Recent content by Dr-NiKoN

Learn ANSI C. Problem solved.

Yup, I was trying to use a latch as a flip-flop. All worked out now :)

I'm trying to create a circuit that has 4 outputs, and loops over all of them so that only 1 output is high at each pulse. Like this: 1 - 1 2 - 0 3 - 0 4 - 0 1 - 0 2 - 1 3 - 0 4 - 0 1 - 0 2 - 0 3 - 1 4 - 0 1 - 0 2 - 0 3 - 0 4 - 1 1 - 1 2 - 0 3 - 0 4 - 0 But...

1 out of 5 italians speak english. 1 out of 5 people in italy are tourists. 1 out 2 tourists speaks english. You meet a english-speaking person in italy, what is the probability that this person is italian. The way I see the "population": P(I) = \frac{2}{10} are italians who speaks english...

I have maybe explained this poorly, but let me try one last time. Say you have a set: A = {1 .. 10} The cryptographic function has A as it's domain and codomain. Now, to create these pairings we have to take elements 'out' of the set. So, if our first pair is: (1, 2), then we are left...

Thanks, I _finally_ get it :)

\binom{\binom{n}{r}}{\frac{r}{2}} ? :) (I know the above isn't correct btw) Is there any way to express this using nPr or nCr?

Ah, yes :) \frac{n!}{(n-r)!} Is to big. But if you take away a factor of r!, you take away all permutations of elements. So, I need to take away (\frac{r}{2})! to take away all permutations of "pairs". So the answer is: \frac{n!}{ (\frac{r}{2})! (n-r)!} ?

Hm, this isn't easy. A set of numbers: A = {1, 2, 3, 4, 5, 6, 7, 8, 9} Permutations of this set will look like: 1 2 3 4 5 6 7 8 9 2 1 3 4 5 6 7 8 9 2 1 4 3 5 6 7 8 9 8 9 1 2 3 4 5 6 7 Let's order these permutations like this: (x_1, x_2), (x_3, x_4), (x_5, x_6), x_7, x_8, x_9 We...

Ah, this may have caused some confusion. I haven't given you the full information. I'll give a third example, that should have all the information: Consider you have 9 numbers(1, 2, 3, 4, 5, 6, 7, 8, 9). You have some sort of crypto-system that can hold 4 pairs of numbers. The...

Let's make it simpler. |A| = 5 The total number of combinations of 2 groups of size 2 is: \frac{5!}{(5-2)!} This should basically be any permutation of 4 elements(2 ordered pairs) where order does matter. But, here the order of the ordered pairs themselves doesn't matter. So, this number...

Ok, let me give another example without "notation". I have 20 numbers. I want combinations of 14 numbers that are ordered as tuples. Some combinations 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 1 4 3 6 5 8 7 10 9 12 11 14 13 20 1 19 2 18 3 17 4 16 5 15 6 14 7 The combination: 3 4 1 2 5 6 7 8 9...

Ok, given a set: A How many distinct x tuples can be created using a subset of A? Example: |A| = 20 I want to know how many combinations of 7 tuples can be made from that set. Example: If A = {1 .. 20} three such combinations could be: {1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}...

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