Complexity of Fibonacci Fast Recursive Program
def fastfib(n, fib_dict = {0: 1, 1: 1}):
if n not in fib_dict:
fib_dict[n] = fastfib(n-1, fib_dict) + fastfib(n-2, fib_dict)
return fib_dict[n]
I think the complexity here is n^2, but I am not sure.
python-3.x recursion fibonacci
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
add a comment |
def fastfib(n, fib_dict = {0: 1, 1: 1}):
if n not in fib_dict:
fib_dict[n] = fastfib(n-1, fib_dict) + fastfib(n-2, fib_dict)
return fib_dict[n]
I think the complexity here is n^2, but I am not sure.
python-3.x recursion fibonacci
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
add a comment |
def fastfib(n, fib_dict = {0: 1, 1: 1}):
if n not in fib_dict:
fib_dict[n] = fastfib(n-1, fib_dict) + fastfib(n-2, fib_dict)
return fib_dict[n]
I think the complexity here is n^2, but I am not sure.
python-3.x recursion fibonacci
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
def fastfib(n, fib_dict = {0: 1, 1: 1}):
if n not in fib_dict:
fib_dict[n] = fastfib(n-1, fib_dict) + fastfib(n-2, fib_dict)
return fib_dict[n]
I think the complexity here is n^2, but I am not sure.
python-3.x recursion fibonacci
python-3.x recursion fibonacci
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
edited Nov 23 '18 at 5:07
Uwe Keim
27.4k31128210
27.4k31128210
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
asked Nov 23 '18 at 5:04
Nourhan Berjawi
364
364
add a comment |
add a comment |
1 Answer
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Since you are filling a dictionary with n values, the lower bound is O(n). However, since you're only doing constant-time operations for each n, (Pythons dictionary lookup operation is O(1), though amortized), this algorithm should be O(n) (amortized). This technique of saving already computed values in a table is called memoization.
answered Nov 23 '18 at 5:34
Tim
1,56541631
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since you are filling a dictionary with n values, the lower bound is O(n). However, since you're only doing constant-time operations for each n, (Pythons dictionary lookup operation is O(1), though amortized), this algorithm should be O(n) (amortized). This technique of saving already computed values in a table is called memoization.
answered Nov 23 '18 at 5:34
Tim
1,56541631
add a comment |
Since you are filling a dictionary with n values, the lower bound is O(n). However, since you're only doing constant-time operations for each n, (Pythons dictionary lookup operation is O(1), though amortized), this algorithm should be O(n) (amortized). This technique of saving already computed values in a table is called memoization.
answered Nov 23 '18 at 5:34
Tim
1,56541631
add a comment |
Since you are filling a dictionary with n values, the lower bound is O(n). However, since you're only doing constant-time operations for each n, (Pythons dictionary lookup operation is O(1), though amortized), this algorithm should be O(n) (amortized). This technique of saving already computed values in a table is called memoization.
answered Nov 23 '18 at 5:34
Tim
1,56541631
Since you are filling a dictionary with n values, the lower bound is O(n). However, since you're only doing constant-time operations for each n, (Pythons dictionary lookup operation is O(1), though amortized), this algorithm should be O(n) (amortized). This technique of saving already computed values in a table is called memoization.
answered Nov 23 '18 at 5:34
Tim
1,56541631
answered Nov 23 '18 at 5:34
Tim
1,56541631
answered Nov 23 '18 at 5:34
Tim
1,56541631
answered Nov 23 '18 at 5:34
Tim
1,56541631
1,56541631
add a comment |
add a comment |
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