Algorithmic efficiency can be measured in terms of 1)time 2)spaceT.pdf
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Algorithmic efficiency can be measured in terms of 1)time 2)space Time:How much amount of time it takes to complete the task space:How much space required to store the data of given task. A)To say a given number is even or odd---O(1) time is enough B)To check whether given number is present in the list or not ---O(n) here n means no of elements present in the list.We need to compare each element in the list with given number whether those two or equal or not C)To find smallest number in a list 2 cases:list with sorted elements---O(1) .just we return the first element of list list with unsorted elements O(n) because we have to traverse all the way in the list. D)O(n2)---because we are checking each element of list1 is present in the list2 or not.for that for each element we have to traverse whole list of list2. Here we are writing two loops.each iterate upto length of list E)O(n)---because elements are sorted.so,just we check the corresponding elements of list1 and list. Solution Algorithmic efficiency can be measured in terms of 1)time 2)space Time:How much amount of time it takes to complete the task space:How much space required to store the data of given task. A)To say a given number is even or odd---O(1) time is enough B)To check whether given number is present in the list or not ---O(n) here n means no of elements present in the list.We need to compare each element in the list with given number whether those two or equal or not C)To find smallest number in a list 2 cases:list with sorted elements---O(1) .just we return the first element of list list with unsorted elements O(n) because we have to traverse all the way in the list. D)O(n2)---because we are checking each element of list1 is present in the list2 or not.for that for each element we have to traverse whole list of list2. Here we are writing two loops.each iterate upto length of list E)O(n)---because elements are sorted.so,just we check the corresponding elements of list1 and list..
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Do the following two problems:
1. Any algorithm that solves the search problem for a 7654 element list by comparing the target
element x to the list items must do at least __________ comparisons in the worst case.
2. Any algorithm that sorts a 7 element list by comparing pairs of items from the list must do at
least __________ comparisons in the worst case.
Solution
If the element is not in the list, the only way to know it is to compare it against every item
present in the list. If there are n elements, then the sequential search requires n comparisons to
discover that the item is not there. In the case where the item is in the list, the analysis is not so
straightforward. There are actually three different scenarios that can occur. In the best case we
will find the item in the first place we look, at the beginning of the list. We will need only one
comparison. In the worst case, we will not discover the element until the very last comparison,
the nth comparison.
So, Any algorithm that solves the search problem for a 7654 element list by comparing the target
element x to the list items must do at least __7654__ comparisons in the worst case.
Next,the algorithm for bubble sort or pair sort requires a pair of nested loops. The outer loop
must iterate once for each element in the data set (of size n) while the inner loop iterates n times
the first time it is entered, n-1 times the second, and so on. Consider the purpose of each loop.
Bubble sort is structured so that on each pass through the list the next largest element of the data
is moved to its proper place. Therefore, to get all n elements in their correct places, the outer
loop must be executed n times.
And on the first iteration of the outer loop, while trying to place the largest element, there have to
be n - 1 comparisons: the first comparison is made between the 1st and 2nd elements, the second
is made between the second and third elements, and so on until the (n-1)th comparison is made
between the (n-1)th and the nth element. On the second iteration of the outer loop, there is no
need to compare the against the last element of the list, because it was put in the correct place on
the previous pass. Therefore, the second iteration requires only n-2 comparisons. This pattern
continues until the second-to-last iteration of the outer loop when only the first two elements of
the list are unsorted so, only one comparison is necessary. The total number of comparisons,
therefore, is (n- 1) + (n- 2)...(2) + (1)
=n(n- 1)/2 orO(n2).
So, any algorithm that sorts a 7 element list by comparing pairs of items from the list must do at
least __21________ comparisons in the worst case..
