check data from user with your database data, your code should be something like that:

based on your desired output, seems like you just wanna do this :

Yes, that's what that means.

I was able to update six by doing a wget https://raw.githubusercontent.com/benjaminp/six/master/six.py in ~/.local/lib/python3.9/site-packages. This solved the problem.

The only way this happens in C is:

• x is an infinity (+∞ or −∞) (equivalently, HUGE_VAL or -HUGE_VAL in implementations that do not support infinities).
• x is a NaN and the C implementation uses non-IEEE-754 behavior for NaNs.
• x is adjacent to an infinity in the floating-point representation and a suitable directed rounding mode is used.
• x+1 and x*2 overflow and the resulting behavior not defined by the C standard happens to report the comparisons as true.
• x is uninitialized and hence indeterminate and takes on different values in the two appearances of x in x+1 == x such that the comparison is true and, in x*2 == x either similarly takes on different values or takes on the value zero.
• x is uninitialized and has automatic storage duration and its address is not taken, and hence the behavior is not defined by the C standard, and the resulting behavior happens to report the comparisons as true.

Proof:

Other than infinity, the statements are mathematically false in real numbers, and therefore this cannot arise from real-number behavior. So it must arise from some non-real-number behavior such as overflow, wrapping, rounding, indeterminate values (which may be different in each use of an object) or uninitialized values. Since * is constrained to have arithmetic operands, we only need to consider arithmetic types. (In C++, one could define a class to make the comparisons true.)

For signed integers, non-real-number behavior with fully defined values occurs for + and * only when there is overflow, so that is a possibility.

For unsigned integers, non-real-number behavior with fully defined values occurs for + and * only when there is wrapping. Then, with wrapping modulo M, we would have x+1 = x+kM for some integer k, so 1 = kM, which is not possible for any M used in wrapping.

For the _Bool type, exhaustive testing of the possible values eliminates them.

For floating-point numbers, non-real-number behavior with fully defined values occurs for + and * only with rounding, underflow, and overflow and with NaNs. NaNs never compare as equal by IEEE-754 rules, so they cannot satisfy this, except for the fact that the C standard does not require IEEE-754 conformance, so an implementation could choose to make the comparisons true.

x*2 will not underflow, since it increases the magnitude. x+1 can be made to underflow in a perverse floating-point format with smaller exponent range than precision, but this will not produce x+1 == x. x+1 == x can be satisfied by rounding for sufficiently large x, but then x*2 must produce a value other than x.

That leaves overflow. If x is the greatest representable finite number (and hence the greatest representable number less than infinity), and the rounding mode is downward (toward −∞) or toward zero, then x+1 will yield x and x*2 will yield x, so the comparisons yield true. Similarly, the greatest negative representable finite number will satisfy the comparisons with rounding upward (toward +∞) or toward zero.

Splitting the selection seems to work...

The download fails because libcurl tries to verify the webservers certificate, but can't.

I can reproduce this on my system: