For Programming Embedded SystemsQ-02 Which structure variable poi.pdf
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For Programming Embedded Systems Q-02 *Which structure variable points to the tick function? A) line 8 B) line 98 C) line 62 D) line 19 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101 VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks) /*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103 switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0 break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0; state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1; tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct 6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1] , state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks [1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137 139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1 Solution A) line 8 is true line 8: int (*TickFct) (int); TickFct is a variable which points to tick function.
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![For Programming Embedded Systems
Q-02 *Which structure variable points to the tick function?
A) line 8
B) line 98
C) line 62
D) line 19 I. #include "rims.h" 2 3 4. typedef struct task state BL LedOff; break; default: int
state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I
Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const
unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned
long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long
tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct
ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum
TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101
VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks)
/*e.g., static int x = 0; */ /*variables for state machine go here. printf ("Period too short "): 103
switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case
TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0
break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0;
state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1;
tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct
6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1]
, state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks
[1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get
Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137
139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct
BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int
X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return
state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
A) line 8 is true
line 8: int (*TickFct) (int);
TickFct is a variable which points to tick function](https://image.slidesharecdn.com/forprogrammingembeddedsystemsq-02whichstructurevariablepoi-230702022910-56315d37/85/For-Programming-Embedded-SystemsQ-02-Which-structure-variable-poi-pdf-1-320.jpg)
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A) line 4
B) line 41
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D) line 30 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int
state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I
Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const
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VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks)
/*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103
switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case
TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0
break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0;
state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1;
tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct
6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1]
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[1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get
Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137
139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct
BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int
X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return
state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
Answer :-
C) line 11
Explanation :
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You can create an array by using the new operator .
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For Programming Embedded SystemsQ-07 Whcih variable and condition.pdf
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Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)?
A) Line 28
B) Line 36
C) Line 25
D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int
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VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks)
/*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103
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TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0
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state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1;
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Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137
139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct
BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int
X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return
state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
c) Line 25 is correct
Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished
tasks.
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A) line 4
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state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
Answer :-
C) line 11
Explanation :
Creating Array :
You can create an array by using the new operator .
Syntax :.
For Programming Embedded SystemsQ-07 Whcih variable and condition.pdf
For Programming Embedded Systems
Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)?
A) Line 28
B) Line 36
C) Line 25
D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int
state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I
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TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0
break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0;
state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1;
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, state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks
[1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get
Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137
139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct
BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int
X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return
state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
c) Line 25 is correct
Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished
tasks.
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You are presented with a patient, named Gerry, who has Down syndrome (trisomy 21). You
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You are interested in the TPTE gene, which is near the centromere of chromosome 21. Virtually
no recombination occurs between the centromere and the TPTE gene. Mary has two different
alleles of the TPTE gene. Would Gerry\'s two chromosome 21s from his mother have the same
or different alleles of the TPTE gene? Or do you not have enough information to decide?
Explain your answer by drawing meiosis I and II for Mary\'s chromosome 21. Use \"A\" for one
allele and \"a\" for the other allele of the TPTE gene. Only draw the chromosomes relevant to
this question (Do not draw all 46 chromosomes).
Solution
a. Nondisjunction produces abnormal gametes. Nondisjunction in meiosis I occurs during
anaphase I when one pair of homologous chromosomes fails to separate.Trisomy is the condition
of having 3 copies of one chromosome type. It is designated as 2n + 1 because the cell has the
normal two sets of each 23 types of chromosomes plus an extra copy of one chromosome.
I think Gerrys two chromosome 21s from his mother have different alleles (A and a) of the TPTE
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Write a template function to fill in a STL list with a given array.
Solution
Hi, I am not able to see the function prototype completely.
But I wrote a function that takes a template array and adding all elements of array in STL list.
Please let me know in case of any issue.
template
void fillArray(T arr[], int size) {
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std::list myList;
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write a method to generate AVL tree of height h wth fewest nodes and what is the running time
of your method
Solution
class Node {
int key, height;
Node left, right;
Node(int d) {
key = d;
height = 1;
}
}
class AVLTree {
static Node root;
// A utility function to get height of the tree
int height(Node N) {
if (N == null) {
return 0;
}
return N.height;
}
// A utility function to get maximum of two integers
int max(int a, int b) {
return (a > b) ? a : b;
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
Node rightRotate(Node y) {
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left), height(y.right)) + 1;
x.height = max(height(x.left), height(x.right)) + 1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
Node leftRotate(Node x) {
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left), height(x.right)) + 1;
y.height = max(height(y.left), height(y.right)) + 1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(Node N) {
if (N == null) {
return 0;
}
return height(N.left) - height(N.right);
}
Node insert(Node node, int key) {
/* 1. Perform the normal BST rotation */
if (node == null) {
return (new Node(key));
}
if (key < node.key) {
node.left = insert(node.left, key);
} else {
node.right = insert(node.right, key);
}
/* 2. Update height of this ancestor node */
node.height = max(height(node.left), height(node.right)) + 1;
/* 3. Get the balance factor of this ancestor node to check whether
this node became unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
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// Right Right Case
if (balance < -1 && key > node.right.key) {
return leftRotate(node);
}
// Left Right Case
if (balance > 1 && key > node.left.key) {
node.left = leftRotate(node.left);
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// Right Left Case
if (balance < -1 && key < node.right.key) {
node.right = rightRotate(node.right);
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if (node != null) {
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preOrder(node.left);
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Which of the following is NOT correct?-Protozoans are fairly susceptible to antimicrobial
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Solution
John Stuart Mill wrote his moral treatise, Utilitarianism, in a period in intellectual history
dominated by Aristotelian virtue and Kantian deontology, as well as the intensely conservative
political and social climate of Victorian England.Utilitarianism was intended to convince the
public of Mill’s first principle. Mill responds to preexisting ethical frameworks, namely virtue
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prominent response in the text is his amalgamation of virtue ethics, which spans all of Chapter 4.
