url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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http://everything2.com/title/invariant | 1,490,404,735,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188717.24/warc/CC-MAIN-20170322212948-00551-ip-10-233-31-227.ec2.internal.warc.gz | 116,075,268 | 7,136 | The two senses of "invariant" are two of the most important concepts in programming and physics, respectively.
The programming sense is a predicate that stays the same as you go through a loop. See sorting example of invariant. Knowing your invariants is a good way to write bug free code.
In physics, invariants are closely tied to conservation properties. I.e. invariance under time and translation implies conservation of momentum.
In software engineering, a explicit logical statement that must evaluate to true at all externally observable times during the lifecycle of a method or component. (Compare with precondition and postcondition). "Externally observable" is an important qualifier, since the component's regular operation may require several steps, between which the invariant is not necessarily satisfied. In the design by contract paradigm, pioneered by the Eiffel language, invariants are embodied as formal assertions that are tested on the entry and exit of a method. Failure of an invariant can denote an error either by the "client" or the "supplier", and places no guarantees that the component still functions as designed.
In complex application software, an invariant represents a known consistent state of an object, or complex data structure of objects. In a database, for example, relational constraints must be maintained, often through transaction management. However, invariants have a role to play in much smaller components. They can be used to guard against possible corruption of an object, and calculations based on invariants may be moved out of iterative code during optimization.
Violations in invariants are often used as detection for race conditions in multithreading environments. Without a suitable synchronization mechanism, one thread of execution operates on the object while in a state where it should not be externally observable. The invariant, and hence the behavior of the object, cannot be guaranteed under these conditions.
In*va"ri*ant (?), n. Math.
An invariable quantity; specifically, a function of the coefficients of one or more forms, which remains unaltered, when these undergo suitable linear transformations.
J. J. Sylvester.
Log in or registerto write something here or to contact authors. | 428 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-13 | latest | en | 0.934308 |
http://www.varsitytutors.com/act_math-help/how-to-find-the-area-of-an-acute-obtuse-isosceles-triangle | 1,484,884,772,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280774.51/warc/CC-MAIN-20170116095120-00535-ip-10-171-10-70.ec2.internal.warc.gz | 757,820,048 | 33,640 | # ACT Math : How to find the area of an acute / obtuse isosceles triangle
## Example Questions
### Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle
What is the area of an isosceles triangle with a vertex of degrees and two sides equal to ?
Explanation:
Based on the description of your triangle, you can draw the following figure:
You can do this because you know:
1. The two equivalent sides are given.
2. Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.
Now, based on the properties of an isosceles triangle, you can draw the following as well:
Based on your standard reference triangle, you know:
Therefore, is .
This means that is and the total base of the triangle is .
Now, the area of the triangle is:
or
### Example Question #151 : Triangles
An isosceles triangle has a height of and a base of . What is its area?
Explanation:
Use the formula for area of a triangle:
### Example Question #152 : Triangles
An isosceles triangle has a base length of and a height that is twice its base length. What is the area of this triangle? | 305 | 1,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.82156 |
https://encyclopedia2.thefreedictionary.com/Pseudo-cylindrical+projection | 1,550,664,864,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494741.0/warc/CC-MAIN-20190220105613-20190220131613-00102.warc.gz | 552,460,682 | 12,887 | # map projection
(redirected from Pseudo-cylindrical projection)
Also found in: Dictionary, Thesaurus.
Related to Pseudo-cylindrical projection: Sanson-Flamsteed projection, Eckert projection
## projection, map:
see map projectionmap projection,
transfer of the features of the surface of the earth or another spherical body onto a flat sheet of paper. Only a globe can represent accurately the shape, orientation, and relative area of the earth's surface features; any projection produces distortion with
.
## map projection,
transfer of the features of the surface of the earth or another spherical body onto a flat sheet of paper. Only a globe can represent accurately the shape, orientation, and relative area of the earth's surface features; any projection produces distortion with regard to some of these characteristics. The particular projection chosen for a given map will depend on the use for which the map is intended. Some projections preserve correct relative distances in all directions from the center of the map (equidistant projection); some show areas equal to (equal-area projection) or shapes similar to (conformal projection) those on a globe of the same scale; some are useful in determining direction. Many map projections can be constructed by the use of a light source to project the features of the globe onto a piece of paper (although in practice one performs the operation mathematically rather than with a light); other projections can be constructed only mathematically. Projections are classified as cylindrical, conic, or azimuthal according to the method of projection with a light source; many projections that can be constructed only mathematically are also classified according to this system.
### Cylindrical Projection
In a typical cylindrical projection, one imagines the paper to be wrapped as a cylinder around the globe, tangent to it along the equator. Light comes from a point source at the center of the globe or, in some cases, from a filament running from pole to pole along the globe's axis. In the former case the poles clearly cannot be shown on the map, as they would be projected along the axis of the cylinder out to infinity. In the latter case the poles become lines forming the top and bottom edges of the map. The Mercator projection, long popular but now less so, is a cylindrical projection of the latter type that can be constructed only mathematically. In all cylindrical projections the meridians of longitude, which on the globe converge at the poles, are parallel to one another. In the Mercator projection, a cylindrical conformal projection, the parallels of latitude, which on the globe are equal distances apart, are drawn with increasing separation as their distance from the equator increases in order to preserve shapes and enable the accurate navigational plotting of courses. However, the price paid for preserving shapes is that areas are exaggerated with increasing distance from the equator. The effect is most pronounced near the poles; e.g., Greenland is shown with enormously exaggerated size, although its shape in small sections is preserved. The poles themselves cannot be shown on the Mercator projection. Students using the Mercator projection obtain an incorrect impression of the relative sizes of the countries of the world.
### Conic Projection
In a conic projection a paper cone is placed on a globe like a hat, tangent to it at some parallel, and a point source of light at the center of the globe projects the surface features onto the cone. The cone is then cut along a convenient meridian and unfolded into a flat surface in the shape of a circle with a sector missing. All parallels are arcs of circles with a pole (the apex of the original cone) as their common center, and meridians appear as straight lines converging toward this same point. Some conic projections are conformal (shape preserving); some are equal-area (size preserving). A polyconic projection uses various cones tangent to the globe at different parallels. Parallels on the map are arcs of circles but are not concentric.
### Azimuthal Projection
In an azimuthal projection a flat sheet of paper is tangent to the globe at one point. The point light source may be located at the globe's center (gnomonic projection), on the globe's surface directly opposite the tangent point (stereographic projection), or at some other point along the line defined by the tangent point and the center of the globe, e.g., at a point infinitely distant (orthographic projection). In all azimuthal projections, the tangent point is the central point of a circular map; all great circles passing through the central point are straight lines, and all directions from the central point are accurate. If the central point is a pole, then the meridians (great circles) radiate from that point and parallels are shown as concentric circles. The gnomonic projection has the useful property that all great circles (not just those that pass through the central point) appear as straight lines; conversely, all straight lines drawn on it are great circles. A navigator taking the shortest route between two points (always part of a great circle) can plot his course on a gnomonic projection by simply drawing a straight line between the two points. Since 1998 the National Geographic Society has used a modified azimuthal projection, the Winkel tripel projection, which produces a less distorted representation of the landmasses near the poles than the Robinson projection (see below).
### Other Projections
Among the other commonly used map projections are the Mollweide homolographic and the sinusoidal, both of which are equal-area projections with horizontal parallels; they are especially useful for world maps. Goode's homolosine projection is a composite using the sinusoidal projection between latitudes 40°N and 40°S and the homolographic projection for the remaining parts. Interruptions, or splits, are often made in the ocean areas in order to show land areas with truer shapes. The Robinson, or orthophanic, projection, devised by A. H. Robinson for Rand McNally and also used for a time by the National Geographic Society, gained acceptance because it accurately represents relative size, Neither a conformal nor an equal-area projection, it is most accurate in the temperate zones.
### Bibliography
See G. P. Kellaway, Map Projections (2d ed. 1970); F. Pearson, Map Projection Methods (1984); J. P. Snyder, Flattening the Earth (1993).
## map projection
[′map prə‚jek·shən]
(mapping)
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Open / Close | 1,344 | 6,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-09 | latest | en | 0.907862 |
http://medical-dictionary.thefreedictionary.com/estimating | 1,532,314,790,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594790.48/warc/CC-MAIN-20180723012644-20180723032644-00473.warc.gz | 225,660,829 | 14,078 | # estimate
(redirected from estimating)
Also found in: Dictionary, Thesaurus, Legal, Financial, Idioms, Encyclopedia.
## estimate
[es´tĭ-ma]
1. a rough calculation or one based on incomplete data.
2. a statistic used to characterize the value of a population parameter. Called also estimator.
3. (es´tĭ-māt) to produce or use such a calculation or statistic.
## es·ti·mate
(es'tĭ-māt),
1. A measurement or a statement about the value of some quantity that is known, believed, or suspected to incorporate some degree of error.
2. The result of applying any estimator to a random sample of data. It is not a random variable but a realization of one, a fixed quantity, and it has no variance although commonly it also furnishes an estimate of what the variance of the estimator is. (Not to be confused with an estimator, which is a prescription for obtaining an estimate.)
[L. aestimo, pp. aestimatum, to appraise]
## estimate
/es·ti·mate/
1. (es´tĭ-mat) a rough calculation or one based on incomplete data.
2. (es´tĭ-mat) a statistic used to characterize the value of a population parameter.
3. (es´tĭ-māt) to produce or use such a calculation or statistic.
## estimate
A popular term for an educated guess about a thing or process. See Cookie cutter estimate, Demand-based estimate, Objective probability estimate, Subjective probability estimate.
## es·ti·mate
(es'ti-măt)
1. A measurement or a statement about the value of some quantity that is known, believed, or suspected to incorporate some degree of error.
2. The result of applying any estimator to a random sample of data. It is not a random variable but a realization of one, a fixed quantity, and it has no variance although commonly it also furnishes an estimate of what the variance of the estimator is. usage note Not to be confused with an estimator, which is a prescription for obtaining an estimate.
[L. aestimo, pp. aestimatum, to appraise]
## es·ti·mate
(es'ti-măt)
1. A measurement or a statement about the value of some quantity that is known, believed, or suspected to incorporate some degree of error.
2. The result of applying any estimator to a random sample of data.
[L. aestimo, pp. aestimatum, to appraise]
## estimate,
n the anticipated fee for dental services to be performed.
## estimate
a measurement which is believed likely to incorporate a degree of error.
Q. Hi friends, I like to estimate my body fat based on my height and weight. Hi friends, I like to estimate my body fat based on my height and weight. When I enquired about this I heard about BMI. Though I understood little about it I want to know more about what is BMI and why is it useful?
A. the BMI is not a very good method...it only helps if you are an average person. you can gain weight if you start training and still get in shape and loose fat. but it is our only cheap method...there are gyms that hold a way of measuring body fat- maybe try going to one of those?
References in periodicals archive ?
Complete flexibility for any type of estimating need.
The implementation and use of in house data retrieved from past projects can offer insight for estimators in estimating their organizations' future demolition projects.
Does Estimating Shrinkage Meet the Best Accounting Practice Requirement?
Chapter 14 - An Introduction To Estimating Web Based Projects
As a practical matter, this part of the test should not create any problems for most taxpayers that are currently estimating shrinkage.
Work performed by Dalesford, an independent Australian specialist who has had direct access to all of the detail of the Ausenco estimating department in Perth, has been determined as professional and convincing.
Xactimate is widely used by insurance carriers, independent adjusters, restoration contractors, and cleaning specialists for estimating the restoration costs on homes and businesses across the U.
Hi-Tech Collision, a leading automotive repair business in Southern California, will implement CCC Pathways(R) Estimating Solution with Digital Imaging, CCC Pathways Enterprise Solution(R), CCC Accumark(TM) Advisor, CCC Connect(TM) and CCC Autoverse(R) Repair Management Dispatch in each of its eleven locations.
There is less room than ever for error in estimating complex IT projects," said Henk Keukenkamp, an experienced project manager who founded SCOPE iT.
Factors that could cause actual results to differ materially include changes in crude oil and natural gas prices; unsuccessful exploratory and development drilling; failure to achieve expected reserve or production levels for existing and future projects due to operating hazards, drilling risks, and the inherent engineering uncertainties in estimating oil and gas reserves; difficulties or cost-overruns in constructing production facilities; potential disruption or interruption of the Company's facilities and operations due to accidents or political events; general domestic and international economic and political conditions; and other matters set forth in the Company's periodic filings with the Securities and Exchange Commission.
announced today that it is supplying free online access to Procedure pages (P-pages) for its print and electronic estimating products.
Owners, developers and construction contractors can now turn their rough ideas into conceptual estimates with near pinpoint accuracy using the latest version of Model Estimating.
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Open / Close | 1,127 | 5,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-30 | latest | en | 0.819495 |
https://msgroups.net/excel/counting-cells-in-filtered-list/252098 | 1,627,986,345,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00411.warc.gz | 417,587,504 | 13,141 | #### counting cells in filtered list
```I have a 3 column spreadsheet, the third column is conditionall
formatted and the spreadsheet has an automatic filter on each column.
I have the following formula at the bottom of column C which gives m
the total count of cells in column C or the total resulting from th
filter applied in Column A =SUBTOTAL(3,C3:C83).
Below that I have a summary
Split No. %
RED 47 58.02%
AMBER 27 33.33%
GREEN 7 8.64%
I need to the results in the No. column of the summary to reflect th
filtered list not the whole spreadsheet - at the moment the formla fo
RED under No is =COUNTIF(C3:C83,"<10") but this counts all cells no
those filtered. Can you help
--
rayda
-----------------------------------------------------------------------
raydaw's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=3298
```
0
3/30/2006 1:36:39 PM
excel 39879 articles. 2 followers.
10 Replies
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Similar Articles
[PageSpeed] 37
```Try
=SUMPRODUCT((C3:C83<10)*(SUBTOTAL(3,
OFFSET(\$A\$2,ROW(\$A\$3:\$A\$83)-ROW(\$A\$1),,1))))
--
HTH
Bob Phillips
(remove nothere from email address if mailing direct)
"raydaw" <raydaw.25hj0m_1143726001.1615@excelforum-nospam.com> wrote in
message news:raydaw.25hj0m_1143726001.1615@excelforum-nospam.com...
>
> I have a 3 column spreadsheet, the third column is conditionally
> formatted and the spreadsheet has an automatic filter on each column.
> I have the following formula at the bottom of column C which gives me
> the total count of cells in column C or the total resulting from the
> filter applied in Column A =SUBTOTAL(3,C3:C83).
>
> Below that I have a summary
> Split No. %
> RED 47 58.02%
> AMBER 27 33.33%
> GREEN 7 8.64%
>
> I need to the results in the No. column of the summary to reflect the
> filtered list not the whole spreadsheet - at the moment the formla for
> RED under No is =COUNTIF(C3:C83,"<10") but this counts all cells not
> those filtered. Can you help?
>
>
> --
> raydaw
> ------------------------------------------------------------------------
> raydaw's Profile:
http://www.excelforum.com/member.php?action=getinfo&userid=32982
>
```
0
bob.phillips1 (6510)
3/30/2006 3:20:16 PM
```Thanks for the help Bob but unfortunately it didn't work. Formula is
still counting all cells in list - even those hidden.
--
raydaw
------------------------------------------------------------------------
raydaw's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=32982
```
0
3/30/2006 4:29:26 PM
```I tested it on my data and it worked, so there must be something in the
data, or my interpretation of it.
Can you post the data, or maybe send me a workbook.
--
HTH
Bob Phillips
(remove nothere from email address if mailing direct)
"raydaw" <raydaw.25hqw0_1143736204.6819@excelforum-nospam.com> wrote in
message news:raydaw.25hqw0_1143736204.6819@excelforum-nospam.com...
>
> Thanks for the help Bob but unfortunately it didn't work. Formula is
> still counting all cells in list - even those hidden.
>
>
> --
> raydaw
> ------------------------------------------------------------------------
> raydaw's Profile:
http://www.excelforum.com/member.php?action=getinfo&userid=32982
>
```
0
bob.phillips1 (6510)
3/30/2006 4:50:37 PM
```Try this
=SUMPRODUCT(--(\$C\$3:\$C\$83<10),SUBTOTAL(3,OFFSET(\$A\$2,ROW(\$A\$3:\$A\$83)-MIN(ROW(\$A\$3:\$A\$83)),,1)))
if that does not work then your values are not what you think, OTOH Bob's
formula should not return all rows. It has a typo in that includes the
ROW(\$A\$3:\$A\$83)-ROW(\$A\$1)
should be
ROW(\$A\$3:\$A\$83)-MIN(ROW(\$A\$3:\$A\$83)
however it should not count all rows
--
Regards,
Peo Sjoblom
http://nwexcelsolutions.com
"raydaw" <raydaw.25hqw0_1143736204.6819@excelforum-nospam.com> wrote in
message news:raydaw.25hqw0_1143736204.6819@excelforum-nospam.com...
>
> Thanks for the help Bob but unfortunately it didn't work. Formula is
> still counting all cells in list - even those hidden.
>
>
> --
> raydaw
> ------------------------------------------------------------------------
> raydaw's Profile:
> http://www.excelforum.com/member.php?action=getinfo&userid=32982
>
```
0
Peo
3/30/2006 5:04:59 PM
```Thank you both for your help. The formula works great now!
However, here is an additional problem for you. I also need to count
the filtered cells in C3:C83 which have values between (and including)
10 and 16. At present the array formula is
=SUM(IF((F3:F83<=16)-(F3:F83<10),1,0)) which works fine but counts the
hidden rows.
I love your formula but don't understand it well enough to adapt (what
is the purpose of the -- at the start?
Sorry to be thick!
--
raydaw
------------------------------------------------------------------------
raydaw's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=32982
```
0
4/3/2006 1:35:00 PM
```In article <raydaw.25oxga_1144071301.3755@excelforum-nospam.com>,
raydaw <raydaw.25oxga_1144071301.3755@excelforum-nospam.com> wrote:
> =SUM(IF((F3:F83<=16)-(F3:F83<10),1,0)) which works fine but counts the
> hidden rows.
Try...
=SUMPRODUCT(SUBTOTAL(3,OFFSET(F3:F83,ROW(F3:F83)-ROW(F3),0,1)),--(F3:F83>
=10),--(F3:F83<=16))
>... what is the purpose of the -- at the start?
Have a look at the following link...
http://www.mcgimpsey.com/excel/formulae/doubleneg.html
Hope this helps!
```
0
domenic22 (716)
4/4/2006 4:11:05 AM
```See http://www.xldynamic.com/source/xld.SUMPRODUCT.html for a detailed
explanation.
--
HTH
Bob Phillips
(remove nothere from email address if mailing direct)
"raydaw" <raydaw.25oxga_1144071301.3755@excelforum-nospam.com> wrote in
message news:raydaw.25oxga_1144071301.3755@excelforum-nospam.com...
>
> Thank you both for your help. The formula works great now!
>
> However, here is an additional problem for you. I also need to count
> the filtered cells in C3:C83 which have values between (and including)
> 10 and 16. At present the array formula is
> =SUM(IF((F3:F83<=16)-(F3:F83<10),1,0)) which works fine but counts the
> hidden rows.
>
> I love your formula but don't understand it well enough to adapt (what
> is the purpose of the -- at the start?
>
> Sorry to be thick!
>
>
> --
> raydaw
> ------------------------------------------------------------------------
> raydaw's Profile:
http://www.excelforum.com/member.php?action=getinfo&userid=32982
>
```
0
bob.phillips1 (6510)
4/4/2006 8:53:23 AM
```HELP - situation is becoming urgent - can anyone tell me how to count
cells with values between 10 and 16 (inclusive) in a filtered list
without including hidden rows
--
raydaw
------------------------------------------------------------------------
raydaw's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=32982
```
0
4/4/2006 9:25:43 AM
```Have you tried the formula I offered?
In article <raydaw.25qgry_1144143000.6184@excelforum-nospam.com>,
raydaw <raydaw.25qgry_1144143000.6184@excelforum-nospam.com> wrote:
> HELP - situation is becoming urgent - can anyone tell me how to count
> cells with values between 10 and 16 (inclusive) in a filtered list
> without including hidden rows
```
0
domenic22 (716)
4/4/2006 9:58:21 AM
```Dominec
Thank you very much for your help - it works!
--
rayda
-----------------------------------------------------------------------
raydaw's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=3298
```
0
4/4/2006 11:00:26 AM
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I am running Exchange 2003 SP1 on Windows 2003. I am using Office 2003. I am hosting email for two different groups in a single domain and I only want users to see people in their respective group. I only need one group to have an Offline Address List. I have removed rights to authenticated users from the Default Global Address List. I have used security groups to give access to the Default Global Address List (for support people). The Default Global Address List contains everyone.. I have created two Global Address List and filtered each one on a custom attribute. Each list shows t...
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I've got a three tier dependent list set up for: Division / Group / Line. (Thanks Debra Dalgliesh) If a user selects a division, group and line there are no issues. However, after selecting all three fields, I've had some users go back and change the contents of the Division. When they do this - it makes the content of Group and Line not make sense. Is there a way to 'blank out' the group and line lists if the division list is entered/changed? or to force the user to re-enter group and line if division was changed. thanks in advance General Question - gen... | 5,167 | 19,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.825746 |
http://www.opengl.org/discussion_boards/showthread.php/177865-Bloxorz-game-how-to-move-the-rectangular-block?p=1238524 | 1,368,869,978,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382185/warc/CC-MAIN-20130516092622-00020-ip-10-60-113-184.ec2.internal.warc.gz | 649,822,446 | 10,793 | ## Bloxorz game . how to move the rectangular block?
hello ,
I am developing the bloxorz game :
see this link :
http://www.albinoblacksheep.com/games/bloxorz
I have some how managed to draw the floor and a block on it. by taking two functions like this :
drawFloor present in Floor class outside main :
void Floor::drawFloor()
{
glRotatef(45.5f,1.0f,1.0f,0.0f);
//Draw floor
for(int i = 1;i<6;i++)
{
for(int j = 1;j<6;j++)
{
//For hole to appear at 3,3
if(i == 3 && j==3)
{
glTranslatef(1.0f,0.0f,0.0f);
continue;
}
glBegin(GL_QUADS);
// Top Face
glColor3f(1.0f,0.0f,0.0f);
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.500f,0.500f,-0.500f); //top left
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.500f,0.500f,0.500f); //bottom left
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.500f,0.50f,0.50f); //bottom right
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f,0.50f,-0.50f);//top right
//Left Face
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.50f, 0.50f, -0.50f); //Top front
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f, 0.30f, -0.50f);
glTexCoord2f(1.0f, 1.0f); glVertex3f(-0.50f, 0.30f, 0.50f);
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.50f, 0.50f, 0.50f);
// Front Face
glTexCoord2f(0.0f, 0.0f); glVertex3f(0.50f,0.50f, 0.50f); //top right
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,0.50f,0.50f); // top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(-0.50f,0.30f, 0.50f); //bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f(0.50f,0.30f, 0.50f); //bottom right
// Bottom Face
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f,0.30f,0.50f); // top right
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.50f,0.30f,0.50f); // top left
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.50f,0.30f,-0.50f); // bottom left
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.50f,0.30f, -0.50f); //bottom right
// Back Face
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,0.50f,-0.50f); // top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(0.50f, 0.50f,-0.50f); //top right
glTexCoord2f(0.0f, 1.0f); glVertex3f( 0.50f, 0.30f,-0.50f); //bottom right
glTexCoord2f(0.0f, 0.0f); glVertex3f( -0.50f,0.30f, -0.50f);//bottom left
// Right face
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.50f, 0.50f, -0.50f); // Top right
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f, 0.50f, 0.50f); // top left
glTexCoord2f(0.0f, 1.0f); glVertex3f( 0.50f, 0.30f, 0.50f); //bottom right
glTexCoord2f(0.0f, 0.0f); glVertex3f( 0.50f, 0.30f, -0.50f); //bottom left
glEnd();
glTranslatef(1.0f,0.0f,0.0f);
}
glTranslatef(-5.0f,0.0f,-1.0f);
}
drawCube present in Cube class :
void Cube::drawCube()
{
glBegin(GL_QUADS);
// Top Face
glColor3f(0.0f,1.0f,0.0f);
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.500f,0.500f,-0.500f); //top left
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.500f,0.500f,0.500f); //bottom left
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.500f,0.50f,0.50f); //bottom right
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f,0.50f,-0.50f);//top right
//Left Face
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.50f, 0.50f, -0.50f); //Top right
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,-0.5f, -0.50f); //top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(-0.50f,-0.5f, 0.50f); //bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.50f, 0.50f, 0.50f); //bottom right
// Front Face
glTexCoord2f(0.0f, 0.0f); glVertex3f(0.50f,0.50f, 0.50f); //top right
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,0.50f,0.50f); // top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(-0.50f,-0.5f, 0.50f); //bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f(0.50f,-0.5f, 0.5f); //bottom right
// Bottom Face
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f,-0.50f,0.50f); // top right
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.50f,-0.50f,0.50f); // top left
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.50f,-0.50f,-0.50f); // bottom left
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.50f,-0.50f, -0.50f); //bottom right
// Back Face
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,0.50f,-0.50f); // top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(0.50f, 0.50f,-0.50f); //top right
glTexCoord2f(0.0f, 1.0f); glVertex3f( 0.50f,-0.50f,-0.50f); //bottom right
glTexCoord2f(0.0f, 0.0f); glVertex3f( -0.50f,-0.50f, -0.50f);//bottom left
// Right face
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.50f, 0.50f, -0.50f); // Top left
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f, 0.50f, 0.50f); // bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f( 0.50f, -0.50f, 0.50f); //bottom right
glTexCoord2f(0.0f, 0.0f); glVertex3f( 0.50f, -0.5f, -0.50f); //top right
glEnd();
}
I have placed the hole at 3,3 position look on drawing floor.
My main problem is moving the block left and right . For that I have taken a structure called
typedef struct
{
public:
float x_pos;
float y_pos;
float z_pos;
float radian;
}cube_param;
I created an object like cube_param cube_pos;
while left key is pressed :
case WM_KEYDOWN: // Is A Key Being Held Down?
{
keys[wParam] = TRUE; // If So, Mark It As TRUE
switch(wParam)
{
case VK_LEFT:
{
cube_pos.x_pos = - 1.0f ;
cube_pos.y_pos = -1.0f;
cube_pos.z_pos = 0.0f;
cube_pos.radian = -90.0f;
cube->drawCube(cube_pos);
}
}
return 0;
}
where drawCube(cube_pos); contains the same code as drawCube() but it has to move it.
Here does the structure cube_pos goes away with values to drawCube(cube_param cube_pos)
Look at it :
// For movement of cube
void Cube::drawCube(cube_param cube_pos)
{
float x_pos = cube_pos.x_pos;
float y_pos = cube_pos.y_pos;
float z_pos = cube_pos.z_pos;
float radian = cube_pos.radian;
//glTranslatef(1.0f,1.0f,0.0f);
glRotatef(radian,x_pos,0.0f,0.0f);
glBegin(GL_QUADS);
// Top Face
glColor3f(0.0f,1.0f,0.0f);
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.5f + x_pos,0.5f + y_pos,-0.5f); //top left
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.5f + x_pos,0.5f + y_pos,0.5f); //bottom left
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.5f + x_pos,0.5f + y_pos,0.5f); //bottom right
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.5f + x_pos,0.5f + y_pos,-0.5f);//top right
//Left Face
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.50f, 0.50f, -0.50f); //Top right
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,-0.5f, -0.50f); //top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(-0.50f,-0.5f, 0.50f); //bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.50f, 0.50f, 0.50f); //bottom right
// Front Face
glTexCoord2f(0.0f, 0.0f); glVertex3f(0.50f,0.50f, 0.50f); //top right
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,0.50f,0.50f); // top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(-0.50f,-0.5f, 0.50f); //bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f(0.50f,-0.5f, 0.5f); //bottom right
// Bottom Face
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f,-0.50f,0.50f); // top right
glTexCoord2f(0.0f, 1.0f); glVertex3f(-0.50f,-0.50f,0.50f); // top left
glTexCoord2f(0.0f, 0.0f); glVertex3f(-0.50f,-0.50f,-0.50f); // bottom left
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.50f,-0.50f, -0.50f); //bottom right
// Back Face
glTexCoord2f(1.0f, 0.0f); glVertex3f(-0.50f,0.50f,-0.50f); // top left
glTexCoord2f(1.0f, 1.0f); glVertex3f(0.50f, 0.50f,-0.50f); //top right
glTexCoord2f(0.0f, 1.0f); glVertex3f( 0.50f,-0.50f,-0.50f); //bottom right
glTexCoord2f(0.0f, 0.0f); glVertex3f( -0.50f,-0.50f, -0.50f);//bottom left
// Right face
glTexCoord2f(1.0f, 0.0f); glVertex3f( 0.50f, 0.50f, -0.50f); // Top left
glTexCoord2f(1.0f, 1.0f); glVertex3f( 0.50f, 0.50f, 0.50f); // bottom left
glTexCoord2f(0.0f, 1.0f); glVertex3f( 0.50f, -0.50f, 0.50f); //bottom right
glTexCoord2f(0.0f, 0.0f); glVertex3f( 0.50f, -0.5f, -0.50f); //top right
glEnd();
}
My idea is when left key is pressed i want to translate each and every vertex by adding or subtracting the x_pos,y_pos,z_pos present in cube_param structure
and then translate and rotate it. will it ultimately work for each and every move ?
If so how to do it. | 3,665 | 7,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-20 | latest | en | 0.553614 |
http://denisemeeks.com/science/notebooks/notebook_ti84.php | 1,623,507,808,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487584018.1/warc/CC-MAIN-20210612132637-20210612162637-00375.warc.gz | 15,658,435 | 23,120 | # TI-84 Functions
Sort by TypeClick to sort functions alphabetically by Type...please wait after clicking... SearchEnter function name in box, do NOT include parentheses, click Search button Clear SearchClick to clear Search box Select AllClick to select printing of all function groups Unselect AllClick to clear printing of all function groups BasicsClick to access basics external website ShortcutsClick to access shortcuts external website ReloadClick to reload the page, clear search and sort CloseClick to close this page Info ContactDr. Denise Meeks, tucsonkosmicgirl@gmail.com with corrections, questions, suggestions. Excel FnGo to Excel Functions Sort by FunctionClick to sort alphabetically by Function...please wait after clicking... TutorialsClick to access YouTube TI-84 tutorials website Print SelectedClick to print selected function groups; if none selected, will print all function groups TipsClick to access tips external website Ref GuideClick to access TI-84 Reference Guide external website Quick GuideClick to access Quick Guide external website ProceduresClick to access Procedures external website Data AnalysisClick to access Data Analysis external website CatalogClick to access Catalog website HomeGo to denisemeeks.com
Type Function or Instruction/Arguments Result Key or Keys/Menu or Screen/Item
complex a+bi Sets the mode to rectangular complex number mode (a+bi). [MODE]
a+bi
memory About Returns the product number and ID number. [2ND][+]
complex abs(complex value) Returns the magnitude of a complex number or list. [MATH]
CPX
5:abs(
math abs(value) Returns the absolute value of a real number, expression, list, or matrix. [MATH]
NUM
1:abs(
list addition list + value Returns list in which value is added to each list element. [+]
list addition listA + listB Returns listA elements plus listB elements. [+]
matrix addition matrixA + matrixB Returns matrixA elements plus matrixB elements. [+]
math addition valueA + valueB Returns valueA plus valueB. [+]
angle angle(value) Returns the polar angle of a complex number or list of complex numbers. [MATH]
CPX
4:angle(
statistics ANOVA(list1,list2 [,list3,...,list20]) Performs a one-way analysis of variance for comparing the means of two to 20 populations. [STAT]
TESTS
H:ANOVA(
math Ans Returns the last answer. [2ND]
[ANS]
memory Archive Moves the specified variables from RAM to the user data archive memory. [2ND][MEM]
5:Archive
program Asm(assemblyprgmname) Executes an assembly language program. [2ND][CATALOG]
Asm(
program AsmComp(prgmASM1,prgmASM2) Compiles an assembly language program written in ASCII and stores the hex version. [2ND][CATALOG]
AsmComp(
program AsmPrgm Must be used as the first line of an assembly language program. [2ND][CATALOG]
AsmPrgm
matrix augment(matrixA,matrixB) Returns a matrix, which is matrixB appended to matrixA as new columns. [2ND][MATRIX]
MATH
7:augment(
list augment(listA,listB) Returns a list, which is listB concatenated to the end of listA. [2ND][LIST]
OPS
9:augment(
format AUTO ANSWERS Displays answers in a similar format as the input. [MODE]
graph AxesOff Turns off the graph axes. [2ND][FORMAT]
AxesOff
graph AxesOn Turns on the graph axes. [2ND][FORMAT]
AxesOn
finance bal(npmt[,roundvalue]) Computes the balance at npmt for an amortization schedule using stored values for PV, I%, and PMT and rounds the computation to roundvalue. [APPS]1:Finance
CALC
9:bal(
probability binomcdf(numtrials,p[,x])
$binomcdf(n,p,k) = \sum_{i=0}^{k} binompdf(n,p,i) = \sum_{i=0}^{k} \left ( {\begin{matrix} n \\ i \end{matrix}} \right ) p^i (1-p)^{n-i}$
Computes a cumulative probability at x for the discrete binomial distribution with the specified numtrials and probability p of success on each trial.
[2ND][DISTR]
B:binomcdf(
probability binompdf(numtrials,p[,x])
$binompdf(n,p,k) = \left ( {\begin{matrix} n \\ p \end{matrix}} \right ) p^k (1-p)^{n-k} = \frac {n!}{k!(n-k)!} p^k (1-p)^{n-k}$
Computes a probability at x for the discrete binomial distribution with the specified numtrials and probability p of success on each trial.
[2ND][DISTR]
A:binompdf(
catalog CATALOG Access the Catalog. [2ND][CATALOG]
[ALPHA]
first letter of function name
time checkTmr(starttime) Returns the number of seconds since you used startTmr to start the timer. The starttime is the value displayed by startTmr. [2ND][CATALOG]
checkTmr(
statistics chi-square
Χ2-Test observedmatrix,expectedmatrix,Calculate/Draw
$\chi ^2 = \sum_{i=0}^{k} \left | {\frac {(O_i - E_i)^2}{E_i} } \right |$
Performs a chi-square test.
[STAT]
TESTS
C:Χ2-Test
statistics chi-square
Χ2cdf(lowerbound,upperbound,df)
$\chi ^2 cdf(a,b,k) = \int_{a}^{b} \chi ^2 pdf(x,k)dx$
Computes the Χ2 distribution probability between lowerbound and upperbound for the specified degrees of freedom df.
[2ND][DISTR]
8:Χ2cdf(
statistics chi-square
Χ2GOF-Test observedlist,expectedlist,df,Calculate/Draw
$\chi ^2_{n-1} = \sum_{i=0}^{n} \left | {\frac {(O_i - E_i)^2}{E_i} } \right |$
Performs a test to confirm that sample data is from a population that conforms to a specified distribution.
[STAT]
TESTS
D:Χ2GOF-Test(
statistics chi-square
Χ2pdf(x,df)
$\chi^2 pdf(x,k) = \frac {(1/2)^{k/2}} {(k/2 -1)!} x^{k/2-1}e^{-x/2}$
Computes the probability density function (pdf) for the Χ2 distribution at a specified x value for the specified degrees of freedom df.
[2ND][DISTR]
7:Χ2pdf(
DRAW
9:Circle(
format CLASSIC Displays inputs and outputs on a single line, such as 1/2+3/4. [MODE]
CLASSIC
memory Clear Entries Clears the contents of the Last Entry storage area. [2ND][MEM]
MEMORY
3:Clear Entries
time ClockOff Turns off the clock display in the mode screen. [2ND][CATALOG]
ClockOff
time ClockOn Turns on the clock display in the mode screen. [2ND][CATALOG]
ClockOn
memory ClrAllLists Sets to 0 the dimension of all lists in memory. [2ND][MEM]
MEMORY
4:ClrAllLists
graph ClrDraw Clears all drawn elements from a graph or drawing. [2ND][DRAW]
DRAW
1:ClrDraw
program ClrHome Clears the home screen within a program. [PRGM]
I/O
8:ClrHome
memory ClrList list1[,list2, ...,listn] Sets to 0 the dimension of one or more lists. [STAT]
EDIT
4:ClrList
program Clr Clears all values from the within a program. [PRGM]
I/O
9:Clr
probability combinations list nCr value
$C(n,r) = \frac{(n!)}{r!(n-r!)}$
Returns a list of the combinations of each element in list taken value at a time.
[MATH]
PRB
3:nCr
probability combinations listA nCr listB
$C(n,r) = \frac{(n!)}{r!(n-r!)}$
Returns a list of the combinations of each element in listA taken each element in listB at a time.
[MATH]
PRB
3:nCr
probability combinations value nCr list
$C(n,r) = \frac{(n!)}{r!(n-r!)}$
Returns a list of the combinations of value taken each element in list at a time.
[MATH]
PRB
3:nCr
probability combinations valueA nCr valueB
$C(n,r) = \frac{(n!)}{r!(n-r!)}$
Returns the number of combinations of valueA taken valueB at a time.
[MATH]
PRB
3:nCr
complex complex value▶Rect Displays complex value or list in rectangular format. [MATH]
CPX
6:Rect
format complex value▶Rect Displays complex value or list in rectangular format. [MATH]
CPX
6:▶Rect
logic compare valueA and valueB Returns 1 if both valueA and valueB are ≠ 0. valueA and valueB can be real numbers, expressions, or lists. [2ND][TEST]
LOGIC
1:and
string concatenation string1 + string2 Concatenates two or more strings. [+]
complex conj(value) Returns the complex conjugate of a complex number or list of complex numbers. [MATH]
CPX
1:conj(
graph CONNECTED Sets connected plotting mode; resets all Y= editor graphstyle settings to \. [MODE]
CONNECTED
graph CoordOff Turns off cursor coordinate value display. [2ND][FORMAT]
CoordOff
graph CoordOn Turns on cursor coordinate value display. [2ND][FORMAT]
Coordon
trigonometry cos(value)
$cos(\theta) = \frac {opp}{adj} = sin \left( \frac {\pi}{2} - \theta \right) = \frac {1}{sec(\theta)}$
Returns cosine of a real number, expression, or list.
[COS]
trigonometry cos-1(value) Returns arccosine of a real number, expression, or list. [2ND][COS-1]
trigonometry cosh(value) Returns hyperbolic cosine of a real number, expression, or list. [2ND][CATALOG]
cosh(
trigonometry cosh-1(value) Returns hyperbolic arccosine of a real number, expression, or list. [2ND][CATALOG]
cosh-1(
math cube value3 Returns the cube of a real or complex number, expression, list, or square matrix. [MATH]
MATH
3:3
math cube root 3√(value) Returns the cube root of a real or complex number, expression, or list. [MATH]
MATH
4:3√(
statistics CubicReg Xlist, Ylist,freqlist, regeq Fits a cubic regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
6:CubicReg
statistics cumSum(list) Returns a list of the cumulative sums of the elements in list, starting with the first element. [2ND][LIST]
OPS
6:cumSum(
matrix cumSum(matrix) Returns a matrix of the cumulative sums of matrix elements. Each element in the returned matrix is a cumulative sum of a matrix column from top to bottom. [2ND][MATRIX]
[MATH]
0:cumSum(
time dayOfWk(year,month,day) Returns an integer from 1 to 7, with each integer representing a day of the week. Use dayOfWk( to determine on which day of the week a particular date would occur. The year must be 4 digits; month and day can be 1 or 2 digit. [2ND][CATALOG]
dayOfWk(
1:Sunday
2:Monday
3:Tuesday...
finance dbd(date1,date2) Calculates the number of days between date1 and date2 using the actual-day-count method. [APPS]1:Finance
CALC
D:dbd(
format decimal value▶Dec Displays a real or complex number, expression, list, or matrix in decimal format. [MATH]
MATH
2:▶Dec
program decrement :DS<(variable,value):commandA:commands Decrements variable by 1; skips commandA if variable < value within a program. [PRGM]
CTL
B:DS<(
angle DEGREE Sets degree angle mode. [MODE]
DEGREE
angle degrees notation: valueo Interprets value as degrees; designates degrees in DMS format. [2ND][ANGLE]
ANGLE
1:o
program DelVar variable Deletes from memory the contents of variable within a program. [PRGM]
CTL
G:DelVar
Depend: Auto
table DependAuto Sets table to generate dependent-variable values automatically. [2ND][TBLSET]
Depend: Auto
matrix det(matrix) Returns determinant of matrix. [2ND][MATRIX]
MATH
1:det(
statistics DiagnosticOff Sets diagnostics-off mode; r, r2, and R2 are not displayed as regression model results. [2ND][CATALOG]
DiagnosticOff
statistics DiagnosticOn Sets diagnostics-on mode; r, r2, and R2 are displayed as regression model results. [2ND][CATALOG]
DiagnosticOn
list dim(list) Returns the dimension of list. [2ND][LIST]
OPS
3:dim(
matrix dim(matrixname) Returns the dimension of matrixname as a list. [2ND][MATRIX]
MATH
3:dim(
list dimension lengthdim(list) Assigns a new dimension (length) to a new or existing list. [2ND][LIST]
OPS
3:dim(
program Disp Displays the home screen within a program. [PRGM]
I/O
3:Disp
program Disp [valueA,valueB,valueC,...,value n] Displays each value within a program. [PRGM]
I/O
3:Disp
program DispGraph Displays the graph within a program. [PRGM]
I/O
4:DispGraph
program DispTable Displays the table within a program. [PRGM]
I/O
5:DispTable
list division list/value Returns list elements divided by value. [÷]
list division listA/listB Returns listA elements divided by listB elements. [÷]
list division value/list Returns value divided by list elements. [÷]
math division valueA/valueB Returns valueA divided by valueB. [÷]
format DMS format value▶DMS Displays value in DMS format. [2ND][ANGLE]
ANGLE
4:▶4DMS
graph DOT Sets dot plotting mode; resets all Y= editor graph-style settings to ⋱. [MODE]
DOT
graph DrawF expression Draws expression (in terms of X) on the graph. [2ND][DRAW]
DRAW
6:DrawF
graph DrawInv expression Draws the inverse of expression by plotting X values on the y-axis and Y values on the x-axis. [2ND][DRAW]
DRAW
8:DrawInv
graph derivative dy/dx Calculates the derivative at a selected point. [2ND][CALC]
6:dy/dx
math e Returns e. [2ND][e]
list e^(list) Returns a list of e raised to a list of powers. [2ND][ex]
math e^(power) Returns e raised to power. [2ND][ex]
finance Eff ▶Eff(nominal rate,compounding periods) Computes the effective interest rate. [APPS]1:Finance
CALC
C:▶Eff(
program Else See If:Then:Else
program End Identifies end of For(, If-Then-Else, Repeat, or While loop. [PRGM]
CTL
7:End
format ENG Sets engineering display mode. [MODE]
ENG
logic equal valueA=valueB Returns 1 if valueA = valueB. Returns 0 if valueA ≠ valueB. valueA and valueB can be real or complex numbers, expressions, lists, or matrices. [2ND][TEST]
TEST
1:=
string Equ▶String(Y= var,Strn) Converts the contents of a Y= var to a string and stores it in Strn. [2ND][CATALOG]
Equ▶String(
math exponent listEexponent Returns list elements times 10 to the exponent. [2ND][EE]
math exponent matrixEexponent Returns matrix elements times 10 to the exponent. [2ND][EE]
math exponent valueEexponent Returns value times 10 to the exponent. [2ND][EE]
string expr(string) Converts string to an expression and executes it. [2ND][CATALOG]
expr(
statistics ExpReg Xlist,Ylist,freqlist,regeq Fits an exponential regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
0:ExpReg
graph ExprOff Turns off the expression display during TRACE. [2ND][FORMAT]
ExprOff
graph ExprOn Turns on the expression display during TRACE. [2ND][FORMAT]
ExprOn
statistics F-test
2-SampFTest Data list1,list2,freqlist1,freqlist2,σ alternative,Calculate/Draw
Performs a two-sample F test. [STAT]
TESTS
E:2-SampFTest
statistics F-test
2-SampFTest Stats Sx1,n1,Sx2,n2,σ alternative,Calculate/Draw
Performs a two-sample F test. [STAT]
TESTS
E:2-SampFTest
math factorial value!
$n! = (n)(n-1)(n-2)...(1)$
Returns factorial of value.
[MATH]
PRB
4:!
math factorial: list!
$n! = (n)(n-1)(n-2)...(1)$
Returns factorial of list elements.
[MATH]
PRB
4:!
statistics Fcdf(lowerbound,upperbound,numerator df,denominator df) Computes the F distribution probability between lowerbound and upperbound for the specified numerator df (degrees of freedom) and denominator df. [2ND][DISTR]
0:Fcdf(
matrix Fill(value,matrixname) Stores value to each element in matrixname. [2ND][MATRIX]
MATH
4:Fill(
list Fill(value,list) Stores value to each element in list. [2ND][LIST]
OPS
4:Fill(
format Fix # Sets fixed-decimal mode for # of decimal places. [MODE]
0123456789
(select one)
format FLOAT Sets floating decimal mode. [MODE]
FLOAT
math fMax(expression,variable,lower,upper[,tolerance]) Returns the value of variable where the local maximum of expression occurs, between lower and upper, with specified tolerance. [MATH]
MATH
7:fMax(
math fMin(expression,variable,lower,upper[,tolerance]) Returns the value of variable where the local minimum of expression occurs, between lower and upper, with specified tolerance. [MATH]
MATH
6:fMin(
math fnInt(expression,variable,lower,upper[,tolerance]) Returns the function integral of expression with respect to variable, between lower and upper, with specified tolerance. [MATH]
MATH
9:fnInt(
graph FnOff [function#,function#,...,function n] Deselects all Y= functions or specified Y= functions. [VARS]
Y-VARS
4:On/Off
2:FnOff
graph FnOn [function#,function#,...,function n] Selects all Y= functions or specified Y= functions. [VARS]
Y-VARS
4:On/Off
1:FnOn
program For :For(variable,begin,end[,increment])
:commands
:End
:commands
Executes commands through End, incrementing variable from begin by increment until variable>end. [PRGM]
CTL
4:For(
math fPart(value) Returns the fractional part or parts of a real or complex number, expression, list, or matrix. [MATH]
NUM
4:fPart(
statistics Fpdf(x,numerator df,denominator df) Computes the F distribution probability between lowerbound and upperbound for the specified numerator df (degrees of freedom) and denominator df. [2ND][DISTR]
9:Fpdf(
format fraction n/d Displays results as a simple fraction. [ALPHA][F1]
1: n/d
or
[MATH]
NUM
D: n/d
format fraction to decimal ▶ F ◀ ▶ D Converts an answer from a fraction to a decimal or from a decimal to a fraction. [ALPHA][F1]
4: ▶ F ◀ ▶ D or
[MATH]
NUM
8: ▶ F ◀ ▶ D
format fraction value▶Frac Displays a real or complex number, expression, list, or matrix as a fraction simplified to its simplest terms. [MATH]
MATH
1:▶Frac
settings FULL Sets full screen mode. [MODE]
FULL
graph FUNC Sets function graphing mode. [MODE]
FUNC
memory GarbageCollect Displays the garbage collection menu to allow cleanup of unused archive memory. [2ND][CATALOG]
GarbageCollect
math gcd(valueA,valueB) Returns the greatest common divisor of valueA and valueB, which can be real numbers or lists. [MATH]
NUM
9:gcd(
probability geometcdf(p,x)
$geometcdf(p,n) = \sum_{i=1}^{n} geometpdf(p,i) = \sum_{i=1}^{n} p\,(1-p)^{i-1}$
Computes a cumulative probability at x, the number of the trial on which the first success occurs, for the discrete geometric distribution with the specified probability of success p.
[2ND][DISTR]
F:geometcdf(
probability geometpdf(p,x)
$geometpdf(p,n) = p(1-p)^{n-1}$
Computes a probability at x, the number of the trial on which the first success occurs, for the discrete geometric distribution with the specified probability of success p.
[2ND][DISTR]
E:geometpdf(
memory Get(variable) Gets data from the CBL 2TM or CBRTM System and stores it in variable. [PRGM]
I/O
A:Get(
program GetCalc(variable[,portflag]) Gets contents of variable on another TI-84 Plus and stores it to variable on the receiving TI-84 Plus. By default, the TI-84 Plus uses the USB port if it is connected. If the USB cable is not connected, it uses the I/O port.
portflag=0 use USB port if connected;
portflag=1 use USB port;
portflag=2 use I/O port. Used within a program
[PRGM]
I/O
0:GetCalc(
date getDate Returns a list giving the date according to the current value of the clock. The list is in {year,month,day} format. [2ND][CATALOG]
getDate
date getDtFmt Returns an integer representing the date format that is currently set on the device.
1 = M/D/Y
2 = D/M/Y
3 = Y/M/D
[2ND][CATALOG]
getDtFmt
date getDtStr(integer) Returns a string of the current date in the format specified by integer, where:
1 = M/D/Y
2 = D/M/Y
3 = Y/M/D
[2ND][CATALOG]
getDtStr(
time getTime Returns a list giving the time according to the current value of the clock. The list is in {hour,minute,second} format. The time is returned in the 24 hour format. [2ND][CATALOG]
getTime
time getTmFmt Returns an integer representing the clock time format that is currently set on the device.
12 = 12 hour format
24 = 24 hour format
[2ND][CATALOG]
getTmFmt
time getTmStr(integer) Returns a string of the current clock time in the format specified by integer, where:
12 = 12 hour format
24 = 24 hour format
[2ND][CATALOG]
getTmStr(
program getKey Returns the key code for the current keystroke, or 0, if no key is pressed. Used within a program. [PRGM]
I/O
7:getKey
graph GOTO FORMAT GRAPH Turns shortcut to the format graph screen on or off. [MODE]
GOTO FORMAT GRAPH
program Goto label Transfers control to label used within a program. [PRGM]
CTL
0:Goto
program GraphStyle(function#,graphstyle#) Sets a graphstyle for function# used within a program. [PRGM]
CTL
H:GraphStyle(
settings graph-table G-T Sets graph-table vertical split-screen mode. [MODE]
G-T
logic greater than valueA>valueB Returns 1 if valueA > valueB. Returns 0 if valueA ≤ valueB. valueA and valueB can be real or complex numbers, expressions, or lists. [2ND][TEST]
TEST
3:>
logic greater than or equal valueA≥valueB Returns 1 if valueA ≥ valueB. Returns 0 if valueA < valueB. valueA and valueB can be real or complex numbers, expressions, or lists. [2ND][TEST]
TEST
4:≥
graph GridOff Turns off grid format. [2ND][FORMAT]
GridOff
graph GridOn Turns on grid format. [2ND][FORMAT]
GridOn
memory Group Groups programs and variables. [2ND][+]
8:Group
settings HORIZ Sets horizontal split-screen mode. [MODE]
HORIZ
graph Horizontal y Draws a horizontal line at y. [2ND][DRAW]
DRAW
3:Horizontal
complex i Returns a complex number. [2ND]i
matrix identity(dimension) Returns the identity matrix of dimension rows x dimension columns. [2ND][MATRIX]
MATH
5:identity(
program If :If condition
:commandA
:commands
If condition = 0 (false), skips commandA. [PRGM]
CTL
1:If
program If :If condition
:Then
:commands
:End
:commands
Executes commands from Then to End if condition = 1 (true). [PRGM]
CTL
2:Then
program If :If condition
:Then
:commands
:Else
:commands
:End
:commands
Executes commands from Then to Else if condition = 1 (true); from Else to End if condition = 0 (false). [PRGM]
CTL
3:Else
complex imag(value) Returns the imaginary (nonreal) part of a complex number or list of complex numbers. [MATH]
CPX
3:imag(
table IndpntAuto Sets table to generate independent-variable values automatically. [2ND][TBLSET]
Indpnt: Auto
program Input Displays graph in a program. [PRGM]
I/O
1:Input
program Input ["text",variable] Prompts for value to store to variable within a program. [PRGM]
I/O
1:Input
program Input [variable] Prompts for value to store to variable within a program. [PRGM]
I/O
1:Input
program Input [Strn,variable] Displays Strn and stores entered value to variable. Used within a program. [PRGM]
I/O
1:Input
string inString(string,substring[,start]) Returns the character position in string of the first character of substring beginning at start. [2ND][CATALOG]
inString(
math int(value) Returns the largest integer ? a real or complex number, expression, list, or matrix. [MATH]
NUM
5:int(
graph integral dy/dx Calculates the integral at a selected point. [2ND][CALC]
7:∫f(x)dx
finance interest ΣInt(pmt1,pmt2[,roundvalue]) Computes the sum, rounded to roundvalue, of the interest amount between pmt1 and pmt2 for an amortization schedule. [APPS]1:Finance
CALC
A:ΣInt(
graph intersection Calculates the intersection of two functions. [2ND][CALC]
5:intersection
list inverse list-1 Returns 1 divided by list elements. [x-1]
matrix inverse matrix-1 Returns matrix inverted. [x-1]
math inverse value-1 Returns 1 divided by a real or complex number or expression. [x-1]
statistics invNorm(area[,m,s]) Computes the inverse cumulative normal distribution function for a given area under the normal distribution curve specified by m and s. [2ND][DISTR]
3:invNorm(
statistics invT(area,df) Computes the inverse cumulative student-t probability function specified by degree of freedom, df for a given area under the curve, used for one-tailed tests. [2ND][DISTR]
4:invT(
math iPart(value) Returns the integer part of a real or complex number, expression, list, or matrix. [MATH]
NUM
3:iPart(
finance irr(CF0,CFList[,CFFreq]) Returns the interest rate at which the net present value of the cash flow is equal to zero. [APPS]1:Finance
CALC
8:irr(
time Is :IS>(variable,value)
:commandA
:commands
Increments variable by 1; skips commandA if variable>value. [PRGM]
CTL
A:IS>(
time isClockOn Identifies if clock is ON or OFF. Returns 1 if the clock is ON. Returns 0 if the clock is OFF. [2ND][CATALOG]
IsClockOn
list L Identifies the next one to five characters as a user-created list name. [2ND][LIST]
OPS
B:L
graph LabelOff Turns off axes labels. [2ND][FORMAT]
LabelOff
graph LabelOn Turns on axes labels. [2ND][FORMAT]
LabelOn
program Lbl label Creates a label of one or two characters within a program. [PRGM]
CTL
9:Lbl
math lcm(valueA,valueB) Returns the least common multiple of valueA and valueB, which can be real numbers or lists. [MATH]
NUM
8:lcm(
string length(string) Returns the number of characters in string. [2ND][CATALOG]
length(
logic less than valueA Returns 1 if valueA < valueB. Returns 0 if valueA ≥ valueB. valueA and valueB can be real or complex numbers, expressions, or lists. [2ND][TEST]
TEST
5:
logic less than or equal valueA≤valueB Returns 1 if valueA ≤ valueB. Returns 0 if valueA > valueB. valueA and valueB can be real or complex numbers, expressions, or lists. [2ND][TEST]
TEST
6:≤
graph Line(X1,Y1,X2,Y2) Draws a line from (X1,Y1) to (X2,Y2). [2ND][DRAW]
2:Line(
graph Line(X1,Y1,X2,Y2,0) Erases a line from (X1,Y1) to (X2,Y2). [2ND][DRAW]
2:Line(
statistics LinReg(a+bx) Xlist,Ylist,freqlist,regeq Fits a linear regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
8:LinReg(a+bx)
statistics LinReg(a+bx) Xlist,Ylist,freqlist,regeq Fits a linear regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
4:LinReg(a+bx)
statistics LinRegTInt Xlist,Ylist,freqlist,c-level,regeq,Calculate Performs a linear regression and computes the t confidence interval for the slope coefficient b. [STAT]
[TESTS]
G:LinRegTInt
statistics LinRegTTest Xlist,Ylist,freqlist,β & ρ,regeq,Calculate Performs a linear regression and a t-test. [STAT]
TESTS
F:LinRegTTest
list List ▶ matr(list1,...,listn,matrixname) Fills matrixname column by column with the elements from each specified list. [2ND][LIST]
OPS
0:List 4 matr(
list list ΔList(list) Returns a list containing the differences between consecutive elements in list. [2ND][LIST]
OPS
7:ΔList(
math ln(value)
$ln(a)^x = xln(a), ln(e^x) = x$
Returns the natural logarithm of a real or complex number, expression, or list.
[LN]
statistics LnReg Xlist,Ylist,freqlist,regeq Fits a logarithmic regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
9:LnReg
math log(value) Returns logarithm of a real or complex number, expression, or list. [LOG]
math logBASE(value, base) Returns the logarithm of a specifed value determined from a specified base: logBASE(value, base). [MATH]
A: logBASE
statistics Logistic Xlist,Ylist,freqlist,regeq Fits a logistic regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
B:Logistic
statistics Manual-Fit equname Fits a linear equation to a scatter plot. [STAT]
CALC
D:Manual-Fit
format MATHPRINT Displays most entries and answers the way they are displayed in textbooks. [MODE]
MATHPRINT
matrix matrix {rows,columns} -> dim(matrixname) Assigns new dimensions to a new or existing matrixname. [2ND][MATRIX]
MATH
3:dim(
list Matr▶list(matrix,listA,...,listn) Fills each list with elements from each column in matrix. [2ND][LIST]
OPS
A:Matr▶list(
matrix Matr▶list(matrix,listA,...,listn) Fills each list with elements from each column in matrix. [2ND][LIST]
OPS
A:Matr▶list(
list Matr▶list(matrix,column#,list) Fills a list with elements from a specified column# in matrix. [2ND][LIST]
OPS
A:▶Matrlist(
matrix Matr▶list(matrix,column#,list) Fills a list with elements from a specified column# in matrix. [2ND][LIST]
OPS
A:▶Matrlist(
list max(list) Returns largest real or complex element in list. [2ND][LIST]
MATH
2:max(
list max(listA,listB) Returns a real or complex list of the larger of each pair of elements in listA and listB. [2ND][LIST]
MATH
2:max(
list max(value,list) Returns a real or complex list of the larger of value or each list element. [2ND][LIST]
MATH
2:max(
math max(valueA,valueB) Returns the larger of valueA and valueB. [MATH]
NUM
7:max(
graph maximum Calculates the maximum of a function. [2ND][CALC]
4:maximum
statistics mean(list[,freqlist]) Returns the mean of list with frequency freqlist. [2ND][LIST]
MATH
3:mean(
statistics median(list[,freqlist]) Returns the median of list with frequency freqlist. [2ND][LIST]
MATH
4:median(
statistics Med-Med Xlist,Ylist,freqlist,regeq Fits a median-median model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
3:Med-Med
memory Mem Mgmt/Del Deletes selected variables, lists, matrices, programs, etc. [2ND][+]
2:Mem Mgmt/Del...
program Menu("title","text1",label1[,...,"text7",label7]) Generates a menu of up to seven items during program execution. [PRGM]
CTL
list min(list) Returns smallest real or complex element in list. [2ND][LIST]
MATH
1:min(
list min(listA,listB) Returns real or complex list of the smaller of each pair of elements in listA and listB. [2ND][LIST]
MATH
1:min(
list min(value,list) Returns a real or complex list of the smaller of value or each list element. [2ND][LIST]
MATH
1:min(
math min(valueA,valueB) Returns smaller of valueA and valueB. [2ND][LIST]
NUM
6:min(
graph minimum Calculates the minimum of a function. [2ND][CALC]
3:minimu
angle minutes notation degreeso minutes' seconds" Interprets minutes angle measurement as minutes. [2ND][ANGLE]
ANGLE
2:'
list multiplication list x value Returns each list element times value. [x]
list multiplication listA x listB Returns listA elements times listB elements. [x]
matrix multiplication matrixA x matrixB Returns matrixA times matrixB. [x]
list multiplication value x list Returns value times each list element. [x]
matrix multiplication value x matrix Returns value times matrix elements. [x]
math multiplication valueA x valueB Returns valueA times valueB. [x]
math nDeriv(expression,variable,value[,H]) Returns approximate numerical derivative of expression with respect to variable at value, with specified H. [MATH]
MATH
8:nDeriv(
format number to fraction ▶ n/d ◀ ▶ Un/d Converts the results from a fraction to mixed number or from a mixed number to a fraction, if applicable. [ALPH][F1]
3: ▶ n/d ◀ ▶ Un/d
or
[MATH]
NUM
A: ▶ n/d ◀ ▶ Un/d
finance nominal interest rate ▶Nom(effective rate,compounding periods) Computes the nominal interest rate. [APPS]1:Finance
CALC
B:▶Nom(
format NORMAL Sets normal display mode. [MODE]
NORMAL
statistics normalcdf(lowerbound,upperbound[,m,s])
$normalcdf(a,b) = \int_{a}^{b} normalpdf(x)dx = \frac {1}{\sqrt{2\pi}} \int_{a}^{b} e^{\frac {-1}{2} x^2}dx$
$normalcdf(a,b,\mu,\sigma) = \left ( {\frac {a-\mu}{\sigma}, \frac {b-\mu}{\sigma}} \right )$
Computes the normal distribution probability between lowerbound and upperbound for the specified m and s.
[2ND][DISTR]
2:normalcdf(
statistics normalpdf(x[,m,s])
$normalpdf(a,b) = \frac {1}{\sqrt{2\pi}} e^{\frac {-1}{2} x^2}$
Computes the probability density function for the normal distribution at a specified x value for the specified m and s.
[2ND]][DISTR]
1:normalpdf(
logic not(value) Negates the input, value can be a real number, expression, or list. [2ND][TEST]
LOGIC
4:not(
logic not equal valueA≠valueB Returns 1 if valueA ≠ valueB. Returns 0 if valueA = valueB. valueA and valueB can be real or complex numbers, expressions, lists, or matrices. [2ND][TEST]
TEST
2:≠
finance npv(interest rate,CF0,CFList[,CFFreq]) Computes the sum of the present values for cash inflows and outflows. [APPS]1:Finance
CALC
7:npv(
logic or valueA or valueB Returns 1 if valueA or valueB is true. valueA and valueB can be real numbers, expressions, or lists. [2ND][TEST]
LOGIC
2:or
program Output(row,column,"text") Displays text beginning at specified row and column within a program. [PRGM]
I/O
6:Output(
program Output(row,column,value) Displays value beginning at specified row and column. [PRGM]
I/O
6:Output(
graph PAR Sets prametric graphing mode. [MODE]
PAR
graph Param Sets parametric graphing mode. [MODE]
PAR
program Pause Suspends program execution until [ENTER] is pressed. [PRGM]
CTL
8:Pause
program Pause [value] Displays value; suspends program execution until [ENTER] is pressed. [PRGM]
CTL
8:Pause
statistics permutations list nPr value
$P(n,r) = \frac{n}{(n-r)!}$
Returns a list of the permutations of each element in list taken value at a time.
[MATH]
PRB
2:nPr
statistics permutations listA nPr listB
$P(n,r) = \frac{n}{(n-r)!}$
Returns a list of the permutations of each element in listA taken each element in listB at a time.
[MATH]
PRB
2:nPr
statistics permutations value nPr list
$P(n,r) = \frac{n}{(n-r)!}$
Returns a list of the permutations of value taken each element in list at a time.
[MATH]
PRB
2:nPr
statistics permutations valueA nPr valueB
$P(n,r) = \frac{n}{(n-r)!}$
Returns the number of permutations of valueA taken valueB at a time.
