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d-musing @ dncroot

Just another technical blog. The next sequence is…. November 27th, 2007. What is the next sequence given:. Hint: Don’t think in terms of mathematics (i.e. this is not a Fibonacci series) but rather in terms of the English language…. Each line describes the previous line. 1 1 - previous line is “one 1” or “a one”. 2 1 - previous line is “two 1s” or “two number 1s”. 1 2 1 1 - previous line is “one 2, followed by one 1”. 1 1 1 2 1 1 - previous line is “one 1, then one 2, then two 1’s”. So, the next line is:.

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d-musing @ dncroot | dncroot.com Reviews

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Just another technical blog. The next sequence is…. November 27th, 2007. What is the next sequence given:. Hint: Don’t think in terms of mathematics (i.e. this is not a Fibonacci series) but rather in terms of the English language…. Each line describes the previous line. 1 1 - previous line is “one 1” or “a one”. 2 1 - previous line is “two 1s” or “two number 1s”. 1 2 1 1 - previous line is “one 2, followed by one 1”. 1 1 1 2 1 1 - previous line is “one 1, then one 2, then two 1’s”. So, the next line is:.

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d musing @ dncroot,solution,posted in puzzles,comments closed,source wikipedia,lights and switches,lightbulb,1 comment,payments of gold,lru caches,posted in java,programming,given two classes,burning rope,the two trains,a move forward,b move forward,thus

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d-musing @ dncroot | dncroot.com Reviews

https://dncroot.com

Just another technical blog. The next sequence is…. November 27th, 2007. What is the next sequence given:. Hint: Don’t think in terms of mathematics (i.e. this is not a Fibonacci series) but rather in terms of the English language…. Each line describes the previous line. 1 1 - previous line is “one 1” or “a one”. 2 1 - previous line is “two 1s” or “two number 1s”. 1 2 1 1 - previous line is “one 2, followed by one 1”. 1 1 1 2 1 1 - previous line is “one 1, then one 2, then two 1’s”. So, the next line is:.

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dncroot.com

1

LRU Caches « d-musing @ dncroot

http://dncroot.com/2007/09/18/lru-caches

Just another technical blog. Laquo; Strings & Concurrency. Payments of Gold ». Implement a LRU (least-recently used) cache. In the Java utilities package, there is a LinkedHashMap object that can be used to implement an LRU cache by providing a flag in the constructor to order keys by access (instead of insertion), and then to override one method. Static MAX ENTRIES = 100; new LinkedHashMap(12, 0.6f, true) { boolean removeEldestEntry(Map.Entry eldest) { return size() MAX ENTRIES; } }.

2

The next sequence is… « d-musing @ dncroot

http://dncroot.com/2007/11/27/the-next-sequence-is

Just another technical blog. Laquo; The Monty Hall Problem. The next sequence is…. What is the next sequence given:. Hint: Don’t think in terms of mathematics (i.e. this is not a Fibonacci series) but rather in terms of the English language…. Each line describes the previous line. 1 1 - previous line is “one 1” or “a one”. 2 1 - previous line is “two 1s” or “two number 1s”. 1 2 1 1 - previous line is “one 2, followed by one 1”. So, the next line is:. Feed Both comments and pings are currently closed.

3

Burning Rope « d-musing @ dncroot

http://dncroot.com/2007/09/12/burning-rope

Just another technical blog. Laquo; The Two Trains. Strings & Concurrency ». You have two lengths of rope and some matches. If you light the rope on fire, it will be completely consumed in 1 hour. However, you cannot assume the rope burns at any constant rate. That means half the rope might burn in 1 minute and the other half in 59 minutes, and various unknown rates in between. Using the ropes and the matches, determine when 45 minutes has elapsed. Feed Both comments and pings are currently closed.

4

Circular Buffers « d-musing @ dncroot

http://dncroot.com/2007/09/11/circular-buffers

Just another technical blog. Laquo; Apples & Oranges. Finding the lowest common parent ». Implement a FIFO circular buffer. The buffer has two methods: insert() returns 0 if the buffer is not empty. If it is empty, return the oldest entry and replace it with the new one. remove() returns the oldest entry or -1 if the buffer is empty. Another way to check empty is when. If ( ( (next 1) %list.length ) = head). But now your list will never contain the max. number of elements specified in the constructor.

5

Finding the lowest common parent « d-musing @ dncroot

http://dncroot.com/2007/09/11/finding-the-lowest-common-parent

Just another technical blog. Laquo; Circular Buffers. The Two Trains ». Finding the lowest common parent. Given a binary tree and any two nodes, which are not a parent of the other, write an algorithm to find the lowest common parent. You must move the lower node up to the level of the other node, then travel up their parents until you find the common node. N2level) { if (n1.level n2.level) { while (n1.level! N2level) { n1.parent = n1 } } } else { while (n2.level!

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d-musing @ dncroot

Just another technical blog. The next sequence is…. November 27th, 2007. What is the next sequence given:. Hint: Don’t think in terms of mathematics (i.e. this is not a Fibonacci series) but rather in terms of the English language…. Each line describes the previous line. 1 1 - previous line is “one 1” or “a one”. 2 1 - previous line is “two 1s” or “two number 1s”. 1 2 1 1 - previous line is “one 2, followed by one 1”. 1 1 1 2 1 1 - previous line is “one 1, then one 2, then two 1’s”. So, the next line is:.

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