c******r
发帖数: 300 | 1 比较走运拿到了几个offer，目前基本在考虑一个大的bank的front office quant
group, fixed income products, 和一个prop trading firm的trading intern，
commodity derivative,
都在nyc，各方面条件都差不多。已经有过bank的经验, 是否换个环境会比较好。prop
firm是个很小的，一共不到20个人，但是和他们聊，感觉每个人都很聪明,而且感觉他
们在扩张。另外还有可能会有一个很大的prop firm的trading intern, 不过还在面试
中。想问一下如果full time直接去这种小的公司是否很好，主要有点担心不稳定；当
然感觉upside也很大，毕竟很小的公司，而且还是trading position.多谢各位有经验
的人多给意见。 |
M*****y
发帖数: 666 | 2 big cong!!
prop
【在 c******r 的大作中提到】
: 比较走运拿到了几个offer，目前基本在考虑一个大的bank的front office quant
: group, fixed income products, 和一个prop trading firm的trading intern，
: commodity derivative,
: 都在nyc，各方面条件都差不多。已经有过bank的经验, 是否换个环境会比较好。prop
: firm是个很小的，一共不到20个人，但是和他们聊，感觉每个人都很聪明,而且感觉他
: 们在扩张。另外还有可能会有一个很大的prop firm的trading intern, 不过还在面试
: 中。想问一下如果full time直接去这种小的公司是否很好，主要有点担心不稳定；当
: 然感觉upside也很大，毕竟很小的公司，而且还是trading position.多谢各位有经验
: 的人多给意见。
|
J*****n
发帖数: 4859 | 3
prop
If trading firm had some senior quant with PhD and experience from big name,
I recommend this trading firm.
【在 c******r 的大作中提到】
: 比较走运拿到了几个offer，目前基本在考虑一个大的bank的front office quant
: group, fixed income products, 和一个prop trading firm的trading intern，
: commodity derivative,
: 都在nyc，各方面条件都差不多。已经有过bank的经验, 是否换个环境会比较好。prop
: firm是个很小的，一共不到20个人，但是和他们聊，感觉每个人都很聪明,而且感觉他
: 们在扩张。另外还有可能会有一个很大的prop firm的trading intern, 不过还在面试
: 中。想问一下如果full time直接去这种小的公司是否很好，主要有点担心不稳定；当
: 然感觉upside也很大，毕竟很小的公司，而且还是trading position.多谢各位有经验
: 的人多给意见。
|
c******r
发帖数: 300 | 4 The firm does have a couple of founding members from a very well-known place
. Thanks then.
Also some technical questions I found to be interesting during the process:
(1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
having HHHHHH, you should obtain the answer very quickly and in the simplest
possible way.
(2) Find the largest number that does not have repeated digits and is
divisible by 7, again you should do it very quickly.
(3) Obtain sqrt(5) using any commonly used
【在 J*****n 的大作中提到】
:
: prop
: If trading firm had some senior quant with PhD and experience from big name,
: I recommend this trading firm.
|
x**y
发帖数: 10012 | 5 lihai...
prop
【在 c******r 的大作中提到】
: 比较走运拿到了几个offer，目前基本在考虑一个大的bank的front office quant
: group, fixed income products, 和一个prop trading firm的trading intern，
: commodity derivative,
: 都在nyc，各方面条件都差不多。已经有过bank的经验, 是否换个环境会比较好。prop
: firm是个很小的，一共不到20个人，但是和他们聊，感觉每个人都很聪明,而且感觉他
: 们在扩张。另外还有可能会有一个很大的prop firm的trading intern, 不过还在面试
: 中。想问一下如果full time直接去这种小的公司是否很好，主要有点担心不稳定；当
: 然感觉upside也很大，毕竟很小的公司，而且还是trading position.多谢各位有经验
: 的人多给意见。
|
t**o
发帖数: 64 | 6 The last one seems interesting. It seems that we need to divide the numbers
into two groups.
place
of
simplest
1s
【在 c******r 的大作中提到】
: The firm does have a couple of founding members from a very well-known place
: . Thanks then.
: Also some technical questions I found to be interesting during the process:
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: (2) Find the largest number that does not have repeated digits and is
: divisible by 7, again you should do it very quickly.
: (3) Obtain sqrt(5) using any commonly used
|
d*s
发帖数: 28 | |
j******n
发帖数: 271 | 8 当然是trading intern， if money is the deciding factor. |
s*******s
发帖数: 1568 | 9 Cong to your offers:)
(1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
having HHHHHH, you should obtain the answer very quickly and in the simplest
possible way.
sorry I made a mistake on calculating the expectation here. By the nature that the two pattern doesn't
share the prefix part, the prob can be calculated as P = Expected tossing to get TT/ (Expected tossing to
get HHHHHH + Expected tossing to get TT) . The result is
1^2 + 2^2/(6 + 1^2+ ...+ 6^2) = 1/22.
