Algorithms Q&A - Recent questions and answers in Divide & Conquer

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

Fri, 22 Dec 2023 18:14:44 +0000

Given sets A, B, C of size n each and integer k, determine if there exist a, b, c (a ∈ A, b ∈ B, c ∈ C) such that

a + b + c = k

Divide and Conquer Algorithm:

Divide If n = 1, directly check if a + b + c = k

If n > 1, split A, B, C into halves: A1, A2 = split(A) B1, B2 = split(B) C1, C2 = split(C)

Conquer Recursively solve 8 subproblems:

T(A1, B1, C1, k)

T(A1, B1, C2, k)

T(A1, B2, C1, k)

T(A1, B2, C2, k)

T(A2, B1, C1, k)

T(A2, B1, C2, k)

T(A2, B2, C1, k)

T(A2, B2, C2, k)

If any returns true, return true. Else return false.

Combine Return result of conquer step.

The recurrence is T(n) = 8T(n/2) + O(n) which solves to O(n^2) by Master's theorem.

As a result, examining all 8 options in a recursive fashion solves the issue in optimal O(n^2) time. Due to halving, the recursion tree depth decreases quickly, resulting in an efficient algorithm.

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

Wed, 20 Dec 2023 08:05:37 +0000

Use the Akra-Bazzi method:

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

Sat, 16 Dec 2023 18:03:35 +0000

Using Master's Theorem:

1.) n^{log _b a}= n ^{log ₂ 2}= n || For instance, f(n) is n*log^kn . Then the Time Complexity of the recurrence relation would be Theta( n*log^k+1n).

2.) n^log _b a= n ^{log ₂ 2}= n || For instance, f(n) is O(n^1-e). Then the Time Complexity for the recurrence relation would be Theta(n).

3.) n^log _b a= n ^{log ₂ 2}= n || For instance, f(n) is omega(n^1+e). Then the Time Complexity for the recurrence relation would be Theta(f(n)).

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

Thu, 14 Dec 2023 01:25:40 +0000

Given, T(n) = 3T(n/2) + n^1.5 log n

Using Master's theorem,

a=3, b=2, f(n) = n^1.5 log^1 n , here k=1.5, p=1

Calculating log a base b = log 3 base 2 = 1.58, k=1.5

Clearly log a base b > k (case 1)

T(n) = theta(n^log a base b)

T(n) = theta(n^1.58)

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

Wed, 13 Dec 2023 00:09:53 +0000

T(n) = T(n-1) + T(n/2) + 1

Using estimation method, we get

R(n) = 2R(n-1) + 1 & S(n) = 2S(n/2) + 1

Solving using the Master's Theorem we get,

R(n) = O(n*2ⁿ) & S(n) = n

Therefore, the time complexity is (n*2ⁿ) in the worst case.

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

Tue, 21 Nov 2023 09:58:14 +0000

Came across an alternative approach to solve this question,

The Akra-Bazzi theorem provides a method for solving certain types of recurrence relations that don't fit directly into the standard forms covered by the master theorem. The Akra-Bazzi theorem is applicable to recurrence relations of the form:

T(n) = g(n) + Σ_{i=1to k} ai *T(bi *n + h_i(n)),

where ai and bi are constants, and hi(n) are functions satisfying certain conditions. In this case, k = 2, a1 = a2 = 1, b1 = 1/3, b2 = 2/3, and h1(n) = h2(n) = 0.

First we check conditions ,as we can see all conditions satisfied from above.

next we check Regularity Condition:

Σ_{i=1to k} |ai/ bi^{-1}|^p = 1.

Σ_{i=1to k} |ai/ bi^{-1}|^p = (1/(1/3)^p) + (1/(2/3)^p) = 3^p + (3/2)^p ~= 1 for all p>1

The Akra-Bazzi theorem states that if the conditions are satisfied, the solution T(n) is equivalent to
Θ (n^p (1 + ∫_{1to n} g(u) / u^{1+p} du)).

Here g(n) = n and p = |ai / bi| *(1 + hi'(n) / n^{1 + hi(n)}) = 3^{-1} + (2/3)^{-1} = 1. [Since, h1(n) = h2(n) = 0]. By substituting the values in above equation stated, we get

Θ (n (1 + ∫_{1to n} u / u2 du)) = Θ (n (1 + ∫_{1to n} 1/u du)) = Θ (n (1 + log n)) = Θ (n + n log n)

= Θ (n log n)

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

Fri, 10 Jul 2020 02:28:31 +0000

We can use divide and conquer for this problem, only because the we observe the squaring in this case to the power of 2 repeated. We also observe a constant c in this problem, which empowers us to us the standard mathematical model big O(1) as a multiplier. Thus the pseudocode for this problem:

power(int z, int n) {
if (n ==1)

return z

//division of integer, return the largest integer value that is smaller than or equal to the number (z)

mid = int n/2

int previous = power (a,mid); // this recursive call else if

if n%2 == 1

return prev * prev * z

else

return prev*prev

//end

}

Recurrence Relation: T(n) = T(n/2) + 1

Time Complexity: O(log n ) time

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

Thu, 02 Jul 2020 02:08:23 +0000

Master theorem:

Using MT: a = 3, b = 4, f(n) = n^0.75

Comparing n^(log_4(3)) to n^0.75

Thus, case 1: Θ(n^log_4(3))

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

Sun, 22 Sep 2019 23:00:00 +0000

Preliminary Work, to build our hypothesis

T(n) = T(n/3) + T(n/5) + T(n/6) + n

We know that n/3 + n/5 + n/6 = n (10/30 + 6/30 + 5/30) = n 21/30 = 7n/10.

