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Median Workbench
/* SQL Server database engine doesn't have a MEDIAN() aggregate function. This is probably because there are several types of median, such as statistical, financial or vector medians. Calculating
Medians are essentially a row-positioning task, since medians are the middle value of an ordered result. Easy to do in SQL? Nope. Joe Celko explains why
The workbench requires SQL Server 2008, but is easily modified to 2005.*/
A Date for an Mock Feud
If you are an older SQL programmer, you will remember back in the 1990’s there were two database magazines on the newsstands, DBMS and DATABASE PROGRAMMING & DESIGN. They started as competitors, but
Miller-Freeman bought DBMS from M&T and kept publishing both titles. Later the two titles are merged into INTELLIGENT ENTERPRISE from CMP Publications, which is more management oriented than the two
original magazines.
I started a regular freelance column in DATABASE PROGRAMMING & DESIGN in 1990 and switched to DBMS in 1992 as an employee. Chris Date took my old slot back at DATABASE PROGRAMMING & DESIGN. There was
a period where we did a version of the famous Fred Allen and Jack Benny mock feud from the classic era of American radio. Chris would run a column with some position on a topic in his column and I
would counter it in my next column. It was great fun, it made people think and (most important) it sold magazines and kept me employed.
One of the topics that bounced around the SQL community in those columns was how to compute the Median with SQL. So Chris and I had to get involved in that and toss code around. Today, the Median is
easy to do, but it was a good programming exercise in its day.
The 'Simple', or financial Median
The Median is not a built-in function in Standard SQL, like AVG(), and it has several different versions. In fact, most descriptive statistics have multiple versions, which is why SQL stayed away
from them and left that job to specialized packages like SAS and SPSS. I will not even mention floating point error handling.
The simple Median is defined as the value for which there are just as many cases with a value below it as above it. If such a value exists in the data set, this value is called the statistical
Median. If no such value exists in the data set, the usual method is to divide the data set into two halves of equal size such that all values in one half are lower than any value in the other half.
The Median is then the average of the highest value in the lower half and the lowest value in the upper half, and is called the financial Median. In English, sort the data, cut it in the middle and
look at the guys on either side of the cut.
The financial Median is the most common term used for this Median, so we will stick to it. Let us use Date's famous Parts table, from several of his textbooks, which has a column for part_wgt in it,
CREATE TABLE Parts
(part_nbr VARCHAR(5) NOT NULL PRIMARY KEY,
part_name VARCHAR(50) NOT NULL,
part_color VARCHAR(50) NOT NULL,
part_wgt INTEGER NOT NULL,
city_name VARCHAR(50) NOT NULL);
INSERT INTO Parts (part_nbr, part_name, part_color, part_wgt, city_name)
VALUES ('p1', 'Nut', 'Red', 12, 'London'),
('p2', 'Bolt', 'Green', 17, 'Paris'),
('p3', 'Cam', 'Blue', 12, 'Paris'),
('p4', 'Screw', 'Red', 14, 'London'),
('p5', 'Cam', 'Blue', 12, 'Paris'),
('p6', 'Cog', 'Red', 19, 'London');
First sort the table by weights and find the three rows in the lower half of the table. The greatest value in the lower half is 12; the smallest value in the upper half is 14; their average, and
therefore the Median, is 13. If the table had an odd number of rows, we would have looked at only one row after the sorting.
Date's First Median
Date proposed two different solutions for the Median. His first solution was based on the fact that if you duplicate every row in a table, the Median will stay the same. The duplication will
guarantee that you always work with a table that has an even number of rows. Since Chris (and myself) are strongly opposed to redundant data, this was an unusual approach for him to use.
The first version that appeared in his column was wrong and drew some mail from me and from others who had different solutions. Here is a corrected version of his first solution:
CREATE VIEW Temp1
AS SELECT part_wgt FROM Parts
SELECT part_wgt FROM Parts;
CREATE VIEW Temp2
AS SELECT part_wgt
FROM Temp1
WHERE (SELECT COUNT(*) FROM Parts)
<= (SELECT COUNT(*)
FROM Temp1 AS T1
WHERE T1.part_wgt >= Temp1.part_wgt)
AND (SELECT COUNT(*) FROM Parts)
<= (SELECT COUNT(*)
FROM Temp1 AS T2
WHERE T2.part_wgt <= Temp1.part_wgt);
SELECT AVG(DISTINCT part_wgt) AS median
FROM Temp2;
This involves the construction of a table of duplicated values, which can be expensive in terms of both time and storage space. The use of AVG(DISTINCT x) is important because leaving it out would
return the simple average instead of the Median. Consider the set of weights (12, 17, 17, 14, 12, 19). The doubled table, Temp1, is then (12, 12, 12, 12, 14, 14, 17, 17, 17, 17, 19, 19). But because
of the duplicated values, Temp2 becomes (14, 14, 17, 17, 17, 17), not just (14, 17). The simple average is (96 / 6.0) = 16; it should be (31/2.0) = 15.5 instead.
Celko's First Median
I found a slight modification of Date's solution will avoid the use of a doubled table, but it depends on a CEILING() function.
SELECT MIN(part_wgt) -- smallest value in upper half
FROM Parts
WHERE part_wgt
IN (SELECT P1.part_wgt
FROM Parts AS P1, Parts AS P2
WHERE P2.part_wgt >= P1.part_wgt
GROUP BY P1.part_wgt
HAVING COUNT(*)
<= (SELECT CEILING(COUNT(*) / 2.0)
FROM Parts)
SELECT MAX(part_wgt) -- largest value in lower half
FROM Parts
WHERE part_wgt IN (SELECT P1.part_wgt
FROM Parts AS P1, Parts AS P2
WHERE P2.part_wgt <= P1.part_wgt
GROUP BY P1.part_wgt
HAVING COUNT(*) <=
(SELECT CEILING(COUNT(*) / 2.0)
FROM Parts)))
--or using the same idea and a CASE expression:
SELECT AVG(DISTINCT CAST(part_wgt AS FLOAT)) AS median
FROM (SELECT MAX(part_wgt)
FROM Parts AS B1
WHERE (SELECT COUNT(*) + 1
FROM Parts
WHERE part_wgt < B1.part_wgt)
<= (SELECT CEILING (COUNT(*) / 2.0)
FROM Parts)
SELECT MAX(part_wgt)
FROM Parts AS B
WHERE (SELECT COUNT(*) + 1
FROM Parts
WHERE part_wgt < B.part_wgt)
<= CASE (SELECT (COUNT(*)% 2)
FROM Parts)
WHEN 0
THEN (SELECT CEILING (COUNT(*)/2.0) + 1
FROM Parts)
ELSE (SELECT CEILING (COUNT(*)/2.0)
FROM Parts)
END) AS medians(part_wgt);
The CEILING() function is to be sure that if there is an odd number of rows in Parts, the two halves will overlap on that value.
Date's Second Median
Now the war was on! Did you notice that we had to use a lot of self-joins in those days? That stinks. Date's second solution was based on Celko's Median, folded into one query:
SELECT AVG(DISTINCT Parts.part_wgt) AS median
FROM Parts
WHERE Parts.part_wgt IN
(SELECT MIN(part_wgt)
FROM Parts
WHERE Parts.part_wgt
IN (SELECT P2.part_wgt
FROM Parts AS P1, Parts AS P2
WHERE P2.part_wgt <= P1.part_wgt
GROUP BY P2.part_wgt
HAVING COUNT(*)
<= (SELECT CEILING(COUNT(*) / 2.0)
FROM Parts))
SELECT MAX(part_wgt)
FROM Parts
WHERE Parts.part_wgt
IN (SELECT P2.part_wgt
FROM Parts AS P1, Parts AS P2
WHERE P2.part_wgt >= P1.part_wgt
GROUP BY P2.part_wgt
HAVING COUNT(*)
<= (SELECT CEILING(COUNT(*) / 2.0)
FROM Parts)));
Date mentions that this solution will return a NULL for an empty table and that it assumes there are no NULLs in the column. But you have to remember Chris Date, unlike Dr. Codd and SQL, does not use
NULLs and thinks they should not be in the Relational Model. If there are NULLs, the COUNT(*) will include them and mess up the computations. You will need to get rid of them with a COUNT(part_nbr)
instead. Play with some sample data and see the differences.
Murchison's Median
Others joined in the battle. Rory Murchison of the Aetna Institute had a solution that modifies Date's first method by concatenating the key to each value to make sure that every value is seen as a
unique entity. Selecting the middle values is then a special case of finding the n-th item in the table.
SELECT AVG(part_wgt)
FROM Parts AS P1
(SELECT COUNT(*)
FROM Parts AS P2
WHERE CAST(part_wgt AS CHAR(5)) + P2.part_nbr >=
CAST(part_wgt AS CHAR(5)) + P1.part_nbr
HAVING COUNT(*) = (SELECT FLOOR(COUNT(*) / 2.0)
FROM Parts)
OR COUNT(*) = (SELECT CEILING((COUNT(*) / 2.0))
FROM Parts));
This method depends on being able to have a HAVING clause without a GROUP BY clause, which is part of the ANSI standard but often missed by new programmers.
Occurrence Count Table
I came back with a working table with the values and a tally of their occurrences from the original table. Now that we have this table, we want to use it to construct a summary table that has the
number of occurrences of each part_wgt and the total number of data elements before and after we add them to the working table.
Philip Vaughan of San Jose, CA proposed a simple Median technique based on a VIEW with unique weights and number of occurrences and then a VIEW of the middle set of weights.
CREATE VIEW ValueSet(part_wgt, occurrence_cnt)
AS SELECT part_wgt, COUNT(*)
FROM Parts
GROUP BY part_wgt;
The MiddleValues VIEW is used to get the Median by taking an average. The clever part of this code is the way that it handles empty result sets in the outermost WHERE clause that result from having
only one value for all weights in the table. Empty sets sum to NULL because there is no element to map the index
CREATE VIEW MiddleValues(part_wgt)
AS SELECT part_wgt
FROM ValueSet AS VS1
WHERE (SELECT SUM(VS2.occurrence_cnt)/2.0 + 0.25
FROM ValueSet AS VS2) >
(SELECT SUM(VS2.occurrence_cnt)
FROM ValueSet AS VS2
WHERE VS1.part_wgt <= VS2.part_wgt) - VS1.occurrence_cnt
AND (SELECT SUM(VS2.occurrence_cnt)/2.0 + 0.25
FROM ValueSet AS VS2) >
(SELECT SUM(VS2.occurrence_cnt)
FROM ValueSet AS VS2
WHERE VS1.part_wgt >= VS2.part_wgt) - VS1.occurrence_cnt;
SELECT AVG(part_wgt) AS median FROM MiddleValues;
Median with Characteristic Function
Anatoly Abramovich, Yelena Alexandrova, and Eugene Birger presented a series of articles in SQL Forum magazine on computing the Median (SQL Forum 1993, 1994). They define a characteristic function,
which they call delta, using the SIGN() function. The delta or characteristic function accepts a Boolean expression as an argument and returns one if it is TRUE and zero if it is FALSE or UNKNOWN. We
can construct the delta function easily with a CASE expression.
The authors also distinguish between the statistical Median, whose value must be a member of the set, and the financial Median, whose value is the average of the middle two members of the set. A
statistical Median exists when there is an odd number of items in the set. If there is an even number of items, you must decide if you want to use the highest value in the lower half (they call this
the left Median) or the lowest value in the upper half (they call this the right Median).
The left statistical Median of a unique column can be found with this query, if you will assume that we have a column called bin_nbr that represents the storage location of a part in sorted order.
SELECT P1.bin_nbr
FROM Parts AS P1, Parts AS P2
GROUP BY P1.bin_nbr
HAVING SUM(CASE WHEN (P2.bin_nbr <= P1.bin_nbr) THEN 1 ELSE 0 END)
= (COUNT(*) / 2.0);
Changing the direction of the test in the HAVING clause will allow you to pick the right statistical Median if a central element does not exist in the set. You will also notice something else about
the Median of a set of unique values: It is usually meaningless. What does the Median bin_nbr number mean, anyway? A good rule of thumb is that if it does not make sense as an average, it does not
make sense as a Median.
The statistical Median of a column with duplicate values can be found with a query based on the same ideas, but you have to adjust the HAVING clause to allow for overlap; thus, the left statistical
Median is found by
SELECT P1.part_wgt
FROM Parts AS P1, Parts AS P2
GROUP BY P1.part_wgt
HAVING SUM(CASE WHEN P2.part_wgt <= P1.part_wgt
THEN 1 ELSE 0 END)
>= (COUNT(*) / 2.0)
AND SUM(CASE WHEN P2.part_wgt >= P1.part_wgt
THEN 1 ELSE 0 END)
>= (COUNT(*) / 2.0);
Should Parts contain an even number of entries with the (COUNT(*)/ 2) entry not repeated, this query may return FLOOR(AVG(DISTINCT part_wgt)), as happens when SQL Server computes an average of
integers. This can be fixed by changing the inner SELECT to:
SELECT (P1.part_wgt * 1.0) AS part_wgt
Notice that here the left and right medians can be the same, so there is no need to pick one over the other in many of the situations where you have an even number of items. Switching the comparison
operators in the two CASE expressions will give you the right statistical Median.
Henderson's Median
Another version of the financial Median, which uses the CASE expression in both of its forms, is:
SELECT CASE COUNT(*) % 2
WHEN 0 -- even sized table
THEN (P1.part_wgt + MIN(CASE WHEN P2.part_wgt > P1.part_wgt
THEN P2.part_wgt
ELSE NULL END))/2.0
ELSE P1.part_wgt --odd sized table
END AS median
FROM Parts AS P1, Parts AS P2
GROUP BY P1.part_wgt
HAVING COUNT(CASE WHEN P1.part_wgt >= P2.part_wgt
THEN 1
ELSE NULL END)
= (COUNT(*) + 1) / 2;
This answer is due to Ken Henderson. The only instance in which [[2]] is correct is when Parts has an odd number of entries and the middle (COUNT(*) / 2 + 1) entry is not repeated in the data.
Celko's Third Median
Another approach involves looking at a picture of a line of sorted values and seeing where the Median would fall. Every value in column part_wgt of the table partitions the table into three sections,
the values that are less than part_wgt, equal to part_wgt or greater than part_wgt. We can get a profile of each value with a tabular subquery expression.
Deriving the query is a great exercise in SQL logic and math. Don't do it unless you just like the exercise. The query finally becomes:
SELECT AVG(DISTINCT part_wgt)
FROM (SELECT P1.part_wgt
FROM Parts AS P1, Parts AS P2
GROUP BY P1.part_nbr, P1.part_wgt
HAVING SUM(CASE WHEN P2.part_wgt = P1.part_wgt
THEN 1 ELSE 0 END)
>= ABS(SUM(CASE WHEN P2.part_wgt < P1.part_wgt THEN 1
WHEN P2.part_wgt > P1.part_wgt THEN -1
ELSE 0 END)))
AS Partitions;
OLAP Median
The new OLAP function in the SQL-99 Standard allow you to replace COUNT() functions with row numberings. Let’s assume we have a table or VIEW named RawData that has only one numeric column. The
obvious solution for data with an odd number of elements is:
WITH SortedData (x, sort_place)
(SELECT x, ROW_NUMBER() OVER(ORDER BY x ASC)
FROM RawData)
SELECT x AS median
FROM SortedData
WHERE sort_place = ((SELECT COUNT(*) FROM RawData)/ 2) + 1;
--A slight modification for an even number of elements:
WITH SortedData (x, sort_place)
(SELECT x, ROW_NUMBER() OVER(ORDER BY x ASC)
FROM RawData)
SELECT AVG(x) AS median
FROM SortedData
WHERE sort_place
IN ((SELECT COUNT(*) FROM RawData)/ 2),
(SELECT COUNT(*) FROM RawData)/ 2) + 1);
You can combine both of these into one query with a CASE expression and I leave that as an exercise for the reader.
But there is another question we have not asked about an even number of rows. Assume the data set look like {1, 2, 2, 3, 3, 3} so the usual median would be AVG({2, 3}) = 2.5 as the measure of central
tendency. But if we consider where the real trend is, then we would use the set-oriented weighted median with AVG({2, 2, 3, 3, 3}) = 13/5 = 2.6 as the measure of central tendency.
This leads to the final query which can handle odd and even counts and do the weighted Median:
WITH SortedData (x, hi, lo)
(SELECT x,
ROW_NUMBER() OVER(ORDER BY x ASC),
ROW_NUMBER() OVER(ORDER BY x DESC)
FROM RawData)
SELECT AVG(x * 1.0) AS median
FROM SortedData
WHERE hi IN (lo, lo+1, lo-1);
This might be slower than you would like, since the CTE might do two sorts to get the "hi" and "lo" values. At the time of this writing, ROW_NUMBER() is new to all SQL products, so not as much work
has gone into optimizing them. A smarter optimizer would map the i-th element of a sorted list of (n) items in ASC order to the (n-i+1)-th element for DESC.
Want to try your own versions?
Have Your Say
Do you have an opinion on this article? Then add your comment below:
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Subject: Hi Joe!
Posted Dave Winters (not signed in)
Posted Friday, April 10, 2009 at 12:44 PM
Message: Well I see you are doing OK!
It amazing that after everything I taught you at EES you are
still messing around with SQL. :)
Dave Winters
Subject: Hi Dave!
Posted Joe Celko (not signed in)
Posted Monday, April 13, 2009 at 8:58 AM
Message: I stick with what I know :) You used to write an SQL Server
column, too!
Subject: Was this content reused from somewhere else?
Posted Adam Machanic (view profile)
Posted Monday, April 13, 2009 at 11:36 AM
Message: You mention characteristic functions, then use a CASE
expression in the actual code example... Which is a good
thing, but it's a bit confusing.
Aside from that, interesting article.
Subject: Re: Content reuse
Posted Phil Factor (view profile)
Posted Monday, April 13, 2009 at 4:14 PM
Message: Some of this was originally in 'SQL for Smarties', which is
one of my favorite books on SQL, along with Ken's 'Guru's
guide'. We all liked the idea of an update done as a
Agreed that 'characteristic Functions' were clever, but made
superfluous by the CASE statement. Joe says in this workbench
'The delta or characteristic function accepts a Boolean
expression as an argument and returns one if it is TRUE and
zero if it is FALSE or UNKNOWN. We can construct the delta
function easily with a CASE expression'.
We'll discuss with Joe a slight re-wording to take out any
possible confusion.
Subject: Hmm...
Posted Adam Machanic (view profile)
Posted Tuesday, April 14, 2009 at 4:25 PM
Message: IMO you should remove the reference altogether. Every six
months or so I run into someone on a forum who just discovered
characteristic functions and believes that because they look
cool and difficult to understand that they're faster than CASE
expressions... I would rather just forget about them
altogether at this point!
Subject: Implementation paranoia
Posted puzsol (view profile)
Posted Thursday, April 16, 2009 at 7:08 AM
Message: I hate to spoil the party of the final solution... but isn't
it slightly (in theory, not in practice) implementation
I mean if the asc and desc are done separately then there is
no gurantee of correlation of the hi and lo values. It (in
theory) would be possible for four (or more) identical values
in the middle to have the data:
x, hi, lo
12, 10, 12
12, 11, 13
12, 12, 10
12, 13, 11
which never matches the in condition, and fails to give a
median value... (AVG(null))
Of course in practice that won't happen as you are more likely
to either get exactly the same indexes (10 to 13 matching 10
to 13 resulting in all similar values being included in the
average) or exact oposite indexes (10 to 13 matching 13 to 10
resulting in the correct middle two being returned) - either
case being the average of a single number and returning the
correct value.
Hey it may be something you dug up from an old column, but it
still has the ability to make you think!
Subject: Your math is wrong ...
Posted Celko (view profile)
Posted Thursday, April 16, 2009 at 4:28 PM
Message: WITH SortedData (x, hi, lo)
(SELECT x,
ROW_NUMBER() OVER(ORDER BY x ASC),
ROW_NUMBER() OVER(ORDER BY x DESC)
FROM RawData)
SELECT AVG(x * 1.0) AS median
FROM SortedData
WHERE hi IN (lo, lo+1, lo-1);
>> I mean if the ASC and DESC are done separately then there
is no guarantee of correlation of the hi and lo values. <<
There is a strong relationship between the hi and lo values. A
good optimizer will look at the first ROW_NUMBER() function,
see that it differs from the second ROW_NUMBER() function only
by direction. The second ROW_NUMBER() function can computed
from the results of the first one -- (COUNT(x)+1 - ROW_NUMBER
() OVER(ORDER BY x ASC)), so there is only one sort.
>> It (in theory) would be possible for four (or more)
identical values in the middle to have the data:<<
No. do the math:
Odd case:
X: 10 11 12 12 12 12 12 13 14
COUNT(*) + 1 = 10
hi 1 2 3 4 5 6 7 8 9
lo 9 8 7 6 5 4 3 2 1
match ^
Even case:
X: 10 11 12 12 12 12 13 14
COUNT(*) + 1 = 9
hi 1 2 3 4 5 6 7 8
lo 8 7 6 5 4 3 2 1
match ^ ^
There is a classic logic problem based on this principle. A
Zen monk walks up a hill to a temple the first day. The next
day, he walks down the hill. Was he ever at one spot on the
trail at the same time on both days? Yes, one and only one
such spot exits. See:
Subject: Wrong?
Posted puzsol (view profile)
Posted Thursday, April 16, 2009 at 7:14 PM
Message: I did say it wouldn't be the outcome of a good optimiser... I
didn't even say it was practical - just theoretical... if you
look at the numbers I put out, there is no match, the hi and
lo values do form a correct sequence (all be it jumbled).
I was merely pointing out that IF two independent sorts are
done, then it COULD depend on the sorting algorithm. Sorting
does not have to be in a straight line (like the Monk), it
just has to look like it's in order, and different algorithms
will do different things when values are said to be equal (to
swap or not to swap?).
Case in point... In SQL Server 2005 create a table (called
Median) with primary key "id" (int) identity field, "val"
(int) as a secondary field, then insert 9 identical values in
the val column (I used "5")... now try these:
select * from Median order by val asc
select * from Median order by val desc
The listings are identical.... ie listed in primary key order
because the secondary values are all the same... so on the
second day did the Monk warp back to the bottom of the hill
climb up the hill again then warp back? (Beam me up
So if these were the median values, you would include them all
by the query! (as the Monk walked an identical path)... Turn
the select AVG into a select * - I had nine values included
from the setup above! (of course it doesn't matter as they are
identical, but it's the principle that I am talking about, not
whether the output happens to be correct.)
My question would be; what happens if identical median values
cross a page boundary, do you get a disjointed sequence (asc
vs desc?)?
val, hi, lo
5, 1, 4
5, 2, 5
5, 3, 6
5, 4, 7
--- page boundary
5, 5, 1
5, 6, 2
5, 7, 3
Don't ask me how the Monk would make that journey!
Subject: puzsol is right
Posted StefanG (view profile)
Posted Friday, April 17, 2009 at 9:29 AM
Message: Joe Celko's last query has exactly the flaws puzsol is talking
It is very easy to demonstrate, we just have to use Joe's own
example data set on SQL server 2005.
Like this:
create table RawData (x int)
insert into RawData (x) values (1)
insert into RawData (x) values (2)
insert into RawData (x) values (2)
insert into RawData (x) values (3)
insert into RawData (x) values (3)
insert into RawData (x) values (3)
WITH SortedData (x, hi, lo)
(SELECT x,
ROW_NUMBER() OVER(ORDER BY x ASC),
ROW_NUMBER() OVER(ORDER BY x DESC)
FROM RawData)
SELECT AVG(x * 1.0) AS median
FROM SortedData
WHERE hi IN (lo, lo+1, lo-1);
This returns null !
The reason is easy to see if we look at the full SortedData
result with:
WITH SortedData (x, hi, lo)
(SELECT x,
ROW_NUMBER() OVER(ORDER BY x ASC),
ROW_NUMBER() OVER(ORDER BY x DESC)
FROM RawData)
SELECT * FROM SortedData
which returns:
x hi lo
--- -------------------- --------------------
None of these rows satisfy the where-condition
This nicely illustrates exactly the problem puzsol is talking
about - the correlation between lo and hi is much weaker than
Joe assumes.
Subject: Single query for even add odd counts
Posted StefanG (view profile)
Posted Friday, April 17, 2009 at 10:15 AM
Message: Joe's last query simply does not work.
A much better single query that really works for both even and
odd number of elements is this:
WITH SortedData (x, sort_place)
(SELECT x, ROW_NUMBER() OVER(ORDER BY x ASC)
FROM RawData)
SELECT AVG(x*1.0) AS median
FROM SortedData
WHERE sort_place BETWEEN ((SELECT COUNT(*) FROM RawData)/ 2.0)
AND ((SELECT COUNT(*)+2 FROM RawData)/ 2.0);
If you really want a set-based median (which I find a strange
concept) you could use this:
WITH SortedData (x, sort_place)
(SELECT x, ROW_NUMBER() OVER(ORDER BY x ASC)
FROM RawData)
SELECT AVG(x*1.0) AS median
FROM SortedData
WHERE x in (
SELECT x FROM SortedData
WHERE sort_place BETWEEN
((SELECT COUNT(*) FROM RawData)/ 2.0) AND
((SELECT COUNT(*)+2 FROM RawData)/ 2.0))
Subject: Slight change to OLAP version
Posted dnoeth (view profile)
Posted Wednesday, May 27, 2009 at 3:57 PM
Message: Why mixing aggregates and OLAP?
I usually prefer to access a table only once, especially if
this might be more complex involving joins/groups/conditions.
This is what i used for years (because my Teradata implemented
OLAP functions since 1999), it works for odd and even counts:
WITH SortedData (x, row_num, cnt)
SELECT x,
ROW_NUMBER() OVER(ORDER BY x ASC),
COUNT(*) OVER ()
FROM RawData
SELECT AVG(x * 1.0) AS median
FROM SortedData
row_num in
((cnt + 1) / 2, (cnt / 2) + 1)
Btw, Puzsol is right, but as Celko already pointed out, the
second ROW_NUMBER is easily replaced by
(COUNT(x)+1 - ROW_NUMBER() OVER(ORDER BY x ASC)):
WITH SortedData (x, hi, lo)
SELECT x,
ROW_NUMBER() OVER(ORDER BY x ASC),
COUNT(*) OVER () + 1
- ROW_NUMBER() OVER(ORDER BY x ASC)
FROM RawData
SELECT AVG(x * 1.0) AS median
FROM SortedData
WHERE hi IN (lo, lo+1, lo-1);
Subject: Median and Weighted Median found here
Posted Peso (view profile)
Posted Saturday, September 19, 2009 at 5:30 AM
Message: http://sqlblog.com/blogs/peter_larsson/archive/2009/09/18/ | {"url":"https://www.simple-talk.com/sql/t-sql-programming/median-workbench/","timestamp":"2014-04-18T06:06:46Z","content_type":null,"content_length":"126792","record_id":"<urn:uuid:d7904263-add1-47b9-8595-22055b9b66a9>","cc-path":"CC-MAIN-2014-15/segments/1398223203235.2/warc/CC-MAIN-20140423032003-00104-ip-10-147-4-33.ec2.internal.warc.gz"} |
MathGroup Archive: January 2001 [00302]
[Date Index] [Thread Index] [Author Index]
Re: Who can help me?
• To: mathgroup at smc.vnet.net
• Subject: [mg26851] Re: [mg26778] Who can help me?
• From: Ranko Bojanic <bojanic at math.ohio-state.edu>
• Date: Thu, 25 Jan 2001 01:13:28 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
Jacqueline Zizi wrote:
>I'm working on this polynomial linked to the truncated icosahedron:
>-17808196677858180 x+
>138982864440593250 x^2-527304830550920588 x^3+
>1301702220253454898 x^4-2358155595920193382 x^5+
>3347791850698681436 x^6-3878279506351645237 x^7+
>3764566420106299695 x^8-3117324712750504866 x^9+
>2229873533973727384 x^10-1390372935143028255 x^11+
>760794705528035032 x^12-367240961907017721 x^13+
>157018216115380477 x^14-59650776196609992 x^15+
>20179153653354540 x^16-6086251542996201 x^17+
>1637007669992780 x^18-392300104078670 x^19+
>83589038962550 x^20-15782712151030 x^21+
>2628070696678 x^22-383466859804 x^23+48618908986 x^24-
>5298021900 x^25+489095520 x^26-37516324 x^27+
>2327268 x^28-112200 x^29+3945 x^30-90 x^31+x^32;
>I'm interested at its value for xÆ2+2 Cos[2[Pi]/7].
>Taking N[] gives 3.2628184 10^7
>But if I simplify first and then take N[] it gives-0.0390625+
>As it is a polynomial with integer coefficients,and 2+2 Cos[2 pi/7]
> is real too,the result should be real.So I prefer the 1st
>solution,but for another reason,I'm not so sure of this result.
>A Plot between 3 and 3.5,does not help me neither to check if the
>value 3.2628184 is good and If I do:polynomial/.xÆ3.2628184
>10^7,it gives 2.7225238332205106`^240
>How could I check the result 3.2628184 10^7?
What you have here is a severe loss of precision of computations.
Let f[x] be the polynomial from the message. Following a Ted Ersek's
recommendation, here is what you have to do in such cases:
g[x_] := Module[{ t }, f[t] /.t -> SetPrecision[x , 60]]
You will have then no problems drawing the graph of g on
any interval you like and you will see that
g[2+2 Cos[2 Pi / 7]] = 0.001080560723438890436215553236443 | {"url":"http://forums.wolfram.com/mathgroup/archive/2001/Jan/msg00302.html","timestamp":"2014-04-16T10:28:55Z","content_type":null,"content_length":"36102","record_id":"<urn:uuid:f2968f51-3d95-4315-bc03-22b137d5498f>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00197-ip-10-147-4-33.ec2.internal.warc.gz"} |
It's the same?, problem with the king of chess
It's the same?, problem with the king of chess
Given an nxn chessboard size, a king is placed in a box
arbitrary coordinate ( x , y) . The problem is to determine the n ^ 2-1
Figure movements so that all board squares are visited once, if there is such a sequence of movements .
The solution to the problem can be expressed as a matrix of dimension nxn representing the chessboard. Each element ( x , y) of this matrix solution
k contain a natural number indicating the order number that has been
accessed box coordinates ( x , y) .
The algorithm works in stages at each stage k deciding where to
moves .
As there are eight possible moves in each step , this will be the
maximum number of children that will be generated by each node .
As for explicit restrictions , the way in which we defined
structure representing the solution ( in this case a two dimensional array of
natural) numbers , we know that its components can be numbers
between zero (indicating that a cell has not been visited yet) n ^ 2 , which is the order of the last possible move . Initially the board is filled with zeros and there is a 1 only in the
original box ( x0 , y0 ) .
Implied restrictions in this case will limit the number of children who
generated from a square by checking the movement does not take the king off the board or on a box previously visited .
Having defined the structure that represents the solution and the restrictions
I will use to implement the algorithm that solves the problem just use the general scheme , obtaining :
ledesma The mov_x and mov_y variables contain the legal moves of a King
(according to the rules of chess), and are initialized at the beginning of the main program by MovimientosPosibles case:
I have implemented well in Mathematica
movx[1] = 0; movx[2] = -1; movx[3] = -1; movx[4] = -1; movx[5] = 0;
movx[6] = 1; movx[7] = 1; movx[8] = 1;
movy[1] = 1; movy[2] = 1; movy[3] = 0; movy[4] = -1; movy[5] = -1;
movy[6] = -1; movy[7] = 0; movy[8] = 1;
rey[k_, x_, y_] := Module[{orden, exito, u, v},
orden = 0;
exito = False;
While[orden < 8 \[Or] exito,
u = x + movx[orden];
v = y + movy[orden];
If[(1 <= u) && (u <= n) && (1 <= v) && (v <=
n) && (tablero[[u, v]] == 0),
tablero[[u, v]] = k;
If[k <= n^2,
rey[k + 1, u, v];
If[! exito, tablero[[u, v]] == 0]
exito = True;
n = 4;
tablero = Array[0 &, {n, n}];
here is the solution according to the code that I showed at the beginning,
that our approach would be to
luis ledesma {{12,1,2,3},{13,11,4,5},{14,10,6,7},{15,16,9,8}} placed in the variable tablero.
my problem is that mathematica, repeat some numbers in the solution, as seen in the example of rey[1,1,1], any idea is welcome, greetings guys
Try replacing
If[! exito, tablero[[u, v]] == 0]
Todd Rowland
If[! exito, tablero[[u, v]] = 0]
It would be interesting to see a Mathematica solution in a more native style.
Thanks Todd for you help, but when performed the change, my program fall in a loop, you believe the conversion of the loop UNTIL is bad?, or the parameters are out of domain, because me not
luis finding the solution, I will work in that, I hope someone can help me. Greetings
Sorry I missed this the first time. It should be
While[orden<8 && !exito,
Todd Rowland which is the translation of do{} while (odern>=8 || exito)
Try using ArrayPlot to visualize the answer. You can also make a more sophisticated graphic with Arrow.
Hello, I've made the changes in the cycle WHILE, but when running the program with the following entry rey[1,1,1], mathematica is very busy and can only be stopped by using the command to
abort , it is the combination ctrl +,.I show you in the picture below
ledesma I hope your instructios for this case, greetings and thanks in advance
To translate the Modula-2 code faithfully, a local variable should be global and a weak inequality should be strict.
Ilian Gachevski
It is amazing!, Ilian Gachevski, I worked very hard in the solution, but cannot find,thanks for your help, is invaluable, greetings everybody
luis ledesma | {"url":"http://community.wolfram.com/groups/-/m/t/187142?p_p_auth=cLXn0OCV","timestamp":"2014-04-16T21:51:39Z","content_type":null,"content_length":"94351","record_id":"<urn:uuid:23a30636-7fc7-48d3-894f-e6c19b9644b5>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00524-ip-10-147-4-33.ec2.internal.warc.gz"} |
Independent Samples t-Test: Chips Ahoy® vs. Supermarket Brand
Independent Samples t-Test: Chips Ahoy® vs. Supermarket Brand
This activity has been undergone anonymous peer review.
This activity was anonymously reviewed by educators with appropriate statistics background according to the CAUSE review criteria for its pedagogic collection.
This page first made public: May 17, 2007
This material is replicated on a number of sites as part of the
SERC Pedagogic Service Project
In this hands-on activity, students count the number of chips in cookies in order to carry out an independent samples t-test to see if Chips Ahoy® cookies have a higher, lower, or different mean
number of chips per cookie than a supermarket brand. First there is a class discussion that can include concepts about random samples, independence of samples, recently covered tests, comparing two
parameters with null and alternative hypotheses, what it means to be a chip in a cookie, how to break up the cookies to count chips, and of course a class consensus on the hypotheses to be tested.
Second the students count the number of chips in a one cookie from each brand, and report their observations to the instructor. Third, the instructor develops the independent sample t-test statistic.
Fourth, the students carry out (individually or as a class) the hypothesis test, checking the assumptions on sample-size/population-shape.
Learning Goals
There are four (4) goals for this lesson:
1. Reinforce concepts of (simple) random samples.
2. Introduce the concept of independent samples (vs. paired samples, if appropriate).
3. Introduce null and alternative hypotheses for comparing two parameters, and the different ways to write them down (i.e., mu1 = mu2 vs. mu1 – mu2 = 0).
4. Address the importance of having uniform guidelines for recording data, and possible, the learning effect.
Context for Use
This activity
1. Can be used at any educational level, or any course, that covers the independent samples t-test (or t- confidence interval).
2. It can be inserted in any existing format for presenting the independent samples t-test.
3. The actual counting of the chips and recording the data (classroom computer) may take 15 minutes, depending on the class size.
4. The activity is effective with classes of 20-25 students, but
1. in small classes students can count chips in multiple cookies, and
2. the number of cookies you need for each sample can be dependent on your texts conditions for the independent samples t-test!
Description and Teaching Materials
This activity
1. Requires the following materials:
1. One package of Chips Ahoy® cookies (approx 44 cookies) and one package of a store brand of chocolate chip cookies (I use Acme; the number of cookies in a bag will vary with the store brand).
2. One, small paper plate per student, to break the cookies over. (Little paper plates usually come 100 to a pack. They require separating before class, because pairs of them stick together.)
3. Wastebasket in the classroom.
4. Napkins are optional.
2. Requires statistical software or a statistically capable graphing calculator for recording the data, making appropriate graphs to check assumptions, and analyzing the data.
3. Works the following way:
1. Use your usual introduction or class discussion of comparing two parameters with null and alternative hypotheses. It can include concepts about random samples, independence of samples, ‘
recently’ covered tests.
2. A more specific discussion of "how should we" gather data in order to compare the mean number of chips per cookies for Chips Ahoy® and the supermarket brand, including, what it means to be a
chip in a cookie, how to break up the cookies to count chips, what the students think the relationship is between the means, and a class consensus on the hypotheses to be tested. If students
think there will be learning going on when they count the chips, then you should randomize the order in which they count chips.
3. Hand out one paper plate to each student.
4. Hand out one of each kind of cookie to each student. If they are not distinguishable, then give out the second cookie only after a student is finished counting the chips in the first. Give
'half' of the students Chips Ahoy® first if the class feels there is a learning effect (counting chips).
5. When a student has the number of chips in each cookie (they can write the numbers on the paper plate), he or she brings the information to the front of the room where you can record it.
6. Getting student help in c), d) & e) makes the process go faster.
7. Return to your usual development of the independent samples t-test, but now with interesting data.
4. More detailed instructions on doing the activity are found in the MSWord file Detailed Instructions for Chips Ahoy (Word) (Microsoft Word 31kB May17 07) and PDF file Detailed Instructions for
Chips Ahoy (PDF) (Acrobat (PDF) 68kB May17 07). It includes the Adaptations and Warnings in the next section.
Teaching Notes and Tips
1. Context. This activity is meant to provide a context for discussing the independent samples t-test. Adapt it to your method of introducing the independent samples t-test, whether you cover the
pooled, approximate, or both.
2. Assumptions/conditions. The concept of comparing chips gives numerous points for discussion. If some assumptions fail, be prepared to "continue for the sake of illustration" and write a caveat.
1. Independent random samples. Ask the students if you can consider the bags of cookies as randomly chosen? Are the two samples independent? What are the populations from which the samples came?
2. Sample size conditions. If the populations of the number of chips are normal, we can use the procedures. Histograms and normal probability plots can be used to verify or eliminate normality.
Of course if the sample sizes are large enough, the sample means are normal, but these conditions differ from text to text! For instance in some books the conditions are more robust, and
depend on the sum of the sample sizes (Moore and McCabe), whereas in others the sample sizes must individually be at least 30 (say). Hence the size of the class needed to make the activity
work well may depend on the text you use.
3. Population variances equal? If you use the pooled t-test the samples should satisfy the conditions. Consider switching to the approximate t-test if you have not already!
3. Warnings.
1. If you teach in a computer lab, you can have the students move outside to break up the cookies. Crumbs in the keyboard may disqualify you from using the lab/classroom again.
2. Be sure to point out that although the sample sizes may be the same for the two samples, this is not a paired t-test.
4. Adaptations.
1. The counting can be done in a take-home fashion. Give the students one of each cookie to take home in little plastic bags, adequately marking the cookies so that the data is not mixed up
(wrap one in a piece of paper or little envelope, use markers, or see below).
2. Because this is not a paired t-test, you can have some students do two of one kind of cookie, so that you get different sample sizes.
3. Having extra cookies (and milk) for consumption is optional, because the counting is a destructive process.
4. If you are independently wealthy or you school has a large budget, you can compare Chips Ahoy??? to a regional bakery's "gourmet" brand. (In at least New Jersey, we could choose Entenmann's!)
This class activity is meant to provide context for discussing hypotheses for comparing two means, as well as the assumptions for the test. Of course its original inspiration was for a change-of-pace
to the standard lecture (relying on its tactile nature of the activity and its food theme). Below are suggestions for formative and summative assessments.
Consider the Chips Ahoy
activity from class, with its discussion of the hypotheses for the independent samples t-test, as well as its assumtions/conditions.
1. Choose one of the following (that best fits your feelings):
1. It helped me learn the independent samples t-test.
2. It hindered me in my learning of the independent samples t-test.
3. It neither helped me nor hindered me in my learning of the independent samples t-test.
2. Explain what one or two aspects of the activity that led you to choose your answer.
(To be administered after discussion of comparing two proportions.) You and a classmate have been discussing the colors of Swedish fish and gummy bears. You think that the proportion of red in the
Swedish fish is greater than the proportion of red in gummy bears, and your classmate thinks that they are the same (for a particular brand (instructors can insert a name)).
1. Define the appropriate parameters, and write down the null and alternative hypotheses that you should be testing.
2. Your classmate has suggested that you go to the supermarket and buy a package of each candy and use the data from the two packages to perform the hypothesis test. Comment on the validity of the
assumptions for the appropriate test.
3. After buying several packages (randomly chosen) of each kind of candy, you find that 23 of 75 Swedish fish are red, but 15 of 60 gummy bears are red. Comment of the validity of the sample size
assumptions for the appropriate test.
4. Complete the appropriate test, being sure to give all of the appropriate conclusions. [Instructor: we are looking for the so called "statistical conclusion" and the "real-world" conclusions.]
References and Resources
To whom it may concern: Checks of DASL, EESEE, CAUSE and MERLOT did not reveal any applications of Chips Ahoy® or chocolate chip cookies for the two sample t-test. However, do check out the
Warner, Brad, and Rutledge, Jim, "Checking the Chips Ahoy! Guarantee," Chance Vol. 12, No. 1, Winter 1999, pp. 10-14. | {"url":"http://serc.carleton.edu/sp/cause/conjecture/examples/18165.html","timestamp":"2014-04-18T14:15:22Z","content_type":null,"content_length":"39190","record_id":"<urn:uuid:5a0fc1db-4cad-4907-9c53-d47a94905d2f>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00362-ip-10-147-4-33.ec2.internal.warc.gz"} |
QSTK Tutorial 1
From Quantwiki
In this tutorial we will take a look at using QSTK code for basic time series analysis of stock price data. Python works well for this because you can do some very powerful analyses in a few lines of
code. There are also convenient libraries for displaying data easily. All of the code you see below is available in QSTK/Examples/Basic/tutorial1.py.
In this tutorial we focus on techniques for looking at equity price data, but you might also want to consult this more general tutorial about NumPy as well.
Acknowledgments: We heavily leverage pandas and NumPy.
Make sure you have installed QSTK correctly. For instructions, visit QSToolKit_Installation_Guide.
Reading Historical Data
In this first section of the code we import several useful libraries. numpy, pylab and matplotlib provide a number of functions to Python that give it MATLAB-like capabilities. datetime helps us
manipulate dates. The qstkutil items are from the QuantSoftware ToolKit
import QSTK.qstkutil.qsdateutil as du
import QSTK.qstkutil.tsutil as tsu
import QSTK.qstkutil.DataAccess as da
import datetime as dt
import matplotlib.pyplot as plt
import pandas as pd
We'll be using historical adjusted close data. QSTK has a DataAccess class designed to quickly read this data into pandas DataFrame object. We must first select which symbols we're interested in, and
for which time periods.
ls_symbols = ["AAPL", "GLD", "GOOG", "$SPX", "XOM"]
dt_start = dt.datetime(2006, 1, 1)
dt_end = dt.datetime(2010, 12, 31)
dt_timeofday = dt.timedelta(hours=16)
ldt_timestamps = du.getNYSEdays(dt_start, dt_end, dt_timeofday)
In the first line above we create a list of 5 equities we're interested to look at. Next we specify a start date and and end date using Python's datetime feature. Finally, we get a list of timestamps
that represent NYSE closing times between the start and end dates. The reason we need to specify 16:00 hours is because we want to read the data that was available to us at the close of the day.
Now we are ready to read the data in:
c_dataobj = da.DataAccess('Yahoo')
ls_keys = ['open', 'high', 'low', 'close', 'volume', 'actual_close']
ldf_data = c_dataobj.get_data(ldt_timestamps, ls_symbols, ls_keys)
d_data = dict(zip(ls_keys, ldf_data))
The first line creates an object that will be ready to read from our Yahoo data source. The second two lines provide the various data types you want to read. The third line creates a list of
dataframe objects which have all the different types of data. The fourth line converts this list into a dictionary and then we can access anytype of data we want easily.
Take a look here for examples of what you can do with them. A DataFrame's rows correspond to points in time, and its columns usually represent specific stocks or equities. The data in each cell of a
DataFrame represents the corresponding item (say closing price) for that stock at that time. You can access the indexes for the rows and the columns as close.index, or close.columns.
The "real" data within a DataFrame can be pulled out and manipulated as a NumPy ndarray. NumPy provides sophisticated indexing and slicing capabilities similar to those that MATLAB allows. For more
details on this take a look at a numpy tutorial, (skip to the indexing and slicing section). We we leverage NumPy features in this tutorial.
First Figure
The figure on the right shows the adjusted close data from the file. In this section we look at the code that generated it. First, we create date objects from the date data:
na_price = d_data['close'].values
plt.plot(ldt_timestamps, na_price)
plt.ylabel('Adjusted Close')
plt.savefig('adjustedclose.pdf', format='pdf')
In the first line we pull out the close prices we cant to plot into a 2D numpy array from a dataframe. In the next line we erase any previous graph.
We can then plot the data using plt.plot. Note that the pyplot plot() command is smart enough to plot several lines at once if it is provided a 2D object. pyplot automatically assigns a color to each
line, but you can, if you like, assign your own colors. We add a legend with the symbol names and also add labels for the axes. Finally, with plt.savefig the figure is written to a file.
Normalized Data
A problem with the previous figure is that the high share price of GOOG dominates. It is difficult to see what's going on with the other equities. Also, as an investor, you really want to see
relative, or normalized price moves. Here's how to normalize the data:
na_normalized_price = na_price / na_price[0, :]
The line of code above nicely illustrates how effective Python (with NumPy) can be. In that single line of code we executed 2000 divide operations. Each row in pricedat was divided by the first row
of pricedat. Thus normalizing the data with respect to the first day's price. The resulting figure is to the right.
Up til now I've shown all of the code. From here forward, I'm only going to show what's important or relevant. So I'm not repeating the code for plotting. You can see it in the source code that
referenced at the beginning of this tutorial.
Daily Returns
It is very often useful to look at the returns by day for individual stocks. The general equation for daily return on day t is:
• ret(t) = (price(t)/price(t-1)) -1
We can compute this in Python all at once as using the builtin QSTK function returnize0:
na_rets = na_normalized_price.copy()
In the figure at right we illustrate the daily returns for SPY and XOM. Observe that they tend to move together but that XOM's moves are more extreme.
Scatter Plots
Scatter plots are a convenient way to assess how similarly two equities move together. In a scatter plot of daily returns the X location of each point represents the return on one day for one stock,
and the Y location is the return for another stock. If the cloud of points is arranged roughly in a line we can infer that the equities move together.
The figures to the right illustrate the visual difference between two equities that move together (top/blue) and two that are less correlated (bottom/red). What do the shapes of the point clouds tell
you about these relationships?
The figure at the top was created with the following line of code:
plt.scatter(na_rets[:, 3], na_rets[:, 1], c='blue')
Exercise: Cumulative Daily Returns
Using the daily returns we can reconstruct cumulative daily returns. Note that in general the cumulative daily return for day t is defined as follows (this is NOT Python code, it is an equation):
daily_cum_ret(t) = daily_cum_ret(t-1) * (1 + daily_ret(t))
I don't provide the code for this, as it is a programming assignment. If you plot the result, it should look exactly like the normalized returns plot above.
Exercise: Combining Daily Returns To Estimate Portfolio Returns
Suppose we want to estimate the performance of a portfolio composed of 75% GLD and 25% SPY? We can approximate* the daily returns in an equation as follows:
portfolio_daily_rets = 0.75 * GLD_daily_rets + 0.25 * SPY_daily_rets
Then, using the equation above for cumulative daily returns, we can plot the performance of the combined portfolio. This also is an assignment in my course, so I don't list the code for it here.
Exercise: Line fit to Daily Returns
Finally, we revisit the scatterplots above that reveal visually how closely correlated (related) the daily movement of two stocks are. It's even better if can quantify this correlation by fitting a
line to them using linear regression. Note the red line in the figure on the right; this was computed using one of NumPy's linear regression tools. The value of the slope of the line is reported as
"corr" which is technically not correct.
Wikipedia has a nice discussion of correlation here: wikipedia
Again, I'm not going to show the code here, but I will tell you that the code is not very complex, and I used the following functions: polyfit(), polyval(), and sort(). | {"url":"http://wiki.quantsoftware.org/index.php?title=QSTK_Tutorial_1","timestamp":"2014-04-19T14:29:13Z","content_type":null,"content_length":"25178","record_id":"<urn:uuid:5d4e6a8a-6d67-4438-a280-6777b84871a1>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00050-ip-10-147-4-33.ec2.internal.warc.gz"} |
Conclusions and discussion
Next: About the choice of Up: Why does the meter Previous: Interplay of pendulum and
Conclusions and discussion
The initial aim of this paper was to rise the question that gives the title, why the semi-period of a meter simple pendulum is approximately equal to one second. In fact, it seems to us that this
question has never been raised in literature, not even at a level of a curiosity (for some references on the subject, see e.g. [29,35,40,10,43]).
We were fully aware that mere coincidences are possible and that speculations based on them can easily drift to the non-scientific domain of numerology, especially in the field of units of measures,
where the large number of units through the ages and around the world cover almost with continuity the range of lengths in the human scale. Therefore, we were searching if there were reasonable
explanations for the close coincidence pointed out in our question.
A physical reason is ruled out. A unit of length defined as a fraction of a planet meridian makes the period of a unitary pendulum only depending on the planet density. But this period has no
connection with the planet rotation period, and then with its 86400th part.
The suspicion remains -- definitely strengthened -- that, among the possible choices of Earth related units, the choice favored the one that approximated a pendulum that beats the second.
Next: About the choice of Up: Why does the meter Previous: Interplay of pendulum and Giulio D'Agostini 2005-01-25 | {"url":"http://www.roma1.infn.it/~dagos/history/sm/node10.html","timestamp":"2014-04-20T15:50:40Z","content_type":null,"content_length":"5114","record_id":"<urn:uuid:18a9d0a6-d677-4e3b-ac31-fbb910dd5656>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00428-ip-10-147-4-33.ec2.internal.warc.gz"} |
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sin[arcsin(1/5)+ arccos x] =1 find the value of x? - Homework Help - eNotes.com
sin[arcsin(1/5)+ arccos x] =1 find the value of x?
This question is best solved in smaller parts. For instance,
sin (theta) = 1
theta = pi/2 or 90 degrees. I'm going to assume all answers for angles will be in radians.
Then, arcsin (1/5) + arccos x equals pi/2, since sin (pi/2)=1
arccos x = pi/2 - arcsin (1/5)
x = cos (pi/2 - arcsin (1/5))
Using the identity: cos (a - b) = cos a cos b + sin a sin b
x = cos (pi/2) cos (arcsin (1/5)) + sin (pi/2) sin (arcsin (1/5))
x = 0 * cos (arcsin (1/5)) + 1 * 1/5
x = 1/5
To solve this problem, you use that sin (pi/2) = 1
Therefore the entire expression on the inside of the sine brackets is equal to pi/2.
Then solve for x by taking the cos of both sides.
Use the difference identity for cosine to evaluate x.
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Atiyah-MacDonald, exercise 2.11
up vote 12 down vote favorite
Let A be a commutative ring with 1 not equal to 0. (The ring A is not necessarily a domain, and is not necessarily noetherian.) Assume we have an injective map of free A-modules A^m -> A^n. Must we
have m <= n?
I believe the answer is yes. For instance, why is there no injective map from A^2 -> A^1? Say it's represented by a matrix (a_1 a_2). Then clearly (a_2, -a_1) is in the kernel. In the A^{n+1} -> A^
{n} case, we can look at the n x (n+1) matrix which represents it; call it M. Let M_i denote the determinant of the matrix obtained by deleting the ith column. Let v be the vector (M_1 -M_2 ... (-1)^
nM_{n+1}). Then v is in the kernel of our map, because the vector M*v^T has ith component the determinant of the (n+1) x (n+1) matrix attained from M by repeating the ith row twice.
That almost finishes the proof, except it is possible that v is the zero vector. I would like to see either this argument finished, or, even better, a nicer proof.
Thank you!
ac.commutative-algebra tag-removed
1 The answer mathoverflow.net/questions/30860/… was given by Robin Chapman in another thread, which has been closed as duplicate. I find Robin's answer very nice, and hope the other thread won't be
deleted. – Pierre-Yves Gaillard Jul 31 '10 at 15:40
add comment
4 Answers
active oldest votes
Define D[i](M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that D[r](M) has no annihilator (i.e. there is no
element a∈A such that aD[r](M)=0); I think r is usually called the McCoy rank of M.
Choose an a∈A such that aD[r+1](M)=0. By assumption, $a$ does not annihilate D[r](M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r
minor. Let M[1], ..., M[r+1] be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. Note, in particular, we know M[r+1] is the determinant
up vote 7 of the upper-left r-by-r minor, so aM[r+1]≠0.
down vote
accepted The claim is that the vector (aM[1],...,aM[r+1],0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row
of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-right (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r,
this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$.
add comment
Here is another solution using only the Cayley-Hamilton Theorem for finitely generated modules (Proposition 2.4. in Atiyah-Macdonald) which, even though looks quite innocent, is a very
powerful statement.
up vote Assume by contradiction that there is an injective map $\phi: A^m \to A^n$ with $m>n$. The first idea is that we regard $A^n$ as a submodule of $A^m$, say the submodule generated by the
21 down first $n$ coordinates. Then, by the Cayley-Hamilton Theorem, $\phi$ satisfies some polynomial equation $$\phi^k + a_{k-1} \phi^{k-1} + \cdots + a_1 \phi + a_0 = 0.$$ Using the injectivity
vote of $\phi$ it is easy to see that if this polynomial has the minimal possible degree, then $a_0 \ne 0$. But then, applying this polynomial of $\phi$ to $(0,\ldots,0, 1)$, the last coordinate
will be $a_0$ which is a contradiction as it should be zero.
Wonderful !!!!! – Pierre-Yves Gaillard Dec 1 '10 at 9:16
1 +1 This is a proof from The Book. – benblumsmith Sep 28 '12 at 14:43
add comment
Dear CJD, if you are still interested in your problem, already solved three weeks ago by Anton, here is another point of view.
Let $M:A^m \to A^n$ be injective. Let $B=\mathbb Z [\ldots,m_{ij},\ldots]$ be the subring of $A$ generated by all the entries of the matrix $M$ ; this $B$ is a noetherian ring and we have
(by restriction) an injective linear map $M:B^m \to B^n$. In other words we may assume that $A$ is noetherian. Now we localize at a prime $\mathfrak p$ of $B$ of height zero and we get an
injective map (localization is exact and thus preserves injections) $L:C^m \to C^n$ (the ring $C$ is the ring $B$ localized at $\mathfrak p$).
up vote 20
down vote Ah, but now $C$ is noetherian of dimension zero, hence artinian and we can talk about lengths. Since lengths are additive in exact sequences (Atiyah-MacDonald, Proposition 6.9) we get
$m.length(C) + length(coker L)= n.length(C)$, hence $m\leq n$.
Friendly greetings, Georges.
I like this proof a lot. I guess it doesn't logically make sense as the solution to the Atiyah-MacDonald exercise, since it uses results that come later in the book, but this proof
definitely requires less book-keeping than Anton's argument using determinants. Thanks for the post! – CJD Oct 28 '09 at 7:24
This is a response to the answer given by Georges Elencwajg posted on Oct 25. His answer is really nice, but there is a small correction to this answer. In the construction, $B=Z[{m_
{ij}}]$ to be replaced with $B=Z(A)[{m_{ij}}]$, where $Z(A)$ is the prime sub ring of A. – N. Kumar Dec 30 '09 at 13:12
I wrote explicitly that B is the subring of A generated by the entries of the matrix M. Any subring of A must, of course, contain the prime subring of A. So your ring is exactly the one
I described. – Georges Elencwajg Jan 2 '10 at 9:14
Sorry to have overlooked at it. – N. Kumar Nov 21 '12 at 14:45
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I posted this question on a different site a couple of years ago. Eventually I found that a book of T.Y. Lam has a very nice treatment. Here is the writeup I posted on the other site:
After paging through several algebra books, I found that T.Y. Lam's GTM Lectures on Rings and Modules has a beautiful treatment of this question.
The above property of a (possibly noncommutative) ring is called the "strong rank condition." It is indeed stronger than the corresponding statement for surjections ("the rank condition")
which is stronger than the isomorphism version "Invariant basis number property". However, in fact it is the case that all commutative rings satisfy the strong rank condition. Lam gives
two proofs [pp. 12-16], and I will now sketch both of them.
First proof:
Step 1: The result holds for (left-) Noetherian rings. For this we establish:
Lemma: Let $M$ and $N$ be (left-) $A$-modules, with $N$ nonzero. If the direct sum $M \oplus N$ can be embedded in $M$, then $M$ is not a Noetherian $A$-module.
Proof: By hypothesis $M$ has a submodule $M_1 \oplus N_1$, with $M_1 \cong M$ and $N_1 \cong N$. But we can also embed $M \oplus N$ in $M_1$, meaning that $M_1$ contains a submodule $M_2 \
oplus N_2$ with $M_2 \cong M$ and $N_2 \cong N$. Continuing in this way we construct an ascending chain of submodules $N_1$, $N_1 \oplus N_2$,..., contradiction.
So if A is (left-) Noetherian, apply the Lemma with $M = A^n$ and $N = A^{m-n}$. $M$ is a Noetherian $A$-module, and we conclude that $A^m$ cannot be embedded in $A^n$.
up vote 8 Step 2: We do the case of a commutative, but not necessarily Noetherian, ring. First observe that, defining linear independent subsets in the usual way, the strong rank condition precisely
down vote asserts that any set of more than $n$ elements in $A^n$ is linearly dependent. Thus a ring $A$ satisfies the strong rank condition iff: for all $m > n$, any homogeneous linear system of
$n$ linear equations and $m$ unknowns has a nonzero solution in $A$.
So, let $MX = 0$ be any homogeneous linear system with coefficient matrix $M = (m_{ij}), \ 1 \leq i \leq n, 1 \leq j \leq m$. We want to show that it has a nonzero solution in $A$. But the
subring $A' = \mathbb{Z}[a_{ij}]$, being a quotient of a polynomial ring in finitely many variables over a Noetherian ring, is Noetherian (by the Hilbert basis theorem), so by Step 1 there
is (even) a nonzero solution $(x_1,...,x_m) \in (A')^m$.
This makes one wonder if it is necessary to consider the Noetherian case separately, and it is not. Lam's second proof comes from Bourbaki's Algebra, Chapter III, §7.9, Prop. 12, page 519.
[Thanks to Georges Elencwajg for tracking down the reference.] It uses the following elegant characterization of linear independence in free modules:
Theorem: A subset $\{u_1,...,u_m\}$ in $M = A^n$ is linearly independent iff: if $a \in A$ is such that $a \cdot (u_1 \wedge \ldots \wedge u_m) = 0$, then $a = 0$.
Here $u_1\wedge \ldots \wedge u_m$ is an element of the exterior power $\Lambda^m(M)$.
(I will omit the proof here; the relevant passage is reproduced on Google books.)
This gives the result right away: if $m > n$, $\Lambda^m(A^n) = 0$.
2 Hi Peter, nice to find you here. I have just checked that the reference to the English edition of Bourbaki is: Algebra, Chapter III, §7.9, Prop.12, page 519. Atiyah-MacDonald's exercise
2.11 is a consequence of Bourbaki's exercise 16, page 641. Friendly greetings, Georges. – Georges Elencwajg Oct 26 '09 at 20:27
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Ohio Masters of Mathematics
Call for Contributions.
Ohio Masters of Mathematics
This collection of biographical sketches, which originated as part of Ohio's Bicentennial celebration, is sponsored by the Ohio Section of the Mathematical Association of America. It is designed to
foster public understanding, education, and appreciation of mathematics as a human endeavor and Ohio's contributions to that enterprise.
Click on each name for biographical article. Listed in order of date of birth. See also Milestones in Mathematics timeline.
1. Jared Mansfield, 1759-1830.
A graduate of Yale, he came to Ohio as Surveyor General in 1803, the year that Ohio became a state. Starting with a principal meridian that marked the boundary of Ohio and Indiana, he set up
coordinates for a system of townships and sections that would eventually spread across the nation. He later served as Professor of Natural Philosophy at the US Military Academy at West Point from
1814 to 1828.
2. Thomas J. Matthews, 1788-1852.
Professor of Mathematics at Miami University from 1845-1852. Earlier he taught at Transylvania University in Kentucky and Woodward High School in Cincinnati. He was the first president of the
Western Literary Institute and College of Professional Teachers, and he is known for his work in surveying the "Matthews Line" between Kentucky and Tennessee and for other civil engineering
3. Joseph Ray, 1807-1855.
Professor of mathematics at Woodward College in Cincinnati and later principal of Woodward High School. He also served a term as President of the Ohio State Teachers Association. Ray is best
known as the author of one of the most popular series of American arithmetic and algebra textbooks of the nineteenth century.
4. Ormsby McKnight Mitchel, 1810-1862.
Grew up in Lebanon, Ohio and graduated from West Point in 1829. In 1835 he accepted a position as professor of mathematics, natural philosophy, and astronomy at the newly revived Cincinnati
College. He became the principal founder of the Cincinnati Astronomical Society in 1842 and embarked on a trip to Germany where he procured a telescope with a lens nearly a foot in diameter.
Construction of the Cincinnati Observatory on Mount Adams began in 1843 with John Quincy Adams delivering an oration at the laying of the cornerstone. Mitchel supervised its construction and
served as its director until 1860. He served as a Union officer in the Civil War and died of yellow fever while on duty in South Carolina.
5. Elias Loomis, 1811-1889.
A prominent astronomer and author of mathematics textbooks who is also known for his meteorological researches and interest in genealogy. Loomis was associated with Western Reserve College in
Ohio from 1837-1844, where he supervised construction of the third college observatory in the United States and collected observations in astronomy, meteorology, and terrestrial magnetism.
6. Eli T. Tappan, 1824-1888
The son of federal judge and U.S. senator Benjamin Tappan. At age 33 he began a new career in mathematics and education, serving as a mathematics teacher in Steubenville and Cincinnati, professor
of mathematics at Ohio University, and president of Kenyon College. He was the author of textbooks on geometry and trigonometry.
7. Robert W. McFarland,1825-1910.
Born in Champaign County, he earned A. B. and A. M degrees from Ohio Wesleyan University. In 1853 he was appointed Professor of Mathematics at Madison College, and then in 1856 at Miami
University, Ohio. During the Civil War, McFarland served in the Union Army, rising to the rank of lieutenant colonel. In 1873 he was appointed as the first Professor of Mathematics and Civil
Engineering at the newly opened Ohio Agricultural and Mechanical College (now Ohio State University). In 1885 he returned to Miami, to serve as president for three years.
8. Edward Olney, 1827-1887.
Grew up in Wood County, Ohio, and taught mathematics in Perrysburg before being appointed professor of mathematics at Kalamazoo College (1853) and the University of Michigan (1863). He was known
as a tough, but fair teacher and the author of a popular series of mathematics textbooks.
9. Aaron Schuyler, 1828-1913.
Grew up in Seneca County, Ohio. After serving as principal of Seneca County Academy for twelve years, he was elected professor of mathematics and philosophy at Baldwin University (now
Baldwin-Wallace College) in Berea and later became president of that institution. In the 1870's he published a series of college mathematics textbooks. He left Ohio to teach at Kansas Wesleyan
University in 1885. Schuyler's able assistant, Ellen H. Warner, may have been the first female professor of mathematics at an American college.
10. E. W. Hyde, 1843-1930.
Educated at Cornell University as a civil engineer, he came to the University of Cincinnati in 1875. There he served as Professor of Mathematics, Dean of the College of Liberal Arts, and
President before being forced out in an academic bloodbath in 1900. He was an associate editor of the Annals of Mathematics, and he published books on advanced subjects when America was still a
mathematical backwater.
11. E. B. Seitz, 1846-1883.
Seitz was also born in Fairfield County, Ohio (see Finkel). He was mainly self-taught in mathematics mastering Ray's texts. He spent one year of general academic study at Ohio Wesleyan in 1870. A
prodigious problem solver, he is the 5th American elected to the London Mathematical Society.
12. Ormond Stone, 1847-1933.
Director of the Cincinnati Observatory from 1875-1882, where he influenced E. H. Moore to study mathematics, and where he was the first to establish standard time for an American city. As
Professor of Astronomy and Mathematics at the University of Virginia, he founded the Annals of Mathematics in 1884.
13. William Hoover, 1850-1938.
Born in Wayne County, he attended both Wittenberg and Oberlin colleges. After teaching high school in Bellefontaine, Wapakoneta, and Dayton, Ohio, and South Bend, Indiana, he was elected
professor of mathematics and astronomy at Ohio University in 1883. He attended the first meeting of the MAA Ohio Section, and he contributed problems and solutions to the American Mathematical
Monthly for over 40 years.
14. Ellen Amanda Hayes, 1851-1930.
Born in Granville, Ohio, and graduated from Oberlin College in 1878. Taught mathematics at Wellesley College from 1879 to 1916 and was appointed head of the mathematics department in 1888. Writer
of several textbooks and in 1891 was elected a member of the New York Mathematical Society (later to become the American Mathematical Society), one of the first six women to join.
15. Elisha S. Loomis, 1852-1940.
Born in Medina County, Ohio, and a graduate of Baldwin University (now Baldwin-Wallace College). Taught mathematics in Ohio at all levels, from the primary grades through college. Known for his
compendium of more than 250 proofs of the Pythagorean theorem.
16. R. D. Bohannan, 1855-1926.
Appointed professor of mathematics and astronomy at The Ohio State University in 1887. He published a number of papers on classical and algebraic geometry and complex function theory. He helped
to organize of the foundational meeting of the Mathematical Association of America and served as chairman of the Ohio Section in 1925-26.
17. C. J. Keyser, 1862-1947.
Born in Rawson, Ohio, and educated at the North West Ohio Normal School (now Ohio Northern University). After holding several school positions in Ohio, Missouri, and New York, he eventually
earned a Ph.D. in mathematics from Columbia University in 1901, and spent the rest of his professional career at that institution. He is best known for his writings in the area of mathematical
18. E.H. Moore, 1862-1932
Born in Marietta, Ohio and graduated from Woodward High School in Cincinnati in 1879. He was a pioneer in the American mathematical research community, and he founded the mathematics department
at the University of Chicago and served as its head from 1896-1931. He was President of the American Mathematical Society and editor of the Transactions of the AMS.
19. Benjamin Finkel, 1865-1947.
Founded The American Mathematical Monthly in 1894, which led to the founding of the Mathematical Association of America in Columbus in 1915. He was born in Fairfield County and received his B.S.
and M.S. degrees from Ohio Normal University in Ada (now named Ohio Northern University, after Finkel's suggestion). He was professor of mathematics and physics at Drury College in Springfield,
Missouri, from 1895 until his death.
20. Harris Hancock, 1867-1944.
A native of Virginia, Hancock obtained a Berlin Ph.D. in 1894 for a thesis on elliptic functions. He taught at the University of Chicago, where E. H. Moore was department head, until he was
appointed Professor at the University of Cincinnati in 1900. A strong proponent of classical education, Hancock was influential in the establishment of Walnut Hills High School in 1920. He
attended the First Annual Meeting of the MAA Ohio Section in 1916 and served as Section Chairman in 1924-25.
21. William D. Cairns, 1871-1955.
Born in Troy, Ohio, Cairns graduated in 1892 from Ohio Wesleyan and earned an A.M. in 1898 from Harvard University. In 1907 he received a Ph.D. degree in mathematics from the University of
Gottingen, where he studied under David Hilbert. Cairns was MAA Secretary-Trasurer from the founding of the MAA in 1915 until 1942. After serving as MAA President in 1943-44, he was made honorary
president for life. He taught at Oberlin College from 1899 until retirement in 1939.
22. Samuel Rasor, 1873-1950.
A native of Ohio, he received his B.S. from Ohio State in 1898 and his M.A. degree in 1902. He then embarked on a career of teaching and service at OSU that would last for nearly fifty years.
Rasor was the author of several mathematical papers and three textbooks. He was chairman of the local organizing committee for the foundational meeting of the Mathematical Association of America
in December 1915, and he served as chairman of the MAA Ohio Section in 1920-21.
23. Charles M. Austin, 1874-1967.
A founder and first president of the National Council of Teachers of Mathemtics. He was born near Waynesville, Ohio, graduated from Ohio Wesleyan University, and taught at Milford High School
before moving to Oak Park, Illinois in 1912.
24. Grace Bareis, 1875-1962.
A charter member of the MAA and the Ohio Section. She received her A.B. degree from Heidelberg College in Tiffin, Ohio, in 1897. In 1909 she became the first person to receive a Ph.D. in
mathematics from The Ohio State University, and she taught at Ohio State until her retirement in 1946.
