Probstat/random word - ประวัติรุ่นแก้ไข

Jittat: /* Character prediction */

2014-09-09T09:47:21Z

Character prediction

Jittat: /* Character prediction */

2014-09-09T09:46:57Z

Character prediction

Jittat: /* Character prediction */

2014-08-21T03:10:52Z

Character prediction

Jittat: /* Character prediction */

2014-08-21T03:07:30Z

Character prediction

Jittat: /* Character prediction */

2014-08-21T03:07:16Z

Character prediction

Jittat: /* Character prediction */

2014-08-21T03:06:50Z

Character prediction

Jittat: /* Character prediction */

2014-08-21T03:02:51Z

Character prediction

Jittat: /* Word prediction */

2014-08-21T02:58:57Z

Word prediction

Jittat เมื่อ 02:49, 21 สิงหาคม 2557

2014-08-21T02:49:11Z

Jittat เมื่อ 02:48, 21 สิงหาคม 2557

2014-08-21T02:48:59Z

@@ แถว 33: / แถว 33: @@
 == Character prediction ==
-: ''This problem was inspired by [https://www.facebook.com/video.php?v=10152584665476690 this fb status].''
+: ''This problem was inspired by [https://www.facebook.com/video.php?v=10152584665476690 this fb status] by Supasate Choochaisri.''
 Here is a simple character prediction.  Let '''K''' be a fixed parameter.  (You will experiment with various values of '''K''').

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 09:46, 9 กันยายน 2557
แถว 33:		แถว 33:

	== Character prediction ==		== Character prediction ==
		+	: ''This problem was inspired by [https://www.facebook.com/video.php?v=10152584665476690 this fb status].''

	Here is a simple character prediction. Let '''K''' be a fixed parameter. (You will experiment with various values of '''K''').		Here is a simple character prediction. Let '''K''' be a fixed parameter. (You will experiment with various values of '''K''').

@@ แถว 47: / แถว 47: @@
 . find character with the highest probability of appearing after  w  in if you pick a word from the list at random.
           if there are more than one character with the highest probability, pick the smallest one  (e.g., if 'd' and 't' get the highest probability, use 'd'.)
 Can you analyze the performance of this prediction?   More precisely, can you compute the probability that the algorithm predict correctly given that we pick a random word of length at least '''K'''.

@@ แถว 50: / แถว 50: @@
 Can you analyze the performance of this prediction?   More precisely, can you compute the probability that the algorithm predict correctly given that we pick a random word of length at least '''K'''.
-Try to answer in the following table:
+Answer in the following table:
 {| class="wikitable"

@@ แถว 52: / แถว 52: @@
 Try to answer in the following table:
-{|
+{| class="wikitable"
 | '''K'''
 | Number of words whose length is at least '''K'''

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 03:06, 21 สิงหาคม 2557
แถว 48:		แถว 48:
	if there are more than one character with the highest probability, pick the smallest one (e.g., if 'd' and 't' get the highest probability, use 'd'.)		if there are more than one character with the highest probability, pick the smallest one (e.g., if 'd' and 't' get the highest probability, use 'd'.)

−	Can you analyze the performance of this prediction?	+	Can you analyze the performance of this prediction? More precisely, can you compute the probability that the algorithm predict correctly given that we pick a random word of length at least '''K'''.
		+
		+	Try to answer in the following table:
		+
		+	{\|
		+	\| '''K'''
		+	\| Number of words whose length is at least '''K'''
		+	\| Probability of correct prediction
		+	\|-
		+	\|2
		+	\|
		+	\|
		+	\|-
		+	\|3
		+	\|
		+	\|
		+	\|-
		+	\|4
		+	\|
		+	\|
		+	\|-
		+	\|5
		+	\|
		+	\|
		+	\|}

