今天是特别的日子，我的他向我求婚。我很开心，我回到家就和朋友分享我给他生日惊喜的过程和他给我惊喜的过程。在分享的过程中，不知不觉谈到10年前的第一份工作的经验。这不是第一次分享但是却发现自己跟多以前从来没察觉的特质。朋友说这些好的经历都因该好好的记载下来，所以我来这里写下来了。
回想起，第一份工作的两年里，我是个无知和不会照顾别人感受的人。我只懂得拼命工作导致效率过高引起不满，在20人当中被排挤。当技术人员教导我们把crystal 放入机器时要刚刚好正中央，我考虑到肉眼来衡量比较困难，于是隔天就用纸皮剪了3 寸和4寸的圆形来方便自己。每次都很准确的放入机器grinding. 几个月后，supervisor就叫人用好的塑料来打造一样的补助平。以后就成为整个程序的一部分。
很快的这个新部门就要开始生产线了，20个人分成2组，我成为leader带领一队，每次前一个shift交棒是都很凌乱，在我这个shift我就把整齐持续的东西交棒回去。过不久我train另个人称为第三个shift的leader， 她也不满意凌乱和埋怨，我还是教导她做整齐来交棒。很快地，分成4个shift了。当我的shift的人有困难我马上出来帮忙解决，训练了几个新人，当每个人忙碌得不可开交就埋怨上个shift的人没有擦地板。我就快点解决手头上的工作然后帮忙擦地板。日子久了，他们也会习惯的去擦地板了。我和他们的距离也小了。新人也不需要我训练了，我的徒弟会训练他们。欣慰的是有一次我听到徒弟说：“这个从Cherry QC 出来的都写到很整齐的，你可以很快的把货物电算完毕然后包裹”。
有一次，生产线必须停顿就因为第三个步骤的其中一个补助工具坏了不能取得样品放入机器做测试。 我却可以用tweezer的尾端来取得样品测试完毕继续让生产线运作。用这个方法很危险，机会只有一次，很容易毁掉整片都拿不到样品。4个shift里面只有我能做好，所以在新的工具来之前，我必须帮忙做完4个shift 的测试工作。
。。。。故事说完了以后，朋友问道：“那么你在两年里面向自己学习到什么？”。 当下我真的想不出来。我说:"我知道自己对人的态度必须改进，当我觉得对的事情就去做，日子久了人们也明白"。 但是现在我有所发现，原来我在很久以前就有的能力，而我居然忽视了它。我有创意能力，......
Wednesday, March 24, 2010
Tuesday, February 23, 2010
生命的他
记得有个朋友和我说过，当她遇到生命的他的时候是那么的自然，就像一颗种子掉入土地之然生长发芽到开花。她觉得痛苦尽力了几十年就是为了遇见他。当时的我真的没有办法去体会她说的一切，但是此时此刻的我却觉得自己感同身受了。回想起与他的相遇，相知到相爱原来都是那么的自然。
感觉很幸福有他在身边。他天天陪着我，我病了他比我还紧张，在车上会拿小枕头给我抱，没饿他替我先饿的帮我买食物。。。。感谢这一切一切，感受这一切一切，珍惜这一切一切。
感觉很幸福有他在身边。他天天陪着我，我病了他比我还紧张，在车上会拿小枕头给我抱，没饿他替我先饿的帮我买食物。。。。感谢这一切一切，感受这一切一切，珍惜这一切一切。
Monday, September 21, 2009
一定可以
祈祷这有个安静的空间，自己的空间，从刚才就祈祷着。大自然的力量真的在身边，当我发出讯号需要盒子般东西的时候，真的有人愿意给与帮助。。。所以我相信这一次我一定可以，可以找到的。。。。。
真奇怪， 把 着 写成这，也许心理也在祈祷着这里也得到平静。心的平静。。
真奇怪， 把 着 写成这，也许心理也在祈祷着这里也得到平静。心的平静。。
Friday, September 04, 2009
Some thoughts
All in the sudden i feel like to write something.
Once upon of time, I would call my ex-boyfriend to help me something and i knew he could not let go our relationship, even after 2 years. One day, one of my friend pointed this issue to me only then i only realized I should play a role to help him let go in which I stopped called him up for helping me. I think this should be be way and the thing I can give him in order to help him start a new venture soon.