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fallacies. In his success, Mill crafted an impartial standard which the moral claims of other
theories can by compared against.
It is in Chapter 4 of Utilitarianism that Mill addresses virtue the most directly. He begins his
argument by stating that virtue can actually be an end to human action besides happiness. Mill
agrees with his opponents that there are people who desire or approve of acts according to virtue
and absence of vice instead of the pleasure and the absence of pain. The reason this is significant
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desirable. If there are people who genuinely value virtue as the criteria of right and wrong, of
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Utilitarian claims. The solution to this problem for utilitarianism is to either swallow the
intuitively unsavory conclusion that virtue is without value, or show that virtue actually has
value belonging to, part of, or from happiness.
Mill states that where opponents make their mistake is in believing that utility is incompatible
with virtue. Mill holds that virtue is not only valuable as a means to pleasure, but in fact as an
end in itself. Virtue and happiness are perceived to be different, but Mill attempts to show that
this is superficial. It is by seeking virtue disinterestedly and for its own sake that virtue can
become part of pleasure, Mill claims.
So not only is Mill successful in bringing virtue ethics into utilitarianism, but he’s successful in
creating a template by which other theories can be synthesized with utility. Mill’s claim is that
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principle is the concept of utility. By providing one specific example, virtue, and showing how it
actually admits of his general concept, utility, Mill is successful crafting an argument which
allows people .
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Bivariate distribution is a two dimensional random variables distribution.
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When the two random variables X and Y are continuous, the joint probability
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Solution
This statement is false. Glucagon stimulates and insulin suppresses the hydrolysis of liver
glycogen, or glycogenolysis. Thus during times of fasting (post absorptive state) , when glucagon
secretion is high and insulin secretion is low, liver glycogen is used as a source of additional
blood glucose..
The principal role of ATP In the cell is A) serving as structural s.pdf
The principal role of ATP In the cell is: A) serving as structural support. B) catalyzing
chemical reactions. C) keeping the phospholipid tails of membranes aligned D) providing
energy. E) functioning as a hormone. Small nonpolar molecules, such as O_2 and CO_2, enter a
cell by: A) vesicle-mediated transport. B) simple diffusion. C) osmosis. D) facilitated
diffusion. E) active transport. Eukaryotic cells differ from prokaryotic cells in that eukaryotic
cells have: A) plasma membrance B) genetic material C) cytoplasm. D) ribosomes. E)
nucleus. If a plant cell is placed in an aqueous solution that is hypertonic to the cell, the cell A)
build up turgor pressure B) build up wall pressure C) undergo plasmolysis D) neither gain nor
lose water E) swell and perhaps burst
Solution
15 . D Providing energy
Atp or adenosine tri phosphate is a energy rich molecule produced by mitochondria of our cell.
These atp molecules are produced by oxidation of organic molecules, (glucose) which is then
utilised by our body for various cellular functions..
The neutral theory of evolution proposed that most genetic mutations .pdf
The neutral theory of evolution proposed that most genetic mutations are functionally neutral.
Neutral mutations are fixed in a population by natural selection random genetic drift
environmental selection asexual reproduction the invariance of DNA
Solution
The neutra theory of evouloution proposed that most genetic mutations are functionally
neutral.Neutral mutations are fixed in a population by-
ans- E. the invariance of DNA
(In a population genetics,fixation is the change in a gene pool with atleast two variant of
particular gene or allele).
The oldest living organisms on Earth are plants. Some bristlecone pi.pdf
The oldest living organisms on Earth are plants. Some bristlecone pines are over 4600 years old,
and a desert creosote bush is known to be 10,000 year old. What special feature of plants
provides for this incredible longevity? How do plant differ from animals in their pattern of
growth and development? (also explain how trees grow upward as too how a nail driven into a
10year old tree will be at the same height 10 years later)
Solution
The oldest living organisms on Earth are plants. Some bristlecone pines are about 4,600 years
old, and a desert creosote bush is known to be 10,000 years old. see actually it is just a matter of
definition. no individual cells last that long time and all the original tissue is typically long dead.
We generally think of it as continuing because the new growth is against the old. growth is
continued and new cells are generated due to plants anaomy structure. they maintain their gene
pool.
trees grow upward as too how a nail driven into a 10year old tree will be at the same height 10
years later because tree grows from the top up.