[MATH]
PRB
2:nPr
graph Plot#(type,Xlist,Ylist,mark) Defines Plot# (1, 2, or 3) of type Scatter or xyLine for Xlist and Ylist using mark. [2ND][STAT PLOT]
STAT PLOTS
1:Plot1-
2:Plot2-
3:Plot3-
graph Plot#(type,Xlist,freqlist) Defines Plot# (1, 2, or 3) of type Histogram or Boxplot for Xlist with frequency freqlist. [2ND][STAT PLOT]
STAT PLOTS
1:Plot1-
2:Plot2-
3:Plot3-
graph Plot#(type,Xlist,freqlist,mark) Defines Plot# (1, 2, or 3) of type ModBoxplot for Xlist with frequency freqlist using mark. [2ND][STAT PLOTS]
STAT PLOTS
1:Plot1-
2:Plot2-
3:Plot3-
graph Plot#(type,datalist,data axis,mark) Defines Plot# (1, 2, or 3) of type NormProbPlot for datalist on data axis using mark. Data axis can be X or Y. [2ND][STAT PLOTS]
STAT PLOTS
1:Plot1-
2:Plot2-
3:Plot3-
graph PlotsOff [1,2,3] Deselects all stat plots or one or more specified stat plots (1, 2, or 3). [2ND][STAT PLOTS]
STAT PLOTS
4:PlotsOff
graph PlotsOn [1,2,3] Selects all stat plots or one or more specified stat plots (1, 2, or 3). [2ND][STAT PLOTS]
STAT PLOTS
5:PlotsOn
finance Pmt_Bgn Specifies an annuity due, where payments occur at the beginning of each payment period. [APPS]1:Finance
CALC
F:Pmt_Bgn
finance Pmt_End Specifies an ordinary annuity, where payments occur at the end of each payment period. [APPS]1:Finance
CALC
E:Pmt_End
probability poissoncdf(m,x) Computes a cumulative probability at x for the discrete Poisson distribution with specified mean m. [2ND][DISTR]
D:poissoncdf(
probability poissonpdf(m,x) Computes a probability at x for the discrete Poisson distribution with the specified mean m. [2ND][DISTR]
C:poissonpdf(
math power of ten 10^(value) Returns 10 raised to the value power. value can be a real or complex number or expression. [2ND][10x]
list power of ten 10^(list) Returns a list of 10 raised to the list power. [2ND][10x]
graph POL
$r = \sqrt{x^2 + y^2} , \; \varphi = atan2(y,x)$
Sets polar graphing mode.
[MODE]
POL
format polar complex value▶Polar Displays complex value in polar format. [MATH]
CPX
7:▶Polar
format PolarGC Sets polar graphing coordinates format. [2ND][FORMAT]
PolarGC
program prgmname Executes the program name. [PRGM]
CTRL
D:prgm
finance principal amount ∑Prn(pmt1,pmt2[,roundvalue]) Computes the sum, rounded to roundvalue, of the principal amount between pmt1 and pmt2 for an amortization schedule. [APPS]1:Finance
CALC
0:GPrn(
list prod(list[,start,end]) Returns product of list elements between start and end. [2ND][LIST]
MATH
6:prod(
program Prompt variableA[,variableB,...,variable n] Prompts for value for variableA, then variableB, etc. Used within a program. [PRGM]
I/O
2:Prompt
graph Pt-Change(x,y) Reverses a point at (x,y). [2ND][DRAW]
POINTS
3:Pt-Change(
graph Pt-Off(x,y[,mark]) Erases a point at (x,y) using mark. [2ND][DRAW]
POINTS
2:Pt-Off(
graph Pt-On(x,y[,mark]) Draws a point at (x,y) using mark. [2ND][DRAW]
POINTS
1:Pt-On(
statistics PwrReg Xlist,Ylist,freqlist,regeq Fits a power regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
A:PwrReg
graph Pxl-Change(row,column) Reverses pixel at (row,column); 0 ≤ row ≤ 62 and 0 ≤ column ≤ 94. [2ND][DRAW]
POINTS
6:Pxl-Change(
graph Pxl-Off(row,column) Erases pixel at (row,column); 0 ≤ row ≤ 62 and 0 ≤ column ≤ 94. [2ND][DRAW]
POINTS
5:Pxl-Off(
graph Pxl-On(row,column) Draws pixel at (row,column); 0 ≤ row ≤ 62 and 0 ≤ column ≤ 94. [2ND][DRAW]
POINTS
4:Pxl-On(
graph pxl-Test(row,column) Returns 1 if pixel (row, column) is on, 0 if it is off; 0 ≤ row ≤ 62 and 0 ≤ column ≤ 94. [2ND][DRAW]
POINTS
7:pxl-Test(
angle P▶Rx(r,θ) Returns X, given polar coordinates r and q or a list of polar coordinates. [2ND][ANGLE]
ANGLE
7:P▶Rx(
angle P▶Ry(r,q) Returns Y, given polar coordinates r and q or a list of polar coordinates. [2ND][ANGLE]
ANGLE
8:P▶Ry(
statistics QuadReg Xlist,Ylist,freqlist,regeq Fits a quadratic regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
statistics QuartReg Xlist,Ylist,freqlist,regeq Fits a quartic regression model to Xlist and Ylist with frequency freqlist, and stores the regression equation to regeq. [STAT]
CALC
7:QuartReg
ANGLE
3:r
probability rand[(numtrials)] Returns a random number between 0 and 1 for a specified number of trials numtrials. [MATH]
PRB
1:rand
probability randBin(numtrials,prob[,numsimulations]) Generates and displays a random real number from a specified binomial distribution. [MATH]
PRB
7:randBin(
probability randInt( lower,upper[,numtrials]) Generates and displays a random integer within a range specified by lower and upper integer bounds for a specified number of trials numtrials. [MATH]
PRB
5:randInt(
probability randIntNoRep(lowerint,upperint) Returns a random ordered list of integers from a lower integer to an upper integer which may include the lower integer and upper integer. [MATH]
PRB
8:randIntNoRep(
matrix randM(rows,columns) Returns a random matrix of rows (1-99) × columns (1-99). [2ND][MATRIX]
MATH
6:randM(
probability randNorm(μ,σ[,numtrials]) Generates and displays a random real number from a specified Normal distribution specified by μ and σ for a specified number of trials numtrials. [MATH]
PRB
6:randNorm(
complex re^θi Sets the mode to polar complex number mode (re^θi). [MODE]
re^θi
format REAL Sets mode to display complex results only when you enter complex numbers. [MODE]
REAL
math real(value) Returns the real part of a complex number or list of complex numbers. [MATH]
CPX
2:real(
graph RecallGDB n Restores all settings stored in the graph database variable GDBn. [2ND][DRAW]
STO
4:RecallGDB
graph RecallPic n Displays the graph and adds the picture stored in Picn. [2ND][DRAW]
STO
2:RecallPic
graph RectGC Sets rectangular graphing coordinates format. [2ND][FORMAT]
RectGC
matrix ref(matrix) Returns the row-echelon form of a matrix. [2ND][MATRIX]
MATH
A:ref(
math remainder(dividend,divisor) Reports the remainder as a whole number from a division of two whole numbers where the divisor is not zero. [MATH]
NUM
0:remainder(
math remainder(dividend, list) Reports the remainder as a whole number from a division of two whole numbers where the divisor is a list. [MATH]
NUM
0:remainder(
math remainder(list, divisor) Reports the remainder as a whole number from a division of two lists where the divisor is not zero. [MATH]
NUM
0:remainder(
math remainder(list, list) Reports the remainder as a whole number from a division of two lists. [MATH]
NUM
0:remainder(
program repeat :Repeat condition
:commands
:End
:commands
Executes commands until condition is true within a program. [PRGM]
CTL
6:Repeat
memory Reset Resets RAM, ARCHIVE, or ALL memory. [2ND][+]
CTL
7:Reset
program Return Returns to the calling program. [PRGM]
CTL
E:Return
math root listxvalue Returns list roots of value. [MATH]
MATH
5:x
math root listAxlistB Returns listA roots of listB. [MATH]
MATH
5:x
math root xthrootxlist Returns the xth root of list elements. [MATH]
MATH
5:x
math root xthrootxvalue Returns the xth root of value. [MATH]
MATH
5:x
math round(value[,#decimals]) Returns a number, expression, list, or matrix rounded to #decimals (≤ 9). [MATH]
NUM
2:round(
matrix row *row(value,matrix,row) Returns a matrix with row of matrix multiplied by value and stored in row. [2ND][MATH]
MATH
E:*row(
matrix row *row+(value,matrix,rowA,rowB) Returns a matrix with rowA of matrix multiplied by value, added to rowB, and stored in rowB. [2ND][MATRIX]
MATH
F:*row+(
matrix row+(matrix,rowA,rowB) Returns a matrix with rowA of matrix added to rowB and stored in rowB. [2ND][MATRIX]
MATH
D:row+(
matrix rowSwap(matrix,rowA,rowB) Returns a matrix with rowA of matrix swapped with rowB. [2ND][MATRIX]
MATH
C:rowSwap(
matrix rref(matrix) Returns the reduced row-echelon form of a matrix. [2ND][MATRIX]
MATH
B:rref(
angle R▶Pr(x,y) Returns R, given rectangular coordinates x and y or a list of rectangular coordinates. [2ND][ANGLE]
ANGLE
5:R4Pr(
angle R▶Pθ(x,y) Returns θ, given rectangular coordinates x and y or a list of rectangular coordinates. [2ND][ANGLE]
ANGLE
6:R▶Pθ(
format SCI Sets scientific notation display mode. [MODE]
SCI
angle seconds notation degreeso minutes' seconds" Interprets seconds angle measurement as seconds. [ALPHA]["]
graph Select(Xlist,Ylist) Selects one or more specific data points from a scatter plot or xyLine plot (only), and then store•s the selected data points to two new lists, Xlist and Ylist. [2ND][LIST]
OPS
8:Select(
list Select(Xlist,Ylist) Selects one or more specific data points from a scatter plot or xyLine plot (only), and then store•s the selected data points to two new lists, Xlist and Ylist. [2ND][LIST]
OPS
8:Select(
program Send(variable) Sends contents of variable to the CBL 2™ or CBR™ System within a program. [PRGM]
I/O
B:Send(
graph SEQ Sets sequence graphing mode. [MODE]
SEQ
list seq(expression,variable,begin,end[,increment]) Returns list created by evaluating expression with regard to variable, from begin to end by increment. [2ND][LIST]
OPS
5:seq(
graph SEQUENTIAL Sets mode to graph functions sequentially. [MODE]
SEQUENTIAL
time SET CLOCK Sets the clock. [MODE]
SET CLOCK
date setDate(year,month,day) Sets the date using a year, month, day format. The year must be 4 digits; month and day can be 1 or 2 digits. [2ND][CATALOG]
setDate(
date setDtFmt(integer) Sets the date format.
1 = M/D/Y
2 = D/M/Y
3 = Y/M/D
[2ND][CATALOG]
setDtFmt(
time setTime(hour,minute,second) Sets the time using an hour, minute, second format. The hour must be in 24 hour format, in which 13 = 1 p.m. [2ND][CATALOG]
setTime(
time setTmFmt(integer) Sets the time format.
12 = 12 hour format
24 = 24 hour format
[2ND][CATALOG]
setTmFmt(
list SetUpEditor Removes all list names from the stat list editor, and then restores list names L1 through L6 to columns 1 through 6. [STAT]
EDIT
5:SetUpEditor
list SetUpEditor list1[,list2,...,list20] Removes all list names from the stat list editor, then sets it up to display one or more lists in the specified order, starting with column 1. [STAT]
EDIT
5:SetUpEditor
graph Shade(lowerfunc,upperfunc[,Xleft,Xright,pattern,patres]) Draws lowerfunc and upperfunc in terms of X on the current graph and uses pattern and patres to shade the area bounded by lowerfunc, upperfunc, Xleft, and Xright. [2ND][DRAW]
DRAW
graph Shadeχ2(lowerbound,upperbound,df) Draws the density function for the c2 distribution specified by degrees of freedom df and shades the area between lowerbound and upperbound. [2ND][DISTR]
DRAW
graph ShadeF(lowerbound,upperbound,numerator df,denominator df) Draws the density function for the F distribution specified by numerator df and denominator df and shades the area between lowerbound and upperbound. [2ND][DISTR]
DRAW
graph ShadeNorm(lowerbound,upperbound[,μ,σ]) Draws the normal density function specified by μ and σ and shades the area between lowerbound and upperbound. [2ND][DISTR]
DRAW
graph Shade_t(lowerbound,upperbound,df) Draws the density function for the Student-t distribution specified by degrees of freedom df, and shades the area between lowerbound and upperbound. [2ND][DISTR]
DRAW
graph SIMUL Sets mode to graph functions simultaneously. [MODE]
SIMUL
trigonometry sin(value)
$sin(\theta) = \frac {opp}{hyp} = cos \left ( \frac {\pi}{2} - \theta \right ) = \frac {1}{csc(\theta)}$
Returns the sine of a real number, expression, or list.
[SIN]
trigonometry sin-1(value) Returns the arcsine of a real number, expression, or list. [2ND][SIN-1]
trigonometry sinh(value) Returns the hyperbolic sine of a real number, expression, or list. [2ND][CATALOG]
sinh(
trigonometry sinh-1(value) Returns the hyperbolic arcsine of a real number, expression, or list. [2ND][CATALOG]
sinh-1
statistics SinReg iterations,Xlist,Ylist,period,regeq Attempts iterations times to fit a sinusoidal regression model to Xlist and Ylist using a period guess, and stores the regression equation to regeq. [STAT]
CALC
C:SinReg
math solve(expression,variable,guess,{lower,upper}) Solves expression for variable, given an initial guess and lower and upper bounds within which the solution is sought. [MATH]
MATH
0:solve(
list SortA(list) Sorts elements of list in ascending order. [2ND][LIST]
OPS
1:SortA(
list SortA(keylist,dependlist1[,dependlist2,...,dependlist n]) Sorts elements of keylist in ascending order, the sorts each dependlist as a dependent list. [2ND][LIST]
OPS
1:SortA(
list SortD(list) Sorts elements of list in descending order. [2ND][LIST]
OPS
2:SortD(
list SortD(keylist,dependlist1[,dependlist2,..., dependlist n]) Sorts elements of keylist in descending order, then sorts each dependlist as a dependent list. [2ND][LIST]
OPS
2:SortD(
math square value2 Returns value multiplied by itself. value can be a real or complex number or expression. [x2]
list square list2 Returns list elements squared. [x2]
matrix square matrix2 Returns matrix multiplied by itself. [x2]
math square root √(value) Returns square root of a real or complex number, expression, or list. [2ND][√]
time startTmr Starts the clock timer. Store or note the displayed value, and use it as the argument for checkTmr( ) to check the elapsed time. [2ND][CATALOG]
startTmr
statistics STAT DIAGNOSTICS Enables or disables wizard syntax help for statistical commands, distributions, and seq(. [MODE]
STAT DIAGNOSTICS
statistics stdDev(list[,freqlist]) Returns the standard deviation of the elements in list with frequency freqlist. [2ND][LIST]
MATH
7:stdDev(
program Stop Ends program execution; returns to home screen. [PRGM]
CTL
F:Stop
memory Store value->variable Stores value in variable. [STO>]
graph StoreGDB n Stores current graph in database GDBn. [2ND][DRAW]
STO
3:StoreGDB
memory StoreGDB n Stores current graph in database GDBn. [2ND][DRAW]
STO
3:StoreGDB
memory StorePic n Stores current picture in picture Picn. [2ND][DRAW]
STO
1:StorePic
string StringEqu(string,Y= var) Converts string into an equation and stores it in Y= var. [2ND][CATALOG]
String▶Equ(
string sub(string,begin,length) Returns a string that is a subset of another string, from begin to length. [2ND][CATALOG]
sub(
list subtraction list - value Subtracts value from list elements. [-]
list subtraction listA - listB Subtracts listB elements from listA elements. [-]
matrix subtraction matrixA - matrixB Subtracts matrixB elements from matrixA elements. [-]
list subtraction value - list Subtracts list elements from value [-]
string subtraction valueA - valueB Subtracts valueB from valueA. [-]
list sum(list[,start,end]) Returns the sum of elements of list from start to end. [2ND][LIST]
MATH
5:sum(
math summation G(expression[,start,end]) Displays the MathPrintTM summation entry template and returns the sum of elements of list from start to end, where start <= end. [MATH]
NUM
0: summation G(
trigonometry tan(value)
$tan(\theta) = \frac {opp}{adj} = \frac {sin(\theta)}{cos(\theta)} = cot \left ( \frac {\pi}{2} - \theta \right ) = \frac {1}{cot(\theta)}$
Returns the tangent of a real number, expression, or list.
[TAN]
trigonometry tan-1(value) Returns the arctangent of a real number, expression, or list. [2ND][TAN-1
graph Tangent(expression,value) Draws a line tangent to expression at X=value. [2ND][DRAW]
DRAW
5:Tangent(
trigonometry Tangent(expression,value) Draws a line tangent to expression at X=value. [2ND][DRAW]
DRAW
5:Tangent(
trigonometry tanh(value) Returns hyperbolic tangent of a real number, expression, or list. [2ND][CATALOG]
tanh(
trigonometry tanh-1(value) Returns the hyperbolic arctangent of a real number, expression, or list. [2ND][CATALOG]
tanh-1(
statistics tcdf(lowerbound,upperbound,df)
$tcdf(-\infty,0,df) = \frac {1}{2}$
$tcdf(-\infty,upbnd,df) = \frac {1}{2} + tcdf(0,upbnd,df)$
$tcdf(-upbnd,upbnd,df) = 2tcdf(0,upbnd,df)$
Computes the Student-t distribution probability between lowerbound and upperbound for the specified degrees of freedom df.
[2ND][DISTR]
6:tcdf(
graph Text(row,column,text1,text2,...,text n) Writes text on graph beginning at pixel (row,column), where 0 ≤ row ≤ 57 and 0 ≤ column ≤ 94. [2ND][DRAW]
DRAW
0:Text(
program Then
See If:Then
Used in a program. [PRGM]
CTL
2:Then
time Time Sets sequence graphs to plot with respect to time. [2ND][FORMAT]
Time
time timeCnv(seconds) Converts seconds to units of time that can be more easily understood for evaluation. The list is in {days,hours,minutes,seconds} format. [2ND][CATALOG]
statistics tpdf(x,df)
$tpdf(t,df) = \frac {\Gamma((df + 1)/2)}{\sqrt{df \pi} \, \Gamma(df/2)}\, \left (1+ \frac {t^2} {df} \right) ^{-\frac {1}{2} (df + 1)}$
Computes the probability density function (pdf) for the Student-t distribution at a specified x value with specified degrees of freedom df.
[2ND][DISTR]
5:tpdf(
graph Trace Displays the graph and enters TRACE mode. [TRACE]
matrix transpose matrixT Returns a matrix in which each element (row,column) is swapped with the corresponding element (column,row) of matrix. [2ND][MATRIX]
MATH
2:T
statistics T-Test Data μ0,list,freqlist,μ alternative,Calculate/Draw
Performs a t test with frequency freqlist. [STAT]
TESTS
2:T-Test
statistics T-Test Stats μ0, x̄,Sx,n,μ alternative,Calculate/Draw
Performs a t test with frequency freqlist. [STAT]
TESTS
2:T-Test
statistics TInterval Data list,freqlist,c-level Computes a t confidence interval. [STAT]
TESTS
8:TInterval
statistics TInterval Stats x̄,Sx,n,c-level Computes a t confidence interval. [STAT]
TESTS
8:TInterval
statistics 2-SampTInt Data list1,list2,freqlist1,freqlist2,c-level,pooled,Calculate Computes a two-sample t confidence interval. [STAT]
TESTS
0:2-SampTInt
statistics 2-SampTInt Stats x̄1,Sx1,n1,x̄2,Sx2,n2,c-level,pooled,Calculate Computes a two-sample t confidence interval. [STAT]
TESTS
0:2-SampTInt
statistics 2-SampTTest Data list1,list2,freqlist1,freqlist2,μ alternative,pooled,Calculate/Draw
Computes a two-sample t test. [STAT]
TESTS
4:2-SampTTest
statistics 2-SampTTest Data x̄1,Sx1,n1,x̄2,Sx2,n2,μ alternative,pooled,Calculate/Draw
Computes a two-sample t test. [STAT]
TESTS
4:2-SampTTest
finance tvm_FV[(N,I%,PV,PMT,P/Y,C/Y)] Computes the future value. [APPS]1:Finance
CALC
6:tvm_FV
finance tvm_I%[(N,PV,PMT,FV,P/Y,C/Y)] Computes the annual interest rate. [APPS]1:Finance
CALC
3:tvm_I%
finance tvm_N[(I%,PV,PMT,FV,P/Y,C/Y)] Computes the number of payment periods. [APPS]1:Finance
CALC
5:tvm_N
finance tvm_Pmt[(N,I%,PV,FV,P/Y,C/Y)] Computes the amount of each payment. [APPS]1:Finance
CALC
2:tvm_Pmt
finance tvm_PV[(N,I%,PMT,FV,P/Y,C/Y)] Computes the present value. [APPS]1:Finance
CALC
4:tvm_PV
memory UnArchive Moves the specified variables from the user data archive memory to RAM. To archive variables, use Archive. [2ND][MEM]
6:UnArchive
format Un/d Displays results as a mixed number, if applicable. [MATH]
NUM
C: Un/d
graph uvAxes Sets sequence graphs to plot u(n) on the x-axis and v(n) on the y-axis. [2ND][FORMAT]
uv
graph uwAxes Sets sequence graphs to plot u(n) on the x-axis and w(n) on the y-axis. [2ND][FORMAT]
uw
graph value Displays the y-coordinate of a given x-coordinate. [2ND][CALC]
1:value
statistics variance(list[,freqlist]) Returns the variance of the elements in list with frequency freqlist. [2ND][LIST]
MATH
8:variance(
statistics var stats 1-Var Stats [Xlist,freqlist] Performs one-variable analysis on the data in Xlist with frequency freqlist. [STAT]
CALC
1:1-Var Stats
statistics var stats 2-Var Stats [Xlist,Ylist,freqlist] Performs two-variable analysis on the data in Xlist and Ylist with frequency freqlist. [STAT]
CALC
2:2-Var Stats
graph Vertical x Draws a vertical line at x. [2ND][DRAW]
DRAW
4:Vertical
graph vwAxes Sets sequence graphs to plot v(n) on the x-axis and w(n) on the y-axis. [2ND][FORMAT]
vw
graph Web Sets sequence graphs to trace as webs. [2ND][FORMAT]
Web
program while :While condition
:commands
:End
:command
Executes commands while condition is true. [PRGM]
CTL
5:While
logic xor valueA xor valueB Returns 1 if only valueA or valueB = 0. valueA and valueB can be real numbers, expressions, or lists. [2ND][TEST]
LOGIC
3:xor
graph ZBox Displays a graph, lets you draw a box that defines a new viewing window, and updates the window. [ZOOM]
ZOOM
1:ZBox
graph ZDecimal Adjusts the viewing window so that ΔX=0.1 and ΔY=0.1, and displays the graph screen with the origin centered on the screen. [ZOOM]
ZOOM
4:ZDecimal
graph zero Calculates the zeroes of a function. [2ND][CALC]
2:zero
graph ZFrac1/2 Sets the window variables so that you can trace in increments of 1/2, if possible. Sets ΔX and ΔY to 1/2. [ZOOM]
ZOOM
B:ZFrac1/2
graph ZFrac1/3 Sets the window variables so that you can trace in increments of 1/3, if possible. Sets ΔX and ΔY to 1/3. [ZOOM]
ZOOM
C:ZFrac1/3
graph ZFrac1/4 Sets the window variables so that you can trace in increments of 1/4, if possible. Sets ΔX and ΔY to 1/4. [ZOOM]
ZOOM
D:ZFrac1/4
graph ZFrac1/5 Sets the window variables so that you can trace in increments of 1/5, if possible. Sets ΔX and ΔY to 1/5. [ZOOM]
ZOOM
E:ZFrac1/5
graph ZFrac1/8 Sets the window variables so that you can trace in increments of 1/8, if possible. Sets ΔX and ΔY to 1/8. [ZOOM]
ZOOM
F:ZFrac1/8
graph ZFrac1/10 Sets the window variables so that you can trace in increments of 1/10, if possible. Sets ΔX and ΔY to 1/10. [ZOOM]
ZOOM
G:ZFrac1/1
graph ZInteger Redefines the viewing window using these dimensions:
ΔX=1 Xscl=10
ΔY=1 Yscl=10
[ZOOM]
ZOOM
8:ZInteger
graph Zoom In Magnifies the part of the graph that surrounds the cursor location. [ZOOM]
ZOOM
2:Zoom In
graph Zoom Out Displays a greater portion of the graph, centered on the cursor location. [ZOOM]
ZOOM
3:Zoom Out
graph ZoomFit Recalculates Ymin and Ymax to include the minimum and maximum Y values, between Xmin and Xmax, of the selected functions and replots the functions. [ZOOM]
ZOOM
0:ZoomFit
graph ZoomRcl Graphs the selected functions in a user-defined viewing window. [ZOOM]
MEMORY
3:ZoomRcl
graph ZoomStat Redefines the viewing window so that all statistical data points are displayed. [ZOOM]
ZOOM
9:ZoomStat
graph ZoomSto Immediately stores the current viewing window. [ZOOM]
MEMORY
2:ZoomSto
graph ZPrevious Replots the graph using the window variables of the graph that was displayed before you executed the last ZOOM instruction. [ZOOM]
MEMORY
1:ZPrevious
graph ZQuadrant1 Displays the portion of the graph that is in quadrant 1. [ZOOM]
ZOOM
graph ZSquare Adjusts the X or Y window settings so that each pixel represents an equal width and height in the coordinate system, and updates the viewing window. [ZOOM]
ZOOM
5:ZSquare
graph ZStandard Replots the functions immediately, updating the window variables to the default values. [ZOOM]
ZOOM
6:ZStandard
statistics Z-test Data μ0,σ,x̄,n,μ alternative,Calculate/Draw
Computes a two-sample z test. [STAT]
TESTS
1:Z-Test(
statistics Z-test Stats σ1,σ2,list,freqlist,alternative,Calculate/Draw Computes a two-sample z test. [STAT]
TESTS
1:Z-Test(
statistics 1-PropZInt x,n,c-level,Calculate Computes a one-proportion z confidence interval [STAT]
TESTS
A:1-PropZInt(
statistics 1-PropZTest p0,x,n,prop alternative,Calculate/Draw Computes a one-proportion z test. [STAT]
TESTS
5:1-PropZTest(
statistics 2-PropZInt x1,n1,x2,n2,c-level,Calculate Computes a two-proportion z interval. [STAT]
TESTS
B:2-PropZInt(
statistics 2-PropZTest x1,n1,x2,n2, p1 alternative,Calculate/Draw Computes a two-proportion z test. [STAT]
TESTS
6:2-PropZTest(
statistics 2-SampZInt Data σ1,σ2,list1,list2,freqlist1,freqlist2,c-level,Calculate Computes a two-sample z confidence interval. [STAT]
TESTS
9:2-SampZInt(
statistics 2-SampZInt Stats σ1,σ2,x̄1,n1,x̄2,n2,c-level,Calculate Computes a two-sample z confidence interval. [STAT]
TESTS
9:2-SampZInt(
statistics 2-SampZTest Data σ1,σ2,list1,list2,freqlist1,freqlist2,μ alternative,Calculate/Draw
Computes a two-sample z test. [STAT]
TESTS
3:2-SampZTest(
statistics 2-SampZTest Stats σ1,σ2,x̄1,n1,x̄2,n2,μ alternative,Calculate/Draw Computes a two-sample z test. [STAT]
TESTS
3:2-SampZTest(
statistics ZInterval Data σ,list,freqlist,c-level Computes a z confidence interval. [STAT]
TESTS
7:ZInterval
statistics ZInterval Stats σ,x̄,n,c-level Computes a z confidence interval. [STAT]
TESTS
7:ZInterval
graph ZTrig Replots the functions immediately, updating the window variables to preset values for plotting trig functions. [ZOOM]
ZOOM
7:ZTrig
Questions, comments, suggestions: Dr. Denise Meeks, email: tucsonkosmicgirl@gmail.com | 16,309 | 56,164 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-25 | longest | en | 0.618124 |
https://www.gamedev.net/forums/topic/32793-is-the-3ds-file-asc-good/ | 1,544,840,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00219.warc.gz | 894,934,814 | 26,965 | #### Archived
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# Is the 3DS file *.asc good?
This topic is 6592 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I'm writing a loader for it (as well as many many other fileypes), and I'm wondering what the ASC file uses. Does it use triangle strips? Triangles only? Something completely different? I just find it wierd that there are 72 faces in a model and 92 points. This means that some of these points overlap, but how do you know which ponts are part of each triangle? -SkyFire Edited by - skyfire360 on 11/27/00 12:00:43 AM
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i havent used the asc format.
but look at the following model
6 points + 4 triangles
X-X-X
|/|/|
X-X-X
http://members.xoom.com/myBollux
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the faces are just indexes pointing onto the vertices/points:
Points: //always x,y,z
1,54,23,364,234,135,543,23,75,253,324,234 // our points
Faces: //always 3 for a face
0,1,2,0,1,3
at the end we have 2 faces:
first with vertices/points 0,1 and 2
([1,54,23],[364,234,135],[543,23,75])
and the second with 0,1, and 3
([1,54,23],[364,234,135],[253,324,234])
so you just have to load the points into an array
VERTEX vertices[verteccount];
and your faces into a facelist:
short int faces[facecount*3]; //always 3 points a face
then you easy can draw the faces like this:
glBegin(GL_TRIANGLES)
{
for(int i=0;i{
glVertex3fv(vertices[faces[i*3+0]]) //grab the index of the current vertex out of faces and send this vertice to glVertex()
glVertex3fv(vertices[faces[i*3+1]]) //for the second vertice of the face, too
glVertex3fv(vertices[faces[i*3+2]]) //and for the last one
}
}glEnd();
every normal 3d format is constructed like this, so i think the ASC, too
we wanna play, not watch the pictures
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I can''t thank you enough dave. I was very confused about what was going on in the ASC file, and now I see it clearly. Thanks Again!
-SkyFire
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× | 659 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-51 | latest | en | 0.892085 |
http://graphmatica.com/help/draw_tan.htm | 1,547,901,148,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583667907.49/warc/CC-MAIN-20190119115530-20190119141530-00399.warc.gz | 98,597,762 | 2,463 | DRAWING TANGENT LINES
At times you may be interested in knowing the slope of a curve at a given point. Graphmatica will provide this information both numerically and graphically about any curve on the screen with just a few clicks of the mouse.
To calculate the slope of a curve and draw the tangent line at a specific point, select Draw Tangent from the Calculus menu. If you have chosen to select the initial point and curve using the mouse, use the mouse or arrow keys to move the crosshairs to any point on a curve on the screen, then click or press enter to select it. The program will draw the tangent line and display the point selected and the slope on the status line as well as in the Printout window if it is on. The tangent line will be displayed only until you hide or delete the equation it belongs to, clear the screen, or draw another tangent line.
If you have chosen to show the Draw Tangent Line dialog box (also true by default), it will appear now if it is not already on screen. This dialog box allows you to choose the equation for the tangent line, plus adjust the x and y coordinates. You can use it to "straighten up" the coordinate produced using the mouse, or to enter an entirely different coordinate. When entering new coordinates manually, you may either:
1. Enter a value for the independent variable (usually x) only, and leave the other one blank.
2. Enter a value for the dependent variable (usually y), and optionally an estimate of the independent variable. The program will solve the equation for the exact independent variable value for you. When the function is not one-to-one, there may be more than one solution for a given value, so you should always include the independent variable estimate to indicate which solution you want to use.
You can only find the exact tangent line for differentiable (i.e. those which do not include the int() or abs() functions) Cartesian and polar equations. At this time, there is no efficient way to produce accurate results for other curves (for parametric equations you will get a rough approximation based on the slope between the two consecutive points nearest where you clicked).
In addition, I do not recommend drawing tangent lines when using logarithmic graph paper (since, of course, straight lines are not straight with a logarithmic scale).
The input method for drawing tangent lines is adjustable using the Tangent Line Options dialog box.
kSoft, Inc. ksoft@graphmatica.com Last updated: Sun 11 Jun 2017 | 521 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-04 | longest | en | 0.883496 |
https://br.pinterest.com/pin/47498971045651570/ | 1,501,119,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426693.21/warc/CC-MAIN-20170727002123-20170727022123-00623.warc.gz | 626,759,493 | 60,205 | Pinterest
# Explore A Ordem, Common Cores e muito mais!
<p>This video that counts to 100 is an interesting visual representation of the process. Play it, then play it again! When your students understand the routine and become familiar with the medium, this can help them recall the order while having a bit of fun.</p>
### Print and Go! St. Patrick's Day Math and Literacy (NO PREP)
Print and Go! St. Patrick's Day Math and Literacy - Save yourself some time and ink with these fun print and go sheets. Perfect for the month of March in kindergarten even if you don't "do" St. Paddy's Day. \$
### Ahoy, Matey! It's Number Time (to 20) (TEKS 1.2C)
*** This activity only goes to 20. This file folder matching activity includes 12 compare and order problems that support the following 1st grade Texas math standard: 1.2C Use objects, pictures, and expanded and standard forms to represent numbers up to 120.
Ejercitación con castillo numérico: 1+, 1-, 10+, 10-
Welcome to Math Playground. Play with numbers and give your brain a workout! Tons of fun math and logic games for kids!
Mr. Nussbaum game - Ice cream mixed numbers 3.3a
### Kansas Standards
Kansas Standards, Kansas State Standards, Kansas Education Standards, Kansas Common Core Standards, Kansas School Standards, Standards Kansa...
Stem and leaf plotter- shodor This site is ABSOLUTELY awesome!! It allows students the opportunity to practice using data in a stem-and-leaf plot to find the mean, median, and mode.
### Ahoy, Matey! It's Compare and Order Time! (TEKS 2.2D)
This file folder matching activity includes 12 compare and order problems that support the following 2nd grade Texas math standard: 2.2D Use place value to compare and order whole numbers up to 1,200 using comparative language, numbers, and symbols (<, >, or =).
### Compare and Order Whole Numbers Up to 1,200 (TEKS 2.2D)
Includes 24 cards that support the following 2nd grade Texas math standard: 2.2D Use place value to compare and order whole numbers up to 1,200 using comparative language, numbers, and symbols (<, >, or =). Also includes a gameboard, an answer
Internet4Classrooms - Helping Students, Teachers and Parents Use the Internet Effectively. Games for NJASK test prep. All grade levels
Weigh the Wangdoodles! Wangdoodles are residents of the planet Algebra-5. Help them test space snacks by weighing them!
Internet4Classrooms - Helping Students, Teachers and Parents Use the Internet Effectively
Integers 7 th Grade Math and Pre-AP Math. What is an Integer? An integer is any whole number or its opposite. Integers are part of a group of numbers.
Common Core Math Centers: Counting Cookies. Students place the cookies on the correct section of a tens frame. The perfect counting center for kindergarten!
This station targets the first grade standards about measuring weight using a balance scale, and then ordering objects from heaviest to lightest. | 695 | 2,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-30 | latest | en | 0.861257 |
https://wiki.alquds.edu/?query=A-law_algorithm | 1,675,393,316,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00685.warc.gz | 625,268,235 | 20,150 | A-law algorithm
Graph of μ-law and A-law algorithms
Plot of F(x) for A-Law for A = 87.6
An A-law algorithm is a standard companding algorithm, used in European 8-bit PCM digital communications systems to optimize, i.e. modify, the dynamic range of an analog signal for digitizing. It is one of two versions of the G.711 standard from ITU-T, the other version being the similar μ-law, used in North America and Japan.
For a given input ${\displaystyle x}$, the equation for A-law encoding is as follows:
${\displaystyle F(x)=\operatorname {sgn}(x){\begin{cases}{\dfrac {A|x|}{1+\ln(A)}},&|x|<{\dfrac {1}{A}},\\[1ex]{\dfrac {1+\ln(A|x|)}{1+\ln(A)}},&{\dfrac {1}{A}}\leq |x|\leq 1,\end{cases}}}$
where ${\displaystyle A}$ is the compression parameter. In Europe, ${\displaystyle A=87.6}$.
A-law expansion is given by the inverse function:
${\displaystyle F^{-1}(y)=\operatorname {sgn}(y){\begin{cases}{\dfrac {|y|(1+\ln(A))}{A}},&|y|<{\dfrac {1}{1+\ln(A)}},\\{\dfrac {e^{-1+|y|(1+\ln(A))}}{A}},&{\dfrac {1}{1+\ln(A)}}\leq |y|<1.\end{cases}}}$
The reason for this encoding is that the wide dynamic range of speech does not lend itself well to efficient linear digital encoding. A-law encoding effectively reduces the dynamic range of the signal, thereby increasing the coding efficiency and resulting in a signal-to-distortion ratio that is superior to that obtained by linear encoding for a given number of bits.
Comparison to μ-law
The μ-law algorithm provides a slightly larger dynamic range than the A-law at the cost of worse proportional distortion for small signals. By convention, A-law is used for an international connection if at least one country uses it. | 476 | 1,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-06 | longest | en | 0.820948 |
https://ww2.mathworks.cn/matlabcentral/answers/285997-plotslice-what-are-the-numbers-below-the-plots?s_tid=prof_contriblnk | 1,596,790,120,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737172.50/warc/CC-MAIN-20200807083754-20200807113754-00010.warc.gz | 552,162,454 | 24,922 | # plotSlice - what are the numbers below the plots?
6 views (last 30 days)
James Russell on 26 May 2016
Answered: jkr on 8 Nov 2017
There is a row of numbers presented in plotSlice, one beneath each variable. What are these numbers? How may they be interpreted?
Tom Lane on 24 Jun 2016
The prediction shown at the left of the plot is the value given by the model when the predictors are set to the numbers shown below the plot.
Also, suppose there are three plots with predictor names x1, x2, and x3. The first plot shows a curve giving the predicted response vs. x1, when x2 and x3 are set to the values below their plots.
jkr on 28 Jun 2016
Thanks. Good explanation of coefficients of a predictive model.
Matt Fetterman on 7 Nov 2017
Hello, Could you explain is the plot showing a probability or the value of the coefficients which could then be converted into a probability.
jkr on 8 Nov 2017
The coefficients in the model represent the impact of each variable on the model. They are not statistical measures. Moreover, they are arbitrary, in the sense that they are directly sensitive to scale. For example, in a model that included voltage as a predictor variable, if the model's coefficient were, say, 2.7 in volts, it would be 2700 in millivolts. For this reason, when as often the variables are in incommensurate units, it can be helpful to normalize each variable (e.g. (x - mean(x))/std(x)). | 346 | 1,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-34 | latest | en | 0.936693 |
https://cracku.in/118-mr-x-finishes-60-of-a-work-in-9-days-and-then-is-j-x-srcc-gbo-2023 | 1,722,873,768,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00861.warc.gz | 142,841,894 | 27,895 | Question 118
# Mr. X finishes 60% of a work in 9 days, and then is joined by Mr. Y and they together finish the next 10% work in 1 day. How long would it take if Y alone is asked to finish the next 5% of the work?
Solution
X can do 60% of the total work in 9 days, this gives us that X can complete the total work in 15 days.
X and Y can together complete 10% of the work in 1 day, this gives us that X and Y can complete the work in 10 days.
We can take the total work to be a multiple of the LCM of these values, let it be 30U.
Now X has an efficiency of 2U units/day.
And X and Y together have an efficiency of 3U units/day.
This gives us that the efficiency of Y to be 1U units/day.
In order for Y to 5% of the total work, that is 3/2 U units of work, Y would require, $$\frac{3}{2\ }=\ 1\ \frac{1}{2}$$ days.
Hence, Option A is the correct answer. | 261 | 858 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.966782 |
www.forexeconomist.com | 1,604,043,091,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00482.warc.gz | 723,502,502 | 8,573 | # Fibonacci Retracement
Fibonacci retracement…
One of the first tools taught everywhere to unsuspecting traders as they begin their journeys.
But what is it?
And just how useful is it, really?
Let’s find out.
## What’s a Fibonacci Retracement?
Leonardo Fibonacci was an Italian mathematician who lived, did his thing, and died in medieval times.
One of his coolest contributions was the Fibonacci sequence.
In this sequence, each number is the sum of the previous two numbers (with the exception of the first two numbers: 0, and 1).
Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, ….
Notice the second 1 is 0 + 1, while 2 = 1 + 1, and 3 = 2 + 1, and so on…
Why is this sequence important?
Turns out this sequence explains many phenomena in nature.
As an example, according to historians, this sequence of numbers determines how a population of rabbits grows.
A memorial statue for a great mathematician.
Even though the sequence packs a bunch of interesting properties, such as that of the golden ratio (we’ll probably touch on that in another page), we will focus on the property used for the Fibonacci retracement.
As you will, if you divide any number in the sequence by the next bigger one, you will get approximately 0.618 or 61.8%.
If you divide any number in the sequence by the second next bigger one, you get approximately 0.382 or 38.2%.
For the third next bigger number in the sequence, the division gets you 0.236 or 23.6%.
Ok, so what?
Well, some freak decided to use this cool-ass ratios in finance.
Does this look familiar? Fibonacci ratios have many applications.
When you expect a move to retrace -in other words, for a trend in price action to reverse- you can use these ratios ahead of time to get a better idea of where you can expect price to fall (or climb back) to.
Wait, what?
You’ll get a clearer picture down below, so read on.
With time, the use of Fibonacci ratios in trading became ubiquitous, which means a good proportion of traders out there started using this tool.
Why is that important?
Because when a good chunk of traders pay attention to the price zones given by the Fibonacci retracement, then those price zones suddenly become much more likely to affect price action -because many orders are placed in those zones.
I still don’t get it…
Ok, have you heard of a self-realizing prophecy?
As in, everyone tells you you’re good at something you didn’t know before.
You start acting like you are good at it, and engage in that activity more often.
Eventually you do become good at it…
It’s the same here.
Since many people placed orders to buy on a support zone signaled by a Fibonacci retracement, then price is much more likely to stall or rebound from that level if it falls to it.
All because a majority (or a good portion) of the market is paying attention to these levels, and acting on them.
What does the crystal ball say about you? It might be more important than you realize.
Great. But how do we use them?
Can I just jump into a chart and draw a random retracement?
No.
The Fibonacci retracement connects two points in your chart (two candlesticks) and calculates the retracement levels based on the vertical distance between those two points.
In order to do this, you have to choose those two points wisely.
Since you’re expecting to see a retracement, you should pick the lowest low as one extreme, and the highest high for the other extreme.
These are called swing low and swing high respectively.
This is important!
Later we will describe what swing trading is, and you will need to know how to identify swing lows and swing highs.
Either way, to use the Fibonacci retracement tool well, you will need to identify the recent swing low and swing high, and use them to get the Fibonacci levels.
Let’s head into an example in the following section.
## Fibonacci Retracement in Forex
So we now know what a Fibonacci retracement is: a forex technical analysis tool which allows us to see potential support or resistance zones for a retracement of price action.
Phew…. What a mouthful.
And you know you have to use swing lows and swing highs.
Well, a swing low is a bearish candlestick for which the previous two candles are higher, and the following two candles are also higher.
What?
It is pretty easy to spot once you’re looking at a chart, so don’t sweat it.
For a swing high… well… it is the reverse!
The highest point in recent price action (a bullish candle where the previous two candlesticks, and the following two candlesticks are lower).
A picture says more than a thousand words:
Notice the lowest and highest points?
There was a higher high later in the chart, but let’s assume you wanted to long GBP/USD before that point appeared, and you were waiting for a retracement.
A good moment to enter could have been on the 61.8% retracement zone you can see on the chart (second yellow line from bottom to top).
Ok, so now that we have them what do we do?
Let’s remember that any technical tool, Fibonacci retracement included, should be used within the context of your overall strategy.
And this overall strategy must absolutely be dictated by fundamentals.
Even if you’re scalping, you need to know what’s happening (or not happening) fundamentally with the currency pair you plan to trade before you start doing your thing.
With that out of the way, let’s assume your fundamental analysis tells you the Euro will gain against the American Dollar.
But you want to enter the trade at a good price, and you believe price will retrace in the short-term due to a momentary break for the USD.
If so, you need to check the latest swing high and swing low.
Start at the swing low and take the tool towards the swing high (I use MT4, this might be slightly different in another platform, but the concept is the same).
Voila!
You got your levels.
If price retraces, good levels to buy EUR/USD might be close to the 38.2%, 50%, or even the 61.8% level (quite a deep retracement, make sure the fundamentals for the long trade have not changed).
Also, to make this more potent, make sure those retracement levels are close to support/resistance zones price action has printed.
If those levels match, congratulations, you likely have some solid levels to buy off-of.
## How useful is the Fibonacci Retracement?
This depends on your strategy.
As I mentioned, using the Fibonacci retracement tool as part of a well-thought out strategy (which takes fundamentals very seriously) is incredibly useful.
And a great example of an application would be swing trading.
But you can use them in many other ways such as scalping, or even day trading the market sentiment.
## Closing
This was a long piece.
However, I think it was very useful.
Wishing you the best,
Emil Christopher,
The Forex Economist
Fill out my online form.
Online contact and registration forms from Wufoo.
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Read More | 1,677 | 7,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-45 | longest | en | 0.950042 |
http://www.jiskha.com/display.cgi?id=1299076045 | 1,498,265,383,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320206.44/warc/CC-MAIN-20170623235306-20170624015306-00187.warc.gz | 562,223,035 | 3,960 | # algebra
posted by .
How many solutions does the following system have:
6x - 3y = 9
2x - y = 3
Explain how to determine the number of solutions without solving the system. Then apply elimination, and interpret the resulting equation.
could you please explain this to me?
• algebra -
number of solutions:no more than the degree of the equation.
In the above, both x,y are degree one, so there is at most one real solution. THere maybe no solutions.
To solve, multiply the lower equation by 3, and you see that they are the same equation, which then has no solution, as the two lines are colinear (the same). It takes graphs that cross each other (intersect) to produce a solution. | 160 | 687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-26 | latest | en | 0.962154 |
https://brainmass.com/economics/utility-demand/consumer-theory-182559 | 1,480,802,023,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541140.30/warc/CC-MAIN-20161202170901-00303-ip-10-31-129-80.ec2.internal.warc.gz | 817,275,353 | 19,080 | Share
Explore BrainMass
# Consumer Theory
Please open the attached word file to view the mathematical notations.
Question
Consider a consumer with utility function
(a) Find the demand for any vector of strictly positive prices, p = (p1,p2)>>0 and any level of income, M>0,. Why does the consumer end up spending her entire income (i.e., choose a consumption point on the budget line)?
(b) Sketch some indifference curves of the utility function u and draw the price consumption curve (PCC), which results when varying the price of the first good. Does the first good have a normal price effect, and how does this show in the PCC?
(c) Is the following statement true or false: A normal good (a good with positive income effect) cannot be a Giffen good. Explain briefly why you arrive at your conclusion.
\$2.19 | 182 | 813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-50 | latest | en | 0.89897 |
http://web.mit.edu/kenta/www/three/prime/twin-primes.html | 1,642,939,426,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00337.warc.gz | 70,494,218 | 3,384 | # Twin Primes
New! See the expanded and extended list: twin-primes.txt
The three pairs of twin primes "greater than" and "less than" a power of 2 are given. For example, the six pairs of twin primes which surround 2^4=16 are {16-12 +/- 1}, {16-10 +/- 1}, {16-4 +/- 1}, {16+2 +/- 1}, {16+14 +/- 1}, {16+26 +/- 1} (that is, {3,5}, {5,7}, {11,13}, {17,19}, {29,31}, {41,43}).
```2^4 [-12,-10,-4,2,14,26]
2^5 [-20,-14,-2,10,28,40]
2^6 [-34,-22,-4,8,38,44]
2^7 [-56,-26,-20,10,22,52]
2^8 [-58,-28,-16,14,26,56]
2^9 [-92,-80,-50,10,58,88]
2^10 [-166,-142,-4,8,26,38]
2^11 [-98,-50,-20,34,40,64]
2^12 [-76,-46,-4,32,62,122]
2^13 [-242,-182,-104,28,40,100]
2^14 [-196,-154,-22,68,248,266]
2^15 [-206,-158,-50,34,64,142]
2^16 [-268,-88,-16,2,44,164]
2^17 [-422,-230,-62,40,178,376]
2^18 [-172,-94,-34,8,206,368]
2^19 [-620,-86,-68,64,100,124]
2^20 [-448,-358,-4,314,554,566]
2^21 [-242,-194,-20,106,136,298]
2^22 [-1054,-754,-502,278,296,374]
2^23 [-800,-650,-158,10,154,454]
2^24 [-316,-244,-76,74,116,464]
2^25 [-662,-464,-422,70,148,208]
2^26 [-826,-664,-196,458,1286,1604]
2^27 [-680,-326,-290,52,202,310]
2^28 [-676,-478,-448,122,632,806]
2^29 [-1082,-350,-74,106,178,808]
2^30 [-1042,-514,-106,8,146,848]
2^31 [-986,-830,-698,220,394,574]
2^32 [-2398,-1624,-1456,92,266,722]
2^33 [-620,-452,-302,298,988,1318]
2^34 [-832,-502,-376,248,1004,1076]
2^35 [-1910,-1190,-1070,322,334,490]
2^36 [-1384,-1006,-346,116,176,764]
2^37 [-1712,-1664,-200,568,976,1036]
2^38 [-2296,-1792,-316,1328,1748,2156]
2^39 [-746,-740,-530,412,562,1030]
2^40 [-1726,-1468,-1378,554,626,794]
2^41 [-1964,-1250,-74,988,1216,1786]
2^42 [-3256,-3142,-1366,326,1298,1448]
2^43 [-1916,-1430,-800,184,1690,1702]
2^44 [-4654,-2704,-118,2066,2864,3584]
2^45 [-1334,-752,-512,2566,2836,3508]
2^46 [-3232,-2434,-886,416,728,1184]
2^47 [-2540,-1820,-1628,880,1870,2164]
2^48 [-1048,-568,-526,242,764,4034]
2^49 [-2912,-2102,-1994,460,646,808]
2^50 [-2692,-2032,-652,644,1244,3158]
2^51 [-3638,-2930,-788,1210,1390,2062]
2^52 [-5128,-3574,-634,392,1046,2534]
2^53 [-4424,-3194,-2450,1156,1546,3706]
2^54 [-1426,-1342,-256,1478,2144,3464]
2^55 [-1586,-1316,-170,1450,2242,3130]
2^56 [-4678,-3448,-2128,176,962,1652]
2^57 [-6074,-3650,-1604,130,598,3538]
2^58 [-3232,-2044,-1966,1046,2234,4646]
2^59 [-2366,-1526,-1058,1042,1714,1954]
2^60 [-4066,-3244,-1504,2732,2876,3272]
2^61 [-2384,-2372,-1922,1918,3298,3430]
2^62 [-5884,-3256,-1702,188,386,1946]
2^63 [-6746,-4238,-1298,1534,4432,5800]
2^64 [-1024,-898,-844,806,2534,6986]
2^65 [-2834,-2444,-674,988,1456,1888]
2^66 [-3994,-1186,-796,986,3308,6728]
2^67 [-8846,-8318,-6830,154,1900,3502]
2^68 [-4354,-3886,-148,1466,3134,5516]
2^69 [-3674,-2252,-92,916,1180,1360]
2^70 [-12682,-7774,-2566,2516,2936,4016]
2^71 [-8318,-4988,-830,790,3064,3514]
2^72 [-2818,-2428,-844,1934,2636,3062]
2^73 [-15710,-10562,-4262,1888,3220,4648]
2^74 [-3052,-2002,-1054,518,1046,8534]
2^75 [-11138,-7418,-2888,592,2284,5470]
2^76 [-12046,-7144,-3856,2222,2276,2306]
2^77 [-11942,-5240,-1820,70,4360,11026]
2^78 [-3772,-1354,-1114,116,4958,8396]
2^79 [-6890,-4808,-4568,1174,3202,3622]
2^80 [-6616,-5224,-1288,1676,8672,11942]
2^81 [-10652,-6782,-3560,226,6820,7648]
2^82 [-3772,-3484,-364,44,2486,4016]
2^83 [-9638,-9536,-2690,1774,7390,7720]
2^84 [-7738,-7636,-214,2852,4052,6236]
2^85 [-23300,-20120,-4820,8458,9208,10678]
2^86 [-8554,-5644,-2524,656,4094,6236]
2^87 [-2528,-2498,-746,4252,6292,6670]
2^88 [-5968,-3736,-2764,2354,3614,5084]
2^89 [-22442,-17942,-5642,1846,2080,2968]
2^90 [-9586,-9244,-2362,10748,11084,13226]
2^91 [-7688,-7328,-956,3424,7240,7990]
2^92 [-13594,-10138,-1948,5246,9542,12776]
2^93 [-3842,-1994,-752,2788,4210,7540]
2^94 [-8194,-7342,-3166,908,1766,4028]
2^95 [-8726,-3590,-998,2050,7462,10774]
2^96 [-13018,-7186,-3394,3362,4034,5972]
2^97 [-12242,-10094,-2342,3838,7366,17158]
2^98 [-5704,-592,-526,3926,5018,12806]
2^99 [-9350,-8846,-1436,4204,6190,22462]
2^100 [-5446,-4324,-2974,5636,8834,10262]
2^101 [-12854,-3770,-1880,2290,3160,19888]
2^102 [-7246,-6634,-3532,8774,8864,13304]
2^103 [-14720,-9158,-2228,3670,5374,7222]
2^104 [-4108,-4006,-1576,3104,3416,20216]
2^105 [-28220,-27710,-18800,1390,5476,7186]
2^106 [-16006,-5824,-1612,3524,16994,20834]
2^107 [-16556,-4118,-3998,4732,8350,9730]
2^108 [-6724,-5974,-694,7802,8192,9122]
2^109 [-17612,-8330,-944,10948,12448,17176]
2^110 [-2932,-2524,-532,10916,23456,30464]
2^111 [-12458,-10490,-3548,220,634,8122]
2^112 [-19768,-2014,-1108,4994,11042,11546]
2^113 [-35060,-21830,-13574,24436,35416,35608]
2^114 [-8776,-3034,-2794,968,2978,46778]
2^115 [-9938,-7298,-5948,4282,6562,15844]
2^116 [-28534,-18328,-4516,1274,3662,12974]
2^117 [-9890,-8492,-5720,2860,7930,12898]
2^118 [-18844,-13624,-1624,4856,7556,8438]
2^119 [-3980,-3656,-800,1774,3004,11632]
2^120 [-14584,-9136,-5608,1856,2636,12854]
2^121 [-25352,-13034,-8642,3778,5140,12376]
2^122 [-10504,-9052,-4666,428,4646,7724]
2^123 [-40988,-38156,-24860,14230,16972,37012]
2^124 [-8866,-5026,-4804,992,7934,10442]
2^125 [-33410,-29072,-23750,9688,22540,23338]
2^126 [-20224,-7984,-7636,3188,6026,13094]
2^127 [-24128,-18410,-16490,994,8020,8164]
2^128 [-15208,-5506,-1408,2432,3302,25796]
``` | 2,630 | 5,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-05 | latest | en | 0.706825 |
http://livny.info/meter-cubed-to-megalitres-2020 | 1,582,858,212,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146940.95/warc/CC-MAIN-20200228012313-20200228042313-00340.warc.gz | 86,447,639 | 5,195 | Meter Cubed To Megalitres 2020 » livny.info
# Cubic meters to liters [m³ to l] conversion tables with.
A cubic metre per second m 3 s −1, m 3 /s, cumecs or cubic meter per second in American English is a derived SI unit of volumetric flow rate equal to that of a stere or cube with sides of one metre ~39.37 in in length exchanged or moving each second. It is popularly used for water flow, especially in rivers and streams, and fractions for. How to convert cubic meters to liters [m³ to l]:. V l = 1 000 × V m³. How many liters in a cubic meter: If V m³ = 1 then V l = 1 000 × 1 = 1 000 l. How many liters in 64 cubic meters: If V m³ = 64 then V l = 1 000 × 64 = 64 000 l. Note: Cubic meter is a metric unit of volume.Liter is a metric unit of volume.
A cubic metre per second m 3 s −1, m 3 /s, cumecs or cubic meter per second in American English is a derived SI unit of volumetric flow rate equal to that of a stere or cube with sides of one metre ~39.37 in in length exchanged or moving each second. Online calculator to convert liters to megaliters L to ML with formulas, examples, and tables. Our conversions provide a quick and easy way to convert between Volume units.
Derivatives of a cubic meter such as cubic centimeters are also used. Megalitres to Cubic Meters villa rosa pizza yardley coupon Kyle's Converter convert m3 s to megalitres day Electric motor pumping convert m3 s to megalitres day up batdorf and bronson gift card to 43 megalitres per day into Old Dam storage.10' dekaa da. Feb 03, 2014 · Converting megaliters to cubic meters only requires the use of one basic equation. Convert megaliters to cubic meters with help from a mathematics educator in this free video clip. Expert: Marija Kero.
Conversion between cubic meter and meter. Cubic meter to Meter Calculator. Volume Conversion table and factors: cubic meter. Quantity: Reference Unit: is equal to: Conversion Factor: Unit: 1. Cubic Meters. A metric unit of volume, commonly used in expressing concentrations of a chemical in a volume of air. One cubic meter equals 35.3 cubic feet or 1.3 cubic yards. One cubic meter also equals 1000 liters or one million cubic centimeters. Cubic Meters to Kiloliters formula.
## Convert M3 S To Megalitres Day.
Cubic Meter value will be converted automatically as you type. The decimals value is the number of digits to be calculated or rounded of the result of liter to cubic meter conversion. You can also check the liter to cubic meter conversion chart below, or go back to liter to cubic meter converter to top. Free online volume conversion. Convert m3 to megalitres cubic meter to Ml. How much is m3 to megalitres? > with much ♥ by CalculatePlus.
### Cubic meter Conversion Table - Volume - Online Unit Converter.
Diferent flow rate units conversion from cubic meter per day to milliliters per day. Between m3/d and mL/d measurements conversion chart page. Convert 1 m3/d into milliliter per day and cubic meters per day to mL/d. The other way around, how many milliliters per day - mL/d are in one cubic meter per day - m3/d unit? Calculate from flow rate into other flow rate unit measures. As this product is often used for billing purposes, it’s important that the meter is read accurately so cost is not over or under estimated. This article outlines the best way to read the capsule in order to ascertain the total water meter reading. The solution. The units of flow commonly used in water meters are read in M3 cubic meters. Mechanical meter recording in Megalitres ML Hundreds of Kilolitres Mechanical meter recording in cubic meters times one hundred m3 x 100 Meter Reading: 345.676 ML or 345 676.5 KL x 10 x 1 x 0.1 Unit of measure is Cubic meters times one hundred m3 x 100 where 1 m3 x 100 = 100 KL Increments of kilolitres 1 KL. | 953 | 3,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.780638 |
https://mathematica.stackexchange.com/questions/183076/mapping-over-a-piecewise-function/183077 | 1,582,010,058,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143635.54/warc/CC-MAIN-20200218055414-20200218085414-00105.warc.gz | 472,413,879 | 31,945 | Mapping over a piecewise function
This is a followup to a question I posted earlier today. Suppose you have a piecewise function such as "snap=max[x,0]= x if x>0, 0 otherwise".
Then you would think that it must be straightworward to apply a function to this, i.e. to obtain "f[x] if x > 0, f[0] otherwise".
My intuition is that I should be able to say "f[snap]", but this doesn't work. The best I can do (after help from AccidentalFourierTransform) is quite complicated:
snap = PiecewiseExpand[Max[0, x]]
MapAt[f[#] &, Map[MapAt[f[#] &, 1], snap, {2}], -1]
This produces the desired result, but it seems strange that I need to do this. Is there a more straight-forward solution?
pwMap1[f_] := MapAt[f, #, {{1, All, 1}, {2}}] &;
pwMap1[f]@snap // TeXForm
$$\begin{cases} f(x) & x>0 \\ f(0) & \text{True} \end{cases}$$
Also:
pwMap2[f_] := InternalToPiecewise[#, f /@ #2] & @@ InternalFromPiecewise[#] &;
pwMap2[f]@snap == pwMap1[f]@snap
True
• Beautiful! Thanks, I've learned a lot from this! – Christoph Oct 4 '18 at 13:33
• @Christoph, thank you for the accept. – kglr Oct 4 '18 at 13:34 | 358 | 1,102 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-10 | latest | en | 0.826745 |
https://www.kast-opmaat.nl/2017_1229/31261_coarse-stone-per-tonne.html | 1,586,077,387,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371576284.74/warc/CC-MAIN-20200405084121-20200405114121-00399.warc.gz | 961,099,020 | 10,792 | Email:[email protected]
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### Calculate #57 Gravel Stone | cubic yards / Tons
Type in inches and feet of your project and calculate the estimated amount of Gravel in cubic yards, cubic feet and Tons, that your need for your project. The Density of #57 Gravel Stone…
### Specifications For CRUSHED AGGREGATE BASE COARSE #2 & …
2. Crushed aggregate base coarse (Compaction Gravel) shall be purchased per standard ton with a standard ton to weight 2,000 pounds for the purpose of this contract. Material shall be ordered in full truck load lots. 3. Contractor shall proceed with delivery of crushed aggregate base coarse (Compaction Gravel) upon 12 hour notice. 4.
### Rip Rap Cost: Your Complete Pricing Guide
Secondly, what you may think you’re saving per ton will be made up for (and then some) in transportation and shipping costs. It’s best to stay regional with your purchases. Size – In most cases, rip rap and other stone materials are sold based upon their weight. It’s only natural, then, that the larger, heavier and denser the rip rap ...
### Crushed Stone: Determing Yards Per Ton - braenstone.com
Crushed stone is quoted at a weight of 2700 pounds per cubic yard. Your stone dealer tells you he has a truck that can deliver 20 tons of stone per load. You need to know who many cubic yards that comes out to. Here is the math: 20 tons X 2000 pounds per ton = 40,000 pounds 40,000 pounds / 2700 pounds per cubic yard = 14.29 cubic yards
### conversion of coarse aggregate cubic metre to metric ton ...
Appendix A: Overview of Portland Cement and Concrete (PDF) artificial rock-like material, basically a mixture of coarse aggregate (gravel or … world total annual production of hydraulic cement is about 2 billion metric tons ….. or raw meal, is prepared, it is fed into a cement kiln and converted into the …. production in the United States was approximately 349 million cubic meters in 2005.
### Building Stone | Walling Stone | Reclaimed Stone Supplies
Price: £85 per square yard. Lovely example of reclaimed Yorkshire Stone – palletised and ready for delivery. Stone matching service available. Coursed stone and backed off stone. Call or email Richard on 01254 233227 / [email protected] with your specific size requirement for our most competitive rates.
### Construction Aggregate Calculator - Vulcan Materials Company
Use this calculator to estimate the quantity of Construction Aggregates needed for your particular job, based on width, length, thickness, and product density.
### Aggregate | Quality Construction Materials | Pennsy Supply
Pennsy Supply is the largest producer of aggregate in central Pennsylvania, producing more than nine million tons every year consisting of primarily crushed stone, sand and gravel. ... It is a mixture of coarse stone and crushed stone dust. Used as a road base, under slab , driveway or walking paths.
### 2019 Pea Gravel Cost | Average Pea Gravel Price Per Ton
How much does pea gravel cost? Learn about the average pea gravel price per ton and other costs associated with installing pea gravel as researched by HomeAdvisor. ... but it’s generally in the range of \$4 to \$6 per bag for plain or neutral pea gravel. More vibrantly colored stone generally comes with a price tag of up to \$8 per bag.
### Coarse L (River Rock) - Northside Mulch & Landscape Supply
53 Stone Crushed limestone that’s dusty and good for compaction that can be used for driveways, bases for pavers, walkways, and patios . 3/8 inch max size per stone Price per Cu Yard
### How Much Does Gravel Cost Per Ton - Gravel Price
Generally gravel is purchased by the ton. A ton is 2,000 pounds. The price per ton of gravel will depend on the type of gravel being purchased. Decorative gravels will be more expensive than the simpler forms. The common white or greyish gravel will average around \$28 per ton whereas the more decorative gravel will average between \$38 and \$72 ...
