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A set of rules or specified instructions required for performing a task or solving a problem. This includes written code as well as other bits of information working together in an attempt to get you what you’re looking for. A social media newsfeed, for example, has an algorithm designed to display content that is interesting to you as the viewer and determines how far your own content reaches when shared.
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The full Question...
Suppose 3 algorithms are used to perform the same task for a certain number of cycles. Algorithm A completes 3 cycles in one minute. Each of Algorithm B and Algorithm C respectively completes 4 and 5 cycles per minute. What is the shortest time required for each Algorithm to complete the same number of cycles?
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Bresham's Mid point circle drawing algorithm.
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algorithm on multiple queues in a single dimensional array
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algorithm to convert a number representing radix r1 to radix r2
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are sahi se ans do dia karo.My ques is how many bits are required for 16-bit system in booth's algorithm?Pls reply me soon
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The easiest way to write a generic algorithm is to simply use the quadratic formula. If it is a computer program, ask the user for the coefficients a, b, and c of the generic equation ax2 + bx + c = 0, then just replace them in the quadratic formula.
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algorithim and flow chart of odd number 1 and 50
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read the value A,b
store the result of subtraction of a,b in continer subtraction
display subtraction
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Algorithms use the language of mathematics, logic and computers. They are independent of any particular natural language like English or Chinese.
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- Hill Cipher is a cryptographic algorithm to encrypt and decrypt an alphabetic text. In this cipher, each letter is represented by a number (eg. A = 0, B = 1, C = 2). To encrypt a message, each block of n letters (considered as an n -component vector) is multiplied by an invertible n × n matrix, against modulus 26.
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move from, to, spare, count:
move from, spare, count-1
single_move from, to
move spare, to, count-1
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int height;
print("Enter height");
height=getString();
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int n = 2 // start at 0 if you really want to
while n < some arbitrary limit
n = n + 2
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sum = 0
for(n = 0; n <= 10; n += 2) sum += n;
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Yes. At least, that is one way of presenting an algorithm. Other ways include a verbal description, and pseudocode.
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It's basically the algorithm to break a number down to digits. Then each digit would be represented as a natural binary number on its own.
To break it down to digits, get a remainder after division with 10 (that's the last digit), then divide the number by 10 to remove that digit. Repeat until there are digits, and store each one into an array (or print them to screen without storing).
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The algorithm for bubble sort, also know as pair exchange...
- Set a swap flag false
- Loop for the first N-1 elements
- Compare each element with the following element
- If the two elements are in the correct order, continue to next loop iteration
- Otherwise, swap the two elements and set the swap flag true
- At loop end, if the swap flag is true, repeat starting at step 1
- Otherwise, sort completed
Bubble sort is so named because out of order elements "bubble" to the end of the array, moving one step per inner loop iteration.
As stated above, the algorithm is slow because it takes a while for a significantly out of order element to reach its final point when it needs to come closer to the beginning of the array. It can be improved by introducing a "distance" parameter, initially set to one half of the array size, and using that distance in step 3 to choose the second element. After the algorithm is completed for that distance, the distance is halved, and we iterate the entire algorithm until the distance is only one. (This variation is more formally known as merge exchange, but it still retains the "bubble" characteristic.)
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One way to begin evaluating problems in algebra and calculus is to test the values for the variables that should not be there. This is known as the "false position method".
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It's an algorithm to find the spanning tree in decision maths. The method is:
- list weights in ascending order from smallest weight to largest weight.
- then starting form the beginning of your list, tick the weights you want to use and cross out the ones which make a cycle. do this till you've reached all the nods then cross out the rest.
- add up the weights you used to give you the total weight of the spanning tree.
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The question is impossible to answer. You cannot determine an algorithm from a given program if the program is not actually given in the question. Please place the program in the discussion section.
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// returns n!
int fact(int n) {
int f_n = 1;
for(int i = n; i > 1; --i) {
f_n *= n;
}
return f_n;
}
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// Author : SAGAR T.U, PESIT
#define TRUE 1
#define FALSE 0
int isStrictBinaryTree (struct tree * n)
{
if( n NULL ) return TRUE;
return FALSE;
}
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transpose(Matrix mat,int rows, int cols ){ //construction step Matrix tmat; for(int i=0;i<rows;i++){ for(int j=0;j<cols;j++){ tmat[j][i] = mat[i][j]; } } }
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Algorithm to Convert Infix to Prefix Form
Suppose A is an arithmetic expression written in infix form. The algorithm finds equivalent prefix expression B.
Step 1. Push ")" onto STACK, and add "(" to end of the A
Step 2. Scan A from right to left and repeat step 3 to 6 for each element of A until the STACK is empty
Step 3. If an operand is encountered add it to B
Step 4. If a right parenthesis is encountered push it onto STACK
Step 5. If an operator is encountered then:
a. Repeatedly pop from STACK and add to B each operator (on the top of STACK) which has same or higher precedence than the operator.
b. Add operator to STACK
Step 6. If left parenthesis is encontered then
a. Repeatedly pop from the STACK and add to B (each operator on top of stack until a left parenthesis is encounterd)
b. Remove the left parenthesis
Step 7. Exit
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1. If TOP = MAXSTK THEN [Stack already filled?]
Print "OVERFLOW"
Go to step 4
Endif
2. TOP = TOP + 1 [Increase TOP by 1]
3. Set ST[STOP] = ITEM [Insert ITEM in new TOP position]
4. End
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- Matrix-Addition(a,b)
- 1 for i =1 to rows [a]
- 2 for j =1 to columns[a]
- 3 Input a[i,j];
- 4 Input b[i,j];
- 5 C[i, j] = A[i, j] + B[i, j];
- 6 Display C[i,j];
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enter the number whose digits are to be added
num is the given value
num=0!
k=num%10
sum=sum=k
k=num/10
num=k
print the sum of the digits
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Please use the discussion area to state your question in English.
