Algorithms Q&A - Recent questions in Divide & Conquer

Solve or estimate this recurrence relation T(n) = 2 T(n/3) + T(n/2) + n

Tue, 01 Jun 2021 21:39:32 +0000

N-th power of complex number z = x + iy in O(log n) time

Sat, 13 Jun 2020 15:43:36 +0000

You are given a complex number z = x + iy. The n-th power of z is given (x + iy)^n. Describe a O(log n) time algorithm to do the same. Recall that that i^2 = -1, and that z^2 = x^2 – y^2 + 2ixy.

Solve this recurrence relation: T(n) = 3 T(n/4) + O(n^0.75)

Sat, 13 Jun 2020 15:40:39 +0000

Solve this recurrence relation: T(n) = T(n-1) + T(n/2) + 1

Sat, 13 Jun 2020 15:39:20 +0000

Solving this recurrence relation: T(n) = 2 T(n/2) + f(n)

Tue, 24 Sep 2019 14:30:02 +0000

How do we solve this recurrence relation generically?

For example, we can say something like: if f(n) = omega(n^1+epsilon), then T(n) = f(n). etc?

Solve this recurrence relation: T(n) = T(n/3) + T(n/5) + T(n/6) + n

Thu, 19 Sep 2019 14:22:51 +0000

Solve this recurrence relation: T(n) = T(n/3) + T(n/5) + T(n/6) + n

The Subarray with the Minimum Peak

Sat, 20 Oct 2018 18:37:58 +0000

Let X[1:n]: be an array of real numbers, and let L be a given fixed positive integer. An L-subarray is any sequence of consecutive elements of the array X, such as X[k: k + L − 1], and is identified simply by the index k of the first element of the subarray. The array has (n − L + 1) L-subarrays. The peak of an L-subarray is the maximum value in that subarray.

Give a divide-and-conquer algorithm that takes as input an array X[1:n] and a positive integer L ≤ n, and returns the L-subarray with the smallest peak. Analyze the time complexity of your algorithm.

[We observe that a simple brute force O(NL) time algorithm is straightforward, simply evaluate each L-subarray. Our interest is in something like O(N log N) or O(N log L) time algorithm.]

Find the smallest domain value where the function becomes negative.

Thu, 27 Sep 2018 01:59:10 +0000

We consider a monotonically decreasing function f : N → Z (monotonically decreasing means that f(i) > f(i + 1) for all i values in the domain). We can evaluate f at any i in constant time, we want to find n = min{i ∈ N such that f(i) ≤ 0}.

Give an O(log n) time to find such an n.

Three Set Sum: Given an integer k and 3 sets A, B and C, find a, b, c such that a + b + c = k

Mon, 24 Jul 2017 05:22:48 +0000

You are given an integer k and 3 sets A, B and C, each containing n numbers. Give an efficient algorithm to determine if there exist 3 integers a, b, c (in sets A, B, C respectively), such that a + b + c = k. The time complexity of your algorithm should be O(n²) or better.

Solve the recurrence relation: T(n)=T(n/2)+T(n/3)+T(n/4) + n

Tue, 14 Feb 2017 22:56:25 +0000

Solve the recurrence relation: T(n)=T(n/2)+T(n/3)+T(n/4) + n

In Quickselect, T(n)<=T(n/5)+T(7n/10)+cn, prove that T(n) = O(n)

Wed, 08 Feb 2017 01:01:38 +0000

Solve the recurrence relation: T(n) = 3 T(n/2) + n^1.5 log n

Thu, 15 Dec 2016 01:24:17 +0000

Solve the recurrence relation: T(n) = 3 T(n/2) + n^1.5 log n

Solve the recurrence Relation: T(n) = T(n/3) + T(2n/3) + n

Wed, 14 Dec 2016 20:25:59 +0000

Solve this recurrence relation: T(n) = T(n/3) + T(2n/3) + n

Rectangles overlap

Mon, 05 Dec 2016 02:21:50 +0000

“Rectangles” A rectangle can be specified in a 2-dimensional plane using the top left (north west) point and the bottom right (south east) point. Given n rectangles (using 2 points each), give an O(n log n) algorithm that tells if any two rectangles from the list overlap. Two rectangles are said to overlap, if there is a common point in both of them. [If a rectangle is entirely contained in another, they are still said to “overlap”, even though their lines don’t cross.]

How to solve this problem?

MVCS using Divide and Conquer

Sat, 03 Dec 2016 13:38:03 +0000

Recall the Maximum Value Contiguous Subsequence problem that we studied in class: “Given an Array A(1..n) of positive and negative values, find a contiguous subsequence A(i..j) such that the sum of the terms in the subarray is maximized”. We studied a linear time O(n) dynamic programming solution that finds the best subsequence.

Now, you need to design a divide and conquer algorithm for the same problem that is also optimal (so it needs to find the maximum value, not something approximate). It is acceptable for the time complexity of your algorithm to be larger than O(n). If you can show an O(n²) algorithm, you get 5+ points. If you show O(n log n) algorithm, you get 10 points. [If your algorithm is w(n²), that is sadly not worth anything. L]

Tree method or PMI?

Wed, 28 Sep 2016 19:24:47 +0000

For some recurrence we can not use MT, we often guess an answer and prove it through PMI. However, I think that the answer is not close.

For example:

http://notexponential.com/293/recurrence-relation-t-n-t-n-3-2-t-2n-3-n

The PMI answer here is T(n) = T(n/3) + 2T(2n/3)+n = O(n²).

In my solution, I assume that the tree could fully extend.

The 1st level is Θ(n);

The 2nd level is Θ((5/3)*n);

The 3rd level is Θ((5/3)²*n);

....

The nth level is Θ((5/3)^n-1*n).

And the height of the left tree is log₃n, because the height of right tree is less than the value. Here I just assume they have the same height. Then the bottom level is n*T(1) since the tree has n leaves.

Thus,

T(n) = n*T(1) + n + (5/3)*n + (5/3)²*n + ... + (5/3)^log₃n^-1*n = n + n*((1 - (5/3)^log₃n-1)/(1 - (5/3)))

= n + (3/2)*n*(5/3)^log₃n-1~ n + (3/2)*n*(5/3)^log₃n= n + (3/2)*n*n^log₃(5/3) ~ n + n^1.5.

Actually, the tree can not fully extend, T(n) < n + n^1.5, T(n) = O(n^1.5).

PMI answer is O(n²) while tree method answer is O(n^1.5). Which one is the right answer?

Can T(n) = 2T(n/2) + nlogn use master theorem?

Wed, 21 Sep 2016 03:51:57 +0000

How to find the "joint" median of 3 sorted lists, in less than linear time?

Sat, 05 Mar 2016 11:27:06 +0000

We are given 3 sorted lists of size: n1, n2 and n3. We want to find the median of the merged list. Can we do that in less than linear time? If so, how?