25. C. N. Moore, 1882-1967.
A native Cincinnatian and graduate of Woodward High School. He earned a Harvard Ph.D. in 1908 before joining the faculty of the University of Cincinnati. He was highly regarded for his research
on convergence factors for infinite series. He attended the first meeting of the MAA Ohio Section in 1916 and served as Section Chairman in 1918-19.
26. Otto Szász, 1884-1952.
A native of Hungary, he came to the United States in 1933 and taught at the University of Cincinnati from 1936-1952, where he was recognized as an important figure in the field of classical
27. Louis Brand, 1885-1971.
A native of Cincinnati. He received three engineering degrees from the University of Cincinnati and a Ph.D. from Harvard. He served on the faculty of his alma mater from 1907-1955 and was well
known for a series of popular textbooks on vector analysis and mechanics. He was Chairman of the MAA Ohio Section in 1941-42.
28. Otto Marcin Nikodym, 1887-1974.
A Polish mathematician, famous for the Radon-Nikodym Theoreom. He was educated at the Universities of Lwow and Warsaw, and the Sorbonne, taught at the Universities of Krakow and Warsaw and at the
High Polytechnical School in Krakow. He came to Ohio in 1948 to join the faculty of Kenyon College, retiring in 1966.
29. J. R. Overman, 1888-1978.
J. Robert Overman was the first faculty member hired at Bowling Green State University, then called Bowling Green Normal College, and served there for 42 years. He earned degrees in mathematics
from Indiana, Columbia, and a Ph D from Michigan. He wrote a series of ten textbooks in school mathematics. He establish many programs st BGSU, including the College of Liberal Arts, serving as
the first dean. He also served as the first librarian and provost.
30. H. C. Christofferson, 1894-1973.
Professor of mathematics and director of secondary education at Miami University from 1928 to 1961. He served as president of the National Council of Teachers of Mathematics (1938-40) and the
Ohio Council of Teachers of Mathematics (1952-54). He was a founder of the latter organization in 1950.
31. I. A. Barnett, 1894-1974.
After earning three degrees at the University of Chicago and serving on the faculties of Washington University, the University of Saskatchewan, and Harvard, Barnett came to the University of
Cincinnati in 1923. Norbert Wiener credited him for suggestiing a problem in the early 1920s that eventually led him to the notion of Wiener measure and its application to Brownian motion.
Barnett served as MAA Ohio Section Chairman in 1933-34 and on the MAA Board of Governors in 1952.
32. Tibor Radó, 1895-1965
One of the most prominent of the Hungarian mathematicians to come to the US in the post-World War I period. He was appointed professor at Ohio State in 1930 in conjunction with the establishment
of a graduate program in mathematics. He served as chairman of the department in the postwar period and was named research professor in 1948. He is known for his solution of Plateau's problem in
33. George R. Stibitz, 1904-1995
A 1926 Graduate of Denison University in applied mathematics, Stibitz is recognized as the father of the modern digital computer. He earned a Ph.D. in physics from Cornell in 1930, and joined
Bell Telephone Laboratories. From the study of the binary circuits controlled by telephone relays, he built a binary adder circuit and then a full-scale calculator in 1939. Several binary
computers of greater sophistication followed.
34. Henry Mann, 1905-2000.
Born and educated in Vienna, Austria, and emigrated to the United States in 1938. He was a member of the faculty at Ohio State from 1946-1964. In 1946 he was awarded the Cole Prize in Number
Theory by the American Mathematical Society for his proof of a conjecture of Schnirelmann and Landau. He authored 80 research papers and three books.
35. Eugene Lukacs, 1906-1987.
A native of Hungary, Lukacs emigrated to the United States before the World War II. He taught at Our Lady of Cincinnati College from 1945-1953 and later helped to found the doctoral program in
mathematics at Bowling Green State University. In between he taught at Catholic University in Washington, DC, where he founded the Statistics Laboratory. His monograph on characteristic functions
in probability theory was the first to present a unified and detailed treatment of the subject.
36. Arnold E. Ross. 1907-2002.
Came to Ohio State as chair of the mathematics department in 1963, having filled that same post at the University of Notre Dame. In 1957 he founded a summer program for talented high school
students, which he taught in and directed until age 94.
37. A. J. Macintyre, 1908-1967.
A native of England, he became Research Professor of Mathematics at the University of Cincinnati in 1959, and Charles Phelps Taft Professor of Mathematics in 1963. Macintyre was married to a
fellow mathematician, Sheila Scott, who died soon after arriving in Ohio.
38. Marshall Hall, 1910-1990.
A member of the faculty of The Ohio State University from 1946-1959. He authored more than 120 research papers on group theory, coding theory, and design theory, and his highly regarded books
Theory of Groups and Combinatorial Theory are classics.
39. Kenneth B. Cummins, 1911-1998.
A high school mathematics teacher and professor of mathematics at Kent State University. He was best known for his courses, institutes, and lectures for mathematics teachers.
40. Hans Zassenhaus, 1912-1991.
A native of Germany, he came to the United States in 1959 and was invited to join the faculty at Ohio State by Arnold Ross in 1964. He worked on a broad range of topics in mathematics and
mathematical physics and was a world famous authority on groups and Lie algebras.
41. D, Ransom Whitney, 1915-2007.
Best known for the famous Mann-Whitney U Statistic, the most widely used non-parametric statistic for two-sample tests. He was born in East Cleveland and served for many years on the faculty of
The Ohio State University.
42. Herbert Ryser, 1923-1985.
A native of Wisconsin, he earned three degrees from the University of Wisconsin. In 1949 he was appointed assistant professor at The Ohio State University and was promoted to the rank of
professor in 1955. He, along with Marshall Hall, established Ohio State's tradition of excellence in combinatorics.
43. James R.C. Leitzel, 1936-1998.
A member of The Ohio State University faculty from 1965-1990. He served as Chairman of the MAA Ohio Section in 1984-85 and was co-founder of the MAA's influential Project NExT (New Experiences in
Teaching), designed to nurture young mathematics faculty and prepare them for the profession.
Other Histories.
Organized by David Kullman (kullmade@miamioh.edu), Miami University, and Thomas Hern (hern@wcnet.org), Bowling Green.
Last changed May 6, 2013.
Return to the Ohio Section Home Page | {"url":"http://sections.maa.org/ohio/ohio_masters/index.html","timestamp":"2014-04-18T15:38:18Z","content_type":null,"content_length":"27851","record_id":"<urn:uuid:f14a8780-b32d-4d5a-971e-2efb7afc8160>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00208-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Limit question()2
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\[\LARGE ^{n}C_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\] where n->infinity
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@terenzreignz @yrelhan4 @shubhamsrg
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Is that a combination?
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summon @ParthKohli
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I'm thinking an approximation is due here?
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what kind of?
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I'm just tossing ideas :D
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yeah I've heard about that :P
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but it sort of resembles a binomial approximation thingy
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So I'm thinking 0.
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TBH,no freaking idea about this one. we can try stirlings
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I have options,but 0 isn't one of them. D:
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That's a shame :D okay, let's see...
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\[\Large _nC_x = \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]
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\[\Large = \frac{\left(\frac{n}e\right)^n}{\left(\frac{n-x}e\right)^{n-x}\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]
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there must be some sort of trick here
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Not that I've heard of :D
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The e's probably cancel out... \[\Large = \frac{\left(n\right)^n}{\left(n-x\right)^{n-x}\left(x\right)^x\sqrt{2\pi n}}\]
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I don't think stirling would work out.
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Then we need a new plan :D
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We need to summon people :O
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Proceed, summoner :)
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I don't want to further torment people :P
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Well, we'll just stare mindlessly at this limit :D
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lol xD
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No idea , sorry!
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Why is \[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]?
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Approximations. It led to a dead end, pay no heed to it (unless you can make use of it, of course :) )
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Why is it \[\sqrt{2 \pi n}\] for the denominator?
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Stirling's approximation for factorials?
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Shouldn't it be\[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi (n-x)}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi x}}\]?
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Whats the question exactly? Is there a summation involved?
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oh right.. typo sorry about that.
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But even then, I'm not sure this was the correct way to do it :)
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@joemath314159 The question is \[\LARGE \lim_{n\rightarrow \infty} \ ^nC_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\]
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@jim_thompson5910 @saifoo
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@saifoo.khan *
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If x is an integer, then the answer is 0. If x is an integer, then that is one term out of the binomial theorem.
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Roly, how did you get your font so big.....
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\[(a+b)^n=\sum_{x=0}^{n}\left(\begin{matrix}n \\ x\end{matrix}\right)a^xb^{n-x}\]let:\[a=\frac{m}{n}, b=1-\frac{m}{n}\]
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Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]
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So if x is an integer, it follows that:\[\left(\frac{m}{n}+\left(1-\frac{m}{n}\right)\right)^n=\sum_{x=0}^n \left(\begin{matrix}n \\ x\end{matrix}\right)\left(\frac{m}{n}\right)^n\left(1-\frac{m}
{n}\right)^{n-x}\]\[1=\sum_{x=0}^n \left(\begin{matrix}n \\ x\end{matrix}\right)\left(\frac{m}{n}\right)^n\left(1-\frac{m}{n}\right)^{n-x}\]Take the limit as n goes to infinity, the summation
converges (since the left hand side is 1), and if a summation converges, then its terms must converges to 0.
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@bookylovingmeh Maybe you'd love to explain why you think it is two?
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Answer is neither 0 nor 2
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my guess (and again this is a guess) is that your original problem is a binomial pdf (see link below) http://en.wikipedia.org/wiki/Binomial_distribution as n---> infinity, it seems like the
binomial distribution is becoming more and more like the normal distribution (based on the central limit theorem) so that makes me guess that as n ---> infinity, the binomial pdf is slowly
approaching a normal pdf (see link below) http://en.wikipedia.org/wiki/Normal_distribution
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does anyone want options btw?
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I think that would be helpful for the people solving the problem to cross-check their end result
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\[\Large \frac{m^x}{x!}.e^-m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]
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its e raise to power -m in the first option sory about that^^
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cool :O
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so it looks like it matches with \[\Large \frac{m^x}{x!}e^{-m}\] as for the "how", not 100% sure on that, but the page will hopefully clear that up
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its A as well as D
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oh yeah same thing
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hmm strangely enough, yeah, A and D are equivalent
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so maybe it's not A afterall
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lol, indeed it is... Where m/n = mu / n = p in that post.
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using \( n- x \approx n\) sould give you \[ \frac{n!}{(n-x)!x!} \left( \frac m n \right )^x \approx \frac{ m^x}{x! n^x} \cdot \frac{ \sqrt{2 \pi n}}{\sqrt{2 \pi (n-x)}} \cdot \frac{\left(\frac{n}
{e} \right )^n}{\left(\frac{n-x}{e} \right )^{n-x}} \\ \approx \frac{ m^x}{x!n^x} \cdot \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{n}{e} \right )^{n-x}} \approx \frac{ m^x}{x!} \cdot \left(\
frac{1}{e} \right )^{x} \] the other part use the definiton of 'e '\[\left( 1 - \frac m n \right )^{n-x} \approx \left( 1 - \frac m {n-x} \right )^{n-x} =\left( 1 - \frac m {n-x} \right )^{\frac
{n-x}{m} \cdot m } = e^{-(m-x)} \] multiply those and get A or D
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Woops!! there is a slight error \[ \left( 1 - \frac m n \right )^{n-x} \approx \left( 1 - \frac m {n-x} \right )^{n-x} =\left( 1 - \frac m {n-x} \right )^{\frac{n-x}{m} \cdot m } = e^{-(m)} \]
And \[ \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{n-x}{e} \right )^{n-x}} = e^{-x} \cdot \frac{n^n}{(n-x)^{n-x}} = e^{-x} \cdot \frac{n^{x}}{ \left(1 - \frac x n \right )^{n-x}} \\ \approx e
^{-x} \cdot \frac{n^x}{\left(1 - \frac x {n-x} \right )^{n-x}} = e^{-x} \cdot \frac{n^x}{e^{-x}} = n^x \] The first part gives \( \frac{m^x }{x!}\) and second part gives \( e^{-x} \)
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Here is the simplified original solution \[\Large \frac{n(n-1)(n-2)....x~factors}{x!}. \frac{m^x}{n^x}\] \[\Large \frac{\frac{n}{n} \frac{(n-1)}{n} \frac{(n-2)}{n}....x~factors}{x!}.{m^x}\] The
other part..which is 1^infinity \[\Large e^{(n-x)(\frac{-m}{n})}=>e^{(\frac{m}{n}-1})\] Combine them.. \[\Large 1.1.1.1------x ~factors . \frac{m^n}{x!}.e^{(-m)}\] @terenzreignz and everyone :D
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some misplaced n with x sorry about that :/
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Jamaica - E-Learning Project
Details of this resource and the teaching philosophy behind it are given below. For first time users, please read this carefully in order to gain as much value as possible from these resources.
Teacher Instructional Material (TIM)
Student Instructional Material (SIM)
This comprehensive programme of work for Secondary school students has been developed specifically for the e-Learning Jamaica Company Limited. It is based on the Mathematics Enhancement Programme
(MEP) resources produced by the
Centre for Innovation in Mathematics Teaching (CIMT)
at the University of Plymouth, UK (http://www.cimt.plymouth.ac.uk) working with the University of Technology, Jamaica (UTech).
The resources are divided into 10 Strands (A-J) plus a final Revision Strand, comprising a total of 40 Units, each with about 4 or 5 sections. Strands and units are allied directly to the CXC CSEC
Syllabus for first examination in Mathematics in 2010. The order in which the strands and units can be taught is not specified, although some suggested sequencing will be given when the project is
complete; it is expected though that teachers will use their preferred order of sequencing.
The previous knowledge required for each unit is given in the Lesson Plans. In some units, extra material, not at present in the CXC/CSEC Mathematics syllabus but considered to be relevant to the
topic, is included. This can be used as extension material for more able students and will offer a challenge to those who are particularly interested in a specific topic. This extra material is
denoted by coloured shading.
We have included many examples and questions in Jamaican contexts; where currency is involved, the $ sign denotes US dollars (students need to be familiar with this as their examination questions
will often use this currency) and J$ denotes Jamaica dollars. Students will need to take care when working through units that involve money as the different currencies might cause confusion.
We recommend that students pay particular attention to both the Revision Tests and the Multiple Choice Questions for each Strand. It is vital that they get used to the style of questions that they
will encounter in their forthcoming CXC examinations.
The resources have been developed with the MEP-style of teaching in mind: this is characterised by interactive, whole-class teaching, with the teacher orchestrating the teaching and learning in the
classroom and with all students involved and on task. The Key Teaching Points can be found here. We do hope that you will read through these suggestions and perhaps adapt your teaching style in the
light of some of the points made. The MEP resources have been designed to be used in this way. It might also be interesting for you to observe other teachers in your school and perhaps to plan a
lesson jointly with several colleagues, observe the lesson being taught and comment constructively on the outcome. This form of Lesson Study has proved to be a very effective form of continuing
professional development to enhance teaching and learning in many countries as it deals with the issues in your classroom.
We hope that you and your students will find these resources stimulating and helpful and that the outcome will be positive for everyone involved.
We are grateful to Cadien Roxborough of Papine High School, Kingston, Jamaica, who has helped with checking the resources, provided useful information regarding Jamaican contexts and made
recommendations for extra material and explanations. Any errors are of course the responsibility of CIMT.
The structure of the resources fall into two catagories; those for use by teachers in delivering the programme and those for use by students in their learning. Access to each set of resources is
available via the links below.
Teacher Instructional Material (TIM)
Student Instructional Material (SIM) | {"url":"http://www.cimt.plymouth.ac.uk/mepjamaica/","timestamp":"2014-04-21T12:09:22Z","content_type":null,"content_length":"5435","record_id":"<urn:uuid:06636576-e94f-4f83-a673-9d51d359d683>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00203-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculates standard deviation based on the entire population given as arguments. The standard deviation is a measure of how widely values are dispersed from the average value (the mean).
Number1, number2, ... are 1 to 30 number arguments corresponding to a population. You can also use a single array or a reference to an array instead of arguments separated by commas.
● STDEVP assumes that its arguments are the entire population. If your data represents a sample of the population, then compute the standard deviation using STDEV.
● For large sample sizes, STDEV and STDEVP return approximately equal values.
● The standard deviation is calculated using the "n" method.
● Arguments can either be numbers or names, arrays, or references that contain numbers.
● Logical values, and text representations of numbers that you type directly into the list of arguments are counted.
● If an argument is an array or reference, only numbers in that array or reference are counted. Empty cells, logical values, text, or error values in the array or reference are ignored.
● Arguments that are error values or text that cannot be translated into numbers cause errors.
● If you want to include logical values and text representations of numbers in a reference as part of the calculation, use the STDEVPA function.
● STDEVP uses the following formula:
where x is the sample mean AVERAGE(number1,number2,…) and n is the sample size.
The example may be easier to understand if you copy it to a blank worksheet.
● Create a blank workbook or worksheet.
● Select the example in the Help topic.
Note Do not select the row or column headers.
Selecting an example from Help
● Press CTRL+C.
● In the worksheet, select cell A1, and press CTRL+V.
● To switch between viewing the results and viewing the formulas that return the results, press CTRL+` (grave accent), or on the Formulas tab, in the Formula Auditing group, click the Show Formulas
Formula Description (Result)
=STDEVP(A2:A11) Standard deviation of breaking strength, assuming only 10 tools are produced (26.05455814) | {"url":"http://office.microsoft.com/en-us/excel-help/stdevp-HP005209281.aspx?CTT=1]","timestamp":"2014-04-18T03:39:16Z","content_type":null,"content_length":"25283","record_id":"<urn:uuid:48be8d1a-b098-48e0-9d68-ca68584ff0df>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00205-ip-10-147-4-33.ec2.internal.warc.gz"} |
Area enclosed by a plane curve... and converse
March 1st 2011, 01:10 PM #1
Mar 2011
Area enclosed by a plane curve... and converse
The area enclosed by a close plane curve ${\bf x}$ is:
$A=-\frac{1}{2}\oint ds\,{\bf x}\cdot{\bf n}$
where ${\bf n}$ is the principal normal vector of the curve.
Is the converse also true? Meaning, if the previous equation holds, does this imply that ${\bf x}$ is planar?
Follow Math Help Forum on Facebook and Google+ | {"url":"http://mathhelpforum.com/differential-geometry/173084-area-enclosed-plane-curve-converse.html","timestamp":"2014-04-19T23:00:53Z","content_type":null,"content_length":"29759","record_id":"<urn:uuid:7dfd6297-ee1d-4e06-8659-7a1617f7c4d2>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00257-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by Rida on Sunday, February 13, 2011 at 5:18am.
for abrar ul haq show the rate of an adult pass is rs 20 more than the children pass. when a company sold 50 adult passess and 200 children passess , the amount collected re. 2250. find the rice of
an adult pass.
• Statistics - PsyDAG, Sunday, February 13, 2011 at 10:09am
Proofread your questions before you post them.
• Ststistic - RAJA WSARMAD, Sunday, March 6, 2011 at 8:10am
let childern passs rate is x
then adult pass price will be (x+20)
as adult price is RS. 20 more than the childern pass price
200 childern passes are sold
200*x =200x
and 50 adult passes are sold
now total price is 2250
now x = childern
so childern pass price is 5
and adult price is 20 more then the childern
adult price is 25
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Nova Nerdplosion
Check out "
Hunting the Hidden Dimension
", with a basic explanation of fractals, the Mandelbrot set, and a large collection of applications (blood vessels! forest health! textiles! eyeballs! heartbeats!). The website has neat stuff now,
and you can watch the whole thing online starting Wednesday, October 29.
I love Nova. Especially Nova episodes about math. I show "
The Proof
", about Andrew Wiles' quest to prove Fermat's Last Theorem, to my classes when we have a down day. It gives them a taste of what professional mathematicians actually do all day. Also when they
complain that a problem takes too long, I can remind them that Wiles kept at the same problem for seven years. | {"url":"http://function-of-time.blogspot.com/2008/10/nova-nerdplosion.html","timestamp":"2014-04-16T04:28:49Z","content_type":null,"content_length":"155310","record_id":"<urn:uuid:bea2acdd-2b7f-4b1a-90c7-5004b7fc2e1e>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00299-ip-10-147-4-33.ec2.internal.warc.gz"} |
What is the feynman diagram for electromagnetic force between an electron and proton?
question dude
I'm kinda in two minds, I suspect that is wrong because wouldn't the fact that they attract each other (instead of repelling) means that the diagram would be drawn differently?
question_dude, there have been a few good answers here, but I think your above question indicates a bit of confusion that hasn't been addressed. It's important not take Feynman diagrams too
literally. The only thing that matters physically is which bits are connected to what. So, when we draw the lowest order diagram for two electrons interacting, the fact that the legs for the two
outgoing electrons are drawn pointing away from each isn't physically significant. So, for two electrons,
diagram (read left to right) is the same however you choose to rotate and twist the different pieces of it. What tells us two electrons repel each other is when we actually use the diagram to
calculate the probability that two electrons with particular initial momenta will scatter off each other into two electrons with two particular final momenta. If we do this, we learn that it is more
likely that by exchanging a photon the electrons will be scattered away from each other than scattered towards each other—a repulsive force.
We can do the same thing with electron-positron scattering. A positron is a type of anti-matter, which means it has the same properties as an electron but opposite charge. Now, when you hear about
anti-matter, you usually just here about it annihilating matter—so you might think the electron-positron interaction just destroys the two particles (and releases energy). That's one possibility, but
they can also just scatter of each other by Coulomb attraction (or, rather, the quantum interaction that creates Coulomb interaction on large scales). The lowest level diagram for that process looks
(read left to right). Notice how similar this looks to the other diagram! In fact, this diagram isn't quite right as the arrows on the two top legs should point the other way. Those arrows are
supposed to tell us whether its an electron or a positron, (
which way its travelling. Unfortunately, this is the best I could find. In any case, the electron and the positron attract each other, even though it doesn't look like it here! To actually see that
they attract, you again would need to use this diagram to calculate the scattering probabilities for different starting and ending momenta. You would see the final momenta corresponding to the two
particles being attracted to each other are the most likely. We can also, by making some approximations and comparing with non-relativistic quantum mechanics, show that these two diagrams give rise
to exactly the same "Coulomb's law" you learn in school for the force between like or different charges.
However, as someone else mentioned, these diagrams aren't the whole story. They're only what are called "lowest order" diagrams (and even then, not all of them). To calculate the total probability
something will happen, we have add up the contribution of every possible way it can happen. These diagrams are just some of the simpler ones. For electron scattering, we also have
diagram (read it from bottom to top—I'm sorry, there's no consistent convention for this): two electrons exchange a photon which along the way spontaneously turns into an electron and positron which
very quickly annihilate back into a photon. There are an infinite number of diagrams with more and more vertices all contributing to the same process—and, in theory, we have to add them all up. This
isn't possible, so instead we just approximate by adding up a few of the diagrams with small numbers of vertices. The higher order diagrams don't contribute much anyways because they have very low
probabilities of happening.
Finally, as other people have said, the diagrams for an electron scattering off a proton are not as straightforward because protons are made of quarks, and
are what the electrons actually interact with. Electron-positron scattering is much simpler and makes the same point. However, if you add up the contributions of the various low order (two vertex)
diagrams for a proton and an electron, you can represent them in a sort of short hand with something looking like the electron-electron and electron-positron diagrams (though instead of arrows on the
lower two legs, which mean electron or positron depending on the direction, you would just have to write 'proton'). Just keep in mind that, unlike the two examples above, this is not an elementary
process but a combination of multiple processes. | {"url":"http://www.physicsforums.com/showthread.php?p=4221537","timestamp":"2014-04-17T03:56:13Z","content_type":null,"content_length":"79520","record_id":"<urn:uuid:637e465e-eaa6-4c64-8dc8-4c7e1666b119>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00188-ip-10-147-4-33.ec2.internal.warc.gz"} |
st: svy commands
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st: svy commands
From "Theodoropoulos, N." <nt18@leicester.ac.uk>
To <statalist@hsphsun2.harvard.edu>
Subject st: svy commands
Date Mon, 21 Oct 2002 19:46:22 +0100
Dear statalisters,
I am using the survey commands in Stata to find some means. I have removed all the stratums in which one psu (primary sampling unit) is detected. I check that there is no stratum with one psu for all my variables by using the svydes command.
I don't have problems to get the invididual means for the individual variables "general" and "producti". However when i want to find the mean of the "general" variable given that "producti" variable is equal to zero i get the message that a stratum with one psu is detected. The "general" variable is continuous, and the "producti" variable is binary.
Any hints will be highly appreciated.
Thanks in advance.
Here is a copy of the output that i get.
svymean general
Survey mean estimation
pweight: empwt_nr Number of obs = 1727
Strata: newstr1 Number of strata = 44
PSU: newpsu1 Number of PSUs = 162
Population size = 1200.4609
Mean | Estimate Std. Err. [95% Conf. Interval] Deff
general | 4.156863 .0044976 4.147957 4.16577 2.432631
. svymean producti
Survey mean estimation
pweight: empwt_nr Number of obs = 1727
Strata: newstr1 Number of strata = 44
PSU: newpsu1 Number of PSUs = 162
Population size = 1200.4609
Mean | Estimate Std. Err. [95% Conf. Interval] Deff
producti | .3359512 .0475784 .2417331 .4301693 17.51394
. svymean general if producti==0
stratum with only one PSU detected
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Beginning Algebra with Applications and Visualization, Second Edition
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[Numpy-discussion] tensor ops in dim > 3?
Konrad Hinsen hinsen at cnrs-orleans.fr
Wed Aug 28 01:03:04 CDT 2002
"Dr. Dmitry Gokhman" <gokhman at sphere.math.utsa.edu> writes:
> I have a rank three n-dim tensor A. For two of the axes I want to perform
> v^t A v (a quadratic form with scalar output, where v is a vector). The
> final output should be a vector. I also need to compute the derivative of
> this with respect to v. This involves symmetrizing and matrix-vector
> multiplication (2 sym(A)v using two axes of A only, which gives a vector)
> with the final result being a matrix.
Whenever dealing with somewhat more complex operations of this time, I think
it's best to go back to the basic NumPy functionality rather then figuring
out if there happens to be a function that magically does it. In this case,
assuming that the first axis of A is the one that is not summed over:
sum( sum(A*v[NewAxis, NewAxis, :], -1) * v[NewAxis, :], -1)
The idea is to align v with one of the dimensions of A, then multiply
elementwise and sum over the common axis. Note that the first (inner)
sum leaves a rank-2 array, so for the second multiplication v gets
extended to rank-2 only.
> PS One more dumb question: I just installed the ScientificPython-2.4.1 rpm
> on my reincarnated Mandrake linux machine running python2.2. Do I need to
> do something to configure it? My scripts aren't finding things (e.g.
> indexing.py).
If you took the binary RPM from my site, they might not work correctly
with Mandrake, as they were made for RedHat. The source RPM should
work with all RPM-based Linux distributions. There is nothing that needs
configuring with an RPM.
Also note that Scientific is a package, so the correct way to import the
indexing module is
import Scientific.indexing
Konrad Hinsen | E-Mail: hinsen at cnrs-orleans.fr
Centre de Biophysique Moleculaire (CNRS) | Tel.: +33-2.38.25.56.24
Rue Charles Sadron | Fax: +33-2.38.63.15.17
45071 Orleans Cedex 2 | Deutsch/Esperanto/English/
France | Nederlands/Francais
More information about the Numpy-discussion mailing list | {"url":"http://mail.scipy.org/pipermail/numpy-discussion/2002-August/013965.html","timestamp":"2014-04-18T01:15:04Z","content_type":null,"content_length":"5037","record_id":"<urn:uuid:8a70a822-c976-45e0-9931-42c2ca644aff>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00020-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Yang-Mills millenium problem solved?
I am an avid reader of Wikipedia as there is always a lot to be learned. Surfing around I have found the article Yang-Mills existence and mass gap and the corresponding discussion page. Well, someone
put out my name but this is not the real matter. A Russian mathematician, Alexander Dynin, presently at Ohio State University, was doing self-promotion on Wikipedia at his paper claiming to have
found a solution to the problem. This is not published material, so Wiki Admins promptly removed it and started a discussion. By my side, I tried to make aware the right person for this and presently
no answer come out. I cannot say if the proof is correct so far but, coming from a colleague, it would be a real pity not to take a look. Waiting for more significant judgments, I will take some time
to read it.
21 Responses to Yang-Mills millenium problem solved?
1. As far as I understand the criteria for this ‘prize’ … one is required to provide a full axiomatic system from which the traditional picture may be derived. Without even looking at the paper, I
suspect that this has not been done.
2. Impressive paper. It certainly looks legit and comes from someone who does not appear to be cranky. If there are flaws in the proof, they are subtle.
3. Hmmm … yes, the paper does look very interesting. This will test my memory, lol.
4. As an alternative to the Dynin paper, I came up with my own (posted much earlier than Dynin’s)
I am saying this only because I just had found his paper.
Nevertheless, even though our papers are done differently, in the end we came to practically the same main results.
5. For everyone who is interested, I would like to say that my paper
is going to be published in August issue of IJGMMP
The task now lies in using these results for solving the problem of the quark confinement. My own preliminary investigation, based on the latest reviews on this subject,e.g. read 2 reviews by
M.Shifman on arxiv.org , indicates that this problem can indeed be solved. Surely, everybody is entitled for his/her opinion whether such an opportunity is realizable.
□ Dear Arkady,
Thank you for your comments pointing out your work. I would like to understand better the content of this work to become aware of the consistency of your conclusions with those I and
Alexander Dynin obtain. Looking at your paper I can see a sequence of equivalences between models reaching in the end a spectrum the same of the Richardson-Gaudin model. So, please correct me
if I am wrong, you conclude that quantum Yang-Mills theory has a mass gap being in the end reducible to such a model that it is a successful model by itself.
Now, the work I have done by solving directly the equations of the Yang-Mills theory, both classical and quantal, and that agrees with the conclusions obtained by different mathematical
techniques by Alexander Dynin, shows that Yang-Mills theory reaches a trivial fixed point at infrared and the spectrum is that of free particles (a harmonic oscillator spectrum) with
superimposed another harmonic oscillator spectrum as these particles would have a kind of internal structure (to be eventually understood with a higher level theory). Richardson-Gaudin model
as discussed by you in your paper does not seem to fit the bill in this way. Please, could you help me to clarify my ideas about as you claimed that you arrive at the same conclusions
obtained by Alexander Dynin (that are the same as mine)?
6. Dear Marco,
the first part of your conclusions is correct. In my work I was driven by ideas of A.Floer
which are nonperturbative in their nature. For some reason or another his ideas were and still in much use in mathematics and much less in physics. Publication of two long papers
http://arxiv.org/abs/hep-th/0610149 and
containing Floer’s ideas, to my knowledge, was left unnoticed by physics community, including you and Dynin. Since my results are nonperturbative, they cannot be compared straightforwardly with
yours. As for Dynin’s results, my own and his are in agreement in the sense that in both cases we had obtained bosonic spectrum which has a gap between ground and first excited state. This
superconducting conclusion is in accord with many-many less rigorous papers advocating the same superconducting picture for Y-M fields. As for my use of Richardson Gaudin model I have to state
that exactly this model was used for description of superconductivity (e.g. read references in my paper) and, furthermore, it is obtainable directly from the combinatorics of the scattering data,
e.g. see my paper
in accord with philosophy advocated originally by Heisenberg:
That is: given the combinatorics of excitation spectrum, restore the underlying microscopic model. I had written a separate paper about this methodology
In addition, my gap paper deals with gravity-Y-M correspondence in the way very different from that known as ADS-CFT correspondence. Because of this, not only the Y-M but gravity is involved in
the gap issues and imposes very stringent constraints on the whole formalism of the Standard Model.
□ Hi Arkady,
The point is that I and Alexander Dynin stipulated that our approaches take to an identical conclusion about the spectrum of the theory, at least in the infrared. These are different ways to
arrive to an identical conclusion that, by itself, permit to learn a lot about Yang-Mills theory.
By your side, what conclusions could be drawn about the spectrum of the theory? While in your paper there is a claim about the gap existence nothing is said about this. I think you should
agree that the work by Alexander goes more in-depth with respect to a simple conclusion of a gap being there.
Please, could you elucidate about?
7. Dear Marco, please, again keep in mind the differences between the perturbative and nonperturbative approaches to the problem. Incidentally, please take a close look at the book by A.Ushveridze
“Quazi-exactly solvable models in quantum mechanics”
It has a very detailed discussion about the Richardson-Gaudin model but, in addition to yours and Dynin’s results which this model is certainly capable of reproducing if the parameters are
assigned properly, it can be (and was already)used in other instances of nuclear and condensed matter physics where combinatorics is analagous to that observed in high energy physics. This
circumstance makes it a universal paradigm for all kinds of exactly and almost exactly solvable quantum mechanical models exibiting gap in their spectrum. Again, please, do not jump off the hook
by ignoring Floer’s approach to Y-M on one hand and combinatorics of the problem on another. Everybody and his brother in mathematics uses Floer’s approach to Y-M fields! Last but not the least,
while my results can be (and shall be)used to solve quark confinement problem (thus relating it to the gap problem as anticipated) I am having a hard time to see how this result will emerge from
your or Dynin’s results. May be you can shed some light on this topic for me.
Warmest regards,
□ Dear Arkady,
The point is quite simple and it appears to me that you are not answering. Dynin uses a non-perturbative technique that, in the proper limit, is able to reproduce my results at least from a
qualitative point of view. If you look at Dynin’s paper you will see that the author does a clearcut statement about the spectrum of the theory that I indeed see in the infrared through
perturbation techniques. Dynin and mine claims are really strong and imply a deep understanding about the way Yang-Mills theory should behave. This is surely better than just claiming to have
a mass gap also by an operational point of view. So, let me state clearly my question about your approach:
What is your conclusion about the full spectrum of the theory? Is this in agreement with lattice data?
8. Dear Marco,
you are really surprising me! Should I doubt about goodness of Dynin’s work, you would not even know about my existence. I would not make a comment on your blog in the first place. Yes, I know
that his work matches with yours and yes, I know about agreement with lattice calculations and so on. But, by doing my work I was not looking at the whole spectrum because I was thinking about
the existence of a gap as such. However, by fiddling with parameters of the Richardson-Gaudin model it should be possible to reproduce the whole spectrum. Hence, please, do not try to say whose
results are deeper and whose are not…After all, I am having now 165 downloads at the ads database
on my gap paper while Dynin’s paper has only 13
while staying much longer at the same database. But, I am not making a big deal out of this fact since, I believe, that we all had made a significant progress in the gap problem and, perhaps, as
the next step, it is of interest to figure out how the obtained results match each other. This is interesting problem for somebody who is thinking about earning his/her PhD, but NOT immediately
for me. I am much more interested currently in the quark confinement problem and I can see already what to do in this regard from my side but I cannot see yet this from yours or Dynin’s side and
you are ignoring my invitation to be a little bit more specific in this regard. In my case, the Abelian reduction causes some uses of the already known results to arrive at final destination….I
suspect, that the number of downloads indicates that my paper points into the right direction by also helping somehow to solve the problem of quark confinement. This problem holds the key to the
complete understanding of hadron physics. I also suspect that the gap problem by itself is interesting and should be considered as solved now but, regrettably, it is mainly of academic interest.
E.g. see how many good results people were able to obtain with or without exact solution of this problem
The confinement is much more subtle issue as far as I can understand things at this stage of my development :-). Please, correct me on this.
Warmest regards,
□ Dear Arkady,
I would like to remember that in your initial comment you claimed (with respect to Dynin’s paper)
Nevertheless, even though our papers are done differently, in the end we came to practically the same main results.
So, we can conclude without any doubt that your original statement was false.
I think you are late also for QCD. It is since 2008 that I partecipate to conferences and publish papers showing the low-energy limit of QCD with my approach.
9. Dear Marco, I had provided my detailed answers to your latest questions. However, my answers were erased. This is not good. Although I hate to do so, but if my reply is not going to be restored,
I will be forced to write it again in about a day or so. If again it is going to be erased, I am not going to participate in this noncence. Sorry.
□ Dear Arkady,
As you may know I live in Italy. This implies that in some hours when you are awake I am sleeping instead. My comment area is moderated. This means that I decide what should appear in my blog
and what should not before publication. This means that when you write your comment this does not appear immediately but wait until I do not moderate it. So, I am convinced that when you
posted your answer, WordPress warned you about this.
I think that before putting out such comments a moment to think and understand is essential. Mostly from a person that showed to be an incompetent about computer science.
10. Yes, this was a computer glitch. Now I know this. No problems. Viva Italia !
11. Dear Marco,
if to have 3 reads on your the most significant paper
is OK then, we are certainly in the different schools of thought. Who is late and for what is late only future might tell. Should your results be useful for others and lead to some next steps,
most likely, your paper would be much better read..It is obvious that you do not care about results of Floer, Donaldson, Atiyah and other mathematicians on Y-M theory.
Warmest regards,
12. I’m sorry, maybe I haven’t followed the latest developments about the Yang Mills Theory, unexpected things in life do surface once in a while.
However, I believe that the Yang Mills Theory has been solved by the use and solution of Feynman diagrams. As far as I know, the theory has not been solved through conics algebra.
If the conics solution is the Millennium Problem then Alexander Dynin solution is not what is required. Moreover, if the Yang Mills Millennium problem is not the problem of conical solution, and
Alexander Dynin solution is correct as per the problem description, then maybe the conics problem should be added to the Millennium problems!
□ If blogs serve any purpose, then, those who write comments should try to stick to facts. For instance, it is being said above that Alexander Dynin is a RUSSIAN PHYSICIST. Well, it takes only
a second to look up into MATH.SCI NET in order to see that he is respectable mathematician, always was and always will be. If the quality of the rest of information on any science blog is of
the same quality, then what is the purpose of blogs: to search for diccidents ;-), to figure out who is your friend? Or what else?
I wonder, why without any noise(“help”) from ANY blogs the gap paper by A. Kholodenko had accumulated 329 downloads as of today, July 24th , 2012
and that by A.Dynin-58
and that by M.Frasca-52
What is the role of blogs in such a process?
People who are sitting in all these respectable places will never ever admit that somebody outside their clan is capable of producing something of value. What else is new about this?! So,
with blogs or without, life goes on as usual……And always will go this way….
In regard to conics algebra. Why not if you are so sure about how the Y-M problem should be solved, you provide us with some exact references for a change! As far as the way you are
expressing yourself: “the Yang Mills Theory has been solved by the use and solution of Feynman diagrams” What is this? Use of Feynman diagrams upfront is good only if one is expanding theory
near trivial vacuum. But why the trivial vacuum is the right initial state? What means “solution of Feynman diagrams”? These are NOT equations, these DIAGRAMS which require some standard
calculation, necessarily perturbative (in such a case you are having an answer BEFOREHAND when you are making such a computation). This surely is not proving a thing….
☆ Dear Arkady,
I can understand your frustration. But let me state a couple of facts about what you claims. Firstly, your links are not good as just the one by Dynin provides the correct diagram of
downloads. Please, note that my paper to consider in this case is this.Secondly, I fear that the number of downloads is not the right indicator for a scientific publication. What really
matters here is the number of citations. In your case this number amounts to 4 that is not such a high number taking also into account a pair of self-citations. This fact is clearly
stated at NASA ADS site and, indeed, people at arXiv avoid always to make this number known to authors and for very good reasons.
In this post I called Alexander a physicist but the reason is that, at that time when I became aware of him through Wikipedia, I did not know his role in the community. This post is
already quite old and I should update it. Please, be aware that I have a lot of papers reviewed on MathSciNet but this does not qualify myself as a mathematician. I am also a reviewer for
this journal but again I do not claim to be a mathematician.
The role of blogs is essential as a vehicle of information for all the community. This blog is read by a lot of my colleagues and also much more authoritative than me. But the dynamics
for a published idea to emerge can follow quite strange ways. A blog can be just one of them.
○ Marco, please, make sure that Math Sci.Net is NOT a journal!
As a member of AMS I am a reviewer for Math.Sci Net for many years.Your record there is
indicates that you are having 25 publications and 6 citations….Mine record is
indicating that I am having 36 publications and 65 citations
Microsoft Corp. had made a ranking of mathematicians using as a guide the ratio of citations/publications WITHOUT COLLABORATORS
I am ranked as 15th from the top for past 10 years Why?
I am letting you to decide and also to find yourself in this list
13. You are right. I have confused it with Mathematical Review that is indexed there.
I think you are somewhat confused. I am not looking for a competition with you about ranking, h-index and all that. This is really silly. I am talking about the paper you claim is the most
downloaded. For this it is important the number of citations as you are claiming a MIllenium Prize I guess. Your paper has just 4 with 2 self-citations. I let you discover about mine.
I will not publish comments by you anymore. | {"url":"http://marcofrasca.wordpress.com/2010/06/17/yang-mills-millenium-problem-solved/","timestamp":"2014-04-17T21:23:41Z","content_type":null,"content_length":"121268","record_id":"<urn:uuid:bd4eaa68-1423-4b96-8034-d118842b2964>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00327-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Generating permutations lazily
up vote 59 down vote favorite
I'm looking for an algorithm to generate permutations of a set in such a way that I could make a lazy list of them in Clojure. i.e. I'd like to iterate over a list of permutations where each
permutation is not calculated until I request it, and all of the permutations don't have to be stored in memory at once.
Alternatively I'm looking for an algorithm where given a certain set, it will return the "next" permutation of that set, in such a way that repeatedly calling the function on its own output will
cycle through all permutations of the original set, in some order (what the order is doesn't matter).
Is there such an algorithm? Most of the permutation-generating algorithms I've seen tend to generate them all at once (usually recursively), which doesn't scale to very large sets. An implementation
in Clojure (or another functional language) would be helpful but I can figure it out from pseudocode.
algorithm functional-programming clojure combinatorics
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6 Answers
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Yes, there is a "next permutation" algorithm, and it's quite simple too. The C++ standard template library (STL) even has a function called next_permutation.
The algorithm actually finds the next permutation -- the lexicographically next one. The idea is this: suppose you are given a sequence, say "32541". What is the next permutation?
If you think about it, you'll see that it is "34125". And your thoughts were probably something this: In "32541",
• there is no way to keep the "32" fixed and find a later permutation in the "541" part, because that permutation is already the last one for 5,4, and 1 -- it is sorted in decreasing
• So you'll have to change the "2" to something bigger -- in fact, to the smallest number bigger than it in the "541" part, namely 4.
• Now, once you've decided that the permutation will start as "34", the rest of the numbers should be in increasing order, so the answer is "34125".
The algorithm is to implement precisely that line of reasoning:
1. Find the longest "tail" that is ordered in decreasing order. (The "541" part.)
2. Change the number just before the tail (the "2") to the smallest number bigger than it in the tail (the 4).
3. Sort the tail in increasing order.
You can do (1.) efficiently by starting at the end and going backwards as long as the previous element is not smaller than the current element. You can do (2.) by just swapping the "4"
with the '2", so you'll have "34521". Once you do this, you can avoid using a sorting algorithm for (3.), because the tail was, and is still (think about this), sorted in decreasing
order, so it only needs to be reversed.
The C++ code does precisely this (look at the source in /usr/include/c++/4.0.0/bits/stl_algo.h on your system, or see this article); it should be simple to translate it to your
language: [Read "BidirectionalIterator" as "pointer", if you're unfamiliar with C++ iterators. The code returns false if there is no next permutation, i.e. we are already in decreasing
template <class BidirectionalIterator>
bool next_permutation(BidirectionalIterator first,
up vote 101 BidirectionalIterator last) {
down vote if (first == last) return false;
accepted BidirectionalIterator i = first;
if (i == last) return false;
i = last;
for(;;) {
BidirectionalIterator ii = i--;
if (*i <*ii) {
BidirectionalIterator j = last;
while (!(*i <*--j));
iter_swap(i, j);
reverse(ii, last);
return true;
if (i == first) {
reverse(first, last);
return false;
It might seem that it can take O(n) time per permutation, but if you think about it more carefully, you can prove that it takes only O(n log n) time for all permutations in total, so
only O(1) -- constant time -- per permutation.
The good thing is that the algorithm works even when you have a sequence with repeated elements: with, say, "232254421", it would find the tail as "54421", swap the "2" and "4" (so
"232454221"), reverse the rest, giving "232412245", which is the next permutation.
2 This will work, assuming you have a total order on the elements. – Chris Conway Dec 9 '08 at 18:16
8 If you start with a set, you can arbitrarily define a total order on the elements; map the elements to distinct numbers. :-) – ShreevatsaR Dec 9 '08 at 18:21
1 This algorithm works very well. Thanks. – Brian Carper Dec 10 '08 at 2:46
2 This answer just doesn't get enough upvotes, but I can only upvote it once... :-) – Daniel C. Sobral Apr 18 '10 at 3:47
@Masse: Not exactly... roughly, you can go from 1 to a larger number. Using the example: Start with 32541. The tail is 541. After doing the necessary steps, the next permutation is
1 34125. Now the tail is just 5. Incrementing 3412 using the 5 and swapping, the next permutation is 34152. Now the tail is 52, of length 2. Then it becomes 34215 (tail length 1),
34251 (tail length 2), 34512 (length 1), 34521 (length 3), 35124 (length 1), etc. You are right that the tail is small most of the time, which is why the algorithm has good
performance over multiple calls. – ShreevatsaR Jul 6 '10 at 7:04
show 12 more comments
Assuming that we're talking about lexicographic order over the values being permuted, there are two general approaches that you can use:
1. transform one permutation of the elements to the next permutation (as ShreevatsaR posted), or
2. directly compute the nth permutation, while counting n from 0 upward.
For those (like me ;-) who don't speak c++ as natives, approach 1 can be implemented from the following pseudo-code, assuming zero-based indexing of an array with index zero on the "left"
(substituting some other structure, such as a list, is "left as an exercise" ;-):
1. scan the array from right-to-left (indices descending from N-1 to 0)
1.1. if the current element is less than its right-hand neighbor,
call the current element the pivot,
and stop scanning
1.2. if the left end is reached without finding a pivot,
reverse the array and return
(the permutation was the lexicographically last, so its time to start over)
2. scan the array from right-to-left again,
to find the rightmost element larger than the pivot
(call that one the successor)
3. swap the pivot and the successor
4. reverse the portion of the array to the right of where the pivot was found
5. return
Here's an example starting with a current permutation of CADB:
1. scanning from the right finds A as the pivot in position 1
2. scanning again finds B as the successor in position 3
3. swapping pivot and successor gives CBDA
4. reversing everything following position 1 (i.e. positions 2..3) gives CBAD
5. CBAD is the next permutation after CADB
up vote 34
down vote For the second approach (direct computation of the nth permutation), remember that there are N! permutations of N elements. Therefore, if you are permuting N elements, the first (N-1)!
permutations must begin with the smallest element, the next (N-1)! permutations must begin with the second smallest, and so on. This leads to the following recursive approach (again in
pseudo-code, numbering the permutations and positions from 0):
To find permutation x of array A, where A has N elements:
0. if A has one element, return it
1. set p to ( x / (N-1)! ) mod N
2. the desired permutation will be A[p] followed by
permutation ( x mod (N-1)! )
of the elements remaining in A after position p is removed
So, for example, the 13th permutation of ABCD is found as follows:
perm 13 of ABCD: {p = (13 / 3!) mod 4 = (13 / 6) mod 4 = 2; ABCD[2] = C}
C followed by perm 1 of ABD {because 13 mod 3! = 13 mod 6 = 1}
perm 1 of ABD: {p = (1 / 2!) mod 3 = (1 / 2) mod 2 = 0; ABD[0] = A}
A followed by perm 1 of BD {because 1 mod 2! = 1 mod 2 = 1}
perm 1 of BD: {p = (1 / 1!) mod 2 = (1 / 1) mod 2 = 1; BD[1] = D}
D followed by perm 0 of B {because 1 mod 1! = 1 mod 1 = 0}
B (because there's only one element)
Incidentally, the "removal" of elements can be represented by a parallel array of booleans which indicates which elements are still available, so it is not necessary to create a new array
on each recursive call.
So, to iterate across the permutations of ABCD, just count from 0 to 23 (4!-1) and directly compute the corresponding permutation.
1 ++ Your answer is underappreciated. Not to take away from the accepted answer, but the second approach is more powerful because it can be generalized to combinations as well. A
complete discussion would show the reverse function from sequence to index. – Die in Sente Feb 15 '09 at 18:46
Indeed. I agree with the previous comment — even though my answer does slightly fewer operations for the specific question asked, this approach is more general, since it works for e.g.
finding the permutation that is K steps away from a given one. – ShreevatsaR May 22 '10 at 15:27
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You should check the Permutations article on wikipeda. Also, there is the concept of Factoradic numbers.
Anyway, the mathematical problem is quite hard.
up vote 3
down vote In C# you can use an iterator, and stop the permutation algorithm using yield. The problem with this is that you cannot go back and forth, or use an index.
4 "Anyway, the mathematical problem is quite hard." No it's not :-) – ShreevatsaR Dec 9 '08 at 17:31
Well, it is.. if you don't know about Factoradic numbers there is no way you could come up with a proper algorithm in an acceptable time. It's like trying to solve a 4th degree
equation without knowing the method. – Bogdan Maxim Dec 9 '08 at 21:19
1 Oh sorry, I thought you were talking about the original problem. I still don't see why you need "Factoradic numbers" anyway... it's pretty simple to assign a number to each of the n!
permutations of a given set, and to construct a permutation from a number. [Just some dynamic programming/counting..] – ShreevatsaR Dec 11 '08 at 9:00
1 In idiomatic C#, an iterator is more correctly referred to as an enumerator. – Drew Noakes Dec 5 '11 at 11:03
@ShreevatsaR: How would you do that short of generating all permutations? E.g. if you needed to generate the n!th permutation. – Jacob Sep 19 '13 at 20:35
show 2 more comments
More examples of permutation algorithms to generate them.
Source: http://www.ddj.com/architect/201200326
1. Uses the Fike's Algorithm, that is the one of fastest known.
2. Uses the Algo to the Lexographic order.
3. Uses the nonlexographic, but runs faster than item 2.
PROGRAM TestFikePerm;
CONST marksize = 5;
marks : ARRAY [1..marksize] OF INTEGER;
ii : INTEGER;
permcount : INTEGER;
PROCEDURE WriteArray;
VAR i : INTEGER;
FOR i := 1 TO marksize
DO Write ;
permcount := permcount + 1;
PROCEDURE FikePerm ;
{Outputs permutations in nonlexicographic order. This is Fike.s algorithm}
{ with tuning by J.S. Rohl. The array marks[1..marksizn] is global. The }
{ procedure WriteArray is global and displays the results. This must be}
{ evoked with FikePerm(2) in the calling procedure.}
dn, dk, temp : INTEGER;
THEN BEGIN { swap the pair }
temp :=marks[marksize];
FOR dn := DOWNTO 1
DO BEGIN
marks[marksize] := marks[dn];
marks [dn] := temp;
marks[dn] := marks[marksize]
marks[marksize] := temp;
END {of bottom level sequence }
ELSE BEGIN
temp := marks[k];
FOR dk := DOWNTO 1
DO BEGIN
marks[k] := marks[dk];
marks[dk][ := temp;
marks[dk] := marks[k];
END; { of loop on dk }
marks[k] := temp;l
END { of sequence for other levels }
END; { of FikePerm procedure }
BEGIN { Main }
FOR ii := 1 TO marksize
DO marks[ii] := ii;
permcount := 0;
WriteLn ;
FikePerm ; { It always starts with 2 }
WriteLn ;
PROGRAM TestLexPerms;
CONST marksize = 5;
marks : ARRAY [1..marksize] OF INTEGER;
ii : INTEGER;
permcount : INTEGER;
PROCEDURE WriteArray;
VAR i : INTEGER;
FOR i := 1 TO marksize
DO Write ;
permcount := permcount + 1;
PROCEDURE LexPerm ;
{ Outputs permutations in lexicographic order. The array marks is global }
{ and has n or fewer marks. The procedure WriteArray () is global and }
{ displays the results. }
work : INTEGER:
mp, hlen, i : INTEGER;
THEN BEGIN { Swap the pair }
work := marks[1];
marks[1] := marks[2];
up vote 3 down vote marks[2] := work;
WriteArray ;
ELSE BEGIN
FOR mp := DOWNTO 1
DO BEGIN
hlen := DIV 2;
FOR i := 1 TO hlen
DO BEGIN { Another swap }
work := marks[i];
marks[i] := marks[n - i];
marks[n - i] := work
work := marks[n]; { More swapping }
marks[n[ := marks[mp];
marks[mp] := work;
BEGIN { Main }
FOR ii := 1 TO marksize
DO marks[ii] := ii;
permcount := 1; { The starting position is permutation }
WriteLn < Starting position: >;
LexPerm ;
WriteLn < PermCount is , permcount>;
PROGRAM TestAllPerms;
CONST marksize = 5;
marks : ARRAY [1..marksize] of INTEGER;
ii : INTEGER;
permcount : INTEGER;
PROCEDURE WriteArray;
VAR i : INTEGER;
FOR i := 1 TO marksize
DO Write ;
permcount := permcount + 1;
PROCEDURE AllPerm (n : INTEGER);
{ Outputs permutations in nonlexicographic order. The array marks is }
{ global and has n or few marks. The procedure WriteArray is global and }
{ displays the results. }
work : INTEGER;
mp, swaptemp : INTEGER;
THEN BEGIN { Swap the pair }
work := marks[1];
marks[1] := marks[2];
marks[2] := work;
ELSE BEGIN
FOR mp := DOWNTO 1
DO BEGIN
ALLPerm<< n - 1>>;
IF >
THEN swaptemp := 1
ELSE swaptemp := mp;
work := marks[n];
marks[n] := marks[swaptemp};
marks[swaptemp} := work;
AllPerm< n-1 >;
BEGIN { Main }
FOR ii := 1 TO marksize
DO marks[ii] := ii
permcount :=1;
WriteLn < Starting position; >;
Allperm < marksize>;
WriteLn < Perm count is , permcount>;
add comment
the permutations function in clojure.contrib.lazy_seqs already claims to do just this.
up vote 2 down vote
Thanks, I wasn't aware of it. It claims to be lazy, but sadly it performs very poorly and overflows the stack easily. – Brian Carper Dec 10 '08 at 2:51
add comment
I provided a solution in C# that produces such permutations lazily.
up vote 0 down vote See my answer here.
add comment
Not the answer you're looking for? Browse other questions tagged algorithm functional-programming clojure combinatorics or ask your own question. | {"url":"http://stackoverflow.com/questions/352203/generating-permutations-lazily/353248","timestamp":"2014-04-20T01:48:49Z","content_type":null,"content_length":"102234","record_id":"<urn:uuid:d40f8273-e40f-4840-821f-2d5c6479f566>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00057-ip-10-147-4-33.ec2.internal.warc.gz"} |
mathematicians rule
There are three men on a train. One of them is an economist, one is a logician and one of them is a mathematician. They have just crossed the border into Scotland and they see a brown cow standing in
a field from the window of the train standing parallel to where they're sitting.
The economist says, 'Look the cows in Scotland are brown'
The logician says, 'No. There are cows in Scotland of which one, at least is brown.'
And the mathematician says, 'No. There is at least one cow in Scotland, of which one side appears to be brown.' | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=35187","timestamp":"2014-04-20T13:40:50Z","content_type":null,"content_length":"12887","record_id":"<urn:uuid:43b39bc7-f2f7-47d9-9fa3-4ba897772abc>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00481-ip-10-147-4-33.ec2.internal.warc.gz"} |
Am I good enough to make it in grad school for Math?
I'm applying to physics programs, so I'm not sure it's exactly the same for math, but I think it would be in your best interests to take the extra year, and I don't think it will look bad to grad
schools (so long as you keep your grades up). I had a similar experience. I added math as a second major in the spring of my junior year and had to take an extra year to fit it all in - and worst of
all has been this semester, because I had to take 6 courses (4 of which were upper level math and physics, one of them a grad class) and it's practically killed me, especially with the physics gre in
the midst of it all. I think it is better to spread it out and learn more from each class, rather than to cram it all in and not get as much out of the courses.
As for learning to write proofs, don't worry, many of us struggled in the beginning. There's a common misconception floating around that you have to be some genius to do math, but that's not true -
if you just spend enough time studying and working at it, you can do it. The extra time would give you a lot of time to work on it. One thing that might help is to try proving the theorems in your
textbook, after you've read each section. Read through the book's proof the first time, then later try to prove them by yourself without looking at the book's proof. That helped me a lot in abstract
algebra and group theory. | {"url":"http://www.physicsforums.com/showthread.php?t=272101","timestamp":"2014-04-21T14:44:24Z","content_type":null,"content_length":"31549","record_id":"<urn:uuid:7d60e8f6-d4e8-4b7a-adc9-75ae0960c5a1>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00329-ip-10-147-4-33.ec2.internal.warc.gz"} |
Digit MATLAB neural network for al identification of the source code, you can ide AI-NN-PR www.pudn.com
] - digital Identification of matlab simulation, cutting through the images, there was a further neural network training, can achieve good results
] - A handwritten numeral recognition source, a complete identification procedures
] - through neural network for digital identification, only increasing the number seven, can increase the number to 10
] - Matlab neural network-based text-letter identification, very good with Oh, you download it
] - this is digit recognition using som
] - I write with bp network digital identification procedures, the number of drawing tools is the
] - MATLAB neural network for digital identification of the source code
] - Matlab source code is the use of network bp realize 0 ~ 9 digital identification system friendly system interface, including the training samples and
] - Based on BP neural network matlab program can achieve several fonts on the figure 0-9 to identify the document training network packet compression
] - | {"url":"http://en.pudn.com/downloads141/sourcecode/math/detail609583_en.html","timestamp":"2014-04-21T14:39:44Z","content_type":null,"content_length":"17682","record_id":"<urn:uuid:b22f8180-2875-4487-bdd2-f0be98525a26>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00198-ip-10-147-4-33.ec2.internal.warc.gz"} |
Milton, MA Prealgebra Tutor
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Milton, MA Trigonometry Tutors | {"url":"http://www.purplemath.com/Milton_MA_Prealgebra_tutors.php","timestamp":"2014-04-18T00:53:41Z","content_type":null,"content_length":"24211","record_id":"<urn:uuid:c284c494-858e-48c3-b8c2-fc536c6554e4>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00482-ip-10-147-4-33.ec2.internal.warc.gz"} |
boolean algebra tutorial
Author Message
vaceniniace Posted: Sunday 19th of Aug 11:35
Hello there I have almost taken the decision to look fora algebra private teacher, because I've been having a lot of problems with algebra homework lately . each time when I come
home from school I waste all my time with my algebra homework, and in the end I still seem to be getting the wrong answers. However I'm also not completely sure whether a math
private teacher is worth it, since it's very costly , and who knows, maybe it's not even so helpful. Does anyone know anything about boolean algebra tutorial that can help me? Or
maybe some explanations about multiplying fractions,quadratic formula or logarithms? Any ideas will be valued.
kfir Posted: Sunday 19th of Aug 20:18
Well of course there is. If you are enthusiastic about learning boolean algebra tutorial, then Algebrator can be of great benefit to you. It is made in such a way that almost
anyone can use it. You don’t need to be a computer expert in order to operate the program.
From: egypt
Bet Posted: Monday 20th of Aug 08:00
Some teachers really don’t know how to explain that well. Luckily, there are programs like Algebrator that makes a great substitute teacher for math subjects. It might even be
better than a real teacher because it’s more accurate and quicker!
From: kµlt øƒ Ø™
Vild Posted: Wednesday 22nd of Aug 08:09
I am a regular user of Algebrator. It not only helps me finish my assignments faster, the detailed explanations given makes understanding the concepts easier. I advise using it to
help improve problem solving skills.
From: Sacramento,
ctevemojrudel Posted: Wednesday 22nd of Aug 10:29
I am so relieved to hear that there is hope for me. Thanks a lot . Why did I not think about this? I would like to begin on this at once . How can I get hold of this program?
Please give me the particulars of where and how I can get this program.
From: Louisville,
TC Posted: Friday 24th of Aug 11:54
http://www.mathpoint.net/math-2374-worksheet.html. There you go. Hopefully you will not have to leave math.
From: Kµlt °ƒ Ø,
working on my time | {"url":"http://www.mathpoint.net/mathproblem/hyperbolas/boolean-algebra-tutorial.html","timestamp":"2014-04-16T13:19:37Z","content_type":null,"content_length":"58563","record_id":"<urn:uuid:7a04b29c-2015-4665-8003-44d02d24633f>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00351-ip-10-147-4-33.ec2.internal.warc.gz"} |
Nova Nerdplosion
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cyclic code
cyclic code
Let $C$ be a linear code over a finite field $A$ of block length $n$. $C$ is called a cyclic code, if for every codeword $c=(c_{1},\ldots,c_{n})$ from $C$, the word $(c_{n},c_{1},\ldots,c_{{n-1}})\in
A^{n}$ obtained by a cyclic right shift of components is also a codeword from $C$.
Sometimes, $C$ is called the $c$-cyclic code, if $C$ is the smallest cyclic code containing $c$, or, in other words, $C$ is the linear code generated by $c$ and all codewords obtained by cyclic
shifts of its components.
For example, if $A=\mathbb{F}_{2}$ and $n=3$, the codewords contained in the $(1,1,0)$-cyclic code are precisely
$(0,0,0),(1,1,0),(0,1,1)\text{ and }(1,0,1).$
Trivial examples of cyclic codes are $A^{n}$ itself and the code containing only the zero codeword.
Mathematics Subject Classification
no label found | {"url":"http://planetmath.org/cycliccode","timestamp":"2014-04-20T15:55:01Z","content_type":null,"content_length":"46766","record_id":"<urn:uuid:0c54763e-31bc-4e6a-820f-856c448acb84>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00414-ip-10-147-4-33.ec2.internal.warc.gz"} |
Fulton, MD Geometry Tutor
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Variation Worded Problem
April 14th 2009, 02:18 AM #1
Apr 2009
The price of painting the outside of a cylindrical tank (the bottom and top are not painted) of radius r and height h varies directly as the total surface area. If r=5 and h=4, the price is $60.
What is the price when r=4 and h=6
If you open up a cylinder into a net then you will observe that it is a rectangle with the top and the buttom being a circle. As the question says the top and the buttom are not painted then this
implies that we only consider the side, which is the rectangle.
The area of a rectangle is width multiplied by length. For the cylinder, the length is the height ( $h$) and the width is the circumference (because this is the length of the circle) this it is
$2\pi r$.
Therefore, the surface area considered is $S = (2\pi r)(h) = 2 \pi h r$.
When $r=4, h=5$, the surface area is $S = 2 \times \pi \times 5 \times 4 = 40\pi$ at which the price is $\ 60$.
Now consider the surface area when $r=4, h=6$, this would equal $S = 2 \times \pi \times 4 \times 6 = 48\pi$.
Now we are able to use general arithmetic (Ratio consideration) to solve for the price. When $r=4, h=6$, therefore the price ( $P$) is $P= \frac{48 \pi \times 60}{40\pi} = ...$
EDIT: I have attached the cylinder diagram and the net considered.
Last edited by Air; April 18th 2009 at 02:58 AM. Reason: EDIT1: Attaching Diagram, EDIT2: Had Put EDIT1 Reason Into Title.
To learn how to set this up as a variation equation, try here.
Once you have learned the basic terms and techniques, the following should make sense:
i) The variation equation must be of the form P = k(SA), where P is the price, k is the constant of variation, and SA is the surface area.
ii) You know that SA = 2(pi)rh, so P = 2k(pi)rh.
iii) You are given that, for r = 5 and h = 4, P = 60. Use this to solve for the constant of variation k.
iv) Now that you have the value of k, plug this into P = 2k(pi)rh. This is your variation equation.
v) Plug in 4 for r and 6 for h. Simplify to find the required value for P.
April 14th 2009, 02:54 AM #2
April 14th 2009, 05:09 AM #3
MHF Contributor
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Re: B1 B2 proof
From: Peter F. Patel-Schneider <pfps@research.bell-labs.com> Date: Wed, 11 Jun 2003 15:51:46 -0400 (EDT) Message-Id: <20030611.155146.92600738.pfps@research.bell-labs.com> To: jjc@hplb.hpl.hp.com Cc:
Here are my comments on Jeremy's proof.
The summary is that there are significant holes in the proof. I don't know
whether they can be fixed.
| We can weaken the constraint on blank nodes corresponding to descriptions
| and dataranges to be that all directed cycles of such nodes must have an
| owl:equivalentClass or owl:disjointWith triple.
| Sketch
| ======
| We assume that Peter shows the correspondence
| theorem without any reuse of bnodes.
| We show that any two identical descriptions are
| equivalent classes.
| The only possible syntactic uses of bnodes
| corresponding to descriptions in OWL DL occur
| either in triples whose OWL Full semantics
| depends only on the class extension of the bnode,
| or on rdf:first triples in lists which are the object
| of owl:unionOf and owl:intersectionOf triples.
| In the first case we can replace a bnode by an
| equivalent one, in the second case, given an equivalent
| bnode we can use the comprehension axioms to construct
| an equivalent list, and then use that since the
| semantics of the owl:unionOf and owl:intersectionOf
| triples only depend on the class extensions.
| These permit us to proceed by induction on the number
| of bnodes reused.
| B1 B2 Proof
| ===========
| Preliminaries:
| We add to the preamble to the mapping rules words like:
| [bnode reuse]
| "When processing an abstract syntax construct corresponding
| to either the description, restriction or dataRange construct
| then, if that exact instance of the construct has already
| occurred then there is at least one blank
| node already corresponding to the construct. In such a case,
| the mapping may nondeterministically use any previous result
| (a blank node) or may apply the mapping rues to the new occurrence."
| Lemma: Simplified EquivalentClass and DisjointClasses construct.
| Given the new mapping rules, if G is a graph corresponding to
| an abstract syntax tree A then G corresponds to A' in which
| all occurrences of EquivalentClass and DisjointClasses have
| at most two descriptions.
| Proof:
| We form A' from A by copying the ontology
| and modifying the EquivalentClass and DisjointClasses
| axioms.
| Take each
| EquivalentClass( d1 .... dn ) n > 2
| and
| if
| T(di) owl:equivalentClass T(dj) . is in G
| add
| EquivalentClassses(di dj)
| and delete each EquivalentClasses( d1 ... dn ) n > 2
| Similary for DisjointWith( d1 ... dn )
| We can take the mapping T:A -> G and use it to map
| A'->G, using the new clause in the preamble of the mapping rules.
This lemma appears to be about optional typing.
| Lemma: Blank description usage.
| If A is an OWL DL abstract syntax tree and G
| is a corresponding RDF graph and b is some blank node
| in G, for which the mapping rule that created b
| was one of those with an optional rdf:type owl:Class
| or rdf:type owl:Restriction or rdf:type owl:DataRange
| triple; suppose t is a triple containing b then t
| matches one of:
| A)
| b rdf:type rdfs:Class .
| b rdf:type owl:DataRange .
| b rdf:type owl:Restriction .
| b rdf:type owl:Class .
| B)
| XX rdf:first b .
| (moreover XX forms part of a list
| where that list is the object of a
| triple with predicate owl:unionOf
| or owl:intersectionOf).
| C)
| b owl:disjointWith XX .
| XX owl:disjointWith b .
| XX owl:equivalentClass b .
| b owl:equivalentClass XX .
| XX owl:someValuesFrom b .
| XX owl:allValuesFrom b .
| XX owl:complementOf b .
| XX rdf:type b .
| D)
| b owl:intersectionOf XX .
| b owl:unionOf XX .
| b owl:complementOf XX .
| b owl:oneOf XX .
| b owl:onProperty XX .
| b owl:cardinality XX .
| b owl:minCardinality XX .
| b owl:maxCardinality XX .
| b owl:someValuesFrom XX .
| b owl:allValuesFrom XX .
| Proof: by inspection
I do not believe that this definition works correctly. It doesn't
correctly take into account the optional type triples, for example.
| Definition: The bnode reuse count BRC or a graph G
| is the amount of bnodes usage more than once each i.e.
| BRC(G) = | { <s, p, b > in G: b blank }|
| + | { <b, owl:equivalentClass, o> in G: b blank }|
| + | { <b, owl:disjointWith, o> in G: b blank }|
| - | { b : b blank and
| <s, p, b> in G or
| <b, owl:equivalentClass, o> in G or
| <b, owl:disjointWith, o> in G }|
| Terminology: we will say that a triple *using* b
| is one that either has object b, or has subject b
| and predicate owl:equivalentClass
| or owl:disjointWith.
| Theorem: to be proved by Peter,
| If graph G corresponds to an
| abstract ontology A
| and the mapping from A to G does not
| involve any bnode reuse, then the correspondence
| theorem holds.
| Lemma: If graph G corresponds to an
| abstract ontology A and BRC(G)=0
| then the correspondence theorem holds.
| Proof:
| Let T be the mapping from A to G.
| All bnodes in G are the object of at most one
| triple. bnode reuse can only occur for
| bnodes corresponding to the EquivalentClass(d1)
| construct, which generates a bnode that can be
| reused that is the object of no triples (or
| alternatively may reuse a bnode without making it
| the object of any triple). Let us call an
| axiom of this form redundant if T(d1) is the
| object of a triple, or if it is not the first
| axiom in the ontology mapping to that bnode.
| Let A' be A with all redundant axioms deleted.
| A' corresponds to G by the restriction of T to A'.
| A' has no bnode reuse so Peter's result holds,
| between A' and A. The vocabulary usage of A' and
| A are identical; and A and A' are logically
| equivalent by the direct semantics.
| Thus the correspondence theorem holds between A
| and G.
| Lemma: identical descriptions.
| If A is an abstract ontology
| and G corresponds to A under T, and d and d'
| are two occurrences of an identical description
| in A with T(d) != T(d') and with the mapping
| of the subdescriptions of d and d' also not sharing
| bnodes then all interpretations
| of G are interpretations of
| T(d) owl:equivalentClass T(d') .
| Proof:
| Let A1 = ontology( EquivalentClass(d) EquivalentClass(d') )
| Let A2 = ontology( EquivalentClass(d d') )
| By the direct semantics A1 entails A2.