So, we could write this recurrence as similar to: R(n) = R(7n/10) + n

n represents 3/10 part of the overall work. So, it APPEARS that T(n) = 10n/3. By "Appears" we mean that this is not a proof, this is just an exercise to reach a good guess. Proof is what comes next.

Proof by Induction

Our hypothesis is: T(s) <= 10s/3 for all s

Base Case

T(1) = C, this works, as long as C <= 10/3. (Sure, we say that it is.)

Induction Hypothesis

Assume that T(s) <= 10s/3 for all s <= n-1

Inductive Step

We now need to prove the hypothesis for n.

So, for n, T(n) = T(n/3) + T(n/5) + T(n/6) + n

Using Induction Hypothesis (which applies for cases n/3, n/5 and n/6), we have that:

T(n) <= 10(n/3)/3 + 10 (n/5)/3 + 10(n/6)/3 + n

= n (10/9 + 10/15 + 10/18 + 1)

= n 10/3

Therefore, T(n) <= 10n/3 for all n.

Therefore, by PMI, our hypothesis is valid.

Now, we can go back to Asymptotic notation, and conclude that:

T(n) = O(n).

Answered: The Subarray with the Minimum Peak

Thu, 02 May 2019 06:08:47 +0000

divide

find the right peak left peak and peak

merge

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

Thu, 02 May 2019 03:10:23 +0000

Change this question to find the medium of n sorted list n1~nn

Answered: Rectangles overlap

Thu, 02 May 2019 00:51:56 +0000

https://algnotes.wordpress.com/2015/05/20/intersection-of-rectangles/

Answered: Find the smallest domain value where the function becomes negative.

Sun, 27 Jan 2019 16:55:20 +0000

First we find a k that makes f(k) < 0

Then we can find n by a simple binary search.

1. Finding k

Make k0 = some small constant for example 1

Check if f(k0) < 0, if not, double the k0's value and repeat.

the final k0 is k.

if f(0) < 0, vice visa, we can find a k that makes f(k) > 0 where k < 0 in the same way.

2. Finding n

since now f(k) * f(0) < 0, we can perform a single binary search to find n.

Time complexity:

Step 1 takes O(log n) time: finding k takes O(log k) time, and since k > n and k / 2 < n, finding the k takes O(log n) time too.

Step2 takes O(log n) time. It's trivial. The proof is left for practicing.

Phase 1 can go faster under certain conditions. For example, if there is a range for n (Such as something a double or long can hold), we can completely skip phase 1. Or you can make that phase faster by changing the k0 in a faster, different way such as i^i in step i. For example, 1, 4, 27, 256... In this case the k0 grows much faster speed.

However it doesn't change the time complexity in phase 2, so the total time complexity is still O(log n).

To reduce the time complexity even more you need a complete different algorithm than binary search. For example, Newton's method to find the root of the equation in O(log n * F(n)) time where F(n) is the time complexity to calculate f'(n). The time complexity seems the same but the constant associated with it is much smaller.

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

Tue, 07 Mar 2017 21:30:07 +0000

How about to solve it this way?

Because T(n/2)+T(n/3)+T(n/4)<=3T(n/2)

So T(n)=O(3T(n/2)+n)

T(n)=O(n^log23)

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

Wed, 08 Feb 2017 01:07:41 +0000

Assume that T(n) <= 10cn, where c is a constant.

Prove it by induction:

First, there is T(1) = O(1), satisfying that T(1) <= 10cn.

Assume that T(m)<=10cm for all m <= n-1

thus, for n, we have T(n) <= T(n/5) + T(7n/10) + cn

since n/5 <= n-1, 7n/10 <= n-1

For n, there is:

T(n) <= 10c * n/5 + 10c * 7n/10 + cn = 2cn + 7cn + cn = 10cn

Thus, T(n) <= 10cn for all n's. So that T(n) = O(n)

Answered: MVCS using Divide and Conquer

Sat, 03 Dec 2016 13:46:21 +0000

We use D&C

Divide the array into two halfs:

Left side recursive call: T(n/2) time to find MVCS in A[1..n/2]

Right side recursive call: T(n/2) time to find MVCS in A[n/2 + 1..n]

If we can solve the "left-right" case in O(n) time, we will have our T(n) = 2 T(n/2) + O(n) recurrence relation, which then leads to T(n) = O(n log n) analysis.

How to solve the boundary case in O(n) time

To solve boundary case, we realize, that it must include both A[n/2] and A[n/2 + 1]. We simply find the maximum in both directions and add those up. For example:

rightMax = MINUS_INFINITY
rightSum = 0
for (j = n/2 + 1 to n) {
rightSum += a[j]
rightMax = max(rightMax, rightSum)
}

This finds the rightMax. Similarly we find leftMax

middleMax is then simply leftMax + rightMax

Overall algorithm can be written as:

MVCS(A,1,n) {
leftOnlyMax = MVCS(A,1,n/2)
rightOnlyMax = MVCS(A,n/2+1,n)
middleMax = as calculated in the merge routine above
overallMax = max(leftOnlyMax, rightOnlyMax, middleMax);
return overallMax
}

Answered: Tree method or PMI?

Wed, 28 Sep 2016 21:31:03 +0000

Each method has its own advantages and disadvantages, but the analysis given in the question is not correct. The time complexity of this recurrence is not O(n^1.5).

[The analysis presented above should be checked again. For example, consider this statement:

Then the bottom level is n*T(1) since the tree has n leaves.

I am not sure this statement is supported by analysis.

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

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