Computational complexity sum of digits square root
$begingroup$
I know that the sum of digits formula is:
$frac{k(k+1)}{2}$
I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:
$frac{k(k+1)}{2}>n$
At the time when the function increases over $n$, the while loop will exit.
Expanding the formula and solving for $k$:
$frac{k^2+k}{2} > n$
This next line I do not understand, can someone explain how this line is achieved?
$k = Oleft(sqrt{n}right)$
Regards
computational-complexity
edited Dec 26 '18 at 10:15
bames
1,9311315
asked Dec 26 '18 at 9:50
fynxfynx
83
$endgroup$
add a comment |
$begingroup$
I know that the sum of digits formula is:
$frac{k(k+1)}{2}$
I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:
$frac{k(k+1)}{2}>n$
At the time when the function increases over $n$, the while loop will exit.
Expanding the formula and solving for $k$:
$frac{k^2+k}{2} > n$
This next line I do not understand, can someone explain how this line is achieved?
$k = Oleft(sqrt{n}right)$
Regards
computational-complexity
edited Dec 26 '18 at 10:15
bames
1,9311315
asked Dec 26 '18 at 9:50
fynxfynx
83
$endgroup$
add a comment |
$begingroup$
I know that the sum of digits formula is:
$frac{k(k+1)}{2}$
I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:
$frac{k(k+1)}{2}>n$
At the time when the function increases over $n$, the while loop will exit.
Expanding the formula and solving for $k$:
$frac{k^2+k}{2} > n$
This next line I do not understand, can someone explain how this line is achieved?
$k = Oleft(sqrt{n}right)$
Regards
computational-complexity
edited Dec 26 '18 at 10:15
bames
1,9311315
asked Dec 26 '18 at 9:50
fynxfynx
83
$endgroup$
I know that the sum of digits formula is:
$frac{k(k+1)}{2}$
I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:
$frac{k(k+1)}{2}>n$
At the time when the function increases over $n$, the while loop will exit.
Expanding the formula and solving for $k$:
$frac{k^2+k}{2} > n$
This next line I do not understand, can someone explain how this line is achieved?
$k = Oleft(sqrt{n}right)$
Regards
computational-complexity
computational-complexity
edited Dec 26 '18 at 10:15
bames
1,9311315
asked Dec 26 '18 at 9:50
fynxfynx
83
edited Dec 26 '18 at 10:15
bames
1,9311315
asked Dec 26 '18 at 9:50
fynxfynx
83
edited Dec 26 '18 at 10:15
bames
1,9311315
edited Dec 26 '18 at 10:15
bames
1,9311315
edited Dec 26 '18 at 10:15
bames
1,9311315
1,9311315
asked Dec 26 '18 at 9:50
fynxfynx
83
asked Dec 26 '18 at 9:50
fynxfynx
83
asked Dec 26 '18 at 9:50
fynxfynx
83
83
add a comment |
add a comment |
0
active
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