### Construction Coarse Aggregates - Cardigan Sand and Gravel ...
Construction Coarse Aggregates. ... Imported Stone Crusher Run (3'') £22.20 per tonne (inc. VAT) This item is currently out of stock once it is back in... Imported Stone Scalpings (1½" scalpings) £22.20 per tonne (inc. VAT) A screened product through oblong mesh to provide a general fill with...
### Auburn Aggregates Pricing & Calculator: Estimate Your Costs!
Auburn Aggregates is transparent about product pricing, delivery fees, and volume discounts. Use our online calculator to estimate your total cost! ... \$2.00 /per ton Fine Stone Dust; \$5.20 /per ton Pure Stone Dust (Clean) \$7.30 /per ton Washed Manufactured Sand; \$7.30 /per ton Underdrain Sand;
### In 1 CFT, how many kg of coarse aggregate is there? - Quora
Nov 04, 2018· It’s depends upon Density of aggregate Density of the aggregate = 1520 to 1680 kg/m^3 & 1m^3 = 35.3147 CFT therefore, Quantity of Aggregates in 1CFT= (1520/35.3147) to (1680/35.3147) => 43.041 kg to 47.572 kg Ans Hope it helps Visit my website:-we...
### Gravel Calculator - how much gravel you need in tons ...
Price per ton \$ Calculate. Calculation results. Gravel volume: 0.444 cubic yards (12 ft 3) Gravel weight: ... Crushed stone is often used as a material for producing concrete. It is also used as a road base, and in some locations it is used for paving roads, for example Russia has over 400,000 km (250,000 mi) of gravel roads. ... Coarse gravel ...
### Products-Coarse-Aggregate - Capitol Aggregates
Coarse Aggregate. Crushed stone is created through the mining and crushing of rock. The rock is crushed into several different graded sizes for use in applications including ready mix concrete, asphalt and decorative stone. Gravel is created by river erosion, resulting in smooth and rounded products.
### Bulk Material Calculator | Contractors Stone Supply
bulk material calculator (sand, gravel, soil and mulch) ... 6” chopped stone will cover 70-75 linear feet per ton 8” chopped stone will cover 45-50 linear feet per ton 10 chopped stone will cover 35-40 linear feet per ton drystack 4x8” will cover 45-50 linear feet per ton
### Cost of Gravel - Estimates and Prices Paid - CostHelper.com
JMJ Landscape Supply Center in Lynchburg, VA charges \$65 a ton for 1/4"-1/2" brown pea stone and \$99.50 a ton for 1/2" earth rose crushed stone; delivery within a 10-mile radius is \$55 per trip. A cubic yard of gravel (weighing roughly 1.2 to 1.7 tons) costs about \$10-\$90 or more if hauled away by the customer, or \$20-\$120 or more
### Gravel and Aggregate Calculator | How much gravel do I need
To use the calculator, select your stone or sand aggregate type from the pull down menu, and enter length, width, and depth requirements in the blanks provided. This calculator will estimate how much stone and/or sand (in tons) that are required for your job …
### Aggregate Volume Calculator | Sand, Stone, Gravel ...
RT @_OSSGA: To build this infrastructure, we need high-quality stone, sand & gravel, sourced from close-to-market pits and quarries across… 68d. Check our newest blog article on the benefits of using our easy to use online tool, the aggregates calculator.…
### How many yards in 1 ton of gravel - answers.com
How many cubic yards in a ton depends on what your material is. For instance, topsoil has a 1:1 ratio. 1 cubic yard of topsoil would be 1 ton, so if you have topsoil, 17 cubic yards would be 17 tons.
### Now just how much are gravel prices per ton? - Driveway Wise
How much are gravel prices per ton????? Irritatingly, it just seems that the price of gravel will just have to be calculated on the basis of 850kgs! But, some aggregate suppliers will give you a price per ton. Just make sure it is a ton. Click below for some options on buying a ton. Stonewarehouse
### Crushed Stone | CR-6 | #57 stone | Rip Rap | Stone Dust ...
The products shown below are available at The Stone Store, a division on Aggtrans, in Hanover, Maryland. Products are available in bulk and bagged in three sizes; 2000#, 250#, 50#. If you require larger quantities, such as: full truck loads (22 Tons) or more, products will be direct shipped.
### Building Stone | Walling Stone | Reclaimed Stone Supplies
Building and Walling Stone We supply Building Stone and Walling Stone for all types of projects. We deal in building stone of all types either suitable for a small garden wall through to the construction of your dream home in the country.
### Calculate 3/4" White River Gravel | cubic yards / Tons
Type in inches and feet of your project and calculate the estimated amount of River Rocks in cubic yards, cubic feet and Tons, that your need for your project. The Density of 3/4" White River Gravel: 2,410 lb/yd³ or 1.21 t/yd³ or 0.8 yd³/t
### AASHTO - # 3 | Laurel Aggregates
This is an excerpt of the official AASHTO Gradations page. The numbers shown are representative of percentages of a sample of stone which will pass through an opening the size of …
### 2019 Crushed Stone Prices - ImproveNet
Crushed stone prices range from \$27.50 to \$64.20 per ton, not including delivery. ... \$27.50 per ton of basic landscaping crushed stone when 23 or more tons are purchased at one time. Costs increase per ton when fewer tons are ordered. ... More often than not, you have to excavate the land, add some coarse sand or landscape gravel and then ...
### PRICE LIST - winstoneaggregates.co.nz
PRODUCT CODE LIST \$ per tonne LIST \$ per m3 CONVERSION General All Passing 10 GAP10 25.00 38.50 1.54 General All Passing 20 GAP20 22.46 37.50 1.67 General All Passing 40 GAP40 21.34 35.00 1.64 General All Passing 65 GAP65 21.43 33.00 1.54 TNZ AP 40 M/4 TNZ40 27.61 45.00 1.63
### Cubic Meters to Metric Ton Using Gravel - OnlineConversion ...
Gravel, dry 1/4 to 2 inch - 1.68 tonne/cubic meter Gravel, wet 1/4 to 2 inch - 2.00 tonne/cubic meter Concrete, Gravel - 2.40 tonne/cubic meter Choose the above density that most closely matches your gravel, and calculate using the formulas below. cubic meters * density = tonnes tonnes / density = cubic meters Thats the metric tonne.
### How many tonnes are in 1 cubic meter of stone aggregate ...
May 21, 2018· Only if it is pure water, 1 cubic meter weighs 1 ton. If you don't mean pure water, you must know the density of.Weight Conversion Tables - Littler Bulk HaulageDry sand coarse 1.6 tonnes per cubic metre. Topsoil (some moisture) 1.44 tonnes per cub...
### Calculating gravel tonnage, cubes, and sand and earth.
Real life stone, gravel and sand densities (weight / volume) vary widely: ... The densities (pounds per square foot, or tons per cubic yard, sometimes called "pounds per foot" or "tons per cube") will also vary according to the season, the saturation levels (dampness) of the piles of medium when you purchase the mediums, the quality of the ...
### 2019 River Rock Prices | Landscaping Stone Costs Per Ton ...
HomeAdvisor's Landscaping Rock Cost Guide provides bulk prices per ton, yard, cubic yard, pallet and truckload for river rocks, riprap, large decorative stones, lava rock, bull rock, riverstone and landscape boulders. Estimate costs to install. | 2,726 | 11,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.883939 |
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# (Solved): d(d)/(S)ubtract the following Polynomial Expressions (x^(3)+2x-10)+(4x^(2)+11x-32) ...
d(d)/(S)ubtract the following Polynomial Expressions (x^(3)+2x-10)+(4x^(2)+11x-32)
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# Revolving Door Power Generator
Generate electricity by a revolving door
(+6, -3) [vote for, against]
A revolving door in a busy place usually spins for hours every day. Much of the energy used to turn the door is wasted. A generator at the door could convert the mechanical energy from the spinning door into electrical energy. It may be a little harder to turn the door, but people will burn more calories. Little kids running in circles in a revolving door will actually be generating lots of energy. Most of the subways in New York City have revolving doors that are used continuously all day. Think of all the energy that could be harnessed from the doors.
— kewldude471, Jul 24 2005
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This would just make it harder to turn the door, so that weak and lazy people would think it was stuck, and would use the 'standard' door instead. Besides, how much energy could the door really produce, even over a day? I suppose you could have a lightbulb over the door that stays alight when people go through.
— dbmag9, Jul 24 2005
//how much energy could the door really produce//
Easy. Suppose that the door needs a hefty push (say, 100N, equivalent to lifting 10kg) over a distance of 1 metre. Then, the energy expended in pushing the door is 100Nm, or 100 Joules.
Now make a guess as to the efficiency with which the energy can be recovered - say 50% (which is very very optimistic). Therefore, 50J of energy is available. This will light a 50W lightbulb for 1 second, or about a third of the time it takes a person to go through.
Now to find the total daily energy yield. This depends on how often the door is used, but let's be generous and assume the door is in continuous rotation at all times. Of course, it might have several people in it sometimes (and they could, collectively, push with more force, if the mechanism were smart enough to increase its resistance), but on the other hand in reality there will be quiet times, so a continuous one-person push seems reasonable. In this case, the average power output of the door over the whole day is about 17 Watts, or about 1.5MJ per day.
I have no idea of the energy involved in building and installing this mechanism. I would imagine that it would be at least a 100MJ (energy to make and press the steel components, to refine the copper for the wires.....). If the guy who delivers and fits it it has to drive for 1hr in total, in a 200hp vehicle, that adds another 540MJ to the energy bill. So, energy-wise, the unit will pay for itself in just over 1 year (assuming no energy-expensive maintenance call- outs), after which you'll be able to run your 17 watt lightbulb whenever the heck you please :-)
— Basepair, Jul 24 2005
would be very useful, but should only be used on revolving doors that are heavily used. anywhere else it won't generate enough power for its installation to be practical.
— FireElf, Jun 11 2006
Aren't revolving doors resistive enough because of the friction from the sealing around the edges for insulating purposes?
— BJS, Jun 11 2006
Problem with any animal-powered generation is that it is a terrible use of a complex energy-consuming "machine" to produce a small amount of energy. How many lightbulbs will be powered by the first lawyer whose client is injured from one of these increased-friction doors? If you really want to power up a few lightbulbs with people exerting the energy, at least go to cooperative participants -- try a fitness club (think treadmills and weight machines).
— thekohser, Jun 12 2006 | 892 | 3,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-30 | latest | en | 0.964997 |
https://bestftxtqnt.web.app/guilbeault84365kit/future-value-of-series-of-payments-excel-zaf.html | 1,642,805,386,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00462.warc.gz | 183,984,243 | 6,779 | ## Future value of series of payments excel
Excel function. Present value. =PV(rate, nper, pmt, fv). Number of periods. = NPER(rate, pmt, pv, fv). Rate of return. =RATE(nper, pmt, pv, fv). Periodic payment. An annuity is a series of equal payments or receipts that PV is the current worth of a future sum of money or stream of present value of cash outflows. Find the Future value at the end of year 5 of Stream A. All payments occur at Please note that there is no build-in function in Excel to calculate Future value of.
Using a block function to find the present worth or internal rate of return for a table of function, PMT, which means payment, but is what we refer to as the Annual Value. There are two Excel functions that work on a series or block of values. Excel function. Present value. =PV(rate, nper, pmt, fv). Number of periods. = NPER(rate, pmt, pv, fv). Rate of return. =RATE(nper, pmt, pv, fv). Periodic payment. An annuity is a series of equal payments or receipts that PV is the current worth of a future sum of money or stream of present value of cash outflows. Find the Future value at the end of year 5 of Stream A. All payments occur at Please note that there is no build-in function in Excel to calculate Future value of.
## Excel (and other spreadsheet programs) is the greatest financial calculator ever made. Solve for annuity payment, PMT, PMT(rate,nper,pv,fv,type) you are dealing with uneven cash flows and there are sign changes in the cash flow stream.
The FV Function is categorized under Excel Financial functions. investments such as certificates of deposit or fixed rate annuities with low interest rates. Microsoft Excel. In the previous section we looked at using the basic time value of money functions to calculate present and future value of annuities (even cash Excel (and other spreadsheet programs) is the greatest financial calculator ever made. Solve for annuity payment, PMT, PMT(rate,nper,pv,fv,type) you are dealing with uneven cash flows and there are sign changes in the cash flow stream. pv – is the present value, or the total amount that a series of future payments is worth now. type – is the number 0 or 1 and indicates when payments are due. If type How to Calculate Future Value Using Excel or a Financial Calculator. posted on 06-07-2019. Future value is one of the most important concepts in finance. Future value is the value of an asset at a specific date. It measures the nominal future sum of For example, when accounting for annuities (annual payments), there is no simple PV to plug into the equation. Either the PV must be calculated Most loans and many investments are annuities, which are payments made at argument would be 10 times 12, or 120 periods. pv is the present value of the
### If you change B9 to 1,000 then the present value (still at a 10% interest rate) will change to \$1,375.72. Reset the interest rate to 12% and B9 to 500 before continuing. Example 3.1 — Future Value of Uneven Cash Flows. Now suppose that we wanted to find the future value of these cash flows instead of the present value.
Annuities. An annuity is a series of payments made at equal intervals. There are Present and future values of annuities. • Perpetuities and Alternatively, we can use the Excel function RATE to calculate the rate of inter- est that equates the A lump sum is a complete payment consisting of a single sum of money, as opposed to a series of payments made over time (such as an annuity). Formula. The Therefore, Equation 1-3 can determine the future value of uniform series of Note that n is the number of time periods that equal series of payments occur. To find the FV of multiple cash flows, sum the FV of each cash flow. Learning Objectives. Calculate the Future Value of Multiple Annuities. Key Takeaways. Key Sep 19, 2007 which I use to calculate the Future Value of a series of future payments that increase at a fixed annual rate and earn interest at a fixed rate. Here it
### The Excel FV function is a financial function that returns the future value of an investment. You can use the FV function to get the future value of an investment assuming periodic, constant payments with a constant interest rate.
An annuity is a series of equal payments or receipts that PV is the current worth of a future sum of money or stream of present value of cash outflows. Find the Future value at the end of year 5 of Stream A. All payments occur at Please note that there is no build-in function in Excel to calculate Future value of. Jul 10, 2019 Learn how to use the Excel NPV function to calculate net present value of a series of cash flows, build your own NPV calculator in Excel and Annuities. An annuity is a series of payments made at equal intervals. There are Present and future values of annuities. • Perpetuities and Alternatively, we can use the Excel function RATE to calculate the rate of inter- est that equates the A lump sum is a complete payment consisting of a single sum of money, as opposed to a series of payments made over time (such as an annuity). Formula. The Therefore, Equation 1-3 can determine the future value of uniform series of Note that n is the number of time periods that equal series of payments occur.
## Calculates a table of the future value and interest of periodic payments.
Therefore, Equation 1-3 can determine the future value of uniform series of Note that n is the number of time periods that equal series of payments occur. To find the FV of multiple cash flows, sum the FV of each cash flow. Learning Objectives. Calculate the Future Value of Multiple Annuities. Key Takeaways. Key Sep 19, 2007 which I use to calculate the Future Value of a series of future payments that increase at a fixed annual rate and earn interest at a fixed rate. Here it nper: total no. of payment period. pmt: amount paid each period. pv - [optional] The present value of future payments must be entered as a negative number. The Microsoft Excel FV function returns the future value of an investment based on an interest rate and a constant payment schedule. The FV function is a built-in I.e. the future value of the investment (rounded to 2 decimal places) is \$12,047.32. Future Value of a Series of Cash Flows (An Annuity) If you want to calculate the future value of an annuity (a series of periodic constant cash flows that earn a fixed interest rate over a specified number of periods), this can be done using the Excel FV function.
MY REQUEST: Trying to solve for interest rate (to debate yay or nay on an annuity) if I need to pay \$234,000 for a five year / 60 month fixed term annuity that will pay out \$4,000 per month over 60 months (i.e. the future value = \$240,000). How can I solve for interest rate (?) Payments made at end of each month after inception. Calculates the future value for a series of constant payments (such as a payroll deduction for a 401K plan), assuming a constant interest rate. For example, you're putting \$500 away for retirement every month for 10 years, with an expected average return of 5% paid monthly. If n is the number of cash flows in the list of values, the formula for NPV is: NPV is similar to the PV function (present value). The primary difference between PV and NPV is that PV allows cash flows to begin either at the end or at the beginning of the period. Unlike the variable NPV cash flow values, Future value is one of the most important concepts in finance. Luckily, once you learn a few tricks, you can calculate it easily using Microsoft Excel or a financial calculator. Let's look at an example to illustrate the process. Future Value of Periodic Payments Calculator. This calculator will show you how much interest. you will earn over a given period of time; at any given interest rate; based on an initial. investment plus a fixed monthly addition. The calculator compounds monthly and assumes. deposits are made at the beginning of each month. Future value is the value of a sum of cash to be paid on a specific date in the future. An ordinary annuity is a series of payments made at the end of each period in the series. Therefore, the formula for the future value of an ordinary annuity refers to the value on a specific future date of a series of periodic payments, where each payment is made at the end of a period. | 1,886 | 8,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.906773 |
https://discourse.julialang.org/t/how-to-get-the-cartesian-indexes-of-the-k-largest-elements-of-a-matrix-2d-array-efficiently/69419 | 1,652,781,721,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00220.warc.gz | 274,043,080 | 7,271 | # How to get the cartesian indexes of the k largest elements of a matrix (2D array) efficiently?
Hi! I would like to find a way to get the cartesian indexes of the `k` largest elements in a matrix (2D array). Do you have suggestions of a computationally efficient method to achieve that?
Example: for `k=3` with the matrix `A` below, the cartesian indexes would be `[(1, 3), (2, 2), (2, 3)]`.
``````julia> A = [1 1 3; 1 3 3]
2×3 Matrix{Int64}:
1 1 3
1 3 3
``````
Thanks!
What Is the result expected for `k=2`?
I guess either ties are broken arbitrarily, or the first element is returned (I am interested in both solutions). By first element, I mean the first reached when going through the linear indexing of the matrix `A` (from 1 to 6 in this example).
Maybe something like
``````c = Vector{CartesianIndex{2}}(undef,k)
for i in lastindex(M):lastindex(M)-k
c[i] = CartesianIndex...
end
``````
(Sorry, on the cell phone…)
I am unsure I understood, but this made me think of 1) flatten the array, 2) sortperm to get the largest elements and 3) retrieve the cartesian indexes from the `k` last elements? This would give something like:
``````function klargest_indexes(m, k)
ci = CartesianIndices(size(m))
p = sortperm(vec(m))[end-k+1:end]
ci[p]
end
k = 2
m = [1 2 3; 4 5 5; 5 4 3]
out = klargest_indexes(m, k)
``````
In such a case, ties yield the last elements in the linear indexing. Is this what you hinted? Is it computationally efficient? I get a `@btime` of `5.374 μs (6 allocations: 3.55 KiB)` for `m = rand(20, 20)`, `k = 2`.
I meant probably something like this:
``````julia> ci = CartesianIndices(size(A))
for i in lastindex(A)-2:lastindex(A)
@show ci[i]
end
ci[i] = CartesianIndex(2, 2)
ci[i] = CartesianIndex(1, 3)
ci[i] = CartesianIndex(2, 3)
``````
(you can by default run over a matrix as if it was a linearly indexed)
``````sortperm(vec(m), 1:k; rev=true)
``````
I think you mean `partialsortperm(vec(m), 1:k; rev=true)`
4 Likes
Thanks! This is indeed more efficient!
Am I missing something, or this is much more efficient?
``````julia> A = [1 1 3; 1 3 3]
2×3 Matrix{Int64}:
1 1 3
1 3 3
julia> function getci(M,k)
ci = CartesianIndices(size(M))
cis = Vector{CartesianIndex{2}}(undef,k)
j = 0
for i in lastindex(M)-k+1:lastindex(M)
j += 1
cis[j] = ci[i]
end
return cis
end
getci (generic function with 1 method)
julia> @btime getci(\$A,3)
32.426 ns (1 allocation: 128 bytes)
3-element Vector{CartesianIndex{2}}:
CartesianIndex(2, 2)
CartesianIndex(1, 3)
CartesianIndex(2, 3)
``````
vs
``````julia> function klargest_indexes(m, k)
ci = CartesianIndices(size(m))
p = partialsortperm(vec(m), 1:k; rev=true)
ci[p]
end
klargest_indexes (generic function with 1 method)
julia> @btime klargest_indexes(\$A,3)
329.935 ns (6 allocations: 384 bytes)
3-element Vector{CartesianIndex{2}}:
CartesianIndex(2, 2)
CartesianIndex(1, 3)
CartesianIndex(2, 3)
``````
I think the sorting is lacking in the proposed `getci` function. Correct me if I am wrong but I believe it to yield the cartesian indexes of the `k` last elements rather than the `k` largest elements.
1 Like
Ah, I see. In the example you posted those are the same, and I misunderstood your question. Thanks.
(I think you can do that without any sorting, just with an auxiliary array of size k, and running once over the elements of the matrix storing the indexes of the k larger)
2 Likes
There you go:
``````julia> function getci!(M,k,cmins,vmins) # inplace
ci = CartesianIndices(size(M))
for i in 1:k
cmins[i] = ci[i]
vmins[i] = M[i]
end
imin = findmin(vmins)[2]
for i in firstindex(M)+k:lastindex(M)
if M[i] > vmins[imin]
cmins[imin] = ci[i]
vmins[imin] = M[i]
imin = findmin(vmins)[2]
end
end
return cmins, vmins
end
function getci(M,k) # allocating
cmins = Vector{CartesianIndex{2}}(undef,k)
vmins = Vector{eltype(M)}(undef,k)
return getci!(M,k,cmins,vmins)
end
getci (generic function with 1 method)
julia> A = [3 1 1
1 3 3]
2×3 Matrix{Int64}:
3 1 1
1 3 3
julia> @btime getci(\$A,3)
78.857 ns (2 allocations: 240 bytes)
(CartesianIndex{2}[CartesianIndex(1, 1), CartesianIndex(2, 2), CartesianIndex(2, 3)], [3, 3, 3])
julia> cmins=Vector{CartesianIndex{2}}(undef,3); vmins=Vector{Int}(undef,3);
julia> @btime getci!(\$A,3,\$cmins,\$vmins)
35.864 ns (0 allocations: 0 bytes)
(CartesianIndex{2}[CartesianIndex(1, 1), CartesianIndex(2, 2), CartesianIndex(2, 3)], [3, 3, 3])
``````
(since here there is an auxiliary array with the associated values, why not return it)
(for `rand(20,20)` and `k=2`, an example you mentioned this is 10 times faster than the alternative:
``````julia> A = rand(20,20); k = 2;
julia> @btime klargest_indexes(\$A,2)
5.540 μs (8 allocations: 3.56 KiB)
2-element Vector{CartesianIndex{2}}:
CartesianIndex(14, 8)
CartesianIndex(10, 10)
julia> @btime getci(\$A,2)
549.492 ns (2 allocations: 208 bytes)
(CartesianIndex{2}[CartesianIndex(10, 10), CartesianIndex(14, 8)], [0.9941985771967297, 0.9992607929792117])
julia> cmins=Vector{CartesianIndex{2}}(undef,k); vmins=Vector{eltype(A)}(undef,k);
julia> @btime getci!(\$A,2,\$cmins,\$vmins);
509.870 ns (0 allocations: 0 bytes)
``````
2 Likes | 1,730 | 5,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-21 | latest | en | 0.868258 |
https://acsdatacommunity.prb.org/discussion-forum/f/forum/877/how-to-interpret-acs-5-year-estimates/3697 | 1,709,638,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00040.warc.gz | 78,481,080 | 22,296 | # How to Interpret ACS 5-Year Estimates
I am not able to find a clear description of how to interpret 5-Year Estimates for ACS data. In particular, I am not clear on what the values in question, in my case, population data (e.g., Population 5 to 9 years and 5 to 9 year olds enrolled in school) represent. I am seeing conflicting information on whether the value (in this case, the population number) is an average over the 60 months, or what exactly it represents.
For example, in the document "Interpretation and Use of American Community Survey Multiyear Estimates" (https://www.census.gov/content/dam/Census/library/working-papers/2012/adrm/rrs2012-03.pdf) , it states, "An important property of ACS...estimates is that they do not represent a single point in time but an average [emphasis added] of the characteristics of a geography over a one-year, three-year, or five-year period, so they are referred to as period estimates. Data collected during the 60 months of five calendar years are combined together to produce estimates for the same levels of geography as did the Census 2000 long form." So, this makes it sound like it is an average.
Meanwhile, in the document "A Compass for Understanding and Using American Community Survey Data: What Researchers Need to Know " (https://www.census.gov/content/dam/Census/library/publications/2009/acs/ACSResearch.pdf), it states, "While one may think of these estimates as representing average characteristics over a single calendar year or multiple calendar years, it must be remembered that the 1-year estimates are not calculated as an average of 12 monthly values and the multiyear estimates are not calculated as the average of either 36 or 60 monthly values. Nor are the multiyear estimates calculated as the average of 3 or 5 single-year estimates [emphasis added]. Rather, the ACS collects survey information continuously nearly every day of the year and then aggregates the results [emphasis added] over a specific time period—1 year, 3 years, or 5 years. The data collection is spread evenly across the entire period represented so as not to over-represent any particular month or year within the period." This is pretty clear that it is not an average, but is an aggregate (whatever that means).
The most recent version (2020) of the document "Understanding and Using American Community Survey Data: What All Data Users Need to Know" (https://www.census.gov/content/dam/Census/library/publications/2020/acs/acs_general_handbook_2020.pdf), only says that, "While an ACS 1-year estimate includes information collected over a 12-month period, an ACS 5-year estimate includes data collected over a 60-month period.", but does not give any more detail on what that means.
Any clarification on how to interpret and write up the 5-Year estimate data would be appreciated.
Parents
• My interpretation of the literature provided is they don't take the average of monthly percents for single year or multiyear estimates, nor do they take the average of yearly percents for multiyear estimates. Doing so would give a different (and incorrect) result than tallying (aggregating) all responses in the positive for a variable over the entire period and determining what percent that is of the total.
I provide an example (Building Permits Survey data, not ACS) in the table below that shows housing units permitted in Raleigh by year during the 5-year period 2018-2022. Averaging the respective yearly percents gives the incorrect answer (27.7) if one wants to know the percentage of all permitted units in the 5 year period that were allowed to be single-family units. The correct answer (25.3) comes from dividing the aggregate (sum) of single-family units in the period by the aggregate of total housing units in the period, and converting that to a percentage by multiplying by 100.
Place Year Total housing units Single-family housing units Percent of units that are single-family units Raleigh 2018 4,211 1,304 31.0 Raleigh 2019 1,207 380 31.5 Raleigh 2020 3,479 1,162 33.4 Raleigh 2021 6,487 1,354 20.9 Raleigh 2022 8,635 1,875 21.7 Total = 24,019 6,075 138.4 Average of the 5 yearly percents = 138.4 / 5 = 27.7 Percent of 5-year aggregate = 6,075 / 24,019*100 = 25.3
• Yes, I agree with both JamiRae & Jeramiah's response. Nice example, Jeramiah! Whenever I am asked to explain this, I use the phrase in the text that Todd posted: the entire 60-month period. | 1,042 | 4,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.90148 |
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Absolute BS Crap Reasonable Nice Amazing
# McDonald's I'M LOVIN IT = Illuminati symbolism
antidote
User ID: 33057717
Bulgaria
08/12/2013 06:43 AM
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McDonald's I'M LOVIN IT = Illuminati symbolism
When you rearange the letters
I'M LOVIN' IT = ILVMINOTI
and if you take out both of the ' you can use them to transform the v in u and the o in a = ILUMINATI
still missing a L though
Anonymous Coward (OP)
User ID: 33057717
Bulgaria
08/12/2013 06:46 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
if you can't get the part with the apostrophes try to write it down on paper, it will make sense
aqmah
User ID: 33394671
United Arab Emirates
08/12/2013 06:47 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
im lovin it = ti nivol mi = the nephilim
Anonymous Coward (OP)
User ID: 33057717
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08/12/2013 06:51 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
im lovin it = ti nivol mi = the nephilim
Quoting: aqmah
do you know what does it mean exactly?
Anonymous Coward
User ID: 45005265
United Kingdom
08/12/2013 07:32 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
When you rearange the letters
I'M LOVIN' IT = ILVMINOTI
and if you take out both of the ' you can use them to transform the v in u and the o in a = ILUMINATI
still missing a L though
Quoting: antidote 33057717
Talk about clutching at straws. You mindless cunts will make anything fit together
Anonymous Coward
User ID: 44536297
United States
08/12/2013 07:35 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
So what does BURGER KING stand for ?
Anonymous Coward
User ID: 33386139
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08/12/2013 07:39 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
So what does BURGER KING stand for ?
Quoting: Anonymous Coward 44536297
Grr! Nuke big.
Anonymous Coward
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United States
08/12/2013 07:41 AM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
bailey-bose
User ID: 22314700
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08/19/2013 12:24 PM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
you could say that the mcdonalds M sign is the horns of baphomet
Anonymous Coward
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08/19/2013 12:35 PM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
tmorais
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08/19/2013 12:36 PM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
you could say that the mcdonalds M sign is the horns of baphomet
Quoting: bailey-bose
Anonymous Coward
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08/19/2013 12:37 PM
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Re: McDonald's I'M LOVIN IT = Illuminati symbolism
Quoting: Anonymous Coward 8780601
:thatsracistjjj:
Shez
User ID: 45260228
United Kingdom
08/19/2013 12:38 PM
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# Accounting for Bonds
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Dear OTA,
Thanks
1) Ghostbusters Corporation issues \$300,000 of 9% bonds, due in 10 years, with interest payable semiannually. At the time of issue, the market rate for such bonds is 10%. Computer the issue price of the bonds.
2) Toy Story Corporation issued \$500,000 of 6% bonds on May 1, 2008. The bonds were dated January 1, 2008, and mature January 1, 2013, with interest payable July 1 and January 1. The bonds were issued at face value plus accrued interest. Prepare Toy Story's journal entries for the May 1 issuance, the July 1 interest payment, and the December 31 adjusting entry.
3) On January 1, 2008, Qix Corporation issued \$400,000 of 7% bonds, due in 10 years. The bonds were issued for \$372,816, and pay interest each July 1 and January 1. Qix uses the effective interest method. Prepare the company's journal entries for the January 1 issuance, the July 1 interest payment, and the December 31 adjusting entry. Assume an effective interest rate of 8%.
4) On January 1,2008, Uncharted Waters Corporation retired \$600,000, of bonds at 99. At the time of retirement, the unamortized premium was \$15,000 and unamortized bond issue costs were \$5,250. Prepare the corporation's journal entry to record the reacquisition of the bonds.
5) The Goofy Company issued \$200,000 of 10% bonds on January 1, 2008. The bonds are due January 1, 2013 with interest payable July 1 and January 1. Assume the bonds were issued at 98. Prepare the journal entries for January 1, July 1, and December 31. The company uses straight line amortization annually on December 31.
#### Solution Preview
1) Ghostbusters Corporation issues \$300,000 of 9% bonds, due in 10 years, with interest payable semiannually. At the time of issue, the market rate for such bonds is 10%. Computer the issue price of the bonds.
The issue price will be the present value of interest and principal. The rate is 5% and the time period is 20. Interest is an annuity and so we use the PVIFA table. For 5% and 20 periods, the factor is 12.46221. For principal we use the PVIF table and the factor for 5% and 20 periods is 0.37689
Present value of the principal
\$300,000 X .37689 \$113,067
Present value of the interest payments
\$13,500 X 12.46221 168,240
Issue price \$281,307
2) Toy Story Corporation issued \$500,000 of 6% bonds on May 1, 2008. The bonds were dated January 1, 2008, and mature January 1, 2013, with interest payable July 1 and January 1. The bonds were issued at face value plus accrued interest. Prepare ...
#### Solution Summary
The solution explains various questions relating to accounting for bond transactions.
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https://www.scribd.com/doc/23362905/Data-Structures | 1,510,997,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804680.40/warc/CC-MAIN-20171118075712-20171118095712-00775.warc.gz | 844,932,955 | 112,825 | # CS 124
Course Notes 1
Spring 2002
An algorithm is a recipe or a well-defined procedure for performing a calculation, or in general, for transforming some input into a desired output. Perhaps the most familiar algorithms are those those for adding and multiplying integers. Here is a multiplication algorithm that is different from the standard algorithm you learned in school: write the multiplier and multiplicand side by side. Repeat the following operations - divide the first number by 2 (throw out any fractions) and multiply the second by 2, until the first number is 1. This results in two columns of numbers. Now cross out all rows in which the first entry is even, and add all entries of the second column that haven’t been crossed out. The result is the product of the two numbers.
75 37 18 9 4 2 1
29 58 116 232 464 928 1856 2175
29 x 1001011 29 58 232 1856 2175
Figure 1.1: A different multiplication algorithm.
1-1
1-2
In this course we will ask a number of basic questions about algorithms:
• Does it halt? The answer for the algorithm given above is clearly yes, provided we are multiplying positive integers. The reason is that for any integer greater than 1, when we divide it by 2 and throw out the fractional part, we always get a smaller integer which is greater than or equal to 1. Hence our first number is eventually reduced to 1 and the process halts. • Is it correct? To see that the algorithm correctly computes the product of the integers, observe that if we write a 0 for each crossed out row, and 1 for each row that is not crossed out, then reading from bottom to top just gives us the first number in binary. Therefore, the algorithm is just doing standard multiplication, with the multiplier written in binary. • Is it fast? It turns out that the above algorithm is about as fast as the standard algorithm you learned in school. Later in the course, we will study a faster algorithm for multiplying integers. • How much memory does it use? The memory used by this algorithm is also about the same as that of standard algorithm.
1-3
The history of algorithms for simple arithmetic is quite fascinating. Although we take these algorithms for granted, their widespread use is surprisingly recent. The key to good algorithms for arithmetic was the positional number system (such as the decimal system). Roman numerals (I, II, III, IV, V, VI, etc) are just the wrong data structure for performing arithmetic efficiently. The positional number system was first invented by the Mayan Indians in Central America about 2000 years ago. They used a base 20 system, and it is unknown whether they had invented algorithms for performing arithmetic, since the Spanish conquerors destroyed most of the Mayan books on science and astronomy. The decimal system that we use today was invented in India in roughly 600 AD. This positional number system, together with algorithms for performing arithmetic, were transmitted to Persia around 750 AD, when several important Indian works were translated into Arabic. Around this time the Persian mathematician Al-Khwarizmi wrote his Arabic textbook on the subject. The word “algorithm” comes from Al-Khwarizmi’s name. Al-Khwarizmi’s work was translated into Latin around 1200 AD, and the positional number system was propagated throughout Europe from 1200 to 1600 AD. The decimal point was not invented until the 10th century AD, by a Syrian mathematician al-Uqlidisi from Damascus. His work was soon forgotten, and five centuries passed before decimal fractions were re-invented by the Persian mathematician al-Kashi. With the invention of computers in this century, the field of algorithms has seen explosive growth. There are a number of major successes in this field:
• Parsing algorithms - these form the basis of the field of programming languages • Fast Fourier transform - the field of digital signal processing is built upon this algorithm. • Linear programming - this algorithm is extensively used in resource scheduling. • Sorting algorithms - until recently, sorting used up the bulk of computer cycles. • String matching algorithms - these are extensively used in computational biology. • Number theoretic algorithms - these algorithms make it possible to implement cryptosystems such as the RSA public key cryptosystem. • Compression algorithms - these algorithms allow us to transmit data more efficiently over, for example, phone lines.
1-4
• Geometric algorithms - displaying images quickly on a screen often makes use of sophisticated algorithmic techniques.
In designing an algorithm, it is often easier and more productive to think of a computer in abstract terms. Of course, we must carefully choose at what level of abstraction to think. For example, we could think of computer operations in terms of a high level computer language such as C or Java, or in terms of an assembly language. We could dip further down, and think of the computer at the level AND and NOT gates. For most algorithm design we undertake in this course, it is generally convenient to work at a fairly high level. We will usually abstract away even the details of the high level programming language, and write our algorithms in ”pseudo-code”, without worrying about implementation details. (Unless, of course, we are dealing with a programming assignment!) Sometimes we have to be careful that we do not abstract away essential features of the problem. To illustrate this, let us consider a simple but enlightening example.
1-5
1.1 Computing the nth Fibonacci number
Remember the famous sequence of numbers invented in the 15th century by the Italian mathematician Leonardo Fibonacci? The sequence is represented as F0 , F1 , F2 . . ., where F0 = 0, F1 = 1, and for all n ≥ 2, Fn is defined as Fn−1 + Fn−2 . The first few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . The value of F30 is greater than a million! It is easy to see that the Fibonacci numbers grow exponentially. As an exercise, try to show that Fn ≥ 2n/2 for sufficiently large n by a simple induction. Here is a simple program to compute Fibonacci numbers that slavishly follows the definition. function F(n: integer): integer if n = 0 then return 0 else if n = 1 then return 1 else return F(n − 1) + F(n − 2)
The program is obviously correct. However, it is woefully slow. As it is a recursive algorithm, we can naturally express its running time on input n with a recurrence equation. In fact, we will simply count the number of addition operations the program uses, which we denote by T (n). To develop a recurrence equation, we express T (n) in terms of smaller values of T . We shall see several such recurrence relations in this class. It is clear that T (0) = 0 and T (1) = 0. Otherwise, for n ≥ 2, we have T (n) = T (n − 1) + T (n − 2) + 1, because to computer F(n) we compute F(n − 1) and F(n − 2) and do one other addition besides. This is (almost) the Fibonacci equation! Hence we can see that the number of addition operations is growing very large; it is at least 2n/2 for n ≥ 4.
1-6
Can we do better? This is the question we shall always ask of our algorithms. The trouble with the naive algorithm the wasteful recursion: the function F is called with the same argument over and over again, exponentially many times (try to see how many times F(1) is called in the computation of F(5)). A simple trick for improving performance is to avoid repeated calculations. In this case, this can be easily done by avoiding recursion and just calculating successive values: function F(n: integer): integer array A[0 . . . n] of integer A[0] = 0; A[1] = 1 for i = 2 to n do: A[i] = A[i − 1] + A[i − 2] return A[n] This algorithm is of course correct. Now, however, we only do n − 1 additions.
1-7
It seems that we have come so far, from exponential to polynomially many operations, that we can stop here. But in the back of our heads, we should be wondering an we do even better? Surprisingly, we can. We rewrite our equations in matrix notation. Then Similarly, and in general, Similarly, F2 F3 = F1 F2 = · = 0 1 1 1 · F0 F1 . 2 · . F0 F1 ,
0 1 1 1 Fn Fn+1
F1 F2
=
0 1 1 1 F0 F1
0 1 1 1
n ·
So, in order to compute Fn , it suffices to raise this 2 by 2 matrix to the nth power. Each matrix multiplication takes 12 arithmetic operations, so the question boils down to the following: how many multiplications does it take to raise a base (matrix, number, anything) to the nth power? The answer is O(log n). To see why, consider the case where n > 1 is a power of 2. To raise X to the nth power, we compute X n/2 and then square it. Hence the number of multiplications T (n) satisfies T (n) = T (n/2) + 1, from which we find T (n) = log n. As an exercise, consider what you have to do when n is not a power of 2. (Hint: consider the connection with the multiplication algorithm of the first section; there too we repeatedly halved a number...) So we have reduced the computation time exponentially again, from n − 1 arithmetic operations to O(log n), a great achievement. Well, not really. We got a little too abstract in our model. In our accounting of the time requirements for all three methods, we have made a grave and common error: we have been too liberal about what constitutes an elementary step. In general, we often assume that each arithmetic step takes unit time, because the numbers involved will be typically small enough that we can reasonably expect them to fit within a computer’s word. Remember, the number n is only log n bits in length. But in the present case, we are doing arithmetic on huge numbers, with about n bits, where n is pretty large. When dealing with such huge numbers, if exact computation is required we have to use sophisticated long integer packages. Such algorithms take O(n) time to add two n-bit numbers. Hence the complexity of the first two methods was larger than we actually thought: not really O(Fn ) and O(n), but instead O(nFn ) and O(n2 ), respectively. The second algorithm is still exponentially faster. What is worse, the third algorithm involves multiplications of O(n)-bit integers. Let M(n) be the time required to multiply two n-bit numbers. Then the running time of the third algorithm is in fact O(M(n)).
1-8
The comparison between the running times of the second and third algorithms boils down to a most important and ancient issue: can we multiply two n-bit integers faster than Ω(n 2 ) ? This would be faster than the method we learn in elementary school or the clever halving method explained in the opening of these notes. As a final consideration, we might consider the mathematicians’ solution to computing the Fibonacci numbers. A mathematician would quickly determine that 1 Fn = √ 5 √ 1+ 5 2
n
√ 1− 5 2
n
.
Using this, how many operations does it take to compute Fn ? Note that this calculation would require floating point arithmetic. Whether in practice that would lead to a faster or slower algorithm than one using just integer arithmetic might depend on the computer system on which you run the algorithm.
CS 124
Lecture 2
In order to discuss algorithms effectively, we need to start with a basic set of tools. Here, we explain these tools and provide a few examples. Rather than spend time honing our use of these tools, we will learn how to use them by applying them in our studies of actual algorithms.
Induction
The standard form of the induction principle is the following: If a statement P(n) holds for n = 1, and if for every n ≥ 1 P(n) implies P(n + 1), then P holds for all n. Let us see an example of this: Claim 2.1 Let S(n) = ∑n i. Then S(n) = i=1
n(n+1) 2 .
Proof: The proof is by induction. Base Case: We show the statement is true for n = 1. As S(1) = 1 = Induction Hypothesis: We assume S(n) = Reduction Step: We show S(n + 1) =
n(n+1) 2 . 1(2) 2 ,
the statement holds.
(n+1)(n+2) . 2
Note that S(n + 1) = S(n) + n + 1. Hence
S(n + 1) = S(n) + n + 1 n(n + 1) +n+1 = 2 n = (n + 1) +1 2 (n + 1)(n + 2) = . 2
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2-2
The proof style is somewhat pedantic, but instructional and easy to read. We break things down to the base case – showing that the statement holds when n = 1; the induction hypothesis – the statement that P(n) is true; and the reduction step – showing that P(n) implies P(n + 1). Induction is one of the most fundamental proof techniques. The idea behind induction is simple: take a large problem (P(n + 1)), and somehow reduce its proof to a proof of a smaller problems (such as P(n); P(n) is smaller in the sense that n < n + 1). If every problem can thereby be broken down to a small number of instances (we keep reducing down to P(1)), these can be checked easily. We will see this idea of reduction, whereby we reduce solving a problem to a solving an easier problem, over and over again throughout the course. As one might imagine, there are other forms of induction besides the specific standard form we gave above. Here’s a different form of induction, called strong induction: If a statement P(n) holds for n = 1, and if for every n ≥ 1 the truth of P(i) for all i ≤ n implies P(n + 1), then P holds for all n. Exercise: show that every number has a unique prime factorization using strong induction.
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O Notation
When measuring, for example, the number of steps an algorithm takes in the worst case, our result will generally be some function T (n) of the input size, n. One might imagine that this function may have some complex form, such as T (n) = 4n2 − 3n log n + n2/3 + log3 n − 4. In very rare cases, one might wish to have such an exact form for the running time, but in general, we are more interested in the rate of growth of T (n) rather than its exact form. The O notation was developed with this in mind. With the O notation, only the fastest growing term is important, and constant factors may be ignored. More formally:
Definition 2.2 We say for non-negative functions f (n) and g(n) that f (n) is O(g(n)) if there exist positive constants c and N such that for all n ≥ N, f (n) ≤ cg(n).
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Let us try some examples. We claim that 2n 3 + 4n2 is O(n3 ). It suffices to show that 2n3 + 4n2 ≤ 6n3 for n ≥ 1, by definition. But this is clearly true as 4n 3 ≥ 4n2 for n ≥ 1. (Exercise: show that 2n3 + 4n2 is O(n4 ).) We claim 10 log 2 n is O(ln n). This follows from the fact that 10 log 2 n ≤ (10 log 2 e) ln n. If T (n) is as above, then T (n) is O(n2 ). This is a bit harder to prove, because of all the extraneous terms. It is, however, easy to see; 4n2 is clearly the fastest growing term, and we can remove the constant with O notation. Note, though, that T (n) is O(n3 ) as well! The O notation is not tight, but more like a ≤ comparison.
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Similarly, there is notation for ≥ and = comparisons. Definition 2.3 We say for non-negative functions f (n) and g(n) that f (n) is is Ω(g(n)) if there exist positive constants c and N such that for all n ≥ N, f (n) ≥ cg(n). We say that f (n) is Θ(g(n)) if both f (n) is O(g(n)) and f (n) is Ω(g(n)). The O notation has several useful properties that are easy to prove. Lemma 2.4 If f 1 (n) is O(g1 (n)) and f 2 (n) is O(g2 (n)) then f 1 (n) + f2 (n) is O(g1 (n) + g2 (n)). Proof: There exist positive constants c 1 , c2 , N1 , and N2 such that f 1 (n) ≤ c1 g1 (n) for n ≥ N1 and f2 (n) ≤ c2 g2 (n) for n ≥ N2 . Hence f1 (n) + f2 (n) ≤ max{c1 , c2 }(g1 (n) + g2 (n)) for n ≥ max{N1 , N2 }. Exercise: Prove similar lemmata for f 1 (n) f2 (n). Prove the lemmata when O is replaced by Ω or Θ.
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Finally, there is a bit for notation corresponding to <<, when one function is (in some sense) much less than another.
Definition 2.5 We say for non-negative functions f (n) and g(n) that f (n) is is o(g(n)) if f (n) = 0. n→∞ g(n) lim Also, f (n) is ω(g(n)) if g(n) is o( f (n)).
We emphasize that the O notation is a tool to help us analyze algorithms. It does not always accurately tell us how fast an algorithm will run in practice. For example, constant factors make a huge difference in practice (imagine increasing your bank account by a factor of 10), and they are ignored in the O notation. Like any other tool, the O notation is only useful if used properly and wisely. Use it as a guide, not as the last word, to judging an algorithm.
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Recurrence Relations
A recurrence relation defines a function using an expression that includes the function itself. For example, the Fibonacci numbers are defined by: F(n) = F(n − 1) + F(n − 2), F(1) = F(2) = 1. This function is well-defined, since we can compute a unique value of F(n) for every positive integer n. Note that recurrence relations are similar in spirit to the idea of induction. The relations defines a function value F(n) in terms of the function values at smaller arguments (in this case, n − 1 and n − 2), effectively reducing the problem of computing F(n) to that of computing F at smaller values. Base cases (the values of F(1) and F(2)) need to be provided. Finding exact solutions for recurrence relations is not an extremely difficult process; however, we will not focus on solution methods for them here. Often a natural thing to do is to try to guess a solution, and then prove it by induction. Alternatively, one can use a symbolic computation program (such as Maple or Mathematica); these programs can often generate solutions. We will occasionally use recurrence relations to describe the running times of algorithms. For our purposes, we often do not need to have an exact solution for the running time, but merely an idea of its asymptotic rate of growth. For example, the relation T (n) = 2T (n/2) + 2n, T (1) = 1 has the exact solution (for n a power of 2) of T (n) = 2n log 2 n + n. (Exercise: Prove this by induction.) But for our purposes, it is generally enough to know that the solution is Θ(n log n).
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The following theorem is extremely useful for such recurrence relations:
Theorem 2.6 The solution to the recurrence relation T (n) = aT (n/b) + cn k , where a ≥ 1 and b ≥ 2 are integers and c and k are positive constants satisfies: O nlogb a T (n) is O nk log n O nk if a > bk if a = bk if a < bk .
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Data Structures
We shall regard integers, real numbers, and bits, as well as more complicated objects such as lists and sets, as primitive data structures. Recall that a list is just an ordered sequence of arbitrary elements. List q := [x1 , x2 , . . . , xn ]. x1 is called the head of the list. xn is called the tail of the list. n = |q| is the size of the list. We denote by ◦ the concatenation operation. Thus q ◦ r is the list that results from concatenating the list q with the list r. The operations on lists that are especially important for our purposes are: head(q) push(q, x) pop(q) inject(q, x) eject(q) size(q) return(x1 ) q := [x] ◦ q q := [x2 , . . . , xn ], return(x1 ) q := q ◦ [x] q := [x1 , x2 , . . . , xn−1 ], return(xn ) return(n)
The head, pop, and eject operations are not defined for empty lists. Appropriate return values (either an error, or an empty symbol) can be designed depending on the implementation. A stack is a list that supports operations head, push, pop. A queue is a list that supports operations head, inject and pop. A deque supports all these operations. Note that we can implement lists either by arrays or using pointers as the usual linked lists. Arrays are often faster in practice, but they are often more complicated to program (especially if there is no implicit limit on the number of items). In either case, each of the above operations can be implemented in a constant number of steps.
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Application: Mergesort
For the rest of the lecture, we will review the procedure mergesort. The input is a list of n numbers, and the output is a list of the given numbers sorted in increasing order. The main data structure used by the algorithm will be a queue. We will assume that each queue operation takes 1 step, and that each comparison (is x > y?) takes 1 step. We will show that mergesort takes O(n log n) steps to sort a sequence of n numbers. The procedure mergesort relies on a function merge which takes as input two sorted (in increasing order) lists of numbers and outputs a single sorted list containing all the given numbers (with repetition).
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function merge (s,t) list s,t if s = [ ] then return t else if t = [ ] then return s else if s(1) ≤ t(1) then u:= pop(s) else u:= pop(t) return push(u, merge(s,t)) end merge
function mergesort (s) list s, q q=[] for x ∈ s inject(q, [x]) rof while size(q) ≥ 2 u := pop(q) v := pop(q) inject(q, merge(u, v)) end if q = [ ] return [ ] else return q(1) end mergesort
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The correctness of the function merge follows from the following fact: the smallest number in the input is either s(1) or t(1), and must be the first number in the output list. The rest of the output list is just the list obtained by merging s and t after deleting that smallest number. The number of steps for each invocation of function merge is O(1) steps. Since each recursive invocation of merge removes an element from either s or t, it follows that function merge halts in O(|s| + |t|) steps. Question: Can you design an iterative (rather than recursive) version of merge? How much time does is take? Which version would be faster in practice– the recursive or the iterative?
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Q : [ [7, 9], [1, 4], [6, 16], [2, 10] ∗ [3, 11, 12, 14], [5, 8, 13, 15] ] Q : [ [6, 16], [2, 10] ∗ [3, 11, 12, 14], [5, 8, 13, 15], [1, 4, 7, 9] ]
Figure 2.1: One step of the mergesort algorithm. The iterative algorithm mergesort uses q as a queue of lists. (Note that it is perfectly acceptable to have lists of lists!) It repeatedly merges together the two lists at the front of the queue, and puts the resulting list at the tail of the queue. The correctness of the algorithm follows easily from the fact that we start with sorted lists (of length 1 each), and merge them in pairs to get longer and longer sorted lists, until only one list remains. To analyze the running time of this algorithm, let us place a special marker ∗ initially at the end of the q. Whenever the marker ∗ reaches the front of q, and is either the first or the second element of q, we move it back to the end of q. Thus the presence of the marker ∗ makes no difference to the actual execution of the algorithm. Its only purpose is to partition the execution of the algorithm into phases: where a phase is the time between two successive visits of the marker ∗ to the end of the q. Then we claim that the total time per phase is O(n). This is because each phase just consists of pairwise merges of disjoint lists in the queue. Each such merge takes time proportional to the sum of the lengths of the lists, and the sum of the lengths of all the lists in q is n. On the other hand, the number of lists is halved in each phase, and therefore the number of phases is at most log n. Therefore the total running time of mergesort is O(n log n).
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An alternative analysis of mergesort depends on a recursive, rather than iterative, description. Suppose we have an operation that takes a list and splits it into two equal-size parts. (We will assume our list size is a power of 2, so that all sublists we ever obtain have even size or are of length 1.) Then a recursive version of mergesort would do the following: function mergesort (s) list s, s1 , s2 if size(s) = 1 then return(s) split(s, s1 , s2 ) s1 = mergesort(s1 ) s2 = mergesort(s2 ) return(merge(s1 , s2 )) end mergesort
Here split splits the list s into two parts of equal length s 1 and s2 . The correctness follows easily from induction. Let T (n) be the number of comparisons mergesort performs on lists of length n. Then T (n) satisfies the recurrence relation T (n) ≤ 2T (n/2) + n − 1. This follows from the fact that to sort lists of length n we sort two sublists of length n/2 and then merge them using (at most) n − 1 comparisons. Using our general theorem on solutions of recurrence relations, we find that T (n) = O(n log n). Question: The iterative version of mergesort uses a queue. Implicitly, the recursive version is using a stack. Explain the implicit stack in the recursive version of mergesort. Question: Solve the recurrence relation T (n) = 2T (n/2) + n − 1 exactly to obtain an upper bound on the number of comparisons performed by the recursive mergesort variation.
CS124
Lecture 3
Spring 2002
Graphs and modeling
Formulating a simple, precise specification of a computational problem is often a prerequisite to writing a computer program for solving the problem. Many computational problems are best stated in terms of graphs. A directed graph G(V, E) consists of a finite set of vertices V and a set of (directed) edges or arcs E. An arc is an ordered pair of vertices (v, w) and is usually indicated by drawing a line between v and w, with an arrow pointing towards w. Stated in mathematical terms, a directed graph G(V, E) is just a binary relation E ⊆ V ×V on a finite set V . Undirected graphs may be regarded as special kinds of directed graphs, such that (u, v) ∈ E ↔ (v, u) ∈ E. Thus, since the directions of the edges are unimportant, an undirected graph G(V, E) consists of a finite set of vertices V , and a set of edges E, each of which is an unordered pair of vertices {u, v}. Graphs model many situations. For example, the vertices of a graph can represent cities, with edges representing highways that connect them. In this case, each edge might also have an associated length. Alternatively, an edge might represent a flight from one city to another, and each edge might have a weight which represents the cost of the flight. A typical problem in this context is to compute shortest paths: given that you wish to travel from city X to city Y, what is the shortest path (or the cheapest flight schedule). We will find very efficient algorithms for solving these problems. A seemingly similar problem is the traveling salesman problem. Supposing that a traveling salesman wishes to visit each city exactly once and return to his starting point, in what order should he visit the cities to minimize the total distance traveled? Unlike the shortest paths problem, however, this problem has no known efficient algorithm. This is an example of an NP-complete problem, and one we will study towards the end of this course.
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A different context in which graphs play a critical modeling role is in networks of pipes or communication links. These can, in general, be modeled by directed graphs with capacities on the edges. A directed edge from u to v with capacity c might represent a cable that can carry a flow of at most c calls per unit time from u to v. A typical problem in this context is the max-flow problem: given a communications network modeled by a directed graph with capacities on the edges, and two special vertices — a source s and a sink t — what is the maximum rate at which calls from s to t can be made? There are ingenious techniques for solving these types of flow problems. In all the cases mentioned above, the vertices and edges of the graph represented something quite concrete such as cities and highways. Often, graphs will be used to represent more abstract relationships. For example, the vertices of a graph might represent tasks, and the edges might represent precedence constraints: a directed edge from u to v says that task u must be completed before v can be started. An important problem in this context is scheduling: in what order should the tasks be scheduled so that all the precedence constraints are satisfied. There are extremely fast algorithms for this problem that we will see shortly.
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Representing graphs on the computer
1 Generally,
we use either n or |V | for the number of nodes in a graph, and m or |E| for the number of edges.
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Depth first search
There are two fundamental algorithms for searching a graph: depth first search and breadth first search. To better understand the need for these procedures, let us imagine the computer’s view of a graph that has been input into it, in the adjacency list representation. The computer’s view is fundamentally local to a specific vertex: it can examine each of the edges adjacent to a vertex in turn, by traversing its adjacency list; it can also mark vertices as visited. One way to think of these operations is to imagine exploring a dark maze with a flashlight and a piece of chalk. You are allowed to illuminate any corridor of the maze emanating from your current position, and you are also allowed to use the chalk to mark your current location in the maze as having been visited. The question is how to find your way around the maze. We now show how the depth first search allows the computer to find its way around the input graph using just these primitives. (We will examine breadth first search shortly.) Depth first search is technique for exploring a graph using a stack as the basic data structure. We start by defining a recursive procedure search (the stack is implicit in the recursive calls of search): search is invoked on a vertex v, and explores all previously unexplored vertices reachable from v. Procedure search(v) vertex v explored(v) := 1 previsit(v) for (v, w) ∈ E if explored(w) = 0 then search(w) rof postvisit(v) end search
Procedure DFS (G(V, E)) graph G(V, E) for each v ∈ V do explored(v) := 0 rof for each v ∈ V do if explored(v) = 0 then search(v) rof end DFS
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By modifying the procedures previsit and postvisit, we can use DFS to solve a number of important problems, as we shall see. It is easy to see that depth first search takes O(|V | + |E|) steps (assuming previsit and postvisit take O(1) time), since it explores from each vertex once, and the exploration involves a constant number of steps per outgoing edge. The procedure search defines a tree in a natural way: each time that search discovers a new vertex, say w, we can incorporate w into the tree by connecting w to the vertex v it was discovered from via the edge (v, w). The remaining edges of the graph can be classified into three types: • Forward edges - these go from a vertex to a descendant (other than child) in the DFS tree. • Back edges - these go from a vertex to an ancestor in the DFS tree. • Cross edges - these go from “right to left”– there is no ancestral relation. Question: Explain why if the graph is undirected, there can be no cross edges. One natural use of previsit and postvisit could each keep a counter that is increased each time one of these routines is accessed; this corresponds naturally to a notion of time. Each routine could assign to each vertex a preorder number (time) and a postorder number (time) based on the counter. If we think of depth first search as using an explicit stack, then the previsit number is assigned when the vertex is first placed on the stack, and the postvisit number is assigned when the vertex is removed from the stack. Note that this implies that the intervals [preorder(u), postorder(u)] and [preorder(v), postorder(v)] are either disjoint, or one contains the other.
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An important property of depth-first search is that the contents of the stack at any time yield a path from the root to some vertex in the depth first search tree. (Why?) This allows us to prove the following property of the postorder numbering: Claim 3.1 If (u, v) ∈ E then postorder(u) < postorder(v) ⇐⇒ (u, v) is a back edge. Proof: If postorder(u) < postorder(v) then v must be pushed on the stack before u. Otherwise, the existence of edge (u, v) ensures that v must be pushed onto the stack before u can be popped, resulting in postorder(v) < postorder(u) — contradiction. Furthermore, since v cannot be popped before u, it must still be on the stack when u is pushed on to it. It follows that v is on the path from the root to u in the depth first search tree, and therefore (u, v) is a back edge. The other direction is trivial. Exercise: What conditions to the preorder and postorder numbers have to satisfy if (u, v) is a forward edge? A cross edge? Claim 3.2 G(V, E) has a cycle iff the DFS of G(V, E) yields a back edge. Proof: If (u, v) is a back edge, then (u, v) together with the path from v to u in the depth first tree form a cycle. Conversely, for any cycle in G(V, E), consider the vertex assigned the smallest postorder number. Then the edge leaving this vertex in the cycle must be a back edge by Claim 3.1, since it goes from a lower postorder number to a higher postorder number.
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A
A
B
E
B
E
C
F
C
F
D
D
Graph is explored in preorder ABCDEF. Postorder is DCBAFE. DB is a back edge. AD is a forward edge. EC is a cross edge.
Figure 3.1: A sample depth-first search.
Application of DFS: Topological sort
We now suggest an algorithm for the scheduling problem described previously. Given a directed graph G(V, E), whose vertices V = {v1 , . . . vn } represent tasks, and whose edges represent precedence constraints: a directed edge from u to v says that task u must be completed before v can be started. The problem of topological sorting asks: in what order should the tasks be scheduled so that all the precedence constraints are satisfied. Note: The graph must be acyclic for this to be possible. (Why?) Directed acyclic graphs appear so frequently they are commonly referred to as DAGs. Claim 3.3 If the tasks are scheduled by decreasing postorder number, then all precedence constraints are satisfied. Proof: If G is acyclic then the DFS of G produces no back edges by Claim 3.2. Therefore by Claim 3.1, (u, v) ∈ G implies postorder(u) > postorder(v). So, if we process the tasks in decreasing order by postorder number, when task v is processed, all tasks with precedence constraints into v (and therefore higher postorder numbers) must already have been processed. There’s another way to think about topologically sorting a DAG. Each DAG has a source, which is a vertex with no incoming edges. Similarly, each DAG has a sink, which is a vertex with no outgoing edges. (Proving this is an exercise.) Another way to topologically order the vertices of a DAG is to repeatedly output a source, remove it from the graph, and repeat until the graph is empty. Why does this work? Similarly, once could repeatedly output sinks, and this gives the reverse of a valid topological order. Again, why?
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Strongly Connected Components
Connectivity in undirected graphs is rather straightforward. A graph that is not connected can naturally be decomposed into several connected components (Figure 3.2). DFS does this handily: each restart of DFS marks a new connected component.
1 3 4 6 7
2 5
12
13
14 8
9
10
11
Figure 3.2: An undirected graph
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In directed graphs, what connectivity means is more subtle. In some primitive sense, the directed graph in Figure 3.3 appears connected, since if it were an undirected graph, it would be connected. But there is no path from vertex 12 to 6, or from 6 to 1, so saying the graph is connected would be misleading. We must begin with a meaningful definition of connectivity in directed graphs. Call two vertices u and v of a directed graph G = (V, E) connected if there is a path from u to v, and one from v to u. This relation between vertices is reflexive, symmetric, and transitive (check!), so it is an equivalence relation on the vertices. As such, it partitions V into disjoint sets, called the strongly connected components (SCC’s) of the graph (in Figure 3.3 there are four SCC’s). Within a strongly connected component, every pair of vertices are connected.
1 2 3 1 2-4-5 3-6
4 5
6
7-8-9-10-11-12 7 8
9
10 12
11
Figure 3.3: A directed graph and its SCC’s
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We now imagine shrinking each SCC into a vertex (a supervertex), and draw an edge (a superedge) from SCC X to SCC Y if there is at least one edge from a vertex in X to a vertex in Y . The resulting directed graph has to be a directed acyclic graph (DAG) – that is to say, it can have no cycles (see Figure 3.3). The reason is simple: a cycle containing several SCC’s would merge to a single SCC, since there would be a path between every pair of vertices in the SCC’s of the cycle. Hence, every directed graph is a DAG of its SCC’s. This important decomposition theorem allows one to think of connectivity information of a directed graph in two levels. At the top level we have a DAG, which has a useful, simple structure. For example, as we have mentioned before, a DAG is guaranteed to have at least one source (a vertex without incoming edges) and a sink (a vertex without outgoing edges). If we want more details, we could look inside a vertex of the DAG to see the full-fledged SCC —a completely connected graph— that lies there. This decomposition is extremely useful and informative; it is thus very fortunate that we have a very efficient algorithm, based on DFS, that finds the strongly connected components in linear time! We motivate this algorithm next. It is based on several interesting and slightly subtle properties of DFS:
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Property 1: If DFS is started at a vertex v, then it will get stuck and restarted precisely when all vertices in the SCC of v, and in all the SCC’s that are reachable from the SCC of v, are visited. Consequently, if DFS is started at a vertex of a sink SCC (a SCC that has no edges leaving it in the DAG of SCC’s), then it will get stuck after it visits precisely the vertices of this SCC. For example, if DFS is started at vertex 11 in Figure 3.3 (a vertex in the only sink SCC in this graph), then it will visit the six vertices in the sink SCC before getting stuck: vertices 12, 10, 9, 7, 8. Property 1 suggests a way of starting a decomposition algorithm, by finding the first SCC: start DFS from a vertex in a sink SCC, and, when stuck, output the vertices that have been visited. They form an SCC! Of course, this leaves us with two problems: (A) How to guess a vertex in a sink SCC, and (B) how to continue our algorithm by outputting the next SCC, and so on.
