To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .

For bottom up approach to satisfy the conditions of red black tree following conditions must be hold:

m[x]=max{m[left[x]]v[left[x]]+p[x]v[left[x]]+p[x]+m[right[x]]

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

NORMALIZATION PROBLEM: Normalization is one of the most basic preprocessing techniques in data analytics. This involves centering and scaling process. Centering means subtracting the data from the mean and scaling means dividing with its standard deviation. Mathematically, normalization can be expressed as: Z = X-x σ In Python, element-wise mean and element-wise standard deviation can be obtained by using .mean() and .std() calls. In this problem, create a random 5 x 5 ndarray and store it to variable X. Normalize X.

Sets are collections (1) without defined order and (2) not allowing duplication. Multisets, also called “bags” are collections without defined order but which permit duplication, i.e., more than one element. We define the function #(a B) to be the number of occurrences of the element a in the bag B. For example, #(1, [1 1 2 3 4 4 5]) is 2 and #(5, [1 1 2 3 4 4 5]) = 1. Bag union and intersection are defined in terms of #. bag-union: List × List -> ListThis function should take as arguments two lists representing bags and should return the list representing their bag-union. bag-intersection : List × List -> ListThis function should take as arguments two lists representing bags and should return the list representing their bag-intersection. Allowed functions. Your code must use only the following functions:1. define, let2. lambda3. cons, car, cdr, list, list?, append, empty?, length, equal?4. and, or, not5. if, cond6. +, -, /, * Racket code only please. Thank you!

Sets are collections (1) without defined order and (2) not allowing duplication. Multisets, also called “bags” are collections without defined order but which permit duplication, i.e., more than one element. We define the function #(a B) to be the number of occurrences of the element a in the bag B. For example, #(1, [1 1 2 3 4 4 5]) is 2 and #(5, [1 1 2 3 4 4 5]) = 1. sum : List × List -> List This function should take as arguments two lists representing bags and should return the list representing the list resulting from simply appending the input lists together. Racket code only please. No loops Allowed: 1. define, let2. lambda, curry3. cons, car, cdr, list, list?, append, empty?, length, equal?4. and, or, not5. if, cond6. map, append-map, andmap, ormap, filter, apply7. +, -, /, * Thank you!

Answer 2

Question

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .

For bottom up approach to satisfy the conditions of red black tree following conditions must be hold:

m[x]=max{m[left[x]]v[left[x]]+p[x]v[left[x]]+p[x]+m[right[x]]

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .

For bottom up approach to satisfy the conditions of red black tree following conditions must be hold:

m[x]=max{m[left[x]]v[left[x]]+p[x]v[left[x]]+p[x]+m[right[x]]

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .

For bottom up approach to satisfy the conditions of red black tree following conditions must be hold:

m[x]=max{m[left[x]]v[left[x]]+p[x]v[left[x]]+p[x]+m[right[x]]

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .

For bottom up approach to satisfy the conditions of red black tree following conditions must be hold:

m[x]=max{m[left[x]]v[left[x]]+p[x]v[left[x]]+p[x]+m[right[x]]

Answer 6

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

Answer 7

Chapter 14, Problem 1P

(a)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given Information: A point of maximum overlap in a set of intervals is a point with the largest number of intervals in the set that overlap it.

Explanation:

Consider that there is no point of maximum overlap in an endpoint of a segment. The maximum overlap occurs in the interior of m segments. Here, the point P is the intersection of those m points.

There must be another point P′ that has the same overlap as P . Therefore, the point P′ is also a point of maximum overlap. Hence, the assumption is not true since there is a point of maximum overlap in an endpoint of maximum overlap.

Hence, it is proved that the there is always a point where maximum overlap has an endpoint of the segment.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .

For bottom up approach to satisfy the conditions of red black tree following conditions must be hold:

m[x]=max{m[left[x]]v[left[x]]+p[x]v[left[x]]+p[x]+m[right[x]]

Answer 13

(b)

Program Plan Intro

To show that there will always be a point where maximum overlap is an endpoint of one of the segments.

Answer 14

(b)

Expert Solution

Explanation of Solution

Explanation:

Consider a balanced binary tree of endpoints. For inserting the interval, it is necessary to insert the endpoints separately. Consider the endpoints as e . For left endpoint e , increase the value of e by 1 and for right endpoint e , decrease the overlap by 1.

For multiple endpoints with same value, insert the left endpoints with the value before the right endpoints with the value.

Consider that e1,e2,…,en be the sequence of the endpoints and s(i,j) represents the sum. Therefore,

p[e1]+p[ei+1]+…+p[ej]

Where 1≤i≤j≤n .

Here, each node x store the new node that includes the endpoints e1[x],...,er[x] . The sum of the values of all nodes is stored as v[x]=s(1[n],r[x]) and the maximum value id obtained by s(1[n],r[x]) .