# Tag Info

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### How long does the Collatz recursion run?

This is Collatz conjecture - still open problem. Conjecture is about proof that this sequence stops for any input, since this is unresolved, we do not know how to solve this runtime recurrence ...
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### Efficient algorithm to compute the $n$th Fibonacci number

You can use matrix powering and the identity $$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix}.$$ In your model of ...

### Solving a recurrence relation with √n as parameter

In your comment you mentioned that you tried substitution but got stuck. Here's a derivation that works. The motivation is that we'd like to get rid of the $\sqrt{n}$ multiplier on the right hand side,...
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### How to solve a recurrence relation with a sum?

Here are several ways to solve your recurrence relation. Guessing Anyone with enough experience in computer science might recognize your recurrence as the one satisfied by $T(n) = 2^n$. Given this ...

### How long does the Collatz recursion run?

You translated the code correctly. There are many methods for solving recurrences. However, it is currently unknown if collatz even halts for all ...
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### Solving or approximating recurrence relations for sequences of numbers

Summations Often one encounters a recurrence of the form $$T(n) = T(n-1) + f(n),$$ where $f(n)$ is monotone. In this case, we can expand $$T(n) = T(c) + \sum_{m=c+1}^n f(m),$$ and so given a ...
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### Big-O proof for a recurrence relation?

As you pointed out, the reason for splitting the term into two pieces is to be able to cancel the $an$ term. If we go directly from $(8/9)cn^2 + an \leq cn^2 + an$, then we get stuck as we cannot do ...

### Why do these recurrences determine the number of ways of tiling a 3xN rectangle with 2x1 dominoes?

The picture should say more than words.

There may be times when you come across a strange recurrence like this: $$T(n) = \begin{cases} c & n < 7\\ 2T\left(\frac{n}{5}\right) + 4T\left(\frac{n}{7}\right) + cn & n\geq 7 \end{... 9 votes Accepted ### Solving recurrence relation with two recursive calls As you mention, the Akra–Bazzi theorem shows that the solution to the recurrence T(n,p) is O(n\log n) for all p \in (0,1). However, this does not reveal the nature of the dependence on p.... 9 votes ### How to solve T(n)=2T(√n)+log n with the master theorem? Let us actually use the master theorem. Define S(n) = T(e^n) for all n. Then$$S(n) = T(e^n) = 2T(\sqrt{e^n}) + \log(e^n) = 2T(e^{n/2}) + n = 2S(n/2) + n$$Now we can apply the second case of ... 8 votes Accepted ### Understanding an algorithm for the gas station problem The problem is in the condition for the first argument to min() in Equation (4) on p. 7. It's currently ... 8 votes ### Solving or approximating recurrence relations for sequences of numbers After checking this post again, I'm surprised this isn't on here yet. Domain Transformation / Change of Variables When dealing with recurrences it's sometimes useful to be able to change your ... 8 votes Accepted ### What is the Big O of T(n)? Yes, all functions f(n) satisfy f(n) \in O(f(n)). The definitions are meaningful even if f(n) isn't the running time of any function. Indeed, this notation comes from number theory, where f(n) ... 8 votes Accepted ### Why do these recurrences determine the number of ways of tiling a 3xN rectangle with 2x1 dominoes? Suppose that the rectangle to be tiled has 3 rows and n columns. □□□...□ □□□...□ □□□...□ 123...n Consider a tiling of this rectangle using 2\times1 dominos. ... 8 votes Accepted ### Solving or estimating the recurrence T(n) = x + T(n-\log_2 n) Assuming an appropriate base case, it is easy to see that T(n) \geq (n/\log_2 n) \cdot x, since at each step we subtract at most \log_2 n, and thus it takes t least n/\log_2 n steps to reach ... 8 votes Accepted ### Recurrence relation of the coin change problem For that recurrence to make sense, V can only be the array that contains the coin values; that is, V=\{C_1, C_2, ..., C_m\}. Whenever confronted with a new dynamic programming problem, you should ... 8 votes Accepted ### Solving recurrence relation with square root The answer cannot be O(\log\log n). Already without applying any recursion we have the inequality T(n) = T(\sqrt{n}) + n \ge n. So the complexity cannot be smaller than O(n). But now to your ... 7 votes Accepted ### Solving the recurrence T(n) = 3T(n-2) with iterative method Note that 3^{n/2-1/2} = \frac{1}{\sqrt{3}} 3^{n/2} = \frac 1{\sqrt{3}} \sqrt{3^n}. So the -\frac 12 indeed becomes a constant factor that is absorbed by the O(), but \frac n2 in the exponent ... 7 votes ### Solving or approximating recurrence relations for sequences of numbers Case 2 of the master theorem, as usually stated, handles only recurrences of the form T(n) = aT(n/b) + f(n) in which f(n) = \Theta(n^{\log_ab}\log^k n) for k \geq 0. The following theorem, taken ... 7 votes ### Changing variables in recurrence relations What S(m) = T(2^m) means is that S and T are two different functions which produce the same result while taking inputs as m and 2^m respectively. Function S can be considered as an ... 7 votes Accepted ### Solution of complex recurrence relation Everything becomes much simpler if when n is odd we stop the sum at \lfloor n/2 \rfloor, and when k is even we discount the term corresponding to k/2 by a half. In that case you can write$$ ...
No it's not always the case that $a=b$, since you might not necessarily use every sub-problem. Consider for example, the binary search algorithm. In the algorithm, you have a sorted array that you ...