Determine a recurrence relation for the number an of ternary strings (made up of 0 s, 1 s, and

Determine a recurrence relation for the number an of ternary...

Key Concepts

Recurrence Relations

A recurrence relation is an equation that expresses each term of a sequence in relation to previous terms. It is a fundamental tool in discrete mathematics and combinatorics for solving problems that involve counting arrangements, allowing one to build a sequence step by step from known initial values.

Linear Homogeneous Recurrence with Constant Coefficients

This type of recurrence relation is characterized by the fact that each term of the sequence is a linear combination of previous terms with constant coefficients, and there is no external term added. These recurrences are solvable by finding a characteristic equation, leading to a closed-form solution that describes the sequence explicitly.

Forbidden Substring Constraints in Combinatorics

In many combinatorial problems, especially those dealing with strings or sequences, certain consecutive patterns or substrings are prohibited. Handling these restrictions often requires modifying standard counting techniques, such as defining recursive relations that carefully avoid the forbidden patterns while counting valid arrangements.

Characteristic Equations

Characteristic equations are algebraic equations derived from linear homogeneous recurrences with constant coefficients. By solving these equations, one can determine the eigenvalues that are then used to express the general term of the sequence in a closed form, providing a direct formula for computing any term.

Initial Conditions

Initial conditions are the specific starting values that anchor a recurrence relation, allowing for the unique determination of the sequence. They are essential when solving recurrences because they provide the necessary information to compute all subsequent terms, ensuring that the final solution accurately reflects the constraints of the problem.

Transcript

00:01 So here we have where we're going to let a sub n represent the number of ternary strings.

00:16 Okay, and so ternary means 0, 1, or 2 rather than the binary, which is just a 0 or 1, that do not contain or are without consecutive zeros or consecutive ones.

00:30 And of course this is of length n.

00:39 Okay, so to encapsulate a sub -n, we'll do it in a few cases.

00:44 So first case is this string could start with a two.

00:49 Okay, then the remaining digits would be n -1 long, and there would be then a n -minus -one ways for that string to be without the double zeros or double -ones.

01:05 Alternatively, if it starts with a zero or a one, we could then consider the case where we have, say, the beginning has n minus 1 minus k, alternating zeros and ones, which are followed by a 2, and then is followed by then n minus k different strings.

01:45 Things.

01:48 Okay.

01:48 And so in this case, there are going to be a to the n minus n minus k ways to get a stream without the double zeros or double ones.

02:13 Okay, but we can reduce this a little bit.

02:15 This is just going to end up being a to the k.

02:19 But we need to consider, this would be the case where it's the zero, but the same thing for when it starts with a one, right? so, would also be a sub k.

02:28 So it's going to end up being two times a subk to represent where it starts as a 0 and alternates, so 0 -1, or the case where it starts with a 1 and alternates 1 -0 -1.

02:44 Okay, so our third case then is where we contain no 2s at all.

02:51 Okay, and in that case, it just alternates.

02:54 So it's either starting with a 0 and then going 1 -0 until n or or it goes one zero one zero all the way until n and there are exactly two ways for that occur which i just enumerated so a so a sub n is going to be equal to the sum of these three cases then right so that that a of n is equal to a of n minus one okay and to be clear this second case runs for k between zero and n minus 2 okay and why it goes up to n minus 2 right we we have put a 2 in here and we are also assuming that this first digit is either a 0 or 1 so that removes two elements from our length n uh so that is the maximum that it could be there okay so if that's the case then a sub n is equal to a and minus 1 plus 2 times a and minus 2 right so to do this for every value of k plus 2 times a and minus 3 all the way up to 2 times a not and then plus our last one here which is just a 2.

04:41 So we go to the next page because while what i wrote is our recurrence relation, it's not written in the most convenient form, right? so this requires us to sum up from every single previous case.

05:02 And if you are looking at large values of n, that could be pretty cumbersome.

05:14 I forgot some of the 2s in there.

05:16 A .0 plus 2.

05:17 Okay, so if we consider for n minus 1, if we just put n minus 1 in the place where you see ends everywhere, right? what we'd get is that a of n minus 1 is equal to a .n minus 2 plus 2 an minus 3.

05:42 So we're going to do a little bit of some sort of wizardry to get this simple, which i don't think.

05:51 Is really obvious.

05:54 So we're going to subtract the two equations from each other, right? so we'll, on the left -hand side, get a of n minus a -n -minus -a -n -minus 1.

06:05 But then on the right -hand side, by subtracting these two equations, we get something very simple...

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