Differences between Plant Growth and Animal Growth:
in plant growth , generally growth involved most of the parts and increase in the number of parts.
growth generally continuw throughout the life of a plant.
growth can be sessional. it depends on the seassion. it can quit the poisonius things during winter
seasons.
Plant growth carries well defined growing regions like stem, leaf.
seedling has no effect in adult plant.
growing pattern differs species to species in case of plant growth.
new parts can be generated around the older ones.
animal growth here growth takes place before the maturity period. after a certain age there is no
growth in body organs.
there is no increase in number of body parts.
growth pattern is absent generally
the young one is generally identical the old or adult one. ( sexual maturity may be an exception).
Suppose there is a person whose total lung capacity (TLC) is 4.90 L, .pdf
Suppose there is a person whose total lung capacity (TLC) is 4.90 L, inspiratory reserve volume
(IRV) is 2.75 L, and tidal volume (TV or V_T) is 0.50 L. Calculate this person\'s functional
residual capacity (FRC). FRC =
Solution
Functinal residual capacity = Total lungs capacity- (Inspiratory reserve volume +Tidal volume)
FRC=4.9- (2.75+0.5)
FRC= 4.9- 3.25
FRC= 1.65 L Answer.
Research the IPv4 ns the IPv6 protocols, then prepare a report that .pdf
Research the IPv4 ns the IPv6 protocols, then prepare a report that explains the likes and
differences of the two protocols and include how and why the transition from IPv4 to IPv6 is to
occur.
Project report body must have 5 pages or more, with Times New Roman12 pt. print double
spaced. Also include a Cover sheet and a Reference page.
Solution
Introductory lines:
The internet protocol or IP specifies and describes the technical details and formats of fragments
of data (Also known as packets.) along with addressing schemes (numerical; actually in binary
but can be interpreted as decimals or hexadecimals assigned to each of the devices connected
through the network) in order to uniquely identify them and create a platform for secure and
reliable communication.
Ipv4 over the years was proved and deployed for communications over the internet. However in
the current days it becomes obsolete and deficient in order to serve the huge and increasing
number of users or network devices.
To make this possible, the IPv6 comes into existence. The IPv6 provides many extra features and
claims to support and serve this increasing number of devices for upcoming coming centuries.
Let’s discuss about the similar and distinct features of both IPv4 and IPv6.
IPv4 in terms of IP addresses, Address allocation, masks and types:
IPv6 in terms of IP addresses, Address allocation, masks and types:
IPv4 in terms of IP header:
Headers in IPv4 are of variable length that varies from 20 to 60 bytes.
IPv4 does not identify packet flow for QOS handling.
It adds checksum filed for identification of transmission errors in the data.
IPv6 in terms of IP header:
Uses fixed length of 40 bytes of headers and there is no concept of option filed in header.
It supports packet flow for QOS by flow level filed.
IPv6 does not implements checksum for error handling.
IPv4 in terms of IP packets ports and security:
IPv4 supports the packet size of 576 bytes.
They also support packet filtering mechanisms through firewalls.
IPv4 supports for port spaces of the range 1 to 65535
IPsec in IPv4 is optional.
IPv6 in terms of IP packets ports and security:
IPv6 supports the packet size of 1280 bytes.
They does not supports packet filtering mechanisms.
The way the portts work both in IPv4 and IPv6 are the same.
IPv6 has inbuilt IPsec support.
IPv4 was successful;;y deployed in 1981 over internet whereas IPv6 was planned to be deployed
from 1999 but could partially deployed till the date.
Why transition from IPv4 to IPv6 must occur ?
The deficiency of IP address of IPv4 type is one of the most eye catching reasons for transition
towards IPv6. Each and every day the Internet is experiencing phenomenal increase of devices
which are being connected and using the internet.
In order to provide services to all the address space of IPv4 seems to be fewer. Hence
a transition from IPv4 to IPv6 is must needed to satisfy such demand of larger address space.
Except this, the other technical features like improve.
Species I Species II Species III Species IV Ancestral sequence IV Sit.pdf
Species I Species II Species III Species IV Ancestral sequence IV Site 2 3 4 C G A T C T T C A
T A C A G T T IV IV
Solution
Phylogenetic tree also called evolutionary tree is a branched chain diagram that indicates
evolutionary relationship among various species and groups of organisms. Phylogenetic trees
represent the hypothetical evolution.
The most parsimonious phylogenetic tree is E.
Solve the differential equation y+y=sin(x). Im thinking I ne.pdf
Solve the differential equation y\'\'+y\'=sin(x). I\'m thinking I need to take the integral of both
sides in order to get it in the proper form, but i really don\'t know. Any help is greatly
appreciated, thanks.
Solution
Ya integrate both sides => dy/dx + y = -cosx again integrate =>y+y^2/2 = -sinx + c.
Please write a non-recursive procedure to print all nodes for a bina.pdf
Please write a non-recursive procedure to print all nodes for a binary tree in a level order
sequence.
