Re^3: OT: Finding Factor Closest To Square Root

Do you really have to do a search for numbers greater than the sqrt(N)?

You right. And now you pointed it out, it seems fairly obvious:

3600: sqrt=60; 59* 61= 3599

10000: sqrt=100; 99*101= 9999

12: sqrt=3.464; 3* 4= 12

Which, in very non-technical terms because my memory doesn't go back to my formal math days, says to me that:

with lo = int( sqrt( N ) ) & hi = int( sqrt( n ) )+1; lo * hi is always less than or equal to N.

If you reduce lo, leaving hi the same, you get further away from N. And if you increase hi, lo will be nearer.

Does that form an explaination? Hmm. Maybe one of our resident math whiz can phrase that correctly?

Examine what is said, not who speaks.

Silence betokens consent.

Love the truth but pardon error.

Comment on Re^3: OT: Finding Factor Closest To Square Root

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Re^4: Closest factor less than sqrt(N) [proof] by sleepingsquirrel (Chaplain) on Feb 21, 2005 at 00:13 UTC
Here's an informal proof of our conjecture "The factor of N closest to sqrt(N) is less than sqrt(N)". Let M be the positive square root of N. (`MM = N, M>0`) The closest factor to M is bewteen 1 and 2M. (i.e. 1 is closer to M than any number greater than twice M. `(M-1 < 2M-M)`) For every pair of numbers whose product is N, one of the pair will be greater than M and the other less than M. (the product of two numbers greater than M is greater than N, and the product of two numbers less than M would be less than N) Take a pair of numbers whose product is N. Call the smaller (`M-X`) and the larger (`M+Y`). (i.e. `X>0, Y>0`) So, `(M-X) (M+Y) = N` Multiplying out, `M^2 + M(Y-X) - XY = N` But `M^2 = N` (by definition) Substituting: `N + M(Y-X) - XY = N` Subtracting N: `M(Y-X) - XY = 0` Since X and Y are positive (by definition, step 4), the term `-XY` is negative If `-XY` is negative, the only way for the entire sum to equal 0 is if the term `M*(Y-X)` is positive But M is positive (step 1), so `(Y-X)` must also be positive, which means that Y is greater than X. Finally, if Y is greater than X, the smaller factor, `(M-X)` is closer to M than `(M+Y)`, so you can limit your factor search to numbers less than sqrt(N). -- All code is 100% tested and functional unless otherwise noted.	[reply]
Re^5: Closest factor less than sqrt(N) [proof] by hv (Prior) on Feb 21, 2005 at 09:45 UTC
Here's an easier way: take `n = m^2`, and a factor greater than the square root `m + d`, then the other factor is: `m^2/(m + d) = m - dm/(m + d) = m - d + d^2/(m + d)` [download] Now `d^2/(m + d)` is greater than zero, so the cofactor is greater than `m - d` and hence closer to the square root. Hugo	[reply] [d/l] [select]
Re^5: Closest factor less than sqrt(N) [proof] by BrowserUk (Patriarch) on Feb 21, 2005 at 00:52 UTC
sleepingsquirrel++ Makes sense to me--but then my version did (to me) too :) Who'd've thunk it--even informality is relative :)	[reply]


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