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 03:02, 21 สิงหาคม 2557
แถว 41:		แถว 41:
	* for '''K = 2''', we will use information from all words whose length is at least 2, to predict the 2nd character of a word after receiving the first character of a random word whose length is at least 2.		* for '''K = 2''', we will use information from all words whose length is at least 2, to predict the 2nd character of a word after receiving the first character of a random word whose length is at least 2.
	* for '''K = 4''', we will use information from all words whose length is at least 4, to predict the 4th character of a word after receiving the first 3 characters of a random word whose length is at least 4.		* for '''K = 4''', we will use information from all words whose length is at least 4, to predict the 4th character of a word after receiving the first 3 characters of a random word whose length is at least 4.
		+
		+	Our prediction algorithm is simple:
		+
		+	procedure PREDICT( w ): // note that the length of w K-1 characters
		+	1. find character with the highest probability of appearing after w in if you pick a word from the list at random.
		+	if there are more than one character with the highest probability, pick the smallest one (e.g., if 'd' and 't' get the highest probability, use 'd'.)
		+
		+	Can you analyze the performance of this prediction?

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 02:58, 21 สิงหาคม 2557
แถว 32:		แถว 32:
	* Given that we pick a word without vowels, which character has the highest probability of occurring? What's the probability?		* Given that we pick a word without vowels, which character has the highest probability of occurring? What's the probability?

−	== ~~Word~~ prediction ==	+	== Character prediction ==
		+
		+	Here is a simple character prediction. Let '''K''' be a fixed parameter. (You will experiment with various values of '''K''').
		+
		+	The parameter '''K''' is important as we will take the information from all the words whose length is at least '''K'''. You can ignore all shorter words. We will use that to predict the '''K'''-th after we get '''K - 1''' characters for a valid words.
		+
		+	For example:
		+	* for '''K = 2''', we will use information from all words whose length is at least 2, to predict the 2nd character of a word after receiving the first character of a random word whose length is at least 2.
		+	* for '''K = 4''', we will use information from all words whose length is at least 4, to predict the 4th character of a word after receiving the first 3 characters of a random word whose length is at least 4.

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 02:49, 21 สิงหาคม 2557
แถว 1:		แถว 1:
−	: ''This is part of [[probstat]] course.'''	+	: ''This is part of [[probstat]] course.''

	Download [http://theory.cpe.ku.ac.th/~jittat/probstat/2014/words/words.txt words.txt] which is a list of common 4325 English words. (I obtain this list from [https://www.englishclub.com/vocabulary/common-words.htm], which took it from [http://www.wordfrequency.info/]. The original list of 5000 words contain duplicates because some word appears in many parts of speech. I also removed words with punctuation marks (e.g., o'clock).)		Download [http://theory.cpe.ku.ac.th/~jittat/probstat/2014/words/words.txt words.txt] which is a list of common 4325 English words. (I obtain this list from [https://www.englishclub.com/vocabulary/common-words.htm], which took it from [http://www.wordfrequency.info/]. The original list of 5000 words contain duplicates because some word appears in many parts of speech. I also removed words with punctuation marks (e.g., o'clock).)

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 02:48, 21 สิงหาคม 2557
แถว 1:		แถว 1:
		+	: ''This is part of [[probstat]] course.'''
		+
	Download [http://theory.cpe.ku.ac.th/~jittat/probstat/2014/words/words.txt words.txt] which is a list of common 4325 English words. (I obtain this list from [https://www.englishclub.com/vocabulary/common-words.htm], which took it from [http://www.wordfrequency.info/]. The original list of 5000 words contain duplicates because some word appears in many parts of speech. I also removed words with punctuation marks (e.g., o'clock).)		Download [http://theory.cpe.ku.ac.th/~jittat/probstat/2014/words/words.txt words.txt] which is a list of common 4325 English words. (I obtain this list from [https://www.englishclub.com/vocabulary/common-words.htm], which took it from [http://www.wordfrequency.info/]. The original list of 5000 words contain duplicates because some word appears in many parts of speech. I also removed words with punctuation marks (e.g., o'clock).)