- because d = d'
| Extending and restricting T we map A1 to G1, a subgraph of G
| and we map {A2} to G1 union
| { <M(T(d)) owl:equivalentClass M(T(d')) > }
^^ ^ ^^ ^
| Since the mapping of T(d) and T(d') do not share bnodes
| both the mapping T:A1 -> G1 and T:A2 -> G2 are covered
| by the correspondence theorem without bnode reuse.
| Thus we have that any interpretation of G1 is an
| interpretation of
| M(T(d)) owl:equivalentClass M(T(d')) .
^^ ^ ^^ ^
| Since G1 is a subgraph of G we are done.
It might be easier to do this directly.
| Theorem: The correspondence theorem holds.
| Proof: by induction on BRC.
| Initial case, BRC(G) = 0, by Peter's result.
| Suppose n > 0, and the result holds for all G with
| BRC(G) < n.
| Suppose BRC(G) = n, with A corresponding to G by mapping T.
| T(A)=G
| w.l.o.g. A does not contain an EquivalentClass
| or DisjointClasses with n > 2 (see lemma).
| Since BRC(G) > 0 we have there is a blank node b
| such that two triples in G use b.
You haven't shown this.
| This must be by use of the bnode reuse
| clause of the preamble. (We have explicitly excluded
| the two mappings rules EquivalentClass, DisjointClasses,
| which introduce multiple copies of bnodes).
| The bnode b hence corresponds to two identical descriptions
| d and d' within A.
| Form A' by adding a final axiom EquivalentClass(d'')
| with d'' = d' = d
| Extend T to T' to map A' by mapping the new axiom
| without invoking the bnode reuse clause.
| Let G' = T'(A'), this is a superset of G.
| Let G'' be G' with one of the triples using b to be
| replaced by a similar triple using T'(d'').
| BRC(G'') = n - 1.
You haven't shown that BRC(G') = BRC(G)
or that BRC(G'') = BRG(G') - 1
| We can construct a mapping T'' of A to G'' by
| not using the bnode reuse clause on one of the
| occassions where we had used it.
I'm not convinced of this.
| By the inductive hypothesis the correspondence
| theorem holds between A and G''. We now show that
| any interpretation of G'' is an interpretation of G
| and we are done.
| By the identical descriptions lemma we have that
| any interpration of G'' is an interpretation of
| G'' union { < M(T'(d'')) owl:equivalentClass M(T(d)) > }
^^ ^ ^^ ^
- because T(d'') and T(d) don't share blank nodes
| The only triple in G which is not in G''
| is the triple using T(d) which has been
| replaced by that using T'(d''). Looking at the
| blank description usage lemma, and the definition
| of 'using' we see that this triple must have the
| form of section (B) or (C) in the blank description
| usage lemma.
| If it is of form (C) then the truth of the triple
| is a function of the class extension of the
| interpretation of the blank node, and does not depend
| on the interpretation of the blank node itself. Since
| the replacement triple in G'' has a blank node with
| identical class extension, then any interpretation
| of G'' is also an interpretation of the replaced triple.
| If it is of form (B), it is the object of an
| rdf:first triple. This rdf:first triple is part of
| a well-formed list, and that list is the object of
| an owl:unionOf or owl:intersectionOf triple, t.
| Under the RDF semantics the blank nodes in the list
| are understood as existentials. By the first comprehension
| axiom for lists, there is such a list and so the set of
| rdf:first & rdf:rest triples forming the list are true.
| The truth of t depends on the semantics of owl:unionOf
| and owl:intersectionOf which have necessary and sufficient
| conditions that depend only on the class extensions
| of the subject and the elements in the list formed
| by the object.
| Given an interpretation of G'' then this shows that there
| is some mapping of the blank nodes making t true in G''.
| There is some other mapping of the blank nodes that makes
| G - { t } true, i.e. which maps the blank nodes
| of the list which is the object of t differently.
| Under this interpretation all the members of the list have
| the same class extension and so t is still true. Thus this
| is an interpreation of G. The proof is complete.
Received on Wednesday, 11 June 2003 15:51:59 GMT | {"url":"http://lists.w3.org/Archives/Public/www-webont-wg/2003Jun/0111.html","timestamp":"2014-04-16T10:23:38Z","content_type":null,"content_length":"18741","record_id":"<urn:uuid:036dc786-4c78-46d6-a854-1524bef5b4f7>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00335-ip-10-147-4-33.ec2.internal.warc.gz"} |
Patent US7577927 - IC design modeling allowing dimension-dependent rule checking
This application is a continuation patent application of U.S. patent application Ser. No. 11/926,289, filed on Oct. 29, 2007, currently pending, which is a continuation of application Ser. No. 10/
708,039, filed on Feb. 4, 2004, now U.S. Pat. No. 7,404,164, currently issued.
1. Technical Field
The present invention relates generally to integrated circuit design, and more particularly, to integrated circuit design modeling that allows for dimension-dependent rule checking.
2. Related Art
Very large scale integrated (VLSI) circuits are designed using computer-implemented design systems that allow a designer to generate and test a circuit design before the more expensive manufacturing
of the integrated circuit (IC). In order to ensure proper design of an IC, each design system and/or IC format includes a set of design rules that each circuit design must meet. That is, each IC
design must pass a design rule check (DRC). One fundamental operation of DRC is dimension-dependent rule checking. Dimension-dependent rule checking ensures that dimensions, namely width and spacing,
of parts of an IC meet a specific dimensional parameter, e.g., a size range. Among the most significant dimension-dependent rule checking are the width-dependent and spacing-dependent rules that
prevent IC parts from being too large/small or too close together/too far apart.
Current DRC tools derive width and spacing of VLSI shapes using one-dimensional visibility or standard “shape-expand” and “shape-shrink” operations. Advancements in IC design have resulted in the
various shapes of IC parts becoming more complex than simple rectangles. As a result, defining what is a “width” of a shape(s), or what is the “spacing” between shapes has become difficult and
inconsistent. In particular, the conventional mathematical definition of “width” of any polygon is the smallest distance between any two parallel lines of support. This definition implies a unique
width for any shape based on its convex hull. (Note that typically an IC non-convex polygon can be much more “narrow” than its associated convex hull, e.g., an L-shape). However, this definition is
not appropriate for DRC among non-convex shapes as it assigns the unique width of the convex hull to the whole shape.
Where a DRC method uses one-dimensional (1D) visibility, the width or spacing near a vertical edge is often determined by other vertical edges that this edge can “see” along the horizontal direction.
This approach is adequate in some cases but it often leads to results that are inconsistent as 1D visibility is not appropriate to characterize the width of a shape. For example, for any edge having
a distance to another edge that varies along its length results in the width being inconsistent or indeterminable. Currently, there is no consistent or satisfactory definition for the “width” of an
arbitrary shape(s) other than a rectangle. The above-described problem applies to spacing-dependent design rules also.
With regard to “expand” and “shrink” operations, most modern shape processing tools have functions that expand or shrink a shape by a given constant amount. The results of these operations are
incrementally used to determine width and spacing. There are several problems with such an approach. First, the expand and shrink operation can give correct width results only for shapes consisting
of axis parallel edges. Once an acute angle is present the results become incorrect. In addition, the expand and shrink operation is simple only in the case of convex shapes. For general non-convex
shapes, even orthogonal ones, the expand and shrink operation is difficult and rather expensive in terms of computational time. Furthermore expand and shrink operations work only for given constant
amounts and, therefore, many calls to these functions are typically required for standard shapes. As a result, the checking of dimension dependent rules becomes very cumbersome and time consuming.
One approach to checking spacing dependent rules is to represent a rule in a discrete form given by a small number of “buckets.” “Buckets” are rule implementations based on predefined numeric values
or ranges of predefined numeric values. For example, buckets for spacing may mandate that a space be greater than a dimension S if at least one line has a width greater than a value W. In another
example, a space is greater than or equal to a dimension S if both lines are greater than a value W. Using the expand/shrink approach to check this type rule is cumbersome and time consuming.
Checking width-dependent spacing rules given in a function form (i.e., non-step or non-bucketed) would be even more cumbersome (if at all possible) by employing such methods. For example, it is not
known how to implement a width-dependent rule that mandates that the spacing between two neighboring shapes must be 2× the maximum width of one of the shapes minus 10 units.
In addition to the above problems, use of buckets leads to IC designs that are hard to manufacture. In particular, there are two problems that result from the IC design rule implementation using
buckets. The first problem is based on the premise that design rules are established prior to the creation of a manufacturable process, and thus, the design rules represent assumptions and
commitments of the process made during design. Typically, however, process constraints usually deviate by some amount upon reaching manufacturing. Therefore, a design rule legal layout may be outside
the process window capability of the manufacturing process. Avoiding this problem infers that all of a specific “bucket” lies within the manufacturing process capability, which leads to the second
problem where productivity is “left on the table.” In particular, the “bucketed” design rules motivate the designer to use the limit provided rather than what is optimal (assuming the process does
not follow the same function as the buckets). In particular, designers actually make calculations based on the bucket specifications to determine which size wires to use rather than using the size
wire that optimizes their functionality and performance requirements. The end results lead to a non-optimal IC design.
Another problem relative to DRC is that design rules capture diverse physical phenomena, each of which may require a slightly different definition of width and spacing as long as the definition can
be treated in a consistent manner. For example in certain cases a shape may need to be evaluated as “thin” as possible, while in other cases it may need to be evaluated as “fat” as possible.
Currently, no mechanism exists for applying different definitions of width and spacing for DRC.
In view of the foregoing, there is a need in the art for a way to model an IC design that allows for dimension-dependent design rule checking regardless of shape. It would also be advantageous to be
able to implement different definitions of width and spacing.
The invention provides a method, system and program product to model an IC design to include dimensions such as a local width and spacing of IC shapes in a consistent fashion. In particular, the
invention uses a core part of Voronoi diagrams to partition edges of a shape into intervals and assigns at least one dimension to each interval such as a local width and spacing. Dimension assignment
can be made as any desirable definition set for width and spacing, e.g., numerical values or continuous dimension-dependent design rules. Design rule checking for dimension-dependent spacing rules
given in any arbitrary functional form of width and spacing is possible. Application of the invention can be made anywhere the width and spacing of VLSI shapes play a role, e.g., relative to a single
edge, a neighboring edges, a neighboring shapes, and/or for edges in more than one layer of the IC design.
A first aspect of the invention is directed to a method of modeling for use with an integrated circuit (IC) design, the method comprising the steps of: partitioning an edge of a shape in the IC
design into a plurality of intervals; and assigning at least one dimension to each interval.
A second aspect of the invention is directed to an integrated circuit (IC) modeling system comprising: means for partitioning an edge of a shape in the IC design into a plurality of intervals; and
means for assigning at least one dimension to each interval.
A third aspect of the invention is directed to a computer program product comprising a computer useable medium having computer readable program code embodied therein for modeling an integrated
circuit, the program product comprising: program code configured to partition an edge of a shape in the IC design into a plurality of intervals; and program code configured to assign at least one
dimension to each interval.
A fourth aspect of the invention is directed to an integrated circuit (IC) check rule evaluation system comprising: means for partitioning an edge of a shape in the IC design into a plurality of
intervals, the partitioning means including: means for generating a core Voronoi diagram for the shape using a first metric, and means for partitioning the edge based on the core Voronoi diagram;
means for assigning at least one dimension to each interval using a second metric; and means for using the at least one dimension to evaluate a check rule.
The foregoing and other features of the invention will be apparent from the following more particular description of embodiments of the invention.
The embodiments of this invention will be described in detail, with reference to the following figures, wherein like designations denote like elements, and wherein:
FIG. 1 shows a simplified example of a Voronoi diagram.
FIG. 2A shows an exterior Voronoi diagram for a set of polygons using the L[∞] metric.
FIG. 2B shows an interior Voronoi diagram for a polygon using the L[∞] metric.
FIG. 3 shows an interior Voronoi diagram for an acute polygon using the Euclidean metric.
FIG. 4 shows an interior Voronoi diagram for the acute polygon of FIG. 3 using the L[∞] metric.
FIGS. 5A-5B show illustrations for description of the L[∞] metric.
FIG. 6 shows a block diagram of a modeling system according to the invention.
FIG. 7 shows a flow diagram of operation of the modeling system of FIG. 6.
FIGS. 8A-8E show illustrative implementations of the method of FIG. 7 relative to the acute polygon of FIG. 3.
FIG. 9A shows illustrative local widths and spacings for the shapes of FIGS. 2A and 2B.
FIGS. 9B-9C show illustrative implementation of the method of FIG. 7 relative to the set of polygons and polygon of FIGS. 2A and 2B, respectively.
FIG. 9D shows a detail of a polygon with a thin neck caused by convcave vertices.
FIGS. 10A-10D show illustrative edge arrangements including non-orthogonal angles.
FIG. 11 shows the shape of FIG. 4 including partitioning according to the L[∞] metric.
FIG. 12 shows a detail of polygon having a concave vertex.
FIG. 13 shows the shape of FIG. 4 partitioned using L[∞] metric and assigned dimensions using the Euclidean metric.
FIGS. 14A-14F show the shape of FIGS. 2B and 9C for narrower partitioning.
For purposes of clarity only, the following description includes the following sections: I. Overview, II. General Definitions, III. System Overview, IV. Methodology, and V. Conclusion.
I. Overview
IC design rules and computer-aided design (CAD) tools normally work with boundaries of IC shapes. The invention leverages this fact by partitioning an edge of a shape into a plurality of intervals.
Each interval is then assigned a dimension, including in one embodiment a local width and a local spacing, based on discrete neighboring geometry as derived from a local Voronoi diagram in the
interior and/or exterior of the shape. In other words, a local width and local spacing can be mapped to an edge of a shape. As a result, every boundary point of a shape may be assigned a unique width
label and a unique spacing label.
A definition set determines how an edge is partitioned and how width and spacing are assigned for an edge. Definition sets may vary depending on design rules. Accordingly, the assignments of width/
spacing can reflect the nature of the design rule of interest and need not be the same for all rules. In addition, each interval can have a constant width/spacing value or it can be assigned a
function such as the distance from another boundary/corner, a factor of another dimension or other function. Once widths and spacings are obtained they can be compared against each other to check any
kind of relation between them, and, in particular, can be checked for compliance with continuous dimension-dependent design rules.
II. General Definitions
As noted above, the invention implements Voronoi diagrams to determine dimensions, i.e., local widths and spacings, of shapes. A “Voronoi diagram” may take a variety of forms depending on the
structure to which it is applied. In the simplest example, referring to FIG. 1, a “Voronoi diagram” 8 for a set of sites (points) 10 a-10 d includes all points that are closer to a corresponding
site, e.g., 10 a, than to any other site 10 b-10 d. The resulting Voronoi diagram 8 includes a collection of regions 12 a-12 d, called “Voronoi cells,” that divide up the plane in which sites 10 a-10
d sit, i.e., the plane of the page. Each Voronoi cell 12 a-12 d is in the shape of a convex polygon (for point sites only) having edges that are bisectors of sites 10 a-10 d. Each Voronoi cell 12 a-
12 d corresponds to, or is owned by, one of sites 10 a-10 d. Again, all points in one cell, e.g., 12 a, are closer to the corresponding site 10 a than to any other site 10 b-10 d.
FIGS. 2A-2B illustrate an exterior Voronoi diagram 20 and an interior Voronoi diagram 30 for orthogonal shapes, respectively. FIG. 2A illustrates a set of disjointed polygonal sites 22A-22F, and FIG.
2B illustrates a polygon 122. FIGS. 3 and 4 illustrate interior Voronoi diagrams 40, 90 for an acute polygon shape 34. As used herein, the term “boundary” shall refer to an outer border or line of a
shape, while an “edge” shall refer to a component of the boundary.
As shown in FIG. 2A, an external Voronoi diagram 20 for a set of disjoined polygonal sites, i.e., polygons 22A-22F, includes a partitioning of the plane containing polygons 22A-22F into Voronoi cells
24A-24F. Each Voronoi cell, e.g, 24A, for a polygon 22A includes a locus of points closer to polygon 22A than to any other polygon 22B-22F. That is, it includes bisectors of adjacent polygons 22B-22F
(others not shown). A Voronoi cell of a site s, e.g., polygon 22A, is denoted as reg(s) and polygon 22A is referred to as the owner of reg(s). A portion 23 of Voronoi diagram 20 that borders two
Voronoi cells 24A and 24B is referred to as a “Voronoi edge,” and includes portions of bisectors between the owners of cells. A point 26 where three or more Voronoi edges 23 meet is called a “Voronoi
vertex.” Dashed lines indicate partitioning of polygons 22A-22F such that each boundary, e.g., 28A, of a polygon 22A has a corresponding portion 23 of Voronoi diagram 20.
Referring to FIGS. 2B, 3 and 4, an interior Voronoi diagram 30, 40, 90 on an interior of a shape 34 (polygon 122 in FIG. 2B) is referred to as a “medial axis.” Formally, the “medial axis” is the
locus of points [q] internal to a polygon such that there are at least two points on the object's boundary that are equidistant from [q] and are closest to [q].
The above-described Voronoi diagrams of FIGS. 1 and 3 are illustrated as based on the Euclidean metric. That is, Voronoi distances are based on the Euclidean metric. The invention, as will be
described further below, may also implement Voronoi diagrams based on an L[∞] (L-infinity) metric. Referring to FIGS. 5A-5B, in the L[∞] metric, the distance between two points (FIG. 5A) p=(x[p], y
[p]) and q=(x[q], y[q]) is the maximum of the horizontal distance and the vertical distance between p and q, i.e.,
d(p,q)=max[|x [p] −x [q] |,|y [p] −y [q]|].
Intuitively, the L[∞] distance is the size of the smallest square touching p and q. The L[∞] distance between any two points is less than or equal to the Euclidean distance between the points.
Further, in the L[∞] metric, the distance between a point p and a line l (FIG. 5B) is d(p,l)=min [d(p,q),∀qεl]. The L[∞] bisector of two polygonal elements (points or lines) is the locus of points at
equal L[∞] distance from the two elements.
Returning to FIG. 4, medial axis 90 of shape 34 is illustrated as based on the L[∞] metric. (Medial axis 30 in FIG. 2B is also based on the L[∞] metric). As easily discerned by comparing FIGS. 3 and
4, the use of the L[∞] metric simplifies Voronoi diagram 90 of polygonal objects and makes it simpler to compute. The L[∞] Voronoi diagrams are a “skeleton” of straight-line segments having linear
combinational complexity. While methodology of the invention will be described based on one or the other metric, it should be recognized that the concepts and functioning of the invention are the
same regardless of the metric used, and the invention should not be limited to any particular metric other than as set forth in the attached claims.
Further explanation of Voronoi diagrams and their application to critical area determination can be found in U.S. Pat. Nos. 6,317,859 and 6,178,539, which are hereby incorporated by reference for all
purposes. See also, E. Papadopoulou et al., “The L[∞] Voronoi Diagram of Segments and VLSI Applications,” International Journal of Computational Geometry and Applications, Vol. 11, No. 5, 2001,
Returning to FIGS. 2A, 2B, 3 and 4, a “core” 50 (indicated by thicker line) of a Voronoi diagram 30, 40, 90 is the portion remaining after excluding “uninteresting portions.” Typically,
“uninteresting portions” (lighter lines) are portions of Voronoi edges coupled to a boundary 52 (FIGS. 2B, 3 and 4) of the shape, e.g., bisectors 54 between adjacent edges. In the case of IC shapes
with acute angles (i.e., concave vertices), as shown in FIGS. 3 and 4, “uninteresting” can be a portion of a bisector 54 coupled to an acute angle 56 at a certain distance from acute angle 56. What
is excluded from core 50 can be flexible and application dependent. For simplicity, core 50 shall include all Voronoi edges 23 that are not coupled to boundary 52 of shape 34, i.e., all bisectors
between non-adjacent edges and non-degenerate bisectors. A “degenerate bisector” is one between two collinear horizontal/vertical edges. However, it should be clear that core 50 is the “interesting”
subset of a Voronoi diagram 30, 40, 90 where the “interesting” classification can vary for each application.
A “spot defect” is caused by particles such as dust or other contaminants in materials and equipment and are classified as two types: “extra material” defects causing a short between different
conducting regions, e.g., 22A-22F of FIG. 2A, and “missing material” defects causing an open, e.g., in polygon 122 of FIG. 2B. In other words, a “short” is a defect creating a bridge with some other
edge in the exterior of a shape while an “open” is a defect creating a void in the interior of a shape. Referring to FIG. 3, a spot defect may be modeled as a “core element” 60 in the form of a
circle (Euclidean metric only) or a square (L[∞] metric in FIGS. 2A, 2B and 4). That is, a “core element” 60 can be represented by a circle or a square depending on the metric used. Since, in
reality, spot defects have any kind of shape, the square (L[∞] metric) defect model is good for most purposes. The term “core disk” will be used to refer to the circles of the Euclidean metric, and
“core square” will be used to refer to the square of the L[∞] metric, and the term “core element” is generic to both.
A “core element” 60 touches a boundary 52 in at least two points 64, and has a center referred to as a “core point” 62, which is a member of core 50 for a shape, e.g., polygon 122 in FIG. 2B. Each
core point 62 is weighted with twice its distance from boundary 52 of a shape, i.e., the diameter of it's core element. That is, each core point 62 is weighted with twice the radius of an associated
core element 60. In the Euclidean metric, there is at most one core disk touching a boundary point except from concave vertices of the shape that can be touched by multiple core elements 60. In
contrast, in the L[∞] metric, there may be more than one core square touching a boundary point. See, for example, point p in FIG. 12. In particular, in the L[∞] metric, where core squares are
axis-parallel squares, boundary points along axis parallel boundary edges can be touched by more than one core square. The “radius of a square” is defined as half the size of its side. The
above-described weighting provides a mechanism to map dimensions (width or spacing) information of a Voronoi diagram (interior or exterior) to the boundary of a shape.
A “core element” 60 may also be called a “minimal open” where core point 62 is in the interior of a shape because any shrinking of core element 60 (i.e., of the wire formed by the shape) of ε>0 makes
it such that core element 60 stops overlapping any boundary point 64, thus ceasing to create an open circuit. ε is a positive number arbitrarily small, and to shrink ε means to move the edges of a
defect towards an interior of the defect by a distance ε. Similarly, referring to FIG. 2A, a “core element” 60 may also be called a “minimal short” where core point 62 is external of a shape, e.g.,
polygons 22A, 22B, because any shrinkage of core element 60 (i.e., spacing between wires) of ε>0 (same definition as above) makes it such that core element 60 stops overlapping any boundary point 64
of a shape 22A, 22B, thus ceasing creation of a short circuit.
A “definition set” is one or more definitions that allow for partitioning of an edge of a shape and assigning of at least one dimension (either as a constant value or a function) to each interval.
Other definitions will be provided, as necessary in the description that follows.
III. System Overview
With reference to the accompanying drawings, FIG. 6 shows a block diagram of a modeling system 100 in accordance with the invention. Modeling system 100 includes a memory 112, a processing unit (PU)
114, input/output devices (I/O) 116 and a bus 118. A database 120 may also be provided for storage of data relative to processing tasks. Memory 112 includes a program product 122 that, when executed
by PU 114, comprises various functional capabilities described in further detail below. Memory 112 (and database 120) may comprise any known type of data storage system and/or transmission media,
including magnetic media, optical media, random access memory (RAM), read only memory (ROM), a data object, etc. Moreover, memory 112 (and database 120) may reside at a single physical location
comprising one or more types of data storage, or be distributed across a plurality of physical systems. PU 114 may likewise comprise a single processing unit, or a plurality of processing units
distributed across one or more locations. I/O 116 may comprise any known type of input/output device including a network system, modem, keyboard, mouse, scanner, voice recognition system, CRT,
printer, disc drives, etc. Additional components, such as cache memory, communication systems, system software, etc., may also be incorporated into system 100.
As shown in FIG. 6, program product 122 may include an interval partitioner 123 having a core Voronoi diagram generator 124 and a partition generator 126, and a dimension assigner 128. At least one
definition set 130 may be provided for use by system 100, as will be described below. A design rule checker 132 may also be provided as part of system 100, or it may be configured as a separate
system. Other system components 134 include any other program product necessary to implement the invention and not otherwise described herein, e.g., user interfaces, communications, etc.
An IC design 140 is provided to modeling system 100 and is modeled according to the following methodology. It should be recognized that modeling system 100 may be provided as a stand-alone system or
as part of a larger IC design system as now known or later developed.
IV. Methodology
Referring to FIG. 7, a flow diagram of operation of modeling system 100 will now be described. The processing of FIG. 7 will be described, first, relative to an (arbitrary acute-polygon) shape 34 as
originally shown in FIGS. 3 and 4, and also shown in FIGS. 8A-8D. Shape 34 is an unlikely occurrence in a VLSI design, but facilitates a good understanding of the overall methodology. To further
facilitate a good understanding, the process description relative to shape 34 will be made relative to the Euclidean metric. Subsequently, the processing of FIG. 7 will be described relative to FIGS.
9A-9C, which mimic FIGS. 2A-2B, respectively, and represent more likely VLSI design possibilities. In this latter description, use of the L[∞] metric will be explained.
In either case, partitioning and assignment of dimensions to a boundary of a shape can be based on a definition set 130 (FIG. 6). A definition set for dimensions (width and spacing) may vary
depending on application. In addition, the definition set may require assignment of an actual figure or be a function dependent on some other dimension or characteristic. In view of the foregoing,
the definition sets discussed below are meant to be illustrative only.
As an overview, the methodology includes: step S1, generate a core Voronoi diagram; step S2, partition edges into intervals (width and/or spacing); step S3, assign dimensions; and step S4, conduct
design rule checking. For purposes of clarity, partitioning and assigning dimensions will be described together in a single section. That section includes sub-sections relating to the different
metrics, shape configuration and special situations.
1. Generate Core Voronoi Diagram
Referring to FIG. 8A, in a first step S1 (FIG. 7), a core 100, 102 for an interior and/or exterior Voronoi diagram 40 (i.e., medial axis) and 92, respectively, are generated for shapes of a given
layer A of an IC by core Voronoi diagram generator 124 (FIG. 6). Step S1 may be segmented into step S1A for generating a Voronoi diagram, and step S1B for determining a core of the Voronoi diagram.
Core Voronoi diagrams 100, 102 are generated as based on the Euclidean metric. Generation of Voronoi diagrams 40, 92 can be completed, for example, according to the algorithm described in U.S. Pat.
No. 6,317,859, previously incorporated by reference. Generation of a core 100, 102 can be accomplished according to a set of user-defined rules to select only interesting portions of the respective
Voronoi diagram. For example, as noted above, “uninteresting portions” (lighter lines) may be portions of Voronoi edges coupled to a boundary 52 of the shape, e.g., bisectors 54 between adjacent
edges. In the case of IC shapes with acute angles, “uninteresting” can be a portion of a bisector 54 coupled to an acute angle 56 at a certain distance from acute angle 56. Once again, what is
excluded from core 100, 102 can be flexible and application dependent.
FIGS. 2A and 2B also show a core 50 for an exterior and interior Voronoi diagram, respectively. However, these Voronoi diagrams are generated based on the L[∞] metric. In terms of FIG. 2B,
“uninteresting portions” (lighter lines) may be portions of Voronoi edges coupled to a boundary 52 of the shape, e.g., bisectors 54 between adjacent edges.
2. Partitioning into Intervals and Assigning Dimensions
A. Arbitrarily (Acute) Shaped Polygon with Euclidean Metric
Turning to FIGS. 8A-8D, in conjunction with FIG. 7, shape 34 of FIG. 3 is shown for description of the process. In addition, another shape 94 is provided for discussion of an exterior Voronoi diagram
For arbitrarily shaped polygons, an illustrative definition set may be:
Definition 1 (Arbitrary case): The width of a point p on the boundary of shape P is the size of the smallest core-disk touching p in the interior of shape P. If there is no core-disk touching p, then
p gets assigned the width of the nearest boundary point touched by a core-disk as the boundary of shape P is followed in a particular direction (clockwise or counterclockwise).
Definition 2 (Arbitrary case): The spacing of a point p on the boundary of shape P is the size of the smallest core-disk touching p in the exterior of shape P. If there is no core-disk touching p
then p gets assigned the spacing of the nearest boundary point touched by a core disk as the boundary of shape P is followed in a particular direction (clockwise or counterclockwise).
In a second step S2 (FIG. 7), each edge of each shape is partitioned into at least one interval by partition generator 126 (FIG. 6). In one embodiment, this step includes partitioning each edge into
width intervals using an interior core Voronoi diagram 100 and spacing intervals using an exterior core Voronoi diagram 102. Referring to FIG. 8B, partitioning into dimension intervals, in this case
in the form of width intervals, w1-w26, is shown.
In one embodiment, partitioning is conducted by projecting (dashed projections 104) (only some shown) for a vertex v of core 100 (hereinafter “core vertex”) to a corresponding boundary projection
point 106 (only some shown) of shape 34. In this case, partitioning rules can be applied to address specific situations. One example set of partitioning rules are as follows:
a) For each core vertex v of a core bisector bisecting edges e1 and e2, project core vertex v to both edges e1 and e2 such that each projection is perpendicular to the respective edge. Referring to
FIG. 8B, an example core vertex of a core bisector 100A bisecting edges 52A, 52B is core vertex v1. Projections 104A, 104B for core vertex v1 to both edges 52A, 52B are shown such that each
projection 104A, 104B is perpendicular to the respective edge 52A, 52B. The result is width intervals w15 and w16 result on edge 52A and intervals w4 and w6 result on edge 52B.
b) For each core vertex v of a core bisector bisecting an edge e1 (or edges) and a concave vertex of a shape, project the core vertex perpendicular to edge e1 (or edges). A “concave vertex” is a
vertex for which a line segment that connects points of edges forming the vertex is entirely outside of the shape, i.e., a vertex forming an angle greater than 180°.
Referring to FIG. 8B, an example core vertex v2 of a core bisector 100B that bisects an edge 52C and a concave vertex w7 of shape 34 is shown. (Note, concave vertices are labeled as intervals because
each concave vertex represents an interval, as will be described in greater detail below.) In this case, a projection 104C is made from core vertex v2 perpendicular to edge 52C and edge 52C is
partitioned accordingly, i.e., into width intervals w12, w13 based on projection 104C. Similarly, a core vertex v3 of a core bisector 100C that bisects an edge 52A and concave vertex w7 is shown.
Projection 104D is made from core vertex v3 perpendicular to edge 52A to form intervals w14 and w15. Assignment of a dimension, as will be described below, for width intervals w12 and w13 are based
on core vertex v2, and for width intervals w14 and w15 is based on the weighting of core vertex v3.
c) Each concave vertex is its own interval. In FIG. 8B, concave vertices are labeled with intervals w5, w7 and w23.
Returning to FIG. 8B, projections 104 define a list of intervals on boundary 52 of the shape. Based on the above projections, intervals w1-w26 result. It should be recognized that the projection
rules can be modified to suit an application, and the above rules are only illustrative.
In an alternative embodiment shown in FIGS. 8C-8D, the partitioning may be made by generating the core-disk 60 centered at each core vertex v, and determining point(s) where the core disk 60 touches
boundary 52 to form points 106. FIG. 8C shows smallest core disks 60 (Euclidean circles) for a variety of core points (not necessarily core vertices) of shape 34. In the Euclidean metric, there is at
most one core-disk 60 touching a boundary point 106 except from concave vertices. (In the L[∞] metric, there is more than one core-disk 60 touching a boundary point. Here, the invention may use the
smallest core-disk 60 such that the smallest possible (narrowest) width/spacing values are assigned. However, a user may choose to use the size of the largest core-disk instead if the local width of
concave vertices is to be ignored.) Referring to FIG. 8D, this embodiment provides the equivalent partitioning as projecting core vertices to boundary 52, as described above.
While the above description has been made relative to width intervals, it should be recognized that this process can be repeated for exterior core Voronoi 92 such that spacing intervals are derived.
That is, as shown in FIG. 8B, projections from core vertices of core Voronoi 92 are made to boundary 52 to derive spacing intervals, e.g., s1. Alternatively, as shown in FIG. 8D, core disks 60 can be
drawn at each core vertex, and the points at which the core disk touch boundary 52 can be used to derive spacing intervals, e.g., s1.
Next, in step S3, assigning dimensions to each interval is conducted by dimension assigner 128 (FIG. 6) according to a selected definition set 130 (FIG. 6). It should be recognized that while the
particular definition set provided above assigns actual dimensions, a definition set may assign dimensions as a function, e.g., width is twice the width of dimension X minus 10 nm. That is, a
definition set may include a set of rules for partitioning and then assign a function. For each point p along an interval, a core point that projects to p, referred to as c(p), defines a dimension.
For example, returning to FIG. 8B, a width for any point in, for example, intervals w9, w10, w111 is assigned a dimension equivalent to the weighting of a respective core point, i.e., vertex v4,
where the weighting equals the diameter of a core disk centered at v4. This is consistent with Definition 1 above. Turning to interval w6, a width assigned to a point p1 along that interval has a
variable width given by the diameter of a core disk centered at core point c(p1). This is also consistent with Definition 1 above. Also according to Definition 1, for a point P4 not touched by a core
disk, the dimension assigned is the width of the nearest boundary point touched by a core-disk as the boundary of shape 34 is followed in a particular direction (clockwise or counterclockwise). Note
that any point along intervals w9, w10, w11, the core point is core vertex v4.
With regard to concave vertices w5, w7 and w23, as noted above, each concave vertex is an interval itself. In this case, assignment of a dimension is based on a number of core points (or weights of
core points), i.e., exactly two core points for each of the two adjacent edges, and potentially a number of core points induced by Voronoi edges bisecting the concave vertex and other concave
vertices (if any). In one embodiment, the dimension assigned is based on the minimum among the weights of these core points. For example, as shown in FIG. 8E, two concave vertices A, B can create a
very thin neck in a shape. For a concave vertex A, determine the core point of minimum weight along the core bisectors induced by core vertex A. For example, in FIG. 8E, a core bisector 100D (between
arrows) is induced by core vertices A and B, and vertices A, B are assigned the weight of the depicted core disk 60.
Assignment of spacing dimensions occurs in a similar fashion. For example, as shown in FIG. 8B, for a point P2 on edge 52A, the spacing of the point would be the weight of a projected core point c(p2
), i.e., the largest core-disk touching point P2, i.e., twice the distance from core point c(p2) to point P2. If no core-disk touched a point, e.g. point P3, then the spacing is assigned the spacing
of the nearest boundary point touched by a core disk.
B. Orthogonally (VLSI) Shaped Polygon with L[∞] metric
A description of the methodology applied to an orthogonal shape using the L[∞] metric will now be made with reference to FIGS. 2A, 2B, 9B, 9C and 9D. The methodology for an orthogonally shaped
polygon such as that shown originally in FIGS. 2A-2B is identical to that for an arbitrary polygon. However, the definition set may be altered to accommodate the different shapes, as will be
described below. FIG. 9A shows with double-ended arrows illustrative important local widths and spacings for the group of shapes 22A-22F and 122 shown in FIGS. 2A and 2B, respectively. FIGS. 2A and
2B illustrate core Voronoi diagrams 50, and FIGS. 9B and 9C illustrate use of core Voronoi diagrams 50 to partition and assign dimensions, as will be described below.
As noted above, the following description will use the L[∞] metric rather than the Euclidean metric. The Voronoi diagram for the L[∞] metric is much easier to compute in practice than a Euclidean
one. Moreover, a vast majority of VLSI shapes include an axis parallel edge or edges of slope +/−1 (ortho-45 shapes). In this case, the L[∞] metric is very appropriate and simple to use. Where a
general IC design contains an edge in more orientations than ortho-45 shapes, an approximation to an ortho-45 shape can be easily made to compute the L[∞] metric Voronoi diagram.
Interval size can be determined based on the definition set that provides a “partitioning profile.” For a larger partitioning, in the case of an orthogonal polygon(s), as shown in FIGS. 9B-9C, a
definition set may be based on a “largest empty disk (square)” partitioning profile. Note that the following definitions are derived from Definition 1 and 2 by taking into account that a core square
touching an orthogonal edge in the L[∞] metric need not be unique as in the Euclidean case. A definition set is as follows:
Definition 3 (Orthogonal case): The width of any boundary point p along an edge e of a shape P is the size of the largest square touching p (along e) that is entirely contained in the interior of P.
Definition 4 (Orthogonal case): The spacing of any boundary point p along an edge e of a shape P is the size of the largest square touching p (along e) that lies entirely in the free space, i.e., it
does not intersect P or any other shape.
Concave corners present a special case similar to the Euclidean case of concave vertices. In particular, concave corners have a width/spacing just as any other point p on the incident edges, and also
have a corner width/spacing. Accordingly, the following definitions are presented to address concave corners:
Definition 5 (Orthogonal case): The corner width of a concave corner point p of a shape P is the size of the largest square cornered at p entirely contained in the interior of P. The width of concave
corner p is the smallest among the widths of the two edges incident top and the corner width of p.
Definition 6 (Orthogonal case): The corner spacing of a concave corner point p of a shape P is the size of the largest square cornered at p entirely contained in the free space. The spacing of
concave corner p is the smallest among the spacings of the two edges incident top and the corner spacing of p.
(It should be recognized that the definition set may be altered to form smaller intervals and smaller dimensions by using a smallest possible core square partitioning profile for each vertex to
determine interval partitioning and dimension assignment. This results in more intervals for each edge.) Returning to the methodology, in second step S2 (FIG. 7), partitioning of each edge of each
shape into dimension intervals is conducted by partition generator 126 (FIG. 6). In one embodiment, this step includes partitioning into width intervals using an interior core Voronoi diagram 50
(FIG. 2B) and partitioning into spacing intervals using an exterior core Voronoi diagram 50 (FIG. 2A).
Returning to FIG. 2A-2B, partitioning into dimension intervals for orthogonally shaped polygons includes evaluating or scanning a Voronoi cell of an edge, e.g., edge e, to identify Voronoi core
vertices v induced by that edge e. For example, referring to FIG. 2B, edge e of polygon 70 has five Voronoi vertices v1, v2, v3, v4, v5 adjacent thereto, and, as shown in FIG. 2A, edge f of polygon
22A has three Voronoi vertices v4, v5, v6 adjacent thereto. For each Voronoi vertex v encountered, as shown in FIGS. 2A, 2B, a core square 60 centered at the Voronoi vertex v is generated. Since the
Voronoi diagrams are generated using the L[∞] metric, core elements are square. The core square has a side length twice the weight of the Voronoi vertex, i.e., twice the weight of the core point
identical to the Voronoi vertex. Note that this step is equivalent to projecting core vertex v on the boundary of a shape along directions +1 and/or −1. Whether the direction +1 or −1 is used depends
on the slope of the boundary edge.
In one embodiment, derivation of intervals can proceed based on a desired partitioning profile. This is similar to the simplified process presented relative to arbitrary shapes. In this embodiment,
for example, partitioning can be conducted such that the intervals are as large as possible (coarser partitioning profile) or as small as possible (narrow/fine partitioning profile). In the former
case, a largest empty core square possible at each vertex is evaluated to determine interval partitioning. For example, referring to FIG. 9B, for edge f, three spacing intervals s1, s2, s3 are
generated because three largest squares 168, 170, 172 along edge f are possible (for core vertices v4, v5, v6 in FIG. 2A). Similarly, referring to FIG. 9C, for edge e, three width intervals w1, w2, w
3 are generated because three largest core squares 162, 164, 166 (for core vertices v1, v3 and v5 in FIG. 2B) along edge e of the shape are possible. In some cases, some core squares may overlap with
each other because a core square touches an axis parallel xboundary edge with its whole side and not only with its corner. See, for example, core square 177 and core square 164 in FIG. 9C. In one
embodiment, for any overlapping portions of a core square, only the portion of largest weighting is maintained, and portions of a smaller weighting are eliminated. Alternatively, other user-defined
rules may be implemented to address overlapping core squares. In FIG. 9C, core squares 177, 178 for vertices v2 and v4 do not provide the largest core squares at the vertices and, hence, they are
eliminated. The above-described process is repeated for both the interior (for width) face and exterior (for spacing) face of each edge.
Next, in step S3, assigning actual or function dimensions to each interval is conducted by dimension assigner 128 (FIG. 6). In one embodiment, step S3 includes implementation of a definition set 130
(FIG. 6) to assign dimensions to each interval. As noted above, according to the invention, a definition set for dimensions (width and spacing) may vary depending on application. In addition, the
definition set may require assignment of an actual figure or be a function dependent on some other dimension or characteristic. It should be recognized that the particular partitioning profile used
to partition each edge into intervals does not require implementation of a corresponding dimension assignment profile. For example, a user may partition an edge based on a largest empty core square
partitioning profile to attain larger intervals, and then assign dimensions based on a definition set that assigns the smallest possible dimension.
Referring to FIG. 9B, for edge f of polygon 22A, any point within the three spacing intervals s1, s2, s3 is each assigned the dimension of their respective three largest squares 168, 170, 172
according to Definition 4. That is, any point p within an interval is assigned the weight of its unique core point c(p), referred to as the core point that projects to p. For any p that is part of an
axis parallel edge, c(p) is the center of the largest core square touching p. Note that the dimensions of squares 168 and 172 are limited by the horizontal dimension, while square 170 is limited by
the vertical dimension. Referring to FIG. 9C, for edge e, any point p within the three width intervals w1, w2, w3 are each assigned the width of the respective largest core square, i.e., core point
that projects to p, according to Definition 3. Note that the dimensions of squares 162 and 166 are limited by the horizontal dimension, while square 164 is limited by the vertical dimension.
With regard to a corner width, as shown in FIG. 9C, the corner width of a point p is the size of largest square that is cornered at p and entirely contained in the interior of polygon according to
Definition 5. In addition, the width of corner p is the smallest among the widths of the two edges g, h incident to p and the corner width of p. With regard to corner spacing, as shown in FIG. 9B,
the corner spacing of a corner point p of polygon 22A is the size of the largest square 178 cornered at p entirely contained in the free space according to Definition 6. In addition, the spacing of
corner p is the smallest among the spacings of the two edges f, i incident to p, i.e., square 172 or 176, and the corner spacing of p, i.e., the size of square 178 at the corner of p. Note that the
largest empty square 178 cornered at a corner p gives the L[∞] nearest neighbor to corner p that may not be the same as the Euclidean nearest neighbor. The difference, however, between the two is
small and negligible in practice. Once the L[∞] nearest neighbor to p is identified, the numeric distance can be reported in the ordinary (Euclidean) manner, e.g., 160A instead of 160B. Another
example of corner width determination is shown in FIG. 9D. In this case, an orthogonal shape includes three core squares 360A-C (dashed boxes) at a concave corner p. In this case, the core square 360
B would be used to define the width of corner p.
It should also be recognized that each interval may be assigned a function that determines its dimension(s) rather than a numerical value based on its associated core square. For example, an interval
s1 in FIG. 9B may be given a function that assigns its width as 2 times the size of the width of polygon 22B minus 10 nm.
1) Handling of Ortho-45° Edges
Referring to FIGS. 10A-10D, various arrangements of widths for shapes containing 45° edges are illustrated. The methodology is the same as described above. Portions of a medial axis 250, 252 not
included in the core are shown with dashed lines 202.
FIG. 10A illustrates core squares 260A-D for core 250 corresponding to a 45° line. In this case, partitioning is based on the largest core squares 260A and 260D along a boundary edge, which results
in two intervals.
As shown in FIG. 10B, for intervals corresponding to two parallel 45° lines, the invention may use the diameter of a core square 200. In the case of two parallel 45° lines, the Euclidean and L[∞]
metric reach their greatest difference in values. To simulate Euclidean distance, the size of a core square, in this case, can be defined to be the length of the diagonal of the core-square. Note
that the length of the diagonal equals the Euclidean distance between the two parallel 45° lines. In all remaining cases, size is determined by the length of the side of the corresponding core
square. FIGS. 10B and 10C also illustrate the advantages of being able to select a core, i.e., based on what is interesting and uninteresting. In FIG. 10B, a bisector 202 of adjacent edges j, k is
excluded from core 250. Thus, the same width w1 may be assigned to the whole 45° edge k, even in the portion close to the acute angle. In FIG. 10C, a portion 204 of bisector 202 incident to the acute
angle is included in core 252, and thus three intervals w1, w2 and w3 are generated, where w3 is assigned the constant value of the small vertical arrow 206. Note also that once intervals w2 and w3
are obtained, any reasonable numeric value approximating the Euclidean metric may be assigned, if desirable.
FIG. 10D illustrates the case of two perpendicular 45° edges forming a 90° angle, which correspond to a rotated rectangle. This case is handled similarly to that of FIG. 10A.
2) Arbitrarily (Acute) Shaped Polygon with L[∞] Metric
Referring to FIG. 11, the arbitrary shape 34 of FIG. 4 is shown including partitioning according to the L[∞] metric. Where the L[∞] metric is applied to non-orthogonal shapes, a definition set may be
as follows:
Definition 7 (Arbitrary case): The width of any boundary point p along a boundary of shape P is the size of the largest core square touching p in the interior of shape P. If there is no core-square
touching p, then p gets assigned the width of the nearest boundary point touched by a core square.
Definition 8 (Arbitrary case): The spacing of any boundary point p along a boundary of shape P is the size of the largest core square touching p in the exterior of shape P. If there is no core-square
touching p, then p gets assigned the spacing of the nearest boundary point touched by a core square.
As noted above, special care must be taken for concave corners. In particular, concave corners represent an interval by themselves. That is, they have a width/spacing just as any other point p in an
edge, and also have a corner width/spacing. Accordingly, the following definitions are presented to address corners:
Definition 9 (Arbitrary case): The corner width of a concave corner point p of a shape P is the size of the smallest among all core square cornered at p.
Referring to FIG. 12, there may be at most 3 core squares cornered at a concave corner p depending on the slopes of the incident edges. In the orthogonal case, there are exactly 3 core squares
cornered at p.
Definition 10 (Arbitrary case): The corner spacing of a concave corner point p of a shape P is the size of the smallest core square cornered at p entirely contained in the free space.
3) Improved Accuracy: Combination L[∞] metric and Euclidean Metric:
Because shapes are in their majority axis parallel and because the L[∞] Voronoi diagram is easier to compute, the invention preferably uses the L[∞] metric. In an alternative embodiment, if more
accuracy is desired, an approximation to the Euclidean metric can be implemented by, as shown in FIG. 13:
First, partitioning the boundary of a shape into intervals according to the L[∞] metric, i.e., the definition based on the largest square centered along the L[∞] core as given in Definitions 5-6.
Second, the Euclidean metric can be used to assign dimension. This procedure can be completed as follows: Consider a core square D centered at core point c(p). Points p1 and p2 are neighbors of point
p, and vice versa. Similarly edges e1 and e2 are neighbors of point p. Instead of assigning the side of core square D as the width of p, Euclidean based metrics may be used. For example, the size of
a circle defined by point p, edge e1 and edge e2 may be used. Alternatively, the size of a circle centered at core point c(p), or the distance between point p and edge e2 (shown by double ended
arrow), or the distance between point p and edge e1 (shown by double ended arrow) may be used.
As another example, as shown in FIG. 13, points p3 and p4 are neighbors. Instead of assigning the side of core square D′ as the width of p3 and p4, the Euclidean based distance between point p4 and
edge e1, or the size of a circle centered at core point c(p4), or the diagonal of core square D′, etc., may be used.
4) Fine and Narrow Partitioning
In certain applications the focus of width and spacing is to evaluate a shape and the spacing around a shape as “narrow” and as “thin” as possible. In particular, in the area of VLSI yield and
critical area calculations, the notion of shorts and opens is widely used. As noted above in the definition section, a “spot defect” causes a short or an open where a “short” is a defect creating a
bridge with some other edge in the exterior of a shape and an “open” is a defect creating a bridge in the interior of a shape. In addition, a defect may be modeled as a core element in the form of a
circle or a square.
In order to implement the invention to partition and assign dimensions in terms of a “narrow” or “thin” evaluation, the following definition set (for L[∞] and Euclidean metric) may be used:
Definition 11: The width of any boundary point p of shape P is the radius of the smallest core element (defect) centered at p causing an open for shape P. The spacing of the boundary point p is the
radius of the smallest core element (defect) centered atp that causes a short.
Definition 12: An open is an element totally covering a core element that is centered on a core point in the interior of a shape P. A short is an element totally covering a core element that is
centered on a core point in the exterior of a shape P.
Returning to the methodology and referring to FIGS. 14A-14E, partitioning of an edge e into width/spacing intervals based on Definition 11 will now be described. FIG. 14A shows an interior Voronoi
cell 124A (also denoted reg(e) and shaded) of shape 122 (from FIGS. 2B and 9C). FIG. 14B shows a second (2^nd) order Voronoi diagram 390 (dashed lines) of shape 122 within Voronoi cell 124A (also
denoted reg(e)) that is the Voronoi diagram of shape 122 truncated within reg(e). The 2nd order Voronoi diagram within Voronoi cell 124A is generated by extending in Voronoi cell 124A all bisectors
54 incident to Voronoi cell 124A except those incident to endpoints 62E of a core 50, i.e., except those induced by an edge adjacent to e. The 2nd order Voronoi diagram for opens within cell 124A is
the 2nd order Voronoi diagram of shape 122 within the cell excluding in addition to edge e the two edges of shape 122 neighboring edge e. For simplicity, the 2nd order Voronoi diagram 390 for opens
may be referred to as the “2nd order Voronoi diagram” or the “2nd order Voronoi partitioning within reg(e).”
The 2nd order Voronoi partitioning within reg(e) partitions edge e into intervals w1, w2, w3. Each interval has an “owner,” i.e., an edge or concave corner or a core Voronoi vertex that induces that
interval. In FIG. 14C, intervals and their owners are indicated by double arrows. The width of any point p along an interval w is simply the distance of p from the owner of the interval. In case the
owner of an interval is a core point as shown in FIG. 14E, the weight of the core point is added to the derivation of the width, i.e., the width is the size of the core disk centered at that core
FIG. 14D shows the 2nd order Voronoi partitioning of each Voronoi cell of shape 122 as well as the width intervals. FIG. 14F shows the actual width values obtained for the 3 intervals of edge e.
Note, the top and bottom interval w1, w3 have constant width while the middle interval w2 has a variable width equal to the distance of any point p along that interval from the horizontal edge e′.
It should be recognized that although the L[∞] metric has been used in FIGS. 14A-14F, the Euclidean metric may be used when reporting distances between a point along an interval and its owner.
In the orthogonal case under the L[∞] metric only the width/spacing definition of this section is equivalent to the one in the previous section Definition 9-10, if we always choose the smallest
core-square touching a point p to derive its width.
3. Design Rule Checking
Returning to FIG. 7, once intervals and associated dimension(s), i.e., width/spacing, have been assigned, in step S4, data can be used to conduct any design rule checking involving width and spacing
using conventional technology. For example, width and/or spacing dimensions for any single edge may be compared to check rules involving the width and/or spacing dimension of the single edge.
Similarly, width and/or spacing dimensions for pairs of neighboring shapes and/or pairs of neighboring edges may be compared to check rules involving pairs of neighboring shapes and/or edges.
It should also be recognized that the above invention is not limited to application within a layer A of an IC. In an alternative embodiment, for check rules involving shapes in more than one layer of
an IC, e.g., a layer A and a layer B, a Voronoi diagram for the combined layers can be used. For example, for each edge of layer A, spacing intervals can be derived with edges in layer B using the
combined Voronoi diagram. The spacing intervals can then be used to compare to check rules involving layer A and layer B.
V. Conclusion
In the previous discussion, it will be understood that the method steps discussed are performed by a processor, such as PU 114 of system 100, executing instructions of program product 122 stored in
memory. It is understood that the various devices, modules, mechanisms and systems described herein may be realized in hardware, software, or a combination of hardware and software, and may be
compartmentalized other than as shown. They may be implemented by any type of computer system or other apparatus adapted for carrying out the methods described herein. A typical combination of
hardware and software could be a general-purpose computer system with a computer program that, when loaded and executed, controls the computer system such that it carries out the methods described
herein. Alternatively, a specific use computer, containing specialized hardware for carrying out one or more of the functional tasks of the invention could be utilized. The present invention can also
be embedded in a computer program product, which comprises all the features enabling the implementation of the methods and functions described herein, and which—when loaded in a computer system—is
able to carry out these methods and functions. Computer program, software program, program, program product, or software, in the present context mean any expression, in any language, code or
notation, of a set of instructions intended to cause a system having an information processing capability to perform a particular function either directly or after the following: (a) conversion to
another language, code or notation; and/or (b) reproduction in a different material form.
While this invention has been described in conjunction with the specific embodiments outlined above, it is evident that many alternatives, modifications and variations will be apparent to those
skilled in the art. Accordingly, the embodiments of the invention as set forth above are intended to be illustrative, not limiting. Various changes may be made without departing from the spirit and
scope of the invention as defined in the following claims. | {"url":"http://www.google.com/patents/US7577927?ie=ISO-8859-1&dq=6373188","timestamp":"2014-04-18T17:32:10Z","content_type":null,"content_length":"129817","record_id":"<urn:uuid:85415dc1-8a12-448d-be7e-52c41903055b>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00337-ip-10-147-4-33.ec2.internal.warc.gz"} |
II. ITEMS BY OTHER AUTHORS
B. PAPERS BY OTHER AUTHORS: I-M
(alphabetical by author)
118. Jones, C.V., "Attributed Graphs, Graph-Grammars, and Structured Modeling," Annals of Operations Research, 38 (1992), pp. 281-324. (Volume on Model Management in Operations Research edited by B.
Shetty, H. Bhargava, and R. Krishnan.)
Chris originated the use of graph grammars as a modeling framework in his dissertation. This paper applies his work on Graph-Based Modeling Systems (GBMS) to SM. In particular, he explains how a
syntax-directed editing environment for SM can be implemented as a GBMS. He has produced a prototype implementation called Networks/SM. This is a step toward the ultimate aim of producing a truly
graph-based SM environment for workstations.
119. Kang, M., G. Wright, R. Chandrasekharan, R. Mookerjee and D. Worobetz, "The Design and Implementation of OR/SM: A Prototype Integrated Modeling Environment," Annals of Operations Research, 72
(1997), pp. 211-240. (Volume on Information Systems and OR, part II, edited by R. Ramesh and H.R. Rao.)
This paper focuses on the computational implementation of the OR/SM (ORACLE/Structured Modeling) system using ORACLE Tools and Database. The functionality of OR/SM is described in the companion
paper by G. Wright, N.D. Worobetz, M. Kang, R. Mookerjee and R. Chandrasekharan listed elsewhere in this Bibliography. Here the architecture is described in some detail, both internal and
interfaces with other systems. More than 50,000 lines of C give this system its rich functionality.
120. Kemp, J., "Users, Needs, and Conceptual Models: the Next Step for Mathematical Programming," Working Paper 10, Dept. of Mathematics and Statistics, Brunel University, UK, 16 pages, 5 April,
This essay on research needs in mathematical programming comments on some of the ideas in four of my papers. There is also a companion paper, "Conceptual Modelling: The Necessary Alternative to
Algebra for MP Practitioners."
121. Kendrick, D., "A Production Model Construction Program: Math Programming to Structured Modeling," in L.F. Pau, J. Motiwalla, Y.H. Pao, and H.H. Teh (eds.), Expert Systems in Economics, Banking
and Management, Elsevier Science Publishers B.V. (North-Holland), 1989, pp. 351-360. See also the companion article by D. Kendrick, "A Production Model Construction System: PM Statement to Math
Programming," J. Economic Dynamics and Control, 14:2 (May 1990), pp. 219-236.
This paper derives from Krishnan's dissertation (listed in Part IIA of this Bibliography). This excerpt from the cited companion article makes the role of both papers clear: "Krishnan's system
consists of (1) a user interface complete with its own modeling language called PM, (2) a model construction program which transforms the PM representation of the model first to mathematical
programming form and then to predicates which are elements of Geoffrion's (1987) Structured Modeling form, and (3) a back end which transforms the Structured Modeling predicates into the
particular format required by Structured Modeling. This system makes heavy use of the concepts and terminology of knowledge-based systems and logic programming... Since many economists and
management scientists have considerable experience with production and distribution modeling but limited familiarity as yet with knowledge-based systems and logic programming, a series of papers
have been prepared to elucidate the elements of Krishnan's system to this community. The first of these papers, Krishnan, Kendrick, and Lee (1988) [listed in this Bibliography] describes the user
interface and the PM language. The current paper [the cited companion article] ... covers the part of the model construction system from the PM representation to the mathematical programming
form. The third paper [the current bibliographic item] ... covers the part from the mathematical programming form to the Structured Modeling form."
122. Kendrick, D., "Parallel Model Representations," Expert Systems with Applications, 1:4 (1990), pp. 383-389.
My 1987 Management Science article argued for a single model representation suitable for managerial communication, mathematical use, and direct computer execution. This article takes the
apparently opposite view that multiple parallel representations should be maintained: graphical, knowledge-base, mathematical, and modeling language. These views are not really opposed, for I had
in mind representations that were produced and kept in synch manually, while David has in mind that each representation would be derivable automatically from the others. I certainly agree that
multiple views of a model are desirable if the modeling environment can ensure consistency and obviate the need for the special skills needed to perform translation. However, there is a possibly
serious difficulty: not all representational formalisms are equally expressive, and so translation may not be invertible. For example, ordinary mathematics is not as expressive in some respects
as mathematically extended semantic data models like SML.
123. Kendrick, D. and R. Krishnan, "A Comparison of Structured Modeling and GAMS," Computer Science in Economics and Management, 2:1 (1989), pp. 17-36. Presented at the Conference on Integrated
Modeling Systems, held at the University of Texas at Austin in October 1986.
The first author is an expert in GAMS, and the second author is versed in SM. The simple transportation example is used to compare SM and GAMS. See also item 26 of this Bibliography.
123A. Kim, J.W., H.D. Kim and S.J. Park, "Multi-facetted Approach to Mathematical Model Representation and Management," Journal of the Korean Operations Research and Management Science Society, 23:2
(1998) [in Korean].
The paper's English abstract says, in part: "In this paper, a multi-facetted modeling approach is proposed as a basis for the development of an integrated modeling environment which provides for
(1) independent management of modeling knowledge from individual models; (2) an object-oriented conceptual blackboard concept; (3) multi-facetted modeling; and (4) declarative representation of
mathematical knowledge".
124. Kim, H.D. and S.J. Park, "Model Formulation Support in an Object-Oriented Structured Modeling Environment," Working Paper, Dept. of Management Science, KAIST, Taejon, Korea, 37 pages, 2/24/94.
The focus of this paper is on partially automating the process of formulating an SML schema. An object-oriented extension of the SM framework is proposed that relies on imposing certain
constraints. Support for the model formulation process focuses mainly on the key roles of attribute elements and generic rules. The approach taken lends itself to a rapid prototyping method based
on the constraint-based metaview approach explained in a Park-Kim paper mentioned later in this Bibliography. An illustration is given based on a corporate strategic planning application at a
large Korean steel company.
125. Kottemann, J.E., "Applying the Jackson System Development and the Geoffrion Structured Modeling to Systems Description," Final Report, College of Business Administration, University of Hawaii,
Honolulu, 9/86.
This work was done under Army contract in the context of combat simulation (cf. Dolk's 11/86 report and Stott's paper, both listed in this Bibliography). It presents an overview of JSD and SM,
points out some of the shortcomings of each, and outlines a way to marry the two approaches. Of particular interest is the argument that SM properly subsumes the entity-attribute-set formalism of
SIMSCRIPT (although the treatment is not rigorous).
The one shortcoming pointed out for SM was that it appears to be designed for "static" rather than dynamic models, and hence does not appear to directly support dynamic simulation. (Various
authors have since examined SM for simulation carefully and fruitfully.)
126. Kottemann, J.E. and D.R. Dolk, "Model Integration and Modeling Languages: A Process Perspective," Information Systems Research, 3:1 (March 1992), pp. 1-16. An earlier version appeared as
"Process-Oriented Model Integration," Proceedings of the Twenty-First Hawaii International Conference on System Sciences, Vol. III, IEEE Computer Society Press, Los Alamitos, CA, pp. 396-402, 1/88.
My work has assumed that languages for model manipulation will be natural extensions of model definition languages like SML, presumably with some simple procedural capability added so that
solvers can be put through their paces more than one step at a time. This article indicates one possible approach to the design of such extensions. It also indicates that such extensions can
accomplish some of the purposes of model integration via model interconnection.
The authors focus on the situation where multiple models (possibly from different modeling paradigms, and possibly written in SML) are used computationally in a coordinated way without having to
integrate the individual model representations. They sketch a model manipulation language for such situations based on the idea of communicating sequential processes from computer science. The
approach has a strongly procedural, rather than declarative, character.
127. Krishnan, R., "Automated Model Construction: A Logic Based Approach," Annals of Operations Research, 21 (1989), pp. 195-226. (Special volume on Linkages with Artificial Intelligence edited by F.
Glover and H. Greenberg.)
Based on Krishnan's Ph.D. thesis. Compared to the Krishnan, Kendrick, and Lee article (see below), this paper has more detail on the PM language, problem-specific inferencing, and model
Krishnan has two other papers in a similar vein. (1) "PDM: A Knowledge-Based Tool for Model Construction," Proceedings of the Twenty-Second Hawaii International Conference on System Sciences,
IEEE Computer Society Press, Los Alamitos, CA, 1/89 and, in somewhat longer form, in Decision Support Systems, 7 (1991), pp. 301-314; and (2) "PM*: A Logic Modeling Language for Model
Construction," which presents an extension of his dissertation research, in Decision Support Systems, 6 (1990), pp. 123-152.
128. Krishnan, R., D.A. Kendrick and R.M. Lee, "A Knowledge-Based System for Production and Distribution Economics," Computer Science in Economics and Management, 1:1 (1988), pp. 53-72.
This article gives an overview of Krishnan's dissertation. His system, called PM and implemented in Prolog, has three main parts. Its front end is a dialog based system that uses axiomatic
knowledge about production-distribution models to guide the user toward a correct specification of the model in PM's internal first-order logic based language. PM's model construction program
uses a two step rule-based procedure to produce an "equational form" of the model from the PM specification. This form, which consists of four kinds of predicates, is transformed syntactically by
PM's back end into SML as it was some years ago. Fragments of the Mexican steel industry model, which happens to be the subject of one of my Informal Notes, is used illustratively in connection
with each of the three parts.
PM can be viewed as a domain-specific system for building structured models. At least three advantages are claimed: (1) a powerful approach for posing queries concerning model structure, with
Prolog as the query processor; (2) greatly facilitated model modification, thanks to explicitly represented knowledge of syntactic interdependencies among elements of the PM language; and (3) the
potential to enable model building by users without technical training in either mathematical programming or modeling languages.
129. LeClaire, B., and E. Suh, "Object-Oriented Model Representation Scheme," College of Business Administration, Oklahoma State University, 25 pages, 1989.
This paper sketches the model representation basics of "object-oriented structured modeling" from a semantic data modeling perspective.
A closely related, subsequent paper is E. Suh, C. Suh and B. LeClaire, "Object-Oriented (O-O) Structured Modeling for an O-O DSS," Technical Report IE-TR-89-08, Dept. of Industrial Engineering,
POSTECH, P.O. Box 125, Pohang, 790-600, Korea, 24 pages, 11/9/89. (See also C. Suh's M.S. Thesis in Part IIA of this Bibliography.)
130. LeClaire, B., R. Sharda and E. Suh, "An Object-Oriented Architecture for Decision Support Systems," Proceedings of the First ISDSS Conference, Austin, TX, September 1990, pp. 567-586. Also
available as a working paper, 31 pages, 2/91; authors respectively at University of Wisconsin-Milwaukee, Oklahoma State Univ., and POSTECH (Korea).
This paper supersedes the previously listed LeClaire-Suh and Suh-Suh-LeClaire papers. It moves from model representation to the broader issues of DSS architecture, still from an O-O perspective.
Emphasis is given to symmetry of treatment between data and models, and a diagrammatic technique based on Chen's entity-relationship diagrams is proposed in connection with so-called
"Object-Oriented Structured Modeling" for model representation. Five classes are proposed for an O-O DSS: metamodels, entities, relationships, models, and relations. A prototype was implemented
in Smalltalk/V 286, and some work has been done to convert to C++.
131. Lenard, M., "Representing Models as Data," Journal of Management Information Systems, 2:4 (1986), pp. 36-48.
This article gives a way to represent structured models in relational database form (cf. Dolk's "Model Management and Structured Modeling" Comm. ACM article). This article also points out the
surprising compatibility between SM and Garth McCormick's theory of factorable functions.
132. Lenard, M., "Fundamentals of Structured Modeling," in G. Mitra (ed.), Mathematical Models for Decision Support, Springer-Verlag, Berlin, 1988. Presented at a NATO Advanced Study Institute in Val
d'Isere, France, 1987.
A tutorial based on my 1987 Mgt. Sci. article, with editorial comment added.
133. Lenard, M., "Structured Model Management," in G. Mitra (ed.), Mathematical Models for Decision Support, Springer-Verlag, Berlin, 1988. Presented at a NATO Advanced Study Institute in Val
d'Isere, France, 1987.
A sequel to an earlier version of Melanie's "An Object-Oriented Approach to Model Management" article listed below in this Bibliography. It includes a progress report on an implementation in
134. Lenard, M., "An Implementation of Structured Modeling Using a Relational Database Management System," presented at the Second Annual Conference on Integrated Modeling Systems, University of
Texas, Austin, 20 pages, October 1987.
This KnowledgeMan prototype targets very simple linear programs. Schemas are specified by filling in several standard relations, rather than by writing a text-based language. Functionality is
135. Lenard, M., "Mathematical Modeling Systems in a Database Environment," Proposal to NSF (DRMS), 51 page Project Description, 9/88.
This proposal, which was funded, focuses mainly on mathematical programming models. SM and the object-oriented paradigm provide the foundation. The proposal states that SM will be extended "so
that it includes a means for expressing the operational (or dynamic) aspects of modeling as well as the representational (or static) aspects. This extension will make SM more compatible with the
object-oriented paradigm by incorporating a key feature of 'Objects', namely that they are a combination of data and procedures." A prototype modeling system will be designed and built based on
the extended framework, and on an object-oriented database language. The model itself -- schema as well as elemental detail -- will be stored in ONTOS, an object-oriented DBMS.
136. Lenard, M., "Extending the Structured Modeling Framework for Discrete-Event Simulation," Proceedings of the Twenty-Fifth Annual Hawaii International Conference on System Sciences, Vol. III, IEEE
Computer Society Press, Los Alamitos, CA, pp. 494-503, 1/92.
This paper sketches 3 proposed SM extensions (new kinds of elements) designed to make it more useful for discrete-event simulation modeling: random attributes (whose values are generated by
probability distributions), actions (which describe state transitions), and transactions (used to describe complex events in terms of a sequence of previously defined actions and transactions).
137. Lenard, M., "An Object-Oriented Approach to Model Management," Decision Support Systems, 9:1 (January 1993), pp. 67-73. An earlier version appeared in Proceedings of the Twentieth Hawaii
International Conference on System Sciences, Vol. I, IEEE Computer Society Press, Los Alamitos, CA, pp. 509-515, 1/87. Presented at the Workshop on Structured Modeling, UCLA, 8/86.
This article was the first to point out that SM has a lot in common with the popular object-oriented programming paradigm. This observation is the basis for a design approach to object-oriented
model management systems. With the addition of ideas from Melanie's earlier article "Representing Models as Data", this leads to the possibility of using a relational database system for O-O
model management (cf. Dolk's "Model Management and Structured Modeling" Comm. ACM article).
138. Lenard, M.L., J.E. Burns, M.V. Kadyan, D.Z. Levy, M.J. Schement, and J.M. Wagner, "Structured Model Management System for Operational Planning: Installation and Users Guide, Prototype Version
1.0," prepared for U.S. Coast Guard Research & Development Center by Crystal Decision Systems, Brookline, MA, 120 pages, 7/1/92. Revised 3/31/93 (Version 2.0) and 10/6/93 (Version 3.0); now 207
This report, prepared under U.S. Coast Guard contract, delivers on the structured modeling extensions proposed by M. Lenard in "Extending the Structured Modeling Framework for Discrete-Event
Simulation" (listed above in this Bibliography). The system addresses discrete-event simulation models, is built on top of a relational DBMS (ORACLE 6.0), and runs under MS-DOS. It makes
extensive use of the ORACLE tools SQL*Forms and SQL*Menu; in particular, most user interaction is though forms selected from a set of pop-down menus. Among the system's features is the ability to
convert extended structured models to SIMSCRIPT II.5 code. The code generation is done entirely in SQL*Plus (ORACLE's version of SQL).