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Answers
a) It is more efficient to delete the last node in an array because we can access the last element in
constant time O(1) and we can delete the value at that position. In case of linked list we need to
traverse the entire list to reach the last node which requires O(n) time and then we delete the
node.
b) It is inefficient to delete the first node in an array because after deleting the first element we
need to shift the remaining elements one position to the left which requires O(n) time. In case of
linked list we can directly delete the first node by changing the pointer to the first node and
freeing the memory which was earlier allocated to the first element. This operation can be done
in O(1) time.
c) In case of an array after deleting the specific element we need to shift the remaining elements
one position to the left and hence it is inefficient. In case of a linked list we need to traverse the
linked list till we reach the node to be deleted and then we can change the pointers to delete the
node.
d) It is more efficient to insert a node at the end of an array because we can access the last index
in O(1) time and then we can insert the element at that position. In case of linked list we need to
traverse till the end of list and then we can insert the element which requires O(n) time and is
inefficient.
e) It is inefficient to insert a node at the front of an array because we need to make room for the
new element to be inserted by shifting all the elements one position to the right and then insert
the element at the first index of array. In case of linked list we can create a new node and insert
the node at first position by modifying the pointers which is efficient.
f) It is inefficient to insert a node in general in case of array because we need to make room for
the element by shifting the element one position to the right and then we can insert the element at
that index position. In case of linked list we need to traverse the list till we reach the specific
position and then we can insert the element by modifying the pointer.
g) It is more efficient to access a node in case of an array because arrays support random access
and we can access any index in constant time. In case of linked list we need to traverse the list in
order to find the element at specified index which is not efficient.
Solution
Answers
a) It is more efficient to delete the last node in an array because we can access the last element in
constant time O(1) and we can delete the value at that position. In case of linked list we need to
traverse the entire list to reach the last node which requires O(n) time and then we delete the
node.
b) It is inefficient to delete the first node in an array because after deleting the first element we
need to shift the remaining elements one position to the left which requires O(n) time. In case of
linked list we can directly delete the first node by changing the pointer to the first node and
freeing the memory which was earlie.
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two singals are expected. one is for CH3, the other is for CH2
Solution
two singals are expected. one is for CH3, the other is for CH2.
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There is in fact a stronger problem, namely 2p C p ? 2 (mod p
Solution
There is in fact a stronger problem, namely 2p C p ? 2 (mod p.
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- Basic searching algorithms like linear search and binary search and how they work.
- The concept of list
Sorting
The document discusses sorting algorithms. It begins by defining sorting and listing common types of sorts like bubble sort, selection sort, insertion sort, merge sort, and radix sort. It then focuses on explaining selection sort, insertion sort, and radix sort in more detail. For each algorithm, it provides an overview of how it works, discusses space and time complexity, and provides examples. Key points made include that selection sort finds the minimum element and swaps it into position each pass, insertion sort maintains a sorted sub-list and inserts new elements, and radix sort sorts based on digits from the least to most significant place value in multiple passes.
Hashing
Hashing is a technique used to store and retrieve data efficiently. It involves using a hash function to map keys to integers that are used as indexes in an array. This improves searching time from O(n) to O(1) on average. However, collisions can occur when different keys map to the same index. Collision resolution techniques like chaining and open addressing are used to handle collisions. Chaining resolves collisions by linking keys together in buckets, while open addressing resolves them by probing to find the next empty index. Both approaches allow basic dictionary operations like insertion and search to be performed in O(1) average time when load factors are low.
L1803016468
The document describes three different searching techniques: linear search, binary search, and DN search. Linear search has a worst case time complexity of O(n) as it searches through each element sequentially. Binary search has a worst case time complexity of O(log n) but requires the list to be sorted. DN search also has a worst case time complexity of O(log n) and can be applied to both sorted and unsorted lists, making it more powerful than the other techniques.
Sorting
Sorting
SORTING PROPERTIES
Bubble sort
Selection Sort
Insertion sort
Shell Sort
Heap Sort
Merge Sort
Quick Sort
Radix Sort
Insertion and Redix Sort
The document describes the insertion sort algorithm. It works by scanning an array from the first element to the last, inserting each element into the proper position in the previously sorted portion of the array. It also describes the quicksort algorithm, which works by partitioning the array around a pivot element and recursively sorting the subarrays.