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Let us first face Problem (A). It turns out that it will be easier not to look for vertices in a sink SCC, but instead look for vertices in a source SCC. In particular: Property 2: The vertex with the highest postorder number in DFS (that is, the vertex where the DFS ends) belongs to a source SCC. The proof is by contradiction. If Property 2 were not not true, and v is the vertex with the highest post-order number, then there would be an incoming edge (u, w) with u not in the SCC of v and w in the SCC of v. If u were searched before v, then u clearly has a higher postorder number. If u were searched after v, then since u does not lie in v’s SCC, it must not be searched until v is popped from the search stack, so again u must have a higher postorder number than v. The reason behind Property 2 is thus not hard to see: if there is an SCC “above” the SCC of the vertex where the DFS ends, then the DFS should have ended in that SCC (reaching it either by restarting or by backtracking). Property 2 provides an indirect solution to Problem (A). Consider a graph G and the reverse graph G R —G with the directions of all edges reversed. G R has precisely the same SCC’s as G (why?). So, if we make a DFS in G R , then the vertex where we end (the one with the highest post-order) belongs to a source SCC of G R —that is to say, a sink SCC of G. We have solved Problem (A).
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Onwards to Problem (B). How does the algorithm continue after the first sink component is output? The solution is clear: delete the SCC just output from G R , and make another DFS in the remaining graph. The only problem is, this would be a quadratic, not linear, algorithm, since we would run an O(m) DFS algorithm for up to each or O(n) vertices. How can we avoid this extra work? The key observation here is that we do not have to make a new DFS in the remaining graph: Property 3: If we make a DFS in a directed graph, and then delete a source SCC of this graph, what remains is a DFS in the remaining graph (the pre-order and post-order numbers may now not be consecutive, but they will be of the right relative magnitude). This is also easy to justify. We just imagine two runs of the DFS algorithm, one with and one without the source SCC. Consider a transcript recording the steps of the DFS algorithm. It is easy to see that the transcript of both runs would be the same (assuming they both made the same choices of what edges to follow at what points), except where the the first went through the source SCC.
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Property 3 allows us to use induction to continue our SCC algorithm. After we output the first SCC, we can use the same DFS information from GR to output the second SCC, the third SCC, and so on. The full algorithm can thus be described as follows: Step 1: Perform DFS on GR . Step 2: Perform DFS on G, processing unsearched vertices in the order of decreasing postorder numbers from the DFS of Step 1. At the beginning and every restart print “New SCC:” When visiting vertex v, print v. This algorithm is linear-time, since the total work is really just two depth-first searches, each of which is linear time. Question: (How does one construct G R from G?) If we run this algorithm on Figure 3.3, Step 1 yields the following order on the vertices (decreasing postorder in G R ’s DFS): 7, 9, 10, 12, 11, 8, 3, 6, 2, 5, 4, 1. Step 2 now produces the following output: New SCC: 7, 8, 10, 9, 11, 12, New SCC: 3, 6, New SCC: 2, 4, 5, New SCC: 1.
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Incidentally, there is more sophisticated connectivity information that one can derive from undirected graphs. An articulation point is a vertex whose deletion increases the number of connected components in the undirected graph. In Figure 3.2 there are 4 articulation points: 3, 6, 8, and 13. Articulation points divide the graph into biconnected components (the pieces of the graph between articulation points) and bridge edges. Biconnected components are maximal edge sets (of at least 2 edges) such that any two edges on the set lie on a common cycle. For example, the large connected component of the graph in Figure 3.2 contains the biconnected components on edges between vertices 1-2-3-4-5-7-8 and 6-9-10. The remaining edges are 3-6 and 8-11 are bridge edges; they disconnect the graph. Not coincidentally, this more sophisticated and subtle connectivity information can also be captured by DFS.
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Putting in Into Practice
Suppose you are debugging your latest huge software program for a major industrial client. The program has hundreds of procedures, each of which must be carefully tested for bugs. You realize that, to save yourself some work, it would be best to analyze the procedures in a particular order. For instance, if procedure Write Check() calls Get Check Number(), you would probably want to test Get Check Number() first. That way, when you look for the bugs in Write Check(), you do not have to worry about checking (or re-checking) Get Check Number(). (Let’s ignore the specious argument that if there are no bugs, you might avoid testing and debugging Get Check Number() altogether by starting with Write Check().) You can easily generate a list of what procedures each procedure calls with a single pass through the code. So here’s the problem: given your program, determine what schedule you should give your testing and debugging team, so that a procedure will be debugged only after anything it calls will be debugged. Go through the program, creating one vertex for each procedure. Introduce a directed edge from vertex A to vertex B if the procedure A calls B. This directed edge represents the fact that A must be debugged before B. We call this graph the procedure graph. If this graph is acyclic, then the topological sort will give you a valid ordering for the debugging. What if the graph is not acyclic? Then your program uses mutual recursion; that is, there is some chain of procedures through which a procedure might end up calling itself. For example, this would be the case if procedure A calls procedure B, procedure B calls procedure C, and procedure C calls procedure A. A topological sort will detect these cycles, but what we really want is a list of them, since instances of mutual recursion are harder to test and debug. In this case, we should use the strongly connected components algorithm on the procedure graph. The SCC algorithm will find all the cycles, showing all instances of mutual recursion. Moreover, if we collapse the cycles in the graph, so that instances of mutual recursion are treated as one large super-procedure, then the SCC algorithm will provide a valid debugging ordering for all the procedures in this modified graph. That is, the SCC algorithm will topologically sort the underlying SCC DAG.
CS124
Lecture 4
Spring 2002
A searching technique with different properties than DFS is Breadth-First Search (BFS). While DFS used an implicit stack, BFS uses an explicit queue structure in determining the order in which vertices are searched. Also, generally one does not restart BFS, because BFS only makes sense in the context of exploring the part of the graph that is reachable from a particular vertex (s in the algorithm below). Procedure BFS (G(V, E), s ∈ V ) graph G(V, E) array[|V |] of integers dist queue q; dist[s] := 0 inject(q, s) while size(q) > 0 v := pop(q) previsit(v) explored(v) := 1 for (v, w) ∈ E if explored(w) = 0 then inject(q, w) dist(w) = dist(v)+1 fi rof end while end BFS
BFS runs, of course, in linear time O(|E|), under the assumption that |E| ≥ |V |. The reason is that BFS visits each edge exactly once, and does a constant amount of work per edge.
4-1
4-2
S
0
1
2
1 2
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2
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Figure 4.1: BFS of a directed graph Although BFS does not have the same subtle properties of DFS, it does provide useful information. BFS visits vertices in order of increasing distance from s. In fact, our BFS algorithm above labels each vertex with the distance from s, or the number of edges in the shortest path from s to the vertex. For example, applied to the graph in Figure 4.1, this algorithm labels the vertices (by the array dist) as shown. Why are we sure that the array dist is the shortest-path distance from s? A simple induction proof suffices. It is certainly true if the distance is zero (this happens only at s). And, if it is true for dist(v) = d, then it can be easily shown to be true for values of dist equal to d + 1 —any vertex that receives this value has an edge from a vertex with dist d, and from no vertex with lower value of dist. Notice that vertices not reachable from s will not be visited or labeled.
4-3
Single-Source Shortest Paths —Nonnegative Lengths
What if each edge (v, w) of our graph has a length, a positive integer denoted length(v, w), and we wish to find the shortest paths from s to all vertices reachable from it? (What if we are interested only in the shortest path from s to a specific node t? As it turns out, all algorithms known for this problem have to compute the shortest path from s to all vertices reachable from it.) BFS offers a possible solution. We can subdivide each edge (u, v) into length(u, v) edges, by inserting length(u, v) − 1 “dummy” nodes, and then apply DFS to the new graph. This algorithm solves
the shortest-path problem in time O( ∑(u,v)∈E length(u, v)). Unfortunately, this can be very large —lengths could be in the thousands or millions. So we need to find a better way. The problem is that this BFS-based algorithm will spend most of its time visiting “dummy” vertices; only occasionally will it do something truly interesting, like visit a vertex of the original graph. What we would like to do is run this algorithm, but only do work for the “interesting” steps.
4-4
To do this, We need to generalize BFS. Instead of using a queue, we will instead use a heap or priority queue of vertices. A heap is an data structure that keeps a set of objects, where each object has an associated value. The operations a heap H implements include the following: deletemin(H) insert(x, y, H) change(x, y, H) return the object with the smallest value insert a new object x/value y pair in the structure if y is smaller than x’s current value, change the value of object x to y We will not distinguish between insert and change, since for our purposes, they are essentially equivalent; changing the value of a vertex will be like re-inserting it. (In all heap implementations we assume that we have an array of pointers that gives, for each vertex, its position in the heap, if any. This allows us to always have at most one copy of each vertex in the heap. Furthermore, it makes changes and inserts essentially equivalent operations.) Each entry in the heap will stand for a projected future “interesting event” of our extended BFS. Each entry will correspond to a vertex, and its value will be the current projected time at which we will reach the vertex. Another way to think of this is to imagine that, each time we reach a new vertex, we can send an explorer down each adjacent edge, and this explorer moves at a rate of 1 unit distance per second. With our heap, we will keep track of when each vertex is due to be reached for the first time by some explorer. Note that the projected time until we reach a vertex can decrease, because the new explorers that arise when we reach a newly explored vertex could reach a vertex first (see node b in Figure 4.2). But one thing is certain: the most imminent future scheduled arrival of an explorer must happen, because there is no other explorer who can reach any vertex faster. The heap conveniently delivers this most imminent event to us.
4-5
As in all shortest path algorithms we shall see, we maintain two arrays indexed by V . The first array, dist[v], will eventually contain the true distance of v from s. The other array, prev[v], will contain the last node before v in the shortest path from s to v. Our algorithm maintains a useful invariant property: at all times dist[v] will contain a conservative over-estimate of the true shortest distance of v from s. Of course dist[s] is initialized to its true value 0, and all other dist’s are initialized to ∞, which is a remarkably conservative overestimate. The algorithm is known as Djikstra’s algorithm, named after the inventor. Algorithm Djikstra (G = (V, E, length); s ∈ V ) v, w: vertices dist: array[V ] of integer prev: array[V ] of vertices H: priority heap of V H := {s : 0} for v ∈ V do dist[v] := ∞, prev[v] :=nil rof dist[s] := 0 / while H = 0 v := deletemin(h) for (v, w) ∈ E if dist[w] > dist[v]+ length(v, w) dist[w] := dist[v] + length(v, w), prev[w] := v, insert(w,dist[w], H) fi rof end while end shortest paths 1
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a2 s0 2 3 6 b4 5
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Figure 4.2: Shortest paths The algorithm, run on the graph in Figure 4.2, will yield the following heap contents (node: dist/priority pairs) at the beginning of the while loop: {s : 0}, {a : 2, b : 6}, {b : 5, c : 3}, {b : 4, e : 7, f : 5}, {e : 7, f : 5, d : 6}, {e : 6, d : 6}, {e : 6}, {}. The distances from s are shown in Figure 2, together with the shortest path tree from s, the rooted tree defined by the pointers prev.
4-7
What is the running time of this algorithm? The algorithm involves |E| insert operations and |V | deletemin operations on H, and so the running time depends on the implementation of the heap H. There are many ways to implement a heap. Even an unsophisticated implementation as a linked list of node/priority pairs yields an interesting time bound, O(|V |2 ) (see first line of the table below). A binary heap would give O(|E| log |V |). Which of the two should we prefer? The answer depends on how dense or sparse our graphs are. In all graphs, |E| is between |V | and |V |2 . If it is Ω(|V |2 ), then we should use the linked list version. If it is anywhere below we should use binary heaps.
|V |2 log |V | ,
heap implementation linked list binary heap d-ary heap Fibonacci heap
deletemin O(|V |) O(log |V |) O(log |V |) O( d log |V | ) log d
insert O(1) O(log |V |) O( log |V | ) log d O(1) amortized
|V |×deletemin+|E|×insert O(|V |2 ) O(|E| log |V |) O((|V | · d + |E|) log |V | log d
O(|V | log |V | + |E|)
A more sophisticated data structure, the d-ary heap, performs even better. A d-ary heap is just like a binary heap, except that the fan-out of the tree is d, instead of 2. (Here d should be at least 2, however!) Since the depth of any such tree with |V | nodes is
log |V | log d ,
it is easy to see that inserts take this amount of time. Deletemins take d times
that, because deletemins go down the tree, and must look at the children of all vertices visited. The complexity of this algorithm is a function of d. We must choose d to minimize it. A natural choice is d=
|E| |V | ,
which is the the average degree! (Note that this is the natural choice because it equalizes the two terms of
|E| + |V | · d. Alternatively, the “exact” value can be found using calculus.) This yields an algorithm that is good for both sparse and dense graphs. For dense graphs, its running time is O(|V | 2 ). For graphs with |E| = O(|V |), it is |V | log |V |. Finally, for graphs with intermediate density, such as |E| = |V | 1+δ , where δ is the density of the graph, the algorithm is linear! The fastest known implementation of Djikstra’s algorithm uses a data structure known as a Fibonacci heap, which we will not cover here. Note that the bounds for the insert operation for Fibonacci heaps are amortized bounds: certain operations may be expensive, but the average cost over a sequence of operations is constant.
4-8
Single-Source Shortest Paths: General Lengths
Our argument of correctness of our shortest path algorithm was based on the “time metaphor:” the most imminent prospective event (arrival of an explorer) must take place, exactly because it is the most imminent. This however would not work if we had negative edges. (Imagine explorers being able to arrive before they left!) If the length of edge (a, b) in Figure 2 were −1, the shortest path from s to b would have value 1, not 4, and our simple algorithm fails. Obviously, with negative lengths we need more involved algorithms, which repeatedly update the values of dist. We can describe a general paradigm for constructing shortest path algorithms with arbitrary edge weights. The algorithms use arrays dist and prev, and again we maintain the invariant that dist is always a conservative overestimate of the true distance from s. (Again, dist is initialized to ∞ for all nodes, except for s for which it is 0). The algorithms maintain dist so that it is always a conservative overestimate; it will only update the a value when a suitable path is discovered to show that the overestimate can be lowered. That is, suppose we find a neighbor w of v, with dist[v] > dist[w] + length(w, v). Then we have found an actual path that shows the distance estimate is too conservative. We therefore repeatedly apply the following update rule.
4-9
procedure update ( (w, v) ) edge (w, v) if dist[v] > dist[w]+ length(w, v) then dist[v] := dist[w] + length(w, v), prev[v] := w
A crucial observation is that this procedure is safe, in that it never invalidates our “invariant” that dist is a conservative overestimate. The key idea is to consider how these updates along edges should occur. In Djikstra’s algorithm, the edges are updated according to the time order of the imaginary explorers. But this only works with positive edge lengths. A second crucial observation concerns how many updates we have to do. Let a = s be a node, and consider the shortest path from s to a, say s, v1 , v2 , . . . , vk = a for some k between 1 and n − 1. If we perform update first on (s, v 1 ), later on (v1 , v2 ), and so on, and finally on (vk−1 , a), then we are sure that dist(a) contains the true distance from s to a, and that the true shortest path is encoded in prev. (Exercise: Prove this, by induction.) We must thus find a sequence of updates that guarantee that these edges are updated in this order. We don’t care if these or other edges are updated several times in between, since all we need is to have a sequence of updates that contains this particular subsequence. There is a very easy way to guarantee this: update all edges |V | − 1 times in a row!
4-10
Algorithm Shortest Paths 2 (G = (V, E, length); s ∈ V ) v, w: vertices dist: array[V ] of integer prev: array[V ] of vertices i: integer for v ∈ V do dist[v] := ∞, prev[v] :=nil rof dist[s] := 0 for i = 1 . . . n − 1 for (w, v) ∈ E update(w, v) end shortest paths 2
This algorithm solves the general single-source shortest path problem in O(|V | · |E|) time.
4-11
Negative Cycles
In fact, there is a further problem that negative edges can cause. Suppose the length of edge (b, a) in Figure 2 were changed to −5. The the graph would have a negative cycle (from a to b and back). On such graphs, it does not make sense to even ask the shortest path question. What is the shortest path from s to c in the modified graph? The one that goes directly from s to a to c (cost: 3), or the one that goes from s to a to b to a to c (cost: 1), or the one that takes the cycle twice (cost: -1)? And so on. The shortest path problem is ill-posed in graphs with negative cycles. It makes no sense and deserves no answer. Our algorithm in the previous section works only in the absence of negative cycles. (Where did we assume no negative cycles in our correctness argument? Answer: When we asserted that a shortest path from s to a exists!) But it would be useful if our algorithm were able to detect whether there is a negative cycle in the graph, and thus to report reliably on the meaningfulness of the shortest path answers it provides. This is easily done. After the |V | − 1 rounds of updates of all edges, do a last update. If any changes occur during this last round of updates, there is a negative cycle. This must be true, because if there were no negative cycles, |V | − 1 rounds of updates would have been sufficient to find the shortest paths.
4-12
Shortest Paths on DAG’s
There are two subclasses of weighted graphs that automatically exclude the possibility of negative cycles: graphs with non-negative weights and DAG’s. We have already seen that there is a fast algorithm when the weights are non-negative. Here we will give a linear algorithm for single-source shortest paths in DAG’s. Our algorithm is based on the same principle as our algorithm for negative weights. We are trying to find a sequence of updates, such that all shortest paths are its subsequences. But in a DAG we know that all shortest paths from s must go in the topological order of the DAG. All we have to do then is first topologically sort the DAG using a DFS, and then visit all edges coming out of nodes in the topological order. This algorithm solves the general single-source shortest path problem for DAG’s in O(m) time.
CS124
Lecture 5
Spring 2002
Minimum Spanning Trees
A tree is an undirected graph which is connected and acyclic. It is easy to show that if graph G(V, E) that satisfies any two of the following properties also satisfies the third, and is therefore a tree: • G(V, E) is connected • G(V, E) is acyclic • |E| = |V | − 1 (Exercise: Show that any two of the above properties implies the third (use induction).) A spanning tree in an undirected graph G(V, E) is a subset of edges T ⊆ E that are acyclic and connect all the vertices in V . It follows from the above conditions that a spanning tree must consist of exactly n − 1 edges. Now weights of its edges; w(T ) = ∑e∈T w(e). The minimum spanning tree in a weighted graph G(V, E) is one which has the smallest weight among all spanning trees in G(V, E). As an example of why one might want to find a minimum spanning tree, consider someone who has to install the wiring to network together a large computer system. The requirement is that all machines be able to reach each other via some sequence of intermediate connections. By representing each machine as a vertex and the cost of wiring two machines together by a weighted edge, the problem of finding the minimum cost wiring scheme reduces to the minimum spanning tree problem. In general, the number of spanning trees in G(V, E) grows exponentially in the number of vertices in G(V, E). (Exercise: Try to determine the number of different spanning trees for a complete graph on n vertices.) Therefore it is infeasible to search through all possible spanning trees to find the lightest one. Luckily it is not necessary to examine all possible spanning trees; minimum spanning trees satisfy a very important property which makes it possible to efficiently zoom in on the answer. suppose that each edge has a weight associated with it: w : E → Z. Say that the weight of a tree T is the sum of the
5-1
Lecture 5
5-2
We shall construct the minimum spanning tree by successively selecting edges to include in the tree. We will guarantee after the inclusion of each new edge that the selected edges, X, form a subset of some minimum spanning tree, T . How can we guarantee this if we don’t yet know any minimum spanning tree in the graph? The following property provides this guarantee: Cut property: Let X ⊆ T where T is a MST in G(V, E). Let S ⊂ V such that no edge in X crosses between S and V − S; i.e. no edge in X has one endpoint in S and one endpoint in V − S. Among edges crossing between S and V − S, let e be an edge of minimum weight. Then X ∪ {e} ⊆ T where T is a MST in G(V, E). The cut property says that we can construct our tree greedily. Our greedy algorithms can simply take the minimum weight edge across two regions not yet connected. Eventually, if we keep acting in this greedy manner, we will arrive at the point where we have a minimum spanning tree. Although the idea of acting greedily at each point may seem quite intuitive, it is very unusual for such a strategy to actually lead to an optimal solution, as we will see when we examine other problems! Proof: Suppose e ∈ T . Adding e into T creates a unique cycle. We will remove a single edge e from this / unique cycle, thus getting T = T ∪ {e} − {e }. It is easy to see that T must be a tree — it is connected and has n − 1 edges. Furthermore, as we shall show below, it is always possible to select an edge e in the cycle such that it crosses between S and V − S. Now, since e is a minimum weight edge crossing between S and V − S, w(e ) ≥ w(e). Therefore w(T ) = w(T ) + w(e) − w(e ) ≤ w(T ). However since T is a MST, it follows that T is also a MST and w(e) = w(e ). Furthermore, since X has no edge crossing between S and V − S, it follows that X ⊆ T and thus X ∪ {e} ⊆ T . How do we know that there is an edge e = e in the unique cycle created by adding e into T , such that e crosses between S and V − S? This is easy to see, because as we trace the cycle, e crosses between S and V − S, and we must cross back along some other edge to return to the starting point.
Lecture 5
5-3
In light of this, the basic outline of our minimum spanning tree algorithms is going to be the following: X := { }. Repeat until |X| = n − 1. Pick a set S ⊆ V such that no edge in X crosses between S and V − S. Let e be a lightest edge in G(V, E) that crosses between S and V − S. X := X ∪ {e}.
The difference between minimum spanning tree algorithms lies in how we pick the set S at each step.
Lecture 5
5-4
Prim’s algorithm:
In the case of Prim’s algorithm, X consists of a single tree, and the set S is the set of vertices of that tree. One way to think of the algorithm is that it grows a single tree, adding a new vertex at each step, until it has the minimum spanning tree. In order to find the lightest edge crossing between S and V − S, Prim’s algorithm maintains a heap containing all those vertices in V − S which are adjacent to some vertex in S. The priority of a vertex v, according to which the heap is ordered, is the weight of its lightest edge to a vertex in S. This is reminiscent of Dijkstra’s algorithm (where distance was used for the heap instead of the edge weight). As in Dijkstra’s algorithm, each vertex v will also have a parent pointer prev(v) which is the other endpoint of the lightest edge from v to a vertex in S. The pseudocode for Prim’s algorithm is almost identical to that for Dijkstra’s algorithm: Procedure Prim(G(V, E), s) v, w: vertices dist: array[V ] of integer prev: array[V ] of vertices S: set of vertices, initially empty H: priority heap of V H := {s : 0} for v ∈ V do dist[v] := ∞, prev[v] :=nil rof dist[s] := 0 / while H = 0 v := deletemin(h) S := S ∪ {v} for (v, w) ∈ E and w ∈ V − S do if dist[w] > length(v, w) dist[w] := length(v, w), prev[w] := v, insert(w,dist[w], H) fi rof end while end Prim
Note that each vertex is “inserted” on the heap at most once; other insert operations simply change the value on the heap. The vertices that are removed from the heap form the set S for the cut property. The set X of edges chosen to be included in the MST are given by the parent pointers of the vertices in the set S. Since the smallest key in the heap at any time gives the lightest edge crossing between S and V − S, Prim’s algorithm follows the generic outline for a MST algorithm presented above, and therefore its correctness follows from the cut property. The running time of Prim’s algorithm is clearly the same as Dijkstra’s algorithm, since the only change is how we prioritize nodes in the heap. Thus, if we use d-heaps, the running time of Prim’s algorithm is O(m log m/n n).
Lecture 5
5-5
Kruskal’s algorithm:
Kruskal’s algorithm uses a different strategy from Prim’s algorithm. Instead of growing a single tree, Kruskal’s algorithm attempts to put the lightest edge possible in the tree at each step. Kruskal’s algorithm starts with the edges sorted in increasing order by weight. Initially X = { }, and each vertex in the graph regarded as a trivial tree (with no edges). Each edge in the sorted list is examined in order, and if its endpoints are in the same tree, then the edge is discarded; otherwise it is included in X and this causes the two trees containing the endpoints of this edge to merge into a single tree. Note that, by this process, we are implicitly choosing a set S ⊆ V with no edge in X crossing between S and V − S, so this fits in our basic outline of a minimum spanning tree algorithm. To implement Kruskal’s algorithm, given a forest of trees, we must decide given two vertices whether they belong to the same tree. For the purposes of this test, each tree in the forest can be represented by a set consisting of the vertices in that tree. We also need to be able to update our data structure to reflect the merging of two trees into a single tree. Thus our data structure will maintain a collection of disjoint sets (disjoint since each vertex is in exactly one tree), and support the following three operations:
• MAKESET(x): Create a new x containing only the element x. • FIND(x): Given an element x, which set does it belong to? • UNION(x,y): replace the set containing x and the set containing y by their union. The pseudocode for Kruskal’s algorithm follows: Function Kruskal(graph G(V, E)) set X X ={} E:= sort E by weight for u ∈ V MAKESET(u) rof for (u, v) ∈ E (in increasing order) do if FIND(u) = FIND(v) do X = X ∪ {(u, v)} UNION(u,v) rof return(X) end Kruskal
Lecture 5
5-6
The correctness of Kruskal’s algorithm follows from the following argument: Kruskal’s algorithm adds an edge e into X only if it connects two trees; let S be the set of vertices in one of these two trees. Then e must be the first edge in the sorted edge list that has one endpoint in S and the other endpoint in V − S, and is therefore the lightest edge that crosses between S and V − S. Thus the cut property of MST implies the correctness of the algorithm. The running time of the algorithm, assuming the edges are given in sorted order, is dominated by the set operations: UNION and FIND. There are n − 1 UNION operations (one corresponding to each edge in the spanning UNION). We will soon show that this is O(m log ∗ n). Note that, if the edges are not initially given in sorted order, then to sort them in the obvious way takes O(m log m) time, and this would be the dominant part of the running time of the algorithm. tree), and 2m FIND operations (2 for each edge). Thus the total time of Kruskal’s algorithm is O(m × FIND + n ×
Lecture 5
5-7
Exchange Property
Actually spanning trees satisfy an even stronger property than the cut property — the exchange property. The exchange property is quite remarkable since it implies that we can “walk” from any spanning tree T to a minimum ˆ spanning tree T by a sequence of exchange moves — each such move consists of throwing an edge out of the current ˆ ˆ tree that is not in T , and adding a new edge into the current tree that is in T . Moreover, each successive tree in the “walk” is guaranteed to weigh no more than its predecessor. Exchange property: Let T and T be spanning trees in G(V, E). Given any e ∈ T − T , there exists an edge e ∈ T − T such that (T − {e}) ∪ {e } is also a spanning tree. The proof is quite similar to that of the cut property. Adding e into T results in a unique cycle. There must be some edge in this cycle that is not in T (since otherwise T must have a cycle). Call this edge e. Then deleting e restores a spanning tree, since connectivity is not affected, and the number of edges is restored to n − 1. To see how one may use this exchange property to “walk” from any spanning tree to a MST: let T be any ˆ spanning tree and let T be a MST in G(V, E). Let e be the lightest edge that is not in both trees. Perform an exchange using this edge. Since the exchange was done with the lightest such edge, the new tree must be lighter than ˆ the old one. Since T is already a MST, it follows that the exchange must have been performed upon T and results in ˆ a lighter spanning tree which has more edges in common with T (if there are several edges of the same weight, then ˆ the new tree might not be lighter, but it still has more edges in common with T ).
Lecture 5
5-8
1 3 4 2 3 5 5
5 2 1 7 6
Figure 5.1: An example of Prim’s algorithm and Kruskal’s algorithm. Which is which?
CS124
Lecture 6
Spring 2002
Disjoint set (Union-Find)
For Kruskal’s algorithm for the minimum spanning tree problem, we found that we needed a data structure for maintaining a collection of disjoint sets. That is, we need a data structure that can handle the following operations:
• MAKESET(x) - create a new set containing the single element x • UNION(x, y) - replace two sets containing x and y by their union. • FIND(x) - return the name of the set containing the element x Naturally, this data structure is useful in other situations, so we shall consider its implementation in some detail. Within our data structure, each set is represented by a tree, so that each element points to a parent in the tree. The root of each tree will point to itself. In fact, we shall use the root of the tree as the name of the set itself; hence the name of each set is given by a canonical element, namely the root of the associated tree. It is convenient to add a fourth operation LINK(x, y) to the above, where we require for LINK that x and y are two roots. LINK changes the parent pointer of one of the roots, say x, and makes it point to y. It returns the root of the now composite tree y. With this addition, we have UNION(x, y) = LINK(FIND(x),FIND(y)), so the main problem is to arrange our data structure so that FIND operations are very efficient.
6-1
Lecture 6
6-2
Notice that the time to do a FIND operation on an element corresponds to its depth in the tree. Hence our goal is to keep the trees short. Two well-known heuristics for keeping trees short in this setting are UNION BY RANK and PATH COMPRESSION. We start with the UNION BY RANK heuristic. The idea of UNION BY RANK is to ensure that when we combine two trees, we try to keep the overall depth of the resulting tree small. This is implemented as follows: the rank of an element x is initialized to 0 by MAKESET. An element’s rank is only updated by the LINK operation. If x and y have the same rank r, then invoking LINK(x, y) causes the parent pointer of x to be updated to point to y, and the rank of y is then updated to r + 1. On the other hand, if x and y have different rank, then when invoking LINK(x, y) the parent point of the element with smaller rank is updated to point to the element with larger rank. The idea is that the rank of the root is associated with the depth of the tree, so this process keeps the depth small. (Exercise: Try some examples by hand with and without using the UNION BY RANK heuristic.) The idea of PATH COMPRESSION is that, once we perform a FIND on some element, we should adjust its parent pointer so that it points directly to the root; that way, if we ever do another FIND on it, we start out much closer to the root. Note that, until we do a FIND on an element, it might not be worth the effort to update its parent pointer, since we may never access it at all. Once we access an item, however, we must walk through every pointer to the root, so modifying the pointers only changes the cost of this walk by a constant factor.
Lecture 6
6-3
procedure MAKESET(x) p(x) := x rank(x) := 0 end
function FIND(x) if x = p(x) then p(x) := FIND(p(x)) return(p(x)) end
function LINK(x, y) if rank(x) > rank(y) then x ↔ y if rank(x) = rank(y) then rank(y) := rank(y) + 1 p(x) := y return(y) end
Lecture 6
6-4
In our analysis, we show that any sequence of m UNION and FIND operations on n elements take at most O((m + n) log∗ n) steps, where log∗ n is the number of times you must iterate the log 2 function on n before getting a number less than or equal to 1. (So log ∗ 4 = 2, log∗ 16 = 3, log ∗ 65536 = 4.) We should note that this is not the tightest analysis possible; however, this analysis is already somewhat complex! Note that we are going to do an amortized analysis here. That is, we are going to consider the cost of the algorithm over a sequence of steps, instead of considering the cost of a single operation. In fact a single UNION or FIND operation could require O(log n) operations. (Exercise: Prove this!) Only by considering an entire sequence of operations at once can obtain the above bound. Our argument will require some interesting accounting to total the cost of a sequence of steps.
Lecture 6
6-5
We first make a few observations about rank.
• if v = p(v) then rank(p(v)) > rank(v) • whenever p(v) is updated, rank(p(v)) increases • the number of elements with rank k is at most
n 2k n 2k−1
• the number of elements with rank at least k is at most
The first two assertions are immediate from the description of the algorithm. The third assertion follows from the fact that the rank of an element v changes only if LINK(v, w) is executed, rank(v) = rank(w), and v remains the root of the combined tree; in this case v’s rank is incremented by 1. A simple induction then yields that when rank(v) is incremented to k, the resulting tree has at least 2 k elements. The last assertion then follows from the third
n assertion, as ∑∞ 2 j = j=k n . 2k−1
Exercise: Show that the maximum rank an item can have is log n.
Lecture 6
6-6
As soon as an element becomes a non-root, its rank is fixed. Let us divide the (non-root) elements into groups according to their ranks. Group i contains all elements whose rank r satisfies log ∗ r = i. For example, elements in group 3 have ranks in the range (4, 16], and the range of ranks associated with group i is (2 i−1 , 22 ). For convenience we shall write this more simply by saying group (k, 2 k ] to mean the group with these ranks. It is easy to establish the following assertions about these groups: • The number of distinct groups is at most log ∗ n. (Use the fact that the maximum rank is log n.) • The number of elements in the group (k, 2 k ] is at most
n . 2k
i−1
Let us assign 2k tokens to each element in group (k, 2 k ]. The total number of tokens assigned to all elements
n from that group is then at most 2k 2k = n, and the total number of groups is at most log ∗ n, so the total number of
tokens given out is n log ∗ n. We use these tokens to account for the work done by FIND operations. Recall that the number of steps for a FIND operation is proportional to the number of pointers that the FIND operation must follow up the tree. We separate the pointers into two groups, depending on the groups of u and p(u) = v, as follows: • Type 1: a pointer is of Type 1 if u and v belong to different groups, or v is the root. • Type 2: a pointer is of Type 2 if u and v belong to the same group. We account for the two Types of pointers in two different ways. Type 1 links are “charged” directly to the FIND operation; Type 2 links are “charged” to u, who “pays” for the operation using one of the tokens. Let us consider these charges more carefully.
Lecture 6
6-7
The number of Type 1 links each FIND operation goes through is at most log ∗ n, since there are only log ∗ n groups, and the group number increases as we move up the tree. What about Type 2 links? We charge these links directly back to u, who is supposed to pay for them with a token. Does u have enough tokens? The point here is that each time a FIND operation goes through an element u, its parent pointer is changed to the current root of the tree (by PATH COMPRESSION), so the rank of its parent increases by at least 1. If u is in the group (k, 2 k ], then the rank of u’s parent can increase fewer than 2 k times before it moves to a higher group. Therefore the 2 k tokens we assign to u are sufficient to pay for all FIND operations that go through u to a parent in the same group.
Lecture 6
6-8
We now count the total number of steps for m UNION and FIND operations. Clearly LINK requires just O(1) steps, and since a UNION operation is just a LINK and 2 FIND operations, it suffices to bound the time for at most 2m FIND OPERATIONS. Each FIND operation is charged at most log ∗ n for a total of O(m log∗ n). The total number of tokens used at most n log ∗ n, and each token pays for a constant number of steps. Therefore the total number of steps is O((m + n) log∗ n). Let us give a more equation-oriented explanation. The total time spent over the course of m UNION and FIND operations is just all FIND ops We split this sum up into two parts:
all FIND ops
(# links in same group) +
all FIND ops
(Technically, the case where a link goes to the root should be handled explicitly; however, this is just O(m) links in total, so we don’t need to worry!) The second term is clearly O(m log ∗ n). The first term can be upper bounded by:
all elements u because each element u can be charged only once for each rank in its group. (Note here that this is because the links to the root count in the second sum!) This last sum is bounded above by
(# ranks in the group of u),
all groups This completes the proof.
log∗ n
(# items in group) · (# ranks in group) ≤
k=1
n k 2 ≤ n log∗ n. 2k
Lecture 6
6-9
x
y UNION(x,y) x
y
a b c FIND(d)
a
b
c
d
d
Figure 6.1: Examples of UNION BY RANK and PATH COMPRESSION.
CS124
Lecture 7
In today’s lecture we will be looking a bit more closely at the Greedy approach to designing algorithms. As we will see, sometimes it works, and sometimes even when it doesn’t, it can provide a useful result.
Horn Formulae
A simple application of the greedy paradigm solves an important special case of the SAT problem. We have already seen that 2SAT can be solved in linear time. Now consider SAT instances where in each clause, there is at most one positive literal. Such formulae are called Horn formulae; for example, this is an instance: (x ∨ y ∨ z ∨ w) ∧ (x ∨ y ∨ w) ∧ (x ∨ z ∨ w) ∧ (x ∨ y) ∧ (x) ∧ (z) ∧ (x ∨ y ∨ w). Given a Horn formula, we can separate its clauses into two parts: the pure negative clauses (those without a positive literal) and the implications (those with a positive literal). We call clauses with a positive literal implications because they can be rewritten suggestively as implications; (x ∨ y ∨ z ∨ w) is equivalent to (y ∧ z ∧ w) → x. Note the trivial clause (x) can be thought of as a trivial implication → x. Hence, in the example above, we have the implications (y ∧ z ∧ w → x), (x ∧ z → w), (x → y), (→ x), (x ∧ y → w) and these two pure negative clauses (x ∨ y ∨ w), (z). We can now develop a greedy algorithm. The idea behind the algorithm is that we start with all variables set to false, and we only set variables to T if an implication forces us to. Recall that an implication is not satisfied if all variables to the left of the arrow are true and the one to the right is false. This algorithm is greedy, in the sense that it (greedily) tries to ensure the pure negative clauses are satisfied, and only changes a variable if absolutely forced.
7-1
Lecture 7
7-2
Algorithm Greedy-Horn(φ: CNF formula with at most one positive literal per clause) Start with the truth assignment t :=FFF· · ·F while there is an implication that is not satisfied do make the implied variable T in t if all pure negatives are satisfied then return t else return “φ is unsatisfiable” Once we have the proposed truth assignment, we look at the pure negatives. If there is a pure negative clause that is not satisfied by the proposed truth assignment, the formula cannot be satisfied. This follows from the fact that all the pure negative clauses will be satisfied if any of their variables are set to F. If such a clause is unsatisfied, all of its variables must be set to T. But we only set a variable to T if we are forced to by the implications. If all the pure negative clauses are satisfied, then we have found a truth assignment. On the example above, Greedy-Horn first flips x to true, forced by the implication → x. Then y gets forced to true (from x → y), and similarly w is forced to true. (Why?) Looking at the pure negative clauses, we find that the first is not satisfied, and hence we conclude the original formula had no truth assignment. Exercise: Show that the Horn-greedy algorithm can be implemented in linear time in the length of the formula (i.e., the total number of appearances of all literals).
Lecture 7
7-3
Huffman Coding
Suppose that you must store the map of a chromosome which consists of a sequence of 130 million symbols of the form A, C, G, or T. To store the sequence efficiently, you represent each character with just 2 bits: A as 00, C as 01, G as 10, and T as 11. Such a representation is called an encoding. With this encoding, the sequence requires 260 megabits to store. Suppose, however, that you know that some symbols appear more frequently than others. For example, suppose A appears 70 million times, C 3 million times, G 20 million times, and T 37 million times. In this case it seems wasteful to use two bits to represent each A. Perhaps a more elaborate encoding assigning a shorter string to A could save space. We restrict ourselves to encodings that satisfy the prefix property: no assigned string is the prefix of another. This property allows us to avoid backtracking while decoding. For an example without the prefix property, suppose we represented A as 1 and C as 101. Then when we read a 1, we would not know whether it was an A or the beginning of a C! Clearly we would like to avoid such problems, so the prefix property is important. You can picture an encoding with the prefix property as a binary tree. For example, the binary tree below corresponds to an optimal encoding in the above situation. (There can be more than one optimal encoding! Just flip the left and right hand sides of the tree.) Here a branch to the left represent a 0, and a branch to the right represents a 1. Therefore A is represented by 1, C by 001, G by 000, and T by 01. This encoding requires only 213 million bits – a 17% improvement over the balanced tree (the encoding 00,01,10,11). (This does not include the bits that might be necessary to store the form of the encoding!)
Lecture 7
7-4
0 (60)
1 A (70)
0 (23)
1 T (37)
0
1
G (20)
C (3)
Figure 7.1: A Huffman tree.
Lecture 7
7-5
Let us note some properties of the binary trees that represent encoding. The symbols must correspond to leaves; an internal node that represents a character would violate the prefix property. The code words are thus given by all root-to-leaf paths. All internal nodes must have exactly two children, as an internal node with only one child could be deleted to yield a better code. Hence if there are n leaves there are n − 1 internal edges. Also, if we assign frequencies to the internal nodes, so that the frequencies of an internal node are the sums of the frequencies of the children, then the total length produced by the encoding is the sum of the frequencies of all nodes except the root. (A one line proof: each edge corresponds to a bit that is written as many times as the frequency of the node to which it leads.) One final property allows us to determine how to build the tree: the two symbols with the smallest frequencies are together at the lowest level of the tree. Otherwise, we could improve the encoding by swapping a more frequently used character at the lowest level up. (This is not a full proof; feel free to complete one.) This tells us how to construct the optimum tree greedily. Take the two symbols with the lowest frequency, delete them from the list of symbols, and replace them with a new meta-character; this new meta-character will lie directly above the two deleted symbols in the tree. Repeat this process until the whole tree is constructed. We can prove by induction that this gives an optimal tree. It works for 2 symbols (base case). We also show that if it works for n letters, it must also work for n + 1 letters. After deleting the two least frequent symbols and replacing them with a meta-character, it as though we have just n symbols. this process yields an optimal tree for these n symbols (by the inductive hypothesis). Expanding the meta-character back into the two deleted nodes must now yield an optimal tree, since otherwise we could have found a better tree for the n symbols.
Lecture 7
7-6
A 60 E 70 I 40 O 50 U 20 Y 30
A 60 E 70 I 40 O 50 [UY] 50
A 60 E 70 O 50 [I[UY]] 90 I
[OA] 110 O E 70 [I[UY]] 90 I A
U
Y
U
Y
U
Y
Figure 7.2: The first few steps of building a Huffman tree.
Lecture 7
7-7
It is important to realize that when we say that a Huffman tree is optimal, this does not mean that it gives the best way to compress a string. It only says that we cannot do better by encoding one symbol at a time. By encoding frequently used blocks of letters (such as, in this section, the block “frequen”) we can obtain much better encodings. (Note that finding the right blocks of letters can be quite complicated.) Given this, one might expect that Huffman coding is rarely used. In fact, many compression schemes use Huffman codes at some point as a basic building block. For example, image and video transmission often use Huffman encoding somewhere along the line. Exercise: Find a Huffman compressor and another compressor, such as a gzip compressor. Test them on some files. Which compresses better? It is straightforward to write code to generate the appropriate tree, and then use this tree to encode and decode messages. For encoding, we simply build a table with a codeword for each sybmol. To decode, we could read bits in one at a time, and walk down the tree in the appropriate manner. When we reach a leaf, we output the appropriate symbol and return to the top of the tree. In practice, however, if we want to use Huffman coding, there are much faster ways to decode than to explicitly walk down the tree one bit at a time. Using an explicit tree is slow, for a variety of reasons. Exercise: Think about this. One approach is to design a system that performs several steps at a time by reading several bits of input and determining what actions to take according to a big lookup table. For example, we could have a table that represents the information, “If you are currently at this point in the tree, and the next 8 bits are 00110101, then output AC and move to this point in the tree.” This lookup table, which might be huge, encapsulates the information needed to handle eight bits at once. Since computers naturally handle eight bit blocks more easily than single bits, and because table lookups are faster than following pointers down a Huffman tree, substantial speed gains are possible. Notice that this gain in speed comes at the expense of the space required for the lookup table. There are other solutions that work particularly well for very large dictionaries. For example, if you were using Huffman codes on a libraray of newspaper articles, you might treat each work as a symbol that can be encoded. In this case, you would have a lot of symbols! We will not go over these other methods here; a useful paper on the subject is “On the Implementation of Minimum-Redundancy Prefix Codes,” by Moffat and Turpin. The key to keep in mind is that while thinking of decoding on the Huffman tree as happening one bit at a time is useful conceptually, good engineering would use more sophisticated methods to increase efficiency.
Lecture 7
7-8
The Set Cover Problem
The inputs to the set cover problem are a finite set X = {x 1 , . . . , xn }, and a collection of subsets S of X such that
S∈S
S = X. The problem is to find the subcollection T ⊆ S such that the sets of T cover X, that is T = X.
T ∈T
Notice that such a cover exists, since S is itself a cover. The greedy heuristic suggests that we build a cover by repeatedly including the set in S that will cover the maximum number of as yet uncovered elements. In this case, the greedy heuristic does not yield an optimal solution. Interestingly, however, we can prove that the greedy solution is a good solution, in the sense that it is not too far from the optimal. This is an example of an approximation algorithm. Loosely speaking, with an approximation algorithm, we settle for a result that is not the correct answer. Instead, however, we try to prove a guarantee on how close the algorithm is to the right answer. As we will see later in the course, sometimes this is the best we can hope to do.
Lecture 7
7-9
Claim 7.1 Let k be the size of the smallest set cover for the instance (X, S ). Then the greedy heuristic finds a set cover of size at most k ln n.
Proof: Let Yi ⊆ X be the set of elements that are still not covered after i sets have been chosen with the greedy heuristic. Clearly Y0 = X. We claim that there must be a set A ∈ S such that |A ∩Yi | ≥ |Yi |/k. To see this, consider the sets in the optimal set cover of X. These sets cover Yi , and there are only k of them, so one of these sets must cover at least a 1/k fraction of Yi . Hence |Yi+1 | ≤ |Yi | − |Yi |/k = (1 − 1/k)|Yi |, and by induction, |Yi | ≤ (1 − 1/k)i |Y0 | = n(1 − 1/k)i < ne−i/k , where the last inequality uses the fact that 1 + x ≤ e x with equality iff x = 0. Hence when i ≥ k ln n we have |Yi | < 1, meaning there are no uncovered elements, and hence the greedy algorithm finds a set cover of size at most k ln n. Exercise: Show that this bound is tight, up to constant factors. That is, give a family of examples where the set cover has size k and the greedy algorithm finds a cover of size Ω(k ln n).
CS124
Lecture 8
Spring 2000
Divide and Conquer
We have seen one general paradigm for finding algorithms: the greedy approach. We now consider another general paradigm, known as divide and conquer. We have already seen an example of divide and conquer algorithms: mergesort. The idea behind mergesort is to take a list, divide it into two smaller sublists, conquer each sublist by sorting it, and then combine the two solutions for the subproblems into a single solution. These three basic steps – divide, conquer, and combine – lie behind most divide and conquer algorithms. With mergesort, we kept dividing the list into halves until there was just one element left. In general, we may divide the problem into smaller problems in any convenient fashion. Also, in practice it may not be best to keep dividing until the instances are completely trivial. Instead, it may be wise to divide until the instances are reasonably small, and then apply an algorithm that is fast on small instances. For example, with mergesort, it might be best to divide lists until there are only four elements, and then sort these small lists quickly by insertion sort.
8-1
Lecture 8
8-2
Maximum/minimum
Suppose we wish to find the minimum and maximum items in a list of numbers. How many comparisons does it take? A natural approach is to try a divide and conquer algorithm. Split the list into two sublists of equal size. (Assume that the initial list size is a power of two.) Find the maxima and minima of the sublists. Two more comparisons then suffice to find the maximum and minimum of the list. Hence, if T (n) is the number of comparisons, then T (n) = 2T (n/2) + 2. (The 2T (n/2) term comes from conquering the two problems into which we divide the original; the 2 term comes from combining these solutions.) Also, clearly T (2) = 1. By induction we find T (n) = (3n/2) − 2, for n a power of 2.
Lecture 8
8-3
Integer Multiplication
The standard multiplication algorithm takes time Θ(n 2 ) to multiply together two n digit numbers. This algorithm is so natural that we may think that no algorithm could be better. Here, we will show that better algorithms exist (at least in terms of asymptotic behavior). Imagine splitting each number x and y into two parts: x = 10 n/2 a + b, y = 10n/2 c + d. Then xy = 10n ac + 10n/2 (ad + bc) + bd. The additions and the multiplications by powers of 10 (which are just shifts!) can all be done in linear time. We have therefore reduced our multiplication problem into four smaller multiplications problems, so the recurrence for the time T (n) to multiply two n-digit numbers becomes T (n) = 4T (n/2) + O(n). The 4T (n/2) term arises from conquering the smaller problems; the O(n) is the time to combine these problems into the final solution (using additions and shifts). Unfortunately, when we solve this recurrence, the running time is still Θ(n2 ), so it seems that we have not gained anything.
Lecture 8
8-4
The key thing to notice here is that four multiplications is too many. Can we somehow reduce it to three? It may not look like it is possible, but it is using a simple trick. The trick is that we do not need to compute ad and bc separately; we only need their sum ad + bc. Now note that (a + b)(c + d) = (ad + bc) + (ac + bd). So if we calculate ac , bd, and (a + b)(c + d), we can compute ad + bc by the subtracting the first two terms from the third! Of course, we have to do a bit more addition, but since the bottleneck to speeding up this multiplication algorithm is the number of smaller multiplications required, that does not matter. The recurrence for T (n) is now T (n) = 3T (n/2) + O(n), and we find that T (n) = nlog2 3 ≈ n1.59 , improving on the quadratic algorithm. If one were to implement this algorithm, it would probably be best not to divide the numbers down to one digit. The conventional algorithm, because it uses fewer additions, is probably more efficient for small values of n. Moreover, on a computer, there would be no reason to continue dividing once the length n is so small that the multiplication can be done in one standard machine multiplication operation! It also turns out that using a more complicated algorithm (based on a similar idea) the asymptotic time for multiplication can be made arbitrarily close to linear– that is, for any ε > 0 there is an algorithm that runs in time O(n1+ε ).
Lecture 8
8-5
Strassen’s algorithm
Divide and conquer algorithms can similarly improve the speed of matrix multiplication. Recall that when multiplying two matrices, A = ai j and B = b jk , the resulting matrix C = cik is given by cik = ∑ ai j b jk .
j
In the case of multiplying together two n by n matrices, this gives us an Θ(n 3 ) algorithm; computing each cik takes Θ(n) time, and there are n2 entries to compute. Let us again try to divide up the problem. We can break each matrix into four submatrices, each of size n/2 by n/2. Multiplying the original matrices can be broken down into eight multiplications of the submatrices, with some additions.
A B C D
E
F
G H
=
AE + BG
AF + BH
CE + DG CF + DH
Letting T (n) be the time to multiply together two n by n matrices by this algorithm, we have T (n) = 8T (n/2) + Θ(n2 ). Unfortunately, this does not improve the running time; it is still Θ(n 3 ).
Lecture 8
8-6
As in the case of multiplying integers, we have to be a little tricky to speed up matrix multiplication. (Strassen deserves a great deal of credit for coming up with this trick!) We compute the following seven products: • P1 = A(F − H) • P2 = (A + B)H • P3 = (C + D)E • P4 = D(G − E) • P5 = (A + D)(E + H) • P6 = (B − D)(G + H) • P7 = (A −C)(E + F) Then we can find the appropriate terms of the product by addition: • AE + BG = P5 + P4 − P2 + P6 • AF + BH = P1 + P2 • CE + DG = P3 + P4 • CF + DH = P5 + P1 − P3 − P7 Now we have T (n) = 7T (n/2) + Θ(n2 ), which give a running time of T (n) = Θ(n log 7 ). Faster algorithms requiring more complex splits exist; however, they are generally too slow to be useful in practice. Strassen’s algorithm, however, can improve the standard matrix multiplication algorithm for reasonably sized matrices, as we will see in our second programming assignment.
CS124
Lecture 9
Spring 2000
9.1 The String reconstruction problem
The greedy approach doesn’t always work, as we have seen. It lacks flexibility; if at some point, it makes a wrong choice, it becomes stuck. For example, consider the problem of string reconstruction. Suppose that all the blank spaces and punctuation marks inadvertantly have been removed from a text file. You would like to reconstruct the file, using a dictionary. (We will assume that all words in the file are standard English.) For example, the string might begin “thesearethereasons”. A greedy algorithm would spot that the first two words were “the” and “sea”, but then it would run into trouble. We could backtrack; we have found that sea is a mistake, so looking more closely, we might find the first three words “the”,“sear”, and “ether”. Again there is trouble. In general, we might end up spending exponential time traveling down false trails. (In practice, since English text strings are so well behaved, we might be able to make this work– but probably not in other contexts, such as reconstructing DNA sequences!)
9-1
Lecture 9
9-2
This problem has a nice structure, however, that we can take advantage of. The problem can be broken down into entirely similar subproblems. For example, we can ask whether the strings “theseare” and “thereasons” both can be reconstructed with a dictionary. If they can, then we can glue the reconstructions together. Notice, however, that this is not a good problem for divide and conquer. The reason is that we do not know where the right dividing point is. In the worst case, we could have to try every possible break! The recurrence would be
n−1
T (n) =
i=1
∑ T (i) + T (n − i).
You can check that the solution to this recurrence grows exponentially. Although divide and conquer directly fails, we still want to make use of the subproblems. The attack we now develop is called dynamic programming. The way to understand dynamic programming is to see that divide and conquer fails because we might recalculate the same thing over and over again. (Much like we saw very early on with the Fibonacci numbers!) If we try divide and conquer, we will repeatedly solve the same subproblems (the case of small substrings) over and over again. The key will be to avoid the recalculations. To avoid recalculations, we use a lookup table.
Lecture 9
9-3
In order for this approach to be effective, we have to think of subproblems as being ordered by size. We solve the subproblems bottom-up, from the smallest to the largest, until we reach the original problem. For this dictionary problem, think of the string as being an array s[1 . . . n]. Then there is a natural subproblem for each substring s[i . . . j]. Consider a two dimensional array D(i, j) that will denote whether s[i . . . j] is the concatenation of words from the dictionary. The size of a subproblem is naturally d = j − i. So now we write a simple loops which solves the subprobelms in order of increasing size: for d := 1 to n − 1 do for i := 1 to n − d do j := i + d; if indict(s[i . . . j]) then D(i, j) := true else for k := i + 1 to j − 1 do if D(i, k) and D(k, j) then D(i, j) := true;
This algorithm runs in time O(n3 ); the three loops each run over at most n values. Pictorially, we can think of the algorithm as filling in the upper diagonal triangle of a two-dimensional array, starting along the main diagonal and moving up, diagonal by diagonal. We need to add a bit to actually find the words. Let F(i, j) be the position of end of the first word in s[i . . . j] when this string is a proper concatenation of dictionary words. Initially all F(i, j) should be set to nil. The value for F(i, j) can be set whenever D(i, j) is set to true. Given the F(i, j), we can reconstruct the words simply by finding the words that make up the string in order. Note also that we can use this to improve the running time; as soon as we find a match for the entire string, we can exit the loop and return success! Further optimizations are possible. Let us highlight the aspects of the dynamic programming approach we used. First, we used a recursive description based on subproblems: D(i, j) is true if D(i, k) and D(k, j) for some k. Second, we built up a table containing the answers of the problems, in some natural bottom-up order. Third, we used this table to find a way to determine the actual solution. Dynamic programming generally involves these three steps.
Lecture 9
9-4
9.2 Edit distance
A problem that arises in biology is to measure the distance between two strings (of DNA). We will examine the problem in English; the ideas are the same. There are many possible meanings for the distance between two strings; here we focus on one natural measure, the edit distance. The edit distance measures the number of editing operations it would be necessary to perform to transform the first string into the second. The possible operations are as follows: • Insert: Insert a character into the first string. • Delete: Delete a character from the first string. • Replace: Replace a character from the first string with another character. Another possibility is to not edit a character, when there is a Match. For example, a transformation from activate to caveat can be represented by D a M c c R t a D i M v v e I M a a M t t D e
The top line represents the operation performed. So the a in activate id deleted, and the t is replaced. The e in caveat is explicitly inserted. The edit distance is the minimal number of edit operations – that is, the number of Inserts, Deletes, or Replaces – necesary to transform one string to the other. Note that Matches do not count. Also, it is possible to have a weighted edit distance, if the different edit operations have different costs. We currently assume all operations have weight 1.
Lecture 9
9-5
We will show how compute the edit distance using dynamic programming. Our first step is to define appropriate subproblems. Let us reprsent our strings by A[1 . . . n] and B[1 . . . m]. Suppose we want to consider what we do with the last character of A. To determine that, we need to know how we might have transformed the first n − 1 characters of A. These n−1 characters might have transformed into any number of symbols of B, up to m. Similarly, to compute how we might have transformed the first n − 1 characters of A into some part of B, it makes sense to consider how we transformed the first n − 2 characters, and so on. This suggests the following submproblems: we will let D(i, j) represent the edit distance between A[1 . . . i] and B[1 . . . j]. We now need a recursive description of the subproblems in order to use dynamic programming. Here the recurrence is: D(i, j) = min[D(i − 1, j) + 1, D(i, j − 1) + 1, D(i − 1, j − 1) + I(i = j)]. In the above, I(i = j) represents the value 1 if i = j and 0 if i = j. We obtain the above expression by considering the possible edit operations available. Suppose our last operation is a Delete, so that we deleted the ith character of A to transform A[1 . . . i] to B[1 . . . j]. Then we must have transformed A[1 . . . i − 1] to B[1 . . . j], and hence the edit distance would be D(i − 1, j) + 1, or the cost of the transformation from A[1 . . . i − 1] to B[1 . . . j] plus one for the cost of the final Delete. Similarly, if the last operation is an Insert, the cost would be D(i, j − 1) + 1. The other possibility is that the last operation is a Replace of the ith character of A with the jth character of B, or a Match between these two characters. If there is a Match, then the two characters must be the same, and the cost is D(i − 1, j − 1). If there is a Replace, then the two characters should be different, and the cost is D(i − 1, j − 1) + 1. We combine these two cases in our formula, using D(i − 1, j − 1) + I(i = j). Our recurrence takes the minimum of all these possibilities, expressing the fact that we want the best possible choice for the final operation!
Lecture 9
9-6
It is worth noticing that our recursive description does not work when i or j is 0. However, these cases are trivial. We have D(i, 0) = i, since the only way to transform the first i characters of A into nothing is to delete them all. Similarly, D(0, j) = j.
Again, it is helpful to think of the computation of the D(i, j) as filling up a two-dimensional array. Here, we begin with the first column and first row filled. We can then fill up the rest of the array in various ways: row by row, column by column, or diagonal by diagonal! Besides computing the distance, we may want to compute the actual transformation. To do this, when we fill the array, we may also picture filling the array with pointers. For example, if the minimal distance for D(i, j) was obtained by a final Delete operation, then the cell (i, j) in the table should have a pointer to (i − 1, j). Note that a cell can have multiple pointers, if the minimum distance could have been achieved in multiple ways. Now any path back from (n, m) to (0, 0) corresponds to a sequence of operations that yields the minimum distance D(n, m), so the transformation can be found by following pointers. The total computation time and space required for this algorithm is O(nm).
Lecture 9
9-7
9.3 All pairs shortest paths
Let G be a graph with positive edge weights. We want to calculate the shortest paths between every pair of nodes. One way to do this is to run Dijkstra’s algorithm several times, once for each node. Here we develop a different dynamic programming solution. Our subproblems will be shortest paths using only nodes 1 . . . k as intermediate nodes. Of course when k equals the number of nodes in the graph, n, we will have solved the original problem. We let the matrix Dk [i. j] represent the length of the shortest path between i and j using intermediate nodes 1 . . . k. Initially, we set a matrix D0 with the direct distances between nodes, given by d i j . Then Dk is easily computed from the subproblems Dk−1 as follows:
Dk [i, j] = min(Dk−1 [i, j], Dk−1 [i, k] + Dk−1 [k, j]). The idea is the shortest path using intermediate nodes 1 . . . k either completely avoids node k, in which case it has the same length as Dk−1 [i, j]; or it goes through k, in which case we can glue together the shortest paths found from i to k and k to j using only intermediate nodes 1 . . . k − 1 to find it. It might seem that we need at least two matrices to code this, but in fact it can all be done in one loop. (Exercise: think about it!)
D = (di j ), distance array, with weights from all i to all j for k = 1 to n do for i = 1 to n do for j = 1 to n do D[i, j] = min(D[i, j], D[i.k] + D[k, j])
Note that again we can keep an auxiliary array to recall the actual paths. We simply keep track of the last intermediate node found on the path from i to j. We reconstruct the path by succesively reconstructing intermediate nodes, until we reach the ends.
Lecture 9
9-8
9.4 Traveling salesman problem
Suppose that you are given n cities and the distances d i j between them. The traveling salesman problem (TSP) is to find the shortest tour that takes you from your home city to all the other cities and back again. As there are (n − 1)! possible paths, this can clearly be done in O(n!) time by trying all possible paths. Of course this is not very efficient. Since the TSP is NP-complete, we cannot really hope to find a polynomial time algorithm. But dynamic programming gives us a much better algorithm than trying all the paths. The key is to define the appropriate subproblem. Suppose that we label our home city by the symbol 1, and other cities are labeled 2, . . . , n. In this case, we use the following: for a subset S of vertices including 1 and at least one other city, let C(S, j) be the shortest path that start at 1, visits all other nodes in S, and ends at j. Note that our subproblems here look slightly different: instead of finding tours, we are simply finding paths. The important point is that the shortest path from i to j through all the vertices in S consists of some shortest path from i to a vertex x, where x ∈ S − { j}, and the additional edge from x to j.
for all j do C({i, j}, j) := d1 j for s = 3 to n do % s is the size of the subset for all subsets S of {1, . . . , n} of size s containing 1 do for all j ∈ S, j = 1 do C(S, j) := mini= j,i∈S [C(S − { j}, i) + di j ] opt := min j=i C({1, . . . , n}, j) + d j1
The idea is to build up paths one node at a time, not worrying (at least temporarily) where they will end up. Once we have paths that go through all the vertices, it is easy to check the tours, since they consists of a shortest path through all the vertices plus an additional edge. The algorithm takes time O(n 2 2n ), as there are O(n2n ) entries in the table (one for each pair of set and city), and each takes O(n) time to fill. Of course we can add in structures so that we can actually find the tour as well. Exercise: Consider how memory-efficient you can make this algorithm.
CS124
Lecture 10
Spring 1999
How many people do there need to be in a room before with probability greater than 1/2 some two of them have the same birthday? (Assume birthdays are distributed uniformly at random.) Surprisingly, only 23. This is easily determined as follows: the probability the first two people have different birthdays is (1 − 1/365). The probability that the third person in the room then has a birthday different from the first two, given the first two people have different birthdays, is (1 − 2/365), and so on. So the probability that all of the first k people have different birthdays is the product of these terms, or (1 − 2 3 k−1 1 ) · (1 − ) · (1 − ) . . . · (1 − ). 365 365 365 365
Determining the right value of k is now a simple exercise.
10-1
Lecture 10
10-2
10.2 Balls into Bins
Mathematically, the birthday paradox is an example of a more general mathematical question, often formulated in terms of balls and bins. Some number of balls n are thrown into some number of bins m. What does the distribution of balls and bins look like? The birthday paradox is focused on the first time a ball lands in a bin with another ball. One might also ask how many of the bins are empty, how many balls are in the most full bin, and other sorts of questions. Let us consider the question of how many bins are empty. Look at the first bin. For it to be empty, it has to be missed by all n balls. Since each ball hits the first bin with probability 1/m, the probability the first bin remains empty is 1 (1 − )n ≈ e−n/m . m Since the same argument holds for all bins, on average a fraction e −n/m of the bins will remain empty. Exercise: Howmany bins have 1 ball? 2?
Lecture 10
10-3
10.3 Hash functions
A hash function is a deterministic mapping from one set into another that appears random. For example, mapping people into their birthdays can be thought of as a hash function. In general, a hash function is a mapping f : {0, . . . , n − 1} → {0, . . . , m − 1}. Generally n >> m; for example, the number of people in the world in much bigger than the number of possible birthdays. There is a great deal of theory behind designing hash functions that “appear random.” We will not go into that theory here, and instead assume that the hash functions we have available are in fact completely random. In other words, we assume that for each i (0 ≤ i ≤ n − 1), the probability that f (i) = j is 1/m (for (0 ≤ j ≤ m − 1). Notice that this does mean that every time we look at f (i), we get a different random answer! The value of f (i) is fixed for all time; it is just equally likely to take on any value in the range. While such completely random hash functions are unavailable in practice, they generally provide a good rough idea of how hashing schemes perform. (An aside: in reality, birthdays are not completely random either. Seasonal distributions skew the calculation. How might this affect the birthday paradox?)
Lecture 10
10-4
h
.