Two prior reports specified the system documented here: (1) Lenard, M., "A Structured Model Management System, Phase I: Concept Study," 31 pages, 12/90; (2) "Extended Structured Modeling Tables
and their Mapping to SIMSCRIPT," 23 pages, 11/91.
139. Lenard, M., "A Prototype Implementation of a Model Management System for Discrete-Event Simulation Models," Proceedings of the 1993 Winter Simulation Conference, IEEE, Piscataway, NJ, 560-568,
This paper describes a database-based prototype implementation of the ideas in "Extending the Structured Modeling Framework for Discrete-Event Simulation" (see above). The database schema is not
model-specific, as in SML, but rather is fixed for all models. Further documentation is provided in reports co-authored with the other implementors.
140. Liang, T., "On Structured Modeling," BEBR Faculty Working Paper No. 1385, College of Commerce and Business Administration, University of Illinois at Urbana-Champaign, 15 pages, 8/87.
"Decomposition" and "integration" can be viewed as opposite poles of a single modeling axis. The first breaks a whole down into its parts and the second puts parts together into larger
structures. This paper comments on my 1987 Management Science article from this perspective, and argues two main points. First, elemental structure may be too extreme a level of decomposition
from the point of view of human cognitive limitations. Second, even higher levels of abstraction than those provided by SM are necessary to support automatic integration.
I could appreciate these points only after reading a prior paper by Liang and Chris Jones. My response to the first point: (a) nitty gritty detail cannot be avoided, (b) a black box model
description is insufficient for many purposes, and (c) generic and modular structure help the modeler keep cognitive control even of a complex model. Concerning the second point: (d) modular
structure already provides an appropriately high level of abstraction, and (e) the aim of automatic integration may be unrealistic. See item 8 of this Bibliography for in-depth material germane
to both points.
141. Liang, T., "Analogical Reasoning and Case-Based Learning in Model Management Systems," Decision Support Systems, 10:2 (September 1993), pp. 137-160. See also "Modeling By Analogy: A Case-Based
Approach to Automated Formulation of Linear Programs," Proceedings of the Twenty-Fourth Annual Hawaii International Conference on System Sciences, Vol. III, IEEE Computer Society Press, Los Alamitos,
CA, pp. 276-283, 1/91. See also "Analogical Formulation of Linear Programs," Working Paper, Krannert School, Purdue University, 41 pages, 9/92 (due for revision).
This highly original work is on a topic that, although almost totally neglected to date by the modeling community, has considerable potential. It develops an analogical approach to new model
formulation (or model revision) based on similarities to existing models. A model representation framework is used that, as the author notes, derives from SM ideas.
142. Liang, T., "SAM: A Language for Analogical Reasoning in Structured Modeling," Krannert School of Management, Purdue University, 28 pages, 5/91. Revision in progress. Related work was presented
at the "New Directions in Structured Modeling" session at the INFORMS International Singapore, June 25-28, 1995.
This paper retargets the author's prior work on analogical modeling directly to SM.
143. Lin, S.E., D. Schuff and R.D. St. Louis, "Subscript-Free Modeling Languages: A Tool for Facilitating the Formulation and Use of Models," to appear in European Journal of Operational Research,
1999. Earlier version presented at the ORSA/TIMS Meeting in Phoenix, 10/93.
This paper is based on Lin's dissertation (see Part IIA of this Bibliography).
143A. Loh, Seow Yick and Gee Kin Yeo, “A Framework for Implementing Structured Modeling,” Technical Report TRB4/97, Dept. of Information Systems & Computer Science, The National University of
Singapore, 18 pages, 4/97. A zipped 54 KB Word 6.0 file is available by ftp here.
This paper presents a general, non-technical discussion of the requirements for a useful implementation of structured modeling.
143B. Loh, Seow Yick and Gee Kin Yeo, “VMS/SM: Implementing the Modeling Process,” Proceedings of IASTED International Conference on Modelling, Simulation and Optimization, Singapore, pp. 66-69, 8/
97. A 179 KB zipped Word 7.0 document is available by ftp here.
This paper explains how various phases of the modeling process are supported and implemented in VMS/SM (see items 62A and 184A of this Bibliography): model construction, model solution (one
highly specialized solver is integrated), solution interpretation, model analysis, and model documentation.
144. Lustosa, L.J., "FW/SM-ANALYZE Integration: Research Proposal for a Prototype Computer-Based Environment for Model Schema and Instance Analyses," Technical Memorandum 09/93, Departmento de
Engenharia Industrial, Pontificia Universidade Católica do Rio de Janeiro, Brazil, 9 pages, revised 1/94.
This project aims to integrate FW/SM with Harvey Greenberg's ANALYZE, a package that focuses almost entirely on LP model instances. The first phase seeks only a "loose" integration. ANALYZE will
be invoked from within FW/SM, and will reformat the information contained in FW/SM-generated files so that most of FW/SM's functions will be immediately available. Therefore, users will always be
in one of the two unmodified environments. In the second phase, an "external" integration will be attempted so that the user perceives a single coherent environment. Possible problems, and some
interesting possibilities like using SML's interpretation sublanguage to generate the ANALYZE "syntax file", are briefly discussed. [Update 9/95: The first phase was finished, but the second
phase was postponed.]
This work is partially supported by Brazil's Conselho Nacional de Desenvolvimento Cientifico e Tecnologico.
144A. Ma, J., Q. Tian and D. Zhou, "An Object-Oriented Approach to Structured Modeling," IRMA '98 Proceedings (1998 Information Resources Management Association International Conference, Boston,
17-20 May 1998), Idea Group Publishing, pp. 406-413. On-line here. See also the following companion paper: Q. Tian, J. Ma, D. Zhou, L. Huang, "Integration of Dynamic Models in Decision Support
Systems," in B. Verma, Z. Liu, A. Sattar, R. Zurawski and J. You (eds.), Proceedings of the Second IEEE International Conference on Intelligent Processing Systems, Gold Coast, Australia, 4-7 August
1998, pp. 281-284. Sponsored by IEEE Industrial Electronics Society. On-line here.
This paper points out that SM and object-oriented modeling are complementary fields, each strong in areas of the other's weakness. In particular, SM could benefit by adapting dynamic models from
object-oriented modeling. The authors propose adapting an "action logic" (first-order and many-sorted) from artificial intelligence to capture system dynamics in conjunction with conventional SM
(which most naturally captures system statics). Their development and examples are in logical notation. It remains to be seen how well this "extension" of SM can be integrated with existing SM
theory, languages, and implementations.
The companion paper focuses more specifically on model integration. It is not clear from the example whether the authors have a general model integration procedure.
145. Magel, K. and L. Shapiro, "JADE: An Object Oriented, Heterogeneous Model Management System," Computer Science Dept., North Dakota State University, 14 pages, 7/14/87.
The aim of this work is to create a model management environment in which users can have access to multiple modeling tools and languages (GAMS, IFPS, SML, etc.). The main integrating device is a
second, canonical representation for each model that combines SM concepts with some ideas from knowledge representation. Encapsulation and peaceful coexistence of tools and models is facilitated
by a consistently object-oriented approach. Some progress was made toward implementation in both the Microsoft Windows and SUN environments; this includes a limited SM interface.
146. Maturana, S., "Comparative Analysis of Mathematical Programming Systems," Working Paper 347, Western Management Science Institute, UCLA, 42 pages, 5/87. Presented at the TIMS/ORSA National
Meeting, New Orleans, May, 1987.
The systems discussed are GAMS, GINO, GXMP, HEQS, and IFPS/OPTIMUM. The analysis criteria are: solver(s) used, system architecture, modeling language features, model management features, data
management features, user interface, intermediate expression handling, and user control. This paper is not directly on SM, but was influenced by early papers on SM.
147. Maturana, S., "Some Extensions of Structured Modeling," Research Paper, John E. Anderson Graduate School of Management, UCLA, 30 pages, 12/23/87.
This paper develops two extensions of the core concepts of SM: (1) genera with a countable infinity of elements, and (2) attribute elements whose values are specified probabilistically. The first
extension is useful, for example, for modeling discrete time with an unbounded horizon, and the second for some types of stochastic modeling and Monte Carlo simulation. Three other possible
extensions are also discussed.
148. Maturana, S., "Integration of a Mathematical Programming Solver into a Modeling Environment," John E. Anderson Graduate School of Management, UCLA, 68 pages, 10/4/88.
Interfacing modeling languages with optimization engines requires converting a given language's description of general model structure, together with instantiating data, into an input file in
some format that is intelligible to an optimizer. This paper lays out the generic problems associated with this task in the context of SML, and describes the design and implementation of software
to accomplish it for FW/SM. The general architecture is that of a compiler. This document also serves as part of the technical documentation of FW/SM.
149. Maturana, S., "Issues in the Design of Modeling Languages for Mathematical Programming," European Journal of Operational Research, 72:2 (Jan 1994), pp. 243-261.
The issues discussed for algebraic modeling languages are: underlying conceptual framework, indexing, recursion, embedded special structures, error checking, and data specification. SML is one of
the languages used for illustration.
150. Maturana, S. and Y. Eterovic, "Vehicle Routing and Production Planning Decision Support Systems: Designing Graphical User Interfaces," International Transactions in Operational Research, 2:3
(1995), pp. 233-247. Presented at IFORS-SPC3 (Digital Technologies/Multimedia: OR/MS in Strategy, Operations and Decision Support), Santa Monica, 1/95. Also available as a working paper, Pontificia
Universidad Católica de Chile, Dept. of Industrial and Systems Engineering, 17 pages, 4/4/95.
This paper falls under the GESCOPP project described in the item by Gazmuri, Maturana, Vicuña and Contesse earlier in this Bibliography. It reviews the general architecture of the implementation,
which uses PowerBuilder for the graphic user interface, WATCOM SQL for the relational database, CPLEX for mixed integer optimization, and some of FW/SM's code for handling SML. The implementation
has been specialized to two industrial applications: one for aggregate production planning for a Chilean home appliance manufacturer, and the other for vehicle routing for a Chilean candy
manufacturer. The emphasis is on the user interface and related matters; more than a dozen screen shots are included.
150A. Maturana, S., P. Gazmuri and C. Villena, "Development and Implementation of a Production Planning DSS for a Manufacturing Firm," working paper, Dept. of Industrial and Systems Engineering,
Pontificia Universidad Católica de Chile, 20 pages, 7/30/98. Revised 3/29/99.
This paper falls under the GESCOPP project described in the item by Gazmuri, Maturana, Vicuña and Contesse earlier in this Bibliography, and follow-on work. It describes in some depth the
authors' practical experiences in developing and applying an optimization-based DSS for a leading Chilean appliance manufacturer. Work continues on a consulting basis with this firm into 1998.
SML was the modeling language, and some of the programs developed for FW/SM were used. The solver was CPLEX 3.0.
150B. Maturana, S., J. Ferrer and F. Barañao, "Design and Implementation of a Generator of Optimization-Based Decision Support Systems," working paper, Dept. of Industrial and Systems Engineering,
Pontificia Universidad Católica de Chile, 26 pages, 6/16/99.
This paper falls under the GESCOPP project described in the item by Gazmuri, Maturana, Vicuña and Contesse earlier in this Bibliography, and follow-on work motivated by real production planning
and truck dispatching applications. The emphasis is well-described by the title. The generator uses SML, PowerBuilder, and CPLEX. Experience with the generator has been very good.
151. Muhanna, W., "An Object-Oriented Framework for Model Management and DSS Development," Decision Support Systems, 9:2 (February, 1993), pp. 217-229. A preliminary version appeared in Proceedings
of the First ISDSS Conference, Austin, TX, September 1990, pp. 553-565.
This article demonstrates the compatibility of the Muhanna-Pick systems framework for modeling-in-the-large with object-oriented ideas, and shows how this framework can coexist with the SM
approach for modeling-in-the-small. The author is extending his earlier prototype model management system, SYMMS, to embody this synthesis [update 9/95: much remains to be done]. Muhanna-Pick
citation: "Composite Models in SYMMS," Proc. Twenty-First Annual Hawaii Intern. Conf. on System Sciences, Vol. III, IEEE Computer Society Press, Los Alamitos, CA, 418-427, 1988.
152. Murphy, F., E.A. Stohr and A. Asthana, "Representation Schemes for Mathematical Programming Models," Management Science, 38:7 (July 1992), pp. 964-991.
This article describes and comments on 7 formalisms for representing mathematical programming models: block-schematic, algebraic, 3 different graphical styles, a database-oriented approach, and
SM. A single running example is used throughout. For a complete SML rendering of that example, including data and a LINDO solution obtained by FW/SM, see FARMER5 in item 36 of this Bibliography.
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quantitative data
Definition of Quantitative Data
● The data that can be counted or measured is called to be Quantitative Data.
More about Quantitative Data
• Quantitative data is also known as numerical data.
• Quantitative data can be analyzed by using statistical method and can also be represented by using graph.
• Discrete and continuous data are quantitative data.
Example of Quantitative Data
● The above bar graph shows the temperature (°F) for the first six days on the month of October in a particular state. The ‘Date’ (on the horizontal axis) is quantitative (discrete data) as it is
measuring the temperature of a particular date. The ‘Temperature (°F)’ (on the vertical axis) is quantitative (continuous data) as the temperature is counted.
Solved Example on Quantitative Data
Identify the quantitative data.
A. Color of jerseys of each team in World Cup Soccer.
B. Name of the players of a soccer team.
C. Number of vehicles sold from a shop in a month.
D. Color of shirts displayed in a sales counter.
Correct Answer: C
Step 1: The data that can be counted or measured is called to be quantitative data.
Step 2: Here, “Number of vehicles sold from a shop in a month” is the quantitative data.
Related Terms for Quantitative Data
• Bar graph
• Circle Graph
• Continuous data
• Data
• Discrete data
• Histogram
• Horizontal Axis
• Line Graph
• Measure
• Vertical Axis | {"url":"http://www.icoachmath.com/math_dictionary/quantitative_data.html","timestamp":"2014-04-20T08:20:49Z","content_type":null,"content_length":"8798","record_id":"<urn:uuid:87755f52-17d2-4e75-aabb-f86f25639d9d>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00428-ip-10-147-4-33.ec2.internal.warc.gz"} |
Understanding Financial Ratios
Anyone learning how to invest in the stock market has probably seen terms like price to earnings and leverage. These are financial ratios, and understanding what they say about a company can mean the
difference between picking a winning stock and a loser.
Financial Ratios
Before beginning the discussion of financial ratios, it's important to lay out some ground rules concerning their use when analyzing stocks. When conducting research, the investor is trying to
understand exactly what the financial ratios reveal about the company. Oftentimes, these ratios will vary greatly by industry. Therefore, it's important to understand how a company is performing
relative to its industry.
It's equally important to understand how a company is performing relative to the entire market. That is why financial ratios are often quoted alongside both an industry benchmark as well as a ratio
for a market index such as the S&P 500.
Stock market analysts like numbers, which means there are a lot of ratios to cover. To help the learning process, they're traditionally separated into six different categories:
• Growth Ratios
• Price Ratios
• Profitability Ratios
• Ratios of Financial Condition
• Investment Ratios
• Management Efficiency Ratios
As the ratios encountered in each of these categories are identified, those the experts believe are the key financial ratios will be in bold type.
Growth Ratios
Growth ratios, or growth rates, tell the analyst just how fast a company is growing. The most important of these ratios include:
• Sales (%): normally stated in terms of a percentage growth from the prior year. Sales is the term used for operating revenues, so it's important to see the sales growth rate as high as possible.
• Net Income (%): growth in net income is even more important than sales because net income tells the investor how much money is left over after all of the operating costs are subtracted from
• Dividends (%): a good indicator of the financial health of a company. Some companies do not pay stock dividends; rather they use these excess profits to reinvest money back into the company to
accelerate growth. The change in dividends (%) should never be negative. That is, once a dividend rate is established, a company needs to have a very good reason to decrease the payout.
Price Ratios
A financial ratio that normalizes a stock's selling price against a measure of the company's profitability is known as a price ratio, and includes:
• Price / Earnings Ratio: the stock's price divided by the latest 12 months of earnings per share. The P/E ratio is an indicator of the premium the investor will be paying for a stock. It is also
an indicator of future growth expectations. In general, low ratios can often mean the market is ignoring the stock for some reason. A relatively high ratio indicates payment of a premium for the
stock's "potential" future earnings.
• Price / Sales Ratio: the stock's price divided by the latest 12 months worth of sales. This measure became popular during the dot com years, when companies showed no earnings or profits. For that
reason, it is not a very useful measure in today's marketplace.
• Price / Book Value Ratio: the stock's price divided by the book value of the company. A company's book value is a measure of the amount by which assets exceed liabilities (in terms of
shareholder's equity). The price / book value rarely falls below 1.0.
• Price / Cash Flow Ratio: identical to price to earnings except that it removes non-cash expenses and other accounting adjustments such as depreciation from that measure.
Profitability Ratios
Also known as profitability margins, these are a measure of a company's operating efficiency. These metrics are normally used to compare a company to its industry (as a benchmark), rather than the
overall stock market, since the profitability of companies can vary greatly by industry.
• Gross Margin (%): a ratio of a company's operating revenue to sales. Operating revenue is the company's sales revenue minus the cost of good sold.
• Pretax Margin (%): a ratio of the company's pretax profits divided by operating revenues.
• Net Profit Margin (%): the ratio of net profits to sales. One of the best indicators of the company's efficiency because net profit takes into consideration all expenses of the company. Investors
would like to see net profit margins as high as possible.
Ratios of Financial Condition
Evaluating the financial condition of a company consists of two related, but distinct, types of measures. The first has to do with how much money the company has borrowed relative to how much money
is "invested" in a company. This concept is known as leverage.
The second measure of financial condition has to do with a company's ability to pay back its creditors in terms of interest payments. The following ratios measure these two aspects of a company's
• Debt / Equity Ratio: a measure of the long-term debt divided by the common stock equity. This financial ratio is a measure of leverage. The lower the debt to equity ratio, the less likely a
company will be affected by a downturn in the economy.
• Current Ratio: a measure of the company's current assets divided by current liabilities based on the most recent quarter. The current ratio is used to provide guidance on a company's immediate
financial health and its ability to meet current obligations.
• Quick Ratio: also known as the acid test, the quick ratio is the sum of cash and receivables divided by the total current liabilities from the most recent quarter. As such, it's another measure
of a company's immediate financial condition.
• Leverage Ratio: calculated by taking the total assets and dividing them by total stock equity. The leverage ratio is an indicator of the company's use of debt. For example, if the ratio is high,
then assets far exceed stock equity (meaning the company has a lot of debt relative to equity), and the company is considered leveraged.
• Interest Coverage: the company's latest 12 months' earnings before interest and taxes (EBIT) divided by the latest 12 months' interest expense. Interest coverage tells the analyst how easily a
company can meet its debt service. The higher the interest coverage value, the easier it is for the company to pay back its debt.
Investment Ratios
This next set of financial ratios is used to understand how efficiently the money invested in a company is providing a return to investors.
• Return on Equity: also known as ROE, this metric is calculated by taking the latest 12 months' net income divided by common stock equity. The return on equity ratio is perhaps the most widely
used, and most valuable, measure of how well a company is performing for its shareholders. The higher the return on equity, the better.
• Return on Capital: also known as ROC, this measure is calculated by taking the latest 12-months' net income divided by invested capital, which is defined as long term debt plus common stock and
preferred equity.
• Return on Assets: also known as ROA, this ratio is determined by taking net income and dividing it by total assets. The metric is used to understand how effectively a company is using their
assets to generate earnings.
Management Efficiency Ratios
This last category of financial ratios is unique in that none would be a direct indicator of a stock's quality. That being said, it is important to understand exactly what these ratios are telling
the investor:
• Revenue per Employee: calculated by taking revenues and dividing them by the total employees of a company. The revenue per employee ratio allows the analyst to understand the labor intensity of a
• Receivables Turnover Ratio: measures how many times accounts receivable have been collected in a given period. Receivables turnover is calculated by taking the last 12 months of sales and
dividing by the average receivables from the latest two quarters.
• Inventory Turnover Ratio: determined by taking the cost of sales for the latest 12 months and dividing by the company's average inventory. The resulting ratio tells the analyst how fast the
company's products are moving on the marketplace. While a relatively low value indicates the company may be having difficulty selling a product, a value that is too high can indicate inadequate
manufacturing capacity.
• Assets Turnover: determined by dividing revenues by sales, this ratio tells the analyst how well a company uses its assets to generate revenue. Companies that require a large infrastructure in
order to produce, or deliver their product, such as electric utilities, require a large asset base to generate sales.
Final Words on Financial Ratios
Eventually these ratios are going to be used to help pick stocks using a screening tool. But there are two more lessons before reaching that point. Up next, there is a discussion of insider trading
About the Author - Understanding Financial Ratios
Copyright © 2006 - 2014 Money-Zine.com | {"url":"http://www.money-zine.com/investing/investing/understanding-financial-ratios/","timestamp":"2014-04-23T19:30:11Z","content_type":null,"content_length":"23601","record_id":"<urn:uuid:28a2a176-d5d3-4d65-ad0c-f37af4745e53>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00525-ip-10-147-4-33.ec2.internal.warc.gz"} |
See also: How do I search for a function?
PDL::Slices -- Indexing, slicing, and dicing
use PDL;
$a = ones(3,3);
$b = $a->slice('-1:0,(1)');
$c = $a->dummy(2);
This package provides many of the powerful PerlDL core index manipulation routines. These routines mostly allow two-way data flow, so you can modify your data in the most convenient representation.
For example, you can make a 1000x1000 unit matrix with
$a = zeroes(1000,1000);
$a->diagonal(0,1) ++;
which is quite efficient. See the PDL::Indexing manpage and the PDL::Tips manpage for more examples.
Slicing is so central to the PDL language that a special compile-time syntax has been introduced to handle it compactly; see the PDL::NiceSlice manpage for details.
PDL indexing and slicing functions usually include two-way data flow, so that you can separate the actions of reshaping your data structures and modifying the data themselves. Two special methods,
copy and sever, help you control the data flow connection between related variables.
$b = $a->slice("1:3"); # Slice maintains a link between $a and $b.
$b += 5; # $a is changed!
If you want to force a physical copy and no data flow, you can copy or sever the slice expression:
$b = $a->slice("1:3")->copy;
$b += 5; # $a is not changed.
$b = $a->slice("1:3")->sever;
$b += 5; # $a is not changed.
The difference between sever and copy is that sever acts on (and returns) its argument, while copy produces a disconnected copy. If you say
$b = $a->slice("1:3");
$c = $b->sever;
then the variables $b and $c point to the same object but with ->copy they would not.
Signature: (P(); C())
Internal vaffine identity function.
s_identity processes bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (a(n); indx ind(); [oca] c())
index, index1d, and index2d provide rudimentary index indirection.
$c = index($source,$ind);
$c = index1d($source,$ind);
$c = index2d($source2,$ind1,$ind2);
use the $ind variables as indices to look up values in $source. The three routines thread slightly differently.
• index uses direct threading for 1-D indexing across the 0 dim of $source. It can thread over source thread dims or index thread dims, but not (easily) both: If $source has more than 1 dimension
and $ind has more than 0 dimensions, they must agree in a threading sense.
• index1d uses a single active dim in $ind to produce a list of indexed values in the 0 dim of the output - it is useful for collapsing $source by indexing with a single row of values along
$source's 0 dimension. The output has the same number of dims as $source. The 0 dim of the output has size 1 if $ind is a scalar, and the same size as the 0 dim of $ind if it is not. If $ind and
$source both have more than 1 dim, then all dims higher than 0 must agree in a threading sense.
• index2d works like index but uses separate piddles for X and Y coordinates. For more general N-dimensional indexing, see the PDL::NiceSlice syntax or PDL::Slices (in particular slice, indexND,
and range).
These functions are two-way, i.e. after
$c = $a->index(pdl[0,5,8]);
$c .= pdl [0,2,4];
the changes in $c will flow back to $a.
index provids simple threading: multiple-dimensioned arrays are treated as collections of 1-D arrays, so that
$a = xvals(10,10)+10*yvals(10,10);
$b = $a->index(3);
$c = $a->index(9-xvals(10));
puts a single column from $a into $b, and puts a single element from each column of $a into $c. If you want to extract multiple columns from an array in one operation, see dice or indexND.
index barfs if any of the index values are bad.
Signature: (a(n); indx ind(m); [oca] c(m))
index, index1d, and index2d provide rudimentary index indirection.
$c = index($source,$ind);
$c = index1d($source,$ind);
$c = index2d($source2,$ind1,$ind2);
use the $ind variables as indices to look up values in $source. The three routines thread slightly differently.
• index uses direct threading for 1-D indexing across the 0 dim of $source. It can thread over source thread dims or index thread dims, but not (easily) both: If $source has more than 1 dimension
and $ind has more than 0 dimensions, they must agree in a threading sense.
• index1d uses a single active dim in $ind to produce a list of indexed values in the 0 dim of the output - it is useful for collapsing $source by indexing with a single row of values along
$source's 0 dimension. The output has the same number of dims as $source. The 0 dim of the output has size 1 if $ind is a scalar, and the same size as the 0 dim of $ind if it is not. If $ind and
$source both have more than 1 dim, then all dims higher than 0 must agree in a threading sense.
• index2d works like index but uses separate piddles for X and Y coordinates. For more general N-dimensional indexing, see the PDL::NiceSlice syntax or PDL::Slices (in particular slice, indexND,
and range).
These functions are two-way, i.e. after
$c = $a->index(pdl[0,5,8]);
$c .= pdl [0,2,4];
the changes in $c will flow back to $a.
index provids simple threading: multiple-dimensioned arrays are treated as collections of 1-D arrays, so that
$a = xvals(10,10)+10*yvals(10,10);
$b = $a->index(3);
$c = $a->index(9-xvals(10));
puts a single column from $a into $b, and puts a single element from each column of $a into $c. If you want to extract multiple columns from an array in one operation, see dice or indexND.
index1d propagates BAD index elements to the output variable.
Signature: (a(na,nb); indx inda(); indx indb(); [oca] c())
index, index1d, and index2d provide rudimentary index indirection.
$c = index($source,$ind);
$c = index1d($source,$ind);
$c = index2d($source2,$ind1,$ind2);
use the $ind variables as indices to look up values in $source. The three routines thread slightly differently.
• index uses direct threading for 1-D indexing across the 0 dim of $source. It can thread over source thread dims or index thread dims, but not (easily) both: If $source has more than 1 dimension
and $ind has more than 0 dimensions, they must agree in a threading sense.
• index1d uses a single active dim in $ind to produce a list of indexed values in the 0 dim of the output - it is useful for collapsing $source by indexing with a single row of values along
$source's 0 dimension. The output has the same number of dims as $source. The 0 dim of the output has size 1 if $ind is a scalar, and the same size as the 0 dim of $ind if it is not. If $ind and
$source both have more than 1 dim, then all dims higher than 0 must agree in a threading sense.
• index2d works like index but uses separate piddles for X and Y coordinates. For more general N-dimensional indexing, see the PDL::NiceSlice syntax or PDL::Slices (in particular slice, indexND,
and range).
These functions are two-way, i.e. after
$c = $a->index(pdl[0,5,8]);
$c .= pdl [0,2,4];
the changes in $c will flow back to $a.
index provids simple threading: multiple-dimensioned arrays are treated as collections of 1-D arrays, so that
$a = xvals(10,10)+10*yvals(10,10);
$b = $a->index(3);
$c = $a->index(9-xvals(10));
puts a single column from $a into $b, and puts a single element from each column of $a into $c. If you want to extract multiple columns from an array in one operation, see dice or indexND.
index2d barfs if either of the index values are bad.
Backwards-compatibility alias for indexND
Find selected elements in an N-D piddle, with optional boundary handling
$out = $source->indexND( $index, [$method] )
$source = 10*xvals(10,10) + yvals(10,10);
$index = pdl([[2,3],[4,5]],[[6,7],[8,9]]);
print $source->indexND( $index );
[23 45]
[67 89]
IndexND collapses $index by lookup into $source. The 0th dimension of $index is treated as coordinates in $source, and the return value has the same dimensions as the rest of $index. The returned
elements are looked up from $source. Dataflow works -- propagated assignment flows back into $source.
IndexND and IndexNDb were originally separate routines but they are both now implemented as a call to range, and have identical syntax to one another.
Signature: (P(); C(); SV *index; SV *size; SV *boundary)
Engine for range
Same calling convention as range, but you must supply all parameters. rangeb is marginally faster as it makes a direct PP call, avoiding the perl argument-parsing step.
Extract selected chunks from a source piddle, with boundary conditions
$out = $source->range($index,[$size,[$boundary]])
Returns elements or rectangular slices of the original piddle, indexed by the $index piddle. $source is an N-dimensional piddle, and $index is a piddle whose first dimension has size up to N. Each
row of $index is treated as coordinates of a single value or chunk from $source, specifying the location(s) to extract.
If you specify a single index location, then range is essentially an expensive slice, with controllable boundary conditions.
$index and $size can be piddles or array refs such as you would feed to zeroes and its ilk. If $index's 0th dimension has size higher than the number of dimensions in $source, then $source is treated
as though it had trivial dummy dimensions of size 1, up to the required size to be indexed by $index -- so if your source array is 1-D and your index array is a list of 3-vectors, you get two dummy
dimensions of size 1 on the end of your source array.
You can extract single elements or N-D rectangular ranges from $source, by setting $size. If $size is undef or zero, then you get a single sample for each row of $index. This behavior is similar to
indexNDb, which is in fact implemented as a call to range.
If $size is positive then you get a range of values from $source at each location, and the output has extra dimensions allocated for them. $size can be a scalar, in which case it applies to all
dimensions, or an N-vector, in which case each element is applied independently to the corresponding dimension in $source. See below for details.
$boundary is a number, string, or list ref indicating the type of boundary conditions to use when ranges reach the edge of $source. If you specify no boundary conditions the default is to forbid
boundary violations on all axes. If you specify exactly one boundary condition, it applies to all axes. If you specify more (as elements of a list ref, or as a packed string, see below), then they
apply to dimensions in the order in which they appear, and the last one applies to all subsequent dimensions. (This is less difficult than it sounds; see the examples below).
1. (synonyms: 'f','forbid') (default)
Ranges are not allowed to cross the boundary of the original PDL. Disallowed ranges throw an error. The errors are thrown at evaluation time, not at the time of the range call (this is the same
behavior as slice).
2. (synonyms: 't','truncate')
Values outside the original piddle get BAD if you've got bad value support compiled into your PDL and set the badflag for the source PDL; or 0 if you haven't (you must set the badflag if you want
BADs for out of bound values, otherwise you get 0). Reverse dataflow works OK for the portion of the child that is in-bounds. The out-of-bounds part of the child is reset to (BAD|0) during each
dataflow operation, but execution continues.
3. (synonyms: 'e','x','extend')
Values that would be outside the original piddle point instead to the nearest allowed value within the piddle. See the CAVEAT below on mappings that are not single valued.
4. (synonyms: 'p','periodic')
Periodic boundary conditions apply: the numbers in $index are applied, strict-modulo the corresponding dimensions of $source. This is equivalent to duplicating the $source piddle throughout N-D
space. See the CAVEAT below about mappings that are not single valued.
5. (synonyms: 'm','mirror')
Mirror-reflection periodic boundary conditions apply. See the CAVEAT below about mappings that are not single valued.
The boundary condition identifiers all begin with unique characters, so you can feed in multiple boundary conditions as either a list ref or a packed string. (The packed string is marginally faster
to run). For example, the four expressions [0,1], ['forbid','truncate'], ['f','t'], and 'ft' all specify that violating the boundary in the 0th dimension throws an error, and all other dimensions get
If you feed in a single string, it is interpreted as a packed boundary array if all of its characters are valid boundary specifiers (e.g. 'pet'), but as a single word-style specifier if they are not
(e.g. 'forbid').
The output threads over both $index and $source. Because implicit threading can happen in a couple of ways, a little thought is needed. The returned dimension list is stacked up like this:
(index thread dims), (index dims (size)), (source thread dims)
The first few dims of the output correspond to the extra dims of $index (beyond the 0 dim). They allow you to pick out individual ranges from a large, threaded collection.
The middle few dims of the output correspond to the size dims specified in $size, and contain the range of values that is extracted at each location in $source. Every nonzero element of $size is
copied to the dimension list here, so that if you feed in (for example) $size = [2,0,1] you get an index dim list of (2,1).
The last few dims of the output correspond to extra dims of $source beyond the number of dims indexed by $index. These dims act like ordinary thread dims, because adding more dims to $source just
tacks extra dims on the end of the output. Each source thread dim ranges over the entire corresponding dim of $source.
Dataflow: Dataflow is bidirectional.
Examples: Here are basic examples of range operation, showing how to get ranges out of a small matrix. The first few examples show extraction and selection of individual chunks. The last example
shows how to mark loci in the original matrix (using dataflow).
pdl> $src = 10*xvals(10,5)+yvals(10,5)
pdl> print $src->range([2,3]) # Cut out a single element
pdl> print $src->range([2,3],1) # Cut out a single 1x1 block
pdl> print $src->range([2,3], [2,1]) # Cut a 2x1 chunk
[23 33]
pdl> print $src->range([[2,3]],[2,1]) # Trivial list of 1 chunk
pdl> print $src->range([[2,3],[0,1]], [2,1]) # two 2x1 chunks
[23 1]
[33 11]
pdl> # A 2x2 collection of 2x1 chunks
pdl> print $src->range([[[1,1],[2,2]],[[2,3],[0,1]]],[2,1])
[11 22]
[23 1]
[21 32]
[33 11]
pdl> $src = xvals(5,3)*10+yvals(5,3)
pdl> print $src->range(3,1) # Thread over y dimension in $src
pdl> $src = zeroes(5,4);
pdl> $src->range(pdl([2,3],[0,1]),pdl(2,1)) .= xvals(2,2,1) + 1
pdl> print $src
[0 0 0 0 0]
[2 2 0 0 0]
[0 0 0 0 0]
[0 0 1 1 0]
CAVEAT: It's quite possible to select multiple ranges that intersect. In that case, modifying the ranges doesn't have a guaranteed result in the original PDL -- the result is an arbitrary choice
among the valid values. For some things that's OK; but for others it's not. In particular, this doesn't work:
pdl> $photon_list = new PDL::RandVar->sample(500)->reshape(2,250)*10
pdl> histogram = zeroes(10,10)
pdl> histogram->range($photon_list,1)++; #not what you wanted
The reason is that if two photons land in the same bin, then that bin doesn't get incremented twice. (That may get fixed in a later version...)
PERMISSIVE RANGING: If $index has too many dimensions compared to $source, then $source is treated as though it had dummy dimensions of size 1, up to the required number of dimensions. These virtual
dummy dimensions have the usual boundary conditions applied to them.
If the 0 dimension of $index is ludicrously large (if its size is more than 5 greater than the number of dims in the source PDL) then range will insist that you specify a size in every dimension, to
make sure that you know what you're doing. That catches a common error with range usage: confusing the initial dim (which is usually small) with another index dim (perhaps of size 1000).
If the index variable is Empty, then range() always returns the Empty PDL. If the index variable is not Empty, indexing it always yields a boundary violation. All non-barfing conditions are treated
as truncation, since there are no actual data to return.
EFFICIENCY: Because range isn't an affine transformation (it involves lookup into a list of N-D indices), it is somewhat memory-inefficient for long lists of ranges, and keeping dataflow open is much
slower than for affine transformations (which don't have to copy data around).
Doing operations on small subfields of a large range is inefficient because the engine must flow the entire range back into the original PDL with every atomic perl operation, even if you only touch a
single element. One way to speed up such code is to sever your range, so that PDL doesn't have to copy the data with each operation, then copy the elements explicitly at the end of your loop. Here's
an example that labels each region in a range sequentially, using many small operations rather than a single xvals assignment:
### How to make a collection of small ops run fast with range...
$a = $data->range($index, $sizes, $bound)->sever;
$aa = $data->range($index, $sizes, $bound);
map { $a($_ - 1) .= $_; } (1..$a->nelem); # Lots of little ops
$aa .= $a;
range is a perl front-end to a PP function, rangeb. Calling rangeb is marginally faster but requires that you include all arguments.
* index thread dimensions are effectively clumped internally. This makes it easier to loop over the index array but a little more brain-bending to tease out the algorithm.
rangeb processes bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (indx a(n); b(n); [o]c(m))
Run-length decode a vector
Given a vector $a of the numbers of instances of values $b, run-length decode to $c.
rld does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (c(n); indx [o]a(n); [o]b(n))
Run-length encode a vector
Given vector $c, generate a vector $a with the number of each element, and a vector $b of the unique values. Only the elements up to the first instance of 0 in $a should be considered.
rle does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); int n1; int n2)
exchange two dimensions
Negative dimension indices count from the end.
The command
$b = $a->xchg(2,3);
creates $b to be like $a except that the dimensions 2 and 3 are exchanged with each other i.e.
$b->at(5,3,2,8) == $a->at(5,3,8,2)
xchg does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Re-orders the dimensions of a PDL based on the supplied list.
Similar to the xchg method, this method re-orders the dimensions of a PDL. While the xchg method swaps the position of two dimensions, the reorder method can change the positions of many dimensions
at once.
# Completely reverse the dimension order of a 6-Dim array.
$reOrderedPDL = $pdl->reorder(5,4,3,2,1,0);
The argument to reorder is an array representing where the current dimensions should go in the new array. In the above usage, the argument to reorder (5,4,3,2,1,0) indicates that the old dimensions
($pdl's dims) should be re-arranged to make the new pdl ($reOrderPDL) according to the following:
Old Position New Position
------------ ------------
You do not need to specify all dimensions, only a complete set starting at position 0. (Extra dimensions are left where they are). This means, for example, that you can reorder() the X and Y
dimensions of an image, and not care whether it is an RGB image with a third dimension running across color plane.
pdl> $a = sequence(5,3,2); # Create a 3-d Array
pdl> p $a
[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]
[20 21 22 23 24]
[25 26 27 28 29]
pdl> p $a->reorder(2,1,0); # Reverse the order of the 3-D PDL
[ 0 15]
[ 5 20]
[10 25]
[ 1 16]
[ 6 21]
[11 26]
[ 2 17]
[ 7 22]
[12 27]
[ 3 18]
[ 8 23]
[13 28]
[ 4 19]
[ 9 24]
[14 29]
The above is a simple example that could be duplicated by calling $a->xchg(0,2), but it demonstrates the basic functionality of reorder.
As this is an index function, any modifications to the result PDL will change the parent.
Signature: (P(); C(); int n1; int n2)
move a dimension to another position
The command
$b = $a->mv(4,1);
creates $b to be like $a except that the dimension 4 is moved to the place 1, so:
$b->at(1,2,3,4,5,6) == $a->at(1,5,2,3,4,6);
The other dimensions are moved accordingly. Negative dimension indices count from the end.
mv does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); int nth; int from; int step; int nsteps)
experimental function - not for public use
$a = oneslice();
This is not for public use currently. See the source if you have to. This function can be used to accomplish run-time changing of transformations i.e. changing the size of some piddle at run-time.
However, the mechanism is not yet finalized and this is just a demonstration.
oneslice does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); char* str)
DEPRECATED: 'oslice' is the original 'slice' routine in pre-2.006_006 versions of PDL. It is left here for reference but will disappear in PDL 3.000
Extract a rectangular slice of a piddle, from a string specifier.
slice was the original Swiss-army-knife PDL indexing routine, but is largely superseded by the NiceSlice source prefilter and its associated nslice method. It is still used as the basic underlying
slicing engine for nslice, and is especially useful in particular niche applications.
$a->slice('1:3'); # return the second to fourth elements of $a
$a->slice('3:1'); # reverse the above
$a->slice('-2:1'); # return last-but-one to second elements of $a
The argument string is a comma-separated list of what to do for each dimension. The current formats include the following, where a, b and c are integers and can take legal array index values
(including -1 etc):
takes the whole dimension intact.
(nothing) is a synonym for ":" (This means that $a->slice(':,3') is equal to $a->slice(',3')).
slices only this value out of the corresponding dimension.
means the same as "a" by itself except that the resulting dimension of length one is deleted (so if $a has dims (3,4,5) then $a->slice(':,(2),:') has dimensions (3,5) whereas $a->slice(':,2,:')
has dimensions (3,1,5)).
slices the range a to b inclusive out of the dimension.
slices the range a to b, with step c (i.e. 3:7:2 gives the indices (3,5,7)). This may be confusing to Matlab users but several other packages already use this syntax.
inserts an extra dimension of width 1 and
inserts an extra (dummy) dimension of width a.
An extension is planned for a later stage allowing $a->slice('(=1),(=1|5:8),3:6(=1),4:6') to express a multidimensional diagonal of $a.
Trivial out-of-bounds slicing is allowed: if you slice a source dimension that doesn't exist, but only index the 0th element, then slice treats the source as if there were a dummy dimension there.
The following are all equivalent:
xvals(5)->dummy(1,1)->slice('(2),0') # Add dummy dim, then slice
xvals(5)->slice('(2),0') # Out-of-bounds slice adds dim.
xvals(5)->slice((2),0) # NiceSlice syntax
xvals(5)->((2))->dummy(0,1) # NiceSlice syntax
This is an error:
xvals(5)->slice('(2),1') # nontrivial out-of-bounds slice dies
Because slicing doesn't directly manipulate the source and destination pdl -- it just sets up a transformation between them -- indexing errors often aren't reported until later. This is either a bug
or a feature, depending on whether you prefer error-reporting clarity or speed of execution.
oslice does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Returns array of column numbers requested
line $pdl->using(1,2);
Plot, as a line, column 1 of $pdl vs. column 2
pdl> $pdl = rcols("file");
pdl> line $pdl->using(1,2);
Signature: (P(); C(); SV *list)
Returns the multidimensional diagonal over the specified dimensions.
The diagonal is placed at the first (by number) dimension that is diagonalized. The other diagonalized dimensions are removed. So if $a has dimensions (5,3,5,4,6,5) then after
$b = $a->diagonal(0,2,5);
the piddle $b has dimensions (5,3,4,6) and $b->at(2,1,0,1) refers to $a->at(2,1,2,0,1,2).
NOTE: diagonal doesn't handle threadids correctly. XXX FIX
diagonalI does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); int nthdim; int step; int n)
Returns a piddle of lags to parent.
$lags = $a->lags($nthdim,$step,$nlags);
I.e. if $a contains
$b = $a->lags(0,2,2);
is a (5,2) matrix
This order of returned indices is kept because the function is called "lags" i.e. the nth lag is n steps behind the original.
$step and $nlags must be positive. $nthdim can be negative and will then be counted from the last dim backwards in the usual way (-1 = last dim).
lags does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); int nthdim; int nsp)
Splits a dimension in the parent piddle (opposite of clump)
$b = $a->splitdim(2,3);
the expression
$b->at(6,4,x,y,3,6) == $a->at(6,4,x+3*y)
is always true (x has to be less than 3).
splitdim does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (x(n); indx shift(); [oca]y(n))
Shift vector elements along with wrap. Flows data back&forth.
rotate does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); int id; SV *list)
Put some dimensions to a threadid.
$b = $a->threadI(0,1,5); # thread over dims 1,5 in id 1
threadI does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C())
A vaffine identity transformation (includes thread_id copying).
Mainly for internal use.
identvaff does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Signature: (P(); C(); int atind)
All threaded dimensions are made real again.
See [TBD Doc] for details and examples.
unthread does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
Dice rows/columns/planes out of a PDL using indexes for each dimension.
This function can be used to extract irregular subsets along many dimension of a PDL, e.g. only certain rows in an image, or planes in a cube. This can of course be done with the usual dimension
tricks but this saves having to figure it out each time!
This method is similar in functionality to the slice method, but slice requires that contiguous ranges or ranges with constant offset be extracted. ( i.e. slice requires ranges of the form 1,2,3,4,5
or 2,4,6,8,10). Because of this restriction, slice is more memory efficient and slightly faster than dice
$slice = $data->dice([0,2,6],[2,1,6]); # Dicing a 2-D array
The arguments to dice are arrays (or 1D PDLs) for each dimension in the PDL. These arrays are used as indexes to which rows/columns/cubes,etc to dice-out (or extract) from the $data PDL.
Use X to select all indices along a given dimension (compare also mslice). As usual (in slicing methods) trailing dimensions can be omitted implying X'es for those.
pdl> $a = sequence(10,4)
pdl> p $a
[ 0 1 2 3 4 5 6 7 8 9]
[10 11 12 13 14 15 16 17 18 19]
[20 21 22 23 24 25 26 27 28 29]
[30 31 32 33 34 35 36 37 38 39]
pdl> p $a->dice([1,2],[0,3]) # Select columns 1,2 and rows 0,3
[ 1 2]
[31 32]
pdl> p $a->dice(X,[0,3])
[ 0 1 2 3 4 5 6 7 8 9]
[30 31 32 33 34 35 36 37 38 39]
pdl> p $a->dice([0,2,5])
[ 0 2 5]
[10 12 15]
[20 22 25]
[30 32 35]
As this is an index function, any modifications to the slice change the parent (use the .= operator).
Dice rows/columns/planes from a single PDL axis (dimension) using index along a specified axis
This function can be used to extract irregular subsets along any dimension, e.g. only certain rows in an image, or planes in a cube. This can of course be done with the usual dimension tricks but
this saves having to figure it out each time!
$slice = $data->dice_axis($axis,$index);
pdl> $a = sequence(10,4)
pdl> $idx = pdl(1,2)
pdl> p $a->dice_axis(0,$idx) # Select columns
[ 1 2]
[11 12]
[21 22]
[31 32]
pdl> $t = $a->dice_axis(1,$idx) # Select rows
pdl> $t.=0
pdl> p $a
[ 0 1 2 3 4 5 6 7 8 9]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[30 31 32 33 34 35 36 37 38 39]
The trick to using this is that the index selects elements along the dimensions specified, so if you have a 2D image axis=0 will select certain X values - i.e. extract columns
As this is an index function, any modifications to the slice change the parent.
$slice = $data->slice([2,3],'x',[2,2,0],"-1:1:-1", "*3");
Extract rectangular slices of a piddle, from a string specifier, an array ref specifier, or a combination.
slice is the main method for extracting regions of PDLs and manipulating their dimensionality. You can call it directly or via he NiceSlice source prefilter that extends Perl syntax o include array
slice can extract regions along each dimension of a source PDL, subsample or reverse those regions, dice each dimension by selecting a list of locations along it, or basic PDL indexing routine. The
selected subfield remains connected to the original PDL via dataflow. In most cases this neither allocates more memory nor slows down subsequent operations on either of the two connected PDLs.
You pass in a list of arguments. Each term in the list controls the disposition of one axis of the source PDL and/or returned PDL. Each term can be a string-format cut specifier, a list ref that
gives the same information without recourse to string manipulation, or a PDL with up to 1 dimension giving indices along that axis that should be selected.
If you want to pass in a single string specifier for the entire operation, you can pass in a comma-delimited list as the first argument. slice detects this condition and splits the string into a
regular argument list. This calling style is fully backwards compatible with slice calls from before PDL 2.006.
If a particular argument to slice is a string, it is parsed as a selection, an affine slice, or a dummy dimension depending on the form. Leading or trailing whitespace in any part of each specifier
is ignored (though it is not ignored within numbers).
The empty string, :, or X cause the entire corresponding dimension to be kept unchanged.
A single number alone causes a single index to be selected from the corresponding dimension. The dimension is kept (and reduced to size 1) in the output.
A single number in parenthesis causes a single index to be selected from the corresponding dimension. The dimension is discarded (completely eliminated) in the output.
Two numbers separated by a colon selects a range of values from the corresponding axis, e.g. 3:4 selects elements 3 and 4 along the corresponding axis, and reduces that axis to size 2 in the
output. Both numbers are regularized so that you can address the last element of the axis with an index of -1 . If, after regularization, the two numbers are the same, then exactly one element
gets selected (just like the <n> case). If, after regulariation, the second number is lower than the first, then the resulting slice counts down rather than up -- e.g. -1:0 will return the entire
axis, in reversed order.
If you include a third parameter, it is the stride of the extracted range. For example, 0:-1:2 will sample every other element across the complete dimension. Specifying a stride of 1 prevents
autoreversal -- so to ensure that your slice is *always* forward you can specify, e.g., 2:$n:1. In that case, an "impossible" slice gets an Empty PDL (with 0 elements along the corresponding
dimension), so you can generate an Empty PDL with a slice of the form 2:1:1.
Dummy dimensions aren't present in the original source and are "mocked up" to match dimensional slots, by repeating the data in the original PDL some number of times. An asterisk followed by a
number produces a dummy dimension in the output, for example *2 will generate a dimension of size 2 at the corresponding location in the output dim list. Omitting the numeber (and using just an
asterisk) inserts a dummy dimension of size 1.
If you feed in an ARRAY ref as a slice term, then it can have 0-3 elements. The first element is the start of the slice along the corresponding dim; the second is the end; and the third is the
stepsize. Different combinations of inputs give the same flexibility as the string syntax.
An empty ARRAY ref keeps the entire corresponding dim
If $n is missing, you get a dummy dim of size 1.
$dex must be a single value. It is used to index the source, and the corresponding dimension is discarded.
In the simple two-number case, you get a slice that runs up or down (as appropriate) to connect $start and $end.
The three-number case works exactly like the three-number string case above.
PDL args for dicing
If you pass in a 0- or 1-D PDL as a slicing argument, the corresponding dimension is "diced" -- you get one position along the corresponding dim, per element of the indexing PDL, e.g. $a->slice( pdl
(3,4,9)) gives you elements 3, 4, and 9 along the 0 dim of $a.
Because dicing is not an affine transformation, it is slower than direct slicing even though the syntax is convenient.
$a->slice('1:3'); # return the second to fourth elements of $a
$a->slice('3:1'); # reverse the above
$a->slice('-2:1'); # return last-but-one to second elements of $a
$a->slice([1,3]); # Same as above three calls, but using array ref syntax
Signature: (P(); C(); SV *args)
info not available
sliceb does not process bad values. It will set the bad-value flag of all output piddles if the flag is set for any of the input piddles.
For the moment, you can't slice one of the zero-length dims of an empty piddle. It is not clear how to implement this in a way that makes sense.
Many types of index errors are reported far from the indexing operation that caused them. This is caused by the underlying architecture: slice() sets up a mapping between variables, but that mapping
isn't tested for correctness until it is used (potentially much later).
Copyright (C) 1997 Tuomas J. Lukka. Contributions by Craig DeForest, deforest@boulder.swri.edu. Documentation contributions by David Mertens. All rights reserved. There is no warranty. You are
allowed to redistribute this software / documentation under certain conditions. For details, see the file COPYING in the PDL distribution. If this file is separated from the PDL distribution, the
copyright notice should be included in the file. | {"url":"http://pdl.perl.org/?docs=Slices&title=indexND","timestamp":"2014-04-16T16:10:01Z","content_type":null,"content_length":"54425","record_id":"<urn:uuid:68759aa6-6856-4b68-9994-41db92190066>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00548-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two particles are moving along two long straight lines, in the same plane, with the same speed: 20 cm/s. The angle... - Homework Help - eNotes.com
Two particles are moving along two long straight lines, in the same plane, with the same speed: 20 cm/s. The angle between the two lines is 60° and their intersection point is O. At a certain moment,
the two particles are located at distances 3 m and 4 m from O and are moving towards O. Subsequesntly the shortest distance between them will be:
A) 50 cm
B) 40 √2 cm
C) 50 √2 cm
D) 50 √3 cm
Two particles are moving along straight lines, in the same plane. Their speed is the same and equal to 20 cm/s. The angle between the two lines is 60 degrees and the intersect at point O.
At a certain moment, the two particles are located at distances 3 m and 4 m from O and are moving towards O.
The displacement D of the particle that is initially at a distance 3 m from O can be divided into two components D*cos 60 is in the same direction as the other particle and D*sin 60 is perpendicular
to the direction in which the particle is moving.
The distance between the particles after t seconds is `S = sqrt((150*sqrt3 - 10*sqrt 3*t)^2 + (250 - 10t)^2)`
The minimum distance between them is at t when S' = 0
=> `(20*t-350)/sqrt(t^2-35*t+325) = 0`
=> t = 17.5
At t = 17.5, S = `50*sqrt 3`
S'' = `(375*sqrt(t^2-35*t+325))/(t^4-70*t^3+1875*t^2-22750*t+105625) `
At t = 17.5, S'' is positive
As S' = 0 and S'' is positive at t = 17.5 this is a point of minimum value.
The minimum distance between the two particles as `50*sqrt 3` cm
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Join eNotes | {"url":"http://www.enotes.com/homework-help/q-two-particles-moving-along-two-long-straight-440916","timestamp":"2014-04-21T03:15:52Z","content_type":null,"content_length":"26891","record_id":"<urn:uuid:d4ce0233-5bbb-4882-8113-7b4b72a97965>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00290-ip-10-147-4-33.ec2.internal.warc.gz"} |
gemv not handling N=0 according to math. conventions
In order to compute a function t = (t₀, t₁,...) ↦ f(t₀ y + ∑ t_i a_i) where y and a_i are vectors, I have used gemv to perform y ← t₀ y + ∑ t_i a_i. However, in case the sum is empty — ∑_{i ∈ ∅}
which translates to N=0 when calling gemv — gemv leaves y unchanged instead of performing the mathematically correct behavior: y ← t₀ y.
Is this broken semantics the intended behavior — or must it be considered as a forgotten corner case? If the former, why? And why hasn't this be fixed — as it seems to me that having the
mathematically correct behavior is definitely better and can be obtained at zero additional (computational) cost.
Re: gemv not handling N=0 according to math. conventions
This dates back from 1988 and the initial BLAS specification of the Level 2
BLAS. The behavior you are describing (which you do not like) is actually
correct and follows the specification from 1988. Interestingly enough, the
behavior was changed for Level 3 BLAS in 1990 and the Level 3 BLAS follows the
behavior you would have liked to see. See below for historical references and
precise quotes.
To address the issue, you could have an IF statement before calling Level 2
BLAS GEMV or you could use the Level 3 BLAS GEMM instead of the Level 2 BLAS
I understand that you probably have found a fix and made this work and your
comment was higher level than just "I want my code to work". Thanks for the
feedback, you are right, this behavior of Level 2 BLAS GEMV may appear quite
surprising / counter-intuitive indeed. I agree with you.
The Level 2 BLAS paper from 1988.
J. J. Dongarra, J. Du Croz, S. Hammarling, and R. J. Hanson, An extended set of
FORTRAN Basic Linear Algebra Subprograms, ACM Trans. Math. Soft., 14 (1988),
pp. 1--17.
page 5 - section 4
The size of the matrix is determined by the arguments M and N for an m-by-n
rectangular matrix [...] Note it is permissible to call the routines with M or
N = 0, in which case the routines exit immediately without referencing their
vector or matrix arguments.
The Level 3 BLAS paper from 1990.
J. J. Dongarra, J. Du Croz, I. S. Duff, and S. Hammarling, A set of Level 3
Basic Linear Algebra Subprograms, ACM Trans. Math. Soft., 16 (1990), pp. 1--17.
page 6 - section 4
The sizes of the matrices are determined by the arguments M, N and K. It is
permissible to call the routines with M or N = 0, in which case the routines
exit immediately without referencing their arguments. If M and N > 0, but K =
0, the operation reduces to C := beta * C [ ... ]
Re: gemv not handling N=0 according to math. conventions
Thanks for your answer. | {"url":"http://icl.cs.utk.edu/lapack-forum/viewtopic.php?f=5&p=10026","timestamp":"2014-04-19T04:34:29Z","content_type":null,"content_length":"20215","record_id":"<urn:uuid:c646b7e4-c112-44c7-81a3-9e2be7871dfb>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00604-ip-10-147-4-33.ec2.internal.warc.gz"} |
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graph theory
involving the study of
s, collections of
s (commonly called
s; this is the sense of
used on
) which are connected by
s (the equivalent of
s on Everything).
A directed graph is one in which the edges are assigned directions, like one-way streets; that is, an edge may go from A to B but not from B to A. Directed graphs may have some bidirectional edges as
A graph is simple if there is only one edge between any two nodes. Otherwise the graph is nonsimple. A nonsimple graph may have some special edges called loops which go from one node back to the same
There are many other types of graphs with special names, but I will only describe one more here. A planar graph is one which can be drawn or embedded in a plane without any of the edges crossing,
other than at nodes. For such graphs, Euler's formula holds, as is the case with convex polyhedra.
In much of the study of graph theory, only the nodes and the connections made by the edges matter, so the graph can be arbitrarily twisted and still remain the same graph; thus graph theory borders
on topology.
-- This node saved by The Nodeshell Rescue Team | {"url":"http://everything2.com/title/graph+theory","timestamp":"2014-04-19T02:10:50Z","content_type":null,"content_length":"34893","record_id":"<urn:uuid:77343ace-0f40-492e-8be5-aa630e004085>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00303-ip-10-147-4-33.ec2.internal.warc.gz"} |
Planarity Testing and Optimal Edge Insertion with Embedding Constraints
Gutwenger, Carsten and Klein, Karsten and Mutzel, Petra (2007) Planarity Testing and Optimal Edge Insertion with Embedding Constraints. In: Graph Drawing 14th International Symposium, GD 2006,
September 18-20, 2006, Karlsruhe, Germany , pp. 126-137 (Official URL: http://dx.doi.org/10.1007/978-3-540-70904-6_14).
Full text not available from this repository.
Many practical applications demand additional restrictions on an admissible planar embedding. In particular, constraints on the permitted (clockwise) order of the edges around a vertex, like
so-called side constraints, abound. In this paper, we introduce a set of hierarchical embedding constraints that also comprises side constraints. We present linear time algorithms for testing if a
graph is ec-planar, i.e., admits a planar embedding satisfying the given embedding constraints, as well as for computing such an embedding. Moreover, we characterize the set of all possible ec-planar
embeddings and consider the problem of finding a planar combinatorial embedding of a planar graph such that an additional edge can be inserted with the minimum number of crossings; we show that this
problem can still be solved in linear time under the additional restrictions of embedding constraints.
Repository Staff Only: item control page | {"url":"http://gdea.informatik.uni-koeln.de/768/","timestamp":"2014-04-20T10:07:22Z","content_type":null,"content_length":"28513","record_id":"<urn:uuid:39750854-9361-42ce-8b49-afeaa8d0d031>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00656-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Hubble Constant from Gravitational Lens Time Delays -
P. Schechter
4.3. The central concentration degeneracy
Equations 10 and 11 tell us that the interpretation of the time delay for PG1115+080 depends crucially upon the internal structure of the galaxy. An error in the radial exponent of plus or minus 0.1
produces a 10% change in the predicted time delay as compared to an isothermal model. Fortunately we can test the isothermality hypothesis. For an isothermal sphere, the angular radius of the
Einstein ring, b, is a directly proportional to the square of the velocity dispersion, ^2:
With an Einstein ring radius b = 1."15, a lens redshift z[L] = 0.31 and a source redshift z[S] = 1.71 we predict [SIS] = 232 km/s. This must be reduced by a factor (1 - [group])^1/2 which amounts to
a 5% correction in the present case. We can test our prediction, but the measurement is a difficult one. The lensing galaxy is crowded by four very much brighter images. [Tonry (1998)] measures a
velocity dispersion of 281 ± 25 km/s, very much larger than predicted. [Treu & Koopmans (2002)] use Tonry's measurement in modeling PG1115+080 and derive a value for H[0] very much larger than under
the isothermal hypothesis. Their value is the one plotted in figure 1.
Why are we so reluctant to abandon the isothermal hypothesis for PG1115+080? First, because velocity dispersion estimates from equation 12 for an ensemble of lensing galaxies are consistent with the
fundamental plane relation for non-lensing ellipticals. PG1115+080 is in no way unusual [(Kochanek et al. 2000)], but would be if we adopted the direct measurement.
A second argument is that the lenses for which we can measure the radial exponent are very nearly isothermal. Lenses with multiple sources, rings or central images all break the central concentration
degeneracy and permit measurement of the radial exponent table 3 (see also WAYTH'S contribution to the present proceedings). While there are differences in the way
A third argument is that a great many nearby galaxies have been studied, and their potentials are consistent with isothermals. This is nicely shown in a figure published by [Romanowsky & Kochanek
(1999)]. For a sample of twenty bright ellipticals, the circular velocity inferred from the velocity dispersion declines only slightly over a factor of 30 in radius. The corresponding decline in
PG1115+080, computed from its measured central dispersion and the Einstein ring radius, is very much larger. It is nonetheless true that the potentials for nearby ellipticals are only very-nearly
isothermal and not perfectly isothermal. The central concentration degeneracy qualifies as another major difficulty associated with time delay estimates of H[0]. | {"url":"http://ned.ipac.caltech.edu/level5/March04/Schechter/Schechter4_3.html","timestamp":"2014-04-20T10:47:33Z","content_type":null,"content_length":"9373","record_id":"<urn:uuid:12274918-67d2-4efc-8a11-810a1ea7eb01>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00208-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Convert bovate to nook - Conversion of Measurement Units
›› Convert bovate to nook
›› More information from the unit converter
How many bovate in 1 nook? The answer is 1.34895213333.
We assume you are converting between bovate and nook.
You can view more details on each measurement unit:
bovate or nook
The SI derived unit for area is the square meter.
1 square meter is equal to 1.66666666667E-5 bovate, or 1.23552691417E-5 nook.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between bovates and nook.
Type in your own numbers in the form to convert the units!
›› Definition: Bovate
A bovate was a measure of land which could be ploughed in one day by one eighth of a plough team with eight oxen, or in other words the measure of land representing one eighth of a carucate. The term
is used in the Domesday Book for places under the Danelaw. The word is derived from the Latin word bo, meaning ox.
›› Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data.
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Predicting GDP With ARIMA Forecasts
Is the U.S. economy headed for a new recession? The risk is clearly elevated these days, in part because the euro crisis rolls on. The sluggish growth rate in the U.S. isn’t helping either. But with
ongoing job growth, albeit at a slow rate, it’s not yet clear that we’ve reached a tipping point. Given all the mixed signals, however, forecasting is unusually tough at the moment. It’s never easy,
of course, but it’s always necessary just the same. But how to proceed? The possibilities are endless, but one useful way to begin is with so-called autoregressive integrated moving averages (ARIMA).
It sounds rather intimidating, but the basic calculation is straightforward and it’s easily performed in a spreadsheet, which helps explain why ARIMA models are so popular in econometrics. A more
compelling reason for this technique’s widespread use: a number of studies report that ARIMA models have a history of making relatively accurate forecasts compared with the more sophisticated
As a simple example of the power of ARIMA forecasting, let’s consider what this statistical tool is telling us about the next quarterly change in nominal GDP for the U.S. Making a few reasonable
assumptions (discussed below), a basic ARIMA forecasting model predicts that fourth quarter 2011 nominal GDP will rise 4.7% at a seasonally adjusted annual rate. For comparison, that’s slightly lower
than the government’s initial estimate of 5.0% growth for the third quarter. (Remember, we’re talking here of nominal GDP growth vs. real growth, which strips out inflation. Real GDP is the more
popularly quoted series.)
For perspective, let’s compare the history of my ARIMA forecast with GDP predictions via the Survey of Professional Forecasters (SPF). SPF data is available on the Philadelphia Fed’s web site and in
this case I focus on nominal GDP. To cut to the chase, ARIMA has a history of dispatching superior forecasts compared with SPF. To be precise, I’m comparing ARIMA forecasts with the mean
quarter-ahead prediction of economists surveyed quarterly in the SPF reports.
Ok, let’s take a closer look at some of the finer points by reviewing a few basic ARIMA concepts. Keep in mind that in the interest of brevity I’m glossing over the details. For a complete discussion
of ARIMA, an introductory econometrics text will suffice. One of many examples: Peter Kennedy’s A Guide to Econometrics
As for ARIMA, the first step is taking the data series (in this case the historical quarterly nominal GDP numbers) and regressing them against a lagged set of the same data. For the analysis here,
I’m regressing each quarterly GDP number against 1) the previous quarter; 2) two quarters previous; and 3) four quarters previous. Next, I ran a multi-regression analysis on this set of lagged data
to estimate the parameters, which tell us how to weight each lagged variable in the formula that spits out the forecast. To check the accuracy of the parameter estimates, I also ran a maximum
likelihood procedure. (As a quick aside, all of this analysis can be easily done in Excel, although more sophisticated software packages are available, such as Matlab and EViews.)
ARIMA’s forecasts are naïve, of course, but based on history it does fairly well compared with SPF. The in-sample forecasting errors (residuals, as statisticians call them) for ARIMA’s average
deviation from the actual reported GDP number is a slight 0.0049% since 1970. That’s a mere fraction of SPF’s 3.06% residual over the past 30 years. There are several additional error tests we can
run, but the simple evaluations above offer a general sense of how a naïve ARIMA model can provide competitive forecasts vs. the expectations of professional economists.
Alas, like all econometric techniques that look backward as a means of looking ahead there’s the risk that sharp and sudden turning points in the trend will surprise an ARIMA model. That’s clear by
looking at recent history, as shown in the chart below. Note that when the actual level of nominal GDP turned down in 2008 as the Great Recession unfolded, ARIMA’s forecasting error rate increased.
But ARIMA fared no worse than the mean SPF predictions. In fact, you can argue that ARIMA did slightly better than SPF, as implied by tallying up the errors for each during 2008 and comparing one to
Errors are inevitable in forecasting. The goal is to keep them to a minimum, a task that ARIMA does quite well. But that assumes a certain amount of trial-and-error testing for building and adjusting
ARIMA models. Still, the future’s always uncertain. But the errors can help improve ARIMA forecasts. By modeling the errors and incorporating their history into ARIMA’s regressions, there’s a
possibility that we can reduce the error in out-of-sample forecasts going forward. That’s because if you design an ARIMA model correctly, the errors should be randomly distributed around a mean of
zero. In other words, errors through time should cancel each other out. In that case, past error terms can be useful for enhancing ARIMA’s forecasting powers, i.e., reduce the magnitude of the errors
in the future.
Yes, we still need an understanding of economic theory to adjust, interpret and design ARIMA models for predicting GDP and other economic and financial data series. But considering the simplicity and
relative reliability of this econometric technique, ARIMA forecasting is a no-brainer. At the very least, it provides a good benchmark for evaluating other forecasts.
2 Responses to Predicting GDP With ARIMA Forecasts
1. I wouldn’t say “so-called” technique for something every econ grad student learns that’s entirely not controversial.
Technically what you did was without the MA components (unless you did something different with the MA part). Anyway, you don’t really address the whole unit root vs. stationarity problem. The
ideal way to perform this estimation is to take the log changes of X and regress that on its lags. Then you project to horizon in terms of changes. This is because X is I(1) and you need to
convert to I(0) by taking the changes, this is the whole integrated part. You can have problems with longer-term forecasts if you do not do this.
I also do not know how you get your forecast error. I took the SPF and did LN(NGDP2/NGDP1) to get the one quarter ahead forecast and compared that to NGDP (post-revisions of course) and got a
standard deviation of 0.6% for SPF and 0.8% for an AR(4) (also tried AR(2), roughly the same) model on log changes in NGDP. This is from Q4 1968 to present. Of course the SPF may incorporate some
data that was not known at the beginning of a quarter or something, but I wouldn’t try to oversell ARIMA.
2. John,
Yes, there’s much, much more to say about ARIMA and this post barely scratches the surface. That said, I did take the second differences in the GDP changes to detrend the data and satisfy the
weakly stationary requirement. Alas, I failed to mention that in the post, although the calculation I report incorporates this adjustment. As for the differences in our error analysis, that may
be due to focusing on different SPF predictions of GDP. There are several survey forecasts in each quarter and I’m looking only at the mean next quarter prediction. In addition, I also ran the
numbers using alternatives for forecast evaluation, including avg absolute error and root avg squared error and ARIMA still comes out on top. There may be differences in our methodologies for
parameter estimation that accounts for the divergent results. That’s another reminder that model design assumptions are critical if you want to be confident with the output. As for your warning
about overselling ARIMA, I agree, as I noted in the post. It’s a useful technique for comparing and contrasting other predictions. But using ARIMA (or any other forecasting methodology) in a
vacuum is asking for trouble. And, of course, we need to have a good understanding of economic theory to interpret ARIMA’s guesstimates. In short, all the usual caveats apply. | {"url":"http://www.capitalspectator.com/predicting-gdp-with-arima-forecasts/","timestamp":"2014-04-19T09:36:13Z","content_type":null,"content_length":"49579","record_id":"<urn:uuid:72d4e5d5-1ddd-46e7-aec0-23bddea23045>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00078-ip-10-147-4-33.ec2.internal.warc.gz"} |
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What is considered a high standard deviation?