Solution
#include
#include
struct node
{
int data;
struct node* left, *right;
};
void printGivenLevel(struct node* root, int level);
int height(struct node* node);
struct node* newNode(int data);
void printLevelOrder(struct node* root)
{
int h = height(root);
int i;
for (i=1; i<=h; i++)
printGivenLevel(root, i);
}
void printGivenLevel(struct node* root, int level)
{
if (root == NULL)
return;
if (level == 1)
printf(\"%d \", root->data);
else if (level > 1)
{
printGivenLevel(root->left, level-1);
printGivenLevel(root->right, level-1);
}
}
int height(struct node* node)
{
if (node==NULL)
return 0;
else
{
int lheight = height(node->left);
int rheight = height(node->right);
if (lheight > rheight)
return(lheight+1);
else return(rheight+1);
}
}
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
printf(\"Level Order traversal of binary tree is \ \");
printLevelOrder(root);
return 0;
}.
Please do C, D, and E For the given f Z right arrow Z, decide wheth.pdf
Please do C, D, and E For the given f: Z right arrow Z, decide whether f is onto and whether it is
one-to-one. Prove that your decisions are correct.
Solution
c) onto as range of f(x) is Z and f(x) = odd intergers when x is even , f(x) = even intergers when
x is odd
Thus even integers + odd integers + 0 (when x is -1) = Z
one to one as no element in x will give repeatable values as functions are increasing functions
d) onto as x/2 covers whole range of intergers
not one to one as as only x/2 covers the whole range (when x is even ) , repeated values will
come when x is odd
e) not onto as only multiples of 3 and 2 are getting in the range
not one to one as it will be get repeated many times ( as in case 6 (for x=2) and 6 (for x=3)).
need help with code I wrote. This code is a maze gui, and i need hel.pdf
need help with code I wrote. This code is a maze gui, and i need help with the method
shorterstpath(); here is my code:
******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private GraphicsContext gc;
// boundary of drawing canvas, 0% border
// private static final double BORDER = 0.05;
private static final double BORDER = 0.00;
private double xmin, ymin, xmax, ymax;
public static void main(String[] args)
{
launch(args);
}
@Override
public void start(Stage primaryStage)
{
Group root = new Group();
Canvas canvas = new Canvas(width, height);
gc = canvas.getGraphicsContext2D();
gc.setLineWidth(2);
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
root.getChildren().add(canvas);
TextInputDialog tid = new TextInputDialog();
tid.setTitle(\"Maze Size\");
tid.setHeaderText(\"Enter maze size between 10 and 50\");
tid.showAndWait();
int size = Integer.parseInt(tid.getResult());
if (size > 50)
size = 50;
if (size < 10)
size = 10;
primaryStage.setTitle(\"Maze Application\");
primaryStage.setScene(new Scene(root));
primaryStage.setResizable(false);
// Make sure that the application goes away when then window is closed
primaryStage.setOnCloseRequest(e -> System.exit(0));
primaryStage.show();
Maze maze = new Maze(this, size);
// Must solve the maze in a separate thread or else
// the GUI wont update until the end.....
Thread solver = new Thread(
new Runnable () {
public void run()
{
while(true)
{
maze.buildAndDrawMaze();
maze.findShortestPath();
try
{
Thread.sleep(5000);
}
catch(Exception e) { }
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
}
}
});
solver.start();
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setPenColor(Color color) {
gc.setStroke(color);
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setFillColor(Color color) {
gc.setFill(color);
}
/**
* Sets the x-scale to the specified range.
*
* @param min the minimum value of the x-scale
* @param max the maximum value of the x-scale
* @throws IllegalArgumentException if {@code (max == min)}
*/
public void setXscale(double min, double max) {
double size = max - min;
if (size == 0.0) {
throw new IllegalArgumentException(\"the min and max are the same\");
}
xmin = min - BORDER * size;
xmax = max + BORDER * size;
}
/**
* Sets the y-scale to the specified range.
*
* @param min the minimum valu.
Microbiology question Which of the following is the most unique a.pdf
Microbiology question :
Which of the following is the most unique aspect of archaeon cell envelopes?
a) The plasma membrane is composed of phospholipids.
b) The cell walls contain a modified type of peptidoglycan.
c) Archaeon capsules can be composed of protein or polysaccharide.
d) Some archaeons have lipid monolayers in their plasma membrane.
Solution
The distinct features is
Some archeons have a lipid monolayer in their plasma memberane..!!!.
Is Event B dependent or independent of Event A A green ball is draw.pdf
Is Event B dependent or independent of Event A? A green ball is drawn from a box with five
balls and placed next to the box. B: A red ball is drawn next and placed next to the green one.
Please provide detail.
Solution
DEPENDENT
because B is an event that depend of A for the text that say is drawn next and placed next to the
green one.
The geography of Taylor Swift - some ideas
Geographical themes connected with Taylor Swift's ERAS tour - coming to the UK in June 2024
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
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You are presented with a patient, named Gerry, who has Down syndrome .pdf
You are presented with a patient, named Gerry, who has Down syndrome (trisomy 21). You
determine that the nondisjunction event occurred in his mother, named Mary, during meiosis I.