Sorting techniques
Insertion sort works by scanning an array from left to right and inserting each element into its sorted position in the previously sorted left portion. Quicksort works by partitioning the array around a pivot value, recursively sorting the left and right partitions, and combining the results. It chooses a pivot element and rearranges the array such that all elements smaller than the pivot are to its left and larger elements to its right.
Master of Computer Application (MCA) – Semester 4 MC0080
This document describes several sorting algorithms and asymptotic analysis techniques. It discusses bubble sort, selection sort, insertion sort, shell sort, heap sort, merge sort, and quick sort as sorting algorithms. It then explains asymptotic notation such as Big-O, Big-Omega, and Theta to describe the time complexity of algorithms. Finally, it asks questions about Fibonacci heaps, binomial heaps, Strassen's matrix multiplication algorithm, and formalizing a greedy algorithm.
searching in data structure.pptx
This document discusses various sorting and searching algorithms. It begins by listing sorting algorithms like selection sort, insertion sort, bubble sort, merge sort, and radix sort. It then discusses searching algorithms like linear/sequential search and binary search. It provides details on the implementation and time complexity of linear search, binary search, bubble sort, insertion sort, selection sort, and merge sort. Key points covered include the divide and conquer approach of merge sort and how binary search recursively halves the search space.
MergeSort.ppt shows how merge sort is done
Merge sort is a divide and conquer sorting algorithm that works by recursively splitting an array into two halves, sorting each subarray, and then merging the sorted subarrays back together. It has three main steps - divide, where the array is split in half; recur, where the sorting happens recursively on the subarrays; and conquer, where the sorted subarrays are merged back together. Merge sort has a time complexity of O(n log n), making it one of the most efficient sorting algorithms.
Do the following two problems1. Any algorithm that solves the sea.pdf
Do the following two problems:
1. Any algorithm that solves the search problem for a 7654 element list by comparing the target
element x to the list items must do at least __________ comparisons in the worst case.
2. Any algorithm that sorts a 7 element list by comparing pairs of items from the list must do at
least __________ comparisons in the worst case.
Solution
If the element is not in the list, the only way to know it is to compare it against every item
present in the list. If there are n elements, then the sequential search requires n comparisons to
discover that the item is not there. In the case where the item is in the list, the analysis is not so
straightforward. There are actually three different scenarios that can occur. In the best case we
will find the item in the first place we look, at the beginning of the list. We will need only one
comparison. In the worst case, we will not discover the element until the very last comparison,
the nth comparison.
So, Any algorithm that solves the search problem for a 7654 element list by comparing the target
element x to the list items must do at least __7654__ comparisons in the worst case.
Next,the algorithm for bubble sort or pair sort requires a pair of nested loops. The outer loop
must iterate once for each element in the data set (of size n) while the inner loop iterates n times
the first time it is entered, n-1 times the second, and so on. Consider the purpose of each loop.
Bubble sort is structured so that on each pass through the list the next largest element of the data
is moved to its proper place. Therefore, to get all n elements in their correct places, the outer
loop must be executed n times.
And on the first iteration of the outer loop, while trying to place the largest element, there have to
be n - 1 comparisons: the first comparison is made between the 1st and 2nd elements, the second
is made between the second and third elements, and so on until the (n-1)th comparison is made
between the (n-1)th and the nth element. On the second iteration of the outer loop, there is no
need to compare the against the last element of the list, because it was put in the correct place on
the previous pass. Therefore, the second iteration requires only n-2 comparisons. This pattern
continues until the second-to-last iteration of the outer loop when only the first two elements of
the list are unsorted so, only one comparison is necessary. The total number of comparisons,
therefore, is (n- 1) + (n- 2)...(2) + (1)
=n(n- 1)/2 orO(n2).