The total space used is merely hm bits. Notice that the Bloom filter sometimes returns the wrong answer – we may reject a proposed password, even though it is not a common password. This sort of error is probably acceptable, as long as it doesn’t happen so frequently as to bother users. Fortunately this error is one-sided; a common password is never accepted. One must set the parameters m and h appropriately to trade off this error probability against space and time requirements. For example, consider a dictionary of 100,000 common passwords, each of which is on average 7 characters long. Uncompressed this would be 700,000 bytes. Compression might reduce it substantially, to around 300,000 bytes. Of course, then one has the problems of searching efficiently on a compressed list. Instead, one could keep a 100,000 byte Bloom filter, consisting of 5 tables of 160,000 bits. The probability of rejecting a reasonable password is just over 2%. The cost for checking a password is at most 5 hashes and 5 lookups into the table.
CS 124
Lecture 11
11.1 Applications: Fingerprinting for pattern matching
Suppose we are trying to find a pattern string P in a long document D. How can we do it quickly and efficiently? Hash the pattern P into say a 16 bit value. Now, run through the file, hashing each set of |P| consecutive characters into a 16 bit value. If we ever get a match for a pattern, we can check to see if it corresponds to an actual pattern match. (In this case, we want to double-check and not report any false matches!) Otherwise we can just move on. We can use more than 16 bits, too; we would like to use enough bits so that we will obtain few false matches. This scheme is efficient, as long as hashing is efficient. Of course hashing can be a very expensive operation, so in order for this approach to work, we need to be able to hash quickly on average. In fact, a simple hashing technique allows us to do so in constant time per operation! The easiest way to picture the process is to think of the file as a sequence of digits, and the pattern as a number. Then we move a pointer in the file one character at a time, seeing if the next |P| digits gives us a number equal to the number corresponding to the pattern. Each time we read a character in the file, the number we are looking at changes is a natural way: the leftmost digit a is removed, and a new rightmost digit b is inserted. Hence, we update an old number N and obtain a new number N by computing N = 10 · (N − 10|P|−1 · a) + b. When dealing with a string, we will be reading characters (bytes) instead of numbers. Also, we will not want to keep the whole pattern as a number. If the pattern is large, then the corresponding number may be too large to do effective comparisons! Instead, we hash all numbers down into say 16 bits, by reducing them modulo some appropriate prime p. We then do all the mathematics (multiplication, addition) modulo p, i.e. N = [10 · (N − 10|P|−1 · a) + b] mod p. All operations mod p can be made quite efficient, so each new hash value takes only constant time to compute! This pattern matching technique is often called fingerprinting. The idea is that the hash of the pattern creates an almost unique identifier for the pattern– like a fingerprint. If we ever find two fingerprints that match, we have a good reason to expect that they must come the same pattern. Of course, unlike real fingerprints, our hashing-based fingerprints do not actually uniquely identify a pattern, so we still need to check for false matches. But since false matches should be rare, the algorithm is very efficient! See Figure 11.1 for an example of fingerprinting.
11-1
Lecture 11
11-2
P = 17935 p = 251 P mod p = 114 6386179357342... 63861 mod p = 107 38617 mod p = 214 86179 mod p = 86 61793 mod p = 47 17935 mod p = 114 79357 mod p = 41 93573 mod p = 201 35734 mod p = 92 57342 mod p = 114
Figure 11.1: A fingerprinting example. The pattern P is a 5 digit number. Note successive calculations take constant time: 38617 mod p = ( (63861 mod p) - (60000 mod p)) · 10 + 7 mod p. Also note that false matches are possible (but unlikely); 57432 = 17935 mod p. One question remains. How should we choose the prime p? We would like the prime we choose to work well, in that it should have few false matches. The problem is that for every prime, there are certainly some bad patterns and documents. If we choose a prime in advance, then someone can try to set up a document and pattern that will cause a lot of false matches, making our fingerprinting algorithm go very slowly. A natural approach is to choose the prime p randomly. This way, nobody can set up a bad pattern and document in advance, since they are not sure what prime we will choose. Let us make this a bit more rigorous. Let π(x) represent the number of primes that are less than or equal to x. It will be helpful to use the following fact: Fact:
x ln x x ≤ π(x) ≤ 1.26 ln x .
Consider any point in the algorithm, where the pattern and document do not match. If our pattern has length |P|, then at that point we are comparing two numbers that are each less than 10 |P| . In particular, their difference (in absolute value) is less than 10|P| . What is the probability that a random prime divides this difference? That is, what is the probability that for the random prime we choose, the two numbers corresponding to the pattern and the current |P| digits in the document are equal modulo p. First, note that there are at most log 2 10|P| distinct primes that divide the difference, since the difference is at most 10|P| (in absoulte value), and each distinct prime divisor is at least 2. Hence, if we choose our prime randomly
Lecture 11
11-3
from all primes up to Z, the probability we have a false match is at most log2 10|P| π(Z). Now the probability that we have a false match anywhere is at most |D| times the probability that we have a false match in any single location, by the union bound. Hence the probability that we have a false match anywhere is at most |D| log2 10|P| π(Z). Exercise: How big should we make Z in order to make the probability of a false match anywhere in the algorithm less than 1/100?
Lecture 11
11-4
How could we improve the probability of a false match? One way is to choose from a larger set of primes. Another way is to choose not just one random prime, but several random primes from Z. This is like choosing several hash functions in the Bloom filter problem. There is a false match only if there is a false match at every random prime we choose. If we choose k primes (with replacement) from the primes up to Z, the probability of a false match at a specific point is at most log2 10|P| π(Z)
k
.
CS124
Lecture 12
12.1 Near duplicate documents1
Suppose we are designing a major search engine. We would like to avoid answering user queries with multiple copies of the same page. That is, there may be several pages with exactly the same text. These duplicates occur for a variety of reasons. Some are mirror sites, some are copies of common pages (such as Unix man pages), some are multiple spam advertisements, etc. Returning just one of the duplicates should be sufficient for the end user; returning all of them will clutter the response page, wasting valuable real estate and frustraing the user. How can we cope with duplicate pages? Determining exact duplicates has a simple solution, based on hashing. Use the text of each page and an appropriate hash function to hash the text into a 64 bit signature. If two documents have the same signature, it is reasonable to assume that they share the same text. (Why? How often is this assumption wrong? Is it a terrible thing if the assumption turns out to be false?) By comparing signatures on the fly, we can avoid returning duplicates. This solution works extremely well if we want to catch exact duplicates. What if, however, we want to capture the idea of “near duplicate” documents, or similar documents. For example, consider two mirror sites on the Web. It may be that the documents share the same text, except that the text corresponding to the links on the page are different, with each referring to the correct mirror site. In this case, the two pages will not yield the same signature, although again, we would not want to return both pages to the end user, because they are so similar. As another example, consider two copies of a newspaper article, one with a proper copyright notice added, and one without. We do not need to return both pages to the user. Again, hashing the document appears to be of no help. Finally, consider the case of advertisers who submit slightly modified versions of their ads over and over again, trying to get more or better spots on the response pages sent back to users. We want to stop their nefarious plans! We will describe a scheme used to detect similar documents efficiently, using a hashing based scheme. Like the Bloom filter solution for password dictionaries, our solution is highly efficient in terms of space and time. The cost for this efficienty is accuracy; our algorithm will sometimes make mistakes, because it uses randomness.
12.2 Set resemblance
We describe a more general problem that will relate to our document similarity problem. Consider two sets of numbers, A and B. For concreteness, we will assume that A and B are subsets of 64 bit numbers. We may define the resemblance of A and B as resemblance(A, B) = R(A, B) = |A ∩ B| . |A ∪ B|
The resemblance is a real number between 0 and 1. Intuitively, the resemblance accurately captures how close the two sets are. Sets and documents will be related, as we will see later.
lecture is based on the work of Andrei Broder, who developed these ideas, and convinced Altavista to use them! (The second feat may have been even more difficult than the first.)
1 This
12-1
Lecture 12
12-2
How quickly can we determine the resemblance of two sets? If the sets are each of size n, the natural approach (compare each element to in A to each element in B) is O(n2 ). We can do better by sorting the sets. Still, these approaches are all rather slow, when we consider that we will have many sets to deal with and hence many pairs of sets to consider. Instead we should ocnsider relaxing the problem. Suppose that we do not need an exact calculation of the resemblance R(A, B). A reasonable estimate or approximation of the resemblance will suffice. Also, since we will be answering a variety of queries over a long period of time, it makes sense to consider algorithms that first do a preprocessing phase, in order to handle the queries much more quickly. That is, we will first do some work, preparing the appropriate data structures and data in a preprocessing phase. The advantage of doing all this work in advance will be that queries regarding resemblance can then be quickly answered. Our estimation process will require a black box that does the following: it produces an effective random permutation on the set of 64 bit numbers. What do we mean by a random permutation? Let us consider just the case of four bit number, of which there are 16. Suppose we write each number on a card. Generating a random permutation is like shuffling this deck of 16 cards and looking at the order at which the numbers appear after ths shuffling. For example, if we find the number 0011 on the first card, then our permutation maps the number 3 to the number 1. We write this as π(3) = 1, where π is a function that represents the permutation. Suppose we have an efficient implemenation of random permutations, which we think of as a black box procedure. That is, when we invoke the black box procedure BB(1, x) on a 64 bit number x, we get out y = π (x) for some 1 fixed, completely random permutation π1 . Similarly, if we invoke the black box BB(2, x), we get out π2 (x) for some different random permutation π2 . (In fact in practice we cannot achieve this black box, but we can get close enough that it is useful to think in these terms for analysis.) Let us use the notation π1 (A) to denote the set of elements obtained by computing BB(1, x) for every x in A. Consider the following procedure: we compute the set π1 (A) and π1 (B), and record the minimum of each set. When does min{π1 (A)} = min{π1 (B)}? This happens only when there is some element x satisfying π1 (x) = min{π1 (A)} = min{π1 (B)}. In other words, the element x that is the minimum element in the set A ∪ B has to be the intersection of the sets A ∩ B. If π1 is a random permutation, then every element in A ∪ B has equal probability of mapping to the minimum element after the permutation is applies. That is, for all x and y in A ∪ B, Pr[π1 (x) = min{π1 (A ∪ B)}] = Pr[π1 (y) = min{π1 (A ∪ B)}]. Thus, for the minimum of π1 (A) and π1 (B) to be the same, the minimum element must lie in π1 (A ∩ B) (see Figure 12.1). Hence |A ∩ B| . Pr[min{π1 (A)} = min{π1 (B)}] = |A ∪ B| But this is just the resemblance R(A, B)! This gives us a way to estimate the resemblance. Instead of taking just one permutation, we take many– say 100. For each set A, we preprocess by computing min{πj (A)} for j = 1 to 100, and store these values. To estimate the resemblance of two sets A and B, we count how often the minima are the same, and divide by 100. It is like each permutation gives us a coin flip, where the probability of a heads (a match) is exactly the resemblance R(A, B) of the two sets.
Lecture 12
12-3
A
B
AIB
Figure 12.1: If the minimum element of π1 (A) and π1 (B) are the same, the minimum element must lie in π1 (A ∩ B).
Four score and seven years ago, our founding Four score and seven score and seven years and seven years ago seven years ago our years ago our founding
Figure 12.2: Shingling: the document is broken up into all segments of k consecutive words; each segment leads to a 64 bit hash value.
12.3 Turning Document Similarity into a Set Resemblance Problem
We now return to the original application. How do we turn document similarity into a set resemblance problem? The key idea is to hash pieces of the document– say every four consecutive words– into 64 bit numbers. This process has been called shingling, and each set of consecutive words is called a shingle. (See Figure 12.2.) Using hashing, the shingles give rise to the resulting numbers for the set resemblance problem, so that for each document D there is a set SD . There are many possible variations and improvements possible. For example, one could modify the number of bits in a shingle or the method for shingling. Similarly, one could throw out all shingles that are not 0 mod 16, say, in order to reduce the number of shingles per document. This approach obscures some important information in the document– such as the order paragraphs appear in, say. However, it seems reasonable to say that if the resulting sets have high resemblance, the documents are reasonably similar. Once we have the shingles for the document, we associate a document sketch with each document. The sketch of a document SD is a list of say 100 numbers: (min{π1 (SD )}, min{π2 (SD )}, min{π3 (SD )}, . . . , min{π100 (SD )}). Now we choose a threshold– for example, we might say that two documents are the similar if 90 out of the 100 entries in the sketch match. Now whenever a user queries the search engine, we check the sketches of the documents we wish to return. If two sketches share 90 entries, we only send one of them. (Alternatively, we could catch the duplicates on the crawling side– we check all the documents as we crawl the Web, and whenever two sketches share more than 90 entries, we assume the associated documents are similar, so that we only need to store one of them!) Recall that our scheme uses random permutations. So, if we set our sketch threshold to 90 out of 100 entries,
Lecture 12
12-4
this does not guarantee that any pair of documents with high resemblance are caught. Also, some pairs of documents that do not have high resemblance may get marked as having high resemblance. How well does this scheme do? We analyze how well the scheme does with the following argument. For each permutation π, the probability i that two documents A and B have the same value in the ith position of the sketch is just the resemblance of the two documents R(A, B) = r. (Here the resemblance R(A, B) of course refers to the resemblance of the sets of numbers obtained by shingling A and B.) Hence, the probability p(r) that at least 90 out of the 100 entries in the sketch match is 100 100 k r (1 − r)100−k . p(r) = ∑ k k=90 What does p(r) look like as a function of r? The graph is shown in Figure 12.3. Notice that p(r) stays very small until r approaches 0.9, and then quickly grows towards 1. This is exactly the property we want our scheme to have– if two documents are not similar, we will rarely mistake them for being similar, and if they are similar, we are likely to catch them! For example, even if the resemablance is 0.8, we will only get 90 matches with probability less than 0.006! −18 When the resemblance is only 0.5, the probability of having 90 entries in the sketch match falls to almost 10 ! If documents are not alike, we will rarely mistake them as being similar. If documents are alike, we will most likely catch them. If the resemblance is 0.95, the documents will have 90 or more entries in common in the sketch with probability greater than .988; if the resemblance is 0.96, the probability jumps to over .997. We are dealing with a very large number of dcouments– most search engines currently index twenty-five to over one hundred million Web pages! So even though the probability of making a mistake is small, it will happen. The worst that happens, though, is that the search engine fails to index a few pages that it should, and it fails to catch a few duplicates that it should. These problems are not a big deal.
Lecture 12
12-5
1
Probability of 90 or more matches
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Resemblance
Figure 12.3: Making the threshold for document similarity 90 out of 100 matches in the sketch leads to the following graph relating resemblance to the probability two documents are considered similar. Notice the sharp change in behavior near where the resemablance is 0.90. Essentially, the procedure behaves like a low pass filter.
CS124
Lecture 13
Hopefully the ideas we saw in our hashing problems have convinced you that randomness is a useful tool in the design and analysis of algorithms. Just to make sure, we will consider several more example of how to use randomness to design algorithms.
13.1 Primality testing
A great deal of modern cryptography is based on the fact that factoring is apparently hard. At least nobody has published a fast way to factor yet. (It is rumored the NSA knows how to factor, and is keeping it a secret. Some of you might well have worked or will work for the NSA, at which point you will be required to keep this secret. Shame on you.) Of course, certain numbers are easy to factor– numbers with small prime factors, for example. So often, for cryptographic purposes, we may want to generate very large prime numbers and multiply them together. How can we find large prime numbers? We are fortunate to find that prime numbers are pretty dense. That is, there’s an awful lot of them. Let π(x) be the number of primes less than or equal to x. Then x , π(x) ≈ ln x or more exactly, π(x) lim x = 1.
x→∞ ln x
This means that on average about one out of every ln x numbers is prime, if we are looking for primes about the size 250 of x. So if we want to find prime numbers of say 250 digits, we would have to check about ln 10 ≈ 576 numbers on average before finding a prime. (We can search smarter, too, throwing out multiples of 2,3,5, etc. in order to check fewer numbers.) Hence, all we need is a good method for testing if a number is prime. With such a test, we can generate large primes easily– just keep generating random large numbers, and test them for primality until we find a suitable prime number. How can we test if a number n is prime? The pedantic way is to try dividing n by all smaller numbers. √ Alternatively, we can try to divide n by all primes up to n. Of course, both of these approaches are quite slow; √ when n is about 10250 , the value of n is still huge! The point is that 10250 has only 250 (or more generally O(log n)) 250 digits, so we’d like the running time of the algorithm to be based on the size 250, not 10 ! How can we quickly test if a number is prime? Let’s start by looking at some ways that work pretty well, but have a few problems. We will use the following result from number theory: Theorem 13.1 If p is a prime and 1 ≤ a < p, then a p−1 = 1 mod p. Proof: There are two nice proofs for this fact. One uses a simple induction to prove the equivalent statement that a p = a mod p. This is clearly true when a = 1. Now (a + 1) p = ∑
p
i=0
p p−i a . i
13-1
Lecture 13
13-2
The coefficient
p i
is divisible by p, unless i = 0 or i = p. Hence (a + 1) p = a p + 1 mod p = a + 1 mod p,
where the last step follows by the induction hypothesis. An alternative proof uses the following idea. Consider the numbers 1, 2, . . . , p − 1. Multiply them all by a, so now we have a, 2a, . . . , (p − 1)a. Each of these number is distinct mod p, and there are p − 1 such numbers, so in fact the sequence a, 2a, . . . , (p − 1)a is the same as the sequence 1, 2, . . . , p − 1 when considered modulo p, except for the order. Hence 1 · 2 · . . . · (p − 1) = a · 2a · . . . · (p − 1)a mod p = ap−1 · 1 · 2 · . . . · (p − 1) mod p. Thus we have ap−1 = 1 mod p. This immediately suggests one way to check if a number n is prime. Compute 2n−1 mod n. If it is not 1, then n is certainly not prime! Note that we can compute 2n−1 mod n quite efficiently, using our previously discussed methods for exponentiation, which require only O(log n) multiplications! Thus this test is efficient. But so far this test is just one-way; if n is composite, we may have that 2n−1 = 1 mod n, so we cannot assume that n is prime just because it passes the test. For example, 2340 = 1 mod 341, and 341 is not prime. Such a number is called a 2-psuedoprime, and unfortunately there are infinitely many of them. (Of course, even though there are infinitely many 2-pseudoprimes, they are not as dense as the primes– that is, there are relatively very few of them. So if we generate a large number n randomly, and see if 2n−1 = 1 mod n, we will most likely be right if we then say n is prime if it passes this test. In practice, this might be good enough! This is not a good primality test, however, if an NSA official you know gives you a number to test for primality, and you think they might be trying to fool you. The NSA might be purposely giving you a 2-pseudoprime. They can be tricky that way.) You might think to try a different base, other than 2. For example, you might choose 3, or a random value of a. Unfortunately, there are infinitely many 3-pseudoprimes. In fact, there are infinitely many composite numbers n such that an−1 = 1 mod n for all a that do not share a factor with n. (That is, for all a such that the greatest common divisor of a and n is 1.) Such n are called Carmichael numbers– the smallest such number is 561. So a test based on this approach is destined to fail for some numbers. There is a way around this problem, due to Rabin. Let n − 1 = 2t u. Suppose we choose a random base a and compute an−1 by first computing au and then repeatedly squaring. Along the way, we will check to see for the values au , a2u , . . . whether they have the following property: a2
i−1 u
= ±1 mod n, a2 u = 1 mod n.
i
That is, suppose we find a non-trivial square root of 1 modulo n. It turns out that only composite numbers have non-trivial square roots – prime numbers don’t. In fact, if we choose a randomly, and n is composite, for at least 3/4 of the values of a, one of two things will happen: we will either find a non-trivial square root of 1 using this process, or we will find that an−1 = 1 mod n. In either case, we know that n is composite! A value of a for which either an−1 = 1 mod n or the computation of an−1 yields a non-trivial square root is called a witness to the compositeness of n. We have said that 3/4 of the possible values of a are witnesses (we will not prove this here!). So if we pick a single value of a randomly, and n is composite, we will determine that n is composite with probability at least 3/4. How can we improve the probability of catching when n is composite? The simplest way is just to repeat the test several times, each time choosing a value of a randomly. (Note that we do not even have to go to the trouble of making sure we try different values of a each time; we can choose values with replacement!) Each time we try this we have a probability of at least 3/4 of catching that n is composite, so if
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k we try the test k times, we will return the wrong answer in the case where n is composite with probability (1/4) . For 50 ; the probability that a random cosmic k = 25, the probability of the algorithm itself making an error is thus (1/2) ray affected your arithmetic unit is probably higher!
This trick comes up again and again with randomized algorithms. If the probability of catching an error on a t single trial is p, the probability of failing to catch an error after t trials is (1 − p) , assuming each trial is independent. By making t sufficiently large, the probability of error can be reduced. Since the probability shrinks exponentially in t, few trials can produce a great deal of security in the answer.
CS 124
Lecture 14
14.1 Cryptography Fundamentals
Cryptography is concerned with the following scenario: two people, Alice and Bob, wish to communicate privately in the presence of an eavesdropper, Eve. In particular, suppose Alice wants to send Bob a message x. (For convenience, we will always assume our message has been converted into a bit string.) Using cryptography, Alice would compute a function e(x), the encoding of x, using some secret key, and transmit e(x) to Bob. Bob receives e(x), and using his own secret key, would compute a function d(e(x)) = x. The function d provides the decoding of the encoding e(x). Eve is presumably unable to recover x from e(x) because she does not have the key – without the key, computing x is either impossible or computationally difficult.
A classical cryptographic method is the one-time pad. A one-time pad is a random string of bits r, equal in length to the message x, that Alice and Bob share and is secret. By random, here we mean that r is equally like to be any bit string of the right length, |r|. Alice compute e(x) = x ⊕ r; Bob computes d(e(x)) = e(x) ⊕ r = x ⊕ r ⊕ r = x. The claim is that Eve gets absolutely no information about the message by seeing e(x). More concretely, we claim Pr(message is x | e(x)) = Pr(message is x); that is, knowing e(x) gives no more information to Eve than she already had. This is a nice exercise in condtional probabilities. Since e(x) provides no information, the one-time pad is completely secure. (Notice that this does not rely on notions of computational difficulty; Eve really obtains no additional information!) There are, however, crucial drawbacks. • The key r has to be as long as x. • The key r can only be used once. (To see this, suppose we use the same key r to encode x and y. The Eve can compute e(x) ⊕ e(y) = x ⊕ y, which might yield useful information!) 14-1
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• The key r has to be exchanged, by some other means. (Private courier?)
14.1.2 DES
The Data Encrytpion Standard, or DES, is a U.S. government sponsored cryptographic method proposed in 1976. It uses a 56 bit key, again shared by Alice and Bob, and it encodes blocks of 64 bits using a complicated sequence of bit operations. Many have suspected that the government engineered the DES standard, so that they could break it easily, but
56 nobody has shown a simpler method for breaking DES other than trying the 2 possible keys. These days, however,
trying even this large number of keys can be accomplished in just a few days with specialized hardware. Hence DES is widely considered no longer secure.
14.1.3 RSA
RSA (named after its inventors, Ron Rivest, Adi Shamir, and Len Adleman) was developed around the same time as DES. RSA is an example of public key cryptography. In public key cryptography, Bob has two keys: a public key, ke , known to everyone, and a private key, kd , known only to Bob. If Alice (or anyone else) wants to send a message x to Bob, she encrypts it as e(x) using the public key; Bob then decrypts it using his private key. For this to be secure, the private key must be hard to compute from the public key, and similarly e(x) must be hard to compute from x. The RSA algorithm depends on some number theory and simple algorithms, which we will consider before describing RSA. We will then describe how RSA is efficient and secure.
14.2 Tools for RSA
14.2.1 Primality
For the time being, we will assume that it is possible to generate large prime numbers. In fact, there are simple and efficient randomized algorithms for generating large primes, that we will consider later in the course.
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14.2.2 Euclid’s Greatest Common Divisor Algorithm
Definition: The greatest common divisor (or gcd) of integers a, b ≥ 0 is the largest integer d ≥ 0 such that d|a and d|b, where d|a denotes that d divides a. Example: gcd(360,84) = 12. One way of computing the gcd is to factor the two numbers, and find the common prime factors (with the right multiplicity). Factoring, however, is a problem for which we do not have general efficient algorithms. The following algorithm, due to Euclid, avoids factoring. Assume a ≥ b ≥ 0. function Euclid(a, b) if b = 0 return(a) return(Euclid(b, a mod b)) end Euclid Euclid’s algorithm relies on the fact that gcd(a, b) = gcd(b, a mod b). You should prove this as an exercise. We need to check that this algorithm is efficient. We will assume that mod operations are efficient (in fact they can be done in O(log2 a) bit operations). How many mod operations must be performed? To analyze this, we notice that in the recursive calls of Euclid’s algorithms, the numbers always get smaller. For the algorithm to be efficient, we’d like to have only about O(log a) recursive calls. This will require the numbers to shrink by a constant factor after a constant number of rounds. In fact, we can show that the larger number shrinks by a factor of 2 every 2 rounds. Claim 1: a mod b ≤ a/2. Proof: The claim is trivially true if b ≤ a/2. If b > a/2, then a mod b = a − b ≤ a/2. Claim 2: On calling Euclid(a, b), after the second recursive call Euclid(a , b ) has a ≤ a/2. Proof: For the second recursive call, we will have a = a mod b.
14.2.3 Extended Euclid’s Algorithm
Euclid’s algorithm can be extended to give not just the greatest common divisor d = gcd(a, b), but also two integers x and y such that ax + by = d. This will prove useful to us subsequently, as we will explain.
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Extended-Euclid(a, b) if b = 0 return(a, 1, 0) Compute k such that a = bk + (a mod b) (d, x, y) = Extended-Euclid(b, a mod b) return((d, y, x − ky)) end Extended-Euclid
Claim 3: The Extended Euclid’s algorithm returns the correct answer. Proof: By induction on a + b. It clearly works if b = 0. (Note the understanding that all numbers divide 0!) If b = 0, then we may assume the recursive call provides the correct answer by induction, as a mod b < a. Hence we have x and y such that bx + (a mod b)y = d. But (a mod b) = a − bk, and hence by substitution we get bx + (a − bk)y = d, or ay + b(x − ky) = d. This shows the algorithm provides the correct output. Note that the Extended Euclid’s algorithm is clearly efficient, as it requires only a few extra arithmetic operations per recursive call over Euclid’s algorithm. The Extended Euclid’s algorithm is useful if we wish to compute the inverse of a number. That is, suppose we wish to find a−1 mod n. The number a has a multiplicative inverse modulo n if and only if the gcd of a and n is 1. Moreover, the Extended Euclid’s algorithm gives us that number. Since in this case computing gcd(a, n) gives x, y such that ax + ny = 1, we have that x = a−1 mod n.
14.2.4 Exponentiation
Suppose we have to compute xy mod z, for integers x, y, z. Multiplying x by itself y times is one possibility, but
2 it is too slow. A more efficient approach is to repeatedly square from x, to get x mod z, x4 mod z, x8 mod z . . .,
x2
log y
mod z. Now xy can be computed by multiplying together modulo z the powers that correspond to ones in the
binary representation of y.
14.3 The RSA Protocol
To create a public key, Bob finds two large primes, p and q, of roughly the same size. (Large should be a few hundred decimal digits. Recently, with a lot of work, 512-bit RSA has been broken; this corresponds to n = pq being 512
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bits long.) Bob computes n = pq, and also computes a random integer e, such that gcd((p − 1)(q − 1), e) = 1. (An alternative to choosing e randomly often used in practice is to choose e = 3, in which case p and q cannot equal 1 modulo 3.)
−1 The pair (n, e) is Bob’s public key, which he announces to the world. Bob’s private key is d = e mod (p −
1)(q − 1), which can be computed by Euclid’s algorithm. More specifically, (p, q, d) is Bob’s private key. Suppose Alice wants to send a message to Bob. We think of the message as being a number x from the range [1, n]. (If the message is too big to be represented by a number this small, it must be broken up into pieces; for example, the message could be broken into bit strings of length log n .) To encode the message, Alice computes and sends to Bob e(x) = xe mod n. Upon receipt, Bob computes d(e(x)) = (e(x))d mod n. To show that this operation decodes correctly, we must prove: Claim 4: d(e(x)) = x. Proof: We use the steps: e(x)d = xde = x1+k(p−1)(q−1) = x mod n.
−1 The first equation recalls the definition of e(x). The second uses the fact that d = e mod (p − 1)(q − 1), and hence
de = 1+ k(p− 1)(q− 1) for some integer k. The last equality is much less trivial. It will help us to have the following lemma: Claim 5: (Fermat’s Little Theorem) If p is prime, then for a = 0 mod p, we have ap−1 = 1 mod p. Proof: Look at the numbers 1, 2, . . . , p − 1. Suppose we multiply them all by a modulo p, to get a · 1 mod p, a · 2 mod p, . . . , a · (p − 1) mod p. We claim that the two sets of numbers are the same! This is because every pair of
−1 numbers in the second group is different; this follows since if a · i = a · j mod p, then by multiplying by a , we
must have i = j mod p. But if all the numbers in the second group are different modulo p, since none of them are 0, they must just be 1, 2, . . . , p − 1. (To get a feel for this, take an example: when p = 7 and a = 5, multiplying a by the numbers {1, 2, 3, 4, 5, 6} yields {5, 3, 1, 6, 4, 2}.) From the above equality of sets of numbers, we conclude 1 · 2 · · · (p − 1) = (a · 1) · (a · 2) · · · (a · (p − 1)) mod p.
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Multiplying both sides by 1−1 , 2−1 , . . . , (p − 1)−1 we have 1 = a p−1 mod p. This proves Claim 5. We now return to the end of Claim 4, where we must prove x1+k(p−1)(q−1) = x mod n. We first claim that x1+k(p−1)(q−1) = x mod p. This is clearly true if x = 0 mod p. If x = 0 mod p, then by Fermat’s Little Theorem, x(p−1) = 1 mod p, and hence xk(p−1)(q−1) = 1 mod p, from which we have x1+k(p−1)(q−1) = x mod p. by the same argument we also have x1+k(p−1)(q−1) = x mod q. But if a number is equal to x both modulo p and modulo q, it is equal to x modulo n = p · q. Hence x1+k(p−1)(q−1) = x mod n, and Claim 4 is proven. We have shown that the RSA protocol allows for correct encoding and decoding. We also should be convinced it is efficient, since it requires only operations that we know to be efficient, such as Euclid’s algorithm and modular exponentiation. One thing we have not yet asked is why the scheme is secure. That is, why can’t the eavesdropper Eve recover the message x also? The answer, unfortunately, is that there is no proof that Eve cannot compute x efficiently from e(x). There is simply a belief that this is a hard problem. It is an unproven assumption that there is no efficient algorithm for computing x from e(x). There is the real but unlikely possibility that someone out there can read all messages sent using RSA! Let us seek some idea of why RSA is believed to be secure. If Eve obtains e(x) = xe mod n, what can she do? She could try all possible values of x to try to find the correct one; this clearly takes too long. Or she could try to factor n and compute d. Factoring, however, is a widely known and well studied problem, and nobody has come up with a polynomial time algorithm for the problem. In fact, it is widely believed that no such algorithm exists. It would be nice if we could make some sort of guarantee. For example, suppose that breaking RSA allowed us to factor n. Then we could say that RSA is as hard as factoring. Unfortunately, this is not the case either. It is possible that RSA could be broken without providing a general factoring algorithm, although it seems that any natural approach for breaking RSA would also provide a way to factor n.
CS124
Lecture 15
15.1 2SAT
We begin by showing yet another possible way to solve the 2SAT problem. Recall that the input to 2SAT is a logical expression that is the conjunction (AND) of a set of clauses, where each clause is the disjunction (OR) of two literals. (A literal is either a Boolean variable or the negation of a Boolean variable.) For example, the following expression is an instance of 2SAT: (x1 ∨ x2 ) ∧ (x1 ∨ x3 ) ∧ (x1 ∨ x2 ) ∧ (x4 ∨ x3 ) ∧ (x4 ∨ x1 ). A solution to an instance of a 2SAT formula is an assignment of the variables to the values T (true) and F (false) so that all the clauses are satisfied– that is, there is at least one true literal in each clause. For example, the assingment x1 = T, x2 = F, x3 = F, x4 = T satisfies the 2SAT formula above. Here is a simple randomized solution to the 2SAT problem. Start with some truth assignment, say by setting all the variables to false. Find some clause that is not yet satisfied. Randomly choose one the variables in that clause, say by flipping a coin, and change its value. Continue this process, until either all clauses are satisfied or you get tired of flipping coins. In the example above, when we begin with all variables set to F, the clause (x1 ∨ x2 ) is not satisfied. So we might randomly choose to set x1 to be T. In this case this would leave the clause (x4 ∨ x1 ) unsatisfied, so we would have to flip a variable in the clause, and so on. Why would this algorithm tend to lead to a solution? Let us suppose that there is a solution, call it S. Suppose we keep track of the number of variables in our current assignment A that match S. Call this number k. We would like to get to the point where k = n, the number of variables in the formula, for then A would match the solution S. How does k evolve over time? At each step, we choose a clause that is unsatisfied. Hence we know that A and S disagree on the value of at least one of the variables in this clause– if they agreed, the clause would have to be satisfied! If they disagree on both, then clearly changing either one of the values will increase k. If they disagree on the value one of the two variables, then with probability 1/2 we choose that variable and make increase k by 1; with probability 1/2 we choose the other variable and decrease k by 1. Hence, in the worse case, k behaves like a random walk– it either goes up or down by 1, randomly. This leaves us with the following question: if we start k at 0, how many steps does it take (on average, or with high probability) for k to stumble all the way up to n, the number of variables?
2 We can check that the average amount of steps to walk (randomly) from 0 to n is just n . In fact, the average amount of time to walk from i to n is n2 − i2 . Note that the time average time T (i) to walk from i to n is given by:
T (n) = 0 T (i − 1) T (i + 1) + + 1, i ≥ 1 T (i) = 2 2 T (0) = T (1) + 1.
15-1
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These equations completely determine T (i), and our solution satisfies these equations! Hence, on average, we will find a solution in at most n2 steps. (We might do better– we might not start with all of our variables wrong, or we might have some moves where we must improve the number of matches!) We can run our algorithm for say 100n2 steps, and report that no solution was found if none was found. This algorithm might return the wrong answer– there may be a truth assignment, and we have just been unlucky. But most of the time it will be right.
CS124
Lecture 16
An introductory example
Suppose that a company that produces three products wishes to decide the level of production of each so as to maximize profits. Let x1 be the amount of Product 1 produced in a month, x2 that of Product 2, and x3 that of Product 3. Each unit of Product 1 yields a profit of 100, each unit of Product 2 a profit of 600, and each unit of Product 3 a profit of 1400. There are limitations on x1 , x2 , and x3 (besides the obvious one, that x1 , x2 , x3 ≥ 0). First, x1 cannot be more than 200, and x2 cannot be more than 300, presumably because of supply limitations. Also, the sum of the three must be, because of labor constraints, at most 400. Finally, it turns out that Products 2 and 3 use the same piece of equipment, with Product 3 using three times as much, and hence we have another constraint x + 3x3 ≤ 600. 2 What are the best levels of production? We represent the situation by a linear program, as follows:
max 100x1 + 600x2 + 1400x3 x1 ≤ 200 x2 ≤ 300 x1 + x2 + x3 ≤ 400 x2 + 3x3 ≤ 600 x1 , x2 , x3 ≥ 0 The set of all feasible solutions of this linear program (that is, all vectors in 3-d space that satisfy all constraints) is precisely the polyhedron shown in Figure 16.1. We wish to maximize the linear function 100x1 + 600x2 + 1400x3 over all points of this polyhedron. Geometrically, the linear equation 100x1 + 600x2 + 1400x3 = c can be represented by a plane parallel to the one determined by the equation 100x1 + 600x2 + 1400x3 = 0. This means that we want to find the plane of this type that touches the polyhedron and is as far towards the positive orthant as possible. Obviously, the optimum solution will be a vertex (or the optimum solution will not be unique, but a vertex will do). Of course, two other possibilities with linear programming are that (a) the optimum solution may be infinity, or (b) that there may be no feasible solution at all. 16-1
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x2
300 opt
200 200
x1
x3
Figure 16.1: The feasible region In this case, an optimal solution exists, and moreover we shall show that it is easy to find.
Linear programs
Linear programs, in general, have the following form: there is an objective function that one seeks to optimize, along with constraints on the variables. The objective function and the constraints are all linear in the variables; that is, all equations have no powers of the variables, nor are the variables multiplied together. As we shall see, almost all problems can be represented by linear programs, and for many problems it is an extremely convenient representation. So once we explain how to solve linear programs, the question then becomes how to reduce other problems to linear programming (LP). There are polynomial time algorithms for solving linear programs. In practice, however, such problems are solved by the simplex method devised by George Dantzig in 1947. The simplex method starts from a vertex (in this
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case the vertex (0, 0, 0)) and repeatedly looks for a vertex that is adjacent, and has better objective value. That is, it is a kind of hill-climbing in the vertices of the polytope. When a vertex is found that has no better neighbor, simplex stops and declares this vertex to be the optimum. For example, in the figure one of the possible paths followed by simplex is shown. No known variant of the simplex algorithm has been proven to take polynomial time, and most of the variations used in practice have been shown to take exponential time on some examples. Fortunately, in practice, bad cases rarely arise, and the simplex algorithm runs extremely quickly. There are now implementations of simplex that solve routinely linear programs with many thousands of variables and constraints. Of course, given a linear program, it is possible either that (a) the optimum solution may be infinity, or (b) that there may be no feasible solution at all. If this is the case, simplex algorithm will discover it.
Reductions between versions of simplex
A general linear programming problem may involve constraints that are equalities or inequalities in either direction. Its variables may be nonnegative, or could be unrestricted in sign. And we may be either minimizing or maximizing a linear function. It turns out that we can easily translate any such version to any other. One such translation that is particularly useful is from the general form to the one required by simplex: minimization, nonnegative variables, and equality constraints. To turn an inequality ∑ ai xi ≤ b into an equality constraint, we introduce a new variable s (the slack variable for this inequality), and rewrite this inequality as ∑ ai xi + s = b, s ≥ 0. Similarly, any inequality ∑ ai xi ≥ b is rewritten as ∑ ai xi − s = b, s ≥ 0; s is now called a surplus variable.
+ We handle an unrestricted variable x as follows: we introduce two nonnegative variables, x and x− , and
replace x by x+ − x− everywhere. The idea is that we let x = x+ − x− , where we may restrict both x+ and x− to be nonnegative. This way, x can take on any value, but there are only nonnegative variables. Finally, to turn a maximization problem into a minimization one, we just multiply the objective function by −1.
A production scheduling example
We have the demand estimates for our product for all months of 1997, d : i = 1, . . . , 12, and they are very i uneven, ranging from 440 to 920. We currently have 30 employees, each of which produce 20 units of the product each month at a salary of 2,000; we have no stock of the product. How can we handle such fluctuations in demand? Three ways:
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• overtime —but this is expensive since it costs 80% more than regular production, and has limitations, as workers can only work 30% overtime. • hire and fire workers —but hiring costs 320, and firing costs 400. • store the surplus production —but this costs 8 per item per month
This rather involved problem can be formulated and solved as a linear program. As in all such reductions, the crucial first step is defining the variables: • Let w0 be the number of workers we have the ith month —we have w0 = 30. • Let xi be the production for month i. • oi is the number of items produced by overtime in month i. • hi and fi are the number of workers hired/fired in the beginning of month i. • si is the amount of product stored after the end of month i.
We now must write the constraints: • xi = 20wi + oi —the amount produced is the one produced by regular production, plus overtime. • wi = wi−1 + hi − fi , wi ≥ 0 —the changing number of workers. • si = si−1 + xi − di ≥ 0 —the amount stored in the end of this month is what we started with, plus the production, minus the demand. • oi ≤ 6wi —only 30% overtime.
Finally, what is the objective function? It is min 2000 ∑ wi + 400 ∑ fi + 320 ∑ hi + 8 ∑ si + 180 ∑ oi , where the summations are from i = 1 to 12.
A Communication Network Problem
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We have a network whose lines have the bandwidth shown in Figure 16.2. We wish to establish three calls: one between A and B (call 1), one between B and C (call 2), and one between A and C (call 3). We must give each call at least 2 units of bandwidth, but possibly more. The link from A to B pays 3 per unit of bandwidth, from B to C pays 2, and from A to C pays 4. Notice that each call can be routed in two ways (the long and the short path), or by a combination (for example, two units of bandwidth via the short route, and three via the long route). Suppose we are a shady network administrator, and our goals is to maximize the network’s income (rather than minimize the overall cost). How do we route these calls to maximize the network’s income?
B 10 13 8 C
Figure 16.2: A communication network This is also a linear program. We have variables for each call and each path (long or short); for example x is 1 the short path for call 1, and x2 the long path for call 2. We demand that (1) no edge bandwidth is exceeded, and (2) each call gets a bandwidth of 2.
6 11 12 A
max 3x1 + 3x1 + 2x2 + 2x2 + 4x3 + 4x3 x1 + x1 + x2 + x2 ≤ 10 x1 + x1 + x3 + x3 ≤ 12 x2 + x2 + x3 + x3 ≤ 8 x1 + x2 + x3 ≤ 6
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x1 + x2 + x3 ≤ 13 x1 + x2 + x3 ≤ 11 x1 + x1 ≥ 2 x2 + x2 ≥ 2 x3 + x3 ≥ 2 x1 , x1 . . . , x3 ≥ 0 The solution, obtained via simplex in a few milliseconds, is the following: x = 0, x1 = 7, x2 = x2 = 1.5, x3 = 1 .5, x3 = 4.5. Question: Suppose that we removed the constraints stating that each call should receive at least two units. Would the optimum change?
Approximate Separation
An interesting last application: Suppose that we have two sets of points in the plane, the black points (x, yi ) : i i = 1, . . . , m and the white points (xi , yi ) : i = m+ 1, . . . , m+ n. We wish to separate them by a straight line ax+ by = c, so that for all black points ax + by ≤ c, and for all white points ax + by ≥ c. In general, this would be impossible. Still, we may want to separate them by a line that minimizes the sum of the “displacement errors” (distance from the boundary) over all misclassified points. Here is the LP that achieves this: min e1 e1 ≥ e2 ≥ . . . em ≥ em+1 ≥ . . . em+n ≥ +e2 + . . . + em + em+1 + . . . + em+n ax1 + by1 − c ax2 + by2 − c axm + bym − c c − axm+1 − bym+1 c − axm+n − bym+n ei ≥ 0
Network Flows
Suppose that we are given the network in top of Figure 16.3, where the numbers indicate capacities, that is, the amount of flow that can go through the edge in unit time. We wish to find the maximum amount of flow that can go through this network, from S to T .
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A 5 S 1
3
C 2
2
1
T
2 B A 5 S 2 1 2 2 2 1 3 3 D C
5
2 T
2 B A 5 S 4 1 2 2 2 2 1 2 2 B A 5 S 4 2 2 B 2 3 D 1 2 2 2 2 1 4 3 3 D C 3 3 D C
5
2 T
5
2
minimum cut, capacity 6 T
5
Figure 16.3: Max flow
Lecture 16
16-8
This problem can also be reduced to linear programming. We have a nonnegative variable for each edge, representing the flow through this edge. These variables are denoted fSA , fSB , . . . We have two kinds of constraints: capacity constraints such as fSA ≤ 5 (a total of 9 such constraints, one for each edge), and flow conservation constraints (one for each node except S and T ), such as fAD + fBD = fDC + fDT (a total of 4 such constraints). We wish to maximize fSA + fSB , the amount of flow that leaves S, subject to these constraints. It is easy to see that this linear program is equivalent to the max-flow problem. The simplex method would correctly solve it. In the case of max-flow, it is very instructive to “simulate” the simplex method, to see what effect its various iterations would have on the given network. Simplex would start with the all-zero flow, and would try to improve it. How can it find a small improvement in the flow? Answer: it finds a path from S to T (say, by depth-first search), and moves flow along this path of total value equal to the minimum capacity of an edge on the path (it can obviously do no better). This is the first iteration of simplex (see Figure 16.3). How would simplex continue? It would look for another path from S to T . Since this time we already partially (or totally) use some of the edges, we should do depth-first search on the edges that have some residual capacity, above and beyond the flow they already carry. Thus, the edge CT would be ignored, as if it were not there. The depth-first search would now find the path S − A − D − T , and augment the flow by two more units, as shown in Figure 16.3. Next, simplex would again try to find a path from S to T . The path is now S − A − B − D − T (the edges C − T and A − D are full are are therefore ignored), and we augment the flow as shown in the bottom of Figure 16.3. Next simplex would again try to find a path. But since edges A − D, C − T , and S − B are full, they must be ignored, and therefore depth-first search would fail to find a path, after marking the nodes S, A,C as reachable from S. Simplex then returns the flow shown, of value 6, as maximum. How can we be sure that it is the maximum? Notice that these reachable nodes define a cut (a set of nodes containing S but not T ), and the capacity of this cut (the sum of the capacities of the edges going out of this set) is 6, the same as the max-flow value. (It must be the same, since this flow passes through this cut.) The existence of this cut establishes that the flow is optimum! There is a complication that we have swept under the rug so far: when we do depth-first search looking for a path, we use not only the edges that are not completely full, but we must also traverse in the opposite direction all edges that already have some non-zero flow. This would have the effect of canceling some flow; canceling may be necessary to achieve optimality, see Figure 16.4. In this figure the only way to augment the current flow is via the path S − B − A − T , which traverses the edge A − B in the reverse direction (a legal traversal, since A − B is carrying
Lecture 16
16-9
non-zero flow).
A 1 S 1 1 T
1 B
Figure 16.4: Flows may have to be canceled
1
In general, a path from the source to the sink along which we can increase the flow is called an augmenting path. We can look for an augmenting path by doing for example a depth first search along the residual network, which we now describe. For an edge (u, v), let c(u, v) be its capacity, and let f (u, v) be the flow across the edge. Note that we adopt the following convention: if 4 units flow from u to v, then f (u, v) = 4, and f (v, u) = −4. That is, we interpret the fact that we could reverse the flow across an edge as being equivalent to a “negative flow”. Then the residual capacity of an edge (u, v) is just c(u, v) − f (u, v). The residual network has the same vertices as the original graph; the edges of the residual network consist of all weighted edges with strictly positive residual capacity. The idea is then if we find a path from the source to the sink in the residual network, we have an augmenting path to increase the flow in the original network. As an exercise, you may want to consider the residual network at each step in Figure 16.3. Suppose we look for a path in the residual network using depth first search. In the case where the capacities are integers, we will always be able to push an integral amount of flow along an augmenting path. Hence, if the maximum flow is f ∗ , the total time to find the maximum flow is O(E f ∗ ), since we may have to do an O(E) depth first search up to f ∗ times. This is not so great. Note that we do not have to do a depth-first search to find an augmenting path in the residual network. In fact, using a breadth-first search each time yields an algorithm that provably runs in O(V E2 ) time, regardless of whether or not the capacities are integers. We will not prove this here. There are also other algorithms and approaches to the
Lecture 16
16-10
max-flow problem as well that improve on this running time. To summarize: the max-flow problem can be easily reduced to linear programming and solved by simplex. But it is easier to understand what simplex would do by following its iterations directly on the network. It repeatedly finds a path from S to T along edges that are not yet full (have non-zero residual capacity), and also along any reverse edges with non-zero flow. If an S − T path is found, we augment the flow along this path, and repeat. When a path cannot be found, the set of nodes reachable from S defines a cut of capacity equal to the max-flow. Thus, the value of the maximum flow is always equal to the capacity of the minimum cut. This is the important max-flow min-cut theorem. One direction (that max-flow≤min-cut) is easy (think about it: any cut is larger than any flow); the other direction is proved by the algorithm just described.
CS124
Lecture 17
A 5 S 1
3
C 2
2
1
T
2 B A 5 S 2 1 2 2 2 1 3 3 D C
5
2 T
2 B A 5 S 4 1 2 2 2 2 1 2 2 B A 5 S 4 2 2 B 2 3 D 1 2 2 2 2 1 4 3 3 D C 3 3 D C
5
2 T
5
2
minimum cut, capacity 6 T
5
Figure 17.1: Max flow
17-1
Lecture 17
17-2
Network Flows
Suppose that we are given the network in top of Figure 17.1, where the numbers indicate capacities, that is, the amount of flow that can go through the edge in unit time. We wish to find the maximum amount of flow that can go through this network, from S to T . This problem can also be reduced to linear programming. We have a nonnegative variable for each edge, rep¡ ¡ ¡ ¢¢¢
resenting the flow through this edge. These variables are denoted fSA fSB capacity constraints such as fSA
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We have two kinds of constraints:
5 (a total of 9 such constraints, one for each edge), and flow conservation con¤ ¥
program is equivalent to the max-flow problem. The simplex method would correctly solve it.
¤
to maximize fSA
fSB , the amount of flow that leaves S, subject to these constraints. It is easy to see that this linear
¤
straints (one for each node except S and T ), such as fAD
fBD
fDC
fDT (a total of 4 such constraints). We wish
Lecture 17
17-3
In the case of max-flow, it is very instructive to “simulate” the simplex method, to see what effect its various iterations would have on the given network. Simplex would start with the all-zero flow, and would try to improve it. How can it find a small improvement in the flow? Answer: it finds a path from S to T (say, by depth-first search), and moves flow along this path of total value equal to the minimum capacity of an edge on the path (it can obviously do no better). This is the first iteration of simplex (see Figure 17.1). How would simplex continue? It would look for another path from S to T . Since this time we already partially (or totally) use some of the edges, we should do depth-first search on the edges that have some residual capacity, above and beyond the flow they already carry. Thus, the edge CT would be ignored, as if it were not there. The
¦ ¦ ¦
depth-first search would now find the path S Figure 17.1.
A
D
T , and augment the flow by two more units, as shown in
ignored, and therefore depth-first search would fail to find a path, after marking the nodes S A C as reachable from S. Simplex then returns the flow shown, of value 6, as maximum.
¦
¦
¦
Next simplex would again try to find a path. But since edges A
¦
and A
D are full are are therefore ignored), and we augment the flow as shown in the bottom of Figure 17.1. D, C T , and S B are full, they must be
¦
¦
¦
¦
¦
Next, simplex would again try to find a path from S to T . The path is now S
A
B
D
T (the edges C
T
Lecture 17
17-4
How can we be sure that it is the maximum? Notice that these reachable nodes define a cut (a set of nodes containing S but not T ), and the capacity of this cut (the sum of the capacities of the edges going out of this set) is 6, the same as the max-flow value. (It must be the same, since this flow passes through this cut.) The existence of this cut establishes that the flow is optimum! There is a complication that we have swept under the rug so far: when we do depth-first search looking for a path, we use not only the edges that are not completely full, but we must also traverse in the opposite direction all edges that already have some non-zero flow. This would have the effect of canceling some flow; canceling may be necessary to achieve optimality, see Figure 17.2. In this figure the only way to augment the current flow is via the
¦ ¦ ¦ ¦ ¦
path S
B
A
T , which traverses the edge A
B in the reverse direction (a legal traversal, since A
B is carrying
non-zero flow).
A 1 S 1 1 T
1 B
Figure 17.2: Flows may have to be canceled
1
Lecture 17
17-5
In general, a path from the source to the sink along which we can increase the flow is called an augmenting path. We can look for an augmenting path by doing for example a depth first search along the residual network, which we now describe. For an edge u v , let c u v be its capacity, and let f u v be the flow across the edge.
¦ ¥ ¨ § ¥ ¨ ¨ ¨ § ¨ § ¨ §
Note that we adopt the following convention: if 4 units flow from u to v, then f u v
§
4, and f v u
4. That is,
we interpret the fact that we could reverse the flow across an edge as being equivalent to a “negative flow”. Then the residual capacity of an edge u v is just
¡ ¨ ¨ § § ¨ ¢ § § ¦ ©¨ § ¨ §
cuv
f uv
The residual network has the same vertices as the original graph; the edges of the residual network consist of all weighted edges with strictly positive residual capacity. The idea is then if we find a path from the source to the sink in the residual network, we have an augmenting path to increase the flow in the original network. As an exercise, you may want to consider the residual network at each step in Figure 17.1. Suppose we look for a path in the residual network using depth first search. In the case where the capacities are integers, we will always be able to push an integral amount of flow along an augmenting path. Hence, if the maximum flow is f , the total time to find the maximum flow is O E f , since we may have to do an O E depth first search up to f times. This is not so great. Note that we do not have to do a depth-first search to find an augmenting path in the residual network. In fact, using a breadth-first search each time yields an algorithm that provably runs in O V E 2 time, regardless of whether or not the capacities are integers. We will not prove this here. There are also other algorithms and approaches to the max-flow problem as well that improve on this running time. To summarize: the max-flow problem can be easily reduced to linear programming and solved by simplex. But it is easier to understand what simplex would do by following its iterations directly on the network. It repeatedly finds a path from S to T along edges that are not yet full (have non-zero residual capacity), and also along any reverse
¦
edges with non-zero flow. If an S
T path is found, we augment the flow along this path, and repeat. When a path
cannot be found, the set of nodes reachable from S defines a cut of capacity equal to the max-flow. Thus, the value of the maximum flow is always equal to the capacity of the minimum cut. This is the important max-flow min-cut theorem. One direction (that max-flow min-cut) is easy (think about it: any cut is larger than any flow); the other direction is proved by the algorithm just described.
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Lecture 17
17-6
Duality
As it turns out, the max-flow min-cut theorem is a special case of a more general phenomenon called duality. Basically, duality means that for each maximization problem there is a corresponding minimizations problem with the property that any feasible solution of the min problem is greater than or equal any feasible solution of the max problem. Furthermore, and more importantly, they have the same optimum. Consider the network shown in Figure 17.3, and the corresponding max-flow problem. We know that it can be written as a linear program as follows:
A 3 S 1 1 T
2 B
Figure 17.3: A simple max-flow problem
3
fSB fAB fAT fBT
¦ ¤ ¦ ¦
¥
fSA
fBT
¥
fSA
fAB fAB
fAT
3 2 1 1 3 0 0 f 0
£ £ £ £
£
¤
max
fSA fSA
fSB
P
Lecture 17
17-7
Consider now the following linear program: min 3ySA ySA
¤
This LP describes the min-cut problem! To see why, suppose that the uA variable is meant to be 1 if A is in the cut with S, and 0 otherwise, and similarly for B (naturally, by the definition of a cut, S will always be with S in the cut, and T will never be with S). Each of the y variables is to be 1 if the corresponding edge contributes to the cut capacity, and 0 otherwise. Then the constraints make sure that these variables behave exactly as they should. For example, the second constraint states that if A is not with S, then SA must be added to the cut. The third one states
¦ ¤
that if A is with S and B is not (this is the only case in which the sum uA
¦
uB becomes 1), then AB must contribute
to the cut. And so on. Although the y and u’s are free to take values larger than one, they will be “slammed” by the minimization down to 1 or 0.
¦
yBT
uB
¦
yAT
¤
¦
yAB
uA uA
¤
ySB
uB uB
¤
¤
¤
¤
2ySB
yAB
yAT
3yBT uA 1 1 0 0 0 y 0
D
Lecture 17
17-8
Let us now make a remarkable observation: these two programs have strikingly symmetric, dual, structure. This structure is most easily seen by putting the linear programs in matrix form. The first program, which we call the primal (P), we write as: max 1 1 0 0 0 0 1 0
1 0 1 0 0 0
¦
0 0 0 1 0 0
¦
0 0 0 0 1 0 1
¦
0 0 0 0 0 1
¥ £ £ £ £ £
3 2 1 1 3 0 0
¥
0 1
1 1
1 1
0
Here we have removed the actual variable names, and we have included an additional row at the bottom denoting that all the variables are non-negative. (An unrestricted variable will be denoted by unr. The second program, which we call the dual (D), we write as: min 3 1 0 0 0 0
2 0 1 0 0 0
1 0 0 1 0 0
1 0 0 0 1 0
unr unr
Each variable of P corresponds to a constraint of D, and vice-versa. Equality constraints correspond to unrestricted variables (the u’s), and inequality constraints to restricted variables. Minimization becomes maximization. The matrices are transpose of one another, and the roles of right-hand side and objective function are interchanged.
¦
1
0
1
¦
0
1
0
¦
0
1
1
0
0
1
3 0
0 1
0 0 1 1 0 0 0
Lecture 17
17-9
Such LP’s are called dual to each other. It is mechanical, given an LP, to form its dual. Suppose we start with a maximization problem. Change all inequality constraints into necessary. Then
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constraints, negating both sides of an equation if
transpose the coefficient matrix
invert maximization to minimization
interchange the roles of the right-hand side and the objective function
introduce a nonnegative variable for each inequality, and an unrestricted one for each equality
for each nonnegative variable introduce a constraint.
constraint, and for each unrestricted variable introduce an equality
straints, we make the dual a maximization, and we change the last step so that each nonnegative variable corresponds to a
£
constraint. Note that it is easy to show from this description that the dual of the dual is the original primal
problem! By the max-flow min-cut theorem, the two LP’s P and D above have the same optimum. In fact, this is true for general dual LP’s! This is the duality theorem, which can be stated as follows (we shall not prove it; the best proof comes from the simplex algorithm, very much as the max-flow min-cut theorem comes from the max-flow algorithm): If an LP has a bounded optimum, then so does its dual, and the two optimal values coincide.
con-
Lecture 17
17-10
Matching
It is often useful to compose reductions. That is, we can reduce a problem A to B, and B to C, and since C we know how to solve, we end up solving A. A good example is the matching problem. Suppose that the bipartite graph shown in Figure 17.4 records the compatibility relation between four boys and four girls. We seek a maximum matching, that is, a set of edges that is as large as possible, and in which no two edges share a node. For example, in Figure 17.4 there is a complete matching (a matching that involves all nodes).
Al
Eve
Bob
Fay
S
T
Charlie
Grace
Dave
Helen
Figure 17.4: Reduction from matching to max-flow (all capacities are 1)
Lecture 17
17-11
To reduce this problem to max-flow, we create a new source and a new sink, connect the source with all boys and all girls with the sinks, and direct all edges of the original bipartite graph from the boys to the girls. All edges have capacity one. It is easy to see that the maximum flow in this network corresponds to the maximum matching. Well, the situation is slightly more complicated than was stated above: what is easy to see is that the optimum integer-valued flow corresponds to the optimum matching. We would be at a loss interpreting as a matching a flow that ships .7 units along the edge Al-Eve! Fortunately, what the algorithm in the previous section establishes is that if the capacities are integers, then the maximum flow is integer. This is because we only deal with integers throughout the algorithm. Hence integrality comes for free in the max-flow problem. Unfortunately, max-flow is about the only problem for which integrality comes for free. It is a very difficult problem to find the optimum solution (or any solution) of a general linear program with the additional constraint that (some or all of) the variables be integers. We will see why in forthcoming lectures.
Lecture 17
17-12
Games
We can represent various situations of conflict in life in terms of matrix games. For example, the game shown below is the rock-paper-scissors game. The Row player chooses a row strategy, the Column player chooses a column strategy, and then Column pays to Row the value at the intersection (if it is negative, Row ends up paying Column). r
¦
p 1
¦
s 1 1 0
r p
¦
0 1 1
0 1
s
Games do not necessarily have to be symmetric (that is, Row and Column have the same strategies, or, in terms of
¦ § ¥
matrices, A
AT ). For example, in the following fictitious Clinton-Dole game the strategies may be the issues on
which a candidate for office may focus (the initials stand for “economy,” “society,” “morality,” and “tax-cut”) and the entries are the number of voters lost by Column. m
¦ §
t 1 1
e
¦
3 2
s
We want to explore how the two players may play “optimally” these games. It is not clear what this means. For example, in the first game there is no such thing as an optimal “pure” strategy (it very much depends on what your opponent does; similarly in the second game). But suppose that you play this game repeatedly. Then it makes sense
¥
to randomize. That is, consider a game given by an m
¨ ¡ ¡ ¡ ¢¢¢
n matrix Gi j ; define a mixed strategy for the row player 1. Intuitively, xi is the probability with which Row plays
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to be a vector x1
xm , such that xi
0, and ∑m 1 xi i
¨
¡ ¡ ¡ ¢¢¢
strategy i. Similarly, a mixed strategy for Column is a vector y1
yn , such that y j
0, and ∑n j
1 yj
1.
Lecture 17
17-13
Suppose that, in the Clinton-Dole game, Row decides to play the mixed strategy 5 5 . What should Column do? The answer is easy: If the xi ’s are given, there is a pure strategy (that is, a mixed strategy with all y j ’s zero except
¡ ¡ ¡ ¢¢¢ ¥ ¨ ¡ ¢ ¡ §
for one) that is optimal. It is found by comparing the n numbers ∑m 1 Gi j xi , for j i
¡
1
n (in the Clinton-Dole
game, Column would compare 5 with 0, and of course choose the smallest —remember, the entries denote what Column pays). That is, if Column knew Row’s mixed strategy, s/he would end up paying the smallest among the
¡ ¡ ¡ ¢¢¢ ¥
n outcomes ∑m 1 Gi j xi , for j i minimum; that is,
1
n. On the other hand, Row will seek the mixed strategy that maximizes this max min ∑ Gi j xi
x j i 1 m
This maximum would be the best possible guarantee about an expected outcome that Row can have by choosing a mixed strategy. Let us call this guarantee z; what Row is trying to do is solve the following LP: 3x1 x1 x1
¤ ¤
Symmetrically, it is easy to see that Column would solve the following LP: min w w w
¦ ¤
The crucial observation now is that these LP’s are dual to each other, and hence have the same optimum, call it V .
¥
¤
¦
¤
3y1 2y1 y1
y2 y2 y2
£
¥
2x2 x2 x2
¦ ¤
£
¦
max z z z
¡
0 0 1
0 0 1
Lecture 17
17-14
Let us summarize: By solving an LP, Row can guarantee an expected income of at least V , and by solving the dual LP, Column can guarantee an expected loss of at most the same value. It follows that this is the uniquely defined optimal play (it was not a priori certain that such a play exists). V is called the value of the game. In this case, the optimum mixed strategy for Row is 3 7 4 7 , and for Column 2 7 5 7 , with a value of 1 7 for the Row player. The existence of mixed strategies that are optimal for both players and achieve the same value is a fundamental result in Game Theory called the min-max theorem. It can be written in equations as follows:
¡ ¨ § ¨ §
x
y
It is surprising, because the left-hand side, in which Column optimizes last, and therefore has presumably an advantage, should be intuitively smaller than the right-hand side, in which Column decides first. Duality equalizes the two, as it does in max-flow min-cut.
¥
max min ∑ xi y j Gi j
min max ∑ xi y j Gi j
y x
Lecture 17
17-15
Circuit Evaluation
OR
AND
AND
NOT
AND
OR
AND
T
F
F
T
Figure 17.5: A Boolean circuit
Lecture 17
17-16
We have seen many interesting and diverse applications of linear programming. In some sense, the next one is the ultimate application. Suppose that we are given a Boolean circuit, that is, a DAG of gates, each of which is either an input gate (indegree zero, and has a value T or F), or an OR gate (indegree two), or an AND gate (indegree two), or a NOT gate (indegree one). One of them is designated as the output gate. We wish to tell if this circuit evaluates (following the laws of Boolean values bottom-up) to T. This is known as the circuit value problem. There is a very simple and automatic way of translating the circuit value problem into an LP: for each gate g
¥
we have a variable xg . For all gates we have 0
¥
xg
Finally, we want to max xo , where o is the output gate. It is easy to see that the optimum value of xo will be 1 if the circuit value if T, and 0 if it is F. This is a rather straight-forward reduction to LP, from a problem that may not seem very interesting or hard at first. However, the circuit value problem is in some sense the most general problem solvable in polynomial time! Here is a justification of this statement: after all, a polynomial time algorithm runs on a computer, and the computer is ultimately a Boolean combinational circuit implemented on a chip. Since the algorithm runs in polynomial time, it can be rendered as a circuit consisting of polynomially many superpositions of the computer’s circuit. Hence, the fact that circuit value problem reduces to LP means that all polynomially solvable problems do! In our next topic, Complexity and NP-completeness, we shall see that a class that contains many hard problems reduces, much the same way, to integer programming.