You are interested in the TPTE gene, which is near the centromere of chromosome 21. Virtually
no recombination occurs between the centromere and the TPTE gene. Mary has two different
alleles of the TPTE gene. Would Gerry\'s two chromosome 21s from his mother have the same
or different alleles of the TPTE gene? Or do you not have enough information to decide?
Explain your answer by drawing meiosis I and II for Mary\'s chromosome 21. Use \"A\" for one
allele and \"a\" for the other allele of the TPTE gene. Only draw the chromosomes relevant to
this question (Do not draw all 46 chromosomes).
Solution
a. Nondisjunction produces abnormal gametes. Nondisjunction in meiosis I occurs during
anaphase I when one pair of homologous chromosomes fails to separate.Trisomy is the condition
of having 3 copies of one chromosome type. It is designated as 2n + 1 because the cell has the
normal two sets of each 23 types of chromosomes plus an extra copy of one chromosome.
I think Gerrys two chromosome 21s from his mother have different alleles (A and a) of the TPTE
gene. But more information could be needed to explain and understand this in a much better way..
Write a template function to fill in a STL list with a given array..pdf
Write a template function to fill in a STL list with a given array.
Solution
Hi, I am not able to see the function prototype completely.
But I wrote a function that takes a template array and adding all elements of array in STL list.
Please let me know in case of any issue.
template
void fillArray(T arr[], int size) {
// creating STL list of type T
std::list myList;
// adding each element from arr in list
for(int i=0; i.
write a method to generate AVL tree of height h wth fewest nodes and.pdf
write a method to generate AVL tree of height h wth fewest nodes and what is the running time
of your method
Solution
class Node {
int key, height;
Node left, right;
Node(int d) {
key = d;
height = 1;
}
}
class AVLTree {
static Node root;
// A utility function to get height of the tree
int height(Node N) {
if (N == null) {
return 0;
}
return N.height;
}
// A utility function to get maximum of two integers
int max(int a, int b) {
return (a > b) ? a : b;
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
Node rightRotate(Node y) {
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left), height(y.right)) + 1;
x.height = max(height(x.left), height(x.right)) + 1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
Node leftRotate(Node x) {
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left), height(x.right)) + 1;
y.height = max(height(y.left), height(y.right)) + 1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(Node N) {
if (N == null) {
return 0;
}
return height(N.left) - height(N.right);
}
Node insert(Node node, int key) {
/* 1. Perform the normal BST rotation */
if (node == null) {
return (new Node(key));
}
if (key < node.key) {
node.left = insert(node.left, key);
} else {
node.right = insert(node.right, key);
}
/* 2. Update height of this ancestor node */
node.height = max(height(node.left), height(node.right)) + 1;
/* 3. Get the balance factor of this ancestor node to check whether
this node became unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && key < node.left.key) {
return rightRotate(node);
}
// Right Right Case
if (balance < -1 && key > node.right.key) {
return leftRotate(node);
}
// Left Right Case
if (balance > 1 && key > node.left.key) {
node.left = leftRotate(node.left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node.right.key) {
node.right = rightRotate(node.right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder traversal of the tree.
// The function also prints height of every node
void preOrder(Node node) {
if (node != null) {
System.out.print(node.key + \" \");
preOrder(node.left);
preOrder(node.right);
}
}
public static void main(String[] args) {
AVLTree tree = new AVLTree();
/* Constructing a tree with random values */
root = tree.insert(root, 10);
root = tree.insert(root, 20);
root = tree.insert(root, 30);
root = tree.insert(root, 40);
root = tree.insert(root, 50);
root = tree.insert(root, 25);
System.out.println(\"The preorder traversal of constructed tree is : \");
tree.preOrder(root);
}
}
The running time of this method is O(h)..
Which of the following is NOT correctWhich of the following is NO.pdf
Which of the following is NOT correct?
Which of the following is NOT correct?-Protozoans are fairly susceptible to antimicrobial
controls, but their cysts are resistant.-Naked viruses are more susceptible to antimicrobial
control agents than are enveloped viruses.-The antimicrobial effect of heat is enhanced by acidic
conditions.-Warm disinfectants generally work better than cool ones.
Solution
The answer is -Naked viruses are more susceptible to antimicrobial control agents than are
enveloped viruses..
What is Mills response to the problem posed by some peoples desi.pdf
What is Mill\'s response to the problem posed by some people\'s desire for virtue for its own
sake?Tthis is except from Utilitarianism.
Solution
John Stuart Mill wrote his moral treatise, Utilitarianism, in a period in intellectual history
dominated by Aristotelian virtue and Kantian deontology, as well as the intensely conservative
political and social climate of Victorian England.Utilitarianism was intended to convince the
public of Mill’s first principle. Mill responds to preexisting ethical frameworks, namely virtue
ethics, deontology, Christian ethics, and egoism. Except for egoism, Mill invariably shows how
other moral theories and values are actually or can become part of his first principle. The most
prominent response in the text is his amalgamation of virtue ethics, which spans all of Chapter 4.