So, any algorithm that sorts a 7 element list by comparing pairs of items from the list must do at
least __21________ comparisons in the worst case..
Answersa) It is more efficient to delete the last node in an array.pdf
Answers
a) It is more efficient to delete the last node in an array because we can access the last element in
constant time O(1) and we can delete the value at that position. In case of linked list we need to
traverse the entire list to reach the last node which requires O(n) time and then we delete the
node.
b) It is inefficient to delete the first node in an array because after deleting the first element we
need to shift the remaining elements one position to the left which requires O(n) time. In case of
linked list we can directly delete the first node by changing the pointer to the first node and
freeing the memory which was earlier allocated to the first element. This operation can be done
in O(1) time.
c) In case of an array after deleting the specific element we need to shift the remaining elements
one position to the left and hence it is inefficient. In case of a linked list we need to traverse the
linked list till we reach the node to be deleted and then we can change the pointers to delete the
node.
d) It is more efficient to insert a node at the end of an array because we can access the last index
in O(1) time and then we can insert the element at that position. In case of linked list we need to
traverse till the end of list and then we can insert the element which requires O(n) time and is
inefficient.
e) It is inefficient to insert a node at the front of an array because we need to make room for the
new element to be inserted by shifting all the elements one position to the right and then insert
the element at the first index of array. In case of linked list we can create a new node and insert
the node at first position by modifying the pointers which is efficient.
f) It is inefficient to insert a node in general in case of array because we need to make room for
the element by shifting the element one position to the right and then we can insert the element at
that index position. In case of linked list we need to traverse the list till we reach the specific
position and then we can insert the element by modifying the pointer.
g) It is more efficient to access a node in case of an array because arrays support random access
and we can access any index in constant time. In case of linked list we need to traverse the list in
order to find the element at specified index which is not efficient.
Solution
Answers
a) It is more efficient to delete the last node in an array because we can access the last element in
constant time O(1) and we can delete the value at that position. In case of linked list we need to
traverse the entire list to reach the last node which requires O(n) time and then we delete the
node.
b) It is inefficient to delete the first node in an array because after deleting the first element we
need to shift the remaining elements one position to the left which requires O(n) time. In case of
linked list we can directly delete the first node by changing the pointer to the first node and
freeing the memory which was earlie.
Similar to Algorithmic efficiency can be measured in terms of 1)time 2)spaceT.pdf (20)
Master of Computer Application (MCA) – Semester 4 MC0080
Master of Computer Application (MCA) – Semester 4 MC0080
Do the following two problems1. Any algorithm that solves the sea.pdf
Do the following two problems1. Any algorithm that solves the sea.pdf
Answersa) It is more efficient to delete the last node in an array.pdf
Answersa) It is more efficient to delete the last node in an array.pdf
More from annaelctronics
two singals are expected. one is for CH3, the oth.pdf
two singals are expected. one is for CH3, the other is for CH2
Solution
two singals are expected. one is for CH3, the other is for CH2.
There is in fact a stronger problem, namely 2p C .pdf
There is in fact a stronger problem, namely 2p C p ? 2 (mod p
Solution
There is in fact a stronger problem, namely 2p C p ? 2 (mod p.