What you
do is compare standard deviations for the behavioral traits of the two species and say that one species shows a larger deviation than the other. By itself, the standard deviation can't be high or
low, without a context. The context is usually provided either by theory (which for something like what you are studying, is hugely inadequate) or by more statistics. If most apecies show a deviation
of about d in some parameter, and a particular species shows a much larger deviation (3d, say), you might be able to say something about that species (eg : they like to screw around
p.s. I haven't taken any real math since high school, so a simple explanation is best. So far, I've read that if you divide the stdev by the mean and get 10% or over, it's considered a high stdev.
That itself must have been written in some specific context. In general, that statement makes little sense. In the standard normal distribution, for instance, the mean is zero. So, by that argument,
any standard normal distribution will have a standard deviation that is too high. | {"url":"http://www.physicsforums.com/showthread.php?t=120767","timestamp":"2014-04-16T22:16:58Z","content_type":null,"content_length":"31570","record_id":"<urn:uuid:2cb9c7b8-ea4e-4ac2-9f53-bec4dce926a8>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00213-ip-10-147-4-33.ec2.internal.warc.gz"} |
On symplectic hypersurfaces
Seminar Room 1, Newton Institute
The Grothendieck-Brieskorn-Slodowy theorem explains a relation between ADE-surface singularities $X$ and simply laced simple Lie algebras $g$ of the same Dynkin type: Let $S$ be a slice in $g$ to the
subregular orbit in the nilpotent cone $N$. Then $X$ is isomorphic to $S\cap N$. Moreover, the restriction of the characteristic map $\chi:g\to g//G$ to $S$ is the semiuniversal deformation of $X$.
We (j.w. Namikawa and Sorger) show that the theorem remains true for all non-regular nilpotent orbits if one considers Poisson deformations only. The situation is more complicated for non-simply
laced Lie algebras. It is expected that holomorphic symplectic hypersurface singularities are rare. Besides the ubiquitous ADE-singularities we describe a four-dimensional series of examples and one
six-dimensional example. They arise from slices to nilpotent orbits in Liealgebras of type $C_n$ and $G_2$. | {"url":"http://www.newton.ac.uk/programmes/MOS/seminars/2011042815301.html","timestamp":"2014-04-19T10:04:11Z","content_type":null,"content_length":"4351","record_id":"<urn:uuid:135432ac-b873-4953-88d0-6f4aada00352>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00168-ip-10-147-4-33.ec2.internal.warc.gz"} |
Tunnel Vision
The most fun you can have on the Internet is to find a beautiful, succinct argument with a conclusion so unexpected it seems like magic. For today’s fun, I am indebted to Michael Lugo, at God Plays
Lugo’s original post is so good it seems almost superflous to paraphrase it, but I can’t resist the temptation.
Drill a tunnel through the earth, from anywhere to anywhere — New York to Maine, or New York to Australia, or wherever else you like. Like so:
Now drop the object of your choice (Lugo suggests a burrito, but you might prefer a gravity-driven train) into the tunnel entrance and wait till it comes out the other side. It’s a standard calculus
problem to calculate how long you’ll have to wait: The answer is 42 minutes, regardless of the length of the tunnel. I’m sure I once found it surprising that the tunnel length doesn’t matter, but
I’ve known it long enough that I now take it in stride. So that’s not how Lugo surprised me.
The surprise is that if you change the size of the earth (while maintaining its density), the answer is still 42 minutes. Whether the earth is the size of a pea or the size of the solar system, it’s
a 42 minute trip from one end of the tunnel to the other. (We’re — quite reasonably — ignoring the effects of relativity here. For an earth that was half the size of the universe, we’d have to make
some corrections.)
Why so? You could, of course, discover this through a direct calculation. But Lugo provides a much slicker argument, namely:
Once you know that the travel time is independent of the tunnel length, it’s clear there are only three things it can depend on: The earth’s density, its radius, and the gravitational constant.
Neither density nor radius have anything to do with time, so we’re really going to have to use that gravitational constant, which is measured in units of
(The first factor is the 32 feet per second squared that you learned about in elementary school; the second factor depends on distance from the center of the earth and the mass of the earth, neither
of which varies in those elementary school problems, so it’s never explicitly mentioned.)
If you multiply that gravitational constant by density (which is pounds per cubic foot), you’ll get a bunch of cancellation, leaving you with
Now you’ve got something that depends only on time! To turn this into an actual number of seconds, you can just invert it and take the square root. At that point, you’re done! (Except maybe for
multiplying by some dimensionless constant.) Any attempt to bring the radius of the earth into the formula is going to introduce units of length, which have no business popping up in a formula for
time. Therefore the earth’s radius must be irrelevant.
Isn’t that cool? And amazing?
(And yes, there are hidden assumptions, e.g. that the solution must be not just a function of mass, length and the gravitational constant, but a rational function — so this is not quite a proof, but
a highly convincing heuristic argument that does give the correct answer. That, I think, in no way detracts from its coolness.)
Lugo’s post is both more concise and more carefully worded, but I thought this slightly longer and less formal version might be helpful to at least a reader or two. Besides, I felt like writing it.
Do you have an equally amazing argument to share?
25 Responses to “Tunnel Vision”
1. 1 1
Thats awesome!
Here is (what I imagine to be) a stupid question.
Surely this only works when the tunnel goes through the middle of the earth? rather than cutting the top as your diagram shows? Wouldn’t gravity drag the item to the side otherwise and it bounce
Also, does this imply something about the speed of a collapsing star? Only density would matter, rather than size?
2. 2 2
Dimensional analysis is a wonderful thing. It should be taught in the opening chapter of every physics textbook.
3. 3 3
@Nick – yes, this assumes you put the burrito on a frictionless rail/elevator or something similar.
As for the collapsing star, the really cool argument starts from the known fact that the time taken through the tunnel is independent of the tunnel length. I’m don’t know, off the top of my head,
any similar amazing symmetry of the dynamics of a collapsing star, but if you do, let’s hear it :-)
4. 4 4
The poem is terrible, but the intent is clear.
5. 5 5
42, eh. Now that rings a bell.
6. 6 6
Interesting outcome, but to nitpick: standard Earth gravity is a bit over 32 ft/sec^2, not 16.
7. 7 7
Unexpected?? It is after all the answer to life, the universe and everything.
8. 8 8
Martin: Fixing this now! Thanks for the catch.
9. 9 9
I don’t understand the argument that the time can’t depend on the radius. With a larger radius, the burrito would move closer to the center of mass more slowly … why could that (intuitively) not
affect the travel time?
10. 10 10
@Phil If you change the radius of the earth, but keep the mass constant, then yes, the burrito moves more slowly. If you change the radius of the earth but keep the *density* constant, the higher
gravity makes the burrito move faster at the start, and it takes the same time even though it has further to go. This all assumes that the earth has constant density throughout.
11. 11 11
Mike H,
The burrito doesn’t necessarily have further to go. It’s (say) 1000 km regardless of the radius of the earth.
By Dr. Landsburg’s exposition, we accept that the time is the same regardless of length, for a fixed radius of earth. Now, how do we intuitively get to the argument that the time is the same for
any radius?
Dr. Landsburg says, “… radius [does not have] anything to do with time.” That’s the part I don’t understand. Why is it so easy to argue that radius has nothing to do with time?
12. 12 12
In SI units g is quite close to 10 (9.8 m/s2), which makes rough calculations simple.
The 42 minutes thing is quite suprising at first, but does make sense – like the period of a pendulum being only dependent on the length and g, however large the swing.
13. 13 13
That’s interesting.
The fact that the earth’s radius is irrelevant can also be seen to be true from the following logic:
1) If, instead of dropping the object from the surface of the earth, you first burrow X miles into the tunnel and drop if from there, the time it takes to reach the point X miles from the end of
the tunnel on the other side must be 42 minutes (i.e. the same as if you did not burrow at all). This is because the object’s motion is simple (simply?) harmonic, and for simple harmonic motion,
the amplitude is irrelevant. E.g. to take another example, if you make a pendulum from a string and a rock, the time it takes the to go from one side to the other is not dependent on how high the
rock swings on either side (roughly speaking, and assuming amplitude is not too high)
2) To modify 1) slightly, once you burrow X miles into the tunnel, you might as well ignore all of the earth’s mass that is farther from the center than you are — this mass is totally irrelevant
to your motion. Thus, you might as well be performing the same experiment on a smaller planet to begin with — and therefore the radius of the planet is irrelevant.
Admittedly this line of reasoning is not as cool as Lugo’s, but perhaps it is the 2nd coolest way of proving the radius is irrelevant.
14. 14 14
> This is because the object’s motion is simple (simply?) harmonic, and for
> simple harmonic motion, the amplitude is irrelevant.
The motion is harmonic because the force on the object is proportional to the distance from the tunnel’s middle point.
15. 15 15
Jonathan Campbell:
Admittedly this line of reasoning is not as cool as Lugo’s,
I’m not sure I agree with that. It’s pretty cool.
16. 16 16
Jonathan Campbell: Outstanding. I like that explanation.
Nick: “Surely this only works when the tunnel goes through the middle of the earth? rather than cutting the top as your diagram shows?”
No, it does not have to go through the center of the earth. That’s what’s so cool about it.
17. 17 17
I think the friction coming from the air would be a far bigger problem. Of course, in an evacuated tube…
18. 18 18
“… radius [does not have] anything to do with time.”
He means, radius is measured in metres/inches/furlongs, time is measured in seconds/years/fortnights.
Radius is a measure of length, with no time units. The only way it could inform a measurement of time is by combining it with another quantity that involves time and length (eg, speed).
19. 19 19
Something from rock climbing that I think is pretty counterintuitive:
A “main anchor” can be built by attaching rope to two independent anchors, and then joining those two pieces of rope together at a point below both anchors. The purpose of this is to share the
load between the two anchors. (Simplified)
The angle at the main anchor determines the percentage of the load that each anchor bears.
-At 0 degrees, each anchor bears 50% of the load.
-At 60 degrees each anchor bears about 58% of the load.
-At 120 degrees each anchor bears 100% of the load, and above 120 degrees each anchor bear GREATER than 100% of the load.
I find it pretty counterintuitive that with a load of X on a main anchor, it is possible to rig the two individual anchors such that they each share a load of greater than X.
Hope that explanation made some sense. If not, check out Wikipedia on climbing anchor equalization.
20. 20 20
Mike H,
If you drop a paperclip, how long it takes to reach the earth depends on the radius of the earth, the density of the earth, and how high you hold the paperclip before dropping it. None of those
have time units either, but the argument “therefore it doesn’t matter how high you hold the paperclip” is invalid.
21. 21 21
However, the gravitational constant does have units of time in it, Phil.
22. 22 22
True! That’s my point. Even though it’s G that has units of time in it, you still need the radius of the earth, density of the earth, and height of the paperclip to determine how long it takes to
So I don’t understand the argument that you don’t need the radius of the earth or density of the earth to know how long it takes for the burrito to travel.
I’m obviously missing something, since everyone agrees the original argument works. I just don’t know what it is that I’m not understanding.
23. 23 23
If the tunnel doesn’t go through the centre of the Earth why won’t the object get stuck at the point closest to the centre? Are you assuming a vacuum?
24. 24 24
Robert Wiblin: Momentum carries it through to the other side. (Ignoring friction.)
25. 25 25
hmm…i wonder if the burrito will still be edible with the heat inside the mantle of the earth XD…
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real analysis limits
January 30th 2009, 12:48 PM #1
Nov 2008
real analysis limits
Can you please help me with these proofs?
(1) Let A on R and let f : A --> R be such that f(x) > 0 for all x on A.
Prove that if lim x -->c f(x) exists and is nonzero, for some c on A, then lim x -->c square root(f(x)) = square root (lim x-->c f(x)).
(2) If f, defined on an interval [a,infinity), is a decreasing function and is bounded below then lim x -->infinity f(x) exists.
Can you please help me with these proofs?
(1) Let A on R and let f : A --> R be such that f(x) > 0 for all x on A.
Prove that if lim x -->c f(x) exists and is nonzero, for some c on A, then lim x -->c square root(f(x)) = square root (lim x-->c f(x)).
(2) If f, defined on an interval [a,infinity), is a decreasing function and is bounded below then lim x -->infinity f(x) exists.
We actualy want to prove:
$\lim_{x\to c}{\sqrt f(x)}= \sqrt m$
For that you must prove: given ε>0 there exists a δ>0 such that:
..........if xεA and 0<|x-c|<δ , then $|\sqrt f(x)-\sqrt m|<\epsilon$. The trick of that case is hiding in the last inequality:
........................... $|\sqrt f(x)-\sqrt m|<\epsilon$.......................1
Ιf we now multiply $|\sqrt f(x)-\sqrt m|$ by $|\sqrt f(x)+\sqrt m|$ and at the same time divide by $|\sqrt f(x)+\sqrt m|$ we get:
$\frac{\ |f(x)-m|}{|\sqrt f(x) + \sqrt m |}$.
Now since f(x)>0 ====> $\sqrt f(x)$>0=====> $\sqrt f(x) +\sqrt m > \sqrt m >0$.
Hence $\frac{\ 1}{|\sqrt f(x) +\sqrt m|}$< $\frac{\ 1}{\sqrt m}$ .......................======>
$\frac{\ |f(x)-m|}{|\sqrt f(x) +\sqrt m|}$< $\frac{\ |f(x)-m|}{\sqrt m}$.
Overall now we have :
$|\sqrt f(x) - \sqrt m|< \frac{\ |f(x) - m|}{|\sqrt f(x) + \sqrt m|} < \frac{\ |f(x) - m|}{\sqrt m}$.................................................. 2
This last inequality is very important to the proof that follows:
Note the work we have done so far can be considered as an aside work.
Now for the proof :
Let ε>0 then $\epsilon.\sqrt m >0$.
But , since $\lim_{x\to c}{f(x)} =m$,then for any E>0 we can find a δ>0 such that :
........................if xεA and 0<|x-c|<δ ,then |f(x)-m|<E...............
Since the above holds for any E>0,it will hold for $\epsilon.\sqrt m=E>0$.
And for that E we can find a δ>0, such that:
...........if xεA and 0<|x-c|< δ,then |f(x)-m|< $\epsilon.\sqrt m$.................................................. .3
Now we are nearly done,so let:
...........xεA and 0<|x-c|<δ.............................................
............then in combination of (2) and (3) we get the desired result:
................................ $|\sqrt f(x) - \sqrt m|<\epsilon$.................................................. .....................
Of course if one can write a more detailed proof the more he/she gets to the inside of the proof,and i think this should be the ultimate goal of every student in analysis,otherwise one may run
into the possibility of ,what you may remember one year you may have forgotten the next.
thanks a lot..this is very helpful..but i still can't solve the 2nd problem ;/
Can you please help me with these proofs?
(1) Let A on R and let f : A --> R be such that f(x) > 0 for all x on A.
Prove that if lim x -->c f(x) exists and is nonzero, for some c on A, then lim x -->c square root(f(x)) = square root (lim x-->c f(x)).
(2) If f, defined on an interval [a,infinity), is a decreasing function and is bounded below then lim x -->infinity f(x) exists.
We want to prove that:
There exists an m such that: $\lim_{ x\to\infty}{f(x)} = m$
Now since f is bounded below and decreasing ,thru the axiom of continuity of real Nos, there exists a real No m such that:
......for all y belonging to the range of f ,yεR(f), $m\leq y$.................................................. ......................................1
......for all ε>0 ,there exists a yεR(f) ,such that $m\leq y< m+\epsilon$.................................................. .......................2
Now let us prove that this m is the limit of f(x) as x goes to infinity.
To prove that,we must prove that:
For all ε>0 there exists a δ>0,such that:
.........if xε[a, $\infty$) and x>δ ,then |f(x)-m|<ε................
So,let ε>0,then according to (2) THERE exists a yεR(f) and $m\leq y< m +\epsilon$.
But since yεR(f) there exists xε[a, $\infty$),and y=f(x).
Here is the trick of the problem.We put for that x=δ,and since x>0 ,δ>0.
(Note we,for simplicity ,consider the case where a>0).
So far we have prove that:
Given ε>0 ,there exists δ>0.It remains to see what happens :
...............if xε[a, $\infty$) and x>δ...........................
But x>δ====> $x\geq\delta$ and since f is decreasing that implies $f(x)\leq f(\delta)$,and according to (2):
...................... $f(x)\leq f(\delta)< m+\epsilon$.................................................. .............................3
Also according to (1) : $m\leq f(x)$,since f(x)εR(f), which implies that m -ε< f(x).............................................. ........................................4
Combining (3) and (4) we get:
.........m-ε<f(x)<m+ε ,which is equivalent to: |f(x) - m|<ε,which is the desired result.
January 31st 2009, 03:52 PM #2
Oct 2008
January 31st 2009, 07:46 PM #3
Nov 2008
February 1st 2009, 04:26 PM #4
Oct 2008
February 1st 2009, 11:07 PM #5
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The Economics of Homebuying
Back To All Lessons
The Economics of Homebuying
A Roof of Your Own
Buying a home, however, is easier and more affordable than you may think. The keys are understanding the procedures and processes of homeownership and developing the skills to reduce the costs of
In this activity, you'll see that there are many programs that help consumers buy a first home. Although some of these programs reduce initial costs, they can raise long-term costs. Next you'll
analyze the economic factors that influence the costs of homeownership. Finally, you will use a mortgage calculator to master the basics of comparing the costs of mortgages. This is important because
the cost of the mortgage is the largest cost of homeownership. Understanding the components of mortgage costs and comparing those costs among several lenders can save you thousands of dollars. Let's
get started.
In this activity, you'll see that there are many programs that help consumers buy a first home. Although some of these programs reduce initial costs, they can raise long-term costs. Next you'll
analyze the economic factors that influence the costs of home ownership. Finally, you will use a mortgage calculator to master the basics of comparing the costs of mortgages. This is important
because the cost of the mortgage is the largest cost of home ownership. Understanding the components of mortgage costs and comparing those costs among several lenders can save you thousands of
dollars. Let's get started.
Read On the Brink. This article on the costs of first-time home-buyers appeared in The Wall Street Journal Classroom Edition in February 1999.
After reading the article, answer the following questions.
Now let's examine the economic factors that influence home sales. You will examine the data in six charts. The scale for the data on these charts is on the left vertical axis. The horizontal axis
tracks these data over the past 29 years. The light blue shaded areas represent economic recessions in the United States. The economic factors that you will examine are the following:
1. Median Household Income. This is the middle level of income. To obtain this number, line up all household incomes in declining order and move down to 50 percent.
2. Median New-Home Prices. Use the same techniques you used for determining median household income. This is the middle level of the price of new homes.
3. Interest Rates. This chart shows the average interest rate on existing homes for each year.
4. Affordability Index. This chart relates median income to home costs. For example, if incomes rise and interest rates fall, homes become more affordable.
5. New-Home Sales. This chart shows the number of new homes sold each year and the number of homes still for sale at the end of the year.
6. Demographics. On this chart, the red line shows the number of people aged 25 to 34, the age when many people start buying homes. The blue line shows the number of people aged 60-69, the age when
many people start selling their homes and moving to smaller apartments or condominiums.
Before analyzing the data in the charts, please read the longer descriptions, which accompany the charts. Now, click here to review the charts.
First let's answer some factual questions.
Now let's analyze the data and see what conclusions we can draw.
Now it's time to shop for a mortgage. There are three keys to reducing mortgage costs. First, the larger the down payment, the lower the principal of the loan. If you borrow less, you will pay less
interest, which is the cost of the loan. Second, the lower the annual interest rate, the less interest you will pay. The annual interest rate is the percentage of the loan (principal) that you pay in
interest each year. Your total payment minus the principal equals the total dollar amount of interest you will pay for the loan. Finally, the faster you pay off the loan, the less interest you will
Let's analyze four different mortgages with different down payments,annual interest rates, and time periods. All of the home buyers are buying a $125,000 house.
These four mortgages are compared in the chart below. Notice that the monthly payment and total payment (principal plus interest) are blank.
You will fill these in, using the mortgage calculator.
│ │ Mortgage 1 │ Mortgage 2 │ Mortgage 3 │ Mortgage 4 │
│ Home price │ $125,000 │ $125,000 │ $125,000 │ $125,000 │
│ Percentage down │ 20% │ 5% │ 20% │ 20% │
│ Total down payment │ 25,000 │ 6,250 │ 25,000 │ 25,000 │
│ Initial principal │ 100,000 │ 118,750 │ 100,000 │ 100,000 │
│ Annual interest rate │ 7% │ 7% │ 6% │ 6.5% │
│ Term │ 30 years │ 30 years │ 30 years │ 15 years │
│ Monthly payment │ │ │ │ │
│ Total payment │ │ │ │ │
Using a mortgage calculator is easy. Just plug in the initial principle,
annual interest rate, and term (in years) for each of the four mortgages. Check monthly payment amounts and total payments. Now go to the Mortgage Calculator and fill in the rest of the chart.
Based on your chart, answer these questions.
Your Housing IQ
Check your Home-buyer IQ Score:
│ 9 - 10 │ Home-buying genius │
│ 7-8 │ Very smart home buyer │
│ 5-6 │ Home maintenance needed │
│ 3-4 │ Have you considered renting? │
│ 0-2 │ Live in your car │
1. How does median income affect home sales? Home prices?
2. Explain the relationship between interest rates and:
A. The affordability index.
B. The prices of new homes.
C. New homes sold.
3. Why do interest rates have this effect on home sales?
4. How do recessions affect home sales? Why?
5. Why did both the number of new homes sold and the prices of new homes sold rise in 1996 and 1997?
6. Compare the number of 25-34-year-olds with the number of 60-69-year-olds in 1989. How will this ratio change in the 2020s? How do you think these changing ratios will affect the housing
1. Kyle and Amber Campbell pay 20 percent down and take out a $100,000 mortgage. It is a 30-year fixed-rate mortgage with an annual interest rate of seven percent.
2. James and Linda Snyder qualify for a special down payment of only five percent. They take out a $118,850 mortgage. It is a 30-year fixed-rate mortgage with an annual interest rate of seven
3. Benny and Silvia Ramirez qualify for a special program that lowers their fixed annual interest rate to six percent. They make a 20 percent down payment and borrow $100,000 for 30 years.
4. Jill Baldwin knows that paying off a mortgage more quickly will save her money. She makes a down payment of 20 percent and borrows $100,000 at a fixed rate of 6.5 percent. Because she will
pay off the mortgage in 15 years, her annual interest rate is lower than she would have paid for a 30-year mortgage.
1. What happens to the monthly payment and total payment for a loan with a smaller down payment?
2. What happens to the monthly payment and total payment for a loan with a lower annual interest rate?
3. What happens to the monthly payment and total payment if the term of the mortgage is 15 years rather than 30 years?
4. What is the trade-off if you get a 15-year mortgage rather than a 30-year mortgage?
5. How can you reduce your cost of buying a home? | {"url":"http://www.econedlink.org/lessons/index.php?lid=121&type=student","timestamp":"2014-04-18T03:09:43Z","content_type":null,"content_length":"24611","record_id":"<urn:uuid:d59c4000-883f-4949-929c-baa284d08732>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00166-ip-10-147-4-33.ec2.internal.warc.gz"} |
Is asymmetric latency common in practice?
up vote 0 down vote favorite
This is a follow-on to a question posted on stack overflow about whether it is possible to determine the one-way latency of a point to point connection without using externally synchronized clocks -
turns out it isn't.
Now the question here is if such asymmetric latencies (where the trip time from A->B is materially different than the trip time B->A) is common in practice? Why?
What about in a typical data center LAN environment? What about across the internet backbone? What about across connections that have asymmetric bandwidth, such as many DSL, satellite or Cable
add comment
4 Answers
active oldest votes
1. Yes.
2. Because asymmetric routing (in it's many and varied forms) as well as asymmetric capacity and asymmetric load, are all common in practice.
3. Not as common, but not unknown.
up vote 2 down vote 4. Very common.
5. Very, very common.
add comment
It's very common if you consider that incoming data may not take the same network route as outgoing to the same server. And you consider that you will nearly always be sharing the network
with other network traffic.
QoS queues can add additional delays in either direction. Other traffic in either direction will affect response times.
up vote 1 Some people still have satellite links that use satellite for download, and modem for upload. It's a very different response time either way.
down vote
On a LAN, not so different latency...
So yes, it's very common in practice. Why did you ask?
I ask because if you want to optimize communication some particular grid/cloud structure of hosts, you'll be able to do it best if you know the one-way latencies of each link. The
one-way latencies are difficult/impossible to calculate in practice, however, so you are left with round-trip times - so I'm curious whether 1/2 the RTT is in practice a good
approximation to the one way latency. – BeeOnRope Dec 23 '09 at 22:38
add comment
To expand on the "Why" question, your request from A -> Z might go:
A -> B -> F -> G -> H -> O -> Z
The next request might go:
A -> B -> E -> I -> J -> O -> Z
up vote 0
down vote And the response may go:
Z -> O -> H -> I -> E -> C -> B -> A
All of the different hops along the way will have different latencies depending on load, outages, etc.
OK, but these just means that latency has some jitter and the whole process has some random deviation on each request - but this also imply that this asymmetry would remain relative
constant (at least in sign) over time? – BeeOnRope Dec 23 '09 at 20:49
Not really. The internet is incredibly fluid, 10 requests sent 0.1 of a second after each other might take 7 different routes to get there, and arrive in a completely different order
than what you sent them. Do it 1 second later and they might arrive in a completely different order – Mark Henderson♦ Dec 23 '09 at 20:53
add comment
It seems to me that it would be possible to determine the one-way latency without the use of externally synchronized clocks if the time deviation between the two hosts is static\constant and
up vote is then factored in to the calculation.
0 down
Can you give details? If you can do it, you'd be able to show up the supposed "proof" on stackoverflow for this very question (link in OP). – BeeOnRope Dec 23 '09 at 20:47
If you "mark" the time on the sender and then "mark" the time on the receiver and then calculate any time difference between the two into your latency measurements then you've measured
your one-way latency without the use of an external clock, everything else being more or less equal. So for instance if the clock on the sender is 5ms faster than the time on the receiver
and you send an ICMP packet from the sender at 00:00:00:00 and the receiver receives it at 00:00:00:20 then simply subtract 5 ms from the receiving end and you've got your one-way latency.
Am I completely off-base? – joeqwerty Dec 23 '09 at 23:23
add comment
Not the answer you're looking for? Browse other questions tagged networking performance tcpip ping latency or ask your own question. | {"url":"http://serverfault.com/questions/96824/is-asymmetric-latency-common-in-practice","timestamp":"2014-04-17T16:14:14Z","content_type":null,"content_length":"78939","record_id":"<urn:uuid:f4e44b7d-6d31-490c-a417-083bc437447a>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00246-ip-10-147-4-33.ec2.internal.warc.gz"} |
Everything you never wanted to know about Jon Watte
Let's assume you're trying to simulate a car driving on an uneven terrain (perhaps some sort of heightmap). Let's assume you know how to measure the distance from the car's chassis to the heightmap
ground at any position (typically, measure the height of the heightmap, and subtract the height of the car at that point in the XZ plane). The car has four wheels at four different positions on the
chassis; these wheels are generally in contact with the ground.
Each contact is a form of physical spring, which affects the car with a force depending on how far the contact is from the chassis. At some length, the force is 0; the wheel lets go of the ground. At
some point, the force is 1/4 the weight of the car times gravity. This is the equilibrium point.
Each spring can be treated as attached to the corners of a box. Each spring, each simulation step, applies force to a simulated rigid body. The trick is that the force is off-center, so it applies
rotational torque to the body.
In general, you can represent a rigid body as:
public class RigidBody {
public Vector3 Position;
public Quaternion Orientation;
public Vector3 LinearVelocity;
public Vector3 AngularVelocity;
public Vector3 Force;
public Vector3 Torque;
public float Mass;
public void Integrate(float dt) {
LinearVelocity += Force / Mass * dt;
Force = Vector3.Zero;
AngularVelocity += Torque / Mass * dt;
Torque = Vector3.Zero;
Position += LinearVelocity * dt;
Orientation += new Quaternion((AngularVelocity * dt), 0) * Orientation;
This assumes that the mass is evenly distributed (like a sphere) in world space; else you'll have to build an inertia tensor for your mass distribution and apply that to the places that now just
divide by Mass.
Now, when you apply an impulse at a (world-space) position to a body, it will apply both force and torque. If I remember the math correctly, the function looks like:
public void ApplyForce(Vector3 forcePosition, Vector3 directionMagnitude) {
float lengthSquared = directionMagnitude.LengthSquared();
if (lengthSquared < 1e-8f) return; // or use your own favorite epsilon
Force += directionMagnitude;
Vector3 distance = forcePosition - Position;
Torque += Vector3.Cross(directionMagnitude, distance);
So, each step, you will run the raycast to detect distance to ground, turn that into a spring force, apply those forces at the positions of the wheels to the body, and then call Integrate() to update
position/orientation of the body. Note that tuning physical simulation systems (spring strength, masses, etc) is notoriously tricky and will probably take many tries. You also usually want to dampen
the springs; in general by adding a counter-force based on the velocity of each of the points in world space. Velocity of a given point is, again if I remember my math right, something like the
public Vector3 PointVelocity(Vector3 worldPoint) {
Vector3 distance = worldPoint - Position;
Vector3 vel = Vector3.Cross(AngularVelocity, distance) + LinearVelocity;
return vel;
The actual rigid body simulation is pretty simple: each update step, add forces/impulses based on the wheel springs (and any collision detection constraints, etc), then integrate the body for the
duration of the time step (typically 1/60th of a second).
However, for a real physics system, you will start doing collision detection, so that the car can't run into walls, or a very hard landing can't push the car underground, etc. At that point, you'll
need collision proxies for the shape of the car (typically, a box), and for the ground (a heightmap or mesh). You'll also formulate the collision constraints using some kind of joint, or contact
constraint. When you start doing that to a number of movable bodies (boxes, traffic cones, and whatnot) you will come to situations where a number of objects are in contact with each other, and
integrating each body in isolation from the others won't be very realistic. At that point, you need a system that solves all the constraints at the same time. Typically, this is formulated as a
linear constraint problem solver, also known as a "big matrix" solver.
Collision detection, and constraint solution, are actually the hard parts of physical simulation. Thus, I recommend you get a good reference book if you want to go down that path. Or just download an
existing engine, and use that; someone else has already done the work of writing and debugging all that gnarly code :-) | {"url":"http://www.enchantedage.com/node/68","timestamp":"2014-04-19T11:56:55Z","content_type":null,"content_length":"17894","record_id":"<urn:uuid:21a69b2c-4b05-4fe8-8614-0164979b6d62>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00274-ip-10-147-4-33.ec2.internal.warc.gz"} |
Examples of Testing Conjectures
• Reasoning About Center and Spread: How do Students Spend Their Time? This activity builds upon the ideas of distribution; shape, center and spread. The lesson begins by having students predict
which daily activities will have a lot and which activities will have a little variation. The activity then has the students examine their choices through the use of computer software. The final
task in the activity has the students use reasoning about distributions to examine graphs and summary statistics to choose variables that have a lot and have a little variability.
• Seeing and Describing the Predictable Pattern: The Central Limit Theorem This activity is designed to develop student understanding of how sampling distributions behave by having them make and
test conjectures about distributions of means from different random samples; from three different theoretical populations (normal, skewed, and multimodal). Students investigate the impact of
sample size and population shape on the shape of the sampling distribution, and learn to distinguish between sample size and number of samples. Students then apply the Empirical Rule (when
appropriate) to estimate the probability of sample means occurring in a specific interval.
• Using Your Hair to Understand Descriptive Statistics: The purpose of this activity is to enhance students' understanding of various descriptive measures in statistics. In particular, students
will gain a visual and hands-on understanding of means, medians, quartiles, and boxplots without doing any computations by completing this activity.
• A ducks story- introducing the idea of testing (statistical) hypotheses: By means of a simple story and a worksheet with questions we pose to the students we guide them from research question to
arriving to a conclusion. The whole process is simply reasoning, no formulas. However we use the reasoning already done by the student to introduce the standard vocabulary of testing statistical
hypotheses (null & alternative hypotheses, p-value, type I and type II error).
• An In-Class Experiment to Estimate Binomial Probabilities: This hands-on activity is appropriate for a lab or discussion section for an introductory statistics class, with 8 to 40 students. Each
student performs a binomial experiment and computes a confidence interval for the true binomial probability. Teams of four students combine their results into one confidence interval, then the
entire class combines results into one confidence interval. Results are displayed graphically on an overhead transparency, much like confidence intervals would be displayed in a meta-analysis.
Results are discussed and generalized to larger issues about estimating binomial proportions/probabilities.
• Independent Samples t-Test: Chips Ahoy??? vs. Supermarket Brand: In this hands-on activity, students count the number of chips in cookies in order to carry out an independent samples t-test to
see if Chips Ahoy??? cookies have a higher, lower, or different mean number of chips per cookie than a supermarket brand. | {"url":"http://serc.carleton.edu/sp/cause/conjecture/examples.html","timestamp":"2014-04-17T21:28:41Z","content_type":null,"content_length":"25789","record_id":"<urn:uuid:9be552f8-3064-46d2-b2e0-a4202246021e>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00508-ip-10-147-4-33.ec2.internal.warc.gz"} |
Life on the lattice
It has been a while since we had any posts with proper content on this blog. Lest my readers become convinced that this blog has become a links-only intellectual wasteland, I hereby want to commence
a new series on algorithms for dynamical fermions (blogging alongside our discussion seminar at DESY Zeuthen/Humboldt University, where we are reading this
review paper
; I hope that is not too lazy to lift this blog above the waste level...).
I will assume that readers are familiar with the most basic ideas of Markov Chain Monte Carlo simulations; essentially, one samples the space of states of a system by generating a chain of states
using a Markov process (a random process where the transition probability to any other state depends only on the current state, not on any of the prior history of the process). If we call the desired
distribution of states Q(x) (which in field theory will be a Boltzmann factor Z
), and the probability that the Markov process takes us to x starting from y P(x,y), we want to require that the Markov process keep Q(x) invariant, i.e. Q(x)=Σ
P(x,y) Q(y). A sufficient, but not necessary condition for this is the the Markov process satisfy the condition of detailed balance: P(y,x)Q(x)=P(x,y)Q(y).
The simplest algorithm that satisfies detailed balance is the Metropolis algorithm: Chose a candidate x at random and accept it with probability P(x,y) = min(1,Q(x)/Q(y)), or else keep the previous
state y as the next state.
Another property that we want our Markov chain to have is that it is ergodic, that is that the probability to go to any state from any other state is non-zero. While in the case of a system with a
state space as huge as in the case of a lattice field theory, it may be hard to design an ergodic Markov step, we can achieve ergodicity by chaining several different non-ergodic Markov steps (such
as first updating site 1, then site 2, etc.) so as to obtain an overall Markov step that is ergodic. As long as each substep has the right fixed-point distribution Q(x), e.g. by satisfying detailed
balance, the overall Markov step will also have Q(x) as its fixed-point distribution, in addition to being ergodic. This justifies generating updates by 'sweeping' through a lattice point by point
with local updates.
Unfortunately, successive states of a Markov chain are not really very independent, but in fact have correlations between them. This of course means that one does not get truly independent
measurements from evaluating an operator on each of those states. To quantify how correlated successive states are, it is useful to introduce the idea of an autocorrelation time.
It is a theorem (which I won't prove here) that any ergodic Markov process has a fixed-point distribution to which it converges. If we consider P(x,y) as a matrix, this means that it has a unique
eigenvalue λ
=1, and all other eigenvalues λ
|) lie in the interior of the unit circle. If we start our process on a state u=Σ
(where v
is the eigenvector belonging to λ
), then P
u = Σ
= c
+ λ
+ ..., and hence the leading deviation from the fixed-point distribution decays exponentially with a characteristic time N
| called the exponential autocorrelation time.
Unfortunately, we cannot readily determine the exponential autocorrelation time in any except the very simplest cases, so we have to look for a more accessible measure of autocorrelation. If we
measure an observable O on each successive state x
, we can define the autocorrelation function of O as the t-average of measurements that are d steps apart: C
, and the integrated autocorrelation time A
(d) gives us a measure of how many additional measurements we will need to iron out the effect of autocorrelations.
With these preliminaries out of the way, in the next post we will look at the Hybrid Monte Carlo algorithm.
Via Jacques Distler: The arXiv now has an API intended to allow web application developers access to all of the arXiv data, search and linking facilities. They have a Blog and a Google group about
it, as well. Anybody wants to guess when we'll see a "My arXiv papers" application for Facebook? | {"url":"http://latticeqcd.blogspot.com/2007_11_01_archive.html","timestamp":"2014-04-18T18:45:45Z","content_type":null,"content_length":"89304","record_id":"<urn:uuid:78b9056a-20e4-4c02-b488-2fbda501d909>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00433-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Introductory and Intermediate Algebra : A Combined Approach
ISBN: 9780201340204 | 0201340208
Format: Paperback
Publisher: Pearson College Div
Pub. Date: 10/1/1999
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Economics Model Answers Two
From Conservapedia
Introductory: 1. In a free market, the price and quantity at which goods are sold are where __________ equals ___________.
supply equals demand
2. Suppose the price demand curve is P = $20 - Q, where P is price and Q is quantity. Also suppose the price supply curve is P = $4 + Q. At what price and quantity will the good be sold?
The good is sold where supply equals demand. That means the supply P must equal the demand P, and the supply Q must equal the demand Q. This can be found by graphing the above two equations and
seeing where they intersect, or by solving them by setting P to the same value in both:
P = $20 − Q
P = $4 + Q
20 − Q = 4 + Q
20 = 4 + 2Q
4 + 2Q = 20
2Q = 16
Q = 8
Plugging this back into the first equation gives P:
P = $20 − Q = $20 − 8 = $12
Checking our work, we plug this P and Q into the second equation:
12 = 4 + 8 = 12
Answer: Q = 8, P = $12
Intermediate: 3. Draw the supply and demand for air. In addition, draw the supply and demand for a good that costs $10000000000000000000000000 trillion dollars.
The demand curve for air must be at a zero price, which is the x-axis (P=0 everywhere, although note that if the quantity of air ever approached zero then the demand price would sharply increase as
people would pay to survive). The supply curve for air must be at fixed quantity, and hence a vertical line. They must intersect at P=0, Q=amount of air in the atmosphere:
Note, however, that air is not a scarce good, and that is why its supply and demand curves are so unusual and nonsensical.
The point where supply equals demand for an extremely expensive good must be at low quantity Q. From there the supply curve would slope upwards, and the demand curve would slope downwards:
(thanks to Kevin for both graphs above)
4. Suppose 1000 persons in a town each have the following weekly demands for gas, and the gas stations have the following weekly supplies:
Gallons Demand Price/gallon Supply Price/gallon
10 $2.50 $.50
20 $2 $.75
30 $1 $1
40 $.75 $1.50
(A) What is the price and overall quantity of gas sold each week?
The price and quantity at which the good (gas) is sold is where supply equals demand. That means that the P for supply equals the P for demand, and the Q for supply equals the Q for demand. Looking
at the above table, where do both the P and Q for supply equal the corresponding P and Q for demand? That occurs on the third line above: Q equals 30 for both supply and demand, and P equals $1 for
both supply and demand. The answer is there: P=$1, Q=30.
(B) Suppose Congress declares war and imposes a price control of $.75 per gallon. At what price and overall quantity will gas sell each week?
When price P is fixed by the government at $.75, then what is the corresponding Q supply and demand? Looking at the table, the corresponding Q in the supply column is 20 gallons, and the
corresponding Q in the demand column is 40 gallons. The lower number is what matters, because nothing can be bought unless it is both supplied and demanded. So the supply is 20 gallons and that is
what is sold at $.75 per gallon. That is 20 times 0.75 = $15 per person, or $15,000 for the whole town. Some may try to buy and sell gas illegally at a higher price, as illegal markets often develop
when there are price controls.
5. Suppose the government limits the supply of Toyota cars that can be imported in 2008 to a certain quota. What effect does this have on the supply curve, and on the equilibrium price? Who is helped
by this import quota, and who is hurt? Be as specific as possible.
Import quotas cause the supply curve to shift upward, and for the equilibrium where supply meets demand to shift to less quantity at a greater price. the consumers are hurt by import quotas, and
everyone pays more for the cars and some people who wanted the car at its lower, free market price, cannot buy it at the higher price. Sellers of Toyota cars are helped because they make bigger
profits per car sold, although they cannot sell as many as before.
6. Suppose you went to see the opening of your favorite new movie, but it is sold out. However, there are four independent scalpers are (illegally) reselling their tickets outside. Four strangers are
each willing to pay the scalpers at most $15, $10, $9 and $8 for a ticket. The scalpers are initially willing to sell each of their tickets for at least $7, $8, $9, and $12. The eight of you get
together and bargain, and then tickets are sold at a common price. What is the price and how many tickets sell at that price? Should scalping be illegal in New Jersey?
One customer (C1) is willing to pay $15. Another customer (C2) is willing to pay $10. Likewise, customers C3 will pay $9 and C4 will pay $8.
One scalper (S1) scalper willing to sell for $7, scalper S2 will sell at $8, S3 will sell at $9, and S4 at $12.
C1, C2, and C3 will all be willing to buy tickets from S1, S2, and S3 at $9. S4 is not willing to sell at the price, and C4 is not willing to buy at the that price, so both choose not to deal. Three
tickets are thus sold at the "common price" or market price of $9 each.
Scalping is a form of free market activity. If there are willing buyers and sellers for the tickets, then why interfere with these sales? Scalping is illegal in New Jersey, but still happens
Honors: Write an essay of about 300 words total on one or more of the following topics: 7. Should government regulations require and monitor honest and full disclosure of information by sellers to
For the buyer to make informed decisions, he must have information. Even "buyer beware" (caveat emptor) assumes a buyer who can find information. The seller has the information, and regulations
requiring it to tell the buyer seem helpful to free enterprise.
8. Should price discrimination be illegal?
For banning price discrimination: a common price encourages more economic activity.
For allowing price discrimination: if buyers and sellers are willing to exchange at different prices, doesn't this maximize overall benefit?
9. What is an economic incentive for excluding homeschooled athletes from high school sports? Should that be legal?
By banning homeschoolers from sports, public and private schools are encouraging students to enroll in their schools. That brings them more money.
10. Describe your view of if or how government should regulate medical prices.
It shouldn't. Price controls, as illustrated by the gasoline problem above, results in shortages.
11. “Economic theory of law and the public domain - When is Piracy Economically Desirable?” See http://lexnet.bravepages.com/media1.html
Piracy of information can increase market activity. Napster caused much additional traffic on the internet in the late 1990s, and this helped the dot-com boom. But copyright violations are illegal in
order to encourage the creation of new works. It is a delicate policy balance between promoting market activity now and promoting creative works for future benefit. | {"url":"http://conservapedia.com/Economics_Model_Answers_Two","timestamp":"2014-04-17T06:44:41Z","content_type":null,"content_length":"20579","record_id":"<urn:uuid:1d5f5615-87e6-457f-b73d-15850761b281>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00348-ip-10-147-4-33.ec2.internal.warc.gz"} |
Help with differential equation
November 24th 2009, 08:22 PM #1
Nov 2009
Help with differential equation
I am having trouble figuring out this problem. I've done differential equation problems before, but this stumps me.
The number of viruses in a bacterium satises the differential equation
= e^(2t)
Solve this equation and sketch some members of the family of solutions for 0 </=t </=1.
What do all the curves have in common (consider their slopes at each point in time)?
Dear jyu484,
Because of the lack of time I attached this file instead of typing the solution. If you have any problems regarding this matter please feel free to ask me.
Hope this helps.
Last edited by Sudharaka; December 29th 2009 at 03:49 AM.
I am having trouble figuring out this problem. I've done differential equation problems before, but this stumps me.
The number of viruses in a bacterium satises the differential equation
= e^(2t)
Solve this equation and sketch some members of the family of solutions for 0 </=t </=1.
What do all the curves have in common (consider their slopes at each point in time)?
Just integrate: $y= \int e^{2t}dt$.
December 28th 2009, 04:04 PM #2
Super Member
Dec 2009
December 29th 2009, 03:59 AM #3
MHF Contributor
Apr 2005 | {"url":"http://mathhelpforum.com/differential-equations/116615-help-differential-equation.html","timestamp":"2014-04-17T16:05:52Z","content_type":null,"content_length":"39931","record_id":"<urn:uuid:f80bd124-5bb6-4ada-a769-d4e2fc7e49cd>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00275-ip-10-147-4-33.ec2.internal.warc.gz"} |
Brief review of polymath1
I don’t have much to say mathematically, or rather I do but now that there is a wiki associated with polymath1, that seems to be the obvious place to summarize the mathematical understanding that
arises in the comments on the various blog posts here and over on Terence Tao’s blog (see blogroll). The reason I am writing a new post now is simply that the 500s thread is about to run out.
So let me quickly make a few comments on how things seem to be going. (At some point in the future I will do so at much greater length.) Not surprisingly, it seems that we have reached a stage that
is noticeably different from how things were right at the beginning. Certain ideas that emerged then have become digested by all the participants and have turned into something like background
knowledge. Meanwhile, the discussion itself has become somewhat fragmented, in the sense that various people, or smaller groups of people, are pursuing different approaches and commenting only
briefly if at all on other people’s approaches. In other words, at the moment the advantage of collaboration is that it is allowing us to do things in parallel, and efficiently because people are
likely to be better at thinking about the aspects of the problem that particularly appeal to them.
Whether there will be a problem with lack of communication I don’t know. But perhaps there are enough of us that it won’t matter. At the moment I feel rather optimistic that we will end up with a new
proof of DHJ(3) (but that is partly because I have a sketch that I have not subjected to appropriately stringent testing, which always makes me feel stupidly optimistic). In fact, what I’d really
like to see is several related new proofs emerging, each drawing on different but overlapping subsets of the ideas that have emerged during the discussion. That would reflect in a nice way the
difference between polymath and more usual papers written by monomath or oligomath.
Finally, a quick word on threading. The largest number of votes (at the time of writing) have gone to allowing full threading, but it is not an absolute majority: those who want unrestricted
threading are outnumbered by those who have voted either for limited threading or for no threading at all. I think that points to limited threading. I’ve allowed the minimum non-zero amount. I can’t
force you to abide by any rules here, but I can at least give my recommendation, which is this. For polymath comments, I’d like to stick as closely as possible to what we’ve got already. So if you
have a genuinely new point to make, then it should come with a new number. However, if you want to give a quick reaction to another comment, then a reply to it is appropriate. If you have a longish
reply, then it should appear as a new comment, but here there is another use of threading that could be very helpful, which is to add replies such as, “I have a counterexample to this conjecture —
see comment 845 below.” In other words, it will allow forward referencing as well as backward referencing. Comments on this post will start at 800, and if yours is the nth reply to comment 8**, then
you should number it 8**.n. Going back to the question of when to reply to a comment, a good rule of thumb is that you should do so only if your reply very much has the feel of a reaction and not of
a full comment in itself.
Also, it isn’t possible to have threading on some posts and not on others, but I’d be quite grateful if we didn’t have threaded comments on any posts earlier than this one. And a quick question: does
anyone know what happens to the threaded comments if you turn the threading feature off again, which is something I might find myself wanting to do?
Ryan O'Donnell Says:
February 23, 2009 at 6:12 pm | Reply
800. Here is a writeup of a Fourier/density-increment argument for Sperner, implementing Terry’s #578 using some of the Fourier calculations Tim and I were doing:
Hope it’s mostly bug-free!
jozsef Says:
February 23, 2009 at 6:14 pm | Reply
Tim – First of all I have to say that I did enjoy to read the blog and to contribute a bit. On the other hand it became less and less comfortable to me posting any comments. You wrote that
“Meanwhile, the discussion itself has become somewhat fragmented, in the sense that various people, or smaller groups of people, are pursuing different approaches and commenting only briefly if at
all on other people’s approaches.”
Let me tell you what is my experience. I’m reading posts using notations from to ergodic theory or Fourier analysis with great interest, but I will rarely make any comment as I’m not expert on that
field. Some post are still accessible to me and some others aren’t. There is no way that I will understand posts referring to noise stability unless I learn something more about it, which I’m not
planning to do in the near future unless it turns out to be crucial for the project. Maybe the blog is still accessible for the majority of readers but I’m finding it more and more difficult to
follow. For the same reasons I feel a bit awkward about posting notes. It’s tempting to work on the problems alone without sharing it with the others but this is clearly against the soul of the
gowers Says:
February 23, 2009 at 6:28 pm | Reply
Metacomment. Jozsef, in the light of what you say, I invite others to give their reactions to how things are going for them. My feeling, as I said in the post, is that we are entering uncharted
territory (or rather, a different uncharted territory from the initial one) and it is not clear that the same rules should still apply. What we have done so far has, in my view, been a wonderful way
of doing very quickly and thoroughly the initial exploration of the problem, and it feels as though the best way of making further progress will be if we start dealing with technicalities and
actually write up some statements formally. I am hoping to do that in the near future, Ryan is doing something along those lines, Terry recently put a Hilbert-spaces lemma on the wiki, etc.
The obvious hazard with this is that people may end up writing things that others do not understand, or can understand only if they are prepared to invest a lot of time and effort. So if we want to
continue to operate as a collective unit, so to speak, it is extremely important for people who write formal proofs to give generous introductions explaining what they are doing, roughly how the
proof goes, why they think it could be useful for DHJ(3) (or something related to DHJ(3)), and so on. This is perhaps an area where people who have been following the discussion without actually
posting comments could be helpful—just by letting us know which bits you find hard to understand. For example, if you see something on the wiki that is clearly meant to be easily comprehensible but
it in fact isn’t, then it would be helpful to know. (However, some of the wiki articles are over-concise simply because we haven’t had time to expand them yet.)
Do others have any thoughts about where we should go from here?
gowers Says:
February 23, 2009 at 6:47 pm | Reply
801. Fourier/Sperner
Ryan, re the remark at the end of the thing you wrote up, if we do indeed have an expression of the form $\mathbb{E}_{p\leq q}\sum_{A\subset[n]}\lambda_{p,q}^{|A|}\hat{f}_p(A)\hat{f}_q(A)$ for the
total weight $\mathbb{E}_{U\subset V}f(U)f(V)$ for $\mathbb{E}_{U\subset V}f(U)f(V)$ when everything is according to equal-slices density, then it seems to me not completely obvious that one couldn’t
prove some kind of positivity result for the contribution from each $A$, especially given your calculation that $\lambda_{p,q}$ splits up as a product. But probably you’ve thought about that—I can
see that it’s a problem that $\lambda_{p,q}$ is not symmetric in p and q, but can one not make it symmetric by talking instead about pairs of disjoint sets?
Or perhaps it’s false that you have positivity and the disjoint-pairs formulation is what you need to get a similar result where it becomes true. I hope this vague comment makes some sense to you.
I’ll see if I can think of a counterexample. In fact, isn’t this a counterexample: let n=1, and let f(0)=-1, f(1)=1. Then the expectation of f(x)f(y) over combinatorial lines (x,y) is -1 and the
expectation of f is 0.
• gowers Says:
February 23, 2009 at 7:25 pm
801.1 Please ignore the last paragraph of this comment — for analytic purposes we need to consider degenerate lines, and as you say the usual proof of Sperner proves positivity. Hmm — not sure
that the threading adds much here.
• Ryan O'Donnell Says:
February 23, 2009 at 8:05 pm
801.2: Yes, it’s not clear to me why $\mathbb{E}_{p \leq q}[\hat{f}_p(A) \hat{f}_q(A)]$ needs to be nonnegative. This is a shame, because: a) we’d be extremely done if we could show this; b) as
$p, q \to 1/2$ the quantity approaches $\hat{f}(A)^2$, which is of course nonnegative. Unfortunately, if you make $p, q$ extremely close to $1/2$ so as to force it nonnegativity, you’ll probably
be swamped with degenerate lines.
Ryan O'Donnell Says:
February 23, 2009 at 8:20 pm | Reply
Metacomment: I sympathise with Jozsef’s position. Luckily, some of the directions we’re working on are in areas I’m familiar with, so I can comment there. But to be honest, I know nothing about
ergodic methods, am extremely shaky on what triangle-removal is, and am only kind of on top of Szemeredi’s regularity lemma. So indeed it takes me quite a while just to read comments on these topics.
That’s fine with me though.
Thing is, even in the areas I’m familiar with it’s awfully tough to keep up with a certain pair of powerhouses who post regularly to the project :) In a perfect world I’d post more definitions, add
more intuition to my comments, update the wiki with stuff about noise sensitivity and so forth — but it already takes me many hours just to quasi-keep up with Tim and Terry and generate
non-nonsensical comments. I’m fine with this too, though.
Finally, I agree that it’s also gotten to the point where it’s sort of impossible to explore certain directions — especially ones with calculations — “online”. You probably won’t be surprised to
learn I spent a fair bit of time working out the pdf document in #800 by myself on paper. I hope this isn’t too much against the spirit of the project, but I couldn’t find any way to do it otherwise.
I would feel like I were massively spamming people if I tried to compute like this online, with all the associated wrong turns, miscalculations, and crazy mistakes I make.
Ryan O'Donnell Says:
February 23, 2009 at 8:25 pm | Reply
802. Sperner/Fourier.
By the way, I’m pretty sure one can also do Roth’s Theorem (or at least, finding “Schur triples” in $\mathbb{F}_2^n$) in this way. It might sound ridiculous to say so, since Roth/Meshulam already
gave a highly elegant density-increment/Fourier-analysis proof.
But the point is that:
. it works in a non-”arithmetic”/”algebraic” way
. it works by doing density increments that restrict just to *combinatorial* subpaces
. it demonstrates that the method can sometimes work in cases where one needs to find *three* points in the set satisfying some relation.
I can supply more details later but right now I have to go to a meeting…
Gil Says:
February 23, 2009 at 8:54 pm | Reply
(If this .1 will automatically creat a threading I will consider it a miracle.)
I think the nature of this collaboration via different people writing blog remarks is somewhat similar to the massive collaboration in ordinary math where people are writing papers partially based on
earlier papers.
Here the time scale is quicker and the nature of contributions is more tentative. Half baked ideas are encourged.
It is not possible to follow everything or even most of the things or even a large fraction of them. (Not to speak about understanding and digesting and remembering). I think it can be useful that
people will not assume that other people read earlier things and from time to time repeat and summarize and overall be clear as possible. (I think comments of the form: “please clarify” can be
useful.) One the other hand, like in usual collaborations people often write things mainly for themselves (e.g. the Hardy-Littlewood rules) and this is very fine.
(A question: is Ryan’s write-up a correct new proof of (weak) Sperner using density incement and Fourier? (The file is called wrong-Sperner but reading it I realize that this means the proof is right
but not yet the “Fourier/Sperner proof from the book” or something like that. Please clarify.)
There are various avenues which I find very interesting even if not directly related to the most promising avenues towards DHJ proof and even some ideas I’d like to share and explore. (One of these
is a non-density-decreasing Fourier strategy.)
One general avenue is various reductions (ususally based on simple combinatorial reasonings) : Joszef mentioned Moser(k=6)-> DHK(k=3) and some reductions to Szemeredi (k=4) were considered.
Summarizing those and pushing for further reductions can be fruitfull.
Also the equivalence of DHJ for various measures (I supose this is also a reduction business) is an interesting aspect worth further study. When there is group acting such equivalences are easy and
of general nature. But for DHJ I am not sure how general they are.
• Ryan O'Donnell Says:
February 23, 2009 at 11:32 pm
Gil wrote: “(The file is called wrong-Sperner but reading it I realize that this means the proof is right but not yet the “Fourier/Sperner proof from the book” or something like that. Please
Yes — I forgot to mention that. I think/hope the writeup’s correct, but I called it “wrong-Sperner” because I still feel there’s something not-from-the-Book about it.
Terence Tao Says:
February 23, 2009 at 10:07 pm | Reply
It certainly does feel that the project is developing into a more mature phase, where it resembles the activity of a highly specialised mathematical subfield, rather than a single large
collaboration, albeit at very different time scales than traditional activity. (The wiki is somewhat analogous to a “Journal of Attempts to Prove Density Hales Jewett”, and the threads are analogous
to things like “4th weekly conference on Density Hales Jewett” :-) .) Also we are hitting the barrier that the number of promising avenues of research seems to exceed the number of people actively
working on the project.
But I think things will focus a bit more (and become more “polymathy”) once we identify a particularly promising approach to the problem (I think we already have a partially assembled skeleton of
such, and a significant fraction of the tools needed have already been sensed). This is what is going on in the 700 thread, by the way; we are focusing largely on one subproblem at a time (right now,
it’s getting a good Moser(3) bound for n=5) and we seem to be using the collaborative environment quite efficiently.
My near-term plan is to digest enough of the ergodic theory proofs that I can communicate in finitary combinatorial language a formal proof of DHJ(2) that follows the ergodic approach, and a very
handwavy proof of DHJ(3). (The Hilbert space lemma I’m working on is a component of the DHJ(2) analysis.) The finitisation of DHJ(2) looks doable, but is already quite messy (more “wrong” than Ryan’s
“wrong” proof of DHJ(2)), and it seems to me that formally finitising the ergodic proof of DHJ(3), while technically possible, is not the most desirable objective here. But there does seem to be some
useful ideas that we should be able to salvage from the ergodic proof that I would like to toss out here once I understand them properly. (For instance, there seems to be an “IP van der Corput lemma”
that lets one “win” when one has 01-pseudorandomness (which roughly means that if one takes a medium dimensional slice of A and then flips one of the fixed digits from 0 to 1, then the pattern of A
on the shifted slice is independent of the pattern of A on the original slice). I would like to understand this lemma better. The other extreme, that of 01-insensitivity, is tractable by Shelah’s
trick of identifying 0 and 1 into a single letter of the alphabet, and the remaining task is to apply a suitable structure theorem to partition arbitrary sets into 01-structured and 01-pseudorandom
components, analogously to how one would apply the regularity lemma to one part of the tripartite graph needed to locate triangles.)
gowers Says:
February 23, 2009 at 10:34 pm | Reply
803. Sperner/Fourier
Re the discussion in 801, this is a proposal for getting positivity. I haven’t checked it, and it could just obviously not work.
The reason I make it is that I just can’t believe the evidence in front of me: the trivial positivity in “physical space” just must be reflected by equally trivial positivity in “frequency space”. I
wonder if the problem, looked at from the random-permutation point of view, is that to count pairs of initial segments it is not completely sensible to count them with the smaller one always first —
after all, the proof of positivity goes by saying that that is actually half the set of pairs of initial segments. So maybe we should look at a sum of the form $\mathbb{E}_{A:B}f(A)f(B)$, where $A:B$
is a hastily invented notation for “$A\subset B$ or $B\subset A$.”
How could we implement this Ryan style? Well, obviously we would average over all pairs of probabilities (p,q). And if we choose a particular pair, then the obvious thing to do would be to choose
independent random variables $(X_1,\dots,X_n)$ and $(Y_1,\dots,Y_n)$ in such a way that the resulting set will give you a probability p of being in A, a probability q of being in B, and a certainty
that A:B. To achieve that in a unified way, for each $i$ we choose a random $t\in [0,1]$ and set $X_i=1$ if and only if $t\leq p$ and $Y_i=1$ if and only if $t\leq q$.
I haven’t even begun to try any calculations here, but I would find it very strange if one didn’t get some kind of positivity over on the Fourier side.
Having said that, the fact that one expands f in a different way for each p does make me not quite as sure of this positivity as I’m pretending to be.
• Ryan O'Donnell Says:
February 23, 2009 at 11:43 pm
803.1 Positivity:
In my experience, obvious positivity on the physical side doesn’t always imply obvious positivity on the Fourier side. Here is an example (although it goes in the reverse direction).
Let $f : \{0,1\}^n \to \mathbb{R}$ be arbitrary and consider $\mathbb{E}[f(x) f(y)]$, where $x$ is uniform random and $y$ is formed by flipping each bit of $x$ with probability $1/5$. (Note that
$(x,y)$ has the same distribution as $(y,x)$, so there’s no asymmetry here.)
Now $x$ and $y$ are typically quite far apart in Hamming distance, and since $f$ can have positive and negative values, big and small, it seems far from obvious that $\mathbb{E}[f(x) f(y)]$
should be nonnegative.
But this is obvious on the Fourier side: it’s precisely $\sum_S (3/5)^{|S|} \hat{f}(S)^2$.
• gowers Says:
February 24, 2009 at 12:05 am
803.2 Positivity
That’s interesting. The non-obvious assertion is that a certain matrix (where the value at (x,y) is the probability that you get y when you start from x and flip) is positive definite. And the
Fourier transform diagonalizes it. And that helps to see why it’s genuinely easier on the Fourier side: given a quadratic form, there is no reason to expect it to be easy to see that it is
positive definite without diagonalizing it. (However, I am tempted to try with your example …)
• gowers Says:
February 24, 2009 at 11:52 am
803.3 Positivity
Got it! Details below in an hour’s time (after my lecture). It will be comment 812 if nobody else has commented by then.
Terence Tao Says:
February 23, 2009 at 11:15 pm | Reply
804. DHJ(2.7)
Randall: I tried to wikify your notes from 593 at
Unfortunately due to a wordpress error, an important portion of your Tex (in particular, part of the statement of DHJ(2.7)) was missing (the portion between a less than sign and a greater than sign –
unfortunately this was interpreted as HTML and thus eaten). So I unfortunately don’t understand the proof. Could you possibly take a look at the page and try to restore the missing portion? Thanks!
• Randall Says:
February 24, 2009 at 5:04 am
Sorry about that. After several missteps, I may have managed to correct it in full….
Terence Tao Says:
February 23, 2009 at 11:22 pm | Reply
805. Sperner
(Previous comment should be 804.)
Ryan’s notes (which, incidentally, might also be migrated to the wiki at some point) seem to imply something close to my Lemma in 578 that had a bogus proof, namely that if f is non-uniform in the
sense that ${\Bbb E} f(x) g(y)$ or ${\Bbb E} g(x) f(y)$ is large for some g, then f has significant Fourier concentration at low modes (i.e. the energy $\sum_{|S| \leq \varepsilon n} |\hat f(S)|^2$
is large), and hence f correlates with a function of low influence (e.g. f correlates with $T_\varepsilon f$, where $T_\varepsilon$ is the Fourier multiplier that multiplies the S fourier coefficient
by $(1-1/\varepsilon n)^{|S|}$).
If this is the case, then (because polynomial combinations of low influence functions are still low influence) one can iterate this in the usual energy-increment manner (or as in the proof of the
regularity lemma) to obtain a decomposition $1_A = f_{U^\perp} + f_U$, where $f_U$ is uniform and $f_{U^\perp}$ has low influence, is non-negative and has the same density as $1_A$. (This is an
oversimplification; there are error terms, and one has to specify the scales at which uniformity or low influence occurs, but ignore these issues for now.)
If this is the case, then the Sperner count ${\Bbb E} 1_A(x) 1_A(y)$ should be approximately equal to ${\Bbb E} f_{U^\perp}(x) f_{U^\perp}(y)$. But if $f_{U^\perp}$ is low influence, this should be
approximately ${\Bbb E} f_{U^\perp}(x)^2$. Meanwhile, we have ${\Bbb E} f_{U^\perp}(x) = \delta$, so by Cauchy-Schwartz we get at least $\delta^2$, as desired.
• Terence Tao Says:
February 23, 2009 at 11:24 pm
p.s. Something very similar goes on in the ergodic proof of DHJ(2). What I call $f_{U^\perp}$ here will be called $P 1_A$, where P is a certain “weak limit” of shift operators. The key point is
that P is going to be an orthogonal projection in L^2 (this is the upshot of a Hilbert space lemma that I am working on on the wiki.)
Ryan O'Donnell Says:
February 24, 2009 at 12:01 am | Reply
806. Sperner.
Re #805, Terry, one thing I had trouble with was — roughly speaking — showing that indeed $\mathbb{E}[f_{U^\bot}(x) f_{U^\bot}(y)]$ is close to $\mathbb{E}[f_{U^\bot}(x)^2]$.
It seemed logical that this would be true but I had technical difficulties actually making the switch from $y$ to $x$. In some sense that’s why I wrote the notes, to really convince myself that one
could do something. Unfortunately, that something was to define $f_{U^\bot}$ all the way down to just the lowest mode; i.e., to look at density rather than energy.
Perhaps we could try to think about, at a technical level, how to really pass from $(x,y)$ to $(x,y)$ for low-influence functions…
• Terence Tao Says:
February 24, 2009 at 1:49 am
806.2 Sperner
Well, I would imagine that $f_{U^\perp}$ would have Fourier transform concentrated in sets S of size $\varepsilon n$, and this should mean that the average influence of $f_{U^\perp}$, i.e. {\Bbb
E} |f_{U^\perp}(x) – f_{U^\perp}(y)|^2, where x and y differ by just one bit, should be small (something like $O(\varepsilon)$). This should then be able to be iterated to extend to x and y being
several bits apart rather than just one bit. (In order to do this properly, one will probably need to have a whole bunch of different scales in play, not just a single scale $\varepsilon n$.
There are various pigeonhole tricks that let one find a whole range of scales at which all statistics are the same [I call this the "finite convergence principle" in my blog], so as a first
approximation let’s pretend that concentration at one scale $\epsilon n$ is the same as concentration at other small scales such as $\epsilon^2 n$. The point is that while the L^2 energy of f
could be significant in the range $\epsilon^2 n \leq |S| \leq \epsilon n$ for a single $\varepsilon$, the pigeonhole principle prevents it from being significant for all such choices of $\
Ryan O'Donnell Says:
February 24, 2009 at 12:08 am | Reply
807. Fourier/density-increment.
In #802, I guess I’m getting the terminology all wrong. I should say that another problem I think the same method will work for is the problem in $\mathbb{F}_2^n$ where you’re looking for “lines” of
length 3 where in the wildcard coordinates you’re allowed either (0,1,1), (1,0,1), or (1,1,0).
I don’t know if this problem has a name. It’s not quite a Schur triple. I personally would call it the “Not-Two” problem because in every coordinate you’re allowed (0,0,0), (1,1,1), (0,1,1), (1,0,1),
or (1,1,0): anything where the number of 0′s is “Not Two”.
• Gil Says:
February 24, 2009 at 12:59 am
A question and 3 remarks:
Q: I remember when we first talked about it there were obvious difficulties for why fourier+increasing density strategy wont work. At the end what is the basic change that make it works is it
changing the measure?
R1: sometimes is is equally convenient (and even more) not to work with $\mu_p$ measures and their associated Fourier transform but to consider functions on [0,1]$n$ and use the Walsh orthonormal
basis even for every [0,1]. So given p we represent our set by a subset of the solid cube and use the “fine” Walsh transform all the time.
R2: If changing the measure can help when small Fourier coefficients do not suffice for quasirandomness, is there hope it will help for case k=4 of Szemeredi?
R3: there are some problems regarding influences threshold phenomena where better understanding of the relation between equal sliced density and specific $\mu_p$ densities are needed. Maybe some
of the tools from here can help.
Gil Says:
February 24, 2009 at 1:13 am | Reply
808. Another Fourier/Sperner approach
Let me mention briefly a non density-increasing approach to Sperner. You first note that not having lines with 1 wild card implies that $\sum \hat f^2 (S)|S| = \hat f^2(S) (n-|S|)$. If you can show
that the non empty fourier coefficients are concentrated (in terms of |S|) then I think you can conclude that the density is small.
You want to control expression of the from $\hat f^2(S) {{n-|S|} \choose {k}}$. Those are related to the number of elements in the set when we fixed the value of n-k variables and look over all the
rest. These numbers are controlled by the density of Sperner families for subsets of {1,2,…,k}. So this looks like circular thing but maybe it allows some bootstrapping.
Ryan O'Donnell Says:
February 24, 2009 at 5:42 am | Reply
809. Response to 806.2 (Sperner).
Terry wrote, “This should then be able to be iterated to extend to x and y being several bits apart rather than just one bit.”
Hmm, but one source of annoyance is that y is not just x with $\epsilon n$ or so bits flipped — it’s x with $\epsilon n$ or so 0′s flipped to 1′s. As you get more and more 1′s in there, the
intermediate strings are less and less distributed as uniformly random strings. So it’s not 100% clear to me that you can keep appealing to average influence, since influence as defined is a
uniform-distribution concept.
Overall I agree that this is probably more of a nuisance then a genuine problem, but it was stymieing me a bit so I thought I’d ask.
• Terence Tao Says:
February 24, 2009 at 6:07 am
809.2 Sperner and influence
Ryan, I think the trusty old triangle inequality should deal with things nicely. (I would need a binomial model of bit-flipping rather than a Poisson model, but I’m sure this makes very little
Suppose $f_{U^\perp}$ has average influence $\mu$; thus flipping a bit from a 0 to 1 or vice versa would only affect things by $\mu$ on the average. Conditioning, we conclude that if we flip just
one bit from a 0 to 1, we expect $f_{U^\perp}$ to change by at most $2 \mu$ on the average. Now we iterate this process $\varepsilon n$ times. As long as $\varepsilon n$ is much less than $\sqrt
{n}$ (as in your notes), there is no significant distortion of the underlying probability distribution and we see that if y differs by $\varepsilon n$ 0->1 bits from x, then $f_{U^\perp}(y)$ and
$f_{U^\perp}(x)$ will differ by $O( \varepsilon n \mu )$ on the average.
