Answer:
d so if he comes back any time soon
Explanation:
The program which would help Eva convert a Lisp file into machine code by interpreting only a single line of code at a time is: A. interpreters
Interpreters can be defined as computer software programs that are designed to translate (interpret) a programming language (code) into machine code, especially by interpreting only a single line of code at a time.
Hence, an interpreter executes the instructions that are written in a programming language (code) one after the other (instruction by instruction), before it translate (interpret) the next line of code.
On the other hand, a compiler takes the entire program (code) and interprets them.
In conclusion, an interpreter can help Eva convert a Lisp file into machine code by interpreting only a single line of code at a time.
Rea d more: brainly.com/question/21130620
Answer:
Write out your birthday in the following format: M/DD/YY. For example, if your birthday is on June 11, 2013, it would be written as 6/11/13.
2. Convert the birthday date to binary format. Using our same example from above, 6/11/13 translated into binary code would be: 110/1011/1101.
3. Select one color of bead to represent “0”, a second color to represent “1” and then the third color to represent a space (/) between the numbers.
4. Layout the beads to represent your birthdate in binary code. Don’t forget the third color for the spaces in between the numbers!
5. Once laid out, string all the beads on to the string or pipe cleaner.
6. Tie a knot around the ends and enjoy your one-of-a kind masterpiece as a piece of jewelry or a bag tag….the options are endless!
Answer: it depends on the font size
Explanation:
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