The term Lewis acid refers to a definition of aci.pdf
The term Lewis acid refers to a definition of acid published by Gilbert N. Lewis in
1923, specifically: An acid substance is one which can employ an electron lone pair from another
molecule in completing the stable group of one of its own atoms.[1] Thus, H+ is a Lewis acid,
since it can accept a lone pair, completing its stable form, which requires two electrons. The
modern-day definition of Lewis acid, as given by IUPAC is a molecular entity (and the
corresponding chemical species) that is an electron-pair acceptor and therefore able to react with
a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis
base.[2] This definition is both more general and more specific—the electron pair need not be a
lone pair (it could be the pair of electrons in a p bond, for example), but the reaction should give
an adduct (and not just be a displacement reaction). Crystal field theory (CFT) is a model that
describes the breaking of degeneracies of electronic orbital states, usually d or f orbitals, due to a
static electric field produced by a surrounding charge distribution (anion neighbors). This theory
has been used to describe various spectroscopies of transition metal coordination complexes, in
particular optical spectra (colours). CFT successfully accounts for some magnetic properties,
colours, hydration enthalpies, and spinel structures of transition metal complexes, but it does not
attempt to describe bonding. CFT was developed by physicists Hans Bethe and John Hasbrouck
van Vleck[1] in the 1930s. CFT was subsequently combined with molecular orbital theory to
form the more realistic and complex ligand field theory (LFT), which delivers insight into the
process of chemical bonding in transition metal complexes. In chemistry, valence bond (VB)
theory is one of two basic theories, along with molecular orbital (MO) theory, that were
developed to use the methods of quantum mechanics to explain chemical bonding. It focuses on
how the atomic orbitals of the dissociated atoms combine to give individual chemical bonds
when a molecule is formed. In contrast, molecular orbital theory has orbitals that cover the whole
molecule.
Solution
The term Lewis acid refers to a definition of acid published by Gilbert N. Lewis in
1923, specifically: An acid substance is one which can employ an electron lone pair from another
molecule in completing the stable group of one of its own atoms.[1] Thus, H+ is a Lewis acid,
since it can accept a lone pair, completing its stable form, which requires two electrons. The
modern-day definition of Lewis acid, as given by IUPAC is a molecular entity (and the
corresponding chemical species) that is an electron-pair acceptor and therefore able to react with
a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis
base.[2] This definition is both more general and more specific—the electron pair need not be a
lone pair (it could be the pair of electrons in a p bond, for example.
MnO2 may act as a catalyst.in the other reaction .pdf
MnO2 may act as a catalyst.in the other reaction of KBrO3 and Sulfuric acid.
ANSWER
Solution
MnO2 may act as a catalyst.in the other reaction of KBrO3 and Sulfuric acid.
ANSWER.
What are P and R hereSolutionWhat are P and R here.pdf
P and R refer to two variables in this context. The document asks what P and R represent as they are not defined within the given text. In just 3 sentences, the document poses a question about the meaning of P and R but does not provide any other context or information to explain what they are.
Intermolecular attractions are attractions betwee.pdf
Intermolecular attractions are attractions between one molecule and a neighbouring
molecule. The forces of attraction which hold an individual molecule together (for example, the
covalent bonds) are known as intramolecular attractions. These two words are so confusingly
similar that it is safer to abandon one of them and never use it. The term \"intramolecular\" won\'t
be used again on this site. All molecules experience intermolecular attractions, although in some
cases those attractions are very weak. Even in a gas like hydrogen, H2, if you slow the molecules
down by cooling the gas, the attractions are large enough for the molecules to stick together
eventually to form a liquid and then a solid. In hydrogen\'s case the attractions are so weak that
the molecules have to be cooled to 21 K (-252°C) before the attractions are enough to condense
the hydrogen as a liquid. Helium\'s intermolecular attractions are even weaker - the molecules
won\'t stick together to form a liquid until the temperature drops to 4 K (-269°C). van der Waals
forces: dispersion forces Dispersion forces (one of the two types of van der Waals force we are
dealing with on this page) are also known as \"London forces\" (named after Fritz London who
first suggested how they might arise). The origin of van der Waals dispersion forces Temporary
fluctuating dipoles Attractions are electrical in nature. In a symmetrical molecule like hydrogen,
however, there doesn\'t seem to be any electrical distortion to produce positive or negative parts.
