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What could be the recurrence relation that function_?
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$begingroup$
float aFunc(nyArray, n)
if ( (n==1)
return nyArray[0];
// let nyArray1,nyArray2,nyArray3,nyArray4 be predefined arrarys
for (i= 0; i<= (n/2) -1 ; i++)
for (j =0; j<= (n/2)-1 ; i++)
nyArray1[i]= nyArray[i];
nyArray2[i]= nyArray[i+j];
nyArray3[i]= nyArray[n/2+j];
nyArray4[i]= nyArray[j];
x1 = aFunc(nyArray1,n/2);
x2 = aFunc(nyArray2,n/2);
x3 = aFunc(nyArray3,n/2);
x4 = aFunc(nyArray4,n/2);
return x1*x2*x3*x4;
I guess that its something like that, but Im not suce.
T(n) = 4T(n/2)+........
complex-analysis algorithms
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
float aFunc(nyArray, n)
if ( (n==1)
return nyArray[0];
// let nyArray1,nyArray2,nyArray3,nyArray4 be predefined arrarys
for (i= 0; i<= (n/2) -1 ; i++)
for (j =0; j<= (n/2)-1 ; i++)
nyArray1[i]= nyArray[i];
nyArray2[i]= nyArray[i+j];
nyArray3[i]= nyArray[n/2+j];
nyArray4[i]= nyArray[j];
x1 = aFunc(nyArray1,n/2);
x2 = aFunc(nyArray2,n/2);
x3 = aFunc(nyArray3,n/2);
x4 = aFunc(nyArray4,n/2);
return x1*x2*x3*x4;
I guess that its something like that, but Im not suce.
T(n) = 4T(n/2)+........
complex-analysis algorithms
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
float aFunc(nyArray, n)
if ( (n==1)
return nyArray[0];
// let nyArray1,nyArray2,nyArray3,nyArray4 be predefined arrarys
for (i= 0; i<= (n/2) -1 ; i++)
for (j =0; j<= (n/2)-1 ; i++)
nyArray1[i]= nyArray[i];
nyArray2[i]= nyArray[i+j];
nyArray3[i]= nyArray[n/2+j];
nyArray4[i]= nyArray[j];
x1 = aFunc(nyArray1,n/2);
x2 = aFunc(nyArray2,n/2);
x3 = aFunc(nyArray3,n/2);
x4 = aFunc(nyArray4,n/2);
return x1*x2*x3*x4;
I guess that its something like that, but Im not suce.
T(n) = 4T(n/2)+........
complex-analysis algorithms
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
float aFunc(nyArray, n)
if ( (n==1)
return nyArray[0];
// let nyArray1,nyArray2,nyArray3,nyArray4 be predefined arrarys
for (i= 0; i<= (n/2) -1 ; i++)
for (j =0; j<= (n/2)-1 ; i++)
nyArray1[i]= nyArray[i];
nyArray2[i]= nyArray[i+j];
nyArray3[i]= nyArray[n/2+j];
nyArray4[i]= nyArray[j];
x1 = aFunc(nyArray1,n/2);
x2 = aFunc(nyArray2,n/2);
x3 = aFunc(nyArray3,n/2);
x4 = aFunc(nyArray4,n/2);
return x1*x2*x3*x4;
I guess that its something like that, but Im not suce.
T(n) = 4T(n/2)+........
complex-analysis algorithms
complex-analysis algorithms
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
asked Mar 28 at 10:54
Harlinton Palacios MosqueraHarlinton Palacios Mosquera
1
1
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Harlinton Palacios Mosquera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right about your equation.
First, let's evaluate the for-loops. The for-loop are nested and run (n/2)-1 times each. Therefore, the run time of the for-loops is ((n/2)-1)^2. We will define this as the function f(n)=((n/2)-1)^2.
The program function is recursively called for times with n/2 as its input. We can now describe the recurrence by the following equation:
T(n)=4T(n/2)+((n/2)-1)^2
This form of equation can be evaluated using the master's theorem. According to the definition of the master's theorem, we use case 2, because n^(log_b(a))=n^(log_2(4))=n^2=Theta(f(n)), which is our test case. The variable a if found by the constant times the recursive function and the variable b is found in the divisor within the recursive call.
By the definition of case 2, the resulting complexity of your algorithm is Theta(n^(log_b(a))*log_2(n))=Theta(n^(log_2(4))*log_2(n))=Theta(n^2*log_2(n)).
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
add a comment |
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1 Answer
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1 Answer
1
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$begingroup$
You are right about your equation.
First, let's evaluate the for-loops. The for-loop are nested and run (n/2)-1 times each. Therefore, the run time of the for-loops is ((n/2)-1)^2. We will define this as the function f(n)=((n/2)-1)^2.
