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Life in Redmond

Monday, August 3, 2015. IMO 2010 Problem 6. This is a problem that I've made a couple of attempts to solve in the past few years. Eventually, perhaps encouraged by recent success of nailing IMO 3/6 problems, it is now solved. Let a(1), a(2), . be a sequence of positive real numbers. Suppose that for some positive integers s, we have. A(n)=max{a(k) a(n-k) 1 =k =n-1}. For all n s. Prove that there exist positive integers p and N, with p =s, such that a(n)=a(n-p) a(p) for all n =N. Posted by Shen-Fu Tsai.

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Life in Redmond | oathbystyx.blogspot.com Reviews

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Monday, August 3, 2015. IMO 2010 Problem 6. This is a problem that I've made a couple of attempts to solve in the past few years. Eventually, perhaps encouraged by recent success of nailing IMO 3/6 problems, it is now solved. Let a(1), a(2), . be a sequence of positive real numbers. Suppose that for some positive integers s, we have. A(n)=max{a(k) a(n-k) 1 =k =n-1}. For all n s. Prove that there exist positive integers p and N, with p =s, such that a(n)=a(n-p) a(p) for all n =N. Posted by Shen-Fu Tsai.

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Life in Redmond | oathbystyx.blogspot.com Reviews

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Monday, August 3, 2015. IMO 2010 Problem 6. This is a problem that I've made a couple of attempts to solve in the past few years. Eventually, perhaps encouraged by recent success of nailing IMO 3/6 problems, it is now solved. Let a(1), a(2), . be a sequence of positive real numbers. Suppose that for some positive integers s, we have. A(n)=max{a(k) a(n-k) 1 =k =n-1}. For all n s. Prove that there exist positive integers p and N, with p =s, such that a(n)=a(n-p) a(p) for all n =N. Posted by Shen-Fu Tsai.

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1

Life in Redmond: Swap and Sum

http://www.oathbystyx.blogspot.com/2015/01/swap-and-sum.html

Tuesday, January 13, 2015. There are two keys to solving the problem Swap and Sum. 1 Treap, which enables O(log N) time for swapping two segments and summing over one segment, because we could attach an addition field 'sum' to each node of Treap. However in this problem there is O(N) element swaps per request so the issue isn't resolved yet. With this, any O(N) swap becomes swapping two segments of same length in the transformed array, and every summation splits to two which is still managable.

2

Life in Redmond: Freedom

http://www.oathbystyx.blogspot.com/2012/12/freedom.html

Sunday, December 2, 2012. Http:/ www.cbsnews.com/8301-18560 162-57556662/north-korean-prison-endured-for-23-brutal-years/. 事實上，在Shin逃出來之前，他曾經舉發他的母親和哥哥試圖越獄的打算，因為隱匿不報也會有嚴重的後果。此外因為沒有家庭的觀念，當時的Shin並不掙扎。直到日後在外頭知道了“家庭”，他才有些後悔。此外他當時也希望藉由舉報能得到飽餐一頓的獎賞。 Posted by Shen-Fu Tsai. Very Good Story. Just like before I start my monastic life, I dont know it is possible to live a life without money. January 13, 2013 at 5:11 AM. Then how did you buy air ticket? January 13, 2013 at 9:44 AM. January 18, 2013 at 11:45 AM.

3

Life in Redmond

http://www.oathbystyx.blogspot.com/2012/03/httpwww.html

Wednesday, March 7, 2012. 很好的文章：當你過世時，你會有這些遺憾嗎？ Http:/ www.mortylefkoe.com/regret dying/#. Posted by Shen-Fu Tsai. Subscribe to: Post Comments (Atom). Books, Music, Movie, etc. Film: Lovers of the arctic circles. Song: You got growing up to do. View my complete profile. 很好的文章：當你過世時，你會有這些遺憾嗎？ http:/ www.mortylefkoe.com/. The moment of truth.

4

Life in Redmond: "The moment of truth"

http://www.oathbystyx.blogspot.com/2012/03/moment-of-truth.html

Sunday, March 4, 2012. The moment of truth". 前幾期的時代雜誌提到一個德國攝影師Rineke Dijkstra，專門拍攝人們的“自然表情“。簡單來說就是在捕捉沒有防備、也沒有為了照相而擺出笑臉的表情。例如剛生完小孩的孕婦，疲憊到不可能刻意擺出表情；沒有拍照經驗的小孩，沒有拍照的標準笑容；對拍過照片的小孩，攝影師就把時間故意拉長，等到他們在鏡頭前等到不耐煩的時候，開始無聊、然後開始想辦法自得其樂，露出自然的神情。比較讓我印象深刻的，是結束鬥牛的鬥牛士，剛經歷過生死邊緣的拼搏，身上沾有鮮血，眼神疲憊而略顯呆滯。攝影師說，她就是在捕捉這種"the moment of truth"，雖然有些表情以今天的標準來看都有點窘。這讓我想起在電影冷山裡，Ada在Inman出發去打仗前，給他一張自己的照片，解釋道："I'm not smiling in it. I don't know how to do that, hold a smile". Posted by Shen-Fu Tsai. Subscribe to: Post Comments (Atom).

5

Life in Redmond: IMO 2010 Problem 6

http://www.oathbystyx.blogspot.com/2015/08/imo-2010-problem-6.html

Monday, August 3, 2015. IMO 2010 Problem 6. This is a problem that I've made a couple of attempts to solve in the past few years. Eventually, perhaps encouraged by recent success of nailing IMO 3/6 problems, it is now solved. Let a(1), a(2), . be a sequence of positive real numbers. Suppose that for some positive integers s, we have. A(n)=max{a(k) a(n-k) 1 =k =n-1}. For all n s. Prove that there exist positive integers p and N, with p =s, such that a(n)=a(n-p) a(p) for all n =N. Posted by Shen-Fu Tsai.

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Life in Redmond

Monday, August 3, 2015. IMO 2010 Problem 6. This is a problem that I've made a couple of attempts to solve in the past few years. Eventually, perhaps encouraged by recent success of nailing IMO 3/6 problems, it is now solved. Let a(1), a(2), . be a sequence of positive real numbers. Suppose that for some positive integers s, we have. A(n)=max{a(k) a(n-k) 1 =k =n-1}. For all n s. Prove that there exist positive integers p and N, with p =s, such that a(n)=a(n-p) a(p) for all n =N. Posted by Shen-Fu Tsai.

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