But that\'s only true on average. The lozenge-shaped diagram represents a small symmetrical
molecule - H2, perhaps, or Br2. The even shading shows that on average there is no electrical
distortion. But the electrons are mobile, and at any one instant they might find themselves
towards one end of the molecule, making that end -. The other end will be temporarily short of
electrons and so becomes +. Note: (read as \"delta\") means \"slightly\" - so + means \"slightly
positive\". An instant later the electrons may well have moved up to the other end, reversing
the polarity of the molecule. This constant \"sloshing around\" of the electrons in the molecule
causes rapidly fluctuating dipoles even in the most symmetrical molecule. It even happens in
monatomic molecules - molecules of noble gases, like helium, which consist of a single atom. If
both the helium electrons happen to be on one side of the atom at the same time, the nucleus is
no longer properly covered by electrons for that instant. How temporary dipoles give rise to
intermolecular attractions I\'m going to use the same lozenge-shaped diagram now to represent
any molecule which could, in fact, be a much more complicated shape. Shape does matter (see
below), but keeping the shape simple makes it a lot easier to both draw the diagrams and
understand what is going on. Imagine a molecule which has a temporary polarity being
approached by one which happens to be entirely non-polar just at .
It is insoluble in water. .pdf
This document discusses a substance that is insoluble in water. The substance does not dissolve or break apart when mixed with water. The document simply states that the substance is insoluble in water and does not provide any additional context or details.
Write it using the ^ symbol. For example 1.34 10^6Solutio.pdf
Write it using the ^ symbol. For example:
1.34 * 10^6
Solution
Write it using the ^ symbol. For example:
1.34 * 10^6.
Xenodiagnosis is the method used to document presesnce of a microorg.pdf
Xenodiagnosis is the method used to document presesnce of a microorganism by exposing the
infected tissue to a vector and than examining the vector for the presence of microorganisms.
Uninfected vectors and detection of pathogen DNA in host cell will be required for
xenodiagnosis.
Solution
Xenodiagnosis is the method used to document presesnce of a microorganism by exposing the
infected tissue to a vector and than examining the vector for the presence of microorganisms.
Uninfected vectors and detection of pathogen DNA in host cell will be required for
xenodiagnosis..
There are many more conditions than just the norms of the culture th.pdf
There are many more conditions than just the norms of the culture that affect and assert the
statement
Solution
There are many more conditions than just the norms of the culture that affect and assert the
statement.
The three layers of smooth muscle that includes mucosa. innermost tu.pdf
The three layers of smooth muscle that includes mucosa. innermost tunic of a ureter; formed by
transitional epithelium that is both stretchable and impermeable to the passage of urine; external
to the transitional epithelium is the lamina propria, composed of dense irregular connective tissue
The correct answer is 2
Solution
The three layers of smooth muscle that includes mucosa. innermost tunic of a ureter; formed by
transitional epithelium that is both stretchable and impermeable to the passage of urine; external
to the transitional epithelium is the lamina propria, composed of dense irregular connective tissue
The correct answer is 2.
The independent variable is the one which changes in each plant. Thi.pdf
The independent variable is the one which changes in each plant. This is the concentration of the
salt solution in each case.Based on the independent variable the changes seen give us the
dependent variable. This is the height of the plants. The heights of the plants[dependent variable]
may be influenced by the salt solutions[independent variable].
Solution
The independent variable is the one which changes in each plant. This is the concentration of the
salt solution in each case.Based on the independent variable the changes seen give us the
dependent variable. This is the height of the plants. The heights of the plants[dependent variable]
may be influenced by the salt solutions[independent variable]..
Species diversity is more in near island as rate of immigration is m.pdf
Species diversity is more in near island as rate of immigration is more.
Also it is more in large islands as more resources are available there and rate kf extinction is low.
Solution
Species diversity is more in near island as rate of immigration is more.
Also it is more in large islands as more resources are available there and rate kf extinction is low..