Mill is successful in his synthesis of virtue ethics, despite philosophical sloppiness and logical
fallacies. In his success, Mill crafted an impartial standard which the moral claims of other
theories can by compared against.
It is in Chapter 4 of Utilitarianism that Mill addresses virtue the most directly. He begins his
argument by stating that virtue can actually be an end to human action besides happiness. Mill
agrees with his opponents that there are people who desire or approve of acts according to virtue
and absence of vice instead of the pleasure and the absence of pain. The reason this is significant
is because if this is a descriptive fact, a truth, then it would pose a problem for Mill’s inductive
proof of Utility, which hinges on the observation that happiness is desired and therefore
desirable. If there are people who genuinely value virtue as the criteria of right and wrong, of
how one should act, then, at very least, happiness is not the only first principle, which is what the
Utilitarian claims. The solution to this problem for utilitarianism is to either swallow the
intuitively unsavory conclusion that virtue is without value, or show that virtue actually has
value belonging to, part of, or from happiness.
Mill states that where opponents make their mistake is in believing that utility is incompatible
with virtue. Mill holds that virtue is not only valuable as a means to pleasure, but in fact as an
end in itself. Virtue and happiness are perceived to be different, but Mill attempts to show that
this is superficial. It is by seeking virtue disinterestedly and for its own sake that virtue can
become part of pleasure, Mill claims.
So not only is Mill successful in bringing virtue ethics into utilitarianism, but he’s successful in
creating a template by which other theories can be synthesized with utility. Mill’s claim is that
all moral sentiments stem from a first principle that does not admit of proof, and that that first
principle is the concept of utility. By providing one specific example, virtue, and showing how it
actually admits of his general concept, utility, Mill is successful crafting an argument which
allows people .
What is a bivariate distributionSolutionBivariate distributio.pdf
What is a bivariate distribution?
Solution
Bivariate distribution is a two dimensional random variables distribution.
When the two random variables X and Y are discrete, the joint probability
mass function is defined as P(x,y) = P(X=x, Y=y).
When the two random variables X and Y are continuous, the joint probability
density function is given by f(x,y).
What are some of the tools you can use to tune or correct a Linux fi.pdf
What are some of the tools you can use to tune or correct a Linux filesystem?
Solution
Some of the tools are Process Activity Command(it gives the real time view of system and we
can keep an eye to factors which are affecting the performance.iptraf which gives the network
statistics and other one could be /proc file system which gives the detailed information about
harware devices connected and kernel information. Nagios is an open source system which give
complete control over all the information like host, network traffic , equipments and it will send
alert when something goes wrong. Another one could be Cacti which is web based monitoring
tool, its a network graphing tool which gives the statistics in the form of graphs.
TrueFalse Insulin spikes in the postabsorptive state.Solution.pdf
True/False: Insulin spikes in the postabsorptive state.
Solution
This statement is false. Glucagon stimulates and insulin suppresses the hydrolysis of liver
glycogen, or glycogenolysis. Thus during times of fasting (post absorptive state) , when glucagon
secretion is high and insulin secretion is low, liver glycogen is used as a source of additional
blood glucose..
The principal role of ATP In the cell is A) serving as structural s.pdf
The principal role of ATP In the cell is: A) serving as structural support. B) catalyzing
chemical reactions. C) keeping the phospholipid tails of membranes aligned D) providing
energy. E) functioning as a hormone. Small nonpolar molecules, such as O_2 and CO_2, enter a
cell by: A) vesicle-mediated transport. B) simple diffusion. C) osmosis. D) facilitated
diffusion. E) active transport. Eukaryotic cells differ from prokaryotic cells in that eukaryotic
cells have: A) plasma membrance B) genetic material C) cytoplasm. D) ribosomes. E)
nucleus. If a plant cell is placed in an aqueous solution that is hypertonic to the cell, the cell A)
build up turgor pressure B) build up wall pressure C) undergo plasmolysis D) neither gain nor
lose water E) swell and perhaps burst
Solution
15 . D Providing energy
Atp or adenosine tri phosphate is a energy rich molecule produced by mitochondria of our cell.
These atp molecules are produced by oxidation of organic molecules, (glucose) which is then
utilised by our body for various cellular functions..
The neutral theory of evolution proposed that most genetic mutations .pdf
The neutral theory of evolution proposed that most genetic mutations are functionally neutral.
Neutral mutations are fixed in a population by natural selection random genetic drift
environmental selection asexual reproduction the invariance of DNA
Solution
The neutra theory of evouloution proposed that most genetic mutations are functionally
neutral.Neutral mutations are fixed in a population by-
ans- E. the invariance of DNA
(In a population genetics,fixation is the change in a gene pool with atleast two variant of
particular gene or allele).
The oldest living organisms on Earth are plants. Some bristlecone pi.pdf
The oldest living organisms on Earth are plants. Some bristlecone pines are over 4600 years old,
and a desert creosote bush is known to be 10,000 year old. What special feature of plants
provides for this incredible longevity? How do plant differ from animals in their pattern of
growth and development? (also explain how trees grow upward as too how a nail driven into a
10year old tree will be at the same height 10 years later)
Solution
The oldest living organisms on Earth are plants. Some bristlecone pines are about 4,600 years
old, and a desert creosote bush is known to be 10,000 years old. see actually it is just a matter of
definition. no individual cells last that long time and all the original tissue is typically long dead.