This will be enough for our purposes as long as the influence $\mu$ is a really small multiple of $1/\varepsilon n$. One can’t quite get this by working at a single scale – one gets $O( 1/\
varepsilon n )$ instead – but one can do so if one works with a whole range of scales and uses the energy increment argument. I’ve sketched out the details on the wiki at
Terence Tao Says:
February 24, 2009 at 6:27 am | Reply
810. DHJ(2.7)
Randall, thanks for cleaning up the file! It is a very nice proof, and I’d like to try to describe it in my own words here.
DHJ(2.7) is the strengthening of DHJ(2.6) that gives us three parallel combinatorial lines (i.e. they have the same wildcard set), the first of which hits the dense set A in the 0 and 1 position, the
second hits it in the 1 and 2 position, and the third in the 2 and 0 position.
To describe Randall’s argument, I’d first like to describe how Randall’s argument gives yet another proof of DHJ(2) which is quite simple (and gives civilised bounds). It uses the density increment
argument: we want to prove DHJ(2) at some density $\delta$ and we assume that we’ve already proven it at any significantly bigger density.
Now let A be a subset of $[2]^n$ of density $\delta$ for some n. We split n into a smallish r and a biggish n-r, thus viewing $[2]^n$ as a whole bunch of (n-r)-dimensional slices, each indexed by a
word in $[2]^r$.
If any of the big slices has density significantly bigger than $\delta$, we are done; so we can assume that all the big slices have density not much larger than $\delta$ (e.g. at most $\delta + \
delta^2 / 2^r$). Because the total density is $\delta$, we can subtract and conclude that all the big slices have density close to $\delta$.
Now we look at the r+1 slices indexed by the words $0^r, 0^{r-1} 1, 0^{r-2} 1^2, \ldots, 0 1^{r-1}, 1^r$. Each of these slices intersects the set A with density about $\delta$. Thus by the pigeonhole
principle, if r is much bigger than $1/\delta$, then two of the A-slices must share a point in common, i.e. there exists $w \in [2]^{n-r}$ and i,j such that $0^i 1^{r-i} w, 0^j 1^{r-j} w$ both lie in
A. Voila, a two-dimensional line.
The same argument gives DHJ(2.7). Now there are three different graphs on r+1 vertices, the 01 graph, the 12 graph, and the 20 graph. i and j are connected on the 01 graph if the $0^i 1^{r-i}$-slice
and $0^j 1^{r-j}$-slice of A have a point in common; similarly define the 12 graph and the 20 graph. Together, this forms an 8-coloured graph on r+1 vertices. By Ramsey’s theorem, if r is big enough
there is a monochromatic subgraph of size $\gg 1/\delta$. But by the preceding pigeonhole argument, we see that none of the 01 graph, the 12 graph, or the 20 graph can vanish completely here, so they
must instead all be complete, and we get three parallel combinatorial lines, each intersecting A in two of the three positions.
ryanworldwide Says:
February 24, 2009 at 6:53 am | Reply
811. Prelude to some more density-increment arguments.
I was hoping to give an illustration of the #578 technique for a problem in ${[3]^n}$. But it got too late for me to finish writing it, so I’ll just give the “background info” I managed to write.
This probably ought to go in the wiki rather than in a post, but having spent a bit of time to get Luca’s converter working for me, I didn’t have the energy to also convert to the third, wiki format.
More on the #578 method tomorrow.
Here, then, some basics on noise and correlated product spaces; for more, see e.g. this paper of Mossel.
Let ${\Omega}$ be a small finite set; for example, ${\Omega = [3]}$. Let ${\mu}$ be a probability distribution on ${\Omega}$. Abusing notation, write also ${\mu = \mu^{\otimes n}}$ for the
corresponding product distribution on ${\Omega^n}$. For ${0 \leq \rho \leq 1}$, define the noise operator ${T_\rho^\mu}$, which acts on functions ${\Omega^n \to {\mathbb R}}$, as follows: $\
displaystyle (T_\rho^\mu f)(x) = \mathop{\bf E}_{{\bf y} \sim^\mu_\rho x}[f({\bf y})],$
where ${{\bf y} \sim^\mu_\rho x}$ denotes that ${{\bf y}}$ is formed from ${x}$ by doing the following, independently for each ${i \in [n]}$: with probability ${\rho}$, set ${{\bf y}_i = x_i}$; with
probability ${1 - \rho}$, draw ${{\bf y}_i}$ from ${\mu}$ (i.e., “rerandomize the coordinate”). (I use boldface for random variables.)
Here are some simple facts about the noise operator:
Fact 1: ${\mathop{\bf E}_{{\bf x} \sim \mu}[T_\rho^\mu f({\bf x})] = \mathop{\bf E}_{{\bf x} \sim \mu}[f({\bf x})]}$.
Fact 2: For ${p \geq 1}$ we have ${\|T_\rho^\mu f\|_{\mu,p} \leq \|f\|_{\mu,p}}$, where ${\|f\|_{\mu,p}}$ denotes ${\mathop{\bf E}_{{\bf x} \sim \mu}[|f({\bf x})|^p]^{1/p}}$.
Fact 3: The ${T^\mu_\rho}$‘s form a semigroup, in the sense that ${T^\mu_{\rho_1} T^\mu_{\rho_2} = T^\mu_{\rho_1 \rho_2}}$.
Fact 4: ${\mathop{\bf E}_{{\bf x} \sim \mu}[f({\bf x}) \cdot T_\rho^\mu f({\bf x})] = \mathop{\bf E}_{{\bf x} \sim \mu}[(T_{\sqrt{\rho}}^\mu f({\bf x}))^2]}$. We use the notation ${\mathbb{S}_\rho^\
mu(f)}$ for this quantity.
Fact 5: ${\mathbb{S}_{1-\gamma}^\mu(f) = \mathop{\bf E}_\mu[(T_{\sqrt{1 - \gamma}}^\mu f)^2] = \mathop{\bf E}_{{\bf V}}[\mathop{\bf E}_\mu[f|_{{\bf V}}]^2]}$, where: a) ${{\bf V}}$ is a randomly
chosen combinatorial subspace, with each coordinate being fixed (from ${\mu}$) with probability ${1 - \gamma}$ and left “free” with probability ${\gamma}$, independently; b) ${f|_{{\bf V}}}$ denotes
the restriction of ${f}$ to this subspace, and ${\mathop{\bf E}_\mu[f|_{{\bf V}}]}$ denotes the mean of this restricted function, under ${\mu}$ (the product distribution on the free coordinates of $
{{{\bf V}}}$).
gowers Says:
February 24, 2009 at 1:32 pm | Reply
812. Positivity
This concerns Ryan’s example of a quantity that is easily seen to be non-negative on the Fourier side, and not so easily seen to be non-negative in physical space. Except that I now have an easy
argument in physical space. I don’t know how relevant this is, but there is some potential motivation for it, which is that perhaps the simple idea behind the proof could help in Ryan’s quest for a
“non-wrong” proof of Sperner.
The problem, laid out in 803.1, was this. Let $f$ be a function defined on $\{0,1\}^n$. Now choose two random points $x,y$ in $\{0,1\}^n$ as follows: $x$ is uniform, and $y$ is obtained from $x$ by
changing each bit of $x$ independently with probability 1/5. Now show that $\mathbb{E}f(x)f(y)$ is non-negative for all real functions $f$.
There are various ways of describing a proof of this. Here is the one I first thought of, and after that I’ll give a slicker version. I’m going to do a different calculation. Again I’ll pick a random
$x$, but this time I’ll create two new vectors $y$ and $z$ out of $x$, which will be independent of each other (given $x$) and obtained from $x$ by flipping coordinates with probability $\lambda$.
And I’ll choose $\lambda$ so that $2\lambda(1-\lambda)=1/5$, which guarantees that the probability that the $i$th coordinates of $y$ and $z$ are equal is $1/5$. Then I’ll work out $\mathbb{E}f(y)f(z)
$. It’s obviously positive because it is equal to $\mathbb{E}_x(\mathbb{E}[f(y)|x])^2$. But also, the joint distribution of $y$ and $z$ is equal to the joint distribution of $x$ and $y$, since $y$ is
uniform and the digits of $z$ are obtained from $y$ by flipping independently with probability $1/5$. (Proof: the events $y_i=z_i$ are independent and have probability 4/5.)
Now for the second way of seeing it. This time I’ll generate $y$ from $x$ in two stages. First I’ll flip each digit with probability $\lambda$. And then I’ll do that again. Now each time you flip
with probability $\lambda$ you are multiplying $f$ by the symmetric matrix $A$ defined by $A_\lambda(x,y)=\binom n{|x-y|}\lambda^{|x-y|}(1-\lambda)^{n-|x-y|}$ (which is the same as convolving $f$ by
the function $w(z)=\binom n{|z|}\lambda^{|z|}(1-\lambda)^{n-|z|}.)$ So we end up with $\langle f,A_\lambda^2f\rangle$, which, since $A_\lambda$ is symmetric, is $\langle A_\lambda f,A_\lambda f\
Note that this proof fails if you flip with probability p greater than 1/2, because then you can’t solve the equation $2\lambda(1-\lambda)=p$. But that’s all to the good, because the result is false
when $p>1/2$. When $p=1/2$ it is “only just true” since flipping with probability 1/2 gives you the uniform distribution for y, so if $f$ averages zero then $\mathbb{E}f(x)f(y)$ is zero too.
I’m sure there’s yet another way of looking at it (and in fact these thoughts led to the above proof) which is to think of the random flipping operation as the exponential of an operation where you
flip with a different probability. (To define that, I would flip N times, each time with probability $c/N$, and take limits.) I would then expect some general nonsense about exponentials to give the
positivity. In the end I have taken a square root instead.
• ryanworldwide Says:
February 24, 2009 at 4:06 pm
812.1. Thanks Tim! I believe this proof is in fact Fact 4 from #811 (with its $\rho = 3/5$).
• ryanworldwide Says:
February 24, 2009 at 4:27 pm
812.2. I was pleased with myself for figuring out the analogous trick in our equal-slices setting… until I realized it was exactly what you said way back in #572, first paragraph. Drat.
Sune Kristian Jakobsen Says:
February 24, 2009 at 1:50 pm | Reply
813. An optimistic conjecture.
The hyper-optimistic conjecture says that $c_n{\mu}\leq \overline{c}_n{\mu}$. Here I would like to suggest an “optimistic conjecture”:
There exist a number $C$ such that $c_n{\mu}\leq C\overline{c}_n{\mu}$ for all $n$.
The hyper-optimistic conjecture implies the optimistic conjecture and the optimistic conjecture implies DHJ.
Let $A_{\epsilon}$ be the set of elements x in A, such that A has measure at least $\epsilon$ in the slice x belong to.
We know that if A is a line-free set, the sum of the measure of A in the slices $\Gamma_{a+r,b,c}, \Gamma_{a,b+r,c}, \Gamma_{a,b,c+r}$ is at most 2. In particular A cannot contain more than 2/3 of
each of three slices in an equilateral triangle. Thus, the slices where A has a density greater than 2/3 forms a triangle-free subset of $\Delta_n:= \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \}$. So, if
the optimistic conjecture is false, we know that for biggest line-free set A, the density of $A_{2/3+\epsilon}$ in A go to 0 as $n\to \infty$.
I think the following would imply DHJ:
For every $\epsilon$ there exist a C such that for every set A with $A=A_{\epsilon}$ (no element in A is “epsilon-lonesome” in its slice) there exist a equilateral triangle-free set $D\subset \
Delta_n$ with more than $\mu(A)/C$ elements.
ryanworldwide Says:
February 24, 2009 at 4:11 pm | Reply
814. Re Sperner & Influences & 809.2.
Hi Terry, not to be ridiculously nitpicky, but could I ask one more question?
It seems to me that:
a) $f_{U^\bot}$ might end up having range $[0,1]$ rather than $\{0,1\}$;
b) average influence is defined in terms of the *squared* change in $f_{U^\bot}$;
c) we don’t have a triangle inequality for $\ell_2^2$. (It wouldn’t be a problem if $f_{U^\bot}$ had range $\{0,1\}$ since $\ell_2^2$ is a metric on $\{0,1\}$.)
• ryanworldwide Says:
February 24, 2009 at 4:33 pm
814.1. Maybe one gets around it by playing with the scales & using triangle inequality for $\ell_2$; I’m thinking about it…
• ryanworldwide Says:
February 24, 2009 at 5:10 pm
814.2. Oops.
That was a dumb question, as it turns out. Assuming $f_{U^\bot}$ is bounded, squared-differences are bounded by 4 times absolute-value-differences, and then use the triangle inequality. Got it.
ryanworldwide Says:
February 24, 2009 at 5:56 pm | Reply
815. Re Sperner.
I think I can answer my own question from #814. I’ll say what Terry was saying in #809.2, being a little bit imprecise. Let ${T \ll \sqrt{n}}$ be an integer. Define random variables ${x_0, \dots,
x_T}$ as follows: ${x_0}$ is a uniformly random string; ${x_{t}}$ is formed from ${x_{t-1}}$ by picking a random coordinate and, if that coordinate is ${0}$ in ${x_{t-1}}$, changing it to a ${1}$.
Now the distribution on ${(x_0, x_T)}$ is pretty much like the distribution on ${(x,y)}$ from before, with ${{\epsilon} = T/n}$. Probably, as Terry says, one should ultimately wrap a Poisson choice
of ${T}$ around this entire process.
Anyway, for fixed ${T}$, by telescoping we have $\displaystyle \mathop{\bf E}[f(x_0)g(x_T)] = \mathop{\bf E}[f(x_0)g(x_0)] + \sum_{t=1}^T \mathop{\bf E}[f(x_0)(g(x_t) - g(x_{t-1}))]. \ \ \ \ \ (1)$
To bound the “error term” here, use Cauchy-Schwarz on each summand. For a given ${1 \leq t \leq T}$ we have $\displaystyle |\mathop{\bf E}[f(x_0)(g(x_t) - g(x_{t-1}))]| \leq \|f\|_2 \sqrt{\mathop{\bf
E}[(g(x_t) - g(x_{t-1}))^2]}. \ \ \ \ \ (2)$
Let ${Mg}$ be the function defined by ${Mg(y) = \mathop{\bf E}_{y'}[(g(y') - g(y))^2]}$, where ${y'}$ is formed from ${y}$ by taking one step in the Markov chain we used in defining the ${x_t}$‘s.
Hence the expression inside the square-root in (2) is ${\mathop{\bf E}[Mg(x_{t-1})]}$. We would like to say that this expectation is close to ${\mathop{\bf E}[Mg(x_0)]}$ because the distributions on
${x_0}$ and ${x_{t-1}}$ are very similar. We can use the argument in my “wrong-Sperner” notes for this. Using ${t \leq T \ll \sqrt{n}}$, I think that’ll give something like $\displaystyle |\mathop{\
bf E}[Mg(x_{t-1})] - \mathop{\bf E}[Mg(x_{0})]| \leq O(t/\sqrt{n}) \|Mg(x_0)\|_2 \leq O(t/\sqrt{n}) \sqrt{\mathop{\bf E}[Mg(x_0)]},$
where on the right I’ve now assumed that ${g}$ is bounded, hence ${Mg}$ is bounded, hence ${(Mg)^2 \leq Mg}$ pointwise.
But ${\mathop{\bf E}[Mg(x_0)]}$ is precisely (well, maybe up to a factor of ${2}$ or something) the “energy” or “average influence” of ${g}$; write it as ${\mathcal{E}(g)}$. So we get $\displaystyle
\mathop{\bf E}[(g(x_t) - g(x_{t-1}))^2] = \mathcal{E}(g) \pm O(t/\sqrt{n}) \sqrt{\mathcal{E}(g)}.$
Let’s assume now that in fact ${T \leq \sqrt{\mathcal{E}(g)} \sqrt{n}}$. Then the error in the above is essentially negligible. Now plugging this into (2) we get that each of the ${T}$ error terms in
(1) is at most ${O(1) \|f\|_2 \sqrt{\mathcal{E}(g)}}$. So overall, we conclude $\displaystyle \mathop{\bf E}[f(x_0)g(x_T)] = \mathop{\bf E}[f(x_0)g(x_0)] \pm O(T \cdot \|f\|_2 \cdot \sqrt{\mathcal{E}
(g)}). \ \ \ \ \ (3)$
Just to repeat, the assumptions needed to deduce (3) were that ${g}$ is bounded and that ${T \leq \sqrt{\mathcal{E}(g)} \sqrt{n}}$.
• Terence Tao Says:
February 24, 2009 at 8:10 pm
815.1 Sperner
Ryan, thanks for fleshing in the details and sorting out the l^1 vs l^2 issue. I just wanted to point out that we may have located the “DHJ(0,2)” problem that Boris brought up a while back –
namely, DHJ(0,2) should be DHJ(2) for “low influence” sets. I think we now have a satisfactory understanding of the DHJ(2) problem from an obstructions-to-uniformity perspective, namely that
arbitrary sets can be viewed as the superposition of a low influence set (or more precisely, a function $f_{U^\perp}$) plus a DHJ(2)-uniform error.
DHJ(1,3) may now need to be tweaked to generalise to functions f which are polynomial combinations of 01-low influence, 12-low influence, and 20-low influence functions, where 01-low influence
means that f(x) and f(y) are close whenever y is formed from x by flipping a random 0 bit to a 1 or vice versa, etc. (With our current definition of DHJ(1,3), we are considering products of
indicator functions which have zero 01-influence, zero 12-influence, and zero 20-influence respectively.)
gowers Says:
February 24, 2009 at 6:47 pm | Reply
816. Obstructions.
I hope I’m going to have time to do some serious averaging arguments this evening. They will be private calculations (though made public as soon as they work), but let me give an outline of what I am
hoping to make rigorous. The broad aim is to prove that a set with no combinatorial lines has a significant local correlation with a set of complexity 1. This is a statement that has been sort of
sketched in various comments already (by Terry and by me and possibly by others too), but I now think that writing it out in a serious way will be a very useful thing to do and should get us
substantially closer to a density-increment proof of DHJ(3). Here is a rough description of the steps.
1. A rather general argument to prove that whenever we feel like restricting to a combinatorial subspace, we can always assume that it has density at most $\delta-\eta$, where $\eta$ is an arbitrary
function of $\delta$. This kind of argument is standard, and has already been mentioned by Terry: basically if you can find a subspace with increased density then you happily pass to that subspace
and you’ve already completed your iteration. If you can’t do that, then the proportion of subspaces (according to any reasonable distribution of your convenience) with substantially smaller density
is tiny. The minor (I hope) technical challenge is to get a version of this principle that is sufficiently general that it can be used easily whenever it is needed.
2. Representing combinatorial lines as (U,V,W), we know that for an average W (each element chosen with probability 1/3 — I’m going for uniform measure here) the density of (U,V) such that (U,V,W) is
in $\mathcal{A}$ is $\delta$.
3. Also, by 1, for almost all W we find that the set of points $(U,V,W\cup W')$ where $|W'|\leq m$ (for some $m$ with $1<<m<<n$) has density almost $\delta$.
4. Combining 2 and 3, we obtain a W such that the density of (U,V) with (U,V,W) in $\mathcal{A}$ is at least $\delta/2$, say, and the density of $(U,V,W\cup W')\in\mathcal{A}$ is almost as large as $
\delta$ (at least).
5. But the points of the latter kind have to avoid those $U$ and $V$ for which the point $(U,V,W)\in\mathcal{A}$, which is a dense complexity-1 obstruction in the “unbalanced” set of $(U,V,W')$ that
have union ${}[n]\setminus W$ and have $W'$ of size at most $m$.
6. Randomly restricting, we can get a similar statement, but this time for a “balanced” set—that is, one where $U',V',W'$ have comparable sizes.
7. From that it is straightforward to get a (local) density increment on a special set of complexity 1 (as defined here).
Now I think I’ve basically shown that a special set of complexity 1 contains large combinatorial subspaces, though I need to check this. But what we actually need is a bit stronger than that: we need
to cover sets of complexity 1 uniformly by large combinatorial subspaces, or perhaps do something else of a similar nature. (This is where the Ajtai-Szemerédi proof could come in very handy.) But if
everything from 1 to 7 works out—I’m not sure how realistic a hope that is but even a failure would be instructive—then we’ll be left with a much smaller-looking problem to solve. I’ll report back
when I either get something working or see where some unexpected difficulty lies.
ryanworldwide Says:
February 24, 2009 at 8:27 pm | Reply
817. Sperner.
Just to clarify, the thing I wrote in #800 says that if you don’t mind restricting to combinatorial subspaces (which we usually don’t, unless we’re really trying to get outstanding quantitatives),
then the decomposition $f = f_{U^\bot} + f_U$ we seek can be achieved trivially: you just take $f_{U^\bot} = \mathbb{E}[f]$.
Terence Tao Says:
February 25, 2009 at 2:35 am | Reply
818. Ergodic proof of DHJ(3)
I managed to digest Randall’s lecture notes on the completion of the Furstenberg-Katznelson proof of DHJ(3) (the focus of the 600 thread) to the point where I now have an informal combinatorial
translation of the argument at
that avoids any reference to infinitary concepts, at the expense of rigour and precision. Interestingly, the argument is morally based on a reduction to something resembling DHJ(1,3), but more
complicated to state. We are trying to get a lower bound for
${\Bbb E} f(\ell(0)) f(\ell(1)) f(\ell(2))$ (1)
where f is non-negative, bounded, and has positive density, and $\ell$ ranges over all lines with “few” wildcards (and I want to be vague about what “few” means). The first reduction is to eliminate
“uniform” or “mixing” components from the second two factors and reduce to
${\Bbb E} f(\ell(0)) f_{01}(\ell(1)) f_{20}(\ell(2))$ (2)
where $f_{01}$, $f_{20}$ are certain “structured” components of f, analogous to $f_{U^\perp}$ from the Sperner theory. They have positive correlation with f, and in fact are positive just about
everywhere that f is positive.
What other properties do $f_{01}, f_{20}$ have? In a perfect world, they would be “complexity 1″ sets, and in particular one would expect $f_{01}$ to be describable as some simple combination of
01-low influence sets and 12-low influence sets (and similarly $f_{20}$ should be some simple combination of 20-low influence sets and 12-low influence sets). Here, ij-low influence means that the
function does not change much if an i is flipped to a j or vice versa.
Unfortunately, it seems (at least from the ergodic approach) that this is not easily attainable. Instead, $f_{01}$ obeys a more complicated (and weaker) property, which I call “01-almost periodicity
relative to 12-low influence”, with $f_{20}$ obeying a similar property. Very roughly speaking, this is a “relative” version of 01-low influence: flipping digits from 0 to 1 makes $f_{01}$ change,
but the way in which it changes is controlled entirely by functions that have low 12-influence. (This is related to the notion of “uniform almost periodicity” which comes up in my paper on the
quantitative ergodic theory proof of Szemeredi’s theorem.)
It is relatively painless (using Cauchy-Schwarz and energy-increment methods) to pass from (1) to (2). To deal with (2) we need some periodicity properties of $f_{01}, f_{20}$ on small
“monochromatic” spaces (the existence of which is ultimately guaranteed by Graham-Rothschild) which effectively let us replace $f_{01}(\ell(1))$ with $f_{01}(\ell(0))$ and $f_{20}(\ell(2))$ with $f_
{20}(\ell(0))$ on a large family of lines $\ell$ (and more importantly, a large 12-low influence family of lines). From this fact, and the previously mentioned fact that $f_{01}, f_{20}$ are large on
f, we can get DHJ(3).
The argument as described on the wiki is far from rigorous at present, but I am hopeful that it can be translated into a rigorous finitary proof (though it is not going to be pleasant – I would have
to deploy a lot of machinery from my quantitative ergodic theory paper). Perhaps a better approach would be to try to export some of the ideas here to the Fourier-type approaches where there is a
better chance of a shorter and more quantitatively effective argument.
• Randall Says:
February 25, 2009 at 5:06 pm
I haven’t read this closely enough to have even an initial impression, however, much of it looks (somewhat) familiar.
First, I notice you removed your discussion of stationarity…instead (tell me if I misread), and in multiple settings, you seem to do a Cesaro average, over lines, rather than employing something
like Graham-Rothschild to get near-convergence along lines restricted to a subspace. Most striking of these are instances of using the so-called IP van der Corput lemma. Looking at its proof,
this does indeed look kosher, but it’s rather surprising to me all the same; assuming I’m understanding this at least somewhat correctly, did you give any thought to whether the ergodic proof
itself could be tidily rewritten to accommodate this averaging method?
Modulo the above, the main part of the argument I still don’t (in principle) understand is how you bound h, the number of functions used to approximate the almost periodic component of f,
independently of n. (This is the part of the argument I got stuck on in my own thoughts.) I see now that you solved this issue in your quantitative ergodic proof of Szemeredi, which I printed
last night as well, though I haven’t read deeply enough yet to see how. Am I to assume that something similar happens here, or is the answer different in the two cases?
• Terence Tao Says:
February 25, 2009 at 8:50 pm
Yes, the argument is distilled from your notes, though as you see I messed around with the notation quite a bit.
The stationarity is sort of automatic if you work with random subspaces of a big space $[3]^n$, and I am implicitly using it all over the place when rewriting one sort of average by another.
I am indeed hoping that Cesaro averaging may be simpler to implement than IP averaging, and may save me from having to use Graham-Rothschild repeatedly. There are a lot of things hidden in the
sketch that may cause this to bubble up. For instance, I am implicitly using the fact that certain shift operators $U_\alpha$ converge (in some Cesaro or IP sense) to an orthogonal projection,
and this may require a certain amount of Graham-Rothschild type trickery (I started writing some separate notes on a finitary Hilbert space IP Khintchine recurrence theorem which will be relevant
I admit I’m a bit sketchy on how to deal with h not blowing up. A key observation here is that of statistical sampling: if one wants to understand an average ${\Bbb E}_{h \in H} c_h$ of bounded
quantities over a very large set H, one can get quite a good approximation to this expression by picking a relatively small number $h_1,\ldots,h_m$ of representatives of H at random and looking
at the local average ${\Bbb E}_{j=1,\ldots,m} c_{h_j}$ instead. (This fact substitutes for the fact used in the Furstenberg approach that Volterra integral operators are compact and hence
approximable by finite rank operators; or more precisely, the Furstenberg approach needs the relative version of this over some factor Y.) I haven’t worked out completely how this trick will mesh
with the IP-systems involved, but I’m hoping that I can throw enough Ramsey theorems at the problem to make it work out.
Perhaps one thing that helps out in the finitary setting that is not immediately available in the ergodic setting is that there are more symmetries available; in particular, the non-commutativity
of the IP systems that makes the ergodic setup so hard seems to be less of an issue in the finitary world (the operations of flipping a random 0 to a 1 and flipping a random 0 to a 2 essentially
commute since there are so many 0s to choose from). There is a price to pay for this, which is that certain Ramsey theorems may break the symmetry and so one may have to choose to forego either
the Ramsey theorem or the symmetry. This could potentially cause a problem in my sketch; as I said, I have not worked out the details (given the progress on the Fourier side of things, I had the
vague hope that maybe just the concepts in the sketch, most notably the concept of almost periodicity relative to a low influence factor, could be useful to assist the other main approach to the
problem, as I am not particularly looking forward to rewriting my quantitative ergodic theory paper again.)
ryanworldwide Says:
February 25, 2009 at 4:53 am | Reply
819. #816 and #818 look quite exciting; I plan to try to digest them soon. Meanwhile, here is a Moser-esque problem I invented for the express purpose of being solvable. I hope it might give a few
tricks we can use (but it might not be of major help due to the PS of #528).
Let’s define a combinatorial bridge in ${[3]^n}$ to be a triple of points ${(x,y,z) \in \{-1,0,1\}^n}$ formed by taking a string with zero or more wildcards and filling in the wildcards with either $
{(0, -1, 0)}$ or ${(0, 1, 0)}$. If there are zero wildcards we call the bridge degenerate. I think I can show, using the ideas from #800, that if ${f : \{-1,0,1\}^n \to \{0,1\}}$ has mean ${\delta >
0}$ and ${n}$ is sufficiently large as function of ${\delta}$, then there is a nondegenerate bridge ${(x,y,z)}$ with ${f(x) = f(y) = f(z) = 1}$.
Roughly, we first use a density-increment argument to reduce to the case when ${f}$ is extremely noise sensitive; i.e., ${\|T_{1-\gamma} f\|_2^2}$ is only a teeny bit bigger than ${\delta^2}$. Here $
{\gamma}$ is something very small to be chosen later. Next, we pick a suitable distribution on combinatorial bridges ${(x,y,z)}$; basically, choose a random one where the wildcard probability is ${{\
epsilon} \ll \sqrt{\delta/n}}$. Now the key is that under this distribution, there is “imperfect correlation” between the random variable ${(x,y)}$ and the random variable ${z}$ — and similarly,
between ${x}$ and ${(y,z)}$. Here I use the term in the sense of the Mossel paper in #811. Because of this (see Mossel’s Lemma 6.2), ${\mathop{\bf E}[f(x)f(y)f(z)]}$ is practically the same as ${\
mathop{\bf E}[T_{1-\gamma}f(x)\cdot T_{1-\gamma}f(y)\cdot T_{1-\gamma}f(z)]}$, when ${\gamma \leq \delta^{O(1)} {\epsilon}}$. But this is extremely close to ${\delta^3}$ because ${\mathop{\bf E}[T_
{1-\gamma}f(x)] = \mathop{\bf E}[T_{1-\gamma}f(y)] = \mathop{\bf E}[T_{1-\gamma}f(z)] = \delta}$ and because the error can be controlled with H\”{o}lder in terms of ${\|T_{1-\gamma} f - \delta\|_2^2}
$, which is teeny by the density-increment reduction.
More details here: http://www.cs.cmu.edu/~odonnell/bridges.pdf. I can try to port this to the wiki soon.
• Terence Tao Says:
February 25, 2009 at 8:38 pm
Ryan, I’m not sure what you mean by “fill the wildcards by either (0,-1,0) or (0,+1,0)”. Wouldn’t this always make x and z equal in a bridge? Perhaps an example would clarify what you mean here.
• ryanworldwide Says:
February 25, 2009 at 10:09 pm
819.2: Er, whoops. You’re right. Even easier then I thought :) To make this problem more interesting, I think it will work if the triples allowed on wildcards are, say: (-,0,-), (-,0,0), (+,0,0),
But: a) this is not as nice-looking, and b) I think it’ll actually take slightly more work.
So, um, never mind for now.
gowers Says:
February 25, 2009 at 6:44 pm | Reply
820. Density increment.
I’ve now finished a wiki write-up that is supposed to establish (and does unless I’ve made a mistake, which is possible) that if $\mathcal{A}$ is a line-free set of uniform density $\delta$ then you
can pass to a combinatorial subspace of dimension $m$, as long as $m=o(n^{1/2})$, and find a special subset of complexity 1 in that subspace of density at least $c\delta^2$, such that the
equal-slices density of $\mathcal{A}$ inside that special subset is at least $\delta+c\delta^2$.
To be less precise about it, I think I’ve shown that if $\mathcal{A}$ contains no combinatorial line, then you get a density increase on a nice set. I start with uniform density and switch to
equal-slices density, but that is deliberate, and explained in the write-up, which, by the way, is here So my attention is now going to turn to trying to copy either the Ajtai-Szemerédi proof, or
Shkredov’s proof, of the corners theorem. I feel optimistic that this can be done, given that special sets of complexity 1 can be shown to contain combinatorial subspaces—though that on its own is
not enough.
• ryanworldwide Says:
February 26, 2009 at 1:02 am
820.1. Hi Tim — nice. I read through the wiki proof and agree that it should be correct. It is quite interesting how the passage between uniform and equal-slices seems necessary. On one hand I
worry a bit that the different measures might get out of control; on the other hand, the optimistic way to look at it is that we may eventually get so proficient at passing between measures that
it’ll be a very useful tool.
• gowers Says:
February 26, 2009 at 10:14 am
One observation that makes me feel slightly less worried about passing from one measure to another is that you can start with equal-slices measure instead. Then as a first step you pass to
uniform measure on a subspace. Then you carry out the argument as given (passing to a subspace of that subspace). And now you’ve ended up with the same measure that you started with.
ryanworldwide Says:
February 25, 2009 at 10:11 pm | Reply
821. Here is (I believe) a simple proof of Terry’s most basic structure theorem. The setting is as follows: ${\Omega}$ is a finite set and ${\mu}$ is a probability distribution on ${\Omega}$. We also
write ${\mu = \mu^{\otimes n}}$ for the product distribution on ${\Omega^n}$. We work in the space of functions ${f : \Omega^n \to {\mathbb R}}$ with inner product ${\langle f, g \rangle = \mathop{\
bf E}_{x \sim \mu}[f(x)g(x)]}$. Henceforth, all probabilities are wrt ${\mu}$ unless specified.
Theorem: Let ${f :\Omega^n \to {\mathbb R}}$ have ${\mathop{\bf Var}[f] \leq 1}$. Let ${0 < {\epsilon} < 1}$. Let ${\frac{1}{n} \leq \sigma_1 \leq \tau_1 \leq \sigma_2 \leq \tau_2 \leq \cdots \leq \
sigma_k \leq \tau_k \leq {\epsilon}}$ be a sequence of “scales” (or “times”) satisfying the following conditions: (a) ${\sigma_{i}/\tau_{i-1} \geq 3\ln(1/{\epsilon})/{\epsilon}}$ for all ${i}$; (b) $
{k \geq 1/{\epsilon}^2}$.
Then there exists ${i}$ such that ${f}$ can be written as ${f = f_{\text{stab}} + f_{\text{sens}}}$, where:
1. ${\mathop{\bf E}[f_{\text{stab}}] = \mathop{\bf E}[f]}$.
2. ${\|f_{\text{stab}}\|_p \leq \|f\|_p}$ for all ${1 \leq p \leq \infty}$; hence ${f_{\text{stab}}}$ has range ${[0,1]}$ if ${f}$ does (and then ${f_{\text{sens}}}$ has range ${[-1,1]}$).
3. ${0 \leq \langle f_{\text{stab}}, f_{\text{sens}} \rangle \leq 3{\epsilon}}$.
4. ${\mathcal{E}(f_{\text{stab}}) \leq 1/(\tau_{i} n)}$, where ${\mathcal{E}(g)}$ denotes the “average influence of ${g}$“: i.e., ${\mathop{\bf E}[(g(x) - g(x'))^2]}$, where ${x \sim \mu}$ and ${x'}$
is formed by rerandomizing a random coordinate of ${x}$.
5. ${\mathcal{S}_{1 - \sigma_{i}}(f_{\text{sens}}) \leq 3{\epsilon}^2}$, where ${\mathcal{S}_{1 - \sigma}(g)}$ denotes the “noise stability of ${g}$ at ${1 - \sigma}$“: i.e., ${\mathop{\bf E}[g(x)g
(x')]}$, where ${x \sim \mu}$ and ${x'}$ is formed by rerandomizing each coordinate of ${x}$ with probability ${\sigma}$.
Proof in next post.
ryanworldwide Says:
February 25, 2009 at 10:12 pm | Reply
822. Proof: For simplicity I present the proof when ${\Omega = \{0,1\}}$ and ${\mu}$ is the uniform distribution; to get the full version, replace the Fourier transform with the “Hoeffding AKA
Efron-Stein orthogonal decomposition”.
For an interval ${J = [a,b]}$ of natural numbers, let ${\mathcal{W}_I(f)}$ denote ${\sum_{a \leq |S| \leq b} \hat{f}(S)^2}$. The condition ${\mathop{\bf Var}[f] \leq 1}$ is equivalent to ${\mathcal
{W}_{[1,n]} \leq 1}$, and we will use this frequently. Now consider the intervals of the form ${J_i = [{\epsilon}/\tau_i, 2\ln(1/{\epsilon})/\sigma_i]}$. By hypothesis (a) these intervals are
disjoint, and by hypothesis (b) there are at least ${1/{\epsilon}^2}$ of them. By Pigeonhole, we can fix a particular ${i}$ such that ${\mathcal{W}_{J_i} \leq {\epsilon}^2}$. Henceforth write ${\
sigma = \sigma_i}$, ${\tau = \tau_i}$.
Now set ${f_{\text{stab}} = T_{1-\tau} f}$ and ${f_{\text{sens}} = f - f_{\text{stab}}}$. Conditions 1 and 2 are satisfied. Condition 4 holds because $\displaystyle n\mathcal{E}(f_{\text{stab}}) = \
sum_S |S|\hat{f}_{\text{stab}}(S)^2 = \sum_S |S| (1-\tau)^{2|S|}\hat{f}(S)^2$
and this is at most ${1/\tau}$ because ${|S| (1-\tau)^{2|S|} \leq 1/\tau}$ for all ${S}$. For Condition 3, $\displaystyle \langle f_{\text{stab}}, f - f_{\text{stab}} \rangle = \sum_S a(|S|) \hat{f}
where $\displaystyle a(s) = (1-\tau)^{s} - (1-\tau)^{2s}.$
We have ${a(0) = 0}$ so we can drop that term. We have ${a(s) \leq s \tau}$ always so we get a contribution of at most ${{\epsilon}}$ from the terms with ${|S| \leq {\epsilon}/\tau}$. We also have $
{a(s) \leq (1-\tau)^s \leq \exp(-\tau s)}$ always so we get a contribution of at most ${{\epsilon}^2}$ from the terms with ${|S| \geq 2\ln(1/{\epsilon})/\tau}$. And the remaining terms have ${|S| \in
[{\epsilon}/\tau, \ln(1/{\epsilon})/\tau] \subset J_i}$, so we get a contribution of at most ${{\epsilon}^2}$ from them.
The proof of Condition 5 is similar. We have $\displaystyle \mathcal{S}_{1 - \sigma}(f_{\text{sens}}) = \sum_S (1-\sigma)^{|S|} \hat{f}_{\text{sens}}(S)^2 = \sum_S b(|S|) \hat{f}(S)^2,$
where $\displaystyle b(s) = (1-\sigma)^{s} (1-(1-\tau)^s)^2.$
Again, ${b(0) = 0}$. We have ${b(s) \leq (\tau s)^2}$ always so we get a contribution of at most ${{\epsilon}^2}$ from the terms with ${|S| \leq {\epsilon}/\tau}$. We also have ${b(s) \leq (1-\sigma)
^s}$ always so we get a contribution of at most ${{\epsilon}^2}$ from the terms with ${|S| \geq 2\ln(1/{\epsilon})/\sigma}$. And the remaining terms have ${|S| \in J_i}$ so we get a contribution of
at most ${{\epsilon}^2}$ from them. $\Box$
• Terence Tao Says:
February 25, 2009 at 10:48 pm
822.1 This looks about right to me. The game is always to first use pigeonhole to find a nice big gap in the energy spectrum, and then use that gap to split f into opposing pieces. (Sometimes one
also needs a (small) third term to deal with the small amount of energy stuck inside the gap.)
One thing that works in your favour here is that the averaging operators $T_{1-\tau}$ are positivity preserving, so if f is positive then so is $f_{stab}$ (and related to this is a useful
comparison principle: if f is pointwise bigger than g, then $f_{stab}$ is pointwise bigger than $g_{stab}$.) Things get more tricky when one doesn’t have this positivity preserving property,
because it is absolutely essential later on that $f_{stab}$ be non-negative. In my paper with Ben we had to introduce the machinery of partitions and conditional expectations to deal with this.
One side effect of this is that it can force the scales to be exponentially separated (e.g. $1/\sigma_i \gg \exp( 1 / \sigma_{i+1} )$) rather than just separated by a large constant depending
only on $\varepsilon$). This leads to the type of tower-exponential bounds which are familiar from the regularity lemma. Once we get to the fancier structure theorems, we may start seeing these
sorts of bounds emerge, but I agree that in the simple case here we don’t have to deal with all that because of the handy $T_{1-\tau}$ operators.
Combining this with your notes that control the Sperner count by the noise stability, it looks like we have a pretty solid Fourier-analytic proof of DHJ(2), which was one of our early objectives
(suggested first, I believe, by Gil).
• ryanworldwide Says:
February 26, 2009 at 1:00 am
822.2: Yes, I think with slightly more work one can get an “Adequate-Sperner” Fourier proof; I think it will require density $\Omega(1/\sqrt{\log n})$. This is still not quite right, but is
Ryan O'Donnell Says:
February 26, 2009 at 6:54 am | Reply
823. Structure Theorem.
A small note: Unless I’m mistaken, the structure theorem around Lemma 1 in Terry’s wiki notes on Furstenberg-Katznelson — which decomposes functions on $[3]^n$ into a 12-stable and a 12-sensitve part
— can be proved in the same way as in #821.
Specifically, employing Tim’s $(U,V,W)$ notation, I think you just need to look at the expectation over $U$ of the Fourier weight of $f_U$ on the various intervals $J_i$. Here $f_U$ denotes the
restricted function, with 0′s fixed into $U$, on binary inputs (1 and 2). Everything seems to go through the same, using the fact that restricting functions with $U$ commutes with flipping 1′s and
2′s, and that $\mathbb{E}_U[\mathrm{Var}[f_U]] \leq \mathrm{Var}[f]$.
• Ryan O'Donnell Says:
February 26, 2009 at 6:56 pm
823.1. The weird $\mathbb{Var}$ at the end here is meant to be “Var”.
Fixed now — Tim.
gowers Says:
February 26, 2009 at 11:48 am | Reply
824. DHJ(3) general strategy
I have a hunch, after looking at Van Vu’s quantitative version of Ajtai-Szemerédi, that it may be possible to strengthen the argument on the wiki to give the whole thing. If this were not a polymath
project I would of course go away and try to make the idea work, keeping quiet if it didn’t and giving full details if it did. But it is polymath, so let me say in advance roughly what I hope will
In the corners problem, it is fairly easy to prove that a corner-free set has a density increment on a Cartesian product of two dense sets. The problem is what to do then. Ajtai and Szemerédi’s
strategy is to use a more sophisticated averaging argument, making heavy use of Szemerédi’s theorem, to obtain a density increment on a highly structured Cartesian product: it is a product of two
arithmetic progressions with the same common difference.
Initially, it looks very discouraging that they have to use Szemerédi’s theorem, but I now believe that that may just be an artificial byproduct of the fact that they are working in the grid and want
to get another grid at the next stage of the induction. But in the Hales-Jewett world, you don’t actually need long arithmetic progressions anywhere, and I think the right analogue will in fact turn
out to be multidimensional subspaces of ${}[2]^n$. My actual idea is slightly more precise than this, but hard to explain. Basically, I want to revisit the argument on the wiki, but aim straight for
a density increase on a subspace, modelling the argument as closely as I can on the Ajtai-Szemerédi argument for the corners problem.
If that works, then it will probably give a tower-type bound. A follow-up project would then be to try to do a more powerful Shkredov-type argument to get a bounded number of exponentials. But I’m
getting way ahead of myself by even discussing this.
Let me also slip in a metacomment. I find it slightly inconvenient that, now that we have modest threading, I can’t just scroll to the bottom of the comments and see what has been added. As Jason
Dyer pointed out, there is a trade-off here. On the one hand, it makes it slightly harder if you are trying to keep up with the discussion in real time, but on the other it probably means that the
comments are better organized if you want to come back to them later. In the end, the second consideration probably trumps the first, because it’s a long-term convenience at the cost of a mild
short-term inconvenience. But I still think that we should be quite careful about when and how much we use the threading.
gowers Says:
February 26, 2009 at 5:32 pm | Reply
825. Ajtai-Szemerédi approach.
No time to justify this, or at least not for a few hours (and possibly not before tomorrow morning) but I am now very optimistic indeed that the Ajtai-Szemerédi approach to corners can be modified to
give DHJ(3). As usual, this surge of optimism may be followed by a realization that there are some serious technical obstacles, but I think that the existing write-up of the complexity-1 correlation
contains methods for dealing with the technicalities that will arise. I plan to write up (i) a sketch proof of Ajtai-Szemerédi on the wiki, (ii) an outline of an argument here on the blog (which will
include the AS-to-DHJ(3) dictionary I will be using), and (iii) a detailed sketch of an argument on the wiki. Those are in decreasing order of probability that I can actually finish them without
getting stuck.
Terence Tao Says:
February 27, 2009 at 12:45 am | Reply
826. Structure theorem
I now think there is a way to do the DHJ(2) structure theory without explicit mention of the Fourier transform (at the cost of worse quantitative bounds), which may be important when trying to extend
the arguments to the DHJ(3) structure theory sketched in my wiki notes.
The point is to replace the self-adjoint operators $T_{1-\tau}$ that Ryan uses by a “one-sided” counterpart. Given a function $f: [2]^n \to {\Bbb R}$, and a parameter $\lambda > 0$, let $S_\lambda f:
[2]^n \to {\Bbb R}$ be defined by $S_\lambda f(x) := {\Bbb E} f(y)$, where y is formed from x by allowing each coordinate of x, with probability $\lambda/n$, of being overwritten by a 1. (This is in
contrast to Ryan’s rerandomisation, which would cause x to be overwritten by a random 0/1 bit rather than by 1.)
Informally, one can think of y as taking about $\lambda/2$ of the 0-bits of x (the exact number basically obeys a Poisson distribution) and flipping them to 1s.
Observe that (x,y) always forms a combinatorial line. Thus lower bounds on ${\Bbb E} f(x) S_\lambda f(x)$ imply lots of combinatorial lines in the support of f.
The key observation is that the $S_\lambda$ are basically a semigroup (for $\lambda$ much smaller than n): $S_\lambda S_{\lambda'} \approx S_{\lambda+\lambda'}$. In particular, if $\lambda \ll \
lambda'$, we have the absorption property $S_\lambda S_{\lambda'}, S_{\lambda'} S_\lambda \approx S_{\lambda'}$. Because of this, one can show that the $S_\lambda$ converge (in the strong operator
topology) to an orthogonal projection P; to see this finitarily, one would inspect the energies $\| S_\lambda f \|_2^2$, which are basically monotone decreasing in $\lambda$, and locate a long gap in
the energy spectrum in which these energies plateau in $\lambda$. I think this gives us the same type of structure/randomness decomposition as before. Note that for large $\lambda$, $S_\lambda f$ is
approximately $S_1$-invariant, which is basically the same thing as having low influence.
• Terence Tao Says:
February 27, 2009 at 12:46 am
correction: (x,y) form a combinatorial line most of the time; there is a degenerate case when x=y (analogous to the case of a degenerate arithmetic progression). But for large lambda, this case
is very rare and can be ignored.
gowers Says:
February 27, 2009 at 1:12 am | Reply
827. Ajtai-Szemerédi
Just to say that the easy step of my three-step plan is now done: an informal write-up of the Ajtai-Szemerédi proof of the corners theorem can now be found on this wiki page. I have to go to bed now
so parts (ii) and (iii) will have to wait till tomorrow.
gowers Says:
February 27, 2009 at 1:44 pm | Reply
828. Progress report
Part (ii) is, perhaps predictably, giving me a bit more of a headache. But it’s an interesting headache in that I am stuck on something that feels as though it shouldn’t be too hard. I don’t know how
much time I’ll have to think about it, so let me say what the problem is.
Call a subset $\mathcal{B}\subset[3]^n$simple if whether or not $x$ belongs to $\mathcal{B}$ depends only on the 1-set of $x$. In set language, there is a set-system $\mathcal{U}$ such that $\mathcal
{B}=\{(U,V,W):U\in\mathcal{U}\}.$ The question I’m struggling with is this. Let $\mathcal{A}$ be a dense subset of $[3]^n$ that correlates with a dense simple set $\mathcal{B}$. Does that imply that
there is a combinatorial subspace with dimension tending to infinity on which $\mathcal{A}$ has a density increment?
I cannot believe that this is a truly hard problem, but I haven’t yet managed to solve it. I think a solution would be very useful. Oh, and I really don’t mind what measure is used in any solution …
• Ryan O'Donnell Says:
February 27, 2009 at 4:26 pm
828.1: What if $\mathcal{A} = \mathcal{B} =$ strings with an even number of 1′s?
• gowers Says:
February 27, 2009 at 4:50 pm
828.2: Then take all sequences x such that $x_1=x_2,$$x_3=x_4,$ etc., and you have a combinatorial subspace of dimension $n/2$ where the density has gone all the way up to 1. Were you imagining
that wildcard sets had to have size 1? That is certainly not intended to be a requirement in my question.
• Terence Tao Says:
February 27, 2009 at 5:29 pm
I can answer your question as “yes”, but in a totally unhelpful way: DHJ(3) implies that any dense set has a density increment on a large combinatorial subspace.
Of course, the whole point is to find a DHJ(3)-free proof of this fact… I don’t currently have time to think about it right now, but I might look at it later.
• Terence Tao Says:
February 27, 2009 at 5:33 pm
Hmm, actually thinking about it a little bit there may be an energy-increment type proof.
Firstly, one should be able to show that a 1-set contains lots of combinatorial subspaces where it is denser, by concatenating the 2 and 3-sets together and applying DHJ(2).
Next, one should be able to apply some sort of structure theorem to decompose an arbitrary set (or function) into a 1-set plus something orthogonal to 1-sets. Actually one needs to mess around
with scales and get a decomposition into a 1-set $f_{U^\perp}$ at some large scale plus something $f_U$ orthogonal to 1-sets at finer scales (in particular, it should have mean zero at finer
scales). (Let me be vague about what “scale” means, but every time you pass to a large combinatorial subspace, you’re supposed to descend one scale.)
Now take the guy $f_{U^\perp}$ which is a 1-set at a large scale and find lots of combinatorial subspaces in it at the finer scale where it is denser. The guy $f_U$ which is orthogonal to 1-sets
at finer scale should not disrupt this density increment.
Have to run, unfortunately… may flesh this out later
• Ryan O'Donnell Says:
February 27, 2009 at 5:52 pm
828.5. Re 828.2:
Ah, no, I guess I was misunderstanding what a combinatorial subspace is. I always think of this as, “fixing some of the coordinates and leaving the rest of them ‘free’.” It had not occurred to me
that one is also allowed to insist on things like “x_i = x_j”, although I now see that there is no reason not to.
gowers Says:
February 27, 2009 at 6:37 pm | Reply
829. Ajtai-Szemerédi
Let me try to explain the motivation for my simple-sets question. As I said earlier, I am trying to model a proof of DHJ(3) on the Ajtai-Szemerédi corners proof (an account of which is given here on
the wiki). I have the following rough dictionary in mind.
1. A vertical line corresponds to a set of sequences (U,V,W) where you fix the 1-set U.
2. A horizontal line corresponds to a set of sequences (U,V,W) where you fix the 2-set V.
3. A line of slope -1, or diagonal, corresponds to a set of sequences (U,V,W) where you fix the 3-set W.
4. An arithmetic progression in [n] with length tending to infinity corresponds to a combinatorial subspace in ${}[2]^n$ with dimension tending to infinity.
5. A subgrid with width tending to infinity corresponds to a combinatorial subspace of ${}[3]^n$ with dimension tending to infinity.
6. A corner corresponds to a combinatorial line.
That’s about it. So now let me say where the first problem arises when one tries to translate the AS proof to the DHJ(3) context. Obviously step 1 is fine: with any sensible model of a random
combinatorial subspace you can assume that the density of A in that subspace is almost always at least $\delta-\eta$, where $\delta$ is the density of A.
Step 2 says that if you have a positive density of vertical lines inside which A has density noticeably less than $\delta$, then you also have a positive density of vertical lines inside which A has
density noticeably greater than $\delta$. If we let $B$ be the set of x-coordinates associated with these vertical lines, then we can use Szemerédi’s theorem to get an arithmetic progression P of
these “over-full” vertical lines, with small common difference. Then we can partition the horizontal lines into arithmetic progressions $Q_i$, each of which is a translate of $P$, and by averaging we
find a subgrid $P\times Q_i$ inside which A has density noticeably larger than $\delta$, which gives us a density increment and we’re done. (Of course, then one has to go on and say what happens if
we don’t have a positive density of sparse vertical lines — this is, as I say, just the first step of the argument.)
Let me gloss over the question of which measure I want to use: I feel that that is just a technicality and the main point is to get the conceptual argument working. If you accept item 1 in the
dictionary, then the obvious first step is to define U to be underfull if the density of pairs (V,W) (out of all pairs such that (U,V,W) is a point in ${}[3]^n$) is noticeably less than $\delta$.
Then if there is a positive density of underfull U, then we have a positive density of overfull U as well. Now I don’t have a problem finding a nice big combinatorial subspace consisting entirely of
overfull Us (which is what item 4 suggests I should be looking for) but I don’t then have a nice analogue of the fact that a Cartesian product of two APs with the same common difference is a grid. Or
rather, maybe I do, but I’m not sure how to get at it. The analogue should be something like that if I have such a combinatorial subspace (by which I mean a system of sets $U_0,U_1,\dots,U_k$ such
that the elements of the system are all sets of the form $\bigcup_{i\in E}U_i$ with 0 belonging to E), then on average the combinatorial subspaces that fix $U_0$ and treat $U_1,\dots,U_k$ as
wildcards should be slightly too dense. But for a fixed combinatorial subspace this doesn’t seem to have to be true, so I need somehow to build a combinatorial subspace such that it is true.
I can feel this explanation getting less clear as I proceed, but perhaps it’s at least clear globally what I want to do: the “simple set” is the set of all (U,V,W) such that U is overfull. So
trivially A intersects it too densely. If we can get a density increment on a subgrid, then we’re done, which means that Step 3 is complete and we may assume that virtually no U are underfull. (Given
the difficulties associated with this step, it’s not altogether clear that this statement is strong enough, but it would be encouraging. I might assume it and see what happens if I try to press on
with the later steps.)
Ryan O'Donnell Says:
February 27, 2009 at 7:50 pm | Reply
830. Combinatorial subspaces; 828.4 vs. 828.2.
It seems to me that an argument along the lines in 828.4 at some point has to find combinatorial subspaces which include these $x_i = x_j$ constraints, as in Tim’s 828.2. If the combinatorial
subspaces are just of the “fixing-some-coordinates” type then I’m not sure (yet) how to get around 828.1.
But will arguments as in 828.4 actually produce such subspaces?
gowers Says:
February 27, 2009 at 8:14 pm | Reply
831. Finding combinatorial subspaces
Ryan, that’s an interesting point. The kind of arguments I had in mind are different though: they are inductive ones where you build a subspace up dimension by dimension, using Sperner’s theorem (or
similar results) at each stage. A prototype for such an argument is the proof I sketched on the wiki of the basic multidimensional Sperner theorem. This argument leads naturally to larger wildcard
Randall Says:
February 27, 2009 at 10:28 pm | Reply
832. 1-sets
I was trying to think about Tim’s question from the ergodic perspective; if I am translating correctly the outlook is not encouraging. Perhaps though I am overgeneralizing.
A 1-set is just a set, membership in which is impervious to switchings of 0s with 2s, right? Okay, so, if I understand Terry’s translations from the ergodic world and back again, the functions that
are measurable with respect to 1-sets would be in some sense analogous to the $\tau\rho^{-1}$ rigid factor…if that’s right, then in order to think about the question ergodically, one should think
whether it would help at all to know that 1_A had non-zero projection onto this factor. And I don’t see how it would. In particular, it isn’t the case that things orthogonal to this factor don’t
contribute to the averages you’re interested in, so I don’t see how decomposing would help. Yes, it is easy to see from DHJ (2) what the positive contribution would be restricted to this rigid
factor, but you lose positivity of the relative AP components; it would not seem an easy task to show that they have non-negative contribution without positivity, especially given that the argument
as it now stands uses positively very heavily (it throws a lot away because it can).
On the other hand, a non-zero projection onto this factor is sufficient to get (what Terry would call, I think) both 01 over 02 and 12 over 02 almost periodic components, and in a special way, so it
might in theory be easier to get a density increment for these guys in the combinatorial world, but at least from an ergodic perspective, I can’t see any strategy that might simplify the proof from
what it already is. Short, that is, of a detailed analysis of what is going on in terms of finite-dimensional modules instead of “almost periodicity”, but that is potentially a step in the direction
of harder, not easier.
Of course, all I say depends on this analogy between 1-sets and the 02-rigid factor being what I suggest it is, and I don’t really see that part well at all (I’m mostly just guessing because I don’t
understand the picture very well from the combinatorial side). (Also of course someone may be posting a proof simultaneously with my posting my doubts.)
• Terence Tao Says:
February 27, 2009 at 10:55 pm
Yeah, a 1-set is what I would call a 02-low influence (in fact, a 02-zero influence) set, and it would correspond to a tau rho^{-1} invariant set in the ergodic theory language. Tim’s approach
actually deals with intersections of 1-sets and 2-sets (which we have been calling “complexity 1 sets”), which would be a combination of a 02-low influence and 01-low influence set, or a set in
the factor generated by both the tau rho^{-1}-invariants and the sigma rho^{-1}-invariants. (One may also want to throw in the 12-invariant sets for good measure.) But never mind this distinction
for now.
Perhaps the place where Tim’s approach and the ergodic approach differ, though, is that Tim can always stop the argument as soon as a density increment is found on a large subspace, whereas this
trick is not available in the ergodic world. So I don’t think Tim needs the strong (and presumably false) claim that “functions orthogonal to complexity 1 sets have negligible impact on the
combinatorial line count”. He only needs the weaker claim that “if a set has no combinatorial lines, then it must have non-trivial correlation with a complexity 1 set”, combined with the (as yet
not fully established) claim that “a set with non-trivial correlation with a complexity 1 set has a density increment on a large combinatorial subspace”.
Hmm. Maybe one thing to put on my “to do” list is to find an ergodic translation of Tim’s argument on the wiki that line-free sets correlate with complexity 1 sets. (If you haven’t already
noticed, I’m a big fan of translating back and forth between these two languages.)
• Randall Says:
February 27, 2009 at 11:06 pm
Actually I just got on here again to retract everything I said (I should think more carefully before posting). I now think, from the ergodic perspective, you trivially get a “measure increment”
if 1_A correlates projects non-trivially onto the 02-rigid factor, for the reason that on fibers, the measure is not constant. Just pick a bunch of fibers on which the measure is too big and then
a subspace on which you have intersection of those fibers (DJH (2)).
• Terence Tao Says:
February 27, 2009 at 11:39 pm
Hmm, I tried my hand at translating Tim’s argument to ergodic theory, and it came out weird… in order to describe it, I have to leave the measure space X and work in a certain extension of it,
vaguely analogous to the Host-Kra parallelopiped spaces $X^{[k]}$ (or the Furstenberg self-joinings used in Tim Austin’s proof of convergence for multiple commuting shifts).
It’s something like this: consider the space of all quadruples $(x, \sigma(\alpha) \rho^{-1}(\alpha) x, \sigma(\alpha \beta) \rho^{-1}(\alpha \beta) x, \sigma(\alpha) \tau(\beta) \rho^{-1}(\alpha
\beta) x)$ in $X^4$, where $\alpha$ is a word to the left of $\beta$ (or maybe to the right; I may have the signs messed up). There should somehow be a natural measure $\tilde \mu$ associated to
these quadruples, which I currently do not know how to define properly; each of the four projections of this measure to X should be the original measure $\mu$.
The last three points of this quadruple form a combinatorial line (I may have messed up the order of operations a bit). So if A is line-free, the function $1_A(x_4)$ on $X^4$ will have strong
negative correlation with $1_A(x_2) 1_A(x_3)$. But the function $1_A(x_2)$ has a “02-invariance property” in the sense that it is insensitive to changes to $\beta$, while the function $1_A(x_3)$
has a certain “01-invariance property” in the sense that it is insensitive to interchanges between $\alpha$ and $\beta$. So A is correlating with a combination of a 02-invariant set and a
01-invariant set, but the correlation only takes place in the extension $(X^4, \tilde \mu)$ of $(X,\mu)$ and the invariance seems to be a bit fancier than just $\tau \rho^{-1}$-invariance or $\
sigma \rho^{-1}$-invariance.
Ryan O'Donnell Says:
February 27, 2009 at 10:36 pm | Reply
832. Following up on 830, perhaps Randall or Terry would be able to answer:
Is there an analogue in the ergodic proof of restrictions to these subspaces with more than one “type” of wildcard? (In other words, subspaces with these “$x_i = x_j$” constraints?) In my limited
understanding of the ergodic proofs, the operators used therein seemed to be more aligned with just the simpler combinatorial subspaces.
• Terence Tao Says:
February 27, 2009 at 10:59 pm
Ryan, I’m not sure I understand your question. A k-dimensional combinatorial subspace is described by a word with k wildcards, but each wildcard can be used more than once. For instance, xxyyzz21
would describe the three-dimensional subspace of $[3]^8$ in which $x_1=x_2, x_3=x_4, x_5=x_6, x_7=2, x_8=1$. A combinatorial line is the case when k=1, so we have one wild card x, which can be
used many times over.
• Jason Dyer Says:
February 27, 2009 at 11:06 pm
Are you meaning like how Moser’s has a wildcard that goes 1-2-3 while the other goes 3-2-1? So, generalizing, there would be six types of wildcards? (123, 132, 213, 312, 231, 321 — obviously
certain combinations would be equivalent to Moser)
• Jason Dyer Says:
February 27, 2009 at 11:26 pm
To clarify further, any pair of distinct wildcards from those six (123, 132, 213, 312, 231, 321) will fix one element and permute the other two, so they are all equivalent to Moser. However, we
could take all the way up to six different wildcards, resulting in lines I believe have never been named. For example, taking x = 123, y = 321 and z= 213, then a line formed with 11xyz would be
11132, 11221, 11313.
• Jason Dyer Says:
February 27, 2009 at 11:29 pm
Argh. There are pairs that don’t fix any of the elements, (123-231 for example) which would also be something different than Moser’s.
• Ryan O'Donnell Says:
February 28, 2009 at 9:34 pm
832.x. Re Terry’s reply: Right, what I’m saying is, in my limited experience, I hadn’t noticed any ergodic-inspired finitary arguments that involved combinatorial subspaces with more than one
type of wildcard.
gowers Says:
February 28, 2009 at 1:40 am | Reply
833. Fourier and Sperner
On a slightly different topic, I still haven’t given up hope of a “right” Fourier proof of Sperner’s theorem. On the wiki I have now written up an argument that almost gets there. The good news is
that it gives a Fourier-type decomposition with the property that a suitable equal-slice-weighted sum of f(A)f(B) over all pairs $A\subset B$ transforms into an expression where the contribution from
every Fourier mode (when this claim is suitably interpreted) is positive. The bad news is that so far the only way I have of proving this positivity, which boils down to proving that a certain
explicitly defined integral kernel is positive definite, is using an inverse Fourier argument, so I can’t quite claim that the proof is a Fourier proof of Sperner. But now that I know that this
kernel is positive definite, I think it will be just a matter of time before one of us comes up with a direct proof, and then there will be a nice clean Fourier proof of Sperner (or at least, I think
it would deserve to be called that).
Terence Tao Says:
February 28, 2009 at 5:07 am | Reply
834. Density increment
I think I can now prove Tim’s claim in 828, namely that a set that correlates with a 1-set has a density increment on a large dimensional subspace.
First observe, by the greedy algorithm and DHJ(2), that if n is large enough depending on m and $\varepsilon$, then any dense subset A of $[2]^n$ can be partitioned into m-dimensional subspaces, plus
a residual set of density at most $\varepsilon$. Indeed one just keeps using DHJ(2) to locate and then delete m-dimensional subspaces from the set. Similarly, the complement of A can also be
partitioned into m-dimensional spaces plus a residual set of density at most density at most $\varepsilon$.
Now, one should be able to pull this statement to $[3]^n$ and show that if A is a dense 1-set of $[3]^n$, then one can partition both A and its complement into m-dimensional subspaces, plus a
residual of small density. I’m not sure what “density” should mean here; equal-slices may be a better bet than uniform at this point.
Now, if B is a dense set which correlates with a dense 1-set A, then by picking a moderately large m, picking $\varepsilon$ tiny compared to 1/m, and then n really large, we should be able to use the
above partition and the pigeonhole principle to get a density increment of B on one of these m-dimensional subspaces.
gowers Says:
February 28, 2009 at 9:35 am | Reply
835. Density increment
Terry, that’s similar to things I was trying, and I’m wondering whether it will run into similar difficulties. The point where I get stuck is where you say, “Now, one should be able to pull this
statement to ${}[3]^n$.” How might one do this? Well, given any combinatorial subspace in ${}[2]^n$ you can easily define an associated combinatorial subspace in ${}[3]^n$ by simply allowing the
wildcards to take value 2 as well. (For convenience I am taking ${}[2]=\{0,1\}$ and ${}[3]=\{0,1,2\}.)$ And the result will be a collection of disjoint subspaces. But they won’t anything like
partition ${}[3]^n$ because so far all the fixed coordinates are 0 or 1. We can correct that by allowing the fixed coordinates that used to be 0 to be either 0 or 2, but I think that doesn’t get us
out of the woods.
What one really needs to answer is the following question: given a sequence $x$ in ${}[3]^n$, to what combinatorial subspace does it belong?
Here’s an example. Suppose $n=5$ and you are presented with $x=110022.$ And suppose that one of the combinatorial subspaces you took in ${}[2]^n$ was $V=aabcbc$. Then you’d be forced to put $x$ into
$V$, since its 1-set is contained in $V$. But if you now allow the wildcards to take the value 2 as well, you can’t get $x$.
• gowers Says:
February 28, 2009 at 11:37 am
835.1 Maybe one could use a variant of the argument where having chosen the first wildcard set you then inductively cover evenly everything else. Sorry — that’s a bit vague but I have no time to
clarify it just yet. In the above example, having chosen aa you might e.g. then partition into things like bcbc, bbcc, bccb, cbbb, etc. Except that even that wouldn’t be right because once you’d
chosen your second wildcard set you’d play the same game with the third, fourth, etc. Sorry, that’s probably still completely unclear.
• Jason Dyer Says:
February 28, 2009 at 4:23 pm
I haven’t had time to test this yet, but would it help to triplicate the original set into a 9-uniform hypergraph (with subgraphs A, B, and C) such that subgraph A uses 0 or 2, B uses 0 or 1, and
C uses 1 or 2?
Then somehow before the last step everything would need to be recombined into the original graph.
• gowers Says:
February 28, 2009 at 5:04 pm
835.3 Jason, I’m not sure I follow what you are saying here.
• Randall Says:
February 28, 2009 at 5:24 pm
Hopefully this issue will be resolved in a simpler way soon anyway, but could the “measure increment” observation I made in my last post be fashioned into a proof? In the regular F-K conversion
they only cared about combinatorial lines, but presumably one could arrange that for any set I of words, if the intersection over I of T_w^{-1}C is non-empty, then you see a subspace-homomorphic
image of I in the original set A you were dealing with. You could do this for 2 sets simultaneously, a 1-set A and a correlated set B. I am guessing that A would wind up associated with a set in
your measure space that would be measurable with respect to the \tau\rho^{-1} rigid factor, and that B would wind up associated with a set that had non-zero projection onto that factor, and you
would then wrap up the proof as indicated above. Of course you’d have to convert this to a combinatorial proof, which presumably Terry is good at (energy incrementation?) But again, one hopes
it’s not that hard, or that something already suggested can be made to work….
• Terence Tao Says:
February 28, 2009 at 7:30 pm
Huh, so the pullback of a combinatorial subspace of $[2]^n$ to $[3]^n$ is not a combinatorial subspace. That’s a bit weird. I agree that my previous comment doesn’t quite work as stated then.
However, it does seem that the pullback of a combinatorial subspace of $[2]^n$ is an “average” of combinatorial subspaces of $[3]^n$ (or what Gil would call a “fractional cover”). Basically,
there is a random way to convert a 2-dimensional subspace e.g. 01aa01bbb10ccc to a 3-dimensional subspace by randomly converting 0s to 0s or 2s, and a wildcard such as a to a, 0, or 2. One has to
carefully select the probability measures here (in particular, to make choices between equal slices and uniform) to make everything work, but perhaps it can be pulled off. (One may also want to
ensure that each wildcard appears a lot of times so that when one pulls back, one (with very high probability) doesn’t accidentally erase all copies of any given wildcard.) As Gil observed, if
you can fractionally cover most of a set A and its complement by large combinatorial subspaces, and B has a large correlation with A, then B has a density increment on one of these subspaces.
• Terence Tao Says:
February 28, 2009 at 7:42 pm
No, that’s rubbish, I take it back; the pullback of, say, aa, is the set {00, 11, 12, 21, 22}, and I can’t cover that space by lines. Strange.
• Jason Dyer Says:
February 28, 2009 at 7:44 pm
835.5 I was meaning based on “allowing the fixed coordinates that used to be 0 to be either 0 or 2″ you have two more subspaces where:
coordinates that used to be a 0 are either 1 or 2, and coordinates that used to be a 0 are a 1
coordinates that used to be a 0 are either 0 or 1, and coordinates that used to be a 1 are a 2.
However, I have not been able to get anything useful out of this.
gowers Says:
February 28, 2009 at 5:24 pm | Reply
836 Density increment
I’m going to return to full polymath mode and think online rather than offline, concentrating on the problem of getting a density increment on a subspace if you already have one on a simple set.