Solution Flies do not have teeth for chewing food.So,like human f.pdf
Solution
:
Flies do not have teeth for chewing food.So,like human flies have some acid in their stomach for
digestion of food.But digestion of food in flies not occur in the stomach.It is occur outside of the
body.Flies vomit acids present in their stomach on solid food and convert it into liquid form.Flies
can intake food in liquid form only.After converting it into liquid form it is consume by the flies..
Program To change this license header, choose License Heade.pdf
Program:
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package chegg;
// A Java program for Dijkstra\'s single source shortest path algorithm.
// The program is for adjacency matrix representation of the graph
import java.util.*;
import java.lang.*;
import java.io.*;
class DijkstraAlgorithem
{
// A utility function to find the vertex with minimum distance value,
// from the set of vertices not yet included in shortest path tree
static final int V=6;
int minDistance(int dist[], Boolean sptSet[])
{
// Initialize min value
int min = Integer.MAX_VALUE, min_index=-1;
for (int v = 0; v < V; v++)
if (sptSet[v] == false && dist[v] <= min)
{
min = dist[v];
min_index = v;
}
return min_index;
}
// A utility function to print the constructed distance array
void print
Solution
(int dist[], int n,int src)
{
System.out.println( \" Here Infinite=\"+Integer.MAX_VALUE);
System.out.println(\" Source Vertex:\"+(src+1));
System.out.println(\"Vertex Distance from Source\");
for (int i = 0; i < V; i++)
System.out.println((i+1)+\" \\t\\t \"+dist[i]);
}
// Funtion that implements Dijkstra\'s single source shortest path
// algorithm for a graph represented using adjacency matrix
// representation
void dijkstra(int graph[][], int src)
{
int dist[] = new int[V]; // The output array. dist[i] will hold
// the shortest distance from src to i
// sptSet[i] will true if vertex i is included in shortest
// path tree or shortest distance from src to i is finalized
Boolean sptSet[] = new Boolean[V];
// Initialize all distances as INFINITE and stpSet[] as false
for (int i = 0; i < V; i++)
{
dist[i] = Integer.MAX_VALUE;
sptSet[i] = false;
}
// Distance of source vertex from itself is always 0
dist[src] = 0;
// Find shortest path for all vertices
for (int count = 0; count < V-1; count++)
{
// Pick the minimum distance vertex from the set of vertices
// not yet processed. u is always equal to src in first
// iteration.
int u = minDistance(dist, sptSet);
// Mark the picked vertex as processed
sptSet[u] = true;
// Update dist value of the adjacent vertices of the
// picked vertex.
for (int v = 0; v < V; v++)
// Update dist[v] only if is not in sptSet, there is an
// edge from u to v, and total weight of path from src to
// v through u is smaller than current value of dist[v]
if (!sptSet[v] && graph[u][v]!=0 &&
dist[u] != Integer.MAX_VALUE &&
dist[u]+graph[u][v] < dist[v])
dist[v] = dist[u] + graph[u][v];
}
// print the constructed distance array
print.
plexusFormed from anterioe rani o these spinal nervesMajor nerve.pdf
plexus
Formed from anterioe rani o these spinal nerves
Major nerves formed from the plexus
Cervical
The cervical plexus is a plexus of the front rami of the initial four cervical spinal nerves which
are situated from C1 to C4 cervical fragment in the neck.
They are found along the side to the transverse procedures between prevertebral muscles from
the average side and vertebral (m. scalenus, m. levator scapulae, m. splenius cervicis) from
horizontal side. There is anastomosis with extra nerve, hypoglossal nerve and thoughtful trun
The cervical plexus has two sorts of branches: cutaneous and strong.