¦
¥
of h and h , we have the inequalities xg
xh , xg
xh (notice the difference). For a NOT gate we say xg
¤
xg
0. If it is an OR gate, say of the gates h and h , then we have the inequality xg
£ £
£
£
£
1. If g is a T input gate, we have the equation xg xh
1; if it is F,
xh . If it is an AND gate 1 xh .
CS124
NP-Completeness Review
Up to this point, we have generally assumed that if we were given a problem, we could find a way to solve it. Unfortunately, as most of you know, there are many fundamental problems for which we have no efficient algorithms. In fact, by classifying these hard problems, we can show that there is a large class of simple problems for which there is (probably) no efficient algorithm– the NP-complete problems. Moreover, if you could design an efficient algorithm for any one of these problems, you could design an algorithm for all of them! It’s an all or none proposition, so if you could solve just one of them, you would become rich and famous overnight. These notes will review the main concepts behind the theory of NP-complete problems. One might ask why it is important to study what problems we cannot solve, instead of focusing on problems we can solve. Especially for an algorithms course. There are several possible responses, but perhaps the best is that if you do not know what is impossible, you might waste a great deal of time trying to solve it, instead of coming to terms with its impossibility and finding suitable alternatives (such as, for example, approximations instead of exact answers).
18-1
Lecture 18
18-2
Polynomial Running Times
The faster the running time, the better. Linear is great, quadratic is all right, cubic is perhaps a bit slow. But how exactly should we classify which problems have efficient algorithms? Where is the cut off point? The choice computer scientists have made is to group together all problems that are solvable in polynomial time. That is, we define a class of problems P as follows: Definition: P is the set of all problems Z with a yes-no answer such that there is an algorithm A and a positive integer k such that A solves Z in O nk steps (on inputs of size n). Let us clarify some points in the definition. The restriction to problems with a yes-no answer is really just a technical convenience. For example, the problem of finding the minimum spanning tree ( on a tree with integer weights) can be recast as the problem of answering the following question: is the size of the minimum spanning tree at least j? If you can answer one question, you can answer the other; considering only yes-no problems proves more convenient. From the definition, all problems with linear, quadratic, or cubic time algorithms are all in P. But so are problems with algorithms that require time Θ n100 . This may seem a little strange; for example, would a problem with an algorithm that runs in time Θ n100 really be said to have an efficient solution? But the main point of defining the class P is to separate these problems from those that require exponential time, or Ω 2n steps (for some ε
Problems that require this much time to solve are clearly asymptotically inefficient, compared with polynomial time algorithms. The class P is also useful because, as we shall see below, it is closed under polynomial time reductions.
¢
¡
¡
¡
¡
ε
0.
Lecture 18
18-3
Reductions
Let A and B be two problems whose instances require a “yes” or “no” answer. (For example, 2SAT is such a problem, as is the question of whether a bipartite graph has a perfect matching.) A (polynomial time) reduction from A to B is a polynomial time algorithm R which transforms an input of problem A into an input for problem B. That is, given an input x to problem A, R will produce an input R x to problem B, such that the answer to x is yes for problem A if and only if the answer for R x is yes for problem B. This idea of reduction should not seem unfamiliar; all along we have seen the idea of reducing one problem to another. (For example, we recently saw how to reduce the matching problem into the max-flow problem, which could be reduced to linear programming.) The only difference is, right now, for convenience we are only considering yes-no type problems. A reduction from A to B, together with a polynomial time algorithm for B, yields a polynomial time algorithm for A. (See Figure 18.1.) Let us explain this in more detail. For any input x of A of size n, the reduction R takes time p n , where p is a polynomial, to produce an input R x for B. This input R x can have size at most p n , since this is the largest input R could possibly construct in p n time! We now submit R x as an input to the algorithm for B, which we assume runs in time q m on inputs of size m, where q is another polynomial. The algorithm for B gives us the right answer for B on R x , and hence also the right answer for A on x. The total time taken was at most pn q p n , which is itself just a polynomial in n!
This idea of reduction explains why the class P is so useful. If we have a problem A in P, and some other problem B reduces to it, then B is in P as well. Hence we say that P is closed under polynomial time reductions. If we can reduce A to B, we are essentially establishing that, give or take a polynomial, A is no harder or B. We can write this as
where here the inequality is represents a fact about the complexities of the two problems. If we know that B is easy,
We can also look at this inequality the other way. If we know that A is hard, then the inequality establishes that B is hard. It is this implication that we will now use, to show that problems are hard. This way of using reductions is very different from the way we have used reductions so far; it is also much more sophisticated and counter-intuitive.
¦
then A
B establishes that B is easy.
§
¦
A
B
¡
¡
¡
¡
¡ ¡
¡ ¡
¡
¡ ¥¡
£ ¤¡
¡
Lecture 18
18-4
x Input for A
Reduction R
R(x) Input for B
Algorithm for B
yes/no Output for B Output for A
Algorithm for A
Figure 18.1: Reductions lead to algorithms.
Lecture 18
18-5
Short Certificates and the Class NP
We will now begin to examine a class of problems that includes several “hard” problems. What we mean by “hard” in this setting is that although nobody has yet shown that there are no polynomial time algorithms to solve these problems, there is overwhelming evidence that this is the case. Recall that the class P is the class of yes-no problems that can be solved in polynomial time. The new class we define, NP, consists of yes-no problems with a different property: if the answer to the problem is yes, then there is a short certificate that can be checked to show that the answer is correct. A bit more formally, a short certificate must have the following properties:
The idea of the short certificate is the following: a problem is in NP if someone else can convince you in polynomial time that the answer is yes when the answer is yes, and they cannot fool you into thinking the answer is yes when the answer is no. Let us move from the abstract to some specific problems. Compositeness: Testing whether a number is composite is in NP, since if somebody wanted to convince you a number is composite, they could give you its factorization (the short certificate). You could then check that the factorization was correct by doing the multiplication, in polynomial time. (Notice you can’t be fooled!) 3SAT: 3SAT is like the 2SAT problem we have seen in the homework, except that there can be up to three literals in each clause. 3SAT is in NP, since if somebody wanted to convince you that a formula is satisfiable, they could give you a satisyfing truth assignment (the short certificate). You could then check the proposed truth assignment in polynomial time by plugging it in and checking each clause. (Again, notice you can’t be fooled!) Finally, note that P is a subset of NP. To see why, note that if a problem is in P, we don’t even need a short certificate; someone can convince themselves of the correct answer just by running the polynomial time algorithm! Now, let us see an example of a problem which does not appear to have short certificates: not-satisfiable-3SAT: This is like 3SAT, but now the answer is yes if there is no satisyfing assignment for the formula. Given a formula with no solution, how can we convince people there is no solution? The obvious way is to list all possible truth assignments, and show that they do not work, but this would not yield a short certificate.
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It must be short: the length of the polynomial is no more than polynomial in the length of the input. It must certify: there is a polynomial time checker (an algorithm!) that takes the input and the short certificate and checks that the certificate is valid.
Lecture 18
18-6
NP-completeness
The “hard” problems we will be looking at will be the hardest problems in NP; we call these problems NPcomplete. An NP-complete problem will have two properties:
problem, it must be at least as hard as any of them! It may seem surprising, that there are problems in NP that have this property. We will start by proving (well, sketching a proof) that an easily stated problem, circuit SAT, is NPcomplete. Once we have a first problem done, it will turn out to be much easier to prove that other problems are NP-complete. This is because once we have one NP-complete problem, it is much easier to prove others:
Claim 18.1 Suppose problem A is NP-complete, problem B is in NP, and problem A reduces to problem B. Then problem B is NP-complete.
and the hardest problems in NP are the NP-complete ones, then B must also be NP-complete. Slightly more formally, we have to show that every problem in NP reduces to B. But we already know that every problem reduces to A, and A reduces to B. By combining reductions, as in the picture below, we have that every problem in NP reduces to B. So once we have one problem, we can start building up “chains” of NP-complete problems easily.
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it is in NP all other problems in NP reduce to it
Thus, our concept of “being the hardest” is based on reductions. If all other problems in NP reduce to a
Intuitively, this must be true because if A reduces to B, then B is at least as hard as A. So as long as B is in NP,
Lecture 18
18-7
x Input for A
Reduction R
R(x) Input for B
Algorithm for B
yes/no Output for B Output for A
Algorithm for A
Figure 18.2: If C reduces to A, and A reduces to B, then C reduces to B. (Transitivity!)
Lecture 18
18-8
Cook’s Theorem
The problem circuit SAT is defined as follows: given a Boolean circuit and the values of some of its inputs, is there a way to set the rest of its inputs so that the output is T? It is easy to show that circuit SAT is in NP.
Claim 18.2 A problem is in NP if and only if it can be reduced to circuit SAT.
This statement is known as Cook’s theorem, and it is one of the most important results in Computer Science. One direction is easy. If a problem A can be reduced to circuit SAT , it can easily be shown to be in NP. A short certificate for an input to problem A consists of the short certificate for the circuit that results from running the reduction from A to circuit SAT on the input. Given this short certificate, a polynomial time algorithm could run the reduction on the input to A to get the appropriate circuit, and then use the short certificate to check the circuit. The other direction is more complicated, so we offer a somewhat informal explanation. Suppose that we have a problem A in NP. We need to show that it reduces to circuit SAT. Since A is in NP, there is a polynomial time algorithm that checks the validity of inputs of A together with the appropriate certificates. But we could program this algorithm on a computer, and this program would really be just a huge Boolean circuit. (After all, computers are just big Boolean circuits themselves!) The input to this circuit is the input to problem A along with a short certificate. Now suppose we are given a specific instance x of A. The question of whether x is a yes instance or no instance is exactly the question of whether there is an appropriate short certificate, which is exactly the same question ask asking if there is some way of setting the rest of the inputs to the Boolean circuit so that the answer is T. Hence, the construction of the circuit we described is the sought reduction from A to circuit SAT!
Lecture 18
18-9
More NP-complete problems
Now that we have proved that circuit SAT is NP-complete, we will build on this to find other NP-complete problems. For example, we will now show that circuit SAT reduces to 3SAT, and since 3SAT is clearly in NP, this shows that 3SAT is NP-complete. Suppose we are given a circuit C with some input gates unset. We must (quickly, in polynomial time) construct from this circuit a 3SAT-formula R C which is satisfiable if and only if there is a satisfying assignment of the circuit inputs. In essence, we want to mimic the actions of the circuit with a suitable formula. The formula R C will have one variable for each gate (that is, each input, and each output of an AND, OR, or NOT), and each gate will also lead to certain clauses, as described below: 1. If x is a T input gate, then add the clause x . 2. If x is a F input gate, then add the clause x . 3. If x is an unknown input gate, then no clauses are added for it. 4. If x is the OR of gates y and z, then add the clauses y x , z x , and x y z . (It is easy to see that the
5. If x is the AND of gates y and z, then add the clauses x y , x z , and y z x . (It is easy to see that the
6. If x is the NOT of gate y, then add the clauses x y and x y . (It is easy to see that the conjunction of these
7. Finally, if gate x is the output gate, add the clause x , expressing the condition that the output gate should be T. The conjuction of all of these clauses yields the formula R C . It should be apparent that this reduction R can be accomplished in polynomial (in fact, in linear) time. To verify it is a valid reduction, we must now show that C has a setting of the unknown input gates that makes the output T if and only if R(C) is satifiable. Suppose C has a valid setting. Then we claim R C can be satisfied by the truth assignment that gives each variable the same value as the appropriate gate when C is run on this valid setting. This truth assignment must satisfy all the clauses of R C , since we constructed R C to compute the same values as the circuit. Note that the output gate is T for C, and hence the final clause listed above is also satisfied.
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Lecture 18
18-10
Conversely (and this is more subtle!), if there is a valid truth assignment for R C , then there is a valid setting for the inputs of C that makes the output T. Just set the unknown input gates in the manner proscribed by the truth assignment for R C . Since R C effectively mimics the computation of the circuit, we know the output gate must be T when these inputs are applied.
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Lecture 18
18-11
From 3SAT to Integer Linear Programming We must take a 3SAT formula and convert it to an integer linear program. This reduction is easy. Restrict all
clearly satisifed if and only if this constraint is; all terms on the left of the equation are either 0 or 1, and there is at least one 1 if and only if one of the literals of the clause is true. It is somewhat strange that linear programming can be solved polynomial time, but when we try to restrict the solutions to be integers, then the problem appears not be solvable in polynomial time (since it is NP-complete).
whole thing to be at least 1. For example, the above clause becomes x
1
y
z
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, a literal x by x, and a literal x by 1
x , and then forcing the
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Lecture 18
18-12
From 3SAT to Independent Set
K such that no two are connected by an edge. The problem is clearly in NP. (Why?) We reduce 3SAT to Independent Set. That is, given a Boolean formula φ with at most 3 literals in each clause, size K or more if and only if the formula φ is satisfiable.
The reduction is illustrated in Figure 18.3. For each clause, we have a group of vertices, one for each literal in the clause, connected by all possible edges. Between groups of vertices, we connect two vertices if they correspond to opposite literals (like x and x). We let K be the number of clauses. This completes the reduction, and it is clear that it can be accomplished in polynomial time. We now show there is a satisfying truth assignment for φ if and only if there is an independent set of size at least K.
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we must (in polynomial time) come up with a graph G
V E and an integer K so that G has an independent set of
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K such that if u v
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E. That is, we are asked to find a set of vertices of size at least
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V E and an integer K. We are asked if there is a set
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Lecture 18
18-13
x+y+z x+y+z x x
x+y x+y+z x y y x
y
z
y
z
z
Figure 18.3: Turning formulae into graphs.
Lecture 18
18-14
If there is a truth assignment for φ, then there is at least one true literal in each clause. Pick just one for each clause in any way. The set I of corresponding vertices must give an independent set of size K. This is because we use only one vertex per clause, so the only way I could not be independent is if it included two opposite literals, which is impossible, because the satisfying assignment cannot set two opposite literals to T. Now suppose G has an independent set I of size K. Since there are K groups, and each group is completely interconnected, there must be one vertex from each group in I. Consider the assignment that sets all literals in the assignment to T, their opposites to F, and any unused variables arbitrarily. It is clear that this is a valid truth assignment (since if a variable is set to T, its opposite must be set to F).
Lecture 18
18-15
From Independent Set to Vertex Cover and Clique
in C. That is, each edge is adjacent to at least one vertex in the vertex cover. The Vertex Cover problem is, given a graph G and a number K, to determine if G has a vertex cover of size at most K. The reduction from Independent Set to Vertex Cover is immediate from the following observation: C is a
and the edge is covered.) So the reduction is trivial; given an instance G K of Independent Set, we produce the
A clique in a graph is a set of fully connected nodes– every possible edge between every pair of the nodes is there. The clique problem asks whether there is a clique of size K or larger in the graph. Again, the reduction from Independent Set is immediate from a simple observation. Let G be the complement of G, which is the graph with the same nodes as G, but the edges of G are precisely those edges that are missing from G. Then C is a clique in
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G
V E if and only if C is an independent set in G. (See Figure 18.4.)
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instance G V
K of Vertex Cover.
and consider some edge u v . Both u and v can’t be in the independent set, so V
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Lecture 18
18-16
Figure 18.4: Independent sets become cliques in the complement.
CS124
Lecture 19
We have defined the class of NP-complete problems, which have the property that if there is a polynomial time algorithm for any one of these problems, there is a polynomial time algorithm for all of them. Unfortunately, nobody has found an algorithm for any NP-complete problem, and it is widely believed that it is impossible to do so. This might seem like a big hint that we should just give up, and go back to solving problems like MAX-FLOW, where we can find a polynomial time solution. Unfortunately, NP-complete show up all the time in the real world, and people want solutions to these problems. What can we do?
19-1
Lecture 19
19-2
What Can We Do?
Actually, there is a great deal we can do. Here are just a few possibilities:
limited. However, these techniques have had some success in practice, and there are arguments in favor of why they are reasonable thing to try for some problems.
Restrict the inputs. NP-completeness refers to the worst case inputs for a problem. Often, inputs are not as bad as those that arise in NP-completeness proofs. For example, although the general SAT problem is hard, we have seen that the cases of 2SAT and Horn formulae have simple polynomial times algorithms. Provide approximations. NP-completeness resutls often arise because we want an exact answer. If we relax the problem so that we only have to return a good answer, then we might be able to develop a polynomial time algorithm. For example, we have seen that a greedy algorithm provides an approximate answer for the SET COVER problem. Develop heuristics. Sometimes we might not be able to make absolute guarantees, but we can develop algorithms that seem to work well in practice, and have arguments suggesting why they should work well. For example, the simplex algorithm for linear programming is exponential in the worst case, but in practice it’s generally the right tool for solving linear programming problems. Use randomness. So far, all our algorithms have been deterministic; they always run the same way on the same input. Perhaps if we let our algorithm do some things randomly, we can avoid the NP-completeness problem? Actually, the question of whether one can use randomness to solve an NP-complete problem is still open, though it appears unlikely. (As is, of course, the problem of whether one can solve an NP-complete problem in polynomial time!) However, randomness proves a useful tool when we try to come up with approximation algorithms and heuristics. Also, if one can assume the input comes from a suitable “random distribution”, then often one can develop an algorithm that works well on average.
To begin, we will look at heuristic methods. The amount we can prove about these methods is (as yet) very
Lecture 19
19-3
Local Search
“Local search” is meant to represent a large class of similar techniques that can be used to find a good solution for a problem. The idea is to think of the solution space as being represented by an undirected graph. That is, each possible solution is a node in the graph. An edge in the graph represents a possible move we can make between solutions. For example, consider the Number Partition problem for the homework assignment. Each possible solution, or division of the set of numbers into two groups, would be a vertex in the graph of all possible solutions. For our possible moves, we could move between solutions by changing the sign associated with a number. So in this case, our graph of all possible solutions, we have an edge between any two possible solutions that differ in only one sign. Of course this graph of all possible solutions is huge; there are 2n possible solutions when there are n numbers in the original problem! We could never hope to even write this graph down. The idea of local search is that we never actually try to write the whole graph down; we just move from one possible solution to a “nearby” possible solution, either for as long as we like, or until we happen to find an optimal solution.
Lecture 19
19-4
To set up a local search algorithm, we need to have the following: 1. A set of possible solutions, which will be the vertices in our local search graph. 2. A notion of what the neighbors of each vertex in the graph are. For each vertex x, we will call the set of adjacent vertices N x . The neighbors must satisfy several properties: N x must be easy to compute from x
sense to represent neighbors as undirected edges), and N x cannot be too big, or more than polynomial in the input size (so that the neighbors of a node are easy to search through). 3. A cost function, from possible solutions to the real numbers. The most basic local search algorithm (say to minimize the cost function) is easily described: 1. Pick a starting point x.
3. Return the final solution.
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N x then x
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Lecture 19
19-5
The Reasoning Behind Local Search
The idea behind local search is clear; if keep getting better and better solutions, we should end up with a good one. Pictorially, if we “project” the state space down to a two dimensional curve, we are hoping that the picture has a sink, or global optimum, and that we will quickly move toward it. See Figure 19.1.
f(x)
x*, global optimum x
Figure 19.1: A very nice state space. There are two possible problems with this line of thinking. First, even if the space does look this way, we might not move quickly enough toward the right solution. For example, for the number partition problem from the homework, it might be that each move improves our solution, but only by improving the residue by 1 each time. If we start with a bad solution, it will take a lot of moves to reach the minimum. Generally, however, this is not much of a problem, as long as the cost function is reasonably simple.
Lecture 19
19-6
The more important problem is that the solution space might not look this way at all. For example, our cost function might not change smoothly when we move from a state to it neighbor. Also, it may be that there are several local optima, in which cas our local search algorithm will hone in a local optimum and get stuck. See Figure 19.2.
f(x)
local optima x*, global optimum
x
Figure 19.2: A state space with many local optima; it will be hard to find the best solution. This second problem, that the solution space might not “look nice”, is crucial, and it underscores the importance of setting up the problem. When we choose the possible moves between solutions – that is, when we construct the mapping that gives us the neighborhood of each node– we are setting up how local search will behave, including how the cost function will change between neighbors, and how many local optima there are. How well local search will work depends tremendously on how smart one is in setting up the right neighborhoods, so that the solution space really does look the way we would like it to.
Lecture 19
19-7
Examples of Neighborhoods
We have already seen an example of a neighborhood for the homework problem. Here are possible neighborhoods for other problems:
MAX3SAT: A possible neighborhood structure is two truth assignments are neighbors if they differ in only one variables. A more extensive neighborhood could make two truth assignments neighbors if they differ in at most two variables; this trades increased flexibility for increase size in the neighborhood. Travelling Salesperson: The k-opt neighborhood of x is given by all tours that differ in at most k edges from x. In practice, using the 3-opt neighborhood seems to perform better than the 2-opt neighborhood, and using 4-opt or larger increases the neighborhood size to a point where it is inefficient.
Lecture 19
19-8
Lots of Room to Experiment
There are several aspects of local search algorithms that we can vary, and all can have an impact on performance. For example: 1. What are the nieghborhoods N x ? 2. How do we choose an inital starting point? 3. How do we choose a neighbor y to move to? (Do we take the first one we find, a random neighbor that improves f , the neighbor that improves f the most, or do we use other criteria?) 4. What if there are ties? There are other practical considerations to keep in mind. Can we re-run the algorithm several times? Can we try several of the algorithms on different machines? Issues like these can have a big impact on actual performance. However, perhaps the most important issue is to think of the right neighborhood structure to begin with; if this is right, then other issues are generally secondary, and if this is wrong, you are likely to fail no matter what you do.
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Lecture 19
19-9
Local Search Variations
There are many variations on the local search technique (below, assume the goal is to minimize the cost function):
swear that genetic algorithms lead to better solutions more quickly than other methods, while others claim that by choosing the right neighborhood functin one can do as well with hill climbing. In the years to come, hopefully more will become understood about all of these methods. If you’re interested, you might try looking for genetic algorithms and simulated annealing in Yahoo. They’re both there.
Hill-climbing – this is the name for the basic variation, where one moves to a vertex of lower (or possibly equal) cost. Metropolis rule – pick a random neighbor, and if the cost is lower, move there. If the cost is higher, move there with some probability (that is usually set to depend on the cost differential). The idea is that possibly moving to a worse state helps avoid getting trapped at local minima. Simulated annealing – this method is similar to the Metropolis rule, except that the probability of going to a higher cost neighbor varies with time. This is analogous to a physical system (such as a chemical polymer) being cooled down over time. Tabu search – this adds some memory to hill climbing. Like with the Metropolis rule and simulated annealing, you can go to worse solutions. A penalty function is added to the cost function to try to prevent cycling and promote searching new areas of the search space. Parallel search (“go with the winners”)– do multiple searches in parallel, occasionally killing off searches that appear less successful and replacing them with copies of searches that appear to be doing better. Genetic algorithms – this trendy area is actually quite related to local search. An important difference is that instead instead of keeping one solution at a time, a group of them (called a population) is kept, and the population changes at each step.
It is still quite unclear what exactly each of these techniques adds to the pot. For example, some people
CS124
Lecture 20
Heuristics can be useful in practice, but sometimes we would like to have guarantees. Approximation algorithms give guarantees. It is worth keeping in mind that sometimes approximation algorithms do not always perform as well as heuristic-based algorithms. Other times they provide insight into the problem, so they can help determine good heuristics. Often when we talk about an approximation algorithm, we give an approximation ratio. The approximation ratio gives the ratio between our solution and the actual solution. The goal is to obtain an approximation ratio as close to 1 as possible. If the problem involves a minimization, the approximation ratio will be greater than 1; if it involves a maximization, the approximation ratio will be less than 1.
20-1
Lecture 20
20-2
Vertex Cover Approximations
In the Vertex Cover problem, we wish to find a set of vertices of minimal size such that every edge is adjacent to some vertex in the cover. That is, given an undirected graph G = (V, E), we wish to find U ⊆ V such that every edge e ∈ E has an endpoint in U. We have seen that Vertex Cover is NP-complete. A natural greedy algorithm for Vertex Cover is to repeatedly choose a vertex with the highest degree, and put it into the cover. When we put the vertex in the cover, we remove the vertex and all its adjacent edges from the graph, and continue. Unfortunately, in this case the greedy algorithm gives us a rather poor aprroximation, as can be seen with the following example:
vertices chosen by greedy vertices in the min cover
Figure 20.1: A bad greedy example. In the example, all edges are connected to the base level; there are m/2 vertices at the next level, m/3 vertices at the next level, and so on. Each vertex at the base level is connected to one vertex at each other level, and the connections are spread as evenly as possible at each level. A greedy algorithm could always choose a rightmost vertex, whereas the optimal cover consists of the leftmost vertices. This example shows that, in general, the greedy approach could be off by a factor of Ω(log n), where n is the number of vertices.
Lecture 20
20-3
A better algorithm for vertex cover is the following: repeatedly choose an edge, and throw both of its endpoints into the cover. Throw the vertices and its adjacent edges out of the graph, and continue. It is easy to show that this second algorithm uses at most twice as many vertices as the optimal vertex cover. This is because each edge that gets chosen during the course of the algorithm must have one of its endpoints in the cover; hence we have merely always thrown two vertices in where we might have gotten away with throwing in 1. Somewhat surprsingly, this simple algorithm is still the best knwon approximation algorithm for the vertex cover problem. That is, no algorithm has been proven to do better than within a factor of 2.
Lecture 20
20-4
Maximum Cut Approximation
We will provide both a randomized and a deterministic approximation algorithm for the MAX CUT problem. The MAX CUT problem is to divide the vertices in a graph into two disjoint sets so that the numbers of edges between vertices in different sets is maximized. This problem is NP-hard. Notice that the MIN CUT problem can be solved in polynomial time by repeated using the min cut-max flow algorithm. (Exercise: Prove this!) The randomized version of the algorithm is as follows: we divide the vertices into two sets, HEADS and TAILS. We decide where each vertex goes by flipping a (fair) coin. What is the probability an edge crosses between the sets of the cut? This will happen only if its two endpoints lie on different sides, which happens 1/2 of the time. (There are 4 possibilities for the two endpoints – HH,HT,TT,TH – and two of these put the vertices on different sides.) So, on average, we expect 1/2 the edges in the graph to cross the cut. Since the most we could have is for all the edges to cross the cut, this random assignment will, on average, be within a factor of 2 of optimal.
Lecture 20
20-5
We now examine a deterministic algorithm with the same “approximation ratio”. (In fact, the two algorithms are intrinsically related– but this is not so easy to see!) The algorithm implements the hill climbing approximation heuristic. We will split the vertices into sets S 1 and S2 . Start with all vertices on one side of the cut. Now, if you can switch a vertex to a different side so that it increases the number of edges across the cut, do so. Repeat this action until the cut can no longer be improved by this simple switch. We switch vertices at most |E| times (since each time, the number of edges across the cut increases). Moreover, when the process finishes we are within a factor of 2 of the optimal, as we shall now show. In fact, when the process finishes, at least |E|/2 edges lie in the cut. We can count the edges in the cut in the following way: consider any vertex v ∈ S 1 . For every vertex w in S2 that it is connected to by an edge, we add 1/2 to a running sum. We do the same for each vertex in S 2 . Note that each edge crossing the cut contributes 1 to the sum– 1/2 for each vertex of the edge. Hence the cut C satisfies C= 1 2
v∈S1
∑ |{w : (v, w) ∈ E, w ∈ S2 }| + ∑ |{w : (v, w) ∈ E, w ∈ S1 }|
v∈S2
.
Since we are using the local search algorithm, at least half the edges from any vertex v must lie in the set opposite from v; otherwise, we could switch what side vertex v is on, and improve the cut! Hence, if vertex v has degree δ(v), then C = ≥ = = 1 2 1 2
v∈S1
∑ |{w : (v, w) ∈ E, w ∈ S2 }| + ∑ |{w : (v, w) ∈ E, w ∈ S1 }|
v∈S2
δ(v) δ(v) +∑ v∈S2 2 v∈S1 2
1 ∑ δ(v) 4 v∈V 1 |E|, 2
where the last equality follows from the fact that if we sum the degree of all vertices, we obtain twice the number of edges, since we have counted each edge twice. In practice, we might expect that hill climbing algorithm would do better than just getting a cut within a factor of 2.
Lecture 20
20-6
Euclidean Travelling Salesperson Problem
In the Euclidean Travelling Salesman Problem, we are given n points (cities) in the x − y plane, and we seek the tour (cycle) of minimum length that travels through all the cities. This problem is NP-complete (showing this is somewhat difficult). Our approximation algorithm involves the following steps: 1. Find a minimum spanning tree T for the points. 2. Create a psuedo tour by walking around the tree. The pseduo tour may visit some vertices twice. 3. Remove repeats from the tour by short-cutting through the repeated vertices. (See Figure 20.2.)
Lecture 20
20-7
X
Minimum spanning tree Constructed pseudo tour Constructed tour
Figure 20.2: Building an approximate tour. Start at X, move in the direction shown, short-cutting repeated vertices.
Lecture 20
20-8
We now show the following inequalities: length of tour ≤ length of pseudo tour ≤ 2(size of T) ≤ 2(length of optimal tour)
Short-cutting edges can only decrease the length of the tour, so the tour given by the algorithm is at most the length of the pseudo tour. The length of our psuedo tour is at most twice the size of the spanning tree, since this pseudo tour consists of walking through each edge of the tree at most twice. Finally, the length of the optimal tour is at least the size of the minimum spanning tree, since any tour contains a spanning tree (plus an edge!). Using a similar idea, one can come up with an approximation algorithm that returns a tour that is within a factor of 3/2 of the optimal. Also, note that this algorithm will work in any setting where short-cutting is effective. More specifically, it will work for any instance of the travelling salesperson problem that satisfies the triangle inequality for distances: that is, if d(x, y) represents the distance between vertices x and y, and d(x, z) ≤ d(x, y) + d(y, z) for all x, y and z.
Lecture 20
20-9
MAX-SAT: Applying Randomness
Consider the MAX-SAT problem. What happens if we do the simplest random thing we can think of– we decide whether each variable should be TRUE or FALSE by flipping a coin.
Theorem 20.1 On average, at least half the clauses will be satisfied if we just flip a coin to decide the value of each variable. Moreover, if each clause has k literals, then on average 1 − 2 −k clauses will be satisfied. The proof is simple. Look at each clause. If it has k literals in it, then each literal could make the clause TRUE with probability 1/2. So the probability the clause is not satisfied is 1 − 2 −k , where k is the number of literals in the clause.
Lecture 20
20-10
Linear Programming Relaxation
The next approach we describe, linear programming relaxation, can often be used as a good heuristic, and in some cases it leads to approximation algorithms with provable guarantees. Again, we will use the MAX-SAT problem as an example of how to use this technique. The idea is simple. Most NP-complete problems can be easily described by a natural Integer Programming problem. (Of course, all NP-complete problems can be transformed into some Integer Programming problem, since Integer Programming is NP-complete; but what we mean here is in many cases the transformation is quite natural.) Even though we cannot solve the related Integer Program, if we pretend it is a linear program, then we can solve it, using (for example) the simplex method. This idea is konwn as relaxation, since we are relaxing the constraints on the solution; we are no longer requiring that we get a solution where the variables take on integer values. If we are extremely lucky, we might find a solution of the linear program where all the variables are integers, in which case we will have solved our original problem. Usually, we will not. In this case we will have to try to somehow take the linear programming solution, and modify it into a solution where all the variables take on integer values. Randomized Rouding is one technique for doing this.
Lecture 20
20-11
MAX-SAT
We may formulate MAX-SAT as an integer programming problem in a straightforward way (in fact, we have seen a similar reduction before, back when we examined reducitons; it is repeated here). Suppose the formula contains variables x1 , x2 , . . . , xn which must be set to TRUE or FALSE, and clauses C1 ,C2 , . . . ,Cm . For each variable xi we associate a variable yi which should be 1 if the variable is TRUE, and 0 if it is FALSE. For each clause C j we have a variable z j which should be 1 if the clause is satisfied and 0 otherwise. We wish to maximize the number of satisfied clauses s, or
j=1
∑ z j.
m
The constraints include that that 0 ≤ y i , z j ≤ 1; since this is an integer program, this forces all these variables to be either 0 or 1. Finally, we need a constraint for each clause saying that its associated variable z j can be 1 if and only if the clause is actually satisfied. If the clause C j is (x2 ∨ x4 ∨ x6 ∨ x8 ), for example, then we need the restriction: y2 + y6 + (1 − y4 ) + (1 − y8 ) ≥ z j . This forces z j to be 0 unless the clause can be satisfied. In general, we replace x i by yi , xi by 1 − yi , ∨ by +, and set the whole thing ≥ z j to get the appropriate constraint. When we solve the linear program, we will get a solution that might have y 1 = 0.7 and z1 = 0.6, for instance. This initially appears to make no sense, since a variable cannot be 0.7 TRUE. But we can still use these values in a reasonable way. If y1 = 0.7, it suggests that we would prefer to set the variable x 1 to TRUE (1). In fact, we could try just rounding each variable up or down to 0 or 1, and use that as a solution! This would be one way to turn the non-integer solution into an integer solution. Unfortunately, there are problems with this method. For example, suppose we have the clause C1 = (x1 ∨x2 ∨x3 ), and y1 = y2 = y3 = 0.4. Then by simple rounding, this clause will not be TRUE, even though it “seems satisfied” to our linear program (that is, z 1 = 1). If we have a lot of these clauses, regular rounding might perform very poorly. It turns out that there an interpretation for 0.7 that suggests a better way than simple rounding. We think of the 0.7 as a probability. That is, we interpret y 1 = 0.7 as meaning that x1 would like to be true with probability 0.7. So we take each variable xi , and independently we set it to 1 with the probability given by y i (and with probability 1 − yi we set xi to 0). This process is known as randomized rounding. One reason randomized rounding is useful is it allows us to prove that the expected number of clauses we satisfy using this rounding is a within a constant factor of the true optimum.
Lecture 20
20-12
First, note that whatever the maximum number of clauses s we can satisfy is, the value found by the linear program, or ∑m z j , is at least as big as s. This is because the linear program could achieve a value of at least s j=1 simply by using as the values for yi the truth assignment that make satisfying s clauses possible. Now consider a clause with k variables; for convenience, suppose the clause is just C 1 = (x1 ∨ x2 . . . ∨ xk ).
Suppose that when we solve the linear program, we find z 1 = β. Then we claim that the probability that this clause is satisfied after the rounding is at least (1 − 1/e)β. This can be checked (using a bit of sophisticated math), but it
follows by noting (with experiments) that the worst possibility is that y 1 = y2 . . . = yk = β/k. In this case, each x1 is FALSE with probability (1 − β/k), and so C1 ends up being unsatisfied with probability (1 − β/k) k . Hence the probability it is satisfied is at least (again using some math) 1 − (1 − β/k) k ≥ (1 − 1/e)β.
maximum number of satisfiable clauses, ∑m z j . Hence we expected to get within a constant factor of the maximum. j=1
after randomized rounding is at least (1 − 1/e) ∑ m z j . This is within a factor of (1 − 1/e) of our upper bound on the j=1
Hence the ith clause is satisfied with probability at least (1 − 1/e)z i , so the expected number of satisfied clauses
Lecture 20
20-13
Combining the Two
Surprisingly, by combining the simple coin flipping algorithm with the randomized rounding algorithm, we can get an even better algorithm. The idea is that the coin flipping algorithm does best on long clauses, since each literal in the clause makes it more likely the clause gets set to TRUE. On the other hand, randomized rounding does best on short clauses; the probability the clause is satisfied (1 − (1 − β/k) k ) decreases with k. It turns out that if we try both algorithms, and take the better result, on average we will satisfy 3/4 of the clauses. We also point out that there are even more sophisticated approximation algorithms for MAX-SAT, with better approximation ratios. However, these algorithms point out some very interesting and useful general techniques.
CS 124
Lecture 21
We now consider a natural problem that arises in many applications, particularly in conjunction with suffix trees, which we will study later. Suppose we have a rooted tree T with n nodes. We would like to be able to answer questions of the following form: what is the least common ancestor of nodes u and v; that is, what is the common ancestor of u and v closest to the root? In this setting, we will not be answering a single questions, but many questions on the same fixed tree T . If we are given the tree T in advance, we can design an appropriate data structure for answering future queries. Our algorithm will therefore be measured on several criteria. Of course one important criterion is the query time, or the time to answer a specific query. However, a second consideration is how much preprocessing time, or time to set up the data structure, is required to answer the questions. A third related aspect to study is the memory required to store the data structure. For example, a trivial algorithm for the problem is to consider each pair of vertices, and compute their least common ancestor by following both paths toward the root until the first shared vertex is found. Then all the the answers can be stored in a table. There are
n 2
pairs of vertices, so our table will require Θ(n 2 ) space. Queries can
be answered by a table lookup, which is constant time. Preprocessing, however, can require Θ(n 3 ) time. The problem of designing an appropriate data structure for this is called the Least Common Ancestor (LCA) Problem. We will show that there is an algorithm for LCA that require only linear preprocessing time and memory, but still answers any query in constant time! This result is as efficient as we could hope for. We will reduce the LCA problem to a seemingly different but in fact quite related problem, called the Range Minimum Query (RMQ) Problem. The RMQ problem applies to an array A of length n of numbers. We would like to be able to answer questions of the following form: given two indices i and j, what is the index of the smallest element in the subarray A[i . . . j]? Again, we may prepocess the array A to derive some alternative data structure to answer the questions quickly. There is a trivial solution for the RMQ problem completely similar to the one above for the LCA problem.
21-1
Lecture 21
21-2
21.1 Reduction: From LCA to RMQ
How to we convert an LCA problem to an RMQ problem? Note that we must do the conversion in linear time, if we are going to totally complete the preprocessing in linear time for the LCA problem. Linear time suggests that we want to do a tree traversal. In fact, the observation we will use is that the LCA of nodes u and v is just the shallowest node encountered between visiting u and v during a depth first search of the tree starting at the root. So let us do a DFS on the tree, and we can record in an array V the nodes we visit. An example is shown in Figure 1. Notice each node can appear multiple times, but the total length of the array is 2n − 1, where n is the number of nodes in the tree. Each of the n − 1 edges yields two values in the array, one when we go down the edge and one when we go up the edge. The first value is the root. Also, from now on we will refer to each node by its number on the DFS search. We will also require two further arrays. The Level Array is derived from V ; L[i] is the distance from the root of V [i]. Adjacent elements in L can only differ by +1 or −1, since adjacent steps in the DFS are connected by an edge. Finally, R[i] is the representative array; R[i] contains the first index of V that contains the value i. (Actually, any occurrence of i can be stored in R[i], but we might as well choose a specific one.) Clearly, to compute LCA(u, v) it suffices to compute RMQ(R[u], R[v]) over the array L. This gives us the index of the shallowest node between u and v, and the array V can be used to determine the actual node from the index.
Lecture 21
21-3
0 1 2 3 4 5 6 7 8 9
V: 0 1 2 1 3 1 0 4 0 5 6 5 7 8 7 5 9 5 0 L: 0 1 2 1 2 1 0 1 0 1 2 1 2 3 2 1 2 1 0 R: 0 1 2 4 7 9 10 12 13 16
Figure 1: Changing an LCA problem into an RMQ problem.
21.2 Solutions for RMQ
We first note that we can do better than the naive Θ(n 3 ) preprocessing time for RMQ on an array A by doing a trivial dynamic programming, using the recurrence RMQ(i, j) = A−1 [min(A[RMQ(i, j − 1)], A[ j])]. Here we are using convenient notation. Clearly min(A[RMQ(i, j − 1)], A[ j]) gives the value A[k], where A[k] is the represent that we want the index of this value; note that if multiple indices have this value, we do not particularly care which index we obtain. Each table entry can be calculated in constant time by building the table in order of ranges [i, j] of increasing size, leading to preprocessing time Θ(n 2 ). In fact, we can reduce our table size and memory using a different dynamic program, and by using a few additional operations per query. Let us create a table M(i, j) such that M(i, j) = A −1 [mink∈[i,i+2 j ) A[k]]. That is, M(i, j) contains the location of the minimum value over the 2 j positions starting from i. This table has size O(n log n), and it can easily be filled in O(n log n) step by using dynamic programming, based on the fact that M(i, j) can be
smallest value that in the subarray A[i . . . j]. However, we want the index of this value. We use the notation A −1 to
Lecture 21
21-4
determined from M(i, j − 1) and M(i + 2 j−1 , j − 1). How do we use the M(i, j) to compute RMQ(i, j), if j is not a power of 2? We may use two overlapping intervals that cover the range [i, j] as follows. Let k = log( j − i + 1) , so that 2 k is the largest power of 2 such that i + 2k ≤ j + 1. Then RMQ(i, j) = A−1 [minA[M(i, k)], A[M( j − 2k , k)]], and this can be computed in constant time from the M. We have shown that we can achieve preprocessing time and memory size Θ(n log n) while maintaining constant query time. Interestingly, this method can be enhanced so as to require preprocessing time and memory size Θ(n log log n) through a recursive construction. (This will be an exercise.) In practice, such a result would probably be good enough – log log n is quite small for reasonable values of n. By continuing the recursive construction for further levels, we could even achieve Θ(n log log log n) preprocessing time and memory size, and so on for any fixed number of logs, while maintaining constant query time. However, this recursive construction would add significant complexity to an actual program, and it still would not lead us to a linear preprocessing time solution.
Lecture 21
21-5
21.3 ±1 RMQ
In order to achieve linear preprocessing, we will use an additional fact about the RMQ problem we obtain from the reduction from LCA. Recall that our RMQ problem is on the Level Array obtained from the LCA problem. The Level Array has one additional property that we are not yet taking advantage of: each entry differs from the previous entry by +1 or −1. We can take advantage of this fact to split the RMQ problem into a different set of small subproblems in such a way that we can avoid some work by doing table look-ups. The split works as follows: partition A into blocks of size
log n 2 .
Let X[1, . . . , 2n/ log n] and Y [1, . . . , 2n/ log n] be
arrays such that X[i] stores the minimum element in the ith block of A, and Y [i] stores the position in the ith block where the element X[i] occurs. Now to answer an RMQ query for indices i and j with i < j on the array A, we can do the following: 1. If i and j are in the same block, we can perform an RMQ on this block. Notice that this requires that each block be preprocessed. 2. If i and j are in different blocks, we have to compute the following values, and take the minimum of them: (a) The minimum from position i to the end of i’s block. (b) The minimum from the beginning of j’s block to position j. (c) The minimum of all blocks between i’s block and j’s block. Steps 2a and 2c also require that we preprocess for RMQ queries on each block. Step 2b requires that we perform an RMQ over the array X. Assuming we have done all this preprocessing, the total query time is still constant. However, if we preprocess each block in order to do RMQ’s, we have not saved on the running time. We need a faster way to deal with preprocessing each block. How can we possibly avoid preprocessing each block separately? We use the following observation. Consider two arrays X and X . Suppose that these two arrays differ by a constant at each position; for example, the arrays might be 1, 2, 3, 4, 3, 2 . . . and 3, 4, 5, 6, 5, 4 . . . and. Then the RMQ answers, which give the index of the minimum element, will be the same for these two arrays. Hence we can “share” the preprocessing used for these two arrays! Another way to explain this is that in the ±1 RMQ problem, the initial value of the array does not matter, only the sequence of +1 and −1 values are necessary to determine the answer. Now, how many different such sequences are there? Since there are only log n/2 elements in a block, there are only (log n/2) − 1 values in the sequence of +1
Lecture 21
21-6
and −1 values. Hence there are only 2 (log n/2)−1 =
n/2 possible sequences. This number is so small, we can afford
to compute and store tables for every possible sequence! Even if we use quadratic preprocessing time and memory, √ √ these tables would take time O( n log2 n) to preprocess and O( n log2 n) memory. For each block in A, we have to determine which table to use; this can easily be done in linear time.
Lecture 21
21-7
21.4 Back to the standard RMQ
We have shown that ±1 RMQ problems can be solved with linear time preprocessing, and therefore we have a linear time preprocessing solution for LCA. What about the general RMQ problem? It turns out that we can also reduce the RMQ problem to the LCA problem in linear time. So we can obtain a linear time solution the general RMQ problem, by turning it into an LCA problem, and solving that as a ±1 RMQ problem! The details of this reduction are omitted here.
CS 124
Lecture 22
Spring 2000
Suffix trees are an old data structure that have become new again, thanks to a recent new linear time algorithm for constructing suffix trees due to Ukkonen that proves more useful for many applications. Here, we will describe a suffix tree and discuss their classical use, pattern matching.
22.1 Definition
A suffix tree T is built for a string S[1 . . . m]. The tree is rooted and directed with m leaves, which are numbered from 1 to m. Each edge is labeled with a nonempty substring of S. The internal nodes of the tree (other than the root) all have at least two outgoing edges, and the labels of all outgoing edges are labeled with different characters. By following the path from the root to leaf i and concatenating the edge labels, one obtains the suffix S[i . . . m]. An example of a suffix tree for the string xyzxzxy\$ is given in Figure 22.1. The figure helps understand some important points about the suffix tree. First, each internal node has two or more children with different starting characters along the edges, since otherwise the node could be removed or moved in order to make this the case. Also, it is important that the last character of the string be a “unique” character, as this guarantees that the suffix tree as defined actually exists. For example, suppose our string was just xyzxzxy. The suffix tree would remain largely the same. In particular, in the not-quite-suffix tree in Figure 22.1 the path for the suffix xy does not end at a leaf, violating the definition. The problem is that the suffix xy is also the prefix of the string. This problem can be avoided by terminating the string with a special character that does not appear elsewhere, since then no suffix can also be a prefix (except for the entire string itself). Hence, from now on, we will assume all strings end with a special character \$. It is also worth noting that a more convenient represenation of the suffix tree does not actually label the edges with characters. Instead, these labels can be represented by a pair of indices; labeling an edge [i, j] represents that the edge label corresponds to characters S[i . . . j]. Besides saving space and ensuring that each edge is conveniently represented by two numbers, this scheme is important for the linear time algorithm for suffix tree construction.
22-1
Lecture 22
22-2
22.2 Construction algorithm
Lecture 22
22-3
Fortunately, there are slightly more complex construction algorithms that require only O(m) time. We will not discuss the algorithm at this point; the details and the subsequent analysis would require a non-trivial amount of time. A reasonable introduction to the algorithm, however, has been written by Mark Nelson and has appeared in Dr. Dobb’s Journal. You can currently find it at http://www.dogma.net/markn/articles/suffixt/suffixt.htm.
Lecture 22
22-4
22.3 Using suffix trees for pattern matching
Once we have constructed our suffix tree, we can use it to efficiently solve pattern matching problems. There are of course other methods for pattern matching, but using suffix trees has an interesting advantage. Once the suffix tree has been constructed, finding all the occurences of any pattern P[1 . . . n] in the string S takes time O(n + k), where k is the number of times that the string S appears in the text. So by incurring a one-time preprocessing charge to establish the suffix trees, we can handle any pattern matching problem after that in time essentially proportional to the length of the pattern, independent of the length of the original string! This is quite powerful, particularly for things like DNA databases, where the underlying database is large and fixed but must be able to deal with lots of queries. Suppose that P lies in the string S; for example, suppose P corresponds to S[i . . . i + n − 1]. Then P is the prefix of the suffix S[i . . . m]. Hence, if we starting matching characters in P against the labels in the suffix tree for S, we will follow part of the path from the root to the leaf vertex labeled i. Hence, to find all occurences of P in S, start at the root, and match down the tree as far as possible. This takes time O(n). If P does not match some path in the tree, then P does not lie in S. If P does match some path in the tree, in matches down to some point z. All the leaves in the subtree below z correspond to suffixes for which P is a prefix, so the labels on these leaves correspond to locations that begin an occurence of P. To find these positions, we just traverse the subtree below z, using for example depth first search. If there are k leaves, the depth first search takes only O(k) time.
Lecture 22
22-5
22.4 Representation
An important point about suffix trees: to make sure everything takes linear time, it is important to use the correct representation. For example, we do not explicitly label each edge with a group of characters– this could take as much as Ω(n2 ) time to just write down! Instead, each edge is labeled with a pair of values, representing characters. For example, an edge labeled [a, b] should be thought of as being labeled by the character S[a] . . . S[b]. Hence each edge is just labeled by two numbers, and only linear space is required.
Lecture 22
22-6
\$ x y \$ 7 zx
8
y zxzxy\$ \$ 1 6
zxy\$ 4
zxzxy\$ zxy\$ 2 3 5 y\$
x
y 7
zx zxzxy zxy 2 3 5 y
y zxzxy 1 6
zxy 4
Figure 22.1: A true suffix tree (top); why we need the \$ character (bottom).
Lecture 22
22-7
22.5 Generalized suffix trees
You may want to put a set {S1 , S2 , . . . , Sk } of strings in a suffix tree data structure. (Note– we assume each string ends with the special character \$.) The structure in this case is called a generalized suffix tree. There are two primary differences. First, now each leaf node may contain multiple pairs of numbers. Each pair of numbers identifies a string Si and a location where the suffix from the root to that leaf starts in S i . Note that multiple strings can have a suffix that share a leaf node! Second, each edge label must be represented by three numbers: a number i and a pair [a, b] represent that the characters on the edge label are S i [a] . . . Si [b]. Construcing a generalized suffix tree can easily be done by extending our quadratic time algorithm. However, the linear time algorithm for suffix trees can also be used to build a generalized suffix tree. Hence if m = ∑k |Si |, i=1 constructing the generalized suffix tree can be done in O(m) time.
Lecture 22
22-8
22.6 Longest common extension
Using generalized suffix trees and the LCA algorithm, we can solve a very general problem called the largest common extension problem. Given strings S 1 and S2 , we wish to pre-process the string so that we can answer questions of the following form: given a pair (i, j), find the longest substring of S 1 that begins at position i that matches a substring of S2 that begins at position j. We will use linear time pre-processing and linear space, after which we can answer queries in constant time. The solution is to build a generalized search tree for S 1 and S2 . When we build this tree, we should also compute the string depth of each node. The string depth of a node is simply the number of characters along the edges from the root to that node. Notice the string depth is not the same as the tree depth. Also, after building the tree, we precompute the information necessary to do LCA queries on the tree. Given a pair (i, j) we compute the least common ancestor u of the leaf nodes corresponding to the suffix beginning at i in S1 and the suffix beginning at j in S2 . The path from the root to u is longest common extension, and hence the string depth of this node is all we need.
Lecture 22
22-9
22.7 Maximal palindromes
A palindrome is a string that reads the same forwards as backwards, such as axbccbxa. A substring U of a string S is a maximal palindrome if and only if it is a palindrome and extending it one character in both directions yields a string that is not a palindrome. Generally we separate even-length maximal palindromes, or even palindromes for short, and odd-length maximal palindromes (odd palindromes) for convenience. For example, in S = axbccbbbaa, the maximal even palindromes are bccb, bb, and aa. The string bbb is a maximal odd palindrome, and we will skip writing the maximal odd palindromes of length 1. Note that every palindrome is contained in a maximal palindrome. Here is a simple way to find all even-length maximal palindromes in linear time. (Finding odd-length maximal palindromes is similar.) Consider S and Sr , the reversal of S. There is a palindrome of length 2k with the middle just after position q if the string of length k starting from position q + 1 of S matches the string of length k starting from position n − q + 1 of Sr . In particular, this palindrome will be maximal if this is the length of the longest match from these positions. Thus, solving the even-length maximal palindrome problem corresponds to computing the longest common extension of (q + 1, n − q + 1) for all possible q. The data stucture can be processed in linear time, and each of the linear number of queries can be answered in constant time, so the total time is linear. | 64,957 | 253,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-47 | latest | en | 0.927225 |
https://www.neetprep.com/questions/55-Physics/688-Kinetic-Theory-Gases?courseId=8&testId=1061963-Past-Year--onward--NTA-Papers-MCQs&questionId=245074-mean-free-path-l-gas-molecule-depends-upon-diameter-d-ofthe-molecule-aslpropto-fracdlpropto-dlpropto-d-lpropto-fracd | 1,719,024,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00754.warc.gz | 793,904,444 | 57,611 | # The mean free path $$l$$ for a gas molecule depends upon the diameter, $$d$$ of the molecule as: 1. $$l\propto \frac{1}{d^2}$$ 2. $$l\propto d$$ 3. $$l\propto d^2$$ 4. $$l\propto \frac{1}{d}$$
Subtopic: Mean Free Path |
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An ideal gas equation can be written as $\mathrm{P}=\frac{\mathrm{\rho RT}}{{\mathrm{M}}_{0}}$ where $\mathrm{\rho }$ and ${\mathrm{M}}_{0}$ are respectively:
1 mass density, the mass of the gas 2 number density, molar mass 3 mass density, molar mass 4 number density, the mass of the gas
Subtopic: Ideal Gas Equation |
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Match Column - I and Column - II and choose the correct match from the given choices.
Column - I Column - II (A) root mean square speed of gas molecules (P) $$\frac13nm\bar v^2$$ (B) the pressure exerted by an ideal gas (Q) $$\sqrt{\frac{3 R T}{M}}$$ (C) the average kinetic energy of a molecule (R) $$\frac{5}{2} R T$$ (D) the total internal energy of 1 mole of a diatomic gas (S) $$\frac32k_BT$$
(A) (B) (C) (D) 1. (Q) (P) (S) (R) 2. (R) (Q) (P) (S) 3. (R) (P) (S) (Q) 4. (Q) (R) (S) (P)
Subtopic: Kinetic Energy of an Ideal Gas |
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The average thermal energy for a mono-atomic gas is:
($$k_B$$ is Boltzmann constant and T absolute temperature)
1. $$\frac{3}{2}k_BT$$
2. $$\frac{5}{2}k_BT$$
3. $$\frac{7}{2}k_BT$$
4. $$\frac{1}{2}k_BT$$
Subtopic: Kinetic Energy of an Ideal Gas |
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A cylinder contains hydrogen gas at a pressure of $$249~\text{kPa}$$ and temperature $$27^\circ~\mathrm{C}.$$ Its density is: ($$R=8.3~\text{J mol}^{-1} \text {K}^{-1}$$)
1. $$0.2~\text{kg/m}^{3}$$
2. $$0.1~\text{kg/m}^{3}$$
3. $$0.02~\text{kg/m}^{3}$$
4. $$0.5~\text{kg/m}^{3}$$
Subtopic: Ideal Gas Equation |
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The mean free path for a gas, with molecular diameter $$d$$ and number density $$n,$$ can be expressed as:
1. $$\frac{1}{\sqrt{2} n \pi \mathrm{d}^2}$$
2. $$\frac{1}{\sqrt{2} n^2 \pi \mathrm{d}^2}$$
3. $$\frac{1}{\sqrt{2} n^2 \pi^2 d^2}$$
4. $$\frac{1}{\sqrt{2} n \pi \mathrm{d}}$$
Subtopic: Mean Free Path |
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The volume occupied by the molecules contained in $$4.5$$ kg water at STP, if the molecular forces vanish away, is:
1. $$5.6$$ m3
2. $$5.6\times10^{6}$$ m3
3. $$5.6\times10^{3}$$ m3
4. $$5.6\times10^{-3}$$ m3
Subtopic: Ideal Gas Equation |
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The temperature at which the rms speed of atoms in neon gas is equal to the rms speed of hydrogen molecules at $$15^{\circ} \mathrm{C}$$ is: (Atomic mass of neon$$=20.2$$ u, molecular mass of hydrogen$$=2$$ u)
1. $$2.9\times10^{3}$$ K
2. $$2.9$$ K
3. $$0.15\times10^{3}$$ K
4. $$0.29\times10^{3}$$ K
Subtopic: Types of Velocities |
74%
From NCERT
NEET - 2022
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NEET 2025 - Target Batch
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NEET 2025 - Target Batch
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains helium (monoatomic), the second contains fluorine (diatomic) and the third contains sulfur hexafluoride (polyatomic). The correct statement, among the following, is:
1 all vessels contain unequal number of respective molecules. 2 the root mean square speed of molecules is same in all the three cases. 3 the root mean square speed of helium is the largest. 4 the root mean square speed of sulfur hexafluoride is the largest.
Subtopic: Types of Velocities |
72%
From NCERT
NEET - 2022
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
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NEET 2025 - Target Batch
The temperature of a gas is $$-50^\circ \mathrm{C}$$. To what temperature the gas should be heated so that the rms speed is increased by $$3$$ times?
1 $$223$$ $$\text{K}$$ 2 $$669^\circ \mathrm{C}$$ 3 $$3295^\circ \mathrm{C}$$ 4 $$3097$$ $$\text{K}$$
Subtopic: Types of Velocities |
51%
From NCERT
NEET - 2023 | 1,748 | 5,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-26 | latest | en | 0.67644 |
http://conversion.org/mass/crith/hundredweight-long | 1,719,221,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00885.warc.gz | 6,840,718 | 7,054 | # crith to hundredweight (long) conversion
Conversion number between crith and hundredweight (long) [long cwt or cwt] is 1.770290312801 × 10-6. This means, that crith is smaller unit than hundredweight (long).
### Contents [show][hide]
Switch to reverse conversion:
from hundredweight (long) to crith conversion
### Enter the number in crith:
Decimal Fraction Exponential Expression
crith
eg.: 10.12345 or 1.123e5
Result in hundredweight (long)
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 crith = (8.99349*10^-05) / (50.80234544) = 1.770290312801 × 10-6 hundredweight (long)
• 1 hundredweight (long) = (50.80234544) / (8.99349*10^-05) = 564879.10077178 crith
• ? crith × (8.99349*10^-05 ("kg"/"crith")) / (50.80234544 ("kg"/"hundredweight (long)")) = ? hundredweight (long)
### High precision conversion
If conversion between crith to kilogram and kilogram to hundredweight (long) is exactly definied, high precision conversion from crith to hundredweight (long) is enabled.
Since definition contain rounded number(s) too, there is no sense for high precision calculation, but if you want, you can enable it. Keep in mind, that converted number will be inaccurate due this rounding error!
### crith to hundredweight (long) conversion chart
Start value: [crith] Step size [crith] How many lines? (max 100)
visual:
crithhundredweight (long)
00
101.770290312801 × 10-5
203.5405806256019 × 10-5
305.3108709384029 × 10-5
407.0811612512038 × 10-5
508.8514515640048 × 10-5
600.00010621741876806
700.00012392032189607
800.00014162322502408
900.00015932612815209
1000.0001770290312801
1100.00019473193440811
Copy to Excel
## Multiple conversion
Enter numbers in crith and click convert button.
One number per line.
Converted numbers in hundredweight (long):
Click to select all
## Details about crith and hundredweight (long) units:
Convert Crith to other unit:
### crith
Definition of crith unit: ≈ 89.9349 mg.
Convert Hundredweight (long) to other unit:
### hundredweight (long)
Definition of hundredweight (long) unit: ≡ 112 lb av . =112 x 0.45359237 kg = 50.80234544 kg
← Back to Mass units | 668 | 2,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.685436 |
https://www.tammysenglishblog.com/2017/07/how-waec-neco-and-nabteb-count-number.html | 1,719,296,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00028.warc.gz | 895,639,403 | 34,992 | # How WAEC, NECO and NABTEB Count the Number of Words of Students' English Essays
This article provides answer(s) to one of the most frequently asked questions on this blog, "How does WAEC, NECO or NABTEB count the number of words in students' English essays?" It is no news that the required number of words of the English essays of these examination bodies (WAEC, NECO and NABTEB) is 450 words. As a result, students are expected to write 450 words while answering any of the essay questions of WAEC, NECO or NABTEB English examination.
It is also important to note that there is no penalty for writing more than 450 words but marks are deducted if a student's essay is less than 450 words.
Having said that, how do you know whether your essay is up to or less than the required number of words? Do you count your words one by one to ascertain this or you just assume your essay has reached the required number? I am cocksure you do either of these but also believe that you will never be a victim of such ignorance after reading this article.
You can easily get the number of words of your English essay using the formula: Number of words per line multiple by Number of lines covered. For example, while writing your essay, if each line of your paper (answer sheet) contains nine (9) words and your essay covers forty two (42) lines, you will get the total number of words by multiplying 9 by 42, which will give you 378 words. This simply means that if you write ten words per line, you need to cover 45 lines to get 450 words. The higher the number of words per line, the lesser the number of lines covered.
The implication is that those with large handwriting will have to cover more lines to get the required number of words unlike those with small handwriting.
Note: While counting the words, English examiners pick the lines at random, and if two out of the lines picked consist of eight words each, it is assumed other lines contain eight words. And the number of words of your essay will be calculated based on that. Therefore, it is very important you ensure that the number of words in each line does not vary. However, if it must vary, the difference shouldn't be much.
With the formula above, you can easily calculate the number of words in your English essay. Meanwhile, you can check the marking scheme for English HERE.
### Tamuno Reuben
Those who seek knowledge seek power because the pen is mightier than the sword. | 539 | 2,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.955165 |
http://en.academic.ru/dic.nsf/cide/192476/vertical | 1,506,153,127,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689572.75/warc/CC-MAIN-20170923070853-20170923090853-00427.warc.gz | 103,366,752 | 14,996 | Vertical
Vertical Ver"ti*cal, a. [Cf. F. vertical. See {Vertex}.] [1913 Webster] 1. Of or pertaining to the vertex; situated at the vertex, or highest point; directly overhead, or in the zenith; perpendicularly above one. [1913 Webster]
Charity . . . is the vertical top of all religion. --Jer. Taylor. [1913 Webster]
2. Perpendicular to the plane of the horizon; upright; plumb; as, a vertical line. [1913 Webster]
{Vertical angle} (Astron. & Geod.), an angle measured on a vertical circle, called an angle of elevation, or altitude, when reckoned from the horizon upward, and of depression when downward below the horizon.
{Vertical anthers} (Bot.), such anthers as stand erect at the top of the filaments.
{Vertical circle} (Astron.), an azimuth circle. See under {Azimuth}.
{Vertical drill}, an drill. See under {Upright}.
{Vertical fire} (Mil.), the fire, as of mortars, at high angles of elevation.
{Vertical leaves} (Bot.), leaves which present their edges to the earth and the sky, and their faces to the horizon, as in the Australian species of Eucalyptus.
{Vertical limb}, a graduated arc attached to an instrument, as a theodolite, for measuring vertical angles.
{Vertical line}. (a) (Dialing) A line perpendicular to the horizon. (b) (Conic Sections) A right line drawn on the vertical plane, and passing through the vertex of the cone. (c) (Surv.) The direction of a plumb line; a line normal to the surface of still water. (d) (Geom., Drawing, etc.) A line parallel to the sides of a page or sheet, in distinction from a horizontal line parallel to the top or bottom.
{Vertical plane}. (a) (Conic Sections) A plane passing through the vertex of a cone, and through its axis. (b) (Projections) Any plane which passes through a vertical line. (c) (Persp.) The plane passing through the point of sight, and perpendicular to the ground plane, and also to the picture.
{Vertical sash}, a sash sliding up and down. Cf. {French sash}, under 3d {Sash}.
{Vertical steam engine}, a steam engine having the crank shaft vertically above or below a vertical cylinder. [1913 Webster]
The Collaborative International Dictionary of English. 2000.