(Recall that I am defining a simple set to be a set $\mathcal{A}$ of sequences x of the form $\{(U,V,W):U\in\mathcal{U}\}.)$ Since I’ll be talking about ${}[2]^n$ quite a bit, it will be natural to
take ${}[3]=\{0,1,2\}$ and ${}[2]=\{0,1\}.$ Set theorists would no doubt approve.
I am aiming for an inductive construction. That is, I want to fix some coordinates and choose a wildcard set in such a way that I am not dead. To say that more precisely, the assumption we are given
to start with is that for every $U\in\mathcal{U}$ the density of points $(U,V,W)\in\mathcal{A}$ is at least $\delta+\eta$. I would now like to find a disjoint quadruple $(U_1,V_1,W_1,X_1)$ that does
not partition ${}[n]$, or even come close to doing so, and I would like the following to be the case. Let $Z_1$ be the complement of $U_1\cup V_1\cup W_1\cup X_1$. Then there are many $U\subset Z_1$
such that the density of $(V,W)$ with $(U_1\cup X_1\cup U,V_1\cup V,W_1\cup W)\in\mathcal{A}$ is at least $\delta+\eta$ (or perhaps very slightly less), and the same is true, with the same $U$, if we
move $X_1$ over to $V_1\cup V$ or $W_1\cup W$.
I’ve got to go in a moment, but it occurs to me that one might be able to get away with less. I’ve tried to choose a 1-dimensional subspace $S_1$ such that for many $U$ the density is good for every
assignment of the wildcards. But perhaps it’s enough just to get the average density good when we set the wildcards, and perhaps that’s easier to prove.
Incidentally, here’s a problem that ought to count as a first exercise. To make sense of it, we need to go for a functional version. So let’s suppose that $f:[3]^n\rightarrow[0,1]$ is a function of
mean $\delta$ and suppose that there is a dense set system $\mathcal{U}$ such that for every $U\in\mathcal{U}$ the average of $f(U,V,W)$ over all $(V,W)$ that make a point with $U$ is at least $\
delta+\eta$. Can we find a combinatorial line on which $f$ averages at least $\delta+\eta'$ (where $\eta'$ is $99\eta/100,$ say)?
Of course, we can do it by quoting DHJ(3) but is there an elementary proof?
That’s all I’ve got time for for now.
• gowers Says:
February 28, 2009 at 5:48 pm
836.1 Just to be clear, to make sense of that last problem, one should either use equal-slices measure or ask for $\mathcal{U}$ to be dense in the probability measure where each element of a set
is chosen independently with probability $1/3$.
Terence Tao Says:
February 28, 2009 at 7:51 pm | Reply
837. Density increment
Randall, I think the problem we’re seeing in the combinatorial world (that the pullback of a $[2]^n$-combinatorial subspace is not a $[3]^n$-combinatorial subspace) is reflected in the ergodic world
that the $\tau \rho^{-1}$ IP-system does not commute with the $\sigma \rho^{-1}$ IP-system, and so DHJ(2) for the $\tau \rho^{-1}$-invariant factor does not seem to quite give us what we want
(applying $\sigma \rho^{-1}(w)$ to a bunch of $\tau \rho^{-1}$-fibers does not give another bunch of $\tau \rho^{-1}$-fibers).
• Terence Tao Says:
February 28, 2009 at 7:52 pm
Ahh, but there’s some sort of “asymptotic commutativity” if one separates the 01 interchange and 02 interchange operations sufficiently. Let me think about this…
• Terence Tao Says:
February 28, 2009 at 8:10 pm
OK, I think I can make Randall’s argument work. In order to exploit asymptotic commutativity, I first need to weaken the notion of a 1-set to a “local 1-set”, which is a 1-set on large slices of
$[3]^n$; more precisely, A is a local 1-set if there exists a small set $I \subset [n]$ of “bad” coordinates such that whenever one fixes those coordinates (thus reducing the n-dimensional cube
to an $n-|I|$-dimensional cube), the slice of A is a 1-set. (Equivalently, A is insensitive to 0-2 interchanges on coordinates outside of I, but could be extremely sensitive to such changes
within I). Every global 1-set is a local 1-set, but not conversely.
Suppose a set B has a density increment with a global 1-set A of non-trivial size. Then by a greedy algorithm, we can find a local 1-set A’ of non-trivial size on which B has a “near-maximal
density increment” with B in the sense that any other local 1-set A” (with perhaps slightly more bad coordinates than A’, and not too much smaller than A’) does not have a significantly higher
B-density than A’ did. (There will be a lot of parameter juggling to sort out to quantify everything here; I will ignore this issue here.)
OK, let’s look at the local 1-set A’, which has a small number of bad coordinates I’. Let m be a medium sized number. Pick m random small wildcard sets $I_1, \ldots, I_m$ (which with high
probability will be disjoint from each other, and from I’).
If we then pick a word w at random from $[3]^n$, then by DHJ(2.5) and the local 1-set nature of A’, with positive probability, the combinatorial space $V(w)$ of $2^m$ elements formed by taking w
and overwriting each of $I_1, \ldots, I_m$ with 0s, 1s, or 2s, will lie in A’. Call the set of w that do this good.
The key point is that the set A” of good w is itself a local 1-set outside of the bad coordinates $I'' := I' \cup I_1 \cup \ldots \cup I_m$. And so by hypothesis, B enjoys essentially the same
density increment on A” that it did on A’. But on the other hand, A” is the union of (parallel) m-dimensional combinatorial subspaces, and so we get an density increment on one of these spaces.
• Terence Tao Says:
February 28, 2009 at 8:18 pm
small correction: V(m) should of course have $3^m$ elements, not $2^m$. (DHJ(2.5) places $2^m$ of the elements of $V(m)$ inside A’, and the local 1-set nature of A’ then automatically extends
this membership of A’ to the rest of $V(m)$.)
Ryan O'Donnell Says:
February 28, 2009 at 10:48 pm | Reply
838. Structure Theorems; specifically, question for Terry re #826:
Terry, could you clarify on your statement therein, “the energies $\| S_\lambda f \|_2^2$… are basically monotone decreasing in $\lambda$“?
Suppose $f : \{0,1\}^n \to \mathbb{R}$ is of the form $f(x) = h(\sum_{i=1}^n x_i)$, where $h : \mathbb{R} \to \mathbb{R}$ is a nonnegative increasing function. Doesn’t it seem as though $\| S_\lambda
f \|_2^2$ will be increasing in $\lambda$?
• Ryan O'Donnell Says:
February 28, 2009 at 10:49 pm
838.1 The nonparsing formula there is $f : \{0,1\}^n \to \mathbb{R}$. Now fixed — Tim
• Terence Tao Says:
February 28, 2009 at 11:03 pm
Ryan, I guess one should specify that f should be bounded within [-1,1], and “basically” means “up to o(1) errors, and with $\lambda$ small compared with $\sqrt{n}$“.
In your example h should also be bounded between [-1,1]. In that case, flipping $\lambda$ 0′s to 1′s will only have an effect of increasing f by $O(\lambda/n) = o(1)$ or so on average.
The reason for the monotone decrease is because of the absorption formula $S_{\lambda'} S_{\lambda} f \approx S_{\lambda+\lambda} f$. Since $S_{\lambda'}$ is basically a contraction for $\lambda'
= o(\sqrt{n})$ (though it does cease to contract for large $\lambda'$, of course), we see that $\| S_{\lambda+\lambda'} f \|_{L^2} \leq \| S_{\lambda} f \|_{L^2} + o(1)$.
I guess the “local” picture (small $\lambda$) is looking a bit different from the “global” picture. Locally, the operation of flipping 0s to 1s is a measure-preserving operation; globally, of
course, it isn’t. (Tragedy of the commons!)
Ryan O'Donnell Says:
February 28, 2009 at 11:04 pm | Reply
839. Fourier proof of Sperner.
I semi-checked to myself that one can prove Sperner by increment-free, purely Fourier arguments. One uses:
. the structure theorem (#821).
. Mossel’s “Lemma 6.2″ to handle the $\langle f_{\text{sens}}, f_{\text{stab}} \rangle$ and $\langle f_{\text{sens}}, f_{\text{sens}} \rangle$ parts.
. the triangle inequality argument (#815) to handle the $\langle f_{\text{stab}}, f_{\text{stab}} \rangle$ parts.
However, this argument is still quite unsatisfactory to me. For one, it requires $\delta \gg 1/\sqrt{\log \log n}$. For another, it requires selecting the parameter called $\epsilon$ in #800
(equivalently, $\lambda/n$ in Terry’s #826) after using the Structure Theorem. And most “wrongly”, this parameter must be $\Theta_\delta(1)/n$.
In particular, the following theorem is true (I’m 99% sure, at least; proof by “pushing shadows around with Kruskal-Katona”) — however, I don’t think we have any Fourier-based proof at all:
Theorem: Let $f : \{0,1\}^n \to \{0,1\}$ have density $\delta > 0$. Fix $p = 1/2 - \epsilon$, $q = 1/2 + \epsilon$, where $\epsilon$ is, say, $1/n^{.75}$. (Or set $p = 1/2$ if you like.) Then $\
mathbb{E}[f(X_{p,q})f(Y_{p,q})] > 0$, where I’m using the notation from Tim’s wiki entry.
Terence Tao Says:
February 28, 2009 at 11:06 pm | Reply
One side effect of the threaded system is that we hit 100 comments long before the number assigned to the comments increases by 100; we’re at 838 now but the thread is already longer than most of the
other threads. Perhaps we may wish to renew this thread well before 899 (e.g. at 850)?
Ryan O'Donnell Says:
February 28, 2009 at 11:23 pm | Reply
840. Sperner.
This is very tiny comment. I just wanted to point out that one can prove the “correct” density-Sperner Theorem under the uniform distribution, in the same way Tim proves it under the equal-slices
Assume $\mathbb{E}[f] = \delta$ under the uniform distribution. Pick a random chain $C$ from (0, 0, …, 0) up to (1, 1, …, 1) and then choose $u, v \sim$ Binomial$(n, 1/2)$, independently. Define $x =
C(u)$, the $u$th string in the chain, and $y = C(v)$. Then $\mathbb{E}[f(x)f(y)] = \mathbb{E}_C[\mathbb{E}_u[f(C(u))]^2] \geq \mathbb{E}_{C,u}[f(C(u))]^2 = \delta^2$, as $x = C(u)$ is uniformly
So we have a distribution on pairs of strings $(x,y)$ such that $x$ and $y$ both have the uniform distribution, and such that they form a nondegenerate “line” with probability at least $\delta^2 - \
PS: If someone wants to express $\mathbb{E}[f(x)f(y)]$ here in terms of Fourier coefficients, I’d be very happy to see it.
• gowers Says:
March 1, 2009 at 11:56 am
840.1 Ryan, I’m not sure how tiny that comment is, and must think about it. Maybe it will turn out that putting the binomial distribution on maximal chains is always more convenient than putting
equal-slices measure on the cube. And clearly this same trick works for ${}[3]^n$. This feels to me like a potentially useful further technique to add to our armoury, and when I get the chance I
think I’ll add something about it to the wiki.
I agree that the Fourier calculation looks more or less compulsory. And something that’s rather nice about this measure is that it combines the uniform distribution on ${}[2]^n$ with a very
natural non-uniform distribution on combinatorial lines, so it provides a potential answer to a question we were trying to answer way back in the Varnavides thread. Indeed, another exercise we
should do is prove a Varnavides-type version of DHJ(3) for this measure on the combinatorial lines (given DHJ(3) itself).
Oh no wait a moment — it’s not obvious how to generalize to DHJ(3) because we don’t have an analogue of maximal chains. This is something else to think about.
Gil Kalai Says:
March 1, 2009 at 7:00 am | Reply
841. A problem where some of the techniques developed here are potentially useful.
Let $f$ be a monotone Boolean function. Let $\psi_k(f)=\int I_k^p(f)$ be the integral of the influence of the k-th variable w.r.t. the probability measure $\mu_p$. ($\psi_k(f)$ is called the “Shapley
value” of $k$.) Let $\epsilon$ be a fixed small real number and let T be the difference between p-q where p is the value for which the probability that $f$ is 1 is $1-\epsilon$ and q is the value for
which the probability that $f$ is 1 is $\epsilon$. (T is called the length of the “threshold interval for $f$.)
It is known that if all $\psi_k(f) < \delta$ then T is small as well. The known bound is $T < 1/C \log \log (1/\delta)$. The proof relies on connecting the influences for different of $\mu_p$s. It
looks that one $\log$ could be eliminated and that a more careful understanding of the relation between influences and other properties of $f$ for different values of $p$ may be useful.
gowers Says:
March 1, 2009 at 12:00 pm | Reply
1. Terry I agree that with threading we need shorter threads, if you’ll excuse the dual use of the word “thread”. Let’s indeed go for 850.
2. The other thing is that I’ve been meaning to say for ages that I loved your notion that the wiki could be thought of as an online journal of density Hales-Jewett studies, and the discussions as
weekly conferences.
gowers Says:
March 1, 2009 at 12:28 pm | Reply
842. General remark
I have a very non-mathematically busy day today, so I’ve got time for just one brief post. First, let me say that I’m quite excited by Terry’s 837.2. I haven’t fully digested it yet but I have
partially digested it enough to see that it is very definitely of the sort of flavour I was hoping for. It’s frustrating not to be able to spend a few hours today playing around with the argument.
And then tomorrow and Tuesday I have two heavy teaching days.
So instead, let me throw out a small thought/question. Recall that in the dictionary I gave in 829, long arithmetic progressions go to combinatorial subspaces in ${}[2]^n$. Therefore, if the general
plan of modelling an argument on Ajtai-Szemerédi works, it will have to give a new proof of the corners theorem that is similar to their proof but avoids the use of Szemerédi’s theorem. As I’ve
already explained, that sounds discouraging at first, but I think it may in fact not be. But I’ve subsequently realized that these thoughts lead inexorably to a suggestion that Jozsef made, way back
in comment 2 (!), that one should look at a generalization of the corners problem in which you don’t have long APs in the two sets of which you take the Cartesian product. Maybe the time is ripe for
having a go at that problem, and in particular seeing whether the Ajtai-Szemerédi approach can be pushed through for it.
Ryan O'Donnell Says:
March 1, 2009 at 5:25 pm | Reply
843. Measures:
One reason I said the comment in #840 is tiny is that I’m finally starting to catch up the fact (well known to the rest of you, I think) that if you don’t mind density increment arguments (and we
certainly don’t!) then the underlying probability measure is quite unimportant. Or rather, you can pass freely between any “reasonable” measures, using the arguments sketched here on the wiki.
In fact, I think “equal-slices” is a good “base measure” to always return to. It has the nice feature that if you have equal-slice density $\delta$, by density-increment arguments you can assume that
you have density very nearly $\delta$ under almost all product measures.
Ryan O'Donnell Says:
March 1, 2009 at 5:29 pm | Reply
844. Density increment / the problem from #828:
What if $\mathcal{A} = \mathcal{B} =$ a random simple set in $[3]^n$ of density $\delta$, meaning that its $\mathcal{U}$ is a random set of density $\delta$ under the $1/3$-biased distribution?
How should we try to find a combinatorial subspace to increment on?
(Perhaps we shouldn’t; perhaps instead we should exploit the fact that this $\mathcal{A}$ is extremely 23-insensitive.)
• Terence Tao Says:
March 1, 2009 at 5:58 pm
Ryan, that’s right; being 20-insensitive, we can invoke DHJ(2.5) and get large subspaces in ${\mathcal A}={\mathcal B}$, which of course is a healthy density increment.
• Ryan O'Donnell Says:
March 1, 2009 at 7:36 pm
844.2. Ah, sorry if you said this already in an earlier comment, Terry; I’m still catching up.
Anonymous Says:
March 1, 2009 at 7:06 pm | Reply
845. Density increment in cubes
Tim, Re:842. In case you are considering corners in cubes, let me mention a few things here. Reading some recent blogs, I thought that you are close to prove a density increment result which would
imply the corner result. I write the statement first and then I will argue that it proves a Moser type result.
Given an $IP_n$ set (or an n-dimensional Hilbert cube if you wish) with $2^n$ elements. Consider the elements as vertices of a graph. The graph is c-dense, i.e. the number of edges is $c2^{2n}$. The
graph contains a huge hole, there is an $\alpha 2^n$ empty subgraph. Prove that then it contains a d-dimensional subcube on the vertices spanning at least $(c+\delta)2^{2d}$ edges. $\delta >0$ might
depend on c and $\alpha$ but not on n, and d grows with n (arbitrary slowly). Now I will try to post the next part as a new thread.
jozsef Says:
March 1, 2009 at 7:11 pm | Reply
oops, I didn’t give my name for the previous note
jozsef Says:
March 1, 2009 at 7:26 pm | Reply
846. A Moser type theorem
845.2 … and apparently I don’t know how to generate a thread.
Still going backwards, let me state the theorem which would follow from 845: Every dense subset of $[3]^n$ contains three elements, formed by taking a string with one or more wildcards $\ast^1$ and $
\ast^2$ in it and replacing the first type wildcards by 1,2 and 3, and the second type wildcards by 4,3, and 2, respectively. For example 123412,
223312, and 323212 is such triple.
jozsef Says:
March 1, 2009 at 7:29 pm | Reply
846.1. In the previous statement we were in $[4]^n$, not in $[3]^n$.
jozsef Says:
March 1, 2009 at 7:56 pm | Reply
846 contd.
With Ron Graham we proved a colouring version of 846. For any $\log n$ colouring of $[4]^n$ there is a monochromatic triple like in 846. The paper is available at
The reasonable bound for the colouring version shows that this problem might be easier to work with than with DHJ k=3. In the next post I will try to prove that 845 implies 846, that is, an
“arithmetic density increment” in a graph with a huge hole implies a Moser type statement.
(Sorry about the number of posts, but I don’t want to write long ones as even the shorter ones are full with typos)
jozsef Says:
March 1, 2009 at 9:29 pm | Reply
847. (part one)
Here we show that a density increment argument implies the Moser-type theorem in 846. It isn’t new that a density increment would give us what we are looking for, the new feature is that it might be
easier to prove density increment in a graph with a huge empty subgraph. For practical reasons we switch to a bipartite settings. Consider a bipartite graph G(A.B), between two copies of the
n-dimensional cube, $A=B=\{0,1\}^n$. G(A,B) has $2^{n+1}$ vertices and, say, $c2^{2n}$ edges. There is a natural one to one mapping between the set of edges and a pointset of $[4]^n$. The two
vertices of an edge are 0-1 sequences. From the two sequences we get one as follows. The i-th position of the new sequence (0,1,2, or 3) is given by $x + 2^y$ where x is the number (0 or 1) in the
i-th position of the vertex in A and y is the number in the i-th position of the vertex in B. We say that two edges span the same cube if substituting all 1-s and 2-s by a wildcard, *, we get the
same sequence. Observe that if (v1,v2) and (w1,w2), two edges of G(A,B), span the same cube and (v1,w2) is in G(A,B), then we are done, there are three points formed by taking a string with one or
more wildcards $\ast^1$ and $\ast^2$ in it and replacing the first type wildcards by 1,2 and 3, and the second type wildcards by 4,3, and 2, respectively. (For example $\ast^1$23$\ast^2$12 gives
123412, 223312, and 323212.) Equivalently, it will give a corner on the Cartesian product of an n-dimensional Hilbert cube. I will continue, but I would like to see if the typing is correct so far or
jozsef Says:
March 1, 2009 at 10:41 pm | Reply
847. (part two)
We said that two edges span the same cube if substituting all 1-s and 2-s by a wildcard we get the same 0,3,* sequence. Similarly we can say that every edge, e, spans a particular subcube; substitute
all 1-s and 2-s with wildcards to get a 0,3,* sequence. Any edge, which differs from e in the * positions only, is in the subcube spanned by e. Our moderate goal here is to show that if there are no
three edges of the form (v1,v2), (w1,w2), and (v1,w2) in G(A,B) that (v1,v2) and (w1,w2) span the same subcube – in which case we were done – then there is a subcube spanned by many edges containing
many other edges (see in post 845). The vertices of the spanning edges provide us the large empty subgraph.
We will follow a simple algorithm. At first consider the set of edges spanning the whole cube. Those are edges without 0 or 3. If there are at least $\alpha n$ such edges, then stop. If there are
less than $\alpha n$ such edges, then select the densest n-1 dimensional subcube. (Such subcubes are spanned by edges having exactly one 0 and no 3-s and by edges with one 3 and no 0-s.) If the
densest subcube is spanned by at least $\alpha (n-1)$ edges then stop, otherwise select the densest n-2 dimensional subcube. If we repeat the algorithm and $\alpha << c$, then it should terminate in
a c-dense d-dimensional subcube spanned by at least $\alpha 2^d$ edges. By the initial assumption there is no edge connecting the end-vertices of the spanning edges. It gives us the huge empty
bipartite graph, $K_{\alpha n/2, \alpha n/2}$. I didn’t do the actual calculations, but it seems correct to me.
gowers Says:
March 2, 2009 at 12:18 am | Reply
848. Density increment
Terry, I’ve tried to understand your argument properly but there are a few points where I’ve had difficulty working out what you mean. I think they all stem from one main point, which comes in the
last paragraph where you say “by hypothesis”. I’m not sure I know what the hypothesis is. You established that it was impossible to get a substantial density increase on a local 1-set, but here you
seem to need that you can’t have a substantial density decrease. Have I misunderstood something? I hope so. (I’m actually writing this offline in my car so by the time I get to post it perhaps this
matter will have been cleared up.)
gowers Says:
March 2, 2009 at 12:19 am | Reply
849. Different measures (also written earlier but offline)
I want to think some more about Ryan’s observation that the permutations trick can be used with the uniform measure. In particular, I want to understand from a non-permutations point of view what the
resulting measure is on pairs $A\subset B$.
So let’s fix two sets $A\subset B$ of sizes $r$ and $s$, respectively. The probability that a random permutation gives both $A$ and $B$ as initial segments is $\binom ns^{-1}\binom sr^{-1}$ and the
probability that we choose those two initial segments (if we binomially choose a random $r$ and a random $s$ without conditioning on their order) is $2^{-2n}\binom nr\binom ns.$ So the probability
that we choose the sets $A$ and $B$ is $2^{-2n}\binom nr\binom sr^{-1},$ which equals $2^{-2n}\frac{n!(s-r)!}{s!(n-r)!},$ which equals
$2^{-2n}\binom ns \binom {n-r}s^{-1}.$
Therefore, the probability of choosing $B$ conditional on having chosen $A$ is $2^{-n}\binom ns \binom {n-r}s^{-1}.$ If $A$ and $B$ were independent, this probability would of course be $2^{-n}$, so
the extra weight is … hmm, I’m not really getting anywhere here.
gowers Says:
March 2, 2009 at 12:20 am | Reply
850. Different measures (again written earlier)
How about the Fourier expansion? Suppose we choose $U$ and $V$ according to this distribution. That is, we choose a random permutation, and then choose initial segments $r$ and $s$ independently with
binomial probabilities on $r$ and $s$. What is the expected value of $w_A(U)w_B(V)$?
To make the calculation friendlier, I’m going to work out something else that I hope will turn out to be the same. It’s more like Ryan’s p-q approach. To pick a random point I’ll pick a number $m$,
binomially distributed with parameters $n$ and 1/2, and I’ll then set $p=m/n$ and choose a point with the measure $\mu_p$. Damn, that doesn’t work, because the probability that I end up choosing the
empty set is strictly greater than $2^{-n}$ (because there’s a chance I’ll pick it if $p$ takes a typical value round 1/2, and also a probability $2^{-n}$ of taking $p=0$). OK, it’s believable that
any average of binomial distributions that ends up with mean $n/2$ is guaranteed to be less concentrated than the binomial distribution itself, so probably this idea was doomed.
OK I don’t at the moment see a conceptual argument for calculating the Fourier expression. Perhaps it’s a question of a brute-force calculation that one hopes will cancel a lot. But it would be
disappointing if the expectation of $w_A(U)w_B(V)$ above were not zero when $Ae B$.
gowers Says:
March 2, 2009 at 12:22 am | Reply
Metacomment: I’m about to start a new thread, which will start at 851. So perhaps further comments on this thread could be limited to a few small replies. | {"url":"http://gowers.wordpress.com/2009/02/23/brief-review-of-polymath1/","timestamp":"2014-04-20T19:18:32Z","content_type":null,"content_length":"372858","record_id":"<urn:uuid:3b25109c-81ee-419f-8a50-86ce67ec7b9f>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00078-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: Re: [mg5303] Euclidean Matrix Norm
Replies: 0
Re: [mg5303] Euclidean Matrix Norm
Posted: Nov 26, 1996 3:41 AM
Armin Gerritsen wrote:
> Does anyone know if Mathematica (2.2.2) can calculate the Euclidean
> matrix-norm of a matrix. (== max |Ax|/|A| , where x a non-zero vector and A
> a square matrix).
The matrix norm of the square matrix A is the square root of the largest
eigenvalue of the matrix (Transpose[A] . A), so put
Norm[A_] := Sqrt[Max[N[Eigenvalues[Transpose[A].A]]]]
Of course, this gives only a numeric approximation. Life is more difficult if
you want the norm symbolically, because Mathematica can't find the Max
--Lou Talman | {"url":"http://mathforum.org/kb/thread.jspa?threadID=224371","timestamp":"2014-04-17T13:12:13Z","content_type":null,"content_length":"14094","record_id":"<urn:uuid:ad7e5e75-57f1-4227-a0e2-33a8dff229d3>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00193-ip-10-147-4-33.ec2.internal.warc.gz"} |
MathGroup Archive: July 1997 [00145]
[Date Index] [Thread Index] [Author Index]
Re: nested * and ** (rules for commutative quantities)
• To: mathgroup at smc.vnet.net
• Subject: [mg7709] Re: nested * and ** (rules for commutative quantities)
• From: Dick Zacher <dick at loc3.tandem.com>
• Date: Wed, 2 Jul 1997 14:21:29 -0400 (EDT)
• Organization: Tandem Computers
• Sender: owner-wri-mathgroup at wolfram.com
Marlies Brinksma wrote:
> I have a set of operators (let's call them Op[index_,arg2_]).
> Two of these operators commute when their indices are different and they are
> noncommutative otherwise. All operators commute with any scalar quantity.
> I would like to define some rules such that expressions like:
> Op[3,a] ** Op[2,v] ** 4 ** Op[6,s] ** Op[2,x] ** Op[1,t]
> will be automatically changed to:
> Times[4, Op[1,t],NonCommutativeMultiply[Op[2,v],Op[2,x]],Op[3,a],Op[6,s]]
> It seems to be very simple but I just cannot come to a set of rules which are
> general enough and don't lead to infinite recursion.
First Observation: There is a rather substantial noncommutative algebra
package, NCAlgebra, available on MathSource, complete with 104 pages of
documentation. I haven't used it, as this is out of my area of interest,
but it looks promising.
Second Observation: This problem is harder than it appears to be!
Below is my own brief solution to the problem you posed. It seems to
give the right behaviour, but it is excruciatingly slow for products
with many factors. For example, a product with 10 factors took 92
seconds on a SPARCstation 20. This has given me a renewed appreciation
for the efficiency of Mathematica's built-in algorithms for dealing with
commutative multiplication.
(* Define a test for whether factors commute *)
(* The first rule is specific to the problem at hand *)
(* The remaining rules for commutingQ are general *)
(* We also need rules for NonCommutativeMultiply *)
{92.41 Second,
FullForm[3*Op[j, y]**Op[j, w]*Op[i, x]**Op[i, t]^2**Op[i, r]*Op[l, e]]
Timing[Op[3,a] ** Op[2,v] ** 4 ** Op[6,s] ** Op[2,x] ** Op[1,t]]
{1.6 Second,4 Op[2,v]**Op[2,x] Op[1,t] Op[3,a] Op[6,s]}
In judging the correctness of the above results, keep in mind that
NonCommutativeMultiply has higher precedence than Times.
Dick Zacher
Performance Engineering Dept., Tandem Computers
zacher_dick at tandem.com phone: 408-285-5746 fax: 408-285-7079 | {"url":"http://forums.wolfram.com/mathgroup/archive/1997/Jul/msg00145.html","timestamp":"2014-04-20T21:38:46Z","content_type":null,"content_length":"37037","record_id":"<urn:uuid:b8c3d721-c361-46ef-b7c4-98d04a8ed8b7>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00361-ip-10-147-4-33.ec2.internal.warc.gz"} |
Array Question
11-15-2012 #16
Registered User
Join Date
Oct 2012
If I do
#include <stdio.h>
float ave(float stu[][3][4]);
float sAve(float stu[][4]);
int main()
int i, j, k;
float stu[2][3][4];
for(i=0; i<2; i++){
for(k=0; k<4; k++){
printf("Enter in the grade for student %d in subject %d", j+1, k+1);
scanf("%d", &stu[i][j][k]);
scanf("%d", &stu[i][j][k]);
printf("All Averages: %f", ave(stu));
printf("Student Averages: %f", sAve(stu));
float ave(float stu[][3][4])
float sum=0;
int i, j, k;
for(i=0; i<2; i++){
for(k=0; k<4; k++){
return (sum/24);
float sAve(float stu[][4])
float sum=0, l=3, m=4;
int i, j;
for(i=0; i<2; i++){
return (sum/12);
C:\Users\J\Desktop\gcc Lab8a.c -o Lab8a
Lab8a.c: In function 'main':
Lab8a.c:29:1: warning: passing argument 1 of 'sAve' from incompatible pointer ty
pe [enabled by default]
Lab8a.c:9:7: note: expected 'float (*)[4]' but argument is of type 'float (*)[3]
It does if you think about it...
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
#include <string.h>
#define MAX_S 10
void print_s(char (*arr2D_ptr)[][MAX_S]);
int main (void)
char arr2D[2][2][MAX_S] = {
"string 1",
"string 2"
} ,
"string 3",
"string 4"
return EXIT_SUCCESS;
void print_s(char (*arr2D_ptr)[][MAX_S])
Last edited by Click_here; 11-15-2012 at 04:20 PM. Reason: Underlined the word think.
Fact - Beethoven wrote his first symphony in C
1. Pass the whole thing to the function and only use the second and third dimension inside the function.
2. Create a temporary two dimensional array, copy the desired elements from the real array, and pass the temporary array to the function.
I think the best that you can do is.
float sAve(float stu[3][4]);
sAve(stu[x]); where x is a number of your choice (0 or 1).
I'm honestly confused, I thought I could just call 2 parts of the array to the function
I'd be tempted to make a subroutine like this:
void get_y_z_slice( int const_x_val, /* The value of x for the yz plain*/
char (*input)[MAX_X][MAX_Y][MAX_Z], /* The input 3D array */
char (*output)[MAX_Y][MAX_Z]); /* Where the resulting plain is stored */
Fact - Beethoven wrote his first symphony in C
Yes I could use two arrays, but that's not the point of doing this
You can like my second string printing example - But if you wanted a dimension (say, z) constant and to vary the other two and store it in another
arr1[x][y] = arr2[x][y][2]
You are going to have to do it manually.
Fact - Beethoven wrote his first symphony in C
Everything in my notes says this should work the way it is without having to use 2 arrays
So if I do say int array[][4][5];
and I want to just take 4 and 5 I would make a prototype
Yes I could use two arrays, but that's not the point of doing this
What is the point of doing this? What are you trying to achieve?
See I tried using pointers but it gives me an error when compiling
I showed you how to send pointers to arrays into a function/subroutine correctly.
I'm sorry, but I just can't gauge what you are after...
Fact - Beethoven wrote his first symphony in C
I thought I made that clear my mistake. The point is to use 1 multi dimensional array, then to call that array to functions to compute different averages
so I want averages of stu[2][3][4] stu[3][4] and stu[4]
Last edited by Sorinx; 11-15-2012 at 05:12 PM.
scanf's with %d need to be float (%f)
main didn't return 0 at the end
Variables m and l and unused
You can NOT pass a multidimensional array to a function/subroutine -> You must use a pointer.
I got your code working changing these things.
Fact - Beethoven wrote his first symphony in C
And by pointer to the arrays, I mean this:
float ave(float (*stu)[][3][4]);
float sAve(float (*stu)[][4]);
Fact - Beethoven wrote his first symphony in C
I get the same exact error...
11-15-2012 #17
TEIAM - problem solved
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11-15-2012 #20
11-15-2012 #21
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11-15-2012 #22
TEIAM - problem solved
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11-15-2012 #23
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search results
Expand all Collapse all Results 51 - 56 of 56
51. CMB 1998 (vol 41 pp. 404)
$L^p$-boundedness of a singular integral operator
Let $b(t)$ be an $L^\infty$ function on $\bR$, $\Omega (\,y')$ be an $H^1$ function on the unit sphere satisfying the mean zero property (1) and $Q_m(t)$ be a real polynomial on $\bR$ of degree $m$
satisfying $Q_m(0)=0$. We prove that the singular integral operator $$ T_{Q_m,b} (\,f) (x)=p.v. \int_\bR^n b(|y|) \Omega(\,y) |y|^{-n} f \left( x-Q_m (|y|) y' \right) \,dy $$ is bounded in $L^p (\
bR^n)$ for $1
Keywords:singular integral, rough kernel, Hardy space
52. CMB 1998 (vol 41 pp. 306)
Oscillatory integrals with nonhomogeneous phase functions related to Schrödinger equations
In this paper we consider solutions to the free Schr\" odinger equation in $n+1$ dimensions. When we restrict the last variable to be a smooth function of the first $n$ variables we find that the
solution, so restricted, is locally in $L^2$, when the initial data is in an appropriate Sobolev space.
Categories:42A25, 42B25
53. CMB 1998 (vol 41 pp. 49)
Stability of weighted darma filters
We study the stability of linear filters associated with certain types of linear difference equations with variable coefficients. We show that stability is determined by the locations of the poles
of a rational transfer function relative to the spectrum of an associated weighted shift operator. The known theory for filters associated with constant-coefficient difference equations is a
special case.
Keywords:Difference equations, adaptive $\DARMA$ filters, weighted shifts,, stability and boundedness, automatic continuity
Categories:47A62, 47B37, 93D25, 42A85, 47N70
54. CMB 1997 (vol 40 pp. 433)
A uniform $L^{\infty}$ estimate of the smoothing operators related to plane curves
In dealing with the spectral synthesis property for a plane curve with nonzero curvature, a key step is to have a uniform $L^{\infty}$ estimate for some smoothing operators related to the curve. In
this paper, we will show that the same $L^{\infty}$ estimate holds true for a plane curve that may have zero curvature.
Categories:42b20, 42b15
55. CMB 1997 (vol 40 pp. 296)
A general approach to Littlewood-Paley theorems for orthogonal families
A general lacunary Littlewood-Paley type theorem is proved, which applies in a variety of settings including Jacobi polynomials in $[0, 1]$, $\su$, and the usual classical trigonometric series in $
[0, 2 \pi)$. The theorem is used to derive new results for $\LP$ multipliers on $\su$ and Jacobi $\LP$ multipliers.
Categories:42B25, 42C10, 43A80
56. CMB 1997 (vol 40 pp. 169)
The class $A^{+}_{\infty}(\lowercase{g})$ and the one-sided reverse Hölder inequality
We give a direct proof that $w$ is an $A^{+}_{\infty}(g)$ weight if and only if $w$ satisfies a one-sided, weighted reverse H\"older inequality.
Keywords:one-sided maximal operator, one-sided $(A_\infty)$, one-sided, reverse Hölder inequality | {"url":"http://cms.math.ca/cmb/msc/42?page=3","timestamp":"2014-04-18T11:20:02Z","content_type":null,"content_length":"33838","record_id":"<urn:uuid:7bc87f90-f01f-4676-b96e-7e312a10de06>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00620-ip-10-147-4-33.ec2.internal.warc.gz"} |
Illustration 16.1
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Click here for help on updating Java and setting Java security.
Illustration 16.1: Representations of Simple Harmonic Motion
Please wait for the animation to completely load.
In 1610 Galileo discovered four moons of Jupiter. Each moon seemed to move back and forth in what we would call simple harmonic motion. What was Galileo really seeing? He was seeing the essentially
uniform circular motion of each moon, but he was looking at the motion edge-on. We can use what Galileo was experiencing to hand-wave some properties of simple harmonic motion by using an analogy
with uniform circular motion. Consider the above animation (position is given in meters and time is given in seconds). Restart.
First, let us look at position as a function of time. The point on the circle marked by the red ball is always at the same radius, R. If we look at the y position as a function of time we see that y
= R cos(ωt), and the x position is x = R sin(ωt). How do we know this? We can decompose the radius vector into components.
What about the velocity? Well, we know it is tangent to the path of the ball and, since the motion is uniform, the magnitude of the velocity is constant and equal to ωR. We can break up the velocity
vector into components. We get that v[y] = ωR sin(ωt) and v[x] = -ωR cos(ωt) and that both are functions of time. Watch the animation to convince yourself that this decomposition is correct as a
function of time. If we know a bit of calculus, we can take the derivative of the position with respect to time. We again get that v[y] = -ωR sin(ωt) and v[x] = ωR cos(ωt).
We also know that the acceleration is a constant, v^2/R, and points toward the center of the circle. We can again decompose this acceleration as a[y] = -ω^2R cos(ωt) and a[x] = -ω^2R sin(ωt). Again,
if we know a bit of calculus, we can take the derivative of the velocity with respect to time. We again find that a[y] = -ω^2R cos(ωt) and a[x] = -ω^2R sin(ωt). Note that since this is simple
harmonic motion there must be a relationship between the position and the force. Since force must be a linear restoring force and, since force is also mass times acceleration, we must have that ma =
- k x or that a(t) = - (k/m) and a(t) = - ω^2 x(t), which is the case if we compare our functions for y(t) and x(t) to a[y](t) and a[x](t).
For simple harmonic motion we change two things, R → A where A is called the amplitude, and we only consider one direction, in this example the y direction. This yields: y = A cos(ωt), v[] = -ωA sin
(ωt), and a = -ω^2A cos(ωt). Simple harmonic motion requires a linear restoring force, an equilibrium position, and a displacement from equilibrium.
next » | {"url":"http://www.compadre.org/Physlets/waves/illustration16_1.cfm","timestamp":"2014-04-17T13:06:15Z","content_type":null,"content_length":"16811","record_id":"<urn:uuid:f33f49b1-cc70-4293-94ec-691f1910f903>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00393-ip-10-147-4-33.ec2.internal.warc.gz"} |
Perpendicular Vectors
August 31st 2009, 05:07 PM
Perpendicular Vectors
If the vectors a=(2,3,4) and b=(10,y,z) are perpendicular, how must y and z be related?
so this is what I did:
(2,3,4) . (10,y,z)=0
y=-20/3 - 4z/3
z=-5 - 3y/4
is this ok? there are no answers in the back of the book, so I am not sure. thanks for your help!
August 31st 2009, 05:13 PM
This looks good to me. | {"url":"http://mathhelpforum.com/calculus/99988-perpendicular-vectors-print.html","timestamp":"2014-04-17T19:36:00Z","content_type":null,"content_length":"4331","record_id":"<urn:uuid:20fe30d5-f5b1-45c8-a4e9-51877d02f313>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00184-ip-10-147-4-33.ec2.internal.warc.gz"} |
Subsystems of Second-Order Arithmetic
Results 1 - 10 of 24
- Transactions of the American Mathematical Society
"... We consider the extent to which one can compute bounds on the rate of convergence of a sequence of ergodic averages. It is not difficult to construct an example of a computable Lebesgue-measure
preserving transformation of [0, 1] and a characteristic function f = χA such that the ergodic averages An ..."
Cited by 27 (4 self)
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We consider the extent to which one can compute bounds on the rate of convergence of a sequence of ergodic averages. It is not difficult to construct an example of a computable Lebesgue-measure
preserving transformation of [0, 1] and a characteristic function f = χA such that the ergodic averages Anf do not converge to a computable element of L2([0,1]). In particular, there is no computable
bound on the rate of convergence for that sequence. On the other hand, we show that, for any nonexpansive linear operator T on a separable Hilbert space, and any element f, it is possible to compute
a bound on the rate of convergence of (Anf) from T, f, and the norm ‖f ∗ ‖ of the limit. In particular, if T is the Koopman operator arising from a computable ergodic measure preserving
transformation of a probability space X and f is any computable element of L2(X), then there is a computable bound on the rate of convergence of the sequence (Anf). The mean ergodic theorem is
equivalent to the assertion that for every function K(n) and every ε> 0, there is an n with the property that the ergodic averages Amf are stable to within ε on the interval [n, K(n)]. Even in
situations where the sequence (Anf) does not have a computable limit, one can give explicit bounds on such n in terms of K and ‖f‖/ε. This tells us how far one has to search to find an n so that the
ergodic averages are “locally stable ” on a large interval. We use these bounds to obtain a similarly explicit version of the pointwise ergodic theorem, and show that our bounds are qualitatively
different from ones that can be obtained using upcrossing inequalities due to Bishop and Ivanov. Finally, we explain how our positive results can be viewed as an application of a body of general
proof-theoretic methods falling under the heading of “proof mining.” 1
- Bulletin of Symbolic Logic
"... Abstract. Propositional proof complexity is the study of the sizes of propositional proofs, and more generally, the resources necessary to certify propositional tautologies. Questions about
proof sizes have connections with computational complexity, theories of arithmetic, and satisfiability algorit ..."
Cited by 21 (0 self)
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Abstract. Propositional proof complexity is the study of the sizes of propositional proofs, and more generally, the resources necessary to certify propositional tautologies. Questions about proof
sizes have connections with computational complexity, theories of arithmetic, and satisfiability algorithms. This is article includes a broad survey of the field, and a technical exposition of some
recently developed techniques for proving lower bounds on proof sizes. Contents
- Philosophia Mathematica , 2003
"... Elementary arithmetic (also known as “elementary function arithmetic”) is a fragment of first-order arithmetic so weak that it cannot prove the totality of an iterated exponential function.
Surprisingly, however, the theory turns out to be remarkably robust. I will discuss formal results that show t ..."
Cited by 17 (5 self)
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Elementary arithmetic (also known as “elementary function arithmetic”) is a fragment of first-order arithmetic so weak that it cannot prove the totality of an iterated exponential function.
Surprisingly, however, the theory turns out to be remarkably robust. I will discuss formal results that show that many theorems of number theory and combinatorics are derivable in elementary
arithmetic, and try to place these results in a broader philosophical context. 1
- Reverse Mathematics , 2001
"... Abstract. A general method of interpreting weak higher-type theories of nonstandard arithmetic in their standard counterparts is presented. In particular, this provides natural nonstandard
conservative extensions of primitive recursive arithmetic, elementary recursive arithmetic, and polynomial-time ..."
Cited by 7 (5 self)
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Abstract. A general method of interpreting weak higher-type theories of nonstandard arithmetic in their standard counterparts is presented. In particular, this provides natural nonstandard
conservative extensions of primitive recursive arithmetic, elementary recursive arithmetic, and polynomial-time computable arithmetic. A means of formalizing basic real analysis in such theories is
sketched. §1. Introduction. Nonstandard analysis, as developed by Abraham Robinson, provides an elegant paradigm for the application of metamathematical ideas in mathematics. The idea is simple: use
model-theoretic methods to build rich extensions of a mathematical structure, like second-order arithmetic or a universe of sets; reason about what is true in these enriched structures;
- Annals of Pure and Applied Logic
"... A notion called Herbrand saturation is shown to provide the modeltheoretic analogue of a proof-theoretic method, Herbrand analysis, yielding uniform model-theoretic proofs of a number of
important conservation theorems. A constructive, algebraic variation of the method is described, providing yet a ..."
Cited by 7 (3 self)
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A notion called Herbrand saturation is shown to provide the modeltheoretic analogue of a proof-theoretic method, Herbrand analysis, yielding uniform model-theoretic proofs of a number of important
conservation theorems. A constructive, algebraic variation of the method is described, providing yet a third approach, which is finitary but retains the semantic flavor of the model-theoretic
version. 1
- BULL SYMB LOGIC , 2004
"... Paul Cohen's method of forcing, together with Saul Kripke's related semantics for modal and intuitionistic logic, has had profound effects on a number of branches of mathematical logic, from set
theory and model theory to constructive and categorical logic. Here, I argue that forcing also has a pla ..."
Cited by 6 (0 self)
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Paul Cohen's method of forcing, together with Saul Kripke's related semantics for modal and intuitionistic logic, has had profound effects on a number of branches of mathematical logic, from set
theory and model theory to constructive and categorical logic. Here, I argue that forcing also has a place in traditional Hilbert-style proof theory, where the goal is to formalize portions of
ordinary mathematics in restricted axiomatic theories, and study those theories in constructive or syntactic terms. I will discuss the aspects of forcing that are useful in this respect, and some
sample applications. The latter include ways of obtaining conservation results for classical and intuitionistic theories, interpreting classical theories in constructive ones, and constructivizing
model-theoretic arguments.
, 2006
"... Abstract. In Das Kontinuum, Weyl showed how a large body of classical mathematics could be developed on a purely predicative foundation. We present a logic-enriched type theory that corresponds
to Weyl’s foundational system. A large part of the mathematics in Weyl’s book — including Weyl’s definitio ..."
Cited by 5 (5 self)
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Abstract. In Das Kontinuum, Weyl showed how a large body of classical mathematics could be developed on a purely predicative foundation. We present a logic-enriched type theory that corresponds to
Weyl’s foundational system. A large part of the mathematics in Weyl’s book — including Weyl’s definition of the cardinality of a set and several results from real analysis — has been formalised,
using the proof assistant Plastic that implements a logical framework. This case study shows how type theory can be used to represent a non-constructive foundation for mathematics. Key words:
logic-enriched type theory, predicativism, formalisation 1
, 2001
"... We discuss the development of metamathematics in the Hilbert school, and Hilbert's proof-theoretic program in particular. We place this program in a broader historical and philosophical context,
especially with respect to nineteenth century developments in mathematics and logic. Finally, we show how ..."
Cited by 5 (2 self)
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We discuss the development of metamathematics in the Hilbert school, and Hilbert's proof-theoretic program in particular. We place this program in a broader historical and philosophical context,
especially with respect to nineteenth century developments in mathematics and logic. Finally, we show how these considerations help frame our understanding of metamathematics and proof theory today.
- THE ANNALS OF PURE AND APPLIED LOGIC , 2009
"... The metamathematical tradition, tracing back to Hilbert, employs syntactic modeling to study the methods of contemporary mathematics. A central goal has been, in particular, to explore the
extent to which infinitary methods can be understood in computational or otherwise explicit terms. Ergodic theo ..."
Cited by 5 (1 self)
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The metamathematical tradition, tracing back to Hilbert, employs syntactic modeling to study the methods of contemporary mathematics. A central goal has been, in particular, to explore the extent to
which infinitary methods can be understood in computational or otherwise explicit terms. Ergodic theory provides rich opportunities for such analysis. Although the field has its origins in
seventeenth century dynamics and nineteenth century statistical mechanics, it employs infinitary, nonconstructive, and structural methods that are characteristically modern. At the same time,
computational concerns and recent applications to combinatorics and number theory force us to reconsider the constructive character of the theory and its methods. This paper surveys some recent
contributions to the metamathematical study of ergodic theory, focusing on the mean and pointwise ergodic theorems and the Furstenberg structure theorem for measure preserving systems. In particular,
I characterize the extent to which these theorems are nonconstructive, and explain how proof-theoretic methods can be used to locate their “constructive content.” | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=241952","timestamp":"2014-04-19T00:29:09Z","content_type":null,"content_length":"35879","record_id":"<urn:uuid:d87b7659-3540-449e-a285-e73ed2d51fae>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00294-ip-10-147-4-33.ec2.internal.warc.gz"} |
Total # Posts: 7
Both car A and car B leave school at the same time, traveling in the same direction. Car A travels at a constant speed of 75 km/h, while gonzalez (gg24935) HW 02 wieman (053305132) 2 car B
travels at a constant speed of 89 km/h. How far is Car A from schoo...
A cyclist maintains a constant velocity of 4.8 m/s headed away from point A. At some initial time, the cyclist is 255 m from point A. What will be his displacement from his starting position after 88
s? Answer in units of m
Ann is driving down a street at 54 km/h. Suddenly a child runs into the street. If it takes Ann 0.722 s to react and apply the brakes, how far will she have moved before she begins to slow down?
Answer in units of m
i divide nd put the answer and tht formula u gave me gave me a wrong answer sir
A jet travels at 435 m/s. How long does it take to travel 253 m? Answer in units of s
A glacier advances at 3.6 × 10−6 cm/s. Howfarwillitmovein4.1y? Answer in units of cm
Light from the sun reaches Earth in 8.3 min. How far is the Earth from the sun? The velocity of light is 3 × 108 m/s. Answer in units of m | {"url":"http://www.jiskha.com/members/profile/posts.cgi?name=gera-pleaseeeee","timestamp":"2014-04-20T23:37:34Z","content_type":null,"content_length":"7377","record_id":"<urn:uuid:32bd7bf8-cd4a-404f-8e1c-b98e296a0498>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00245-ip-10-147-4-33.ec2.internal.warc.gz"} |
Improper Integrals
February 8th 2010, 05:09 PM #1
Nov 2009
Improper Integrals
1. Determine whether the improper integral converges or diverges.
Lower limit: 20
Upper limit: 22
First i let u = x-2, so du=dx. and i calculated what u would be as a result of the substitution: when x=22, u=20, and when x=20, u =18. I'm not sure whether I have to split and do the integral
separated into two terms.
1. Determine whether the improper integral converges or diverges.
Lower limit: 20
Upper limit: 22
First i let u = x-2, so du=dx. and i calculated what u would be as a result of the substitution: when x=22, u=20, and when x=20, u =18. I'm not sure whether I have to split and do the integral
separated into two terms.
as you have it written, this is not an improper integral.
also, given the limits of integration, the absolute value is not needed either.
recheck the problem.
Yes it is.
It should be 1/(abs(x-21))^(1/3)
I.e. Cube root
February 8th 2010, 05:23 PM #2
February 8th 2010, 05:26 PM #3
Nov 2009 | {"url":"http://mathhelpforum.com/calculus/127870-improper-integrals.html","timestamp":"2014-04-17T23:22:39Z","content_type":null,"content_length":"35474","record_id":"<urn:uuid:7499188d-6f73-4492-b0cd-063d0dcabaee>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00201-ip-10-147-4-33.ec2.internal.warc.gz"} |
Sheep Tags
This is from the Rocky Mountain Bighorn Society. Anyone interested in hunting bighorns in Colorado needs to join this group.
LiveCloud Login
"So you are wondering what's the deal with the weighted points system in place for sheep and goats in Colorado.
First off I think there is a little misunderstanding as to how the CDOW weighted draw works. Your name does not go in the hat more times it a lot more complex than that.
Here's how it was explained to me by some folks at the CDOW
You send in your application and it is assigned a number (10 digiits) After all the apps are in the CDOW picks 10 numbers from 0-9 in random order through a drawing. That number can be any
combination of 0-9, or 9-0. Then they match your assigned number to the drawn number. Example the committe draws the random number of: 8617290453, the draw sequence is 0123456789. Then, the draw
takes each application number and inverts it. Lets say your assigned number is 372016, it becomes 610273.
This allows for the numbers to be random and not have any bearing on when the application was recieved.
Then the inverted number is matched to the random draw numbers. In this case the 6 becomes a zero, because the normal 6 position is now a 0 (look at the number 8617290453, the 0 is where the 6
usually is), the 1 becomes a 6, the 0 becomes a 8, etc. the new number is then 068147. That is your assigned draw number.
Then the numbers will be put in order of smallest to largest. The smallest number has the best chance of drawing, the biggest has the worst. This totally random. Basically the lower you number the
better chance you will draw. The CDOW actually gives the license to the lowest # then keeps going down the list of increasing #s until all of the alloted tags are spoken for.
Here's and example off how an unweighted draw would work using random #s arranged low to high. there are 4 applicants with 3 points they are
In a sheep unit with 2 tags applicants 012345 and 123456 get tags. They have the lowest numbers.
Here is how the weighted pref. points work with sheep and goats and moose. You go through all the steps above.
If you have one weighted pt, the CDOW divides your random draw number by 2, if you are weighted twice, (your 5th year in the draw without a tag) they divide your number by 3.... and so on down the
line. This makes your random # smaller for each weighted point.
Here's and example using the same numbers from above but with weighted points factored in.
012345 0 weight 123456 0 weight 234567 1 weighted pt 345678 3 weighted pts
After dividing for weighted points the numbers are 12345 123456 117284 86420
The first and last applicants would get tags. In this case a lucky guy with just 3 pts no weight and a guy with 3 pts + 3 weight would draw.
It is important to remember it is possible to have smallest numbers drawn first with out dividing it with weighted points.
It is still pure luck, if you get a bad original (high) number ie 9876543210 you are still at a disadvantage to a a guy who just gets in on his 4th year may just get great numbers ie 0123456789.
Hope this helps with the issue. Understand, with the weighted, your name doesn't go into the "hat" more times, your probability of having a better number increases.
Is your head spinning yet?
The truth is once you have 3 points you could draw. The longer your in the better chance you have.
GOOD LUCK to all in the draws.
All around sheep freak"
Thanks to Sandbrew | {"url":"http://www.longrangehunting.com/forums/f92/sheep-tags-41599/index3.html","timestamp":"2014-04-18T01:36:39Z","content_type":null,"content_length":"76681","record_id":"<urn:uuid:7ee04327-2fd7-4705-99da-2072288075ff>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00134-ip-10-147-4-33.ec2.internal.warc.gz"} |
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AP Calculus AB Summer Homework Help
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Post subject: AP Calculus AB Summer Homework Help
Katniss Posted:
Tue Aug 09, 2011 1:17 pm
I have AP Calculus AB summer work, and I'm having trouble working out some of the problems. I'll probably be posting a few others as well. Any help would be
Joined: Wed Feb 17, 2010 7:33 appreciated!
x 0 1 2 3 4 5 6
f(x) -10 -8 -1 0 5 7 2
Using the table of values given above, find the average of change for f on the interval (2,5) and write an equation for the secant line passing through the
corresponding points.
Selling White Dino Star
Tony Fisher wrote:
By midnight Rox will have revealed all her secrets, told everyone that she loves us and end up banning herself tomorrow when she reads it all back.
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Post subject: Re: AP Calculus AB Summer Homework Help
KelvinS Posted:
Tue Aug 09, 2011 1:20 pm
Average of change for f on the interval (2,5) = df/dx = [7-(-1)]/(5-2) = 8/3
Joined: Mon Mar 30, 2009 5:13 pm
That's then the gradient of your line, m=8/3.
Then just solve for y=mx+c (derive c) based on the two points, y=7 when x=5, and y=-1 when x=2.
gives c= y-mx = 7-8/3*5 = ... (based on the first point)
Or, c = y-mx = -1-8/3*2 = ... (based on the second point)
If you want something you’ve never had, you’ve got to do something you’ve never done - Thomas Jefferson
Last edited by KelvinS on Tue Aug 09, 2011 1:31 pm, edited 1 time in total.
Post subject: Re: AP Calculus AB Summer Homework Help
Katniss Posted:
Tue Aug 09, 2011 1:31 pm
Kelvin Stott wrote:
Joined: Wed Feb 17, 2010 7:33 pm
Average of change for f on the interval (2,5) = df/dx = [7-(-1)]/(5-2) = 8/3
Thanks for the quick reply, but can you explain what you did there?
Selling White Dino Star
Tony Fisher wrote:
By midnight Rox will have revealed all her secrets, told everyone that she loves us and end up banning herself tomorrow when she reads it all back.
Transsexual and Proud!
Post subject: Re: AP Calculus AB Summer Homework Help
KelvinS Posted:
Tue Aug 09, 2011 1:33 pm
portal1920 wrote:
Joined: Mon Mar 30, 2009 5:13 pm
Kelvin Stott wrote:
Average of change for f on the interval (2,5) = df/dx = [7-(-1)]/(5-2) = 8/3
Thanks for the quick reply, but can you explain what you did there?
Yep, you're basically just calculating the gradient between the two points where x=2 (f=-1) and x=5 (f=7), so you can ignore all the other points/data.
I've also added some hints on the next part, above.
If you want something you’ve never had, you’ve got to do something you’ve never done - Thomas Jefferson
Post subject: Re: AP Calculus AB Summer Homework Help
Katniss Posted:
Tue Aug 09, 2011 2:18 pm
Kelvin Stott wrote:
Joined: Wed Feb 17, 2010 7:33 pm
Then just solve for y=mx+c (derive c) based on the two points, y=7 when x=5, and y=-1 when x=2.
gives c= y-mx = 7-8/3*5 = ... (based on the first point)
Or, c = y-mx = -1-8/3*2 = ... (based on the second point)
So, would the line be y=(8/3)x+(19/3)?
Selling White Dino Star
Tony Fisher wrote:
By midnight Rox will have revealed all her secrets, told everyone that she loves us and end up banning herself tomorrow when she reads it all back.
Transsexual and Proud!
Post subject: Re: AP Calculus AB Summer Homework Help
Katniss Posted:
Tue Aug 09, 2011 2:49 pm
Just came across another one.
Joined: Wed Feb 17, 2010 7:33 pm
Given that h(x) = (tan^-1)[1-(pi)(x)] is a composite function of the form h(x)=f(g(x)), find f and g.
[I think may be referring to the previous problem where f(x)=x/(x-1), and g(x)=1/(x-1), but I have no idea.]
Selling White Dino Star
Tony Fisher wrote:
By midnight Rox will have revealed all her secrets, told everyone that she loves us and end up banning herself tomorrow when she reads it all back.
Transsexual and Proud!
Post subject: Re: AP Calculus AB Summer Homework Help
KelvinS Posted:
Tue Aug 09, 2011 2:51 pm
portal1920 wrote:
Joined: Mon Mar 30, 2009
5:13 pm Kelvin Stott wrote:
Then just solve for y=mx+c (derive c) based on the two points, y=7 when x=5, and y=-1 when x=2.
gives c= y-mx = 7-8/3*5 = ... (based on the first point)
Or, c = y-mx = -1-8/3*2 = ... (based on the second point)
So, would the line be y=(8/3)x+(19/3)?
Just put x=2 and then x=5 back into your equation to make sure the line runs through those two points. Sorry for not giving you a direct answer, but you should do some
of the thinking for yourself.
If you want something you’ve never had, you’ve got to do something you’ve never done - Thomas Jefferson
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Reverse Polish Notation
Reverse Polish Notation is a way of expressing arithmetic expressions that avoids the use of brackets to define priorities for evaluation of operators. In ordinary notation, one might write
(3 + 5) * (7 – 2)
and the brackets tell us that we have to add 3 to 5, then subtract 2 from 7, and multiply the two results together. In RPN, the numbers and operators are listed one after another, and an operator
always acts on the most recent numbers in the list. The numbers can be thought of as forming a stack, like a pile of plates. The most recent number goes on the top of the stack. An operator takes the
appropriate number of arguments from the top of the stack and replaces them by the result of the operation.
In this notation the above expression would be
3 5 + 7 2 – *
Reading from left to right, this is interpreted as follows:
• Push 3 onto the stack.
• Push 5 onto the stack. Reading from the bottom, the stack now contains (3, 5).
• Apply the + operation: take the top two numbers off the stack, add them together, and put the result back on the stack. The stack now contains just the number 8.
• Push 7 onto the stack.
• Push 2 onto the stack. It now contains (8, 7, 2).
• Apply the – operation: take the top two numbers off the stack, subtract the top one from the one below, and put the result back on the stack. The stack now contains (8, 5).
• Apply the * operation: take the top two numbers off the stack, multiply them together, and put the result back on the stack. The stack now contains just the number 40. | {"url":"http://www-stone.ch.cam.ac.uk/documentation/rrf/rpn.html","timestamp":"2014-04-20T13:42:15Z","content_type":null,"content_length":"3813","record_id":"<urn:uuid:d94eb7ce-47e4-40f2-a1a3-9c4f15251d55>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00444-ip-10-147-4-33.ec2.internal.warc.gz"} |
Inductive Inference, DFAs, and Computational Complexity,” in
Results 1 - 10 of 66
, 2001
"... The paper investigates techniques for extracting data from HTML sites through the use of automatically generated wrappers. To automate the wrapper generation and the data extraction process, the
paper develops a novel technique to compare HTML pages and generate a wrapper based on their similarities ..."
Cited by 296 (7 self)
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The paper investigates techniques for extracting data from HTML sites through the use of automatically generated wrappers. To automate the wrapper generation and the data extraction process, the
paper develops a novel technique to compare HTML pages and generate a wrapper based on their similarities and di#erences. Experimental results on real-life data-intensive Web sites confirm the
feasibility of the approach. 1
- IEEE Transactions on Knowledge and Data Engineering , 2003
"... Contemporary workflow management systems are driven by explicit process models, i.e., a completely specified workflow design is required in order to enact a given workflow process. Creating a
workflow design is a complicated time-consuming process and typically there are discrepancies between the ac ..."
Cited by 233 (39 self)
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Contemporary workflow management systems are driven by explicit process models, i.e., a completely specified workflow design is required in order to enact a given workflow process. Creating a
workflow design is a complicated time-consuming process and typically there are discrepancies between the actual workflow processes and the processes as perceived by the management. TherefS3A we have
developed techniques fi discovering workflow models. Starting pointfS such techniques is a so-called "workflow log" containinginfg3SfiHfl" about the workflow process as it is actually being executed.
We present a new algorithm to extract a process modelf3q such a log and represent it in terms of a Petri net. However, we will also demonstrate that it is not possible to discover arbitrary workflow
processes. In this paper we explore a classof workflow processes that can be discovered. We show that the #-algorithm can successfqFS mine any workflow represented by a so-called SWF-net. Key words:
Workflow mining, Workflow management, Data mining, Petri nets. 1
- ACM Transactions on Software Engineering and Methodology , 1998
"... this article we describe a Markov method that we developed specifically for process discovery, as well as describe two additional methods that we adopted from other domains and augmented for our
purposes. The three methods range from the purely algorithmic to the purely statistical. We compare the m ..."
Cited by 230 (7 self)
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this article we describe a Markov method that we developed specifically for process discovery, as well as describe two additional methods that we adopted from other domains and augmented for our
purposes. The three methods range from the purely algorithmic to the purely statistical. We compare the methods and discuss their application in an industrial case study.
, 1999
"... XML is rapidly emerging as the new standard for data representation and exchange on the Web. An XML document can be accompanied by a Document Type Descriptor (DTD) which plays the role of a
schema for an XML data collection. DTDs contain valuable information on the structure of documents and thus ha ..."
Cited by 101 (4 self)
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XML is rapidly emerging as the new standard for data representation and exchange on the Web. An XML document can be accompanied by a Document Type Descriptor (DTD) which plays the role of a schema
for an XML data collection. DTDs contain valuable information on the structure of documents and thus have a crucial role in the efficient storage of XML data, as well as the effective formulation and
optimization of XML queries. In this paper, we propose XTRACT, a novel system for inferring a DTD schema for a database of XML documents. Since the DTD syntax incorporates the full expressive power
of regular expressions, naive approaches typically fail to produce concise and intuitive DTDs. Instead, the XTRACT inference algorithms employ a sequence of sophisticated steps that involve: (1)
finding patterns in the input sequences and replacing them with regular expressions to generate “general ” candidate DTDs, (2) factoring candidate DTDs using adaptations of algorithms from the logic
optimization literature, and (3) applying the Minimum Description Length (MDL) principle to find the best DTD among the candidates. The results of our experiments with real-life and synthetic DTDs
demonstrate the effectiveness of XTRACT’s approach in inferring concise and semantically meaningful DTD schemas for XML databases. 1
, 1996
"... Agents that operate in a multi-agent system need an efficient strategy to handle their encounters with other agents involved. Searching for an optimal interactive strategy is a hard problem
because it depends mostly on the behavior of the others. In this work, interaction among agents is represented ..."
Cited by 84 (2 self)
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Agents that operate in a multi-agent system need an efficient strategy to handle their encounters with other agents involved. Searching for an optimal interactive strategy is a hard problem because
it depends mostly on the behavior of the others. In this work, interaction among agents is represented as a repeated two-player game, where the agents' objective is to look for a strategy that
maximizes their expected sum of rewards in the game. We assume that agents' strategies can be modeled as finite automata. A model-based approach is presented as a possible method for learning an
effective interactive strategy. First, we describe how an agent should find an optimal strategy against a given model. Second, we present an unsupervised algorithm that infers a model of the
opponent's automaton from its input/output behavior. A set of experiments that show the potential merit of the algorithm is reported as well. Introduction In recent years, a major research effort has
been invested in desi...
- Journal of ACM , 1994
"... Abstract. We present new procedures for inferring the structure of a finite-state automaton (FSA) from its input \ output behavior, using access to the automaton to perform experiments. Our
procedures use a new representation for finite automata, based on the notion of equivalence between tesfs. We ..."
Cited by 73 (1 self)
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Abstract. We present new procedures for inferring the structure of a finite-state automaton (FSA) from its input \ output behavior, using access to the automaton to perform experiments. Our
procedures use a new representation for finite automata, based on the notion of equivalence between tesfs. We call the number of such equivalence classes the diLersL@of the automaton; the diversity
may be as small as the logarithm of the number of states of the automaton. For the special class of pennatatton aatornata, we describe an inference procedure that runs in time polynomial in the
diversity and log(l/6), where 8 is a given upper bound on the probability that our procedure returns an incorrect result. (Since our procedure uses randomization to perform experiments, there is a
certain controllable chance that it will return an erroneous result.) We also discuss techniques for handling more general automata. We present evidence for the practical efficiency of our approach.
For example, our procedure is able to infer the structure of an automaton based on Rubik’s Cube (which has approximately 10 lY states) in about 2 minutes on a DEC MicroVax. This automaton is many
orders of magnitude larger than possible with previous techniques, which would require time proportional at least to the number of global states. (Note that in this example, only a small fraction
(10-14, of the global
- Proceedings of the Eighth National Conference on Artificial Intelligence , 1990
"... This paper surveys some recent theoretical results on the efficiency of machine learning algorithms. The main tool described is the notion of Probably Approximately Correct (PAC) learning,
introduced by Valiant. We define this learning model and then look at some of the results obtained in it. We th ..."
Cited by 40 (1 self)
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This paper surveys some recent theoretical results on the efficiency of machine learning algorithms. The main tool described is the notion of Probably Approximately Correct (PAC) learning, introduced
by Valiant. We define this learning model and then look at some of the results obtained in it. We then consider some criticisms of the PAC model and the extensions proposed to address these
criticisms. Finally, we look briefly at other models recently proposed in computational learning theory. 2 Introduction It's a dangerous thing to try to formalize an enterprise as complex and varied
as machine learning so that it can be subjected to rigorous mathematical analysis. To be tractable, a formal model must be simple. Thus, inevitably, most people will feel that important aspects of
the activity have been left out of the theory. Of course, they will be right. Therefore, it is not advisable to present a theory of machine learning as having reduced the entire field to its bare
essentials. All ...
- Lecture note in AI, 1042: Adaptation and Learning in Multi-agent Systems, Lecture Notes in Artificial Intelligence , 1995
"... Agents that operate in a multi-agent system need an efficient strategy to handle their encounters with other agents involved in that system. Searching for an optimal interactive strategy is a
hard problem because it depends mostly on the behavior of the others. In this work, interaction among agents ..."
Cited by 34 (0 self)
Add to MetaCart
Agents that operate in a multi-agent system need an efficient strategy to handle their encounters with other agents involved in that system. Searching for an optimal interactive strategy is a hard
problem because it depends mostly on the behavior of the others. In this work, interaction among agents is represented as a repeated two-player game, where an agents' objective is to look for a
strategy that maximizes their expected sum of rewards in the game. We assume that agents' strategies can be modeled as finite automata. A model based reasoning approach is presented as a possible
method for learning an efficient interactive strategy. First, we describe how an agent should find an optimal strategy against a given model. Second, we present a heuristic algorithm that infers a
model of the opponent's automata from its input/output behavior. A set of experiments that show the potential merit of the algorithm is reported as well. Keywords: Opponent modeling, Model based
reasoning, Finite au...
- In Proceedings of the Seventh Annual ACM Conference on Computational Learning Theory , 1994
"... In this paper we propose a new formal model for studying reinforcement learning, based on Valiant's PAC framework. In our model the learner does not have direct access to every state of the
environment. Instead, every sequence of experiments starts in a fixed initial state and the learner is provide ..."
Cited by 32 (3 self)
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In this paper we propose a new formal model for studying reinforcement learning, based on Valiant's PAC framework. In our model the learner does not have direct access to every state of the
environment. Instead, every sequence of experiments starts in a fixed initial state and the learner is provided with a "reset" operation that interrupts the current sequence of experiments and starts
a new one (from the initial state). We do not require the agent to learn the optimal policy but only a good approximation of it with high probability. More precisely, we require the learner to
produce a policy whose expected value from the initial state is "-close to that of the optimal policy, with probability no less than 1 \Gamma ffi . For this model, we describe an algorithm that
produces such an (",ffi)-optimal policy, for any environment, in time polynomial in N , K, 1=", 1=ffi, 1=(1 \Gamma fi) and r max , where N is the number of states of the environment, K is the maximum
number of actions in a...