I have this thought is because recently i have found a gal keeps calling her ex-boyfriend even she left him 2 years ago. Maybe she was me, as a mirror that last time of me, so selfish to somehow 'using' the other party? I don't know......Just I can't express out, feel so uneasy........Maybe i could find out my own answer during my silent time in the morning for the coming days. ^^
Once upon of time, I would call my ex-boyfriend to help me something and i knew he could not let go our relationship, even after 2 years. One day, one of my friend pointed this issue to me only then i only realized I should play a role to help him let go in which I stopped called him up for helping me. I think this should be be way and the thing I can give him in order to help him start a new venture soon.
I have this thought is because recently i have found a gal keeps calling her ex-boyfriend even she left him 2 years ago. Maybe she was me, as a mirror that last time of me, so selfish to somehow 'using' the other party? I don't know......Just I can't express out, feel so uneasy........Maybe i could find out my own answer during my silent time in the morning for the coming days. ^^
Tuesday, September 01, 2009
Restaurant City Facebook
- Truffle is ? Fungi
- Tomato is ? Fruit
- Rump found at? Back
- Grapes grow ? Vines
- Homous? chickpeas
- Chinese xxx? not mango
- Saffron is a ? spices
- Calamari is fried? squid
Friday, August 14, 2009
Pohlig-Hellman Part 1 - method 2
Method 2:
We should always use this method first, if the factoring gives us more
than one prime.
let us define some of the general terms,
5^{x} congruent 3 mod 2017
so , a = 5, b =3, n = 2017
n-1 = 2016 = 2^{5} x 3^{2} x 7
so, we have 3 prime, p_{1} =2, p_{2} = 3, p_{3} =7
steps : now, we still deal with prime 3, in order to let us see
the constrast
p_{2} = 3, x congruent n_{0} mod 9 .............we
have to find n0
Let x = 9q + r .........we form this equation for
multiply 9, r is remainder
Deal with both sides
5^{x} = 3 mod 2017.........subsitute x = 9q + r
5^{9q + r} = 3 mod 2017.........raise both side power
to the rest of the prime factor
( 5^{9q + r} )^{2^5 x 7 } = 3 ^ {2^5 x 7 } mod
2017.........by doing this is to let us to cancel off all the
numbers with power of (n-1) to 1 , as theorem
a ^ {n - 1} mod n = 1, so we can cancel off
the q and to solve r
5 ^ {2016q + 224r} = 3 ^ {224} mod 2017
5 ^ { 224r} = 3 ^ {224} mod 2017
calculation:
5 ^ {5} = 1108 mod n
5 ^ {10} = 1328
5 ^ {20} = 726
5 ^ {40} = 639
5 ^ {80} = 887
5 ^ {160} = 139
5 ^ {320} = 1168
5 ^ {640} = 732
calculation:
3 ^ {7} = 170 mod n
3 ^ {14} = 662mod n
3 ^ {28} = 555 mod n
3 ^ {56} = 1441 mod n
3 ^ {112} = 988 mod n
3 ^ {224} = 1933 mod n
3 ^ {448} = 1005 mod n
3 ^ {896} = 1525 mod n
5 ^ { 224r} = 3 ^ {224} mod 2017
576 ^{ r } = 1933 mod n .............as you can see
here, r is the remainder form with multiply
of 9,
so the possible r value could be 0,1,2,3,4,5,6,7,8
Since right not giving us 1 mod n, we will cry if the
multiply is 81, the we have to test each of them
the worst is 81 times.
so, now subsitue r for possible number
when r = 2, 576 ^{ 2 } = 988
when r = 3, 576 ^{ 3 } = 294
when r = 4, 576 ^{ 4 } = 1933 ...........bingo, stop
the loop
conclude r =4, so our equation for prime 3
x congruent 4 mod 9
We should always use this method first, if the factoring gives us more
than one prime.