Synonyms:
(to the horizon), ,
### Look at other dictionaries:
• vertical — vertical, ale, aux [ vɛrtikal, o ] adj. et n. • 1587; point vertical 1545; bas lat. verticalis, de vertex, verticis « sommet » I ♦ Adj. 1 ♦ Qui suit la direction de la pesanteur, du fil à plomb en un lieu; perpendiculaire à un plan horizontal.… … Encyclopédie Universelle
• vertical — ver‧ti‧cal [ˈvɜːtɪkl ǁ ˈvɜːr ] adjective 1. HUMAN RESOURCES a vertical organization, system etc is one in which decisions and rules are passed on to employees through several different levels of management: • Our team has abandoned the vertical… … Financial and business terms
• vertical — VERTICÁL, Ă, verticali, e, adj. Care este orientat perpendicular pe un plan orizontal; care are direcţia căderii corpurilor; (sens curent) care este orientat drept (în sus). ♢ Dreaptă verticală = dreaptă care uneşte un punct de pe pământ cu… … Dicționar Român
• vertical — vertical, perpendicular, plumb can mean situated at right angles to the plane of the horizon or extending from that plane at such an angle. Vertical suggests a relation to the vertex or topmost point (see APEX 1); it is used most often when the… … New Dictionary of Synonyms
• vertical — vertical, ale (vèr ti kal, ka l ) adj. 1° Qui est placé haut au dessus de la tête, du vertex. • Lorsque le soleil, au Capricorne, échauffe pendant trois semaines l île de France de ses feux verticaux...., BERN. DE ST PIERRE Paul et Virg..… … Dictionnaire de la Langue Française d'Émile Littré
• Vertical — Vertical, adj. & adv. welches aus dem Latein. verticalis entlehnet ist, scheitelrecht, so wie perpendiculär durch senkrecht gegeben wird. Einen Mörser vertical richten. Daher die Vertical Fläche, welche auf der horizontalen Fläche senkrecht… … Grammatisch-kritisches Wörterbuch der Hochdeutschen Mundart
• Vertical — Ver ti*cal, n. 1. Vertical position; zenith. [R.] [1913 Webster] 2. (Math.) A vertical line, plane, or circle. [1913 Webster] {Prime vertical}, {Prime vertical dial}. See under {Prime}, a. [1913 Webster] … The Collaborative International Dictionary of English
• vertical — (Del lat. verticālis). 1. adj. Geom. Dicho de una recta o de un plano: Que es perpendicular a otra recta o plano horizontal. 2. Que tiene la dirección de una plomada. U. t. c. s. f.) 3. Que, en figuras, dibujos, escritos, impresos, etc., va de la … Diccionario de la lengua española
• Vêrtical — Datos generales Origen Lima, Perú Información artística … Wikipedia Español
• vertical — adj. 2 g. 1. Que forma um ângulo reto com o plano do horizonte (ex.: plano vertical). 2. Que está colocado no vértice. 3. Direito, aprumado. 4. Que está organizado segundo um esquema hierárquico (ex.: estruturas verticais de um organismo). • s. f … Dicionário da Língua Portuguesa | 1,413 | 4,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-39 | latest | en | 0.864573 |
https://www.kodytools.com/units/potential/from/esupotential/to/kvolt | 1,726,338,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00877.warc.gz | 799,603,454 | 13,921 | # ESU of Electric Potential to Kilovolt Converter
1 ESU of Electric Potential = 0.299792544 Kilovolts
## One ESU of Electric Potential is Equal to How Many Kilovolts?
The answer is one ESU of Electric Potential is equal to 0.299792544 Kilovolts and that means we can also write it as 1 ESU of Electric Potential = 0.299792544 Kilovolts. Feel free to use our online unit conversion calculator to convert the unit from ESU of Electric Potential to Kilovolt. Just simply enter value 1 in ESU of Electric Potential and see the result in Kilovolt.
Manually converting ESU of Electric Potential to Kilovolt can be time-consuming,especially when you don’t have enough knowledge about Electrical Potential units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online ESU of Electric Potential to Kilovolt converter tool to get the job done as soon as possible.
We have so many online tools available to convert ESU of Electric Potential to Kilovolt, but not every online tool gives an accurate result and that is why we have created this online ESU of Electric Potential to Kilovolt converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert ESU of Electric Potential to Kilovolt (ESU to kV)
By using our ESU of Electric Potential to Kilovolt conversion tool, you know that one ESU of Electric Potential is equivalent to 0.299792544 Kilovolt. Hence, to convert ESU of Electric Potential to Kilovolt, we just need to multiply the number by 0.299792544. We are going to use very simple ESU of Electric Potential to Kilovolt conversion formula for that. Pleas see the calculation example given below.
$$\text{1 ESU of Electric Potential} = 1 \times 0.299792544 = \text{0.299792544 Kilovolts}$$
## What Unit of Measure is ESU of Electric Potential?
ESU is a unit of measurement for electric potential. ESU stands for electrostatic unit. One ESU of electric potential is equal to 299.79 volts.
## What is the Symbol of ESU of Electric Potential?
The symbol of ESU of Electric Potential is ESU. This means you can also write one ESU of Electric Potential as 1 ESU.
## What Unit of Measure is Kilovolt?
Kilovolt is a unit of measurement for electric potential. Kilovolt is multiple of electric potential unit volt. One kilovolt is equal to 1000 volts.
## What is the Symbol of Kilovolt?
The symbol of Kilovolt is kV. This means you can also write one Kilovolt as 1 kV.
## How to Use ESU of Electric Potential to Kilovolt Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select ESU of Electric Potential and in the first input field, enter a value.
• From the second dropdown, select Kilovolt.
• Instantly, the tool will convert the value from ESU of Electric Potential to Kilovolt and display the result in the second input field.
## Example of ESU of Electric Potential to Kilovolt Converter Tool
ESU of Electric Potential
1
Kilovolt
0.299792544
# ESU of Electric Potential to Kilovolt Conversion Table
ESU of Electric Potential [ESU]Kilovolt [kV]Description
1 ESU of Electric Potential0.299792544 Kilovolt1 ESU of Electric Potential = 0.299792544 Kilovolt
2 ESU of Electric Potential0.599585088 Kilovolt2 ESU of Electric Potential = 0.599585088 Kilovolt
3 ESU of Electric Potential0.899377632 Kilovolt3 ESU of Electric Potential = 0.899377632 Kilovolt
4 ESU of Electric Potential1.2 Kilovolt4 ESU of Electric Potential = 1.2 Kilovolt
5 ESU of Electric Potential1.5 Kilovolt5 ESU of Electric Potential = 1.5 Kilovolt
6 ESU of Electric Potential1.8 Kilovolt6 ESU of Electric Potential = 1.8 Kilovolt
7 ESU of Electric Potential2.1 Kilovolt7 ESU of Electric Potential = 2.1 Kilovolt
8 ESU of Electric Potential2.4 Kilovolt8 ESU of Electric Potential = 2.4 Kilovolt
9 ESU of Electric Potential2.7 Kilovolt9 ESU of Electric Potential = 2.7 Kilovolt
10 ESU of Electric Potential3 Kilovolt10 ESU of Electric Potential = 3 Kilovolt
100 ESU of Electric Potential29.98 Kilovolt100 ESU of Electric Potential = 29.98 Kilovolt
1000 ESU of Electric Potential299.79 Kilovolt1000 ESU of Electric Potential = 299.79 Kilovolt
# ESU of Electric Potential to Other Units Conversion Table
ConversionDescription
1 ESU of Electric Potential = 299.79 Volt1 ESU of Electric Potential in Volt is equal to 299.79
1 ESU of Electric Potential = 299.79 Watt/Ampere1 ESU of Electric Potential in Watt/Ampere is equal to 299.79
1 ESU of Electric Potential = 0.299792544 Kilovolt1 ESU of Electric Potential in Kilovolt is equal to 0.299792544
1 ESU of Electric Potential = 0.000299792544 Megavolt1 ESU of Electric Potential in Megavolt is equal to 0.000299792544
1 ESU of Electric Potential = 299792.54 Millivolt1 ESU of Electric Potential in Millivolt is equal to 299792.54
1 ESU of Electric Potential = 299792544 Microvolt1 ESU of Electric Potential in Microvolt is equal to 299792544
1 ESU of Electric Potential = 29979254400 Abvolt1 ESU of Electric Potential in Abvolt is equal to 29979254400
1 ESU of Electric Potential = 29979254400 EMU of Electric Potential1 ESU of Electric Potential in EMU of Electric Potential is equal to 29979254400
1 ESU of Electric Potential = 1 Statvolt1 ESU of Electric Potential in Statvolt is equal to 1
1 ESU of Electric Potential = 299792544000 Nanovolt1 ESU of Electric Potential in Nanovolt is equal to 299792544000
1 ESU of Electric Potential = 299792544000000 Picovolt1 ESU of Electric Potential in Picovolt is equal to 299792544000000
1 ESU of Electric Potential = 2.99792544e-7 Gigavolt1 ESU of Electric Potential in Gigavolt is equal to 2.99792544e-7
1 ESU of Electric Potential = 2.99792544e-10 Teravolt1 ESU of Electric Potential in Teravolt is equal to 2.99792544e-10
1 ESU of Electric Potential = 2.8744670789587e-25 Planck Voltage1 ESU of Electric Potential in Planck Voltage is equal to 2.8744670789587e-25 | 1,617 | 5,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.906227 |
https://www.mql5.com/en/forum/158491 | 1,508,568,714,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00500.warc.gz | 948,530,841 | 19,275 | 35
Sometimes the function OrderOpenPrice return number with more digits than it should be and that's make problem because when i use that number inside of if statement like "Ask==OrderOpenPrice+50*pips" it doesn't make true the condition to execute the code at the brackets of if statement.I try the fuction "NormalizeDouble" but the problem remain .
Below is the code to see the price that open the position (OrderOpenPrice) and x is variable to see that price after using the NormalizeDouble with 1 digit.
``` if (OrderSelect(OrdersTotal()-1, SELECT_BY_POS, MODE_TRADES))
{
if (OrderType() == OP_SELL && OrderMagicNumber()==MagicNumber)
{
double x = NormalizeDouble(OrderOpenPrice(),1);
Comment(OrderOpenPrice()+" "+x);
}
}```
Sometimes the results of the Comment function are like : 85.31100000000001 85.3
Sometimes are : 86.59 86.59999999999999
I attached photos for these examples
it's like sometimes the fuction OrderOpenPrice doesn't work and sometimes the NormalizeDouble doesn't work.
it doesn't make any sense.
Thanks !
16096
dionikolo:
Sometimes the function OrderOpenPrice return number with more digits than it should be and that's make problem because when i use that number inside of if statement like
I attached photos for these examples
I try the fuction "NormalizeDouble"
1. No it does not, Floating point has infinite number of digits. It's your not understanding floating point and that some numbers can't be represented exactly. (like 1/10.) Double-precision floating-point format - Wikipedia, the free encyclopedia
2. You can't compare doubles for equality and price could easily move from below to above and never be equal. The == operand. - MQL4 forum
3. Don't attach a image, insert the image
4. Do NOT use NormalizeDouble, EVER. For ANY Reason. It's a kludge, don't use it. It's use is always wrong
19297
Vasyl Nosal
DoubleToString()
35
WHRoeder:
1. No it does not, Floating point has infinite number of digits. It's your not understanding floating point and that some numbers can't be represented exactly. (like 1/10.) Double-precision floating-point format - Wikipedia, the free encyclopedia
2. You can't compare doubles for equality and price could easily move from below to above and never be equal. The == operand. - MQL4 forum
3. Don't attach a image, insert the image
4. Do NOT use NormalizeDouble, EVER. For ANY Reason. It's a kludge, don't use it. It's use is always wrong
Ok..Thank you WHRoeder! Just to know all that need to work my code right is to change all these "if statements" Ask==OrderOpenPrice+something with Ask>=OrderOpenPrice+something or Ask<=OrderOpenPrice+something because of the choppy price and the Floating point (Correct me if i am wrong, i am not a pro Programmer). This forum is perfect for newbie like me! I appreciate all that Information ! Thanks again !
16096 | 705 | 2,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-43 | latest | en | 0.830371 |
https://cs.stackexchange.com/questions/115518/sampling-a-uniform-distribution-of-fixed-size-strings-containing-no-forbidden-su/115520#115520 | 1,642,472,377,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00356.warc.gz | 260,152,424 | 35,216 | # Sampling a uniform distribution of fixed size strings containing no forbidden substrings
Given a list of "forbidden" words (substrings), an alphabet, and a desired output string length, how would I efficiently sample output strings containing no forbidden word?
For short output strings with few forbidden words, I would use simple rejection sampling. Pick a string (uniformly) with the specified alphabet and length, return that string if it contains no element of the forbidden list, try again otherwise.
If I use that algorithm for output lengths several times larger than the typical forbidden word, then the probability of rejection will be higher. (Most words are 2 or 3 characters long.)
Assume the requested output length is too long to enumerate and store every possible value. My alphabet size would be 16 to 36 characters, but solutions to large alphabets would be interesting to think about. (In which case I would call these things random sentences, forbidden n-grams, and dictionary words.)
My forbidden word list will have one hundred to one thousand strings. I would like to avoid solutions requiring expensive precomputation or lots of memory.
My first idea was to try to build a random string incrementally, in contrast to the all-or-nothing approach of straightforward rejection sampling. I doubt that my algorithm produces each possible output with equal probability.
The algorithm idea follows:
1. Initialize a char buffer long enough to fit outlen characters.
2. Pick a random letter of the alphabet and append it to the buffer.
3. If the buffer ends with a forbidden word of length k, then remove the last k letters from the char buffer and go to 2.
4. Otherwise, go to 2 if the buffer has less than outlen characters.
5. Return the contents of the buffer if it is full.
Step 3 serves to rewind the algorithm, returning the char buffer to a previous legal state.
I understand that clearing the whole buffer in step 3 definitely would produce uniform output just like the straightforward rejection sampling method. However, the average number of rejections before the first valid output is generated will be the same.
I've gotten stuck trying to determine if my proposed algorithm is uniform. I have had no luck finding alternative algorithms either. I haven't yet looked at how this algorithm's performance would compare to basic rejection sampling.
• The method you are proposing is not uniform. One counter-example is if the alphabet is abc, you want to sample a string of length 3 and forbidden words are "ab" and "aa". In this case there are 17 valid words, but valid words that starts with "a" ("aca", "acb", "acc") will be sampled less frequently than $\frac{3}{17}$ ($\frac{3}{21}$ to be more precise). Oct 6 '19 at 23:15
• Moreover it is not well defined your algorithm when there are forbidden words which are suffixes of other words (which will be removed?), and whatever option you pick it'll introduce some more bias toward certain words. Oct 6 '19 at 23:20
• You can use rejection sampling in a more efficient way. Create the string adding one char on each step, but sample from the set of chars that don't introduce a forbidden word. This have a known bias (you know the probability that sampled word appears). Accept/Reject this new word according to the output of a Bernoulli sampled variable. (See the wikipedia article to understand the method). Oct 7 '19 at 0:18
## 1 Answer
Suppose the alphabet is $$\{a,b\}$$, and you have one forbidden word, $$aa$$. Suppose we are trying to generate a word of length 3. The first two letters will be distributed uniformly over $$ab,ba,bb$$. Hence the first letter has the following distribution: $$a$$ with probability $$1/3$$, $$b$$ with probability $$2/3$$. In contrast, the allowable words are $$aba,abb,bab,bba,bbb.$$ So the first letter should have the distribution $$a$$ with probability $$2/5$$, $$b$$ with probability $$3/5$$.
Here is an algorithm that does work. Construct a DFA (or UFA) for your language. For each state $$q$$, using dynamic programming you can count how many words of length $$m$$ are accepted when the automaton is started at $$q$$. Let us denote this by $$c(q,m)$$.
The correct distribution of the first letter $$\sigma_1$$ of a word of length $$n$$ in the language is $$\Pr[\sigma_1 = \sigma] = \frac{c(\delta(q_0,\sigma),n-1)}{c(q_0,n)}.$$ More generally, given the first $$\ell$$ letters $$\sigma_1 \ldots \sigma_\ell$$, the following letter has the distribution $$\Pr[\sigma_{\ell+1} = \sigma \mid \sigma_1 \ldots \sigma_\ell] = \frac{c(\delta(q_0,\sigma_1\ldots\sigma_\ell\sigma),n-\ell-1)}{c(\delta(q_0,\sigma_1\ldots\sigma_\ell),n-\ell)}.$$ Ignoring the cost of arithmetic, you can implement this scheme in roughly $$O(|Q|n)$$, where $$Q$$ is the set of states, or in $$O(|\Sigma|n^2)$$. (The former, assuming that $$|Q| = \Omega(|\Sigma|)$$.)
As an example, let us consider the counterexample above. We construct a DFA containing two states (we can omit the sink state, obtaining a UFA) $$q_0,q_1$$. The transition function is $$\delta(q_0,a) = q_1$$, $$\delta(q_0,b) = q_0$$, $$\delta(q_1,b) = q_0$$. The relevant values of $$c$$ are $$\begin{array}{c|cc} n & c(q_0,n) & c(q_1,n) \\\hline 0 & 1 & 1 \\ 1 & 2 & 1 \\ 2 & 3 & 2 \\ 3 & 5 & 3 \end{array}$$ These are computed by the recurrences $$c(q_0,n) = c(q_0,n-1) + c(q_1,n-1)$$ and $$c(q_1,n) = c(q_0,n-1)$$, with base case $$c(q,0) = 1$$.
Since $$\delta(q_0,a) = q_1$$ and $$\delta(q_0,b) = q_0$$, we see that (for $$n = 3$$) $$\Pr[\sigma_1 = a] = c(q_1,2)/c(q_0,3) = 2/5$$ and $$\Pr[\sigma_1 = b] = c(q_0,2)/c(q_0,3) = 3/5$$. | 1,502 | 5,616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-05 | latest | en | 0.882176 |
https://blueggsee.com/category/type-of-math-skill/measurement/speed/ | 1,674,781,325,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00765.warc.gz | 157,179,995 | 18,795 | # How far can an Ornate box turtle move in 10 minutes?
Math skills: Speed
Related Environmental issues: Biodiversity / Deforestation
Ornate box turtles are endangered in northern United States including Wisconsin.
The average walking speed of a box turtle about 0.2mph. How far can a box turtle move in 10 minutes?
1 mile = 5280 feet
0.2 mile = 5280 x 0.2 = 1056 feet
1056feet/60min=xfeet/10min
1056/6=176feet
### FACTFUL ENVIRONMENTAL MATH PROBLEMS (2020)
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# Asian rhinos are threatened — Who can run faster, Rhinos or Usain Bolt?
Math skills: Speed
Related Environmental issues: Biodiversity / Deforestation
Javan rhinos, or Asian rhinos, are one of the most endangered animals with only about 60 indivisuals.
Rhinos can run at the speed of 50 km per hour. Usain Bolt ran 100 meters at 9.58 seconds. Who can run faster? Explain.
50 km = 50,000 m
50,000 / 100 = 500
Usain Bolt runs 50km in
[9.58 x 500 = 4,790 seconds] if he keeps running at the same speed. The running speed of rhinos is 50km per 3,600 seconds (=1hour) while that of Usain Bolt is 50km per 4,790 seconds. Rhino can run about 1.3 times faster than Usain Bolt.
(Usain Bolt’s running speed is 37.5 km per hour.)
“Once the most widespread of Asian rhinos, Javan rhinos are now listed as critically endangered. With only one known population in the wild, it is one of the world’s rarest large mammals.
There are between 58 and 68 in the wild, with none living in captivity. The rhinos are often poached for their horns, although loss of habitat, especially resulting from the Vietnam War, has also contributed to their decline.
The only population of Javan rhinos can be found in Ujung Kulon National Park on the south-western tip of Java, Indonesia. The only other population, in Vietnam, was wiped out in 2010.”
(Source: Wanderlust, 2020 : https://www.wanderlust.co.uk/content/10-of-the-worlds-most-endangered-animals/)
### Factful Environmental Math Problems (2020)
Just published and now available from Amazon (paperback) | 669 | 2,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | longest | en | 0.876392 |
http://forums.cervenka.cz/viewtopic.php?pid=4387 | 1,582,383,951,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00277.warc.gz | 57,459,756 | 4,309 | #### Topic: Internal Mesh Connectivity
Hello,
I´m using Atena 3D to model a reinforced beam. The concretes-structure-mesh is made out of structured hexahedral elements.
Now I have to get sure that the upper and lower reinforcement bars are NOT getting connected with the stirrups.
So I was modelling the concretes mesh in that way, that there is one hexahedral element between the reinforcement bars and the stirrups.
Now to may question.
I want to know, how the reinforcement notes (and also the macro joints of the reinforcement) in generel are connected to the surrounding hexahedral elements.
For example, how this connections from the reinforcement notes to the notes of the hexahedral elements are weighted.
Or if this connections could be shown in the program.
Hopefully someone could help me in this case.
Thanks!
#### Re: Internal Mesh Connectivity
Dear Max,
the element in between is not necessary. The bars would be connected only in case of sharing macro node.
Regarding the connection of bar nodes in hexahedra elements, when no bond is set to bar the position of the bar node is calculated as linear combination of hexahedra nodes.
If there is also the bond law then the inner iterative process starts and the slipping may occur.
You can also check the paper that presents this topic:
https://www.researchgate.net/publicatio … _with_bond
With best regards,
Michaela Vaitova
#### Re: Internal Mesh Connectivity
Dear Max, for the basic information related to connecting reinforcement to the surrounding elements, please see ATENA
Troubleshooting, 2.1.21 When I model the reinforcement, do the bars and stirrups have to be in geometric connection (with common intersection points) or they can be modeled according to their axis so that they do not touch one the other in geometric sense?
I understand you include the layer of volume elements between the longitudinal bar elements and the stirrups to avoid both being connected to the same nodes, even just with small weights. If that least to very small elements (possibly breaking the assumptions for concrete as continuum), it may be better to use bond law(s) for one or both of the reinforcements (as Michaela suggests). If you are interested primarily in other cracking than what can be captured as slip along the bar axes, please explain (see also Troubleshooting, 2.1.1).
Regards. | 501 | 2,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-10 | longest | en | 0.931093 |
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Question From class 9 Chapter POLYNOMIALS
Find the remainder when the polynomial is divided by .
Solution :
. <br> By the remiander theorem ,we know that when p (x) is divided by then the remainder is p(2). <br> Now , <br> . <br> Hence , the required remainder is 21.
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The polynomial when divided by x+1 leaves remainder 19. Also, find the remainder when p(x) is divided by x+2.
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posted by .
What is the definition of the slope of a curve at a point x=a?
Formal definition of a derivative?
or slope between two points?
• math -
the slope at a point a is the slope of the line tangent to the curve at x=a.
To get that, pick a point number b, near a, and think of the slope between two points on the curve at f(a) and f(b).
As b gets closer and closer to a, the line between the two points gets closer and closer to the tangent line.
Finally, when b=a, the two points are the same point, and the line is tangent to the curve.
That's all the formal definition of the derivative is. It's the limit of the slope of a line joining two points on the curve, as the two points get closer and closer to each other.
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https://foreach.id/EN/fluids/flowratemolar/mol%7Cday-to-megamol%7Csecond.html | 1,624,300,734,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00500.warc.gz | 240,675,640 | 8,579 | # Convert mol/day to megamol/second (mol/d to Mmol/s)
Batch Convert
• megamol/second [Mmol/s]
• mol/day [mol/d]
Copy
_
Copy
• megamol/second [Mmol/s]
• mol/day [mol/d]
## Mol/day to Megamol/second (mol/d to Mmol/s)
### Mol/day (Symbol or Abbreviation: mol/d)
Mol/day is one of molar flow rate units. Mol/day abbreviated or symbolized by mol/d. The value of 1 mol/day is equal to 0.000011574 mol/second. In its relation with megamol/second, 1 mol/day is equal to 1.1574e-11 megamol/second.
#### Relation with other units
1 mol/day equals to 0.000011574 mol/second
1 mol/day equals to 1.1574e-23 examol/second
1 mol/day equals to 1.1574e-20 petamol/second
1 mol/day equals to 1.1574e-17 teramol/second
1 mol/day equals to 1.1574e-14 gigamol/second
1 mol/day equals to 1.1574e-11 megamol/second
1 mol/day equals to 1.1574e-8 kilomol/second
1 mol/day equals to 1.1574e-7 hectomol/second
1 mol/day equals to 0.0000011574 dekamol/second
1 mol/day equals to 0.00011574 decimol/second
1 mol/day equals to 0.0011574 centimol/second
1 mol/day equals to 0.011574 millimol/second
1 mol/day equals to 11.574 micromol/second
1 mol/day equals to 11,574 nanomol/second
1 mol/day equals to 11,574,000 picomol/second
1 mol/day equals to 11,574,000,000 femtomol/second
1 mol/day equals to 11,574,000,000,000 attomol/second
1 mol/day equals to 0.00069444 mol/minute
1 mol/day equals to 0.041667 mol/hour
1 mol/day equals to 0.69444 millimol/minute
1 mol/day equals to 41.667 millimol/hour
1 mol/day equals to 1,000 millimol/day
1 mol/day equals to 6.9444e-7 kilomol/minute
1 mol/day equals to 0.000041667 kilomol/hour
1 mol/day equals to 0.001 kilomol/day
### Megamol/second (Symbol or Abbreviation: Mmol/s)
Megamol/second is one of molar flow rate units. Megamol/second abbreviated or symbolized by Mmol/s. The value of 1 megamol/second is equal to 1000000 mol/second. In its relation with mol/day, 1 megamol/second is equal to 86400000000 mol/day.
#### Relation with other units
1 megamol/second equals to 1,000,000 mol/second
1 megamol/second equals to 1e-12 examol/second
1 megamol/second equals to 1e-9 petamol/second
1 megamol/second equals to 0.000001 teramol/second
1 megamol/second equals to 0.001 gigamol/second
1 megamol/second equals to 1,000 kilomol/second
1 megamol/second equals to 10,000 hectomol/second
1 megamol/second equals to 100,000 dekamol/second
1 megamol/second equals to 10,000,000 decimol/second
1 megamol/second equals to 100,000,000 centimol/second
1 megamol/second equals to 1,000,000,000 millimol/second
1 megamol/second equals to 1,000,000,000,000 micromol/second
1 megamol/second equals to 1,000,000,000,000,000 nanomol/second
1 megamol/second equals to 1,000,000,000,000,000,000 picomol/second
1 megamol/second equals to 1e+21 femtomol/second
1 megamol/second equals to 1e+24 attomol/second
1 megamol/second equals to 60,000,000 mol/minute
1 megamol/second equals to 3,600,000,000 mol/hour
1 megamol/second equals to 86,400,000,000 mol/day
1 megamol/second equals to 60,000,000,000 millimol/minute
1 megamol/second equals to 3,600,000,000,000 millimol/hour
1 megamol/second equals to 86,400,000,000,000 millimol/day
1 megamol/second equals to 60,000 kilomol/minute
1 megamol/second equals to 3,600,000 kilomol/hour
1 megamol/second equals to 86,400,000 kilomol/day
### How to convert Mol/day to Megamol/second (mol/d to Mmol/s):
#### Conversion Table for Mol/day to Megamol/second (mol/d to Mmol/s)
mol/day (mol/d) megamol/second (Mmol/s)
0.01 mol/d 1.1574e-13 Mmol/s
0.1 mol/d 1.1574e-12 Mmol/s
1 mol/d 1.1574e-11 Mmol/s
2 mol/d 2.3148e-11 Mmol/s
3 mol/d 3.4722e-11 Mmol/s
4 mol/d 4.6296e-11 Mmol/s
5 mol/d 5.787e-11 Mmol/s
6 mol/d 6.9444e-11 Mmol/s
7 mol/d 8.1019e-11 Mmol/s
8 mol/d 9.2593e-11 Mmol/s
9 mol/d 1.0417e-10 Mmol/s
10 mol/d 1.1574e-10 Mmol/s
20 mol/d 2.3148e-10 Mmol/s
25 mol/d 2.8935e-10 Mmol/s
50 mol/d 5.787e-10 Mmol/s
75 mol/d 8.6806e-10 Mmol/s
100 mol/d 1.1574e-9 Mmol/s
250 mol/d 2.8935e-9 Mmol/s
500 mol/d 5.787e-9 Mmol/s
750 mol/d 8.6806e-9 Mmol/s
1,000 mol/d 1.1574e-8 Mmol/s
100,000 mol/d 0.0000011574 Mmol/s
1,000,000,000 mol/d 0.011574 Mmol/s
1,000,000,000,000 mol/d 11.574 Mmol/s
#### Conversion Table for Megamol/second to Mol/day (Mmol/s to mol/d)
megamol/second (Mmol/s) mol/day (mol/d)
0.01 Mmol/s 864,000,000 mol/d
0.1 Mmol/s 8,640,000,000 mol/d
1 Mmol/s 86,400,000,000 mol/d
2 Mmol/s 172,800,000,000 mol/d
3 Mmol/s 259,200,000,000 mol/d
4 Mmol/s 345,600,000,000 mol/d
5 Mmol/s 432,000,000,000 mol/d
6 Mmol/s 518,400,000,000 mol/d
7 Mmol/s 604,800,000,000 mol/d
8 Mmol/s 691,200,000,000 mol/d
9 Mmol/s 777,600,000,000 mol/d
10 Mmol/s 864,000,000,000 mol/d
20 Mmol/s 1,728,000,000,000 mol/d
25 Mmol/s 2,160,000,000,000 mol/d
50 Mmol/s 4,320,000,000,000 mol/d
75 Mmol/s 6,480,000,000,000 mol/d
100 Mmol/s 8,640,000,000,000 mol/d
250 Mmol/s 21,600,000,000,000 mol/d
500 Mmol/s 43,200,000,000,000 mol/d
750 Mmol/s 64,800,000,000,000 mol/d
1,000 Mmol/s 86,400,000,000,000 mol/d
100,000 Mmol/s 8,640,000,000,000,000 mol/d
1,000,000,000 Mmol/s 86,400,000,000,000,000,000 mol/d
1,000,000,000,000 Mmol/s 8.64e+22 mol/d
#### Steps to Convert Mol/day to Megamol/second (mol/d to Mmol/s)
1. Example: Convert 414 mol/day to megamol/second (414 mol/d to Mmol/s).
2. 1 mol/day is equivalent to 1.1574e-11 megamol/second (1 mol/d is equivalent to 1.1574e-11 Mmol/s).
3. 414 mol/day (mol/d) is equivalent to 414 times 1.1574e-11 megamol/second (Mmol/s).
4. Retrieved 414 mol/day is equivalent to 4.7917e-9 megamol/second (414 mol/d is equivalent to 4.7917e-9 Mmol/s).
▸▸ | 2,237 | 5,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.779783 |
https://www.matrix.edu.au/2017-hsc-maths-ext1-paper-solutions/ | 1,540,064,310,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513384.95/warc/CC-MAIN-20181020184612-20181020210112-00544.warc.gz | 989,920,782 | 58,947 | # 2017 HSC Maths Ext 1 Exam Paper Solutions
In this post, we give you the solutions to the 2017 Maths Extension 1 paper.
Have you seen the 2017 HSC Mathematics Extension 1 Paper, yet? In this post, we will work our way through the 2017 HSC Maths Extension 1 Exam Paper and give you the solutions, written by our leading teacher Jonathan Le, to all of the questions.
Read on, to see how to answer all of the 2017 questions.
## 2017 HSC Mathematics Extension 1 Exam Paper Solutions
This paper has two sections
• Section I consists of 10 Multiple Choice Questions
• Section II consists of 4 Extended Response Questions
## Section 1: 10 Multiple Choice Questions (10 Marks)
##### 1
A Use factor theorem, it is clear that, $$P(2)=0$$
so $$(x-2)$$ is a factor.
##### 2
B We have $$3\log_a(2)=1.893$$, so calculate $$\log_a(4)=2\log_a(2)$$.
##### 3
B $$\angle BAD = 180^\circ – 110^\circ = 70^\circ$$ (Opposite angles in a cyclic quadrilateral) and $$\angle TAB=65^\circ$$ (alternate segment theorem).
##### 4
C Standard Auxiliary angle formula problem.
##### 5
D As $$x\rightarrow \infty$$, $$y\rightarrow -2$$ The only graph with horizontal asymptote $$y=-2$$ is (D)
##### 6
D The normal is usually $$x+py=p^3+2p$$ for a point with co-ordinate $$(2p,p^2)$$ (Set $$a=1$$ in the formula in the reference sheet). Now replace $$p$$ with $$\frac{1}{p}$$ and re-arrange.
##### 7
A The domain is found by solving $$-1\leq \frac{3}{x} \leq 1$$. The direction of the arrows can be found quickly by drawing $$y=\frac{3}{x}$$ and the lines $$y=\pm 1$$
##### 8
C Use the chain rule $$\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}=2\pi r \times 5$$, then substitute in $$r=15$$
##### 9
C Find the power of $$x$$ of each general term. Option (C) yields $$7r-28$$, and is actually the only one that yields an integer value of $$r$$.
##### 10
B The total number of ways to cross 2 squares out of 9 is $$^9C_3$$. Since the grid is small, we can manually count the number of possible ways to get three in a row, which will be 8. Hence, the probability is $$\frac{8}{^9C_3}$$.
## Section 2: Extended Response Questions (60 Marks)
### Question 11 (15 marks)
(a)
\begin{equation*}
x_p=\frac{3(-4)+2(1)}{2+3}=-2
\end{equation*}
(b)
\begin{align*}
\frac{d}{dx}\big(\tan^{-1}(x^3)\big)&=\frac{3x^2}{1+(x^3)^2}\\
&=\frac{3x^2}{1+x^6}
\end{align*}
(c)
Multiply both sides by $$(x+1)^2$$.
\begin{align*}
\frac{2x}{x+1} &> 1\\
2x(x+1) &> (x+1)^2\\
(x+1)^2-2x(x+1)&<0\\
(x+1)[x+1-2x]&<0\\
(x+1)(1-x)&<0\\
(x+1)(x-1)&>0
\end{align*}
Hence, $$x>1$$ or $$x<-1$$
(d)
Domain: $$-1\leq x \leq 1$$
Range: $$0 \leq y \leq 2\pi$$
(e)
Compute the differential term first.
\begin{align*}
x&=u^2-1\\
dx&=2u~du
\end{align*}
Compute the new limits, choosing u to be positive.
\begin{align*}
x=3 ~~~\rightarrow ~~~& u=2\\
x=0 ~~~\rightarrow ~~~& u=1\\
\end{align*}
Substitute.
\begin{align*}
\int_0^3 \frac{x}{\sqrt{x+1}}dx&=\int_1^2\frac{u^2-1}{u}\times 2u~du\\
&= 2\int_1^2 u^2-1 du\\
& = 2\Big[\frac{u^3}{3}-u\Big]_1^2\\
&= \frac{8}{3}
\end{align*}
(f)
Use the fact that $$\int f^n(x)f'(x)dx=\frac{1}{n+1}f^{n+1}(x)+C$$
\begin{equation*} \int \sin^2x\cos x dx=\frac{1}{3} \sin^3 x + C\end{equation*}
(g) (i)
\begin{equation*}P(3 \text{ red})=\binom{8}{3}\Big(\frac{1}{5}\Big)^3\Big(\frac{4}{5}\Big)^5\end{equation*}
(g) (ii)
\begin{equation*}P(\text{no flowers})=\Big(\frac{4}{5}\Big)^8\end{equation*}
(g) (iii)
\begin{align*}
P(\text{at least one red}) &= 1-P(\text{no red})\\
& = 1-\Big(\frac{4}{5}\Big)^8
\end{align*}
### Question 12 (15 marks)
(a)
Let $$D$$ be any point on the major arc $$AC$$.
\begin{align*}
\angle ADC &= 50^\circ~~~(\angle \text{ at centre } =2\times \angle \text{ at circumference})\\
\\
\angle ABC&= 180^\circ – 50^\circ~~~(\text{opposite angles of a cyclic quadrilateral are supplementary})\\
& = 30^\circ
\end{align*}
(b) (i)
The graph of $$y=|x+1|$$ is the shape of $$y=|x|$$ but shifted to the left by 1 unit.
The graph of $$y=3-|x-2|$$ is the shape of $$y=|x|$$ but upside-down, shifted to the right by 1 unit and shifted up by 3 units, so the vertex is at $$(2,3)$$. But observe that $$(2,3)$$ lies on the right branch of $$y=|x+1|$$.
(b) (ii)
We are asked to solve $$|x+1| + |x-2| = 3$$, but this is equivalent to $$|x+1|=3-|x-2|$$, which in turn is equivalent to solving $$y=|x+1|$$ and $$y=3-|x-2|$$ simultaneously. Since we have the graphs of the two equations above, we can find the solutions geometrically.
From the diagram above, the two equations are equal over the domain $$-1\leq x \leq 2$$, so therefore that must be the solution set to the original equation $$|x+1|+|x-2|=3$$.
(c) (i)
First, find the ratio of the left part to the right part.
\begin{align*}
\frac{V_{left}}{V_{right}}&=\frac{\pi\int_{-1}^h 1-x^2~dx}{\pi\int_h^1 1-x^2~dx}\\
&=\dfrac{\Big[x-\dfrac{x^3}{3}\Big]_{-1}^h}{\Big[x-\dfrac{x^3}{3}\Big]_h^1}\\
&= \frac{h-\frac{h^3}{3}+\frac{2}{3}}{\frac{2}{3}-h+\frac{h^3}{3}}\\
& = \frac{3h-h^3+2}{2-3h+h^3}
\end{align*}
But the ratio is given to be $$2:1$$, so this fraction must be equal to 2.
\begin{align*}
\frac{3h-h^3+2}{2-3h+h^3}&=2\\
3h-h^3+2&=2(2-3h+h^3)\\
3h-h^3+2&=4-6h+2h^3
\end{align*}
And hence the required result $$3h^3-9h+2=0$$.
(c) (ii)
Define $$f(h)=3h^3-9h+2$$, so $$f'(h)=9h^2-9$$.
\begin{align*}
h_2&=h_1-\frac{f(h_1)}{f'(h_1)}\\
&=h_1 – \frac{3h_1^3-9h_1+2}{9h_1^2-9}\\
&=\frac{2}{9}~~~(\text{since } h_0=1)
\end{align*}
(d)
Differentiate $$t$$ with respect to $$x$$ and notice that $$\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{v}$$.
\begin{align*}
\frac{dt}{dx}&=2e^{-2x}\\
\frac{1}{v}&= 2e^{-2x}\\
v&=\frac{1}{2}e^{2x}\\
v^2&=\frac{1}{4}e^{4x}
\end{align*}
But $$a=\frac{d}{dx}\Big(\frac{1}{2}v^2\Big)$$.
\begin{align*}
a&=\frac{d}{dx}\big(\frac{1}{8}e^{4x}\big)\\
&=\frac{1}{2}e^{4x}
\end{align*}
(e)
\begin{align*}
\lim_{x \rightarrow 0} \Big(\frac{1-\cos2x}{x^2}\Big)&=\lim_{x \rightarrow 0} \Big(\frac{2\sin^2x}{x^2}\Big)\\
&=2\times \lim_{x \rightarrow 0} \Big(\frac{\sin x}{x} \times \frac{\sin x}{x}\Big)\\
&=2\times \lim_{x \rightarrow 0} \Big(\frac{\sin x}{x} \Big)\times \lim_{x \rightarrow 0} \Big(\frac{\sin x}{x} \Big)\\
&=2\times 1 \times 1\\
&=2
\end{align*}
### Question 13 (15 marks)
(a)
The particle satisfies the equation for simple harmonic motion $$v^2=n^2(a^2-x^2)$$.
Substitute in the given data and solve simultaneously.
\begin{align*}
x=2,v=4 ~~&\rightarrow~~ 16=n^2(a^2-4)\\
x=5,v=3 ~~&\rightarrow~~ 9=n^2(a^2-25)
\end{align*}
Divide these equations.
\begin{align*}
\frac{16}{9} &= \frac{a^2-4}{a^2-25}\\
16a^2-400&=9a^2-36\\
7a^2&=364\\
a^2&=52
\end{align*}
Substitute back in to find $$n$$.
\begin{align*}
9&=n^2(52-25)\\
n^2&=\frac{9}{27}\\
n^2&=\frac{1}{3}\\
n&=\frac{1}{\sqrt{3}}~~~(\text{taking } n>0)
\end{align*}
Hence, the period is $$T=\frac{2\pi}{n}=2\pi\sqrt{3}$$.
(b) (i)
$$(1+x)^n+(1-x)^n=\Big[\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n\Big] +\Big[\binom{n}{0}-\binom{n}{1}x+\binom{n}{2}x^2 – \cdots + (-1)^n\binom{n}{n}x^n\Big]$$
Notice that all the odd powered terms cancel and that we have two of each even powered term. Furthermore, the last term does not cancel because $$n$$ is even.
Hence, we have the expansion
\begin{align*}
(1+x)^n+(1-x)^n &= 2\binom{n}{0}+2\binom{n}{2}x^2+2\binom{n}{4}x^4+\cdots + 2\binom{n}{n}x^n\\
&= 2\Big[\binom{n}{0}+\binom{n}{2}x^2+\binom{n}{4}x^4+\cdots + \binom{n}{n}x^n\Big]
\end{align*}
(b) (ii)
Differentiate both sides of (i) with respect to $$x$$.
\begin{equation*}n(1+x)^{n-1}-n(1-x)^{n-1}=2\Big[2\binom{n}{2}x+4\binom{n}{4}x^3+\cdots + n\binom{n}{n}x^{n-1}\Big]\end{equation*}
(b) (iii)
Let $$x=1$$ and then re-arrange.
\begin{align*}
2\Big[2\binom{n}{2}+4\binom{n}{4}+ \cdots + n\binom{n}{n}\Big]&=n2^{n-1}\\
2\binom{n}{2}+4\binom{n}{4}+ \cdots + n\binom{n}{n}=n2^{n-2}\\
\binom{n}{2}+2\binom{n}{4}+ \cdots + \frac{n}{2}\binom{n}{n}=n2^{n-3}
\end{align*}
(c) (i)
Let $$y=0$$ to find the time of flight.
\begin{align*}
Vt\sin\theta-\frac{1}{2}gt^2&=0\\
V\sin\theta-\frac{1}{2}gt&=0\\
t&=\frac{2V\sin\theta}{g}
\end{align*}
Substitute back into $$x$$ to find horizontal range.
\begin{align*}
x&= V\Big(\frac{2V\sin\theta}{g}\Big)\cos\theta\\
&= \frac{2V^2\sin\theta\cos\theta}{g}\\
&= \frac{V^2\sin2\theta}{g}\\
\end{align*}
(c) (ii)
Substitute the given inequality for $$V^2$$ into (i).
\begin{align*}
x & = \frac{V^2\sin2\theta}{g}\\
&< \frac{100g\sin2\theta}{g}\\
&=100\sin2\theta\\
&\leq100 ~~~ (\text{ since } \sin2\theta\leq1)
\end{align*}
Hence we have the bound $$x<100.$$
(c) (iii)
We are given that $$\frac{V^2\sin2\theta}{g}\geq 100$$
Let $$V^2=200g$$.
\begin{align*}
\frac{200g\sin2\theta}{g}&\geq 100\\
200\sin2\theta &\geq 100\\
\sin2\theta \geq \frac{1}{2}
\end{align*}
The boundary of the inequality is obtained by solving $$\sin2\theta = \frac{1}{2}$$
\begin{align*}
\sin2\theta&=\frac{1}{2}\\
2\theta &= \frac{\pi}{6},\frac{5\pi}{6}\\
\theta &= \frac{\pi}{12},\frac{5\pi}{12}
\end{align*}
A rough sketch shows that the values of $$\theta$$ must lie between these two boundaries, since we want the sine curve to be above the line $$y=\frac{1}{2}$$. Hence, we have the result $$\frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}$$
(c) (iv)
The greatest height is achieved when $$\dot{y}=0.$$
\begin{align*}
V\sin\theta-gt&=0\\
t &= \frac{V\sin\theta}{g}
\end{align*}
For the actual maximum height, substitute this into $$y$$.
\begin{align*}
y_{max}&=Vt\sin\theta-\frac{1}{2}gt^2\\
& = V\Big(\frac{V\sin\theta}{g}\Big)\sin\theta-\frac{1}{2}g\Big(\frac{V\sin\theta}{g}\Big)^2\\
& = \frac{V^2\sin^2\theta}{g}-\frac{V^2\sin^2\theta}{2g}\\
& = \frac{V^2\sin^2\theta}{2g}\\
& = \frac{200g\sin^2\theta}{2g} ~~~~ (\text{since } V^2=200g)\\
& = 100\sin^2\theta
\end{align*}
But note that $$\frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12}$$, so the maximum height is attained when $$\theta=\frac{5\pi}{12}$$ (the curve is still increasing at this stage and has not yet dipped down).
Hence the maximum height is
\begin{align*}
y_{max} &=100\sin^2\big(\frac{5\pi}{12}\big)\\
& = 50 – 50\cos\Big(\frac{5\pi}{6}\Big)\\
& = 50 + 25 \sqrt{3}
\end{align*}
Note: Given that this part was only worth 2 marks, it most likely wasn’t necessary to go as far as finding a nice exact form for $$y_{max}$$. Full marks most likely awarded for finding that $$y_{max}=100\sin^2\Big(\frac{5\pi}{12}\Big)$$.
### Question 14 (15 Marks)
(a)
Base Case: test $$n=1$$
\begin{equation*}8^3+6^1=518=7\times 74\end{equation*}
Hence, the result is true for $$n=1$$
Inductive Hypothesis: assume true for $$n=k$$
$$8^{2k+1}+6^{2k-1}=7M~~~ (\text{where } M \text{ is an integer})$$
Inductive Step: prove true for $$n=k+1$$
\begin{align*}
8^{2(k+1)+1}+6^{2(k+1)-1}&=8^{2k+3}+6^{2k+1}\\
&=8^2\times 8^{2k+1}+6^2\times 6^{2k-1}\\
& = 8^2\times (7M-6^{2k-1})+6^2\times 6^{2k-1}\\
& = 64 \times 7M – 64(6^{2k-1})+36(6^{2k-1})\\
& = 64 \times 7M – 28(6^{2k-1})\\
& = 7\big[64M-4(6^{2k-1})\big]
\end{align*}
And hence, the result is true by the principle of mathematical induction.
(b) (i)
The equation of the tangent from $$P$$ is $$y=px-p^2$$ (reference sheet), noting that $$a=1$$. Solve simultaneously with the bottom parabola $$x^2=-4ay.$$
\begin{align*}
x^2&=-4a(px-p^2)\\
&=-4apx+4ap^2\\
x^2+4apx-4ap^2&=0
\end{align*}
(b) (ii)
Note that the roots of the equation in (i) are the $$x$$ coordinates of where the tangent intersects $$Q$$ and $$R$$.
\begin{align*}
x_M &= \frac{x_R+x_Q}{2}\\
&= \frac{\text{Sum of roots of the quadratic in (i)}}{2}\\
& = \frac{-4ap}{2}\\
& = -2ap
\end{align*}
Substitute back into the linear function to find the $$y$$ coordinate.
\begin{align*}
y_M&=p(-2ap)-p^2\\
&=-2ap^2-p^2\\
&=-p^2(2a+1)
\end{align*}
Hence, $$M$$ has coordinates $$M(-2ap,-p^2(2a+1))$$.
(b) (iii)
If $$M$$ lies on the parabola $$x^2=-4y$$, then it must satisfy that equation. So we substitute the coordinates of $$M$$ into the equation, then solve for $$a$$.
\begin{align*}
(-2ap)^2 &= -4(-p^2(2a+1))\\
4a^2p^2&=4p^2(2a+1)\\
a^2 &= 2a+1\\
a^2-2a-1&=0\\
a&=1\pm\sqrt{2}
\end{align*}
Since $$a>0$$, the value of $$a$$ is $$a=1+\sqrt{2}$$
(c) (i)
\begin{align*}
\frac{d}{dt}(F(t)e^{0.4t})&=F'(t)e^{0.4t}+0.4e^{0.4t}F(t)\\
&=(50e^{-0.5t}-0.4F(t))e^{0.4t}+0.4e^{0.4t}F(t)\\
&= 50e^{-0.1t}-0.4e^{0.4t}F(t)+0.4e^{0.4t}F(t)\\
&=50e^{-0.1t}
\end{align*}
(c) (ii)
Integrate the result from (i) with respect to $$t$$.
\begin{equation*}F(t)e^{0.4t}=-500e^{-0.1t}+C\end{equation*}
Substitute in the fact that $$F(0)=0$$ (given).
\begin{align*}
0\times e^0 &= -500e^0+c\\
C &= 500
\end{align*}
Hence
\begin{align*}
F(t)e^{0.4t}&=-500e^{-0.1t}+500\\
F(t)&=-500e^{-0.5t}+500e^{-0.4t}\\
&= 500(e^{-0.4t}-e^{-0.5t})
\end{align*}
(c) (iii)
\begin{align*}
F'(t)&=500(-0.4e^{-0.4t}+0.5e^{-0.5t})\\
&= -200e^{-0.4t}+250e^{-0.5t}\\
&= 0 ~~~~(\text{for max/min value of }F)
\end{align*}
Solve for $$t$$:
\begin{align*}
200e^{-0.4t} &=250e^{-0.5t}\\
200e^{0.1t}&= 250\\
e^{0.1t}&=\frac{5}{4}\\
t&=10\ln\Big(\frac{5}{4}\Big)
\end{align*} | 5,696 | 12,970 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2018-43 | longest | en | 0.682604 |
https://www.gamedev.net/forums/topic/512661-why-i-need-the-numerical-integrator/ | 1,513,491,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948593526.80/warc/CC-MAIN-20171217054825-20171217080825-00274.warc.gz | 770,351,380 | 35,743 | # Why I need the numerical integrator?
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If given the velocity and the acceleration,I can immediately get Distance=1/2 a t^2 + v t what is the numerical integrator used for?
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Original post by DingOunanIf given the velocity and the acceleration,I can immediately getDistance=1/2 a t^2 + v t
But what happens if the acceleration is constantly varying (i.e. the player is changing direction or speed)? You could get down and dirty with a bunch of calculus, but that could be very complicated (especially if acceleration is not controlled by a known function). Numerical integration is a very simple approach which approximates the (arbitrarily complicated) calculus required for this task.
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Thank you very much for the quick reply.
because the newtonian dynamic equation contains not only the external forces(applied by user),but also the constraining forces(like the force that attach the mass in a pendulum system)
the constraining force is difficult to calculate,so we use the lagrange dynamic equation,which is not easy to be integrated symboly.then we need the numerical integrator.
Am I right?
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Quote:
Original post by DingOunanbecause the newtonian dynamic equation contains not only the external forces(applied by user),but also the constraining forces(like the force that attach the mass in a pendulum system) the constraining force is difficult to calculate,so we use the lagrange dynamic equation,which is not easy to be integrated symboly.then we need the numerical integrator.
You are correct, numerical integration can be used anytime we need to approximate a difficult integral.
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As an addendum to what swiftcoder said, numerical integration is sometimes necessary because we don't actually know what function is being integrated.
If we're updating the character's position, for example, and the player can apply forces to the character with the keyboard, we can't predict what forces the player will choose to apply. Therefore, we don't know how the acceleration of the character will change with time except at the present moment and any moments in the past where we chose to remember the state of the system.
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Another reason to use numerical integration is that many functions don't have primitives that can be expressed in closed formulas in terms of well-known functions, so analytical integration is not possible.
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I like the question. ;-)
The reasons people use numerical integration are that (1) it's easier than solving ODEs, and we are lazy (or would rather do other things), (2) accuracy doesn't matter much in games; it just has to 'look right,' and (3) since we need to know the value of the state every [small timestep] anyway, numerical integration can actually be faster than evaluating a complicated expression (the analytic solution to the differential equation) at all of those points.
But let me play devil's advocate for a second. I would assert that game physics, between collisions, can be evaluated symbolically in most cases. Because most of the time, what are you doing? Rigid body dynamics with collisions, and nothing more. (How often do you see lots of constraint forces in games, besides those that arise from collisions?) And what's more, evaluating the analytic solution to these differential equations is very cheap, so argument (3) above doesn't actually work.
I also disagree with the "unpredictable player input" argument, nilkn (Sorry!). But you see, over any time interval, the player is either hitting a key, or not hitting a key. The way you will probably interpret this -- if you say that keystrokes equate to forces (which I admit they usually don't; usually we just use kinematic models. But those are even simpler) is that over any time interval, the force exerted by the player is a constant. And we know the solution to the ODEs when the input is a constant!!
The only complication comes from handling collisions, but really the way that you do this won't be much different whether you're using analytic solutions or numerical integration. In fact, it might be faster when you use the analytic solution, because whereas when you use numerical integration, often what I've seen people do is use binary search to accurately determine the moment of collision, if you have the analytic solution then you have access to all the derivatives of your state trajectory and can use something that converges faster than binary search, like Newton-Raphson.
Plus, doing things this way will be more accurate, and avoid different-integrator-timestep issues, such as those that plagued Quake3 (players with different framerates could jump different heights. Naturally, you can also solve this problem with a constant physics rate, but this has its own problems).
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Emergent: I think your argument breaks down if you introduce some sort of drag, especially if you have random wind.
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Quote:
Original post by alvaroEmergent: I think your argument breaks down if you introduce some sort of drag, especially if you have random wind.
True -- if the drag in nonlinear (say, quadratic); otherwise it can be solved easily. Random wind also isn't necessarily a problem (if the ODE is linear), since you end up with a stochastic differential equation, which you can again solve.
But I see where you're going, and I think you're right: The real argument, I suppose, is that numerical integration is more flexible. If I have linear drag and then want to switch to a quadratic drag, I can, easily, if I'm using numerical integration. Whereas if I'm using analytic solutions then everything breaks -- I simply can't (since an analytic solution doesn't exist).
Perhaps the "perfect" way to deal with these things is to abstract away the dynamics, so that to the rest of the code -- collision detection, etc -- it doesn't matter whether what's going on under the hood is numerical integration or the evaluation of analytic expressions (I know this partially contradicts what I said before about Newton-Raphson vs. binary search, but in truth, if you really want to, you can get derivatives out of your numerical integrator). But if you're not going to do that abstraction, then numerical integration is more flexible.
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Original post by EmergentBut you see, over any time interval, the player is either hitting a key, or not hitting a key. The way you will probably interpret this -- if you say that keystrokes equate to forces (which I admit they usually don't; usually we just use kinematic models. But those are even simpler) is that over any time interval, the force exerted by the player is a constant. And we know the solution to the ODEs when the input is a constant!!
The point I was trying to make is that the ODE you're talking about is itself an approximation that is only valid locally (i.e., it's only valid for that specified time interval where you know exactly which keys are pressed).
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Original post by DingOunanIf given the velocity and the acceleration,I can immediately getDistance=1/2 a t^2 + v twhat is the numerical integrator used for?
You are asking wrong question. A programmer invents new algorithms, he is not supposed to try to find a use for a some strange mathematical function, because he is not mathematician.
An integrator is used by people that don't know calculus, and by people who, while they know calculus, prefer simpler methods.
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Original post by RagharA programmer invents new algorithms
Since we appear to be arguing semantics, a programmer writes code - nothing more, nothing less.
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he is not supposed to try to find a use for a some strange mathematical function, because he is not mathematician.
A computer scientist, on the other hand, is both a programmer and a mathematician, and as such he must spend a great deal of time messing around with 'strange mathematical functions'.
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An integrator is used by people that don't know calculus, and by people who, while they know calculus, prefer simpler methods.
Or, more commonly, to solve a problem so complex as to either defy an analytical approach or render it too expensive to be calculated in the time allotted.
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Original post by nilknThe point I was trying to make is that the ODE you're talking about is itself an approximation that is only valid locally (i.e., it's only valid for that specified time interval where you know exactly which keys are pressed).
It's not an approximation. It is exact. But let me be explicit so we're not talking past each other. Here's an example:
Consider the ODE,
(d^2/dt^2) x = u
where 'u' is the input, and 'x' is the state (in this case a scalar). And for simplicity let's say that you can hit one of two keys, UP or DOWN, so that u can take one of three values:
- UP key pressed <==> u = +1
- DOWN key pressed <==> u = -1
- No key pressed <==> u = 0
[For completeness, I suppose we should also lump "Both UP and DOWN pressed" in with "No key pressed," (as in, "they cancel each other out"), but this isn't terribly important.]
Then, let's say that your input as a function of time is,
{ +1 if 0 <= t < 1u(t) = { 0 if 1 <= t < 2 { -1 if 2 <= t < 3
Then the exact solution for x(t), (assuming x(0)=0) is,
{ t^2 if 0 <= t < 1x(t) = { 1 if 1 <= t < 2 { 1 - t^2 if 2 <= t < 3
See what I mean? The solution is exact. If you can solve your ODE for a constant input, then you've also solved it for piecewise constant inputs.
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Original post by swiftcoder
Quote:
Original post by RagharA programmer invents new algorithms
Since we appear to be arguing semantics, a programmer writes code - nothing more, nothing less.
A programmer is supposed to create a program, if he would be unable to create a new algorithm, he'd be unable to do his work. The term programmer is also used by SW engineers as a non arrogant description of theirs capabilities. What you described is code monkey, or coder.
My point was a programmer (SW engineer, computer "scientist" with ability to create a working program) isn't seeking a use for an algorithm, or math equation, he is creating/modifying algorithms that are relevant to his code.
While history knows situations where an author of an algorithm desperately tried to find at least some use for his algorithm, majority of reliable programs were created by people who threw part of design specifications out of window, and rather created theirs own solutions.
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he is not supposed to try to find a use for a some strange mathematical function, because he is not mathematician.
A computer scientist, on the other hand, is both a programmer and a mathematician, and as such he must spend a great deal of time messing around with 'strange mathematical functions'.
Wow, I'd call that pink glasses. It reminded me of Peter Molyneux's speeches. In the real world, a common computer scientist ends in the strange world of string manipulations. ~_^
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An integrator is used by people that don't know calculus, and by people who, while they know calculus, prefer simpler methods.
Or, more commonly, to solve a problem so complex as to either defy an analytical approach or render it too expensive to be calculated in the time allotted.
How is that different from "prefer simpler methods?
While few first terms of the Taylor series can be used as an approximation (and sometimes they work well), there is also problem called frame lag. A player don't see what happens now, he sees what happened while ago.
D = v0 + a/2 works, but D = v0 + a works better. The whole system responds more readily, and feels more snappy. There are also pesky problems with limited accuracy on computers, which in better situations only lose energy, in worst situation they make the whole system explode. Integrator offers greater flexibility than analytic solution. Is system gaining energy? Add limiter to that small part where it matters. Is system losing energy? Bump it up, or bias integrating function a little. Analytic solutions are more pesky to work with.
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Original post by Raghar...
Fair enough. However, when I go to sort a list of integers, I find an existing sorting method which has been proven by someone else. If I develop an entirely new algorithm, then the onus is on me to provide a proof (in the mathematical sense), that my algorithm functions as advertised - thus to create a new algorithm I must wield at least some portion of the mathematician's skill.
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An integrator is used by people that don't know calculus, and by people who, while they know calculus, prefer simpler methods.
Or, more commonly, to solve a problem so complex as to either defy an analytical approach or render it too expensive to be calculated in the time allotted.
How is that different from "prefer simpler methods?
I took mild offence at your implication that the only reason to use numerical integration is a lack of facility with mathematics.
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Original post by Emergent
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Original post by nilknThe point I was trying to make is that the ODE you're talking about is itself an approximation that is only valid locally (i.e., it's only valid for that specified time interval where you know exactly which keys are pressed).
It's not an approximation. It is exact. But let me be explicit so we're not talking past each other. Here's an example:
Consider the ODE,
(d^2/dt^2) x = u
where 'u' is the input, and 'x' is the state (in this case a scalar). And for simplicity let's say that you can hit one of two keys, UP or DOWN, so that u can take one of three values:
- UP key pressed <==> u = +1
- DOWN key pressed <==> u = -1
- No key pressed <==> u = 0
[For completeness, I suppose we should also lump "Both UP and DOWN pressed" in with "No key pressed," (as in, "they cancel each other out"), but this isn't terribly important.]
Then, let's say that your input as a function of time is,
{ +1 if 0 <= t < 1u(t) = { 0 if 1 <= t < 2 { -1 if 2 <= t < 3
Then the exact solution for x(t), (assuming x(0)=0) is,
{ t^2 if 0 <= t < 1x(t) = { 1 if 1 <= t < 2 { 1 - t^2 if 2 <= t < 3
See what I mean? The solution is exact. If you can solve your ODE for a constant input, then you've also solved it for piecewise constant inputs.
Ah, I see much more clearly what you mean now.
You present a convincing argument, so I stand corrected. [smile]
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This topic is 3337 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. | 3,376 | 15,042 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-51 | longest | en | 0.902158 |
https://www.coursehero.com/tutors-problems/Math/11514724-1-Producers-will-supply-units-of-a-certain-commodity-to-the-mark/ | 1,544,999,529,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827998.66/warc/CC-MAIN-20181216213120-20181216235120-00488.warc.gz | 848,293,118 | 22,131 | View the step-by-step solution to:
# ommodity to the market when the price is ???? = ????(????) dollars per unit, and the consumers will demand (buy) ???? units when the price is ???? = ?...
1) Producers will supply ???? units of a certain commodity to the market when the price is ???? = ????(????) dollars per unit, and the consumers will demand (buy) ???? units when the price is ???? = ????(????) dollars per unit where ????(????) = ???????? + ???? and ????(????) = ???????? + ???? for constants ????, ????, ???? and ????. Express the equilibrium production level and the equilibrium price in terms of the coefficients ????, ????, ???? and ????.
2) A manufacturer of a product sells all that is produced. The total revenue is given by ???????????? = 16????, and the total cost is given by ???????????? = 4???? + 14400, where ???? represents the number of units produced and sold. (a) Find the level of production at the break-even point, and draw the break-even chart. (b) Find the level of production at the break-even point if the total cost increases by 5%?
3) ???????????? = 0.5???? 2 + 12???? represents total revenue in dollars and ???????????? = 7???? + 352 represents total cost in dollars for a manufacturer. If ???? represents both the number of units produced and the number of units sold, find the break-even price.
4) A manufacturer of dining-room sets produced two styles: early American and contemporary. From past experience, management has determined that 22% more of the early American styles can be sold than the contemporary styles. A profit of \$150 is made on each early American set sold, whereas a profit of \$250 is made on each contemporary set. If, in the forthcoming year, management desires a total profit of \$4060, how many units of each style must be sold?
5) Supply and demand equations for a certain product are 3???? − 200???? + 1800 = 0 and 3???? + 100???? − 1800 = 0 respectively, where p represents the price per unit in dollars and q represents the number of units sold per time period. (a) Find the equilibrium price algebraically, and show the system in the coordinate plane. (b) Find the equilibrium price when a tax of 27 cents per unit is imposed on the supplier.
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Ask a Question | 648 | 2,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-51 | latest | en | 0.884816 |
https://quantumcomputing.stackexchange.com/questions/8381/understanding-the-action-of-operators-on-vectors-in-tensor-product-spaces | 1,713,382,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00217.warc.gz | 438,710,569 | 41,245 | # Understanding the action of operators on vectors in tensor product spaces
I'm studying Quantum Computing: A Gentle Introduction. On page 33, Section 3.1.2, after defining tensor product with 3 properties (distribution over addition on both left and right, scalar on both sides), it says all element of $$V \otimes W$$ have form $$|v_1\rangle \otimes |w_1\rangle +|v_2\rangle \otimes |w_2\rangle +\ldots+|v_k\rangle \otimes |w_k\rangle$$, where $$k$$ is the minimum of the dimensions of $$V$$ and $$W$$.
Assume $$V$$ has $$k+1$$ dimensions (and $$W$$ has $$k$$ dimensions), why the basis $$|v_{k+1}\rangle$$ is not used?
Similarly, in the classic Quantum computing and quantum information textbook, section 2.1.7, formula 2.46, it also uses a single index $$i$$ $$(A\otimes B)(\sum_i a_i|v_i\rangle \otimes |w_i\rangle).$$ I thought it should be $$(A\otimes B)(\sum_ {i,j} a_{i,j}|v_i\rangle \otimes |w_j\rangle).$$
Is there anything deeper to explain that a single index with a subset of bases is sufficient?