, 2000
"... Learning from positive data constitutes an important topic in Grammatical Inference since it is believed that the acquisition of grammar by children only needs syntactically correct (i.e.
positive) instances. However, classical learning models provide no way to avoid the problem of over-generalizati ..."
Cited by 23 (0 self)
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Learning from positive data constitutes an important topic in Grammatical Inference since it is believed that the acquisition of grammar by children only needs syntactically correct (i.e. positive)
instances. However, classical learning models provide no way to avoid the problem of over-generalization. In order to overcome this problem, we use here a learning model from simple examples, where
the notion of simplicity is defined with the help of Kolmogorov complexity. We show that a general and natural heuristic which allows learning from simple positive examples can be developed in this
model. Our main result is that the class of regular languages is probably exactly learnable from simple positive examples. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=161339","timestamp":"2014-04-20T05:03:50Z","content_type":null,"content_length":"40513","record_id":"<urn:uuid:1ec6afe9-140d-4547-9750-69a246e0e063>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00116-ip-10-147-4-33.ec2.internal.warc.gz"} |
Continuous iteration of fractals
Continuous iteration of fractals.
(Extra picture for search engine related reasons)
Like many others, I was intrigued when I first saw pictures of Julia sets and Mandelbrod sets, and I had lots of fun generating them on a computer.
But does this have anything to do with physics? Is there a dynamic system whose behavior is modeled by the startlingly simple equation that so surprisingly generates all those pretty pictures:
z -> z^2 + c
To get a continuous time evolution, we need the “Nth iterative root” of this map, a map f(z), such that:
z->f(z)->f2(z)->f3(z)-> ... ->fN(z) = z^2 + c
For large N, we would start to get something like a dynamic system in continuous time.
Taking the Nth iterative root of a function is apparently a difficult subject. I’ll just list a few cases where the answer is simple.
The map
z-> z + c
has as Nth root
z-> z+c/N
The map
z-> az
has as Nth root
z-> (a^(1/N)) z
Note that already, we come across multi-valuedness: there are N Nth roots of a complex number. We will come back to multi-valuedness later.
Mobius transformations :
z-> (az+b)/(cz+d)
form a group, and they are a representation of the group of 2X2 matrices
( a b )
( c d )
So taking the Nth root of this matrix, will produce coefficients for an Nth root Mobius transformation.
Actually, finding a group of matrices which have the same group structure as functions under composition, is an interesting approach. An online article on this is by
Aldrovandi and Freitas
Consider a non-linear dynamical system with 1 variable:
dz/dt = f(z)
Expand f(z) into a Taylor polynomial:
f(z) = a0z^0 + a1z^1 + a2z^2 + ....
Using some fairly easy differentiation rules, we can write this in a matrix form:
(z^0) ( 0 0 0 0 ...........) (z^0)
(z^1) (a0 a1 a2 a3............) (z^1)
d/dt (z^2) = ( 0 2a0 2a1 2a2 2a3 .......) (z^2)
(z^3) ( 0 0 3a0 3a1 3a2 3a3 ...) (z^3)
(...) (...........................) (...)
In a slightly modified form this matrix is called the Bell matrix of the function f.
It turns out that Bell matrices form a group, and that compositions of functions correspond to multiplications of their Bell matrices and vice versa.
So all we need to do is find the Nth root of the Bell matrix, and we will have the required Nth iterative root!
The problem is, that to truncate the infinite Bell matrix into a finite one, we need the higher columns and rows to get progressively less important. But in our case, they don’t.
At this point, I found the web site of Markus Mueller . Apparently, he had succeeded in making continuous iterations of the real case (Verhulst dynamics). He has some nice pictures of it on his web
page, and he was kind enough to explain his method to me.
First, he reverse-iterates a point a number of times by using the inverse iteration: z->sqrt(z-c). It turns out the reverse iteration has an attractive fixed point (z0). Near this fixed point, it
becomes increasingly well approximated by linear dynamics:
(z-z0) -> lambda * (z-z0)
So near this point, we can find continuous iterates. The nice idea that Markus Mueller had was to then simply forward iterate back to the original point, using “ordinary” iteration.
Because Markus Mueller uses real numbers, the multi-valuedness of the square roots and the Nth roots of lambda are not so worrying. Each choice of either the positive square root or the negative
square root leads to a different fixed point. We choose a combination with a small real-valued Nth root of lambda, since that’s the only one that will give smooth real-valued continuous iterates.
This method appears to work quite well numerically.
However, the time evolution generated in this way apparently cannot be that of a dynamical system in 1 variable. This can be seen by the fact that for example it crosses the line z=0 may times in
time, but each time follows a different trajectory afterwards: The value of z(t) itself does not define a state from which z(t+dt) is uniquely determined.
I decided to try the method on complex numbers instead of just reals. After all, it is the complex numbers which give the nice Julia pictures. Again, we run into a multi-valuedness problem. To
produce some pictures, I initially just ignored the multi-value problem, by simply choosing roots. Again, the method works numerically, and soon I was having Lissajous-like curves on my screen, that
winded through the complex plane. But unfortunately, there were usually self intersections. These intersections are bad, because they means that we do not have a dynamical system uniquely defined by
After some thought, I realized that he self-intersections were in fact linked to the multi-valuedness: If all maps and their inverses the story were single-valued, then we should not get self
There is a way to get rid of multi-valuedness: Use a Riemann surface! In our case, we need a Riemann surface that I will call “unwrapped-C”. It is basically the Riemann surface of the complex
In unwrapped-C, each number (u) can be written as 2 real numbers u=(R,theta). These are just polar coordinates of the complex plane, except that theta is no longer limited to a 2pi interval.
To multiply in unwrapped-C, we use
u1*u2 = (R1*R2,theta1+theta2)
This is the same as in ordinary C, but we retain the extra multiples of 2pi.
There is now a unique Nth root, or pth power:
u1^p = (R1^p, p*theta)
The tricky part is now to add a constant (c). On the ordinary complex plane, adding a constant is just a translation. If we construct the Riemann surface of unwrapped-C, we make a number of copies of
C, cut them open along a line from infinity to the origin, and the glue them together like multi-story spiraling parking garage. On this surface, we can do a translation by a constant (c) just as in
C. But look at the line extending from –c to the origin. If we take a small region that contains part of this line, and translate it by c, we see that the small region gets sliced in 2: each half
ending in a different floor of the parking garage.
We will just accept the fact there is a discontinuity in this map, and define z->z+c as a translation in the local copy of C within unwrapped-C.
OK, so we now have a map z->z^2+c whose inverse is single valued, at the cost of a discontinuity. So lets make pictures!
The pictures I made use a coordinate system in which the horizontal axis is theta, and the vertical axis is log(R). If we draw a Julia set on it, we will get a periodic wall paper of deformed
ordinary Julia sets. Since continuous iterations will involve back-iterating to the fixed point, I thought we might as well start near the fixed point. The fixed points are small red dots. Around the
fixed point, I make a small circle of starting M points. I then iterate these points N times, using the iterative Nth root of a “normal” iteration. These N points are joined in a line. The M lines of
N points are then normally iterated. On each line, I also put 1 small rectangle, so that you can see what a “normal” iteration does: it jumps from rectangle to rectangle along a curve.
The blue lines can be interpreted as time evolutions of the stateof the system in continuous time.
I also insert red lines going from –c to the origin. If a curve hits this line, it will “jump”.
So what do we get?
The first example is c=0, so z-> z^2.
(Click to enlarge)
This case can also be solved exactly. There is a fixed point at z=1, and a fixed point at z=0. The point z=0 is grotesquely deformed at -infinity (of the vertical scale), because of the use of
logarithm. The point z=1 is unstable, the state will evolve away from it after an infinitely small perturbation. In our coordinates, the dynamics looks like an explosion: The system goes away from z=
1 at ever increasing rate. That is, using logarithmic scale. If we used linear scale, we would see that the lines in the black part of the Julia picture all move to z=0.
Next we try to slightly move c. We use c=-0.2.
(Click to enlarge)
We again get a seemingly well behaved picture, at least for a certain region in the plane. One thing that was surprising to me, is that the border between the interior and exterior is not a
separatrix. This might have been a problem, because the border is an infinitely long line (like the coast of Britain). But we get orbits that cross the border, and points in the exterior are supposed
to go to infinity, while those on the interior attract to a fixed point! The clever way that the system fulfills both, is by becoming an oscillation of infinite amplitude, but sampled at a certain
sample rate, can appear to be always finite at each snapshot!! In other words, an infinite oscillation that looks finite when viewed through a stroboscope. Check that rectangles on lines that cross
the border are either all outside or all inside the interior.
Next is an example with c = -1.
(Click to enlarge)
This time, we are getting some trouble from the red lines attached to copies of –c. But there are still quite large regions with smooth dynamics, even some regions that appear to stay smooth for all
time. Not only the lines 0-(-c) are discontinuous, also image of those lines under z->z^2+c. So the discontinuities get worse and worse, although certain regions appear to be safe from them.
Finally an example with c = 0.012+i0.66
(Click to enlarge)
This value of c lies outside the Mandelbrod set. We therefore know that images of the point c will end up in infinity. So images of the discontinuities across our red lines will presumably also end
up at infinity. It seems that in this case, there may be no safe place from discontinuities.
It seems that the dynamical systems we find, are pretty weird. An extraordinary properly is "stroboscopy", infinite oscillation that appears finite when viewed through a stroboscope.
It is possible that a better picture of the dynamics can be found by finding a more suitable Riemann surface to plot it on. | {"url":"http://westy31.home.xs4all.nl/ContFract/Continuous_iteration_of_fractals.html","timestamp":"2014-04-20T21:03:30Z","content_type":null,"content_length":"13757","record_id":"<urn:uuid:944bfb8a-4064-4f33-83f1-51977f6f65d9>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00027-ip-10-147-4-33.ec2.internal.warc.gz"} |
[SciPy-dev] code for weave / blitz function wrapper generation
Hoyt Koepke hoytak@gmail....
Tue Apr 22 18:10:44 CDT 2008
I recently wrote a general purpose function for automatically creating
wrappers to C++ functions using weave, and I thought others might find
it useful as well. In particular, I think a good place for it would
be as a method in the weave ext_module class if others agree. I'm
also looking for thoughts, suggestions, and for more people to test
it. All the heavy lifting is done by weave, so hopefully the subtle
errors are minimal.
I've attached the code and an example (with setup.py) for people to
look at. To try it out, run ./setup.py build and then
./test_wrapper.py. The interesting function is in
Here's a brief overview:
To create a function module, you specify the python name, the cpp
function's call name, a list of argument types as strings, and whether
it returns a value. It supports argument types of (from the doc
"double" (python float)
"float" (python float)
"int" (python int)
"long int" (python int)
Blitz Arrays (converted from numpy arrays), 1 <= dim <= 11
"Array<double, dim>", (numpy dtype=float64)
"Array<float, dim>", (numpy dtype=float32)
"Array<long, dim>" (numpy dtype=int32)
"Array<long int, dim>" (numpy dtype=int32)
"Array<int, dim>" (numpy dtype=int32)
"Array<short int, dim>" (numpy dtype=int16)
"Array<short, dim>" (numpy dtype=int16)
"Array<unsigned long int, dim>" (numpy dtype=uint32)
"Array<unsigned int, dim>" (numpy dtype=uint32)
"Array<unsigned, dim>" (numpy dtype=uint32)
"Array<size_t, dim>" (numpy dtype=uint32)
"Array<unsigned short int, dim>" (numpy dtype=uint16)
C++ types:
"string" (python string)
Thus, for example, if you have the following c++ function definition:
string printData(Array<float, 2>& array2d,
Array<long int, 1>& idxarray,
const string& msg,
double a_double,
int an_int);
You could wrap it by calling
add_function_wrapper(wrappermod, "printData", "example::printData",
arglist = ["Array<float, 2>",
"Array<long int, 1>",
"int"], returns_value=True)
This would add a function to the ext_module wrappermod called
printData which simply converts everything and calls the function.
Anyway, I hope this might be useful to someone! I'll welcome any feedback.
Hoyt Koepke
UBC Department of Computer Science
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Results 1 - 10 of 228
, 2009
"... We develop fast algorithms for estimation of generalized linear models with convex penalties. The models include linear regression, twoclass logistic regression, and multinomial regression
problems while the penalties include ℓ1 (the lasso), ℓ2 (ridge regression) and mixtures of the two (the elastic ..."
Cited by 192 (6 self)
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We develop fast algorithms for estimation of generalized linear models with convex penalties. The models include linear regression, twoclass logistic regression, and multinomial regression problems
while the penalties include ℓ1 (the lasso), ℓ2 (ridge regression) and mixtures of the two (the elastic net). The algorithms use cyclical coordinate descent, computed along a regularization path. The
methods can handle large problems and can also deal efficiently with sparse features. In comparative timings we find that the new algorithms are considerably faster than competing methods.
- Journal of Machine Learning Research , 2007
"... Logistic regression with ℓ1 regularization has been proposed as a promising method for feature selection in classification problems. In this paper we describe an efficient interior-point method
for solving large-scale ℓ1-regularized logistic regression problems. Small problems with up to a thousand ..."
Cited by 153 (6 self)
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Logistic regression with ℓ1 regularization has been proposed as a promising method for feature selection in classification problems. In this paper we describe an efficient interior-point method for
solving large-scale ℓ1-regularized logistic regression problems. Small problems with up to a thousand or so features and examples can be solved in seconds on a PC; medium sized problems, with tens of
thousands of features and examples, can be solved in tens of seconds (assuming some sparsity in the data). A variation on the basic method, that uses a preconditioned conjugate gradient method to
compute the search step, can solve very large problems, with a million features and examples (e.g., the 20 Newsgroups data set), in a few minutes, on a PC. Using warm-start techniques, a good
approximation of the entire regularization path can be computed much more efficiently than by solving a family of problems independently.
- ANNALS OF STATISTICS , 2009
"... The Lasso is an attractive technique for regularization and variable selection for high-dimensional data, where the number of predictor variables pn is potentially much larger than the number of
samples n. However, it was recently discovered that the sparsity pattern of the Lasso estimator can only ..."
Cited by 122 (9 self)
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The Lasso is an attractive technique for regularization and variable selection for high-dimensional data, where the number of predictor variables pn is potentially much larger than the number of
samples n. However, it was recently discovered that the sparsity pattern of the Lasso estimator can only be asymptotically identical to the true sparsity pattern if the design matrix satisfies the
so-called irrepresentable condition. The latter condition can easily be violated in the presence of highly correlated variables. Here we examine the behavior of the Lasso estimators if the
irrepresentable condition is relaxed. Even though the Lasso cannot recover the correct sparsity pattern, we show that the estimator is still consistent in the ℓ2-norm sense for fixed designs under
conditions on (a) the number sn of nonzero components of the vector βn and (b) the minimal singular values of design matrices that are induced by selecting small subsets of variables. Furthermore, a
rate of convergence result is obtained on the ℓ2 error with an appropriate choice of the smoothing parameter. The rate is shown to be
, 904
"... We consider the empirical risk minimization problem for linear supervised learning, with regularization by structured sparsity-inducing norms. These are defined as sums of Euclidean norms on
certain subsets of variables, extending the usual ℓ1-norm and the group ℓ1-norm by allowing the subsets to ov ..."
Cited by 97 (15 self)
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We consider the empirical risk minimization problem for linear supervised learning, with regularization by structured sparsity-inducing norms. These are defined as sums of Euclidean norms on certain
subsets of variables, extending the usual ℓ1-norm and the group ℓ1-norm by allowing the subsets to overlap. This leads to a specific set of allowed nonzero patterns for the solutions of such
problems. We first explore the relationship between the groups defining the norm and the resulting nonzero patterns, providing both forward and backward algorithms to go back and forth from groups to
patterns. This allows the design of norms adapted to specific prior knowledge expressed in terms of nonzero patterns. We also present an efficient active set algorithm, and analyze the consistency of
variable selection for least-squares linear regression in low and high-dimensional settings.
, 2006
"... Variable selection plays an important role in high dimensional statistical modeling which nowa-days appears in many areas and is key to various scientific discoveries. For problems of large
scale or dimensionality p, estimation accuracy and computational cost are two top concerns. In a recent paper, ..."
Cited by 90 (12 self)
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Variable selection plays an important role in high dimensional statistical modeling which nowa-days appears in many areas and is key to various scientific discoveries. For problems of large scale or
dimensionality p, estimation accuracy and computational cost are two top concerns. In a recent paper, Candes and Tao (2007) propose the Dantzig selector using L1 regularization and show that it
achieves the ideal risk up to a logarithmic factor log p. Their innovative procedure and remarkable result are challenged when the dimensionality is ultra high as the factor log p can be large and
their uniform uncertainty principle can fail. Motivated by these concerns, we introduce the concept of sure screening and propose a sure screening method based on a correlation learning, called the
Sure Independence Screening (SIS), to reduce dimensionality from high to a moderate scale that is below sample size. In a fairly general asymptotic framework, the SIS is shown to have the sure
screening property for even exponentially growing dimensionality. As a methodological extension, an iterative SIS (ISIS) is also proposed to enhance its finite sample performance. With dimension
reduced accurately from high to below sample size, variable selection can be improved on both speed and accuracy, and can then be ac-
, 2008
"... For supervised and unsupervised learning, positive definite kernels allow to use large and potentially infinite dimensional feature spaces with a computational cost that only depends on the
number of observations. This is usually done through the penalization of predictor functions by Euclidean or H ..."
Cited by 77 (20 self)
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For supervised and unsupervised learning, positive definite kernels allow to use large and potentially infinite dimensional feature spaces with a computational cost that only depends on the number of
observations. This is usually done through the penalization of predictor functions by Euclidean or Hilbertian norms. In this paper, we explore penalizing by sparsity-inducing norms such as the ℓ
1-norm or the block ℓ 1-norm. We assume that the kernel decomposes into a large sum of individual basis kernels which can be embedded in a directed acyclic graph; we show that it is then possible to
perform kernel selection through a hierarchical multiple kernel learning framework, in polynomial time in the number of selected kernels. This framework is naturally applied to non linear variable
selection; our extensive simulations on synthetic datasets and datasets from the UCI repository show that efficiently exploring the large feature space through sparsity-inducing norms leads to
state-of-the-art predictive performance. 1
, 2007
"... It is now well understood that (1) it is possible to reconstruct sparse signals exactly from what appear to be highly incomplete sets of linear measurements and (2) that this can be done by
constrained ℓ1 minimization. In this paper, we study a novel method for sparse signal recovery that in many si ..."
Cited by 76 (5 self)
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It is now well understood that (1) it is possible to reconstruct sparse signals exactly from what appear to be highly incomplete sets of linear measurements and (2) that this can be done by
constrained ℓ1 minimization. In this paper, we study a novel method for sparse signal recovery that in many situations outperforms ℓ1 minimization in the sense that substantially fewer measurements
are needed for exact recovery. The algorithm consists of solving a sequence of weighted ℓ1-minimization problems where the weights used for the next iteration are computed from the value of the
current solution. We present a series of experiments demonstrating the remarkable performance and broad applicability of this algorithm in the areas of sparse signal recovery, statistical estimation,
error correction and image processing. Interestingly, superior gains are also achieved when our method is applied to recover signals with assumed near-sparsity in overcomplete representations—not by
reweighting the ℓ1 norm of the coefficient sequence as is common, but by reweighting the ℓ1 norm of the transformed object. An immediate consequence is the possibility of highly efficient data
acquisition protocols by improving on a technique known as compressed sensing.
"... Proofs subject to correction. Not to be reproduced without permission. Contributions to the discussion must not exceed 400 words. Contributions longer than 400 words will be cut by the editor. 1
2 ..."
Cited by 60 (2 self)
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Proofs subject to correction. Not to be reproduced without permission. Contributions to the discussion must not exceed 400 words. Contributions longer than 400 words will be cut by the editor. 1 2
- JOURNAL OF MACHINE LEARNING RESEARCH , 2007
"... Recently, a lot of attention has been paid to ℓ1-regularization based methods for sparse signal reconstruction (e.g., basis pursuit denoising and compressed sensing) and feature selection (e.g.,
the Lasso algorithm) in signal processing, statistics, and related fields. These problems can be cast as ..."
Cited by 57 (4 self)
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Recently, a lot of attention has been paid to ℓ1-regularization based methods for sparse signal reconstruction (e.g., basis pursuit denoising and compressed sensing) and feature selection (e.g., the
Lasso algorithm) in signal processing, statistics, and related fields. These problems can be cast as ℓ1-regularized least-squares programs (LSPs), which can be reformulated as convex quadratic
programs, and then solved by several standard methods such as interior-point methods, at least for small and medium size problems. In this paper, we describe a specialized interior-point method for
solving large-scale ℓ1-regularized LSPs that uses the preconditioned conjugate gradients algorithm to compute the search direction. The interior-point method can solve large sparse problems, with a
million variables and observations, in a few tens of minutes on a PC. It can efficiently solve large dense problems, that arise in sparse signal recovery with orthogonal transforms, by exploiting
fast algorithms for these transforms. The method is illustrated on a magnetic resonance imaging data set.
, 2009
"... Techniques from sparse signal representation are beginning to see significant impact in computer vision, often on non-traditional applications where the goal is not just to obtain a compact
high-fidelity representation of the observed signal, but also to extract semantic information. The choice of ..."
Cited by 44 (1 self)
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Techniques from sparse signal representation are beginning to see significant impact in computer vision, often on non-traditional applications where the goal is not just to obtain a compact
high-fidelity representation of the observed signal, but also to extract semantic information. The choice of dictionary plays a key role in bridging this gap: unconventional dictionaries consisting
of, or learned from, the training samples themselves provide the key to obtaining state-of-theart results and to attaching semantic meaning to sparse signal representations. Understanding the good
performance of such unconventional dictionaries in turn demands new algorithmic and analytical techniques. This review paper highlights a few representative examples of how the interaction between
sparse signal representation and computer vision can enrich both fields, and raises a number of open questions for further study. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=526567","timestamp":"2014-04-24T06:46:56Z","content_type":null,"content_length":"39355","record_id":"<urn:uuid:5c3bab5b-d3ce-42c9-931f-896bd7d67e9d>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00612-ip-10-147-4-33.ec2.internal.warc.gz"} |
the first resource for mathematics
London Mathematical Society Monographs Series 33. Princeton, NJ: Princeton University Press (ISBN 978-0-691-12437-7/hbk). xiv, 362 p. $ 65.00; £ 38.95 (2007).
This monograph describes both the theory of the small sieve (principally in the case of sieving dimension one) and its applications to prime numbers. In the early days of the sieve it had been
believed that the methods were incapable of proving the existence of primes, because of the “parity phenomenon”, but this idea is long out of date, as the book amply proves.
We start with chapters on the history of sieve ideas, on the Vaughan identity and its generalizations, on the Rosser–Iwaniec sieve, and on the “Alternative Sieve”. This last chapter introduces a very
fruitful technique. If one wants to find primes in a difficult set $𝒜$ one looks for a simpler set $ℬ$ for which one can calculate the sifting functions which occur. Then, if the sieve data $#{𝒜}_{d}
$ and $#{ℬ}_{d}$ match up for a suitable set of values of $d$, one will be able to estimate sifting functions for $𝒜$ in terms of the corresponding functions for $ℬ$. A simple example is discussed,
in which $𝒜$ is the Piatetski-Shapiro sequence $\left[{n}^{\gamma }\right]$ and $ℬ$ is the set of integers in an interval $\left(x,2x\right]$.
These ideas are then put into practice for a simple upper bound sieve, which is fed into Chebyshev’s method to show that any interval $\left(x,x+{x}^{1/2}\right]$ with $x$ sufficiently large contains
an integer whose largest prime factor exceeds ${x}^{0·74}$.
The subsequent chapters present further applications, developing these ideas. They concern primes in the interval $\left[x,x+{x}^{\theta }\right]$, the Brun–Titchmarsh Theorem on average, and primes
in almost-all short intervals. The vector sieve is then introduced, with an application to Goldbach numbers in short intervals.
The last three substantive chapters show how sieve methods can be used in algebraic number fields. The first of these presents results on the distribution of Gaussian primes in narrow sectors and
small discs. The second describes the work of Fouvry and Iwaniec on primes ${a}^{2}+{p}^{2}$, and of Friedlander and Iwaniec on primes ${a}^{2}+{b}^{4}$. Finally, there is a discussion of the
reviewer’s work on primes of the form ${a}^{3}+2{b}^{3}$. The book ends with an epilogue, five appendices covering basic analytic techniques, and a bibliography of 168 items.
This volume takes the reader right up to the leading edge of current research. The area is one in which multiple case-by-case analyses and detailed calculations are sometimes unavoidable. The book is
therefore unsuitable for lazy students. However those who want to make a serious study of the area will appreciate the author’s unified approach to the methods which have been employed, and the
wealth of applications described. The book is to be recommended to anyone with an interest in sieves, from beginning PhD students to established researchers.
11N36 Applications of sieve methods
11N35 Sieves
11N05 Distribution of primes
11N25 Distribution of integers with specified multiplicative constraints
11-02 Research monographs (number theory) | {"url":"http://zbmath.org/?q=an:1220.11118&format=complete","timestamp":"2014-04-17T06:44:54Z","content_type":null,"content_length":"25342","record_id":"<urn:uuid:6260ae17-fc57-4740-82d6-84711fc106c7>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00004-ip-10-147-4-33.ec2.internal.warc.gz"} |
Inductor Spice Model
Spice model of inductor including stray capacitances:
Red=input voltage
Green=input current
Blue=load voltage
Yellow=load current
This system (including reasonable leakage capacitance) representing a 50 ohm source driving a 100uH inductor with stray capacitance of 9pF and a 50pF load (small whip), shows input and output
currents are essentially equal, while output voltage greatly increases at resonance.
In the following SPICE plot, increasing stray capacitance shows expected departure from equal input and output currents. Distributed capacitance to ground is now at 45pF, nearly equal to the 50pF
antenna capacitance:
Again despite stray capacitance nearly equaling load capacitance, current levels at input and output are reasonably similar:
Green=source current
Yellow=load current
Red=source voltage
Blue=load voltage
If we multiply output voltage times current, the apparent power is many dozens of times the input power. The reason is simple, phase relationship between voltage and current along the inductor
changes even while current remains essentially uniform.
Although casual experimenters often assume a loading coil "replaces" the missing length in an antenna and has sinking voltage with rising current, this is not the case unless stray capacitance is
extreme or the inductor is incorrectly and inefficiently operated at or near self-(series) resonance. The loading inductor really just corrects power factor, and bringing voltage and current into
phase at resonance. It does replace missing length by simulating an antenna.
Linear loading is no different, it simply behaves like a very lossy low-Q coil and does not add radiation to the system. Unless of course we are talking about heat radiation. | {"url":"http://www.w8ji.com/inductor_spice_model.htm","timestamp":"2014-04-17T00:49:25Z","content_type":null,"content_length":"3270","record_id":"<urn:uuid:f4974dde-b76c-4152-9865-f355528374e6>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00610-ip-10-147-4-33.ec2.internal.warc.gz"} |
Fraction Bars Author Resume
Vitae and Publications
Name of institutions, Dates of attendance, Degrees, and year received
Maine Maritime Academy 1951-1954 B.S. Marine Science (1954)
University of Maine 1956-1958 B.S. Education (1958)
University of Maine 1958-1959 M.A. Mathematics (1959)
University of Michigan 1963-1966 D. Ed. Mathematics (1966)
Maine Maritime Academy - Valedictorian
University of Maine - Graduate Scholarship
University of Michigan - two years of National Science Foundation full support University of Michigan - one year teaching fellowship
Maine State Department of Education, Assistant Mathematics Television Teacher, 1959-61
University of Maine at Portland-Gorham, Assistant Professor of Mathematics, 1961-63 and 1966-67
University of New Hampshire, Assistant Professor of Mathematics 1967-72
University of Oregon, Visiting Associate Professor of Mathematics 1974-75
University of New Hampshire, Associate Professor of Mathematics, 1972-87
University of New Hampshire, Professor of Mathematics Education, 1987-2009
University of New Hampshire, Emertius Professor of Mathematics Education
Bennett, Albert B. Jr., and Musser, Gary L., "A Concrete Approach to Integer Addition and Subtraction", Arithmetic Teacher, (May 1976), 332-36.
Bennett, Albert B., Jr., "Star Patterns", Arithmetic Teacher (January 1978), 12-14.
Bennett, Albert B. Jr. and Schiot, Peter, "Fraction games - versions and revisions", Teaching Mathematics: Strategies That Work, Northeast Regional Exchange, Inc., Chelmsford, MA, 1985.
Bennett, Albert B. Jr., Decimal Squares, Scott Resources, Inc., Fort Collins, CO, 1982, 2nd Edition 1992, 3rd Edition 2003. (A program of games, activities, tests and manipulative materials.
Includes: Decimal Squares Step by Step Teacher's Guide and Decimal Squares Step by Step Workbook).
Bennett, Albert B. Jr., Eugene Maier, and L. Ted Nelson, Math and the Mind's Eye, Math Learning Center, Salem, Oregon, 1987.
Bennett, Albert B. Jr., "Visual Thinking and Number Relationshiops". Mathematics Teacher, (April 1988) 267-72.
Bennett, Albert B. Jr., "Visualizing the Geometric Series". Mathematics Teacher, (February 1989) 130-36.
Bennett, Albert B. Jr., and Foreman, Linda, Mathematics Alive, Math Learning Center, Salem, Oregon, 1989, 1995.
Bennett, Albert B. Jr., "Fraction Patterns - Visual and Numerical." Mathematics Teacher, 82 (April 1989) 254-59.
Bennett, Albert B. Jr., Laurie J. Burton and Leonard T. Nelson, Mathematics For Elementary Teachers: A Conceptual Approach, McGraw Hill Publishers, New York, NY, 9^th Edition, 2012.
Bennett, Albert B. Jr., Laurie J. Burton and Leonard T. Nelson, Mathematics For Elementary Teachers: An Activity Approach, McGraw Hill Publishers, New York, NY, 9^th Edition, 2012.
Bennett, Albert B. Jr. and Davidson, Patricia S., Fraction Bars, Scott Resources, Inc., Fort Collins, CO, 1973, 1st Edition, 1973. (A program of games, activities, tests and manipulative materials.
Includes: Fraction Bars Step by Step Teacher's Guide and Fraction Bars Step by Step Workbook.)
Bennett, Albert B. Jr.., Fraction Bars, Scott Resources, Inc., Fort Collins, CO, 2009, 5th Edition, 2009. (A program of games, activities, tests and manipulative materials. Includes: Fraction Bars
Teacher's Guides Grades 1-2, 3-4 and 5-8, and Fraction Bars Classroom Centers Grades 1-2, 3-4, and 5-8.)
Bennett, Albert B. Jr., and Nelson, Leonard T., "A Conceptual Model for Solving Percent Problems", Mathematics Teaching in the Middle School (April 1994), 20-25.
Bennett, Albert B. Jr., and Nelson, Leonard T. "Divisibility Tests: So Right for Discoveries," Mathematics Teaching in the Middle School (April 2002), 460-464.
Computer Software
Bennett, Albert B. Jr., Fraction Bars Online Games Set 2, 2007.
Bennett, Albert B. Jr. and Bennett, Albert B. III, Fraction Bars Online Games Set 1, 2002.
Bennett, Albert B. Jr., Decimal Squares Online Games, 2009.
Bennett, Albert B. Jr. and Bennett, Albert B., III, The Mathematics Investigator, software to accompany Mathematics for Elementary Teachers: A Conceptual Approach, McGraw Hill Publishers, New York,
NY, 1998. | {"url":"http://fractionbars.com/Author.html","timestamp":"2014-04-17T04:03:17Z","content_type":null,"content_length":"17399","record_id":"<urn:uuid:197febce-5676-4459-9260-c4d6092bbb17>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00413-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Researchers find a way to keep quantum memory and logic in synch
A quantum computer, like any other computer, requires a way to store and retrieve information. In other words, some sort of memory. But because of the rich quantum entanglement gooey center of
quantum computing, the memory and the logic need to be linked in a manner that's very different from that in classical computing: the magic of entanglement.
Physicists have been crowing about how they can create entangled states for a while now. Unfortunately, quantum computing requires something more: a memory state should last for a long time,
independently of the logic parts, while the logic part should be accessible and fast. This makes coupling these two elements together in a useful way difficult... until now, that is.
Its the decoherence, stupid
Quantum computing, at a practical level, is all about how predictably things change with time, called coherence. If you give a swing a push, then, after a few measurements of the swing's position,
you can use that data to predict the swing's position at any time in the future. This is because the eddies in the air motion around the swing simply don't have much effect on the swing's motion. In
the quantum case, the qubit is in a superposition of two states, and the relative probability of measuring the qubit in one of them changes predictably with time (and, in fact, changes very much like
the position of the swing with time).
Now imagine that some obnoxious shot-putter is throwing large metal balls at the swing after you give it a push. Every time one hits the swing, it noticeably changes the swing's motion and, in a very
short time, you can't use previous measurements to predict the future.
The quantum world is filled with tiny swings that we can set into motion. Unfortunately, it is also filled with obnoxious, swing-hating shot-putters. We may set a qubit into motion, but the
predictability of its behavior decays very, very quickly.
Fortunately, it doesn't have to be this way. What we can do is give the qubit an additional push every now and again to keep its motion predictable. We simply need to make sure that the pushes are
always at the right time in the cycle (just like pushing a swing) and occur often enough that our pushes are more significant than the random lurches due to the shot-putter.
This technique allows one to extend the coherence of a qubit by quite a bit, and is well known in physics (your local MRI scanner will likely use the same technique).
Physicists have known how to extend the coherence of a qubit for a long time. But in quantum computing, you may want to store two entangled qubits that evolve at different rates. Then, extending the
coherence of one is likely to destroy the coupling between the two qubits. In other words, if we preserve the coherence of our logic qubit, we cannot connect it to the memory qubit, and, if we don't
extend the coherence, it will destroy itself before any memory operations can be completed.
All you coherence are belong to us
To solve this problem, researchers from a number of institutions took a more careful approach to the preservation of coherence. They realized that, even though the two qubits evolve at their own
pace, there are periodic moments where they are at the same point in their cycle. If they give qubit a push at just that moment, then they still preserve the coherence of the logic qubit, but do not
destroy its entanglement with the memory qubit.
You can think of this in terms of music. If you take two strings on a guitar and tune them so that they are nearly—but not quite—the same frequency, then you will not only hear both tones, but they
will oscillate in volume. This oscillation is the frequency difference between the two vibrations, and creates what are called beat notes. Our quantum system has the same beat notes, and if we choose
to apply a push to the qubit at the same point in this beat note oscillation, we will preserve the coherence of the fast decaying qubit (logic) without altering the state of the slow decaying qubit
In this case, the researchers were working with nitrogen vacancy centers in diamond—not guitars. Diamond consists of carbon atoms, arranged such that each carbon atom is connected to four others. But
when nitrogen is introduced, one carbon atom is spurned by the nitrogen, leaving an electron that has a set of unique levels.
The spin—the direction the electron's internal magnetic field is pointing—provides a nice pair of states that can be set and read using light. Although it's great for computation, the electron is
very sensitive to the environment and its coherence never lasts very long (typically a microsecond or so), so microwave pulses are used to flip the spin and preserve the coherence.
The nitrogen nuclear spin can also be used as a qubit. Because the nuclear magnetic moment is much smaller, it is much less sensitive to the environment, so it stays around for milliseconds—just what
we want for memory. However, it takes a few microseconds to perform an operation on a nuclear spin qubit, by which time the electronic qubit has long since gone off to do other things.
Nevertheless, the electron's coherence lasts long enough that the nuclear spin undergoes many oscillations before it decays away. So, the researchers identified times in those oscillations when the
microwave pulses would have no influence on the nuclear spin—and hit the electron with spin flips that coincide with those points. That is, they hit the beat note between the slow nuclear spin and
the fast electron spin.
They demonstrated that they could perform basic logic operations, and performed one of the classic "look, we made a quantum computer" demonstrations: a search of a database containing four items.
Although that sounds a bit cynical, this is a significant step, because it provides a convenient and natural link between a qubit that is near-perfect for logic operations with a qubit that is
near-perfect for memory operations.
Nature, 2012, DOI: 10.1038/nature10900 | {"url":"http://arstechnica.com/science/2012/04/pushing-two-diamond-swings-to-make-a-better-quantum-logic-gate/","timestamp":"2014-04-17T15:43:43Z","content_type":null,"content_length":"43198","record_id":"<urn:uuid:00cb3211-957e-47ef-9731-ddfa9e19bbff>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00270-ip-10-147-4-33.ec2.internal.warc.gz"} |
matrix proving problem
August 4th 2010, 08:12 AM #1
Aug 2010
Suppose u and v are solutions to the linear system Ax=b. Show that if scalars α and β satisfy α+β=1, then αu+βv is also a solution to the linear system Ax=b.
Just plug $\alpha u+\beta v$ for $x$ into the left side of $Ax=b$ and use the linearity of applying $A$ to that vector to simplify everything to just $b$
$Ax = A(\alpha u+\beta v)=\alpha (Au)+\beta (Av)=\ldots = b$
August 4th 2010, 10:29 AM #2 | {"url":"http://mathhelpforum.com/advanced-algebra/152775-matrix-proving-problem.html","timestamp":"2014-04-19T08:37:32Z","content_type":null,"content_length":"33598","record_id":"<urn:uuid:15a0e64d-40f8-45a8-aee7-06cb1677d6b0>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00241-ip-10-147-4-33.ec2.internal.warc.gz"} |
Multiple choice questions from Chi-Square test
1. 198114
Multiple choice questions from Chi-Square test
1. In a chi-square test, the sample data are called observed frequencies.
2. One advantage of the chi-square tests is that they can be used when the data are measured on a nominal scale.
3. A chi-square test for goodness of fit is used to evaluate a hypothesis about how a population is distributed across three categories. If the researcher uses a sample of n = 100 participants, then
the chi-square test will have df = 99.
4. The chi-square test for independence requires that each individual be categorized on two separate variables.
5. For the chi-square test for goodness of fit, what is the value of df for a test with four categories and a sample of n = 100?
6. The chi-square distribution is
positively skewed, with no values less than zero.
negatively skewed, with no values greater than zero.
symmetrical, centered at a value determined by the degrees of freedom.
symmetrical, centered at a value of zero.
7. A researcher obtains a value of -8.50 for a chi-square statistic. What can you conclude because the value is negative?
the observed frequencies are consistently larger than the expected frequencies
The researcher made a mistake. The value of chi-square cannot be negative.
the expected frequencies are consistently larger than the observed frequencies
there are large differences between the observed and expected frequencies
8. A chi-square test for independence is used to evaluate the relationship between two variables. If one variable is classified into 4 categories and the other variable is classified into 2
categories, then the chi-square statistic will have
df = 6
cannot determine the value of df from the information provided
df = 3
df = 8
9. The sample data for a chi-square test are called ________.
observed proportions
observed frequencies
expected proportions
expected frequencies
10. In the chi-square test for independence, the null hypothesis states that
there is no relationship between the two variables being examined.
there is a relationship between the two variables being examined.
there is no difference between the two variables being examined.
there is a difference between the two variab
Answers to Multiple choice questions from Chi-Square test is given in the answer. | {"url":"https://brainmass.com/statistics/chi-squared-test/198114","timestamp":"2014-04-21T00:21:36Z","content_type":null,"content_length":"28391","record_id":"<urn:uuid:eec70a02-5e2b-40d5-ae69-1a84a926a343>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00240-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mathematical modelling of boundary layer flow and heat transfer in forced convection
Raja Ismail, Raja Mohd. Taufika (2006) Mathematical modelling of boundary layer flow and heat transfer in forced convection. Masters thesis, Universiti Teknologi Malaysia, Faculty of Science.
PDF (Full text)
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A mathematical model for the boundary layer flow and heat transfer in forced convection is developed. Boundary layer is a narrow region of thin layer that exists adjacent to the surface of a solid
body where the effects of viscosity are obvious, manifested by large flow velocity and temperature gradient. The concept of boundary layer was first introduced by Ludwig Prandtl (1875-1953) in 1905.
The derivation of both velocity and temperature boundary layer equations for flow past a horizontal flat plate and semi-infinite wedge are discussed. The velocity and temperature boundary layer
equations are first transformed into ordinary differential equations via a similarity transformation. The differential equations corresponding to the flow past a horizontal flat plate and a
semi-infinite wedge are nonlinear and known respectively as the Blasius and the Falkner-Skan equation. The approximate solutions of these equations are obtained analytically using a series expansion,
namely the Blasius series and an improved perturbation series using the Shanks transformation. The solutions presented include the velocity and temperature profiles, the skin friction and the heat
transfer coefficient.
Item Type: Thesis (Masters)
Additional Information: Thesis (Master of Science (Mathematics)) - Universiti Teknologi Malaysia, 2006; Supervisor : Prof. Dr. Norsarahaida S. Amin
Uncontrolled Keywords: Mathematical model; boundary layer flow; heat transfer; forced convection
Subjects: Q Science > QA Mathematics
Divisions: Science
ID Code: 3715
Deposited By: Ms Zalinda Shuratman
Deposited On: 26 Jun 2007 03:19
Last Modified: 05 Jul 2012 00:08
Repository Staff Only: item control page | {"url":"http://eprints.utm.my/3715/","timestamp":"2014-04-17T09:47:57Z","content_type":null,"content_length":"20487","record_id":"<urn:uuid:b61bc915-ec09-40ea-a47d-73bfccfd11bd>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00504-ip-10-147-4-33.ec2.internal.warc.gz"} |
The counting principle: permuations and combinations
Permutation and Combinations Handout
Section 1: the counting principle
7. How many sequences of two letters each can be formed?
8. How many three-digit numerals can be written using the symbols 6, 7, and 8?
9. How many different ways can ten questions on a true-false test be answered?
10. An automobile manufacturer produces 7 models, each available in 6 different colors. In addition, the buyer can choose one of 4 different upholstery fabrics and one of 5 different colors for the
interior. How many varieties of cars can be ordered from the manufacturer?
11. How many different telephone numbers can be formed from (a) two different letters and five digits if the first digit cannot be 0; (b) seven digits if the third cannot be O?
12. In long-distance direct dialing, the area code consists of three digits and the local number of seven digits. Taking the area code into consideration, how many telephone numbers can be formed if
the sixth digit cannot be O?
13. Each row of a four-rowed signal device contains a red and a green light. If at most one light can be lit in any one row, how many different signals can be sent by this device?
14. A witness to a holdup reports that the license of the getaway car consisted of 6 different digits. He remembers the first three but has forgotten the remainder. How many licenses do the police
have to check?
15. In how many ways can you write four-digit numerals, using the digits 3, 4, 5, 6, and 7, if you may use a digit as many times as desired in any one numeral?
16. How many three-letter arrangements can be made of the letters P, R, I, M, and E, if any letter may be repeated?
17. How many positive odd integers whose numerals contain three digits can be formed, using the digits 1,2,3,4, and 5? (Hint: Fill the units place with an odd digit; then fill the remaining places.)
18. How many positive even integers of three digits can be formed from the digits 1,2,3,4, 5, and 6? (Hint: Fill the units place with an even digit; then fill the remaining places.)
19. How many positive odd integers less than 70,000 can be represented, using the digits 2,3,4, 5, and 6?
20. How many positive integers < 8000 can be represented, using 3, 5, 6, 7?
21. How many auto license plates of four symbols can be made in which at least two of the symbols are letters and the rest are digits?
22. How many three-letter code words can be made if at least one of the letters must be one of the vowels: A, E, I, 0, or U?
Section 2: permutations
1. In how many ways can the letters in the word PHOENIX be arranged if each letter is used only once in each arrangement?
2. Seven salesmen are to be assigned to seven different counters in a department store. How many ways can the assignment be made?
3. A school has six sections of first-year algebra. In how many ways can a pair of twins be assigned to algebra classes if their parents have requested that they be placed in different classes?
4. A symphony concert is to consist of five works: 2 modern, 2 of the classical period, and a piano concerto for soloist and orchestra. If the concerto is the last work on the program, how many ways
can the program be arranged?
5. How many 4-letter radio station call letters can be made if the first letter must be a W or a K and no letter may be repeated?
6. A beauty salon has 4 assistants who wash hair, 2 stylists who cut hair, and a manicurist. By how many different arrangements of the personnel in the salon may a woman have her hair washed, cut,
and then her nails manicured?
7. A business school gives courses in typing, shorthand, transcription, business English, technical writing, and accounting. In how many ways may a student arrange his program if he has three courses
a day?
8. A sandlot baseball team has 13 players. If there are three pitchers, two catchers, two center-fielders, and one of each other position, how many 9-man starting teams can be fielded?
9. How many permutations of the letters A N S W E R end in a vowel?
10. How many permutations of the letters E Q U I N 0 X end in a consonant?
11. How many permutations of the letters M O D E R N have consonants in the second and third positions?
12. How many even integers whose numerals contain three digits can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated in a numeral?
13. How many odd natural numbers are there having a 4-digit numeral in which no digit is repeated?
14. How many numbers divisible by 5 can be formed from the digits 1,2, 3, 4, 5, 6, using each digit exactly once in each numeral?
15. To lead a certain cheer, the seven cheerleaders form a circle, each facing the center. In how many orders can they arrange themselves?
16. A milliner wants to arrange six different flowers around the brim of a hat. In how many orders can she place them?
17. In how many ways can five keys be arranged on a key ring?
18. In how many ways can a girl arrange nine charms on a bracelet?
19. In how many ways can five men and five girls be seated at a round table so that each girl is between two men?
20. In how many ways can four young couples be seated as a round table so that girls and boys are seated alternately?
21. Show that 6P4 = 6(5P3). 23. Show that 5Pr = 5(4Pr-1)
22. Show that nP4 = n([n-1]P[3]) 24. Show that 5P3 - 5P2 = 2(5P2)
Section 3: counting subsets: combinations
1. How many combinations can be formed from the letters D R E A M, taking two at a time? Show the combinations.
2. How many combinations can be formed from the letters N I L E, taking two at a time? Write the combinations.
3. How many straight lines can be formed by joining any two of five points, no three of which are in a straight line?
4. Seven points lie on the circumference of a circle. How many chords can be drawn joining them?
5. In how many ways can a student choose to answer five questions out of eight on an examination, if the order of his answers is of no importance?
6. Eva's Hamburger Haven sells hamburgers with cheese, lettuce, tomato; relish, ketchup, or mustard. How many different hamburgers can be made, choosing any three of the "extras"?
7. Julie owes letters to her grandmother, her uncle, her cousin, and two school friends. Tuesday night she decides to write to two of them. From how many combinations can she choose?
8. Local 352 is holding an election of four officers. In how many ways can they be chosen from a membership of 75? (Disregard order.)
9. A quality control engineer has to inspect a sample of 5 fuses from a box of 100. How many different samples can he choose?
10. The twelve engineers in the instrumentation department of Ajax Missile Corp. are to divide into project groups of four persons each. How many possible groups are there?
11. Six points lie on the circumference or a circle. How many inscribed triangles can be drawn having these points as vertices?
12. Eleven points lie on the circumference of a circle. How many inscribed hexagons can be drawn having these points as vertices?
13. If 6 students are to be chosen from 12 to participate in an honors section of a course, in how, many ways can this group be selected? In how many ways can the group not chosen to participate be
14. How many ways can a chemistry student choose 5 experiments to perform out of 10? In how many ways can 10 things be divided into 2 equal groups?
15. How many different five-card hands can be drawn from a pack of 52 cards? (Set up solution and estimate answer.)
16. How many different thirteen-card hands can be dealt from the 52 cards in a pack? (Set up solution and estimate answer.)
17. Find n, given nC2 = 100C98.
18. Find n if nC5 = nC3.
19. Prove nCr = nC(n-r) by using the formula nCr = n! / [(n - r)!r!]
20. Show that the total number of subsets of a set with n elements is 2^n. Hint. Each member of the set either is or is not selected in forming a subset.
Section 4: combination and permutation applications
1. Mrs. Henry McGrath has five hats, nine winter dresses, three handbags, and six pairs of shoes. How many different winter outfits does she have?
2. On a geometry test, each student may select one theorem to prove from three which are given and two constructions to perform from three which are given. In how many ways can a student make his
3. Students in an English class are to write a report on two books read outside of class. For the first, they may choose from four different books, and for the second, from three different books. How
many different choices of reports are there? (Disregard order.)
4. There are ten men and six women in a repertory theatre group. Four of the men can play male leads, and the others play supporting roles. Three of the women play female leads and the others play
supporting roles. In how many ways can a play with male and female leads and two male and three female supporting roles be cast?
5. A chef interested in using up some leftover meats and vegetables decides to make a stew consisting of three kinds of meat and four vegetables. If there are five different meats and seven different
vegetables available, how many different kinds of stew can the chef make?
6. Seven boys and seven girls were nominated for homecoming king and queen. How many ways can a king, a queen, and her court of two girls be chosen?
7. A bridge deck has thirteen cards of each suit. How many 13-card hands having seven spades are there? How many hands having exactly seven spades, three hearts, two diamonds, and one club are there?
8. A deck of cards has thirteen cards of each suit. How many 10-card hands having three cards of one kind and two of another are there? How many hands having three 10s and two queens are there?
9. A department-store window designer has eight spring dresses, four shorts-and-shirt outfits, and seven bathing suits from which to choose three dresses, two shorts-and-shirt outfits, and two
bathing suits. How many ways can she arrange these on seven manikins?
10. The executive division of a business organization proposed three speakers for the annual banquet, the sales department proposed two, the production department proposed two, and the accounting
department proposed one, If there is to be one speaker from each department, how many ways can they be arranged on one side of the head table?
11. How many five-letter arrangements of the letters R E G I 0 N A L consisting of three consonants and two vowels can be formed if no letter is repeated?
Section 1: 7. 676 8. 27 9. 1024 10. 840 11. a. 5.85 x 10^7 b. 9 x 106 12. 9 x 10^9 13. 80
14. 210 15. 625 16. 125 17. 75 18. 108 19. 1562 20. 340 21. 1,565,616 22. 8315
Section 2: 1. 5040 2. 5040 3. 30 4. 24 5. 27,600 6. 8 7. 120 8. 12 9. 240
10.2160 11. 288 12. 24 13. 2240 14. 120 15. 720 16. 120 17. 12 18. 20,160
19. 2880 20. 144
Section 3: 1. 10 2. 6 3. 10 4. 21 5. 56 6. 20 7. 10 8. 1,215,450 9. 75,287,520
10. 495 11. 20 12. 462 13. 924; 924 14. 252; 126 15. approx. 2.6 x 10^6
16. approx. 7 x 10^11 17. n = 100 18. n = 8
Section 4: 1. 810 2. 9 3. 12 4. 180 5. 350 6. 735 7. approx. 5.6 x 10^9; approx. 5 x 10^8
8. 4.1 x 10^9; 2.6 x 10^7 9. approx. 3.6 x 10^7 10. 288 11. 2880 | {"url":"http://members.bellatlantic.net/~vze4hzs6/mq6/combperm.htm","timestamp":"2014-04-17T10:16:03Z","content_type":null,"content_length":"15034","record_id":"<urn:uuid:c291b188-a4b1-4a39-a430-6dff1ea63232>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00201-ip-10-147-4-33.ec2.internal.warc.gz"} |
Class jnt.linear_algebra.LU
All Packages Class Hierarchy This Package Previous Next Index
Class jnt.linear_algebra.LU
public class LU
extends Object
LU matrix factorization. (Based on TNT implementation.) Decomposes a matrix A into a triangular lower triangular factor (L) and an upper triangular factor (U) such that A = L*U. By convnetion, the
main diagonal of L consists of 1's so that L and U can be stored compactly in a NxN matrix.
Initalize LU factorization from matrix.
LU factorization (in place).
Returns a copy of the compact LU factorization.
Returns a copy of the pivot vector.
Solve a linear system, with given factorization.
Solve a linear system, using a prefactored matrix in LU form.
public LU(double A[][])
Initalize LU factorization from matrix.
A - (in) the matrix to associate with this factorization.
public double[][] getLU()
Returns a copy of the compact LU factorization. (useful mainly for debugging.)
the compact LU factorization. The U factor is stored in the upper triangular portion, and the L factor is stored in the lower triangular portion. The main diagonal of L consists (by
convention) of ones, and is not explicitly stored.
public int[] getPivot()
Returns a copy of the pivot vector.
the pivot vector used in obtaining the LU factorzation. Subsequent solutions must permute the right-hand side by this vector.
public double[] solve(double b[])
Solve a linear system, with given factorization.
b - (in) the right-hand side.
solution vector.
public static int factor(double A[][],
int pivot[])
LU factorization (in place).
A - (in/out) On input, the matrix to be factored. On output, the compact LU factorization.
pivit - (out) The pivot vector records the reordering of the rows of A during factorization.
0, if OK, nozero value, othewise.
public static void solve(double LU[][],
int pvt[],
double b[])
Solve a linear system, using a prefactored matrix in LU form.
LU - (in) the factored matrix in LU form.
pivot - (in) the pivot vector which lists the reordering used during the factorization stage.
b - (in/out) On input, the right-hand side. On output, the solution vector.
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La Salle University: School of Arts and Sciences: Mathematics and Computer Science
4 credits
The use of technology as a tool for solving problems in mathematics, learning mathematics and building mathematical conjectures; electronic spreadsheets, a Computer Algebra System (CAS), and a
graphing calculator; the use of these tools, programming within all three environments, including spreadsheet macros, structured CAS programming, and calculator programming. A TI-89 calculator is
required for mathematics majors; a TI graphing calculator is required for other majors.
3 credits
Algebraic operations; linear and quadratic equations; exponents and radicals; elementary functions; graphs; and systems of linear equations. Students who have other college credits in mathematics
must obtain permission of the department chair to enroll in this course. NOTE: Not to be taken to fulfill major requirements.
4 credits
Review of algebra; simultaneous equations; trigonometry; functions and graphs; properties of logarithmic, exponential, and trigonometric functions; problem-solving and modeling. A TI graphing
calculator is required.
4 credits
Introduction to functions and modeling; differentiation. There will be a particular focus on mathematical modeling and business applications. Applications include: break-even analysis; compound
interest; elasticity; inventory and lot size; income streams; and supply and demand curves. The course will include the frequent use of Microsoft Excel. A TI-84 or TI-83 graphing calculator is
required. Co-requisite: CSC 151. Prerequisite: MTH 101 or its equivalent.
4 credits
Functions of various types: rational, trigonometric, exponential, logarithmic; limits and continuity; the derivative of a function and its interpretation; applications of derivatives including maxima
and minima and curve sketching; antiderivatives, the definite integral and approximations; the fundamental theorem of calculus; integration using substitution. A TI-89 calculator is required for
mathematics majors; a TI graphing calculator is required for other majors. Prerequisite: MTH 113 or its equivalent.
3 credits
Overview of mathematical concepts that are essential tools in navigating life as an informed and contributing citizen; logical reasoning, uses and abuses of percentages, interpreting statistical
studies and graphs, the basics of probability, descriptive statistics, and exponential growth. Applications of these topics include population statistics, opinion polling, voting and apportionment,
statistics in disease diagnoses and health care, lotteries and games of chance, and financial mathematics.
4 credits
Differentiation and integration of inverse trigonometric and hyperbolic functions; applications of integration, including area, volume, and arc length; techniques of integration, including
integration by parts, partial fraction decomposition, and trigonometric substitution; L’Hopital’s Rule; improper integrals; infinite series and convergence tests; Taylor series; parametric equations;
polar coordinates; and conic sections. A TI graphing calculator is required for other majors. Prerequisite: MTH 120.
4 credits
Three-dimensional geometry including equations of lines and planes in space, vectors. An introduction to multi-variable calculus including vector-valued functions, partial differentiation,
optimization, and multiple integration. Applications of partial differentiation and multiple integration. A TI-89 calculator is required. Prerequisite: MTH 221.
4 credits
Systems of linear equations; matrices; determinants; real vector spaces; basis and dimension; linear transformations; eigenvalues and eigenvectors; orthogonality; applications in mathematics,
computer science, the natural sciences, and economics. Prerequisite: MTH 221.
3 credits
This course is the first half of a two-semester course in discrete mathematics. The intended audience of the course consists of computer science majors (both B.A. and B.S.) and IT majors. Topics in
the course include logic, sets, functions, relations and equivalence relations, graphs, and trees. There will be an emphasis on applications to computer science.
3 credits
This course is the second half of a two-semester course in discrete mathematics. The intended audience of the course consists of computer science majors (both B.A. and B.S.) and IT majors. Topics in
the course include number theory, matrix arithmetic, induction, counting, discrete probability, recurrence relations, and Boolean algebra. There will be an emphasis on applications to computer
science. Prerequisite: MTH 260.
3 credits
Propositional logic; methods of proof; sets and cardinality; basic properties of integers; elementary number theory; structure of the real numbers; sequences; functions and relations. Prerequisite:
MTH 221.
3 credits
Theory behind calculus topics such as continuity, differentiation, integration, and sequences and series (both of numbers and of functions); basic topology, Fourier Series. Prerequisites: MTH 222 and
4 credits
Analytical, graphical, and numerical techniques for first and higher order differential equations; Laplace transform methods; systems of coupled linear differential equations; phase portraits and
stability; applications in the natural and social sciences. Prerequisite: MTH 221.
3 credits
Topics from Euclidean geometry including motions and similarities, collinearity and concurrence theorems, compass and straightedge constructions; the classical non-Euclidean geometries; finite
geometries. Prerequisite: MTH 240.
3 credits
Sets and mappings; groups, rings, fields, and integral domains; substructures and quotient structures; homomorphisms and isomorphisms; abelian and cyclic groups; symmetric and alternating groups;
polynomial rings. Prerequisite: MTH 302.
3 credits
Permutations and combinations; generating functions; recurrence relations and difference equations; inclusion/exclusion principle; derangements; other counting techniques, including cycle indexing
and Polya’s method of enumeration. Prerequisite: MTH 221.
3 credits
An introduction to specialized areas of mathematics. The subject matter will vary from term to term. Prerequisite: junior Mathematics standing.
3 credits
In-depth historical development of arithmetic, algebra, geometry, trigonometry, and calculus in Western mathematics (Europe and Near East) from ancient times into the 1700s; highlights from the
mathematical work of such figures as Hippocrates, Euclid, Archimedes, Heron, Diophantus, Fibonacci, Cardano, Napier, Descartes, Fermat, Newton, and Leibniz; non-Euclidean geometry (1800s); important
contributions of Euler and Gauss; the advent of computers. Prerequisite: MTH 302.
3 credits
Sample spaces and probability measures; descriptive statistics; combinatorics, conditional probability and independence; random variables, joint densities and distributions; conditional
distributions; functions of a random variable; expected value and variance; Chebyshev’s inequality; correlation coefficient; laws of large numbers; the Central Limit Theorem; various distribution
models; introduction to confidence intervals. Prerequisite: MTH 222.
3 credits
Measures of central tendency and variability; random sampling from normal and non-normal populations; estimation of parameters; maximum likelihood estimates; confidence intervals and hypothesis
testing; normal, chi-square, Student’s t and F distributions; analysis of variance; randomized block design; correlation and regression; goodness of fit; contingency tables. Prerequisite: MTH 410.
4 credits
Basic concepts; interpolation and approximations; summation and finite differences; numerical differentiation and integration; roots of equations. Prerequisite: MTH 222
3 credits
Analytic functions; Cauchy-Riemann equations; Cauchy’s integral theorem; power series; infinite series; calculus of residues; contour integration; conformal mapping. Prerequisite: MTH 222 or
permission of the instructor.
3 credits
Uses of mathematical methods to model real-world situations, including energy management, assembly-line control, inventory problems, population growth, predator-prey models. Other topics include:
least squares, optimization methods interpolation, interactive dynamic systems, and simulation modeling. Prerequisite: MTH 221.
3 credits
Topological spaces; subspaces; product spaces, quotient spaces; connectedness; compactness; metric spaces; applications to analysis. Prerequisite: MTH 302.
3 credits
An introduction to specialized research, concentrating on one particular aspect of mathematics. The subject matter will vary from term to term. Prerequisite: senior Mathematics standing. | {"url":"http://www.lasalle.edu/schools/sas/mathcomp/index.php?section=math&page=course_descriptions&prog=43","timestamp":"2014-04-20T14:19:06Z","content_type":null,"content_length":"37027","record_id":"<urn:uuid:90a73e2f-53db-4e1e-af49-e278c63f7279>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00531-ip-10-147-4-33.ec2.internal.warc.gz"} |
Truss Analysis - Sections Method
I was analyzing a truss and when I checked my answer there was a "new" point where the moment was taken about that was not in the original diagram. After reading the text I could not find anything
that stated why, when, or how to do this? I've attached the image of the answer.
Can I figure out the forces without creating this alternate point? If I were to create an additional point, how would I know what lengths to use? | {"url":"http://www.physicsforums.com/showthread.php?t=650180","timestamp":"2014-04-16T22:13:04Z","content_type":null,"content_length":"26501","record_id":"<urn:uuid:53686436-7383-4118-a1f5-447f9c8f816e>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00140-ip-10-147-4-33.ec2.internal.warc.gz"} |
Prove the existence of the square root of 2
October 16th 2011, 10:18 AM #1
Oct 2011
Prove the existence of the square root of 2
I'm currently in Calculus I Honors at my university and so far, after 1.5 months of instruction, we've yet to touch calculus... But in the mean time, my professor has gone deep into number
theory, and we spent the first 2 weeks proving the existence of real numbers using a Cauchy Sequence. For our first exam, he prompted us with the extra-credit:
Prove the existence of the square root of 2 using a Cauchy Sequence of rationals. (Also, feel free to re-explain Cauchy Sequences to me........)
THANK YOU SO MUCH!
Re: Prove the existence of the square root of 2
consider the following sequence:
$x_0 = 1$
$x_{n+1} = \frac{1}{2}\left(x_n+\frac{2}{x_n}\right)$
is this sequence cauchy? a cauchy sequence is one for which:
$\forall \epsilon > 0,\ \exists N \in \mathbb{Z}^+$ such that
$|x_m - x_n| < \epsilon,\ \forall \ m,n > N$
October 17th 2011, 01:50 AM #2
MHF Contributor
Mar 2011 | {"url":"http://mathhelpforum.com/number-theory/190518-prove-existence-square-root-2-a.html","timestamp":"2014-04-18T08:17:49Z","content_type":null,"content_length":"33541","record_id":"<urn:uuid:98447c29-7563-4833-9079-b872cea7d313>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00368-ip-10-147-4-33.ec2.internal.warc.gz"} |
Sinusoidal Function
December 1st 2012, 08:58 PM
Sinusoidal Function
A bug has landed on the rim of a jelly jar and is moving around the rim. The location where the bug initially lands is described and its angular speed is given. Impose a coordinate system with
the origin at the center of the circle of motion. For each of the scenarios below, answer the two questions.
Attachment 26014
Both coordinates of P(t) = (x(t), y(t)) are sinusoidal functions in the variable t. Sketch a rough graph of the functions
and y(t)
on the domain 0≤ t ≤ 11
Attachment 26015
Attachment 26016
Use the graph sketches to help you find the the amplitude, mean, period and phase shift for each function. Write x(t)
and y(t)
in standard sinusoidal form. (Round phase shifts to four decimal places.)
Amplitude: 2
Mean: 0
Period: 5.5
Phase Shift: -2.337
Amplitude: 2
Mean: 0
Period: 5.5
Phase Shift: -0.9629
I am having a hard time figuring out how to get these phase shift numbers...can someone explain how to get them? I understand how to get the rest, it is just that the phase shifts I don't
understand. Thank you! | {"url":"http://mathhelpforum.com/trigonometry/208861-sinusoidal-function-print.html","timestamp":"2014-04-16T08:42:04Z","content_type":null,"content_length":"6142","record_id":"<urn:uuid:35e47c86-5d65-449c-a80b-b340b6d32e9a>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00260-ip-10-147-4-33.ec2.internal.warc.gz"} |
Possible Answer
This MATLAB function produces the reduced row echelon form of A using Gauss Jordan elimination with partial pivoting. - read more
This MATLAB function computes the reduced row echelon form of the symbolic matrix A. - read more
Share your answer: how to do reduced row echelon form in matlab?
Question Analizer
how to do reduced row echelon form in matlab resources | {"url":"http://www.askives.com/how-to-do-reduced-row-echelon-form-in-matlab.html","timestamp":"2014-04-16T22:56:18Z","content_type":null,"content_length":"35772","record_id":"<urn:uuid:4d91cbca-81e5-4e5e-beb2-a5d2cadeb3a9>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00617-ip-10-147-4-33.ec2.internal.warc.gz"} |
FOM: category theory, cohomology, group theory, and f.o.m.
Stephen G Simpson simpson at math.psu.edu
Tue Feb 22 00:21:24 EST 2000
Reply to Andrej Bauer's posting of Feb 21, 2000.
You asked me to help you find the earlier FOM discussions of category
theory. I think November 1997 to February 1998 was the most
extensive. The conclusion was that category theory is no good as
a global foundational setup, because (i) it is more complicated than
set theory, (ii) it depends on set theory, (iii) it has no underlying
or motivating foundational picture. There was also a lot of category
theory discussion on FOM around April-May 1999, about small vs large
categories, the set-theoretic basis of category theory, etc. Let's
not go over all this ground again, unless you have some new point to
> sets, classes and operations can be explained in terms of objects
> and morphisms.
No, they can't. This was well covered in the earlier FOM discussion.
> 3. Disjoint sum is left adjoint of the diagonal functor.
But the disjoint sum of two sets can be explained to 5-year old
children much better and more easily without category theory, in terms
of sets of marbles, etc. This is part of why sets are fundamental to
math but adjoint functors are not.
> 4. Cartesian product is the right adjoint of the diagonal functor.
Same remark as above for disjoint sum, except you might need a 12-year
old child, to explain Cartesian product in terms of a rectangular
array or table.
> 6. Existential quantification is the left adjoint to the inverse image
> functor.
> 7. Universal quantification is the right adjoint to the inverse image
> functor.
As a non-category-theorist and a human being, I of course find this
way of viewing quantifiers somewhat unnatural. But putting that
aside, don't you agree with me that this alleged definition of
quantifiers in terms of adjoint functors is circular? Quantification
has to be understood *before* you can even define what you mean be a
category, let alone a functor and a left adjoint. After all, a
category is defined as a certain kind of algebraic structure where
*every* pair of morphisms of a certain kind has a composition, and
*for every* object *there exists* an identify morphism, etc etc etc.
This illustrates why logic is more fundamental than category theory.
> You may not *know* that you are using adjoint functors all the time,
> but that's hardly an argument for anything.
Instead of casually dismissing the fact that the vast majority of
mathematicians don't know and don't care about adjoint functors, why
not ponder this fact and try to learn something from it? For
instance, you could contrast it dramatically to the situation
regarding truly fundamental mathematical concepts such as set,
function, number, etc.
> And yes, people's ignorance of adjoint functors greatly hinders
> research in algebra, logic, and geometry.
Probably you can give some examples illustrating this by showing that,
with hindsight, somebody could have perhaps discovered some theorem
more quickly using adjoint functors. Isn't hindsight wonderful? :-)
And of course there are lots of examples illustrating the same point
for lots of other specialized concepts, where knowledge of those
concepts might be thought to help in unexpected ways.
But none of this proves that these specialized concepts are part of
> You sound like a physicist of a few decades ago who would advise
> against teaching Lie groups to nuclear physicists.
I don't advise against teaching category theory to mathematicians.
But I think category theorists tend to overrate its value, especially
for f.o.m.
> Category theorists have exposed and made precise certain vague
> analogies in disparate branches of mathematics.
But the mathematics in question was perfectly rigorous and valid and
had a pefectly good foundation already, without category-theoretic
language. Contrast this to the truly foundational work of Cauchy and
Weierstrass, where they were systematically replacing non-rigorous
math by rigorous math.
> It turned out that these concepts (specifically "adjoint functors")
> permeate almost every branch of mathematics, including algebra,
> geometry, topology, and logic. And so, I claim, this puts certain
> aspects of category theory at the foundations of mathematics.
The concept of ``group'' also permeates almost every branch of
mathematics. Do you think this puts group theory at the foundation of
mathematics? If not, ask yourself why not.
> However, as should be clear from the examples above, adjoint functors
> are much, much, more wide-spread than groups.
Most core mathematicians could give a much longer list of places in
mathematics where groups come up in highly non-trivial ways. Think of
Klein's Erlanger Programm, etc. Yet surely you would deny that group
theory is part of f.o.m.
> Every time you use an implication or form a cartesian product you
> have an adjoint functor at hand. Is that pervasive enough?
Define ``at hand''. I can assure you that adjoint functors are the
farthest thing from my hand and my mind in these situations.
> I just want to defend the position that "adjoint functors" are of a
> general interest to f.o.m.
The pervasive interest and fundamental nature of sets and classes in
f.o.m. is extremely well established. The interest of adjoint
functors in f.o.m. is much, much less well established, or maybe not
established at all.
As a mathematician, I am somewhat interested and curious to understand
why adjoint functors arise in a variety of contexts and explain a
number of analogies. Similarly, I am interested and curious to
understand why groups arise in a variety of contexts and explain a
number of things. But this doesn't convince me that adjoint functors
or groups are part of f.o.m.
-- Steve
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