let us define some of the general terms,
5^{x} congruent 3 mod 2017
so , a = 5, b =3, n = 2017
n-1 = 2016 = 2^{5} x 3^{2} x 7
so, we have 3 prime, p_{1} =2, p_{2} = 3, p_{3} =7
steps : now, we still deal with prime 3, in order to let us see
the constrast
p_{2} = 3, x congruent n_{0} mod 9 .............we
have to find n0
Let x = 9q + r .........we form this equation for
multiply 9, r is remainder
Deal with both sides
5^{x} = 3 mod 2017.........subsitute x = 9q + r
5^{9q + r} = 3 mod 2017.........raise both side power
to the rest of the prime factor
( 5^{9q + r} )^{2^5 x 7 } = 3 ^ {2^5 x 7 } mod
2017.........by doing this is to let us to cancel off all the
numbers with power of (n-1) to 1 , as theorem
a ^ {n - 1} mod n = 1, so we can cancel off
the q and to solve r
5 ^ {2016q + 224r} = 3 ^ {224} mod 2017
5 ^ { 224r} = 3 ^ {224} mod 2017
calculation:
5 ^ {5} = 1108 mod n
5 ^ {10} = 1328
5 ^ {20} = 726
5 ^ {40} = 639
5 ^ {80} = 887
5 ^ {160} = 139
5 ^ {320} = 1168
5 ^ {640} = 732
calculation:
3 ^ {7} = 170 mod n
3 ^ {14} = 662mod n
3 ^ {28} = 555 mod n
3 ^ {56} = 1441 mod n
3 ^ {112} = 988 mod n
3 ^ {224} = 1933 mod n
3 ^ {448} = 1005 mod n
3 ^ {896} = 1525 mod n
5 ^ { 224r} = 3 ^ {224} mod 2017
576 ^{ r } = 1933 mod n .............as you can see
here, r is the remainder form with multiply
of 9,
so the possible r value could be 0,1,2,3,4,5,6,7,8
Since right not giving us 1 mod n, we will cry if the
multiply is 81, the we have to test each of them
the worst is 81 times.
so, now subsitue r for possible number
when r = 2, 576 ^{ 2 } = 988
when r = 3, 576 ^{ 3 } = 294
when r = 4, 576 ^{ 4 } = 1933 ...........bingo, stop
the loop
conclude r =4, so our equation for prime 3
x congruent 4 mod 9
Pohlig-Hellman Part 1 - method 1 Options
i was struggling for one whole day to figure out the technique to use it.
Hence, I will write a few topic about this algorithmn.
1st of all, I have found there are 2 main methods to solve this kind of question.
So now, let us explore this 2 methods. We should always use method 2
first because it is fast, if it doesn't give us ans then only look for
method 1 to solve.
let us define some of the general terms,
5^{x} congruent 3 mod 2017
so , a = 5, b =3, n = 2017
n-1 = 2016 = 2^{5} x 3^{2} x 7
so, we have 3 prime, p_{1} =2, p_{2} = 3, p_{3} =7
Method 1:
Normally this is called baby steps, which as it named very tedious with a lot of steps to perform. This method for sure will let u find the answer but a lot of steps. We have to choose method wisely based on question.
We choose this method if the power of factoring prime (p) very large and method 2 does'nt help us to find the asnwer if the exponent is big.
When the fatoring only give us one prime also use this method.
step 1: Remeber formula to deal with right hand side:-
b ^ {(n-1)/p_{i}}........p is prime, raise the power of prime from 1 until the last constants.
Example above i take prime = 3 to solve
my p_{2} = 3,
FIRST, come out the equation first, raise the power of 3 less than the maximun.
this example, max exponent for 3 is 2, so we raise until 1.
And sure we will have the constants need to be solved.
let constants be c_{i}
x = c_{0}3^{0} + c_{1}3^{1}
.......see, we have constants needed to be solved,
if the power of prime 3 larger, the
longer the steps i need to deal in order to
find all my constants.
step 2: Start to solve 1st constant 0
c0 = ?, b ^ {(n-1)/p_{i}} = 3 ^ {(2016)/3} = 3 ^ {672}
calculation:
3 ^ {7} = 170 mod n
3 ^ {14} = 662mod n
3 ^ {28} = 555 mod n
3 ^ {56} = 1441 mod n
3 ^ {112} = 988 mod n
3 ^ {224} = 1933 mod n
3 ^ {448} = 1005 mod n
3 ^ {896} = 1525 mod n
3 ^ {672} = 294 mod n...............
oh no, if we see 1 mod n here ,
we are done, else we have to continue
to deal with left hand side.
