Jumbo Conforming Loan的Limit要从$729K降了 - 未名空间MITBBS历史存档

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Jumbo Conforming Loan的Limit要从$729K降了

Jumbo Conforming Loan的Limit要从$729K降了# Living

r*y2011-05-15 07:05

1 楼

givn m sets. some sets may be subsets of others. If one set is a subset of
another set, then this smaller set will be removed. Finally we have the sets
from which we can not remove any set. Assume the elements in all the sets
are
same type and only "==" operator is supported on these elements.
Is there any good algorithm to do this job ? Thanks.

b*d2011-05-15 07:05

2 楼

具体降的幅度要看County，很多地方要降到$625,500。
http://finance.yahoo.com/news/Federal-Retreat-on-Bigger-nytimes

g*y2011-05-15 07:05

3 楼

基本想法：把Element映射到数字，然后用Bitmap来计算包容关系。
pesudo-code =>
foreach set in sets {
Bitmap b1 = transform(set);
boolean add = true;
foreach b2 in map.keySet() {
Bitmap b3 = b1.and(b2);
if (b3.equals(b2)) {
bitmaps.remove(b2);
}
else if (b3.equals(b1)) {
add = false;
}
}
if (add) map.put(b1, set);
}
return map.values();

c*o2011-05-15 07:05

4 楼

为这个好消息得好好喝一盅

r*y2011-05-15 07:05

5 楼

So it is using hashing. But how to do hashing here is also a big deal, right
?
Thanks

【在 g**********y 的大作中提到】

: 基本想法：把Element映射到数字，然后用Bitmap来计算包容关系。
: pesudo-code =>
: foreach set in sets {
: Bitmap b1 = transform(set);
: boolean add = true;
: foreach b2 in map.keySet() {
: Bitmap b3 = b1.and(b2);
: if (b3.equals(b2)) {
: bitmaps.remove(b2);
: }

r*n2011-05-15 07:05

6 楼

这个对高房价区感觉杀伤力很大啊

【在 b*d 的大作中提到】

: 具体降的幅度要看County，很多地方要降到$625,500。
: http://finance.yahoo.com/news/Federal-Retreat-on-Bigger-nytimes

g*y2011-05-15 07:05

7 楼

I am not sure what's your question. But as example, here is a simplified
Java implementation, assume sets no more than 32. If more than that, you can
use BigInteger or write your own class.
public class Subset {
public Set[] merge(Set[] sets) {
HashMap