(2) Fin
【在 c******r 的大作中提到】
: The firm does have a couple of founding members from a very well-known place
: . Thanks then.
: Also some technical questions I found to be interesting during the process:
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: (2) Find the largest number that does not have repeated digits and is
: divisible by 7, again you should do it very quickly.
: (3) Obtain sqrt(5) using any commonly used
|
f****e
发帖数: 590 | 10 赞，第一题的思路是？
of
simplest
1s
【在 s*******s 的大作中提到】
: Cong to your offers:)
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: sorry I made a mistake on calculating the expectation here. By the nature that the two pattern doesn't
: share the prefix part, the prob can be calculated as P = Expected tossing to get TT/ (Expected tossing to
: get HHHHHH + Expected tossing to get TT) . The result is
: 1^2 + 2^2/(6 + 1^2+ ...+ 6^2) = 1/22.
: (2) Fin
|
|
|
d*j
发帖数: 13780 | 11 co ask
【在 f****e 的大作中提到】
: 赞，第一题的思路是？
:
: of
: simplest
: 1s
|
A*******u
发帖数: 66 | 12 Big cong !
prop
【在 c******r 的大作中提到】
: 比较走运拿到了几个offer，目前基本在考虑一个大的bank的front office quant
: group, fixed income products, 和一个prop trading firm的trading intern，
: commodity derivative,
: 都在nyc，各方面条件都差不多。已经有过bank的经验, 是否换个环境会比较好。prop
: firm是个很小的，一共不到20个人，但是和他们聊，感觉每个人都很聪明,而且感觉他
: 们在扩张。另外还有可能会有一个很大的prop firm的trading intern, 不过还在面试
: 中。想问一下如果full time直接去这种小的公司是否很好，主要有点担心不稳定；当
: 然感觉upside也很大，毕竟很小的公司，而且还是trading position.多谢各位有经验
: 的人多给意见。
|
B*****t
发帖数: 335 | 13 第三题IBM的原题是用2个"2"，而不是2个"1".
-log2(ln(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt((exp2))))))))
place
of
simplest
1s
【在 c******r 的大作中提到】
: The firm does have a couple of founding members from a very well-known place
: . Thanks then.
: Also some technical questions I found to be interesting during the process:
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: (2) Find the largest number that does not have repeated digits and is
: divisible by 7, again you should do it very quickly.
: (3) Obtain sqrt(5) using any commonly used
|
a*******h
发帖数: 123 | 14 (3) 如果允许用三角函数的话。。。是 sec(arctan(1+1))
of
simplest
1s
【在 s*******s 的大作中提到】
: Cong to your offers:)
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: sorry I made a mistake on calculating the expectation here. By the nature that the two pattern doesn't
: share the prefix part, the prob can be calculated as P = Expected tossing to get TT/ (Expected tossing to
: get HHHHHH + Expected tossing to get TT) . The result is
: 1^2 + 2^2/(6 + 1^2+ ...+ 6^2) = 1/22.
: (2) Fin
|
B*****t
发帖数: 335 | 15 第4个应该是个NPC问题。
第一题又算了一便, 结果应该是3/66=1/22,这回应该没错了。
算的比较麻烦，有简单的思路么？
place
of
simplest
1s
【在 c******r 的大作中提到】
: The firm does have a couple of founding members from a very well-known place
: . Thanks then.
: Also some technical questions I found to be interesting during the process:
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: (2) Find the largest number that does not have repeated digits and is
: divisible by 7, again you should do it very quickly.
: (3) Obtain sqrt(5) using any commonly used
|
c******r
发帖数: 300 | 16 If the numbers are ordered, then you can solve this problem by dynamic
programming with O(N^2)
complexity. Otherwise I thought it is a NP-Complete problem. and of course
by the nature of this problem,
there are many ways to reduce the complexity.
Why you need the number to be ordered if the dp already costs O(N^2)? It
seems you can always sort it in O(Nlog(N)). I am also not quite sure why
this is a NPC problems, the bound 1.5 is somehow crucial here. Otherwise,
you probably are right, it must b
【在 s*******s 的大作中提到】
: Cong to your offers:)
: (1) Toss a fair coin until you get HHHHHH or TT, what is the probability of
: having HHHHHH, you should obtain the answer very quickly and in the simplest
: possible way.