We generally think of it as continuing because the new growth is against the old. growth is
continued and new cells are generated due to plants anaomy structure. they maintain their gene
pool.
trees grow upward as too how a nail driven into a 10year old tree will be at the same height 10
years later because tree grows from the top up.
Differences between Plant Growth and Animal Growth:
in plant growth , generally growth involved most of the parts and increase in the number of parts.
growth generally continuw throughout the life of a plant.
growth can be sessional. it depends on the seassion. it can quit the poisonius things during winter
seasons.
Plant growth carries well defined growing regions like stem, leaf.
seedling has no effect in adult plant.
growing pattern differs species to species in case of plant growth.
new parts can be generated around the older ones.
animal growth here growth takes place before the maturity period. after a certain age there is no
growth in body organs.
there is no increase in number of body parts.
growth pattern is absent generally
the young one is generally identical the old or adult one. ( sexual maturity may be an exception).
Suppose there is a person whose total lung capacity (TLC) is 4.90 L, .pdf
Suppose there is a person whose total lung capacity (TLC) is 4.90 L, inspiratory reserve volume
(IRV) is 2.75 L, and tidal volume (TV or V_T) is 0.50 L. Calculate this person\'s functional
residual capacity (FRC). FRC =
Solution
Functinal residual capacity = Total lungs capacity- (Inspiratory reserve volume +Tidal volume)
FRC=4.9- (2.75+0.5)
FRC= 4.9- 3.25
FRC= 1.65 L Answer.
Research the IPv4 ns the IPv6 protocols, then prepare a report that .pdf
Research the IPv4 ns the IPv6 protocols, then prepare a report that explains the likes and
differences of the two protocols and include how and why the transition from IPv4 to IPv6 is to
occur.
Project report body must have 5 pages or more, with Times New Roman12 pt. print double
spaced. Also include a Cover sheet and a Reference page.
Solution
Introductory lines:
The internet protocol or IP specifies and describes the technical details and formats of fragments
of data (Also known as packets.) along with addressing schemes (numerical; actually in binary
but can be interpreted as decimals or hexadecimals assigned to each of the devices connected
through the network) in order to uniquely identify them and create a platform for secure and
reliable communication.
Ipv4 over the years was proved and deployed for communications over the internet. However in
the current days it becomes obsolete and deficient in order to serve the huge and increasing
number of users or network devices.
To make this possible, the IPv6 comes into existence. The IPv6 provides many extra features and
claims to support and serve this increasing number of devices for upcoming coming centuries.
Let’s discuss about the similar and distinct features of both IPv4 and IPv6.
IPv4 in terms of IP addresses, Address allocation, masks and types:
IPv6 in terms of IP addresses, Address allocation, masks and types:
IPv4 in terms of IP header:
Headers in IPv4 are of variable length that varies from 20 to 60 bytes.
IPv4 does not identify packet flow for QOS handling.
It adds checksum filed for identification of transmission errors in the data.
IPv6 in terms of IP header:
Uses fixed length of 40 bytes of headers and there is no concept of option filed in header.
It supports packet flow for QOS by flow level filed.
IPv6 does not implements checksum for error handling.
IPv4 in terms of IP packets ports and security:
IPv4 supports the packet size of 576 bytes.
They also support packet filtering mechanisms through firewalls.
IPv4 supports for port spaces of the range 1 to 65535
IPsec in IPv4 is optional.
IPv6 in terms of IP packets ports and security:
IPv6 supports the packet size of 1280 bytes.
They does not supports packet filtering mechanisms.
The way the portts work both in IPv4 and IPv6 are the same.
IPv6 has inbuilt IPsec support.
IPv4 was successful;;y deployed in 1981 over internet whereas IPv6 was planned to be deployed
from 1999 but could partially deployed till the date.
Why transition from IPv4 to IPv6 must occur ?
The deficiency of IP address of IPv4 type is one of the most eye catching reasons for transition
towards IPv6. Each and every day the Internet is experiencing phenomenal increase of devices
which are being connected and using the internet.
In order to provide services to all the address space of IPv4 seems to be fewer. Hence
a transition from IPv4 to IPv6 is must needed to satisfy such demand of larger address space.
Except this, the other technical features like improve.
Species I Species II Species III Species IV Ancestral sequence IV Sit.pdf
Species I Species II Species III Species IV Ancestral sequence IV Site 2 3 4 C G A T C T T C A
T A C A G T T IV IV
Solution
Phylogenetic tree also called evolutionary tree is a branched chain diagram that indicates
evolutionary relationship among various species and groups of organisms. Phylogenetic trees
represent the hypothetical evolution.
The most parsimonious phylogenetic tree is E.
Solve the differential equation y+y=sin(x). Im thinking I ne.pdf
Solve the differential equation y\'\'+y\'=sin(x). I\'m thinking I need to take the integral of both
sides in order to get it in the proper form, but i really don\'t know. Any help is greatly
appreciated, thanks.