Bec a^{0} = 1 mod n
step 3: Now we have to calclate left hand in order to solve constant 0
5^{x} = 3 mod 2017 ....................see, originally
3 is no power, but in step 2 we power it,
so left hand also need to power with the same
thing
(5 ^ {x}) ^ {672} ....................replace x
with equation we have formend in step 1,
x = c_
{0}3^{0} + c_{1}3^{1} = c0 + 3c1
(5 ^ {c0 + 3c1}) ^ {672} .............multiply 672
with each of the items, as u will find all the numbers
give us
(n-1)t except the current constant we want to solve.
As we
know whatever number power of (n-1)t mod n will give us 1.
Hence,
we can cancel it bec of 1. t is any integer number.
(5 ^ {672c0 + 2016c1})
5 ^ {672c0} x 5 ^{2016c1}
5 ^ {672c0} x 1
( 5 ^ {672} ) ^ {c0}
calculation:
5 ^ {5} = 1108 mod n
5 ^ {10} = 1328
5 ^ {20} = 726
5 ^ {40} = 639
5 ^ {80} = 887
5 ^ {160} = 139
5 ^ {320} = 1168
5 ^ {640} = 732
( 5 ^ {672} ) ^ {c0} = 294 ^ {c0}
NOW, put both sides to gether
294 ^ {c0} = 294 mod n, as very clear the constant
should be 1 to make both side tally.
NOTE: in baby steps
method, every time we solve the constants,
the
constants range sure less the the current prime
so, 0 <= c_{i} < 3..........current prime is 3
ans : c0 = 1
step 4: Continue to solve constant 1
c1 = ?, b ^ {(n-1)/p_{i}} = 3 ^ {(2016)/
9} ..........as u can see i raise the prime power ( bottom)
= 3 ^ {224}
= 1933 mod
n ..........since not 1 mod n, so proceed to deal left hand
step 5: Continue to solve constant 1 - left hand
5^{x} = 3 mod 2017 ....................see, originally
3 is no power, but in step 2 we power it, so left hand
also
need to power with the same thing
(5 ^ {x}) ^ {224} ....................replace x
with equation we have formend in step 1 & step 3
x = c_
{0}3^{0} + c_{1}3^{1} = 1 + 3c1
(5 ^ { 1 + 3c1}) ^ {224}
(5 ^ { 224 + 672c1})
5 ^ {224} x 5 ^{672c1}
576 x 294^ {c1}
now put both sides together
576 x 294^ {c1} = 1933 mod n .........we have to
subsitue all the possible constanst to find
the answer, c1 can be 0,1,2
when c1=1, 576 x 294 = 1933 mod n, hence c1=
1
step 6: Calculate the equation
now, subsitue all the constants into step 1 equation.
x = c_{0}3^{0} + c_{1}3^{1} = 1 + 3c1
x = 1 + 1 x 3 = 4
Hence, the equation for prime 3 is:-
x congruent 4 mod 9
Method 2: please look at another topic
Hence, I will write a few topic about this algorithmn.
1st of all, I have found there are 2 main methods to solve this kind of question.
So now, let us explore this 2 methods. We should always use method 2
first because it is fast, if it doesn't give us ans then only look for
method 1 to solve.
let us define some of the general terms,
5^{x} congruent 3 mod 2017
so , a = 5, b =3, n = 2017
n-1 = 2016 = 2^{5} x 3^{2} x 7
so, we have 3 prime, p_{1} =2, p_{2} = 3, p_{3} =7
Method 1:
Normally this is called baby steps, which as it named very tedious with a lot of steps to perform. This method for sure will let u find the answer but a lot of steps. We have to choose method wisely based on question.
We choose this method if the power of factoring prime (p) very large and method 2 does'nt help us to find the asnwer if the exponent is big.
When the fatoring only give us one prime also use this method.
step 1: Remeber formula to deal with right hand side:-
b ^ {(n-1)/p_{i}}........p is prime, raise the power of prime from 1 until the last constants.