: sorry I made a mistake on calculating the expectation here. By the nature that the two pattern doesn't
: share the prefix part, the prob can be calculated as P = Expected tossing to get TT/ (Expected tossing to
: get HHHHHH + Expected tossing to get TT) . The result is
: 1^2 + 2^2/(6 + 1^2+ ...+ 6^2) = 1/22.
: (2) Fin
|
B*****t
发帖数: 335 | 17 这里ordered的意思不是要把number排序，而是每次取数的顺序是不能改变的。
【在 c******r 的大作中提到】
: If the numbers are ordered, then you can solve this problem by dynamic
: programming with O(N^2)
: complexity. Otherwise I thought it is a NP-Complete problem. and of course
: by the nature of this problem,
: there are many ways to reduce the complexity.
: Why you need the number to be ordered if the dp already costs O(N^2)? It
: seems you can always sort it in O(Nlog(N)). I am also not quite sure why
: this is a NPC problems, the bound 1.5 is somehow crucial here. Otherwise,
: you probably are right, it must b
|
c******r
发帖数: 300 | 18 Sorry, I misunderstood the previous post. I think you do not have this assumption, you can pick any two numbers at a time. It should not be a NP hard problem. But I am now curious whether it will be NP hard if we decrease the lower bound 1.5 to 0 (or 1).
【在 B*****t 的大作中提到】
: 这里ordered的意思不是要把number排序，而是每次取数的顺序是不能改变的。
|
B*****t
发帖数: 335 | 19 I don't think it's NP-Hard. It should be NPC.
assumption, you can pick any two numbers at a time. It should not be a NP
hard problem. But I am now curious whether it will be NP hard if we decrease
the lower bound 1.5 to 0 (or 1).
【在 c******r 的大作中提到】
: Sorry, I misunderstood the previous post. I think you do not have this assumption, you can pick any two numbers at a time. It should not be a NP hard problem. But I am now curious whether it will be NP hard if we decrease the lower bound 1.5 to 0 (or 1).
|
c******r
发帖数: 300 | 20 Sorry, I am not an expert on this NP staffs, not a computer scientist by
training. But I guess what I wanted to say is the original problem (with the
lower bound) can be sovled by a polynomial algorithm. Think about how the
final number can be represented using the input definitely helps.
decrease
【在 B*****t 的大作中提到】
: I don't think it's NP-Hard. It should be NPC.
:
: assumption, you can pick any two numbers at a time. It should not be a NP
: hard problem. But I am now curious whether it will be NP hard if we decrease
: the lower bound 1.5 to 0 (or 1).
|
|
|
c**g
发帖数: 45 | 21 NPC 是 NP-Hard的子集吧
I don't think it's NP-Hard. It should be NPC.
assumption, you can pick any two numbers at a time. It should not be a NP
hard problem. But I am now curious whether it will be NP hard if we decrease
the lower bound 1.5 to 0 (or 1).
【在 B*****t 的大作中提到】
: I don't think it's NP-Hard. It should be NPC.
:
: assumption, you can pick any two numbers at a time. It should not be a NP
: hard problem. But I am now curious whether it will be NP hard if we decrease
: the lower bound 1.5 to 0 (or 1).
|
B*****t
发帖数: 335 | 22 您说的没错，我只是想把范围缩小一下。
decrease
【在 c**g 的大作中提到】
: NPC 是 NP-Hard的子集吧
:
: I don't think it's NP-Hard. It should be NPC.
: assumption, you can pick any two numbers at a time. It should not be a NP
: hard problem. But I am now curious whether it will be NP hard if we decrease
: the lower bound 1.5 to 0 (or 1).
|
w*********l
发帖数: 1337 | 23 npc难道不imply NP-hard?
decrease
【在 B*****t 的大作中提到】
: I don't think it's NP-Hard. It should be NPC.
:
: assumption, you can pick any two numbers at a time. It should not be a NP
: hard problem. But I am now curious whether it will be NP hard if we decrease
: the lower bound 1.5 to 0 (or 1).
|
r****1
发帖数: 328 | |
z****i
发帖数: 406 | 25 I guess the first one is 1/22.
HHHHH and TT are disjoint, so the probability that HHHHHH appears first is
the E(TT)/(E(TT)+E(HHHHHH)), where E represents number of tosses until some
pattern appears for the first time.
We know E(TT) = 2+2^2 = 6, and E(HHHHHH) = 2+2^2+2^3+2^4+2^5+2^6 = 126.
So the answer is 6/(6+126) = 1/22.
Please let me know if it's wrong, or if there is an easier way. |
f*k
发帖数: 95 | 26 prob #1. i got 1/22 as well.
prob #4. is interesting.