Solution
Ya integrate both sides => dy/dx + y = -cosx again integrate =>y+y^2/2 = -sinx + c.
Please write a non-recursive procedure to print all nodes for a bina.pdf
Please write a non-recursive procedure to print all nodes for a binary tree in a level order
sequence.
Solution
#include
#include
struct node
{
int data;
struct node* left, *right;
};
void printGivenLevel(struct node* root, int level);
int height(struct node* node);
struct node* newNode(int data);
void printLevelOrder(struct node* root)
{
int h = height(root);
int i;
for (i=1; i<=h; i++)
printGivenLevel(root, i);
}
void printGivenLevel(struct node* root, int level)
{
if (root == NULL)
return;
if (level == 1)
printf(\"%d \", root->data);
else if (level > 1)
{
printGivenLevel(root->left, level-1);
printGivenLevel(root->right, level-1);
}
}
int height(struct node* node)
{
if (node==NULL)
return 0;
else
{
int lheight = height(node->left);
int rheight = height(node->right);
if (lheight > rheight)
return(lheight+1);
else return(rheight+1);
}
}
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
printf(\"Level Order traversal of binary tree is \ \");
printLevelOrder(root);
return 0;
}.
Please do C, D, and E For the given f Z right arrow Z, decide wheth.pdf
Please do C, D, and E For the given f: Z right arrow Z, decide whether f is onto and whether it is
one-to-one. Prove that your decisions are correct.
Solution
c) onto as range of f(x) is Z and f(x) = odd intergers when x is even , f(x) = even intergers when
x is odd
Thus even integers + odd integers + 0 (when x is -1) = Z
one to one as no element in x will give repeatable values as functions are increasing functions
d) onto as x/2 covers whole range of intergers
not one to one as as only x/2 covers the whole range (when x is even ) , repeated values will
come when x is odd
e) not onto as only multiples of 3 and 2 are getting in the range
not one to one as it will be get repeated many times ( as in case 6 (for x=2) and 6 (for x=3)).
need help with code I wrote. This code is a maze gui, and i need hel.pdf
need help with code I wrote. This code is a maze gui, and i need help with the method
shorterstpath(); here is my code:
******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private GraphicsContext gc;
// boundary of drawing canvas, 0% border
// private static final double BORDER = 0.05;
private static final double BORDER = 0.00;
private double xmin, ymin, xmax, ymax;
public static void main(String[] args)
{
launch(args);
}
@Override
public void start(Stage primaryStage)
{
Group root = new Group();
Canvas canvas = new Canvas(width, height);
gc = canvas.getGraphicsContext2D();
gc.setLineWidth(2);
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
root.getChildren().add(canvas);
TextInputDialog tid = new TextInputDialog();
tid.setTitle(\"Maze Size\");
tid.setHeaderText(\"Enter maze size between 10 and 50\");
tid.showAndWait();
int size = Integer.parseInt(tid.getResult());
if (size > 50)
size = 50;
if (size < 10)
size = 10;
primaryStage.setTitle(\"Maze Application\");
primaryStage.setScene(new Scene(root));
primaryStage.setResizable(false);
// Make sure that the application goes away when then window is closed
primaryStage.setOnCloseRequest(e -> System.exit(0));
primaryStage.show();
Maze maze = new Maze(this, size);
// Must solve the maze in a separate thread or else
// the GUI wont update until the end.....
Thread solver = new Thread(
new Runnable () {
public void run()
{
while(true)
{
maze.buildAndDrawMaze();
maze.findShortestPath();
try
{
Thread.sleep(5000);
}
catch(Exception e) { }
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
}
}
});
solver.start();
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setPenColor(Color color) {
gc.setStroke(color);
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setFillColor(Color color) {
gc.setFill(color);
}
/**
* Sets the x-scale to the specified range.
*
* @param min the minimum value of the x-scale
* @param max the maximum value of the x-scale
* @throws IllegalArgumentException if {@code (max == min)}
*/
public void setXscale(double min, double max) {
double size = max - min;
if (size == 0.0) {
throw new IllegalArgumentException(\"the min and max are the same\");
}
xmin = min - BORDER * size;
xmax = max + BORDER * size;
}
/**
* Sets the y-scale to the specified range.
*
* @param min the minimum valu.
Microbiology question Which of the following is the most unique a.pdf
Microbiology question :
Which of the following is the most unique aspect of archaeon cell envelopes?
a) The plasma membrane is composed of phospholipids.
b) The cell walls contain a modified type of peptidoglycan.
c) Archaeon capsules can be composed of protein or polysaccharide.
d) Some archaeons have lipid monolayers in their plasma membrane.
Solution
The distinct features is
Some archeons have a lipid monolayer in their plasma memberane..!!!.
Is Event B dependent or independent of Event A A green ball is draw.pdf
Is Event B dependent or independent of Event A? A green ball is drawn from a box with five
balls and placed next to the box. B: A red ball is drawn next and placed next to the green one.
Please provide detail.
Solution
DEPENDENT
because B is an event that depend of A for the text that say is drawn next and placed next to the
green one.
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