| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1617.1 | we want easy problems! please ! :-) | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Tue Jun 02 1992 03:18 | 25 |
|
x x
floor(---) = floor(---) ------------- (1)
n+y n
when (1) is true then:
-xy + y (x mod n)
n = ---------------------- -----------------(2)
x mod(n+y) - x mod n
so the problem is : find n such that { -xy + y (x mod n) } is minimum
subject to condition (1) .
one observation is found, denominator of (2) is 1 for all values
of n such as (1) is true, i tried it with
x=12,y=.5, n=5 and n=4 , these values meet condition (1), but n=4
is the minimum, in both cases you get denominator of (2) as 1 (ignoring
the sign, it cancels with nominator of (2) ).
is this problem fit for linear programming (some kind of simplex
method?, which i have not studied...)
/nasser
|
| 1617.2 | | TRACE::GILBERT | Ownership Obligates | Tue Jun 02 1992 18:51 | 21 |
| > -xy + y (x mod n)
> n = ---------------------- -----------------(2)
> x mod(n+y) - x mod n
> so the problem is : find n such that { -xy + y (x mod n) } is minimum
> subject to condition (1) .
Huh? I don't follow that. Besides minimizing the (fractional part of!?!)
the numerator, shouldn't some consideration be given to the denominator,
which also depends on n?
> is this problem fit for linear programming (some kind of simplex
> method?, which i have not studied...)
Perhaps, but I don't think so.
> we want easy problems! please ! :-)
Is this problem not easy?
For reference: This problem appeared in Fibonacci Quarterly, problem H-296.
I haven't seen the solution, if any.
|
| 1617.3 | | GUESS::DERAMO | Dan D'Eramo, zfc::deramo | Tue Jun 02 1992 19:09 | 26 |
| > If x and y are positive real numbers, what is the smallest positive
> integer n such that
>
> x x
> floor(---) = floor(---) ?
> n+y n
Well, I'll solve two thirds of the cases. :-)
x < y
Since we are given that 0 < x, 0 < y, and 1 <= n,
then if x < y it follows that 0 < x/(n+y) < 1 and
so the first floor is 0. Therefore the second
floor is 0, so x < n. The smallest positive n
for which this is true, given 0 < x, is floor(x)+1.
x = y
Here x/(n+y) <= x/(1+x) < 1, so both floor's must
be 0 and so again n is floor(x)+1.
It remains to solve the problem for 0 < y < x. :-)
Dan
|
| 1617.4 | late night rampling.. | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Wed Jun 03 1992 01:31 | 26 |
| >> -xy + y (x mod n)
>> n = ---------------------- -----------------(2)
>> x mod(n+y) - x mod n
>> so the problem is : find n such that { -xy + y (x mod n) } is minimum
>> subject to condition (1) .
>
>Huh? I don't follow that. Besides minimizing the (fractional part of!?!)
>the numerator, shouldn't some consideration be given to the denominator,
>which also depends on n?
as i said, it seemd that denominator is always 1. ( i only did few
tries), that is why i ignored denomenator,... i must be wrong.
>> we want easy problems! please ! :-)
>Is this problem not easy?
>
>For reference: This problem appeared in Fibonacci Quarterly, problem H-296.
>I haven't seen the solution, if any.
uhmm, i guess a problem is easy if one can solve it, else it is hard..
any way, whish i have more time to work on these problems you post, they
seem fun, but i got to go and struggle with my homeworks.. :-(
/Nasser
|