1.5 is kinda threshold.
Looking at 1.5*x >= 1.5+x ==> x = 3.0
That means, if we have smth >= 3.0, then multiplication is always better.
And, since every number >= 1.5, we can make sure the first operation's
result is >= 3.0
So, either (x1+x2)*x3*x4....xN or x1*x2*x3...xN (x1 and x2 are two smallest
elements in the set) |
s*******s
发帖数: 1568 | 27 I made a mistake, hehe, sorry
【在 B*****t 的大作中提到】
: 第4个应该是个NPC问题。
: 第一题又算了一便, 结果应该是3/66=1/22,这回应该没错了。
: 算的比较麻烦，有简单的思路么？
:
: place
: of
: simplest
: 1s
|
s*******s
发帖数: 1568 | 28 try the case, 1.5 1.5 1.5 1.5
smallest
【在 f*k 的大作中提到】
: prob #1. i got 1/22 as well.
: prob #4. is interesting.
: 1.5 is kinda threshold.
: Looking at 1.5*x >= 1.5+x ==> x = 3.0
: That means, if we have smth >= 3.0, then multiplication is always better.
: And, since every number >= 1.5, we can make sure the first operation's
: result is >= 3.0
: So, either (x1+x2)*x3*x4....xN or x1*x2*x3...xN (x1 and x2 are two smallest
: elements in the set)
|
n******r
发帖数: 1247 | 29 NP-hard is the usual way to put it as it includes NPC
if you want to claim NPC, the problem has to be a decision problem and you
need to proof it belongs to the NPC equivalent class (can be reduced to and
from other NPC problems)
【在 B*****t 的大作中提到】
: 您说的没错，我只是想把范围缩小一下。
:
: decrease
|
f*k
发帖数: 95 | 30 Thanks. i overlooked this case.
but we may still use greedy method here
(X1+Xm)*(X2+Xm-1)*(X3+Xm-2)....*(Xm/2+Xm/2+1)*Xm+1*Xm+2*Xm+3*.....XN
【在 s*******s 的大作中提到】
: try the case, 1.5 1.5 1.5 1.5
:
: smallest
|
|
|
s*******s
发帖数: 1568 | 31 mmm, what is m ?
when X < 3??
then try case 1.5,1.5,2.5,2.5
【在 f*k 的大作中提到】
: Thanks. i overlooked this case.
: but we may still use greedy method here
: (X1+Xm)*(X2+Xm-1)*(X3+Xm-2)....*(Xm/2+Xm/2+1)*Xm+1*Xm+2*Xm+3*.....XN
|
g******r
发帖数: 29 | 32 这个 E(TT)/(E(TT)+E(HHHHHH)) 可以用optional stopping theorem或其他的定理简单
说明吗
some
【在 z****i 的大作中提到】
: I guess the first one is 1/22.
: HHHHH and TT are disjoint, so the probability that HHHHHH appears first is
: the E(TT)/(E(TT)+E(HHHHHH)), where E represents number of tosses until some
: pattern appears for the first time.
: We know E(TT) = 2+2^2 = 6, and E(HHHHHH) = 2+2^2+2^3+2^4+2^5+2^6 = 126.
: So the answer is 6/(6+126) = 1/22.
: Please let me know if it's wrong, or if there is an easier way.
|
c**********e
发帖数: 2007 | 33 Congratulations!
银行更稳定一些，如果没有绿卡，还是银行好。小公司的钱途更好一些。
银行也看哪个啦，高盛放简历上比花旗好看多了。其实主要是个人喜好。
很多人觉得新手从银行做起更好一些。
prop
【在 c******r 的大作中提到】
: 比较走运拿到了几个offer，目前基本在考虑一个大的bank的front office quant
: group, fixed income products, 和一个prop trading firm的trading intern，
: commodity derivative,
: 都在nyc，各方面条件都差不多。已经有过bank的经验, 是否换个环境会比较好。prop
: firm是个很小的，一共不到20个人，但是和他们聊，感觉每个人都很聪明,而且感觉他
: 们在扩张。另外还有可能会有一个很大的prop firm的trading intern, 不过还在面试
: 中。想问一下如果full time直接去这种小的公司是否很好，主要有点担心不稳定；当
: 然感觉upside也很大，毕竟很小的公司，而且还是trading position.多谢各位有经验
: 的人多给意见。
|
f*k
发帖数: 95 | 34 we can iterate to find m.
(1.5+1.5)*(2.5*2.5)
【在 s*******s 的大作中提到】
: mmm, what is m ?
: when X < 3??
: then try case 1.5,1.5,2.5,2.5
|