Cutaneous (4 branches):
Incredible auricular nerve - innervates skin close concha auricle (external ear) and outer acoustic
meatus (ear waterway) (C2&C3)
Transverse cervical nerve - innervates front locale of neck (C2&C3)
Lesser occipital - innervates the skin and the scalp posterosuperior to the auricle (C2)
Supraclavicular nerves - innervate the skin above and underneath the clavicle (C3,C4)
Muscular:
Ansa cervicalis (circle framed from C1-C3), and so forth (geniohyoid (C1 just), thyrohyoid (C1
just), sternothyroid, sternohyoid, omohyoid)
Phrenic (C3-C5 (fundamentally C4))- innervates stomach and the pericardium
Segmental branches (C1-C4)- innervates front and center scalenes
Brachial
The brachial plexus is a system of nerves shaped by the foremost rami of the lower four cervical
nerves and first thoracic nerve (C5, C6, C7, C8, and T1). This plexus stretches out from the
spinal line, through the cervicoaxillary trench in the neck, over the primary rib, and into the
armpit. It supplies afferent and efferent nerve strands to the mid-section, shoulder, arm and hand.
The Brachial Plexus
1 Roots.
2 Trunks.
3 Divisions.
4 Cords.
5 Major Branches. 5.1 Musculocutaneous Nerve. 5.2 Axillary Nerve. 5.3 Median Nerve. 5.4
Radial Nerve. 5.5 Ulnar Nerve. 5.6 Practical Relevance: Dissecting the Brachial Plexus.
6 Minor Branches.
lumbosacral
The anterior divisions of the lumbar nerves, sacral nerves, and coccygeal nerve form the
lumbosacral plexus, the first lumbar nerve being frequently joined by a branch from the twelfth
thoracic. For descriptive purposes this plexus is usually divided into three parts:
lumbar plexus
sacral plexus
pudendal plexus
lumbar plexus
sacral plexus
pudendal plexus
plexus
Formed from anterioe rani o these spinal nerves
Major nerves formed from the plexus
Cervical
The cervical plexus is a plexus of the front rami of the initial four cervical spinal nerves which
are situated from C1 to C4 cervical fragment in the neck.
They are found along the side to the transverse procedures between prevertebral muscles from
the average side and vertebral (m. scalenus, m. levator scapulae, m. splenius cervicis) from
horizontal side. There is anastomosis with extra nerve, hypoglossal nerve and thoughtful trun
The cervical plexus has two sorts of branches: cutaneous and strong.
Cutaneous (4 branches):
Incredible auricular nerve - innervates skin close con.
Ok, so the number of double bond equivalents in the compound is [C4H.pdf
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that\'s left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2)
Solution
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that\'s left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2).
order of overlap of atomic orbitals from highest extent of overlap t.pdf
order of overlap of atomic orbitals from highest extent of overlap to lowest is,
I2 > IBr > ICl > IF
Because,
I2-is homo diatomic molecule and having good overlapping between the atomic orbitals of two
iodine atoms.
IBr-is a hetero di-atomic molecule and having almost good overlapping between the atomic
orbitals of iodine and bromine atoms even though these two atoms belongs to different periods.
ICl-is a hetero di-atomic molecule and having moderate overlapping between the atomic orbitals
of iodine and chlorine atoms.Because, these two atoms belongs to different periods and size
different is more.
IF-is a hetero di-atomic molecule and having poor overlapping between the atomic orbitals of
iodine and fluorine atoms.Because, these two atoms belongs to different periods and size
different is too more.
Solution
order of overlap of atomic orbitals from highest extent of overlap to lowest is,
I2 > IBr > ICl > IF
Because,
I2-is homo diatomic molecule and having good overlapping between the atomic orbitals of two
iodine atoms.
IBr-is a hetero di-atomic molecule and having almost good overlapping between the atomic
orbitals of iodine and bromine atoms even though these two atoms belongs to different periods.
ICl-is a hetero di-atomic molecule and having moderate overlapping between the atomic orbitals
of iodine and chlorine atoms.Because, these two atoms belongs to different periods and size
different is more.
IF-is a hetero di-atomic molecule and having poor overlapping between the atomic orbitals of
iodine and fluorine atoms.Because, these two atoms belongs to different periods and size
different is too more..
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