The program function is recursively called for times with n/2 as its input. We can now describe the recurrence by the following equation:
T(n)=4T(n/2)+((n/2)-1)^2
This form of equation can be evaluated using the master's theorem. According to the definition of the master's theorem, we use case 2, because n^(log_b(a))=n^(log_2(4))=n^2=Theta(f(n)), which is our test case. The variable a if found by the constant times the recursive function and the variable b is found in the divisor within the recursive call.
By the definition of case 2, the resulting complexity of your algorithm is Theta(n^(log_b(a))*log_2(n))=Theta(n^(log_2(4))*log_2(n))=Theta(n^2*log_2(n)).
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
add a comment |
$begingroup$
You are right about your equation.
First, let's evaluate the for-loops. The for-loop are nested and run (n/2)-1 times each. Therefore, the run time of the for-loops is ((n/2)-1)^2. We will define this as the function f(n)=((n/2)-1)^2.
The program function is recursively called for times with n/2 as its input. We can now describe the recurrence by the following equation:
T(n)=4T(n/2)+((n/2)-1)^2
This form of equation can be evaluated using the master's theorem. According to the definition of the master's theorem, we use case 2, because n^(log_b(a))=n^(log_2(4))=n^2=Theta(f(n)), which is our test case. The variable a if found by the constant times the recursive function and the variable b is found in the divisor within the recursive call.
By the definition of case 2, the resulting complexity of your algorithm is Theta(n^(log_b(a))*log_2(n))=Theta(n^(log_2(4))*log_2(n))=Theta(n^2*log_2(n)).
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
add a comment |
$begingroup$
You are right about your equation.
First, let's evaluate the for-loops. The for-loop are nested and run (n/2)-1 times each. Therefore, the run time of the for-loops is ((n/2)-1)^2. We will define this as the function f(n)=((n/2)-1)^2.
The program function is recursively called for times with n/2 as its input. We can now describe the recurrence by the following equation:
T(n)=4T(n/2)+((n/2)-1)^2
This form of equation can be evaluated using the master's theorem. According to the definition of the master's theorem, we use case 2, because n^(log_b(a))=n^(log_2(4))=n^2=Theta(f(n)), which is our test case. The variable a if found by the constant times the recursive function and the variable b is found in the divisor within the recursive call.
By the definition of case 2, the resulting complexity of your algorithm is Theta(n^(log_b(a))*log_2(n))=Theta(n^(log_2(4))*log_2(n))=Theta(n^2*log_2(n)).
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You are right about your equation.
First, let's evaluate the for-loops. The for-loop are nested and run (n/2)-1 times each. Therefore, the run time of the for-loops is ((n/2)-1)^2. We will define this as the function f(n)=((n/2)-1)^2.
The program function is recursively called for times with n/2 as its input. We can now describe the recurrence by the following equation:
T(n)=4T(n/2)+((n/2)-1)^2
This form of equation can be evaluated using the master's theorem. According to the definition of the master's theorem, we use case 2, because n^(log_b(a))=n^(log_2(4))=n^2=Theta(f(n)), which is our test case. The variable a if found by the constant times the recursive function and the variable b is found in the divisor within the recursive call.
By the definition of case 2, the resulting complexity of your algorithm is Theta(n^(log_b(a))*log_2(n))=Theta(n^(log_2(4))*log_2(n))=Theta(n^2*log_2(n)).
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 19:55
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
answered Mar 28 at 11:18
Martin PekárMartin Pekár
11
11
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Martin Pekár is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
add a comment |
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Thank you for your feedback, if run time of the for-loops is ((n/2)-1)^2 and the function is f(n)=((n/2)-1)^2. The T(n) with recursive function time should be T(n)= T(n)=4T(n/2)+((n/2)-1)^2 ?
$endgroup$
– Harlinton Palacios Mosquera
Mar 28 at 12:19
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Yes, you are right! I will update it. Also, note I changed the result from using case 1 to use case 2 because of the mistake.
$endgroup$
– Martin Pekár
Mar 28 at 19:56
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
$begingroup$
Thank you so much.
$endgroup$
– Harlinton Palacios Mosquera
Mar 29 at 8:27
add a comment |
Harlinton Palacios Mosquera is a new contributor. Be nice, and check out our Code of Conduct.
Harlinton Palacios Mosquera is a new contributor. Be nice, and check out our Code of Conduct.
Harlinton Palacios Mosquera is a new contributor. Be nice, and check out our Code of Conduct.
Harlinton Palacios Mosquera is a new contributor. Be nice, and check out our Code of Conduct.
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