1,998 reputation
1330
bio website chaoxuprime.com
location Urbana, IL
age 24
visits member for 4 years, 1 month
seen Aug 14 at 21:05

PhD student in CS theory at UIUC.

I'm interested in algorithms, computational geometry, theory of computation and problem solving in general.


Aug
14
answered Is this function submodular?
Aug
14
revised Is this function submodular?
update proof to remove the requirement of symmetric
Aug
13
comment Find taxicab numbers in $O(n)$ time
so this would be similar to sorting X+Y cs.smith.edu/~orourke/TOPP/P41.html
Aug
13
asked Is this function submodular?
Jul
29
awarded  Notable Question
Jul
20
awarded  Yearling
Jul
2
awarded  Popular Question
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Apr
19
awarded  Good Question
Mar
8
answered Convert curves to monotone curves without introduce intersections
Mar
3
comment Convert curves to monotone curves without introduce intersections
$f_x(t)=t$ for $t=0$ and $t=1$, not all $t \in [0,1]$.
Mar
3
comment Convert curves to monotone curves without introduce intersections
h(t)=t does not imply g*=g, neither does it imply f*=f.
Mar
2
accepted Convert curves to monotone curves without introduce intersections
Mar
1
comment Convert curves to monotone curves without introduce intersections
$f(I)\cap g(I)=\emptyset$ does not imply $f_y(I)\cap g_y(I) = \emptyset$.
Feb
27
comment Convert curves to monotone curves without introduce intersections
$g^*$ and $f^*$ are just two complete curves defined in terms of $f,g$ and $h$. $h(t)=t$ doesn't always work as there might be a intersection. For example this graph. i.imgur.com/HF1FNgk.png We can parametrize $f$ and $g$ so at time $t$, both $f_y(t) = g_y(t)$(with different $x$ coordinate). If $h(t)=t$, then $g^*(t)=f^*(t)$, so there is a intersection.
Feb
21
asked Convert curves to monotone curves without introduce intersections
Jan
18
accepted Union of two matchings that have same number of edges in each component
Jan
18
comment Union of two matchings that have same number of edges in each component
"every component contains edges from both matching" not true. Consider $M$ is a empty matching. However, the proof still holds, and it looks like this can be generalized to all graphs contains an almost perfect matching.
Jan
14
asked Union of two matchings that have same number of edges in each component