Example above i take prime = 3 to solve
my p_{2} = 3,
FIRST, come out the equation first, raise the power of 3 less than the maximun.
this example, max exponent for 3 is 2, so we raise until 1.
And sure we will have the constants need to be solved.
let constants be c_{i}
x = c_{0}3^{0} + c_{1}3^{1}
.......see, we have constants needed to be solved,
if the power of prime 3 larger, the
longer the steps i need to deal in order to
find all my constants.
step 2: Start to solve 1st constant 0
c0 = ?, b ^ {(n-1)/p_{i}} = 3 ^ {(2016)/3} = 3 ^ {672}
calculation:
3 ^ {7} = 170 mod n
3 ^ {14} = 662mod n
3 ^ {28} = 555 mod n
3 ^ {56} = 1441 mod n
3 ^ {112} = 988 mod n
3 ^ {224} = 1933 mod n
3 ^ {448} = 1005 mod n
3 ^ {896} = 1525 mod n
3 ^ {672} = 294 mod n...............
oh no, if we see 1 mod n here ,
we are done, else we have to continue
to deal with left hand side.
Bec a^{0} = 1 mod n
step 3: Now we have to calclate left hand in order to solve constant 0
5^{x} = 3 mod 2017 ....................see, originally
3 is no power, but in step 2 we power it,
so left hand also need to power with the same
thing
(5 ^ {x}) ^ {672} ....................replace x
with equation we have formend in step 1,
x = c_
{0}3^{0} + c_{1}3^{1} = c0 + 3c1
(5 ^ {c0 + 3c1}) ^ {672} .............multiply 672
with each of the items, as u will find all the numbers
give us
(n-1)t except the current constant we want to solve.
As we
know whatever number power of (n-1)t mod n will give us 1.
Hence,
we can cancel it bec of 1. t is any integer number.
(5 ^ {672c0 + 2016c1})
5 ^ {672c0} x 5 ^{2016c1}
5 ^ {672c0} x 1
( 5 ^ {672} ) ^ {c0}
calculation:
5 ^ {5} = 1108 mod n
5 ^ {10} = 1328
5 ^ {20} = 726
5 ^ {40} = 639
5 ^ {80} = 887
5 ^ {160} = 139
5 ^ {320} = 1168
5 ^ {640} = 732
( 5 ^ {672} ) ^ {c0} = 294 ^ {c0}
NOW, put both sides to gether
294 ^ {c0} = 294 mod n, as very clear the constant
should be 1 to make both side tally.
NOTE: in baby steps
method, every time we solve the constants,
the
constants range sure less the the current prime
so, 0 <= c_{i} < 3..........current prime is 3
ans : c0 = 1
step 4: Continue to solve constant 1
c1 = ?, b ^ {(n-1)/p_{i}} = 3 ^ {(2016)/
9} ..........as u can see i raise the prime power ( bottom)
= 3 ^ {224}
= 1933 mod
n ..........since not 1 mod n, so proceed to deal left hand
step 5: Continue to solve constant 1 - left hand
5^{x} = 3 mod 2017 ....................see, originally
3 is no power, but in step 2 we power it, so left hand
also
need to power with the same thing
(5 ^ {x}) ^ {224} ....................replace x
with equation we have formend in step 1 & step 3
x = c_
{0}3^{0} + c_{1}3^{1} = 1 + 3c1
(5 ^ { 1 + 3c1}) ^ {224}
(5 ^ { 224 + 672c1})
5 ^ {224} x 5 ^{672c1}
576 x 294^ {c1}
now put both sides together
576 x 294^ {c1} = 1933 mod n .........we have to
subsitue all the possible constanst to find
the answer, c1 can be 0,1,2
when c1=1, 576 x 294 = 1933 mod n, hence c1=
1
step 6: Calculate the equation
now, subsitue all the constants into step 1 equation.
x = c_{0}3^{0} + c_{1}3^{1} = 1 + 3c1
x = 1 + 1 x 3 = 4
Hence, the equation for prime 3 is:-
x congruent 4 mod 9
Method 2: please look at another topic
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