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1.start
2.n=1,s=0
3 compute s=s+n
4 compute n=n+1
5 check n<=10 go to step3 else go to step 7
6 display s
7 stop
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The only way to delete objects in an object oriented programming language (unless they were created in heap memory) is for the object to go out of scope. If the object is declared in the heap, in c++ you would use delete[] ptr; or delete ptr; where ptr is a pointer to your object.
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You sort a doubly linked list the same way you sort any other kind of list or array.
You implement a procedure to sort the list or array, and that procedure calls the appropriate insert, delete, or move methods of the list or array.
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Lxi h,xx50
mov a,m
inx h
mov b,m
mvi ,ffh
inr c
sub b
jnc xx08
add b
inx h
mov m,c
inx h
mov m,a
hlt
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def digits(x):
""" Return amount of digits of x. """
y = math.log10(x)
if y % 1 0:
return int(y)
else:
return int(math.floor(y) + 1)
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step 1:start
step 2:input the values of a,b,c
step 3:if(a>b&&a>c) max=a
step 4:else if(b>c) max=b
step 5:else max=c
step 6:output:max
step 7:stop
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I am assuming you mean positive integer in decimal notation.
Here is an algorithm that lists the binary representation backwards, that is
from right to left, least significant digit first.
N is the number decimal number
While N > 0
print N rem 2 // output the remainder
N = N/2 // replace N by n divided by 2
end loop
Example: trace for n = 101
n output
101
50 1
25 0
12 1
6 0
3 0
1 1 -- Last one had 0; 3%2 is 1
0 1
Output would be 1010011 ---- from left to right would be 1100101
Check: 64 + 32 + 0 + 0 + 4 + 1 = 101.
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factorial using recursion style in c++ is
unsigned int fact(unsigned int a)
{
if (a<=1)
return 1;
else
{
f*=fact(a-1);
return a;
}
}
when using looping structure factorial is
unsigned int fact (unsigned int n)
{
unsigned int i,f=1;
for(i=1;i<=n;i++)
f*=i ;
return f;
}
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Algorithm
Step1: Read A, B, C
Step2: If A > B is True, then check whether A > C, if yes then A is greatest otherwise C is greatest
Step3: If A > B is False, then check whether B > C, if yes then B is greatest otherwise C is greatest
Following these steps flowchart can be made.
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The belt-and-braces technique is easy enough:
>
> prefix_to_infix(stream, stack)
> if stack is not empty
> pop a node off the stack
> if this node represents an operator
> write an opening parenthesis to stream
> prefix_to_infix(stream, stack)
> write operator to stream
> prefix_to_infix(stream, stack)
> write a closing parenthesis to stream
> else
> write value to stream
> endif
> endif
> endfunc
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#include<stdio.h>
#include<conio.h>
int fib(int a);
main()
{
int a;
clrscr();
scanf("%d",&a);
for(int i=0;i<a;i++)
printf("%d\n",fib(i));
}
int fib(int a)
{
if(a==0)
return 0;
if(a==1)
return 1;
else
return (fib(a-1)+fib(a-2));
}
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To delete an node in a linked list...
- If the head node is empty, stop, and return failed.
- Interate thorugh each member of the list.
- Remember at each iteration, the address of the privious node.
- If the current node is the head node or not the desired node, continue to the next iteration.
- Move the current node's pointer value to the previous node's pointer.
- Delete the current node, and return success.
- Return failed..
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The hexadecimal notation is base-16, so for small numbers, arrange them right to left as powers of 16 (units 0 to 9 and A through F representing 10, 11, 12, 13, 14, and 15).
Examples: 21 = 15 hex (16+5) and 31 = 1F (16+15)
For larger numbers, see the process at the related link below.
Here is a program:
#include
#include
void main()
{
int n;
clrscr();
printf("Enter Decimal Number: ");
scanf("%d",&n);
printf("Hexadecimal value is: %x",n);
getch();
}
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What is FIFO algorithm?
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int linearSearch(int a[], int first, int last, int key) { // function: // Searches a[first]..a[last] for key. // returns: index of the matching element if it finds key, // otherwise -1. // parameters: // a in array of (possibly unsorted) values. // first, last in lower and upper subscript bounds // key in value to search for. // returns: // index of key, or -1 if key is not in the array. for (int i=first; i<=last; i++) { if (key == a[i]) { return i; } } return -1; // failed to find key }
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Delete Front----
DQDELETE_FRONT(QUEUE, FRONT, REAR, ITEM)
1. [check for queue underflow]
If FRONT<0, Then: Print: "Queue is empty", and Return.
2. ITEM = QUEUE[FRONT];
3. Set FRONT = FRONT + 1.
4. Return.
Delete Rear----
DQDELETE_REAR(QUEUE, REAR, FRONT, ITEM)
1. [check for queue underflow]
If REAR<0, Then: Print: "Queue is empty", and Return.
2. ITEM = QUEUE[REAR].
3. Set REAR = REAR - 1.
4.Return.
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There are many ways of checking for a complete binary tree. Here is one method:
1. Do a level order traversal of the tree and store the data in an array
2. If you encounter a nullnode, store a special flag value.
3. Keep track of the last non-null node data stored in the array - lastvalue
4. Now after the level order traversal, traverse this array up to the index lastvalue and check whether the flag value is encountered. If yes, then it is not a complete binary tree, otherwise it is a complete binary tree.
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