Thanks
The reason is relatively straightforward. Consider an $$m$$ dimensional vector space $$V$$ with basis $$\lbrace \vert v_1 \rangle,...,\vert v_m \rangle \rbrace$$, and an $$n$$ dimensional vector space $$W$$ with basis $$\lbrace \vert w_1 \rangle,...,\vert w_n \rangle \rbrace$$. As your intuition suggests, we can naturally express any element $$A \in V \otimes W$$ in the form $$A = \sum \limits_{j=1}^{m} \sum \limits_{k=1}^n \lambda_{jk} \vert v_j \rangle \otimes \vert w_k \rangle,$$ where $$\lambda_{jk}$$ are scalar coefficients.
The reason we can express $$A$$ in $$\text{min}(m,n)$$ terms is that we can group the set of $$m$$ vectors $$\lambda_{jk} \vert v_j \rangle$$ into a new set of $$n$$ vectors $$\vert a_k \rangle$$ given by $$\vert a_k \rangle=\sum \limits_{j=1}^m \lambda_{jk} \vert v_j \rangle,$$ which gives an expression for $$A$$ with one index in $$n$$ terms as $$A=\sum \limits_{k=1}^n \vert a_k \rangle \otimes \vert w_k \rangle.$$ We can do the same to express $$A$$ in $$m$$ terms by $$A = \sum \limits_{j=1}^m \vert v_j \rangle \otimes \vert b_j \rangle, \;\;\;\; \vert b_j \rangle = \sum \limits_{k=1}^n \lambda_{jk} \vert w_k \rangle,$$ showing that $$A$$ can always be expressed in $$\text{min}(m,n)$$ terms.
However, to show that the $$mn$$ elements $$\lbrace \vert v_j \rangle \otimes \vert w_k \rangle \rbrace$$ form a basis in $$V \otimes W$$, we still need to show that these elements are linearly independent. That proof is not so easy (or concise). I would refer you to Linear Algebra via Exterior Products by Winitzki, section 1.7.3, if you want that level of rigor.
• What do you mean by "the set of $min(m,n)$ terms form a basis in $V \otimes W$"? This space has dimension $mn$. Oct 2, 2019 at 12:03
• Thank you for pointing that out. That was a mistake. Edited to fix. Oct 2, 2019 at 12:49
• what is the difficulty in proving that $\{|v_j\rangle\otimes|w_k\rangle\}$ are linearly independent? They are obviously orthogonal to each other (essentially by definition of the inner product in the tensor product space), and therefore linearly independent, no?
– glS
Oct 4, 2019 at 18:03
• If $\{|v_j\rangle \otimes |w_k\rangle \}$ span the whole $V \otimes W$ (and they do) then they have to be linearly independent. If they are linearly dependent then they can span only subspace of dimension less than $mn$. Oct 6, 2019 at 20:41
• I see, in 1.7.3 we don't know yet that dimension of tensor product is $mn$. Anyway, in the field $\mathbb{R}$ or $\mathbb{C}$ it's much easier since we can introduce the scalar product. Oct 7, 2019 at 6:39 | 1,106 | 3,602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.755554 |
https://www.physicsforums.com/threads/calculating-impact-force.366172/ | 1,521,935,607,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651465.90/warc/CC-MAIN-20180324225928-20180325005928-00678.warc.gz | 863,084,772 | 15,583 | # Calculating impact force?
1. Dec 29, 2009
### alexa23
calculating impact force??
I need help calculating impact force. Here's all the information I have:
the object weighs 180 pounds
being dropped from a height of 11 feet
I'm not completely sure if the force can be determined from that. I'm having a hard time figuring it out for some reason.
2. Dec 30, 2009
### Tominator
Re: calculating impact force??
Hi Alexa
I recommend you to read forum rules before posting because your thread seems like a homework and homeworks are discussed in other forum.
By the way welcome to PF
3. Jan 6, 2010
### waynexk8
Re: calculating impact force??
Hi all,
I would like to ask the same thing, and first to prove its not homework, first I am 48, and second, we were wondering how mush force the muscles would be taking on the transition from negative to positive, like in strength training, and repping a barbell up and down. We call this transition MMMs {Momentary Maximum Muscle Tensions}
These tensions or forces are as you imagine the highest there is in the whole rep. {repetition} As when you are on the eccentric of the rep, you are still controlling it down, but when a given load is lowed quite fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.
Would it be possible to work out the force of ??? Let’s say a 90kg weight, dropping at .5 of a second for 1m, please ???
Wayne
4. Jan 6, 2010
### Redbelly98
Staff Emeritus
Re: calculating impact force??
More information is needed. Specifically, either how long or over what distance is the object brought to a stop? Note, this is different than the time or distance that the object is falling.
5. Jan 7, 2010
### waynexk8
Re: calculating impact force??
Hmm, as you can see this is a bit tricky to say, as when you are repping up and down with a barbell it quite hard to tell.
Therefore, what we say the last fifth of the eccentric, and that would also be one fifth of .5 of a second.
And thx for you time.
Wayne | 511 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-13 | longest | en | 0.942151 |
http://www.decodedscience.com/a-brief-introduction-to-prime-numbers/4084 | 1,440,905,546,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064869.18/warc/CC-MAIN-20150827025424-00200-ip-10-171-96-226.ec2.internal.warc.gz | 383,005,792 | 14,203 | # A Brief Introduction to Prime Numbers
Selection of Prime Numbers – Image by Decoded Science
Would you like a simple introduction to primes? This article introduces a series about prime numbers by answering the following basic questions: What are prime numbers? Are primes important? How many prime numbers are there?
## What is a Prime Number?
Prime numbers are those Natural numbers that are greater than one and can only be evenly divided by the numbers one and itself. As a set, P={all ‘i’ in N such that “i>1″ and, for ‘j’ in N, where “i/j” is in N only if ‘j’ equals either one or ‘i’}.
The set of prime numbers begins as {2, 3, 5, 7, 11, 13, 17…}.
## What are Non-Prime Numbers Called?
Non-prime numbers greater than 2 are called “composite” numbers. These are positive integers that are “composed of” the product of primes. The set of composite numbers begins as {4, 6, 8, 9, 10…}.
## Why are Zero and One Commonly Excluded From Lists of Prime Numbers?
Zero: Since zero divided by anything is still zero, with no remainder, one could say that zero is not a prime number. Zero may be excluded from lists of prime numbers because no-one ever found an acceptable reason to include it in a discussion of primes.
One: The number one is not a composite number, so why is it excluded? This controversy raged in the 20th century; at least, there were disagreements. One can say, “There is no Natural number other than 1 and itself that can divide 1 evenly; therefore 1 is a prime number.” Several mathematicians did indeed consider ‘1’ to be the first Prime number.
However, the number ‘1’ requires exceptional handling in a variety of “primal” situations. The first example is the “fundamental theorem of arithmetic.” There seems to be consensus that the number one is simply defined as “neither prime nor composite.”
## What is the “Fundamental Theorem of Arithmetic?”
The Fundamental Theorem of Arithmetic states that any Natural number that is greater than one is the product of a unique set of prime numbers. The sequence of numbers can be rearranged in the list, but an ordered set would be unique.
Two examples should make this concept clear. “7 = 7″, since 7 is prime. “84 = 2*2*3*7″ is the unique decomposition of 84 into primes, ordered by increasing value of the primes. One could also write “84 = (2^2)*3*7″.
If the number ‘1’ were included in the set of prime numbers, then this theorem would have a problem. “7 = 7 = 7*1 = 7*1*1 = 7*1*1*1 = …”. The theorem would have to exclude ‘1’ or stipulate that ‘1’ may only be used once.
The more descriptive name for this theorem is the “Unique Factorization Theorem“, as each divisor of a Natural number is called a “factor”.
Sieve for Seven – Image by Scot Nelson
## The Infinite Universe of Prime Numbers
Euclid, an ancient Greek philosopher, proved that the number of prime numbers is infinite.
Another ancient Greek, Eratosthenes, described a mathematical “sieve” to filter out composite numbers that are the products of the first primes found. These two Greeks allow us to say, without being too rigorous, that the number of prime numbers must be a “countable” infinity.
This introductory article will not delve into Euclid’s proof or a more detailed description of the sieve of Eratosthenes. An earlier article, “Cantor Defeated Galileo in the Battle of Infinite Numbers“, explains countable versus uncountable infinities.
Leonhard Euler by Johann Georg Brucker: Image by Haham hanuka
## The Future of Prime Numbers
Future articles in this series may include specific types of prime numbers, and ways in which prime numbers are used. Leonhard Euler is a modern mathematician who contributed to the study of prime numbers.
One use for prime numbers familiar to many in the fields of cryptography and of computer security is to multiply a pair of large prime numbers for data encryption. Because it is difficult to determine factors of a large number, this is a fairly secure way of encrypting computer data. Whether faster algorithms or “quantum computing” will ever make this approach obsolete are questions that remain unsolved.
External Resources:
Weisstein, Eric W. MathWorld (Wolfram). Fundamental Theorem of Arithmetic, Mersenne Prime, and Prime Number. Accessed Oct. 17, 2011. | 998 | 4,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2015-35 | latest | en | 0.959333 |
https://brainmass.com/engineering/electrical-engineering/single-turn-circular-coils-and-resultant-magnetic-field-2199 | 1,643,177,889,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00489.warc.gz | 203,012,795 | 75,354 | Explore BrainMass
# Single turn circular coils and resultant magnetic field
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Two single turn circular coils share a common horizontal axis and are separated by a horizontal distance of 2.5m. The radius of one of the coils is 50 cm and it carries a current of 5 A. The radius of the other coil is 25 cm and it carries a current I in such a direction that the magnetic field which it produces in the opposition to that produced by the current in the first coil. Determine the value of I in order that the resultant magnetic field strength at the mid-point on the axis between the coils shall be zero.
Please view the attachment for the complete question and the diagram.
https://brainmass.com/engineering/electrical-engineering/single-turn-circular-coils-and-resultant-magnetic-field-2199
#### Solution Preview
The first step in finding the current I is to find H for the 5A coil. ...
#### Solution Summary
Solution also mentions how to approach the problem.
\$2.49 | 250 | 1,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-05 | longest | en | 0.917813 |
http://www.notimeforflashcards.com/tag/math-for-kids/page/2 | 1,436,102,422,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097475.46/warc/CC-MAIN-20150627031817-00106-ip-10-179-60-89.ec2.internal.warc.gz | 676,670,129 | 47,919 | ## Monster Math Tray – Learning After School
Math is probably my son’s favorite subject right now and I am running with it. This monster math activity took 2 minutes to set up and could be used over and over . No need to buy anything other than paper and googly eyes! I like making simple tray activities like this that I can have ready for him at the table in the playroom to do after school. They are educational and appealing but not so long that he gets overwhelmed after a long day of school. As you will see my toddler demanded she get in on the action too, luckily this was easy to adapt to her level as well. For a fun variation check out how you can use dice for even more monster math over at Inner Child Learning.
1. Gather your materials. You will need some bright paper, googly eyes, a pencil ( if you want to use the monsters more than once) , and scissors. I also had a small cup and cookie tray to keep everything contained.
2. Start by folding your paper in half and cutting out the shape of a monster. Best part is that they can be detailed or a blob , not need for extraordinary artistic skills!
3. Next write simple math equations on the monster. If your child isn’t up to equations yet just do numbers. You can also do shapes and have your child place the eyes on the shapes. I used markers for the photos but If you want to use this more than once you can use pencil, or laminate the monsters and use dry erase markers.
4. All ready – now add a math wizard! He loves this . I love that while doing math he is also working on fine motor skills that he needs for writing.
5. After he did addition we flipped it over and did some subtraction. He loved it too!
6. ” I do it TOO!!!” To make it toddler friendly I only used the largest eyes, and wrote simple numbers on the monsters. My daughter still needed a little help as I thought she would but she was ecstatic to be doing big girl math with her big brother. All I know is she begged to do math. Let’s keep that spirit going right?
## Monster Books
Check out our round up of monster books for some reading after your math!
## Haunted House Math Activity
Using holidays like Halloween as a theme for great learning activities is a sure fire hit in my house. My daughter who is 2 is always excited to do any project but my almost 6 year old is a lot more picky. This Halloween math activity was such a hit that when I asked my son to rate it 1-100 he gave it a 100 without hesitation! Better yet it’s pretty easy to make , adapt for various levels and frugal too.
1. Gather your materials. You will need some craft paper , markers , white card stock ( or paper plates !), scissors, painter’s tape and something to attach the house to a wall. I used push pins but more painter’s tape would work too.
2. Cut out simple ghost shapes from the paper plates / card stock. Add faces and numbers. I did 1-10 but you can write whatever numbers your child is working on.
3. Draw a haunted house on craft paper. Mine took 3 tries the first was so bad I should have taken a picture to make you all laugh. The other ones became coloring paper for my toddler.
4. Write out simple equations , number words or even just numbers to match up. You will see further down that for a toddler like my daughter you don’t even need anything to match. Just play with the numbers on the ghosts.
5. Add painter’s tape to the ghosts and on the haunted house where you will place the equations.
6. Add the equations to the house, put the ghosts next to it ready to be put in the house and call your little mathematician.
7. As soon as my son saw the activity he said it was too easy and it probably was. I grabbed my iPhone and asked him if he wanted me to time him. His face lit up. I don’t suggest timing children who don’t want to be timed or who will feel negatively pressured . Matching the words with the numbers on the ghosts was an easy task for my son but he has a competitive spirit and timing him made it more fun because it made it challenging.
8. He flew through it. Placing the ghosts on top of the matching words.
9. Next I switched the words on the house to simple equations. These were not going to be as easy and I told him for this time we would not be timing it. I think that if I’d done the harder task first he would have gotten frustrated when a few of the harder equations didn’t come to him immediately.
10. After my son was done I removed all the tape and equations so the house was clear , and put the ghosts back on the wall. Then invited my daughter who is 2 to come and put the ghosts in the haunted house. It was perfect for her. She grabbed the ghosts and named the numbers she knew and asked me to confirm the numbers she didn’t. She was very specific about where they should be. I was thrilled that they both had fun with math at their own levels of learning!
## Ghosts In The House
Ghosts in the House! by Kazuno Kohara is on my must buy list! A little girl moves into house and soon finds out it is haunted. Luckily she is a witch and knows just what to do. The ghosts in the story seem mischievous but never scary and even when she washes them in the washing machine, they are still smiling. My son loved this book, the text was the perfect length for a 3 year old, short but still descriptive. I loved the simple black and orange colors and had to look at the copyright twice because I was certain this was written sometime in the 30s, nope 2008. The simplicity of the book and colors is balanced so well with the little details like the little girl’s constant companion , a white cat that puts on a black costume when the little witch pops on her hat. This detail had my son in stitches, “Cats don’t wear clothes , silly cat!” . Absolutely a perfect Halloween book for children not yet ready to be scared for fun!
This post contains an affiliate link.
## Cheeseball Math
by Kim
I got the inspiration for this activity from Disney’s Family Fun magazine June/July 2012 issue. The idea they shared in the magazine was to cover shower caps with shaving cream and toss cheese puffs at each other to see how many puffs you could get to stick. I quickly saw potential for this to turn into some fun learning.
Grab some cardboard, shaving cream, post-it notes, a marker, and some cheeseballs. We are about to get cheesy. Yes, I know how dorky how I am.
I drew numbers on some post-it notes and affixed them to the cardboard for my preschooler. You could write directly on the cardboard, but this way I could easily change the numbers to allow a wider spectrum for practice.
I then wrote some math equations for my five year old son on post-it notes and affixed those to a different piece of cardboard.
My son read the equations and figured them out.
He threw the corresponding number of cheeseballs at the targets that equaled the answer.
My three year old starting throwing the number of cheeseballs at the corresponding numbers written above the targets.
Littlest sister couldn’t resist how much fun everyone was having and had to join in. She helped count as we threw and then did some throwing herself when it was her turn.
There you have it, fun math! It combines throwing, aiming, math equations, number recognition, counting, and even sensory fun by spreading the shaving cream. We had a blast!
.
Kim is a contributing writer for No Time For Flash Cards, a mom to a toddler, a preschooler, and a foster parent, too. She juggles her day by trying out fun activities and crafts with the kids. After all, she is just a big kid herself. See what she has been up to over at Mom Tried It. | 1,680 | 7,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2015-27 | latest | en | 0.967163 |
https://www.codeproject.com/Questions/788490/How-find-shortest-path-in-d-Array | 1,716,060,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00166.warc.gz | 625,899,419 | 33,202 | 15,900,907 members
See more:
I have 2d array. I want find shortest path from left top value to right down. I want use Deikstra algho, but it take much memory.
```0 2 2 2 2 2
1 0 1 0 0 1
2 2 2 0 2 2
2 2 2 0 1 0```
Thats is path i want found. Does anybode have c# code?
0 2 2 2 2 2
1 0 1 0 0 1
2 2 2 0 2 2
2 2 2 0 1 0
i make it in pascal, but cannot make in c#
Delphi
```const maxn = 40;
var n, m : integer;
a : array [1..maxn, 1..maxn] of integer;
dist : array [1..maxn * maxn] of integer;
bool : array [1..maxn * maxn] of boolean;
var i, j : integer;
begin
for i := 1 to n do
for j := 1 to m do
end;
function num (x, y : integer) : integer;
begin
num := (x - 1) * m + y;
end;
procedure solve;
var i, cur, min : integer;
x, y : integer;
begin
fillchar (bool, sizeof (bool), false);
fillchar (dist, sizeof (dist), \$FF);
bool[1] := true;
dist[1] := a[1, 1];
cur := 1;
while cur <> n * m do
begin
x := cur div m + 1;
y := cur mod m;
if cur mod m = 0 then
begin
x := x - 1;
y := m;
end;
if (y > 1) and ((dist[num (x, y - 1)] = -1) or (dist[num (x, y - 1)] > dist[num (x, y)] + a[x, y - 1])) then
dist[num (x, y - 1)] := dist[num (x, y)] + a[x, y - 1];
if (y < m) and ((dist[num (x, y + 1)] = -1) or (dist[num (x, y + 1)] > dist[num (x, y)] + a[x, y + 1])) then
dist[num (x, y + 1)] := dist[num (x, y)] + a[x, y + 1];
if (x > 1) and ((dist[num (x - 1, y)] = -1) or (dist[num (x - 1, y)] > dist[num (x, y)] + a[x - 1, y])) then
dist[num (x - 1, y)] := dist[num (x, y)] + a[x - 1, y];
if (x < n) and ((dist[num (x + 1, y)] = -1) or (dist[num (x + 1, y)] > dist[num (x, y)] + a[x + 1, y])) then
dist[num (x + 1, y)] := dist[num (x, y)] + a[x + 1, y];
min := 10000;
for i := 1 to n * m do
if (not ((dist[i] = -1) or (bool[i]))) and (dist[i] < min) then
begin
min := dist[i];
cur := i;
end;
bool[cur] := true;
end;
writeln (dist[n * m]);
end;
begin
assign (input, 'input.txt'); reset (input);
assign (output, 'output.txt'); rewrite (output);
solve;
close (input);
close (output);
end.```
Posted
Updated 20-Jun-14 22:51pm
v3
OriginalGriff 21-Jun-14 4:42am
And?
What have you tried?
Where are you stuck?
What help do you need?
[no name] 21-Jun-14 4:55am
i have tried)
phil.o 21-Jun-14 8:13am
"What have you tried?
- I have tried."
How funny! | 945 | 2,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.486822 |
http://www.wikihow.com/Calculate-Kilowatts-Used-by-Light-Bulbs | 1,474,954,075,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660966.51/warc/CC-MAIN-20160924173740-00059-ip-10-143-35-109.ec2.internal.warc.gz | 788,330,692 | 52,054 | Edit Article
# wikiHow to Calculate Kilowatts Used by Light Bulbs
Ever wonder how much that light bulb is costing you? Is it really worth switching to compact fluorescent or LED bulbs? All you need to find out is the bulb's wattage, and the cost of electricity in your building. Replacing your incandescent bulbs with energy-efficient options typically saves a few dollars in the first year, and more over a longer period of time.
### Part 1 Kilowatts and Kilowatt-Hours
1. 1
Find the wattage rating of the bulb. The wattage is often printed directly on the bulb as a number followed by a W. If you don't see it there, check the packaging the bulb was sold in. The watt is a unit of power, measuring the energy the bulb uses each second.
• Ignore phrases like "100-watt equivalent," which are used to compare brightness. You want the actual number of watts the bulb uses.
2. 2
Divide this number by one thousand. This converts the number from watts to kilowatts. An easy way to divide by one thousand is to move the decimal point three places to the left.
• Example 1: A typical incandescent bulb draws 60 watts of power, or 60 / 1000 = 0.06 kilowatts.
• Example 2: A typical fluorescent bulb uses 15 watts, or 15 / 1000 = 0.015 kW. This bulb only uses ¼ as much power as the bulb in example 1, since 15 / 60 = ¼.
3. 3
Estimate the number hours the bulb is on per month. To calculate your utility bill, you'll need to know how much you use your bulb. Assuming you receive monthly utility bills, count up the number of hours the bulb is on in a typical month.
• Example 1: Your 0.06 kW bulb is turned on for 6 hours a day, every day. In a 30-day month, that's a total of (30 days/month * 6 hrs/day) = 180 hours per month.
• Example 2: Your 0.015 kW fluorescent bulb is on for 3 hours a day, 3 days a week. In one month, it will be on for roughly (3 hours/day * 3 days/wk * 4 wks/month) = 28 hours per month.
4. 4
Multiply the kilowatt use by the number of hours. Your energy company charges you for each "kilowatt-hour" (kWh), or each kilowatt of power in use for one hour. To find the kilowatt-hours your light bulb consumes per month, multiply the kilowatt use by the number of hours it's on each month.
• Example 1: The incandescent bulb uses 0.06 kW of power and is on for 180 hours a month. Its energy usage is (0.06 kW * 180 hours/month) = 10.8 kilowatt-hours per month.
• Example 2: The fluorescent bulb uses 0.015 kW and is on for 28 hours a month. Its energy usage is (0.015 kW * 28 hours/month) = 0.42 kilowatt-hours per month.
### Part 2 Calculating Cost
1. 1
Calculate the cost of running your light bulb. Check your utility bill for the cost of each kilowatt-hour of electricity. (Average costs are roughly \$0.12 per kWh in the US, or €0.20 per kWh in Europe.[1]) Multiply this by the number of kWh your bulb uses each month to estimate how much you pay to power that bulb.
• Example 1: Your power company charges 10 US cents per kWh, or \$0.10. The incandescent bulb uses 10.8 kWh/month, so running it is costing you (\$0.10/kWh * 10.8 kWh/mo.) = \$1.08 per month.
• Example 2: At the same cost of \$0.10 per month, the low-use fluorescent bulb costs you (\$0.10/kWh * 0.42 kWh/mo.) = \$0.042 per month, or about four cents.
2. 2
Save on lighting costs. Light bulbs account for about 5% of the electricity bill of the average US household.[2] Although other energy savers will have a larger effect, replacing incandescent bulbs is always worth it in the long term:[3][4]
• Replacing a traditional incandescent bulb with a compact fluorescent light (CFL) will typically pay for itself within 9 months. It will also last nine times as long before burning out, saving you even more money in the long run.
• Light-emitting diodes (LEDs) are even more efficient and have a lifespan of 50,000 hours (almost six years of constant use). Over their whole lifespan, they save about \$7 US per year.[5]
3. 3
Choose the right replacement. Consider the following when shopping for energy-efficient lighting:[6][7][8]
• Poorly made CFL bulbs can burn out quickly. The best options have an EnergyStar logo in the US, or a rating of A+ or higher on the European Union energy label.
• If you're lucky, the packaging will list "lumens," a measure of brightness. If not, use this estimate: a 60 watt incandescent bulb, 15 watt CFL, or 10 watt LED are all roughly the same brightness.
• Look for a color descriptor. "Warm white" is closer to the yellow glow of an incandescent. "Cool white" heightens the contrast, which may feel harsh in living spaces.[9]
• "Directional" LED lights focus light on a small area instead of illuminating a whole room.
## Community Q&A
Search
• If 100V is applied across a 200V, 100W bulb, what will be the amount of power consumed?
wikiHow Contributor
This formula is complicated for light bulbs, since the amount of heat produced changes the properties of the light bulb filament. From a practical standpoint, the bulb will be extremely dim. It's best to use a circuit as close to the bulb's rated voltage as possible. Using a higher voltage is even worse, since the bulb could overheat or explode.
• How many KWH does it take to turn on a 60-watt light bulb?
wikiHow Contributor
A 60-watt incandescent lamp will require 60 watts : 120 volts = .5 amperes after turning it on, but at first, it will require 12 - 16 times that amount, about 12 x .5 amperes = 6 amperes for a fraction of a second. To convert it to KWH will be 6 x 120 = 720 watts x about .1 seconds, which will give you .072 KWH.
• What does it means when a light bulb says "9 watts 2000 hours"?
wikiHow Contributor
The number of hours is an estimate of the bulb's lifespan. It does not affect the kilowatt-hour calculations.
• Is 1.5 volts more or less than 20 watts?
wikiHow Contributor
They measure two different things, so you cannot compare them directly. The watts are a measure of power, and that's what you need to calculate kilowatt-hours and electricity costs. Voltage is a property of the battery or household electrical circuit. If "1.5 volts" is printed on a bulb, that bulb is designed to be run off a 1.5 volt battery.
• If I am using a 1.5 volt battery to power a 20 watt lightbulb, what is the amperage?
wikiHow Contributor
Amps = Watts / Volts (or Current = Power / Voltage), so the current in your circuit is equal to (20 watts) / (1.5 volts) = about 13.3 amps. Note that you do not need to know the current for any of the calculations in this article.
• How much electrical energy is consumed by a 30 W bulb in 5 minutes?
wikiHow Contributor
A one watt bulb consumes 1 joule of elecrical energy every second. To calculate the total amount of energy consumed in your example, multiply (1 joule/second) x (30 watts) x (5 minutes) x (60 seconds/minute) = 9000 joules. Joules are a very small unit of energy, useful for problems like this one with a small unit of energy. Energy companies use a unit called the kilowatt-hour instead. 9000 joules = 0.0025 kilowatt hours.
• How do I calculate and compare the energy use of two different watt bulbs over the same period of time?
wikiHow Contributor
1 KWh = 1000 W over a period of one hour. A 60 W light bulb over a period of one hour = 0.06 KWh. 100 W light bulb over a period of one hour = 0.1 KWh.
## Tips
• Watts are a measure of power, not brightness. A 15W fluorescent bulb can be as bright as a 60W incandescent bulb, since fluorescent bulbs are more efficient. LED light bulbs are even more efficient, and can produce the same brightness for less than 8 watts of power.[10]
• Do not believe the myth that leaving fluorescent lights on saves money. Although turning on the light does take a small amount of extra energy, leaving the light on between uses will use much more.[11]
## Warnings
• Check your light fixture label before switching to a higher wattage light bulb. Each fixture has a maximum wattage. Using a light bulb that draws more than the maximum wattage can lead to short circuits and other damage.
• A light bulb made for a higher voltage than your socket will use less wattage than stated on the label. This will slightly reduce the kilowatt-hours used, but the light output will be dimmer and yellower.[12] For example, a 60 watt, 130V bulb powered by a typical U.S. 120V household circuit will draw fewer than 60 watts, and produce dimmer, yellower light than a bulb labeled 60W and 120V.
## Article Info
Categories: Electrical Maintenance
In other languages:
Thanks to all authors for creating a page that has been read 216,088 times. | 2,204 | 8,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-40 | longest | en | 0.927784 |
https://www.mathematik.rwth-aachen.de/cms/mathematik/Forschung/Publikationen/Bibliographie/~ckus/Einzelansicht/?lidx=1&file=229817 | 1,656,157,436,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00717.warc.gz | 966,928,564 | 9,457 | # Borcherds lifts and Maaß lifts on the paramodular group of level 3
• Borcherds Lifts und Maaß Lifts zur paramodularen Gruppe der Stufe 3
Kreuzer, Judith; Krieg, Aloys (Thesis advisor)
Aachen (2014)
Dissertation / PhD Thesis
Aachen, Techn. Hochsch., Diss., 2014
Abstract
The main goal of this thesis is to give a full characterization of those modular forms on the paramodular group Gamma_3^+ of level 3 that are Maaß lifts and Borcherds lifts at the same time. The Maaß lift is an additive lift that yields a Fourier expansion. Moreover, we obtain a certain relation among the Fourier coefficients. In contrast, the Borcherds lift yields an infinite product expansion of the modular form. Thus, it is multiplicative and we can determine the divisors of a Borcherds lift quite easily. By construction, we already know certain properties of a modular form obtained by a Maaß lift or a Borcherds lift. But these properties differ a lot in structure, which makes it difficult to compare the two lifts. The question if a modular form is a Maaß lift and a Borcherds lift at the same time has hardly been examined until now. A first systematical approach to this question was taken by B. Heim and A. Murase in the case of Siegel modular forms of degree 2 without character. In answering the question of simultaneous lifts for the paramodular group Gamma_3^+ of level 3, it turned out that it is not possible to give an analogous proof to the Siegel case. Thus, a new proof was developed. This new approach is presented first by characterizing simultaneous lifts in the case of Siegel modular forms with characters. Afterwards, the same line of argument will be used for the paramodular case of level 3. An important step in describing simultaneous lifts lies in finding properties of Maaß lifts and Borcherds lifts on Gamma_3^+. Some of these characteristics turn out to be interesting in their own right. For example, a Maaß lift on Gamma_3^+ can have only certain characters, and a method to construct Borcherds lifts with any given divisor is presented. Moreover, it is possible to give structural statements on these constructed Borcherds lifts without knowing them in concrete terms; e.g. if it is a cusp form or if it has trivial character. In addition to that, one can show that every non-cuspidal Borcherds lift on Gamma_3^+ must have trivial character. Finally, it turns out that -up to a multiplicative constant- there are exactly six modular forms on Gamma_3^+ that are Maaß lifts and Borcherds lifts at the same time. | 598 | 2,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-27 | longest | en | 0.901 |
http://www.chronicle.com/blognetwork/castingoutnines/2011/10/17/math-monday-tv-lawyers-solve-np-complete-problem-part-1/ | 1,495,954,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609605.31/warc/CC-MAIN-20170528062748-20170528082748-00395.warc.gz | 579,609,335 | 14,838 | Previous The pedagogy of phlebotomy Next Article about the flipped MATLAB class
# Math Monday: TV Lawyers Solve NP-Complete Problem (part 1)
October 17, 2011, 7:30 am
For the next couple of weeks, Math Monday here at the blog will feature a guest blogger. Ed Aboufadel is Professor of Mathematics and chair of the Mathematics Department at Grand Valley State University, where I work. He’ll be writing a two-part series on a neat appearance of an NP-complete problem on network TV, adding yet another data point that mathematics is indeed everywhere. Thanks in advance, Ed!
On the new USA-network TV series Suits [1], Harvey Specter is a senior partner at the law firm of Pearson Hardman, and Mike Ross is his new associate. Mike never went to law school, but he combines a photographic, elephantine memory with near-genius intelligence to fake it well. Harvey is in on the deception, but none of the other partners know. During the eighth episode of the first season of Suits (broadcast August 11, 2011), Harvey and Mike, working with Louis Litt, a socially-challenged junior partner, solved an NP-complete problem in a short amount of time in order to rescue over $150 million for a client. Were they just lucky, smarter than the rest of us, or did they do something practically impossible? In the story, Lucille Jackson, a client of Pearson Hardman, needed help to deal with an investment firm led by Tony Maslow. Lucille had trusted the firm with over$150 million that belonged to a non-profit organization similar to Habitat for Humanity. The investment firm claimed to have lost the money due to poor investments, and due to the malfeasance of one of Tony’s partners who conveniently dies early in the episode. Eventually, Harvey, Mike, and Louis figure out that Maslow had hidden the money overseas, and with the help of biology student and hacker extraordinaire Lola Jensen, they pull together the following information:
• The exact amount that was embezzled was $152,375,242.18. • On one day, all of the money was wired to one or more banks in Liechtenstein, but they don’t know which banks were involved. • There are sixteen banks in Liechtenstein. • For the day the money was transferred, the lawyers have a list of every deposit made to every account for every bank in Liechtenstein (but no information about where the deposits came from). • The deposits at eleven of the sixteen banks can be ignored for various legal reasons. In order to confront Tony with proof of the embezzlement, Harvey, Mike, and Louis determine that they need to identify a collection of deposits to the five banks that add up to exactly$152,375,242.18. Louis remarks that this is “simple mathematics”. It certainly is a simple problem to state, but it is not clear if it is a simple problem to solve.
The problem they needed to solve is called subset-sum: given a set of integers $$S$$ and a target $$k$$, is there a subset of $$S$$ whose elements add up exactly to $$k$$? (Since we are dealing with money, multiply everything by 100 so that we have integers.) Stated this way, subset-sum is a yes-no decision problem, but asking for the elements in the subset is an equivalent problem [2].
Subset-sum is related to other decision problems such as the traveling salesman problem and the knapsack problem, and games like FreeCell. These problems are known as NP-complete problems which arise in theoretical computer science. “Brute force” solutions of these problems involve generating a large, finite collection of all possible solutions, and then checking each element of the collection until an actual solution is found. Checking a potential solution is quick (what computer scientists refer to as “polynomial time”) but generating the complete collection of solutions can take a long time. For example, in the subset-sum problem where there are 25 elements in $$S$$, then $$S$$ has $$2^{25}$$, or over 33 million, subsets. It would take a long time to write down or generate all of these subsets, but not very long to determine if the elements of a specific subset summed to a target $$k$$.
Generally, the time needed to solve an NP-complete problem grows based on the size of the inputs into the problem, due to the time needed to generate all possible solutions. In the case of subset-sum, if $$S$$ has $$n$$ elements, then there are $$2^n$$ subsets to generate. Now a solver might get lucky and generates the subset that adds up to $$k$$ early in the solution process. But theoretically, it may be necessary to inspect all $$2^n$$ subsets before finding the subset that adds up to the target or to determine that the subset does not exist.
All NP-complete problems can be solved using this “brute force” approach. At the same time, solution methods have been developed for NP-complete problems that are fast most of the time (which are appreciated by industries such as airlines and telecommunications), but they are not perfect methods. The algorithms might be designed to solve the problem quickly for a typical instance of the problem, but these methods are vulnerable to rare instances of requiring the inspection of all possible solutions. It also the case that there are fast methods that find a nearly- optimal solution rather than an optimal one, which is often sufficient in real applications. (With subset-sum, an example would be to find a subset that sums up to within 1% of the target $$k$$.)
Another fascinating fact about NP-complete problems is that a solution method to one of the problems can, in theory, be modified to solve any other NP-complete problem. This theoretical result comes from reducing any NP-complete problem to a Boolean logic problem known as satisfiability.
Returning to Harvey, Mike, and Louis, although no details are provided on the show, they solve their subset-sum problem, finding a set of seven deposits from the five banks that add up to exactly \$152,375,242.18. The lawyers didn’t have a lot of time to solve the problem. In the story, when they realize the “simple” mathematics problem that they needed to solve, it was nighttime on what was a warm, sunny day in New York City. On the following day, clean, showered, and refreshed, they have the details about the seven deposits ready for a meeting with Tony and his lawyer. After confronted with the deposits, Tony returns all of the money to the house-building charity, and once again, mathematics saved the day!
So, were they lucky? Did they happen upon a subset of the set of deposits in a short period of time? Did they have a solution method handy that solved this particular instance quickly? Perhaps near-genius Mike had an insight after reading all the deposits into his photographic memory. In Part II, we will explore the likelihood that they could solve the problem given the time they had.
References for Part I:
[1] http://en.wikipedia.org/wiki/Suits_%28TV_series%29
[2] Cormen, T. et al, Introduction to Algorithms, 2nd edition, the MIT Press, 2001.
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# How do you solve the linear equation $7x-\left( x-3 \right)=3\left( x+10 \right)$ ?
Last updated date: 04th Aug 2024
Total views: 387.9k
Views today: 7.87k
Answer
Verified
387.9k+ views
Hint: From the question given, we have been asked to solve $7x-\left( x-3 \right)=3\left( x+10 \right)$ for getting the value of $x$ . For that we need to use basic arithmetic simplifications like addition and transformations like shifting from right hand side to left hand side and reduce it. We can get the solution of the equation.
Complete step-by-step solution:
Now considering from the question we have been asked to solve the given equation $7x-\left( x-3 \right)=3\left( x+10 \right)$ for getting the value of $x$.
For doing that we need to use basic arithmetic simplifications like addition and transformations like shifting from right hand side to left hand side and reduce it.
Firstly we will simplify the given equation as follows
\begin{align} & 7x-\left( x-3 \right)=3\left( x+10 \right) \\ & \Rightarrow 7x-x+3=3x+30 \\ & \Rightarrow 6x+3=3x+30 \\ \end{align}
Here we will shift all the terms except the term involving $x$ in it to one side and keep that respective term in one side. For that here we will shift $3x$ from right hand to left hand side. After doing that we will have
\begin{align} & \Rightarrow 6x-3x+3=30 \\ & \Rightarrow 6x-3x=30-3 \\ \end{align} .
Now as we need the value of $x$ we will perform the basic simple arithmetic subtraction on the both sides after that we will have $\Rightarrow 3x=27$ .
Now we will divide with $3$on both sides after that we will have
\begin{align} & \Rightarrow \dfrac{3x}{3}=\dfrac{27}{3} \\ & \Rightarrow x=9 \\ \end{align}
Therefore we can conclude that the value of $x$ obtained from the given equation $7x-\left( x-3 \right)=3\left( x+10 \right)$ is $9$.
Note: We should be very careful while doing the calculation in this problem. Also, we should be very careful while solving the equation. Also, we should be very careful while transforming the equation. Also, we should do exact transformations to the given equation to obtain the perfect answer. This type of questions are very easy and do not require much calculations so the possibility of mistakes in questions of this type are very less. Similarly we can solve any other equations like for the equation $5x-4=3x-6$ the value of $y$ will be given as $\Rightarrow 2x=-2\Rightarrow x=-1$ . | 681 | 2,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-33 | latest | en | 0.867697 |
https://communities.sas.com/t5/SAS-Statistical-Procedures/How-to-output-or-calculate-Studentized-Residuals-in-Proc-NLMIXED/td-p/165507?nobounce | 1,516,418,921,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00327.warc.gz | 660,680,127 | 33,873 | Solved
Contributor
Posts: 45
# How to output or calculate Studentized Residuals in Proc NLMIXED?
I am struggling to output or calculate Studentized Residuals in Proc NLMIXED.
Thank you,
Marcio
Accepted Solutions
Solution
07-03-2014 03:03 PM
Contributor
Posts: 45
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
Thank you Steve, I ended up get this working and here is how.
As always thanks for your input,
Marcio
predict L + U*(z1) out=ppp;
run;
data ppp; set ppp;
resid=y-pred;
resid_t=resid; run;
proc standard data=ppp std =1 out =pred_std;
var resid_t;
run;
title;
proc print data=ppp; run;
proc print data=pred_std;
run;
symbol1 value=dot color=blue i=none;
proc gplot data=pred_std;
plot resid_t*x resid_t*pred/vref=(0);
run;
All Replies
Posts: 2,655
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
I don't know, and the person who I think does is seldom found around here anymore. However, take a look at Dale McLerran's post:
https://communities.sas.com/message/55521#55521
where he calculates individual deviances for a Poisson. I believe you could take the square root of the individual deviance and use it as a denominator in calculating the studentized residual. to me, the key statement is the PREDICT statement with the OUT= option.
Steve Denham
Solution
07-03-2014 03:03 PM
Contributor
Posts: 45
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
Thank you Steve, I ended up get this working and here is how.
As always thanks for your input,
Marcio
predict L + U*(z1) out=ppp;
run;
data ppp; set ppp;
resid=y-pred;
resid_t=resid; run;
proc standard data=ppp std =1 out =pred_std;
var resid_t;
run;
title;
proc print data=ppp; run;
proc print data=pred_std;
run;
symbol1 value=dot color=blue i=none;
proc gplot data=pred_std;
plot resid_t*x resid_t*pred/vref=(0);
run;
Posts: 2,655
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
You should give yourself a "Correct answer". The use of STANDARD is something I overlook way too often.
Steve Denham
Valued Guide
Posts: 684
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
proc standard would not be giving you the standardized or studentized residual. You need to obtain the estimated standard error for EACH residual, which is not directly given by NLMIXED.The standard procedure is just scaling by the the variation in the raw residuals (not the same thing). I think getting standardized residuals from NLMIXED will require some coding (post-model-fitting) which would depend on the chosen conditional distribution. It would require the "hat" matrix.
Contributor
Posts: 45
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
lvm, thanks. This is what I have used.
data ppp; set ppp;
resid=y-pred;
resid_t=resid; run;
proc standard data=ppp std =1 out =pred_std;
var resid_t;
run;
Valued Guide
Posts: 684
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
proc standard will not give you a standardized or studentized residual. Each residual has a unique variance, var(r_i), and you cannot recover this from proc standard (even though I saw a website that says that you can).
Posts: 2,655
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
I think I saw the same thing, hence my excitementat using PROC STANDARD. I think an IML solution may be the only way out now.
Steve Denham
Contributor
Posts: 45
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
Hi Steve, do you have any directions on how I could get that?
Posts: 2,655
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
Search 's blog: The DO Loop. You are going to need the entire X'VX matrix from NLMIXED, and the Y vector, to calculate the residual, and the variance for that residual. The last step is the only easy one for me--residual/sqrt(var(r_i)). I need to learn how to do more things "long hand".
Steve Denham
Contributor
Posts: 45
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
Hi Steve, thanks!!! I was not able to find nothing related to NLMIXED or 'studentized' in his blog, any idea what I should look at?
Thanks,
Marcio
Posts: 2,655
## Re: How to output or calculate Studentized Residuals in Proc NLMIXED?
It wouldn't be specific, but more as background on how to use IML to calculate what you need from the various matrixes, following 's note regarding the individual variances of the residual.
Now it seems to me that a residual is a sum of two non-independent random variables, and the estimate and the Y value are correlated, so the key is getting the covariance between the two to estimate the variance of the residual. Surely someplace on the web has a matrix representation. Then it is just a matter of of following the right examples in the documentation and Rick's blog.
Steve Denham
🔒 This topic is solved and locked. | 1,335 | 5,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-05 | latest | en | 0.738387 |
http://homeenergypros.org/forum/topics/calculating-attic-ventilation?xg_source=activity&id=6069565%3ATopic%3A139566&page=3 | 1,508,474,142,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823630.63/warc/CC-MAIN-20171020025810-20171020045810-00396.warc.gz | 142,801,902 | 21,000 | # Calculating attic ventilation
All too often an existing home has far too little attic ventilation and some difficult choices for adding more. We all know the existing, although somewhat old, method of one ft² of net free vent area (NFA) for every 300 ft² of attic floor space and double that if the ceiling plane has not been well air sealed. Whatever that number comes out to be should be roughly divided half high and half low. 60/40 and 40/60 cold country vs ac country are options. But when you are short on vent area and cash, here is a detail not often mentioned.
The NFA is only part of the ventilation equation. The other half is the pressure across those vents. A 10 ft attic will have twice the attic stack pressure of a 5 ft attic. Now, I would have to do a bunch of reading to find out what height attic those old calculations were based upon, but a simple fudge factor would be, short attics need more and tall attics may get by with less.
Feel free to post any references that show how to adjust for attic height, fudge is good but real numbers taste better.
Bud
Tags: attic, ventilation
Views: 3111
### Replies to This Discussion
<We use it because we are stuck with it!> I agree that's what we are doing, but that doesn't make it right.
<No one likes change!> As you know, I've already encountered that! But it would be hypocritical of us to complain about other contractors following old habits while we have in front of us a standard from the same "hand me down" origin.
<No one really understands why attic ventilation needs to happen.> Although home owners may not understand, we (all auditors and related contractors) should.
<How to make it work correctly … with incorrect work... > That is often the challenge with a retrofit and add to that, the solution will often have a very limited budget. Thus it is good to know what is needed in each situation and whether there exists any wiggle room.
<It has gotten entrenched into code and we are stuck.> Stuck is not stopped, it just means we are going to have to dig a little to get out, and get a little muddy in the process.
There are few problems that a bunch of money can't fix, but very few owners will pop for those expensive solutions. If a house needs more ventilation, the owners deserve to know if the estimated solution is a wag or one based upon their home and their location. The 1:300 is a wag and we can do better. I was hoping someone had.
Bud
Bud, I know you believe every house is a unique structure and worth individual attention.
If you have a 900 sf ranch with a full basement, or 1800 sf conditioned space. ca 1965. No vaults etc. The 1:300 is 3 sf of attic ventilation. That would be NFA. Ridge Vent is 11 NFA inches per foot. Turtles are rated at 50 NFA for a 10 inch diameter (I don't buy that one, but I'm not rating them.) A 6 x 16 soffit rates out at 64 NFA.
3 sf = 432 sq in NFA. Does not take much.
The 1:300 was about 2x of the 1:576, which Rowley said was OK. It is half the 1:186 that he thought was too much. He was also talking about total ventilation - Gable to Gable. No high / low.
Food for thought.
John, to be honest, I don't think we can consider any of the Rowely work as relevant, 1939, cross ventilation only, and all done in a lab. One could say we have had 70 years of field work to add credibility to his conclusions, however, there isn't a one size fits all solution.
If we look at just the cross ventilation limitations, as compared to what we experience with a good ridge and soffit combination, there is no way equal vent areas will produce the same effective air flow. That alone should allow one to install substantially less NFA for a ridge/soffit install.
Then, 1939 is even older than I am. A lot has changed since then and our understanding of attic issues has improved, I hope.
Your example is for 1:300, which assumes an air sealed ceiling, but that is not always the case and certainly wasn't way back then. So air sealed or not is a factor of two, height can be another factor of two and other variables can probably push that 3 ft² to anywhere between 10 ft² to zero. I'm exaggerating, but there is a cost that goes with ventilation and we need a better way to pick a number.
Bud
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9 hours ago | 1,423 | 5,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-43 | latest | en | 0.957436 |
https://calculator-online.org/equation/e/7x_to_pow_two_pls_6x_minus_nine_equal_minus_x_to_pow_two_pls_fourteen_minus_three | 1,721,598,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00546.warc.gz | 130,101,394 | 15,566 | Mister Exam
7x^2+6x-9=-x^2+14-3 equation
The teacher will be very surprised to see your correct solution 😉
v
Numerical solution:
Do search numerical solution at [, ]
The solution
You have entered [src]
2 2
7*x + 6*x - 9 = - x + 14 - 3
$$\left(7 x^{2} + 6 x\right) - 9 = \left(14 - x^{2}\right) - 3$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(7 x^{2} + 6 x\right) - 9 = \left(14 - x^{2}\right) - 3$$
to
$$\left(\left(x^{2} - 14\right) + 3\right) + \left(\left(7 x^{2} + 6 x\right) - 9\right) = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 8$$
$$b = 6$$
$$c = -20$$
, then
D = b^2 - 4 * a * c =
(6)^2 - 4 * (8) * (-20) = 676
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{5}{4}$$
$$x_{2} = -2$$
Vieta's Theorem
rewrite the equation
$$\left(7 x^{2} + 6 x\right) - 9 = \left(14 - x^{2}\right) - 3$$
of
$$a x^{2} + b x + c = 0$$
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{3 x}{4} - \frac{5}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{3}{4}$$
$$q = \frac{c}{a}$$
$$q = - \frac{5}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{3}{4}$$
$$x_{1} x_{2} = - \frac{5}{2}$$
The graph
Rapid solution [src]
x1 = -2
$$x_{1} = -2$$
x2 = 5/4
$$x_{2} = \frac{5}{4}$$
x2 = 5/4
Sum and product of roots [src]
sum
-2 + 5/4
$$-2 + \frac{5}{4}$$
=
-3/4
$$- \frac{3}{4}$$
product
-2*5
----
4
$$- \frac{5}{2}$$
=
-5/2
$$- \frac{5}{2}$$
-5/2
x1 = -2.0
x2 = 1.25
x2 = 1.25 | 887 | 1,865 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-30 | latest | en | 0.580084 |
http://www.numbersaplenty.com/190507422848 | 1,601,167,211,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00719.warc.gz | 202,652,944 | 4,363 | Search a number
190507422848 = 2717322232
BaseRepresentation
bin1011000101101100100…
…0001001010010000000
3200012201210022210122212
42301123020021102000
511110124400012343
6223303511121252
716522644242651
oct2613310112200
9605653283585
10190507422848
1173880523516
1230b08673228
1314c707b5612
1493134c7528
154e4ede8c18
hex2c5b209480
190507422848 has 72 divisors (see below), whose sum is σ = 383452465545. Its totient is φ = 94278434304.
The previous prime is 190507422839. The next prime is 190507422851. The reversal of 190507422848 is 848224705091.
Multipling 190507422848 by its sum of digits (50), we get a square (9525371142400 = 30863202).
It is a powerful number, because all its prime factors have an exponent greater than 1 and also an Achilles number because it is not a perfect power.
It can be written as a sum of positive squares in 2 ways, for example, as 149868088384 + 40639334464 = 387128^2 + 201592^2 .
It is an ABA number since it can be written as A⋅BA, here for A=2, B=308632.
It is a super-2 number, since 2×1905074228482 (a number of 23 digits) contains 22 as substring.
It is a Smith number, since the sum of its digits (50) coincides with the sum of the digits of its prime factors.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 190507422796 and 190507422805.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 8 ways as a sum of consecutive naturals, for example, 854293265 + ... + 854293487.
Almost surely, 2190507422848 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 190507422848
190507422848 is an abundant number, since it is smaller than the sum of its proper divisors (192945042697).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
190507422848 is an frugal number, since it uses more digits than its factorization.
190507422848 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 806 (or 398 counting only the distinct ones).
The product of its (nonzero) digits is 1290240, while the sum is 50.
The spelling of 190507422848 in words is "one hundred ninety billion, five hundred seven million, four hundred twenty-two thousand, eight hundred forty-eight". | 719 | 2,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-40 | latest | en | 0.854794 |
https://math.stackexchange.com/questions/3458497/how-much-bigger-is-the-power-set-explicitly | 1,721,219,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00699.warc.gz | 328,446,156 | 37,001 | # How much bigger is the power set -- explicitly
Recently, I gave a talk to highschool pupils about cardinality and explained them that, for any set $$X$$, there is no surjective map $$f:X\to \mathcal P(X)$$ because $$R=\{x\in X: x\notin f(x)\}$$ is not in the range of $$f$$.
Apparently, the pupils liked the interpretation that every member of $$X$$ organizes a party $$f(x)\subseteq X$$ (the invited people). Some people spoil every party they attend and, in order to keep their chances for the best party award, they do not invite themselves. Then, $$R$$ is the consolation party where exactly those poor spoilers are invited.
The question: Can one explicitly write down other parties which (given the map $$f$$) are certainly not organized by some $$x\in X$$?
• I think the question is fairly clear, but just in case someone doesn't: the question can be formalized as whether there is a first-order formula $\varphi(x,y)$ in the language of set theory such that for any set $X$ with at least two elements and any $f\!:X\to\mathcal P(X)$, and setting $A=\{x\in X: \varphi(x,f)\}$, the following two conditions hold: 1) $A\notin\mathrm{ran}(f)$, and 2) $A\ne\{x\in X: x\ne f(x)\}$. Commented Dec 1, 2019 at 19:35
I think Zwicker's survival game construction comes close to what you are looking for. It will not always yield a party different from the "consolation party", but it often will and is based on a rather different idea.
Once you have your $$X$$ and your $$f$$, for each $$x_0\in X$$ define a "survival game" as follows : you start at position $$x_0$$, and then you have to jump to a position $$x_1\in f(x_0)$$ (if $$f(x_0)$$ is empty you have lost from the start), then to a position in $$f(x_1)$$, etc. If you find a way to go on jumping forever (aka a cycle in mathematical parlance), you win ; otherwise you lose.
Clearly, there are winning and losing initial positions. Let $$A$$ be the set of losing initial positions. If $$A=f(a)$$ for some $$a\in X$$, then $$a$$ must be a losing position since all the positions immediately accessible from it are, so $$a\in A=f(a)$$, but then we can win by jumping and staying on $$a$$ forever, which is a contradiction.
Update 05/12/2019: here is a construction of a $$R'$$ that will always be different from your $$R$$.
As explained in Andres Caicedo's comment to the OP, we assume that $$X$$ has at least two elements ; call them $$x_1$$ and $$x_2$$. Let $$Y=X \setminus \lbrace x_1,x_2 \rbrace$$ ($$Y$$ may be empty).
Let $$R_Y=\lbrace y \in Y | y \not\in f(y) \rbrace$$. Then the four sets $$S_1=R_Y$$, $$S_2=R_Y \cup \lbrace x_1 \rbrace$$, $$S_3=R_Y\cup \lbrace x_2 \rbrace$$, and $$S_4=R_Y\cup \lbrace x_1,x_2 \rbrace$$ are all distinct. It follows that they cannot all lie in $$Z=\lbrace f(x_1),f(x_2),R \rbrace$$ since this latter set has at most three elements. So there one of the $$S_k$$'s (call it $$R'$$) which is not in $$Z$$. This $$R'$$ satisfies the requirements by construction.
• The solution given in the upadate is formally correct but somehow disappointing. Your first solution is just great. Commented Dec 6, 2019 at 8:27
• @Jochen I completely agree, the second solution is not very interesting, but I posted it all the same because it's not obvious, and therefore informative (however disappointing it may be). The brilliantly simple idea in the first is Zwicker's not mine. Commented Dec 6, 2019 at 10:43 | 965 | 3,398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-30 | latest | en | 0.937219 |
https://www.coursehero.com/file/ps1akr/The-principle-of-Corresponding-States-PCS-as-stated-by-van-der-Waals-Substances/ | 1,548,114,232,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00055.warc.gz | 763,526,381 | 388,421 | # The principle of corresponding states pcs as stated
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The principle of Corresponding States (PCS) as stated by van der Waals : “Substances behave alike at the same reduced states. Substances at same reduced states are at corresponding states.” That is, “Substances at corresponding states behave alike.”
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9/22/2010 20 Generalized Compressibility Chart ME200 Therm I Lecture 13, Prof. Mongia
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 21 p R 1.0
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9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 22 p R 10.0
Generalized Compressibility Chart Sonntag, Borgnakke, and Van Wylen, Fundamentals of Thermodynamics, 5 th Edition, page 763. 23 Z c =[Pv/RT] c Table A-1 or A-1E Z c = 0.22-0.304
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Generalized Compressibility Chart (cont.) Calculate two reduced properties: p R , T R and v R p R = p/p c T R = T/T c v R = v/ ( RT c /p c ) (pseudoreduced specific volume) Determine Z If p R << 1, then ideal gas at all temperatures If T R >> 2, then ideal gas at all pressures (except when p R >> 1) Near critical point and for saturated vapor, real gas behavior 9/22/2010 24 ME200 Therm I Lecture 13, Prof. Mongia
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 25 3.96: Butane (C 4 H 10 ) in a piston-cylinder assembly undergoes an isothermal compression at 173 o C from p 1 = 1.9 MPa to p 2 = 2.5 MPa. Determine the work , in kJ/kg. From Table A-1 (p. 816): T c = 425 K, p c = 38 bar M = 58.12, R=0.143kJ/kg.K
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9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 26 T R = 446 K/425 K = 1.05 p R1 = 19 bar/38 bar = 0.5 p R2 = 25 bar/38 bar = 0.66 From Fig. A-1, p. 911, Z 1 = 0.835, Z 2 = 0.775 Z av =0.805
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 27
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9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 28
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http://mathhelpforum.com/calculus/110255-trigonometry.html | 1,524,493,224,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00559.warc.gz | 195,099,520 | 9,868 | 1. ## trigonometry
hi!
Can someone teach me how to solce arccos(cos(7Pi/ 6))?
i tried to change 7pi/ 6 into the principal angle but what i thought of was:
cos (7pi/6)= -cos (pi/6)
although it is in the principal angle, i do not know how to continue after that.
thanks!
2. Originally Posted by alexandrabel90
hi!
Can someone teach me how to solce arccos(cos(7Pi/ 6))?
i tried to change 7pi/ 6 into the principal angle but what i thought of was:
cos (7pi/6)= -cos (pi/6)
although it is in the principal angle, i do not know how to continue after that.
thanks!
HI
it would be $\displaystyle \frac{5\pi}{6}$
cos (7pi/6)= -cos (pi/6)=cos (5pi/6)
Lets say
$\displaystyle \cos^{-1} \cos (\frac{5\pi}{6})=x$
$\displaystyle \cos (\frac{5\pi}{6})=\cos x$
So now , cancel the 'cos' , thus $\displaystyle x=\frac{5\pi}{6}$
Note that its different when its $\displaystyle \cos \cos^{-1}$ something
3. but i thought that way that you did only applies to when the angle is in the principal range of 0 to 180 degrees?
so i was trying to convert 7pi/6 into the principal angle of cos x where x is in the principal range so that i can use the method that you have just used..
4. Originally Posted by alexandrabel90
but i thought that way that you did only applies to when the angle is in the principal range of 0 to 180 degrees?
so i was trying to convert 7pi/6 into the principal angle of cos x where x is in the principal range so that i can use the method that you have just used..
yeah you are right , i just edited . | 447 | 1,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-17 | latest | en | 0.921916 |
http://www.topperlearning.com/forums/ask-experts-19/calculate-the-number-of-electrons-that-will-constitute-a-flo-chemistry-acids-bases-and-salts-50117/reply | 1,493,326,777,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122621.35/warc/CC-MAIN-20170423031202-00489-ip-10-145-167-34.ec2.internal.warc.gz | 706,813,740 | 37,099 | Question
Sun October 02, 2011
# calculate the number of electrons that will constitute a flow of charge of 4c
Mon October 03, 2011
Since we know that Q = ne
Q=charge,
n=number of electrons,
e=charge of 1 electron
n = Q/e
n = 4/ (1.6×10-19) = 2.5×1019
Related Questions
Thu December 15, 2016
# Testing by tech team. Please ignore.
Sun November 13, 2016 | 124 | 364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-17 | latest | en | 0.934242 |
https://www.jiskha.com/display.cgi?id=1295230131 | 1,516,631,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891377.59/warc/CC-MAIN-20180122133636-20180122153636-00244.warc.gz | 909,774,259 | 3,412 | # college math 209
posted by .
solve the equation
(x+4)^2=2x^+19x+46
• college math 209 -
(x+4)^2 = 2x^2 + 19x + 46
(x + 4)(x + 4) = 2x^2 + 19x + 46
x^2 + 8x + 16 = 2x^2 + 19x + 46
collect like terms, then solve
you try and post back if you need help
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More Similar Questions | 440 | 1,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-05 | latest | en | 0.810942 |
https://math.stackexchange.com/questions/753354/prove-that-there-are-two-frogs-in-one-square | 1,718,843,165,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00579.warc.gz | 335,978,691 | 39,119 | # Prove that there are two frogs in one square.
A certain chessboard is infinite in size. There is a frog sitting in the center of every square. After a certain time, all the frogs jump such that
1. They may jump to any possible square in the infinite chessboard
2. They may jump and land at the same square again
Prove that its possible for all the frogs to jump simultaneously such that there are exactly two frogs per square after the jump.
Let there be n frogs sitting on n squares where $n\rightarrow \infty$
Lets choose our king frog who's sitting at a particular square, now
CASE I: King frog wants to stay at his original square, then its probability is $\frac { 1 }{ n }$. Then we are left with n-1 frogs who have n squares left. I am stuck at this point, I know the probability of remaining frogs to choose king's square is $\frac { n-1 }{ n }$ but how to proceed after that?
CASE II: King frog doesn't want to stay at his original square, then he has $n-1$ ways to go, also the probability of remaining of the n-1 frogs to land at king's square is $\frac { n-1 }{ n }$. So we have
$$\lim _{ n\rightarrow \infty }{ \frac { n(n-1) }{ n } }$$
• I'm fairly sure this basic problem has been raised before. What I don't understand is what your question is, and whether it might be different from those previously raised. It looks as though you should be looking for a systematic process rather than a random one. Commented Apr 14, 2014 at 14:18
• Choose a definite orientation of the board so as to have all the frogs "facing" to the same direction (say, "the front"). Now, what about having each frog on a white square jumping onto its same square, whereas each frog on a black square jumping precisely to the black square in front of it? Commented Apr 14, 2014 at 14:23
• I thought a chessboard has just two dimensions? What would be a chessboard that is infinite in dimensions? Or do you just mean infinite in size? Fun aside, in an infinite chess board the ranks wouldn't be numbered 1 to 8, but rank numbers would be all integers. If a frog is on rank 2k it jumps to rank k with the same file; if a frog is on rank 2k+1 it also jumps to rank k with the same file. Two frogs on every square. Commented Apr 14, 2014 at 15:21
• It seems to me that the introduction of probability makes this a more interesting problem than posters seem to think, if we interpret "it is possible" as "there is a positive probability". Of course we could have the situation where ranks $2k+1$ and $2k$ all jump to rank $k$, but the probability of this particular outcome happening by chance is 0. (However, I'm inclined to suspect that the probability of ending up with 2 frogs on every square is also 0.) Commented Apr 14, 2014 at 16:17
• Actually, now that I read it again, if we want to intepret it probabilistically, we need to specify the probability distribution with which frogs choose a new square. There's no uniform distribution on an inifinite lattice. Commented Apr 14, 2014 at 16:19
Suppose you describe the squares of the board $B$ using integer coordinates: $$B = \{(x,y) \mid x,y \in \mathbb{N}\}.$$ Use the following map, which I will describe in steps: first map $$(x,0),(x,1) \mapsto (x,0) \quad \forall x \geq 0;$$ this takes care of the first row, each square of which now has exactly $2$ frogs occupying it. Next map $$(x,2),(x,3) \mapsto (x,1) \quad \forall x \geq 0;$$ this takes care of the second row, each square of which now has exactly $2$ frogs occupying it. In general, the map is $$(x,2i),(x,2i+1) \mapsto (x,i) \quad \forall x \geq 0, i \geq 0.$$ I think you can see this will have exactly the property you want!
Think about doing it for one layer of the chess board. It may be helpful to split the layer into two parts. I.e consider the split line being the natural numbers. Tell 0 to jump and 1 to move 1 left. Tell 2 to move 1 left and 3 to move 2 left.... tell $2n$ to move n left and $2n+1$to move $n+1$ left.
Find a similar algorithm for $\{-1,-2,-3,\ldots\}$. And then just order that this rule should be followed on all levels of the chess board. | 1,096 | 4,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.95993 |
http://forums.devshed.com/python-programming-11/generating-integers-functions-949240.html | 1,539,872,272,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00434.warc.gz | 137,445,523 | 9,670 | ### Thread: Generating integers with functions
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#### Generating integers with functions
I want to define a function A that creates a list of N random integers where each integer is a random number between 1 and 10 and N is given to the function as an argument. Then define another function to output A. Define a function that takes A as argument and returns the average of all elements, as well as the maximum deviation
Why doesn't this work so far?
Code:
```import random
def N():
N = int(raw_input("Enter a value for N:"))
return N
N()
def A(N):
list = []
for i in range(N):
list.append(random.randrange(1,11))
return list
print list```
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Variables defined within functions are local to those functions. You need to use the return values. Also, list is a builtin. Don't use it as a variable name.
python Code:
```import random
def how_many():
N = int(raw_input("Enter a value for N:"))
return N
def make_range(N):
mylist = []
for i in range(N):
mylist.append(random.randrange(1,11))
return mylist
def main():
N = how_many()
print(make_range(N))
main()```
-Mek
• lunapt agrees
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thank you!! is it possible to output mylist if i wanted to use it as an argument for main()
something similar to this?
Code:
```def main(mylist):
N=how_many()
print (make_range(N))
main(make_range(N))``` | 429 | 1,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-43 | latest | en | 0.730623 |
http://what-when-how.com/Tutorial/topic-656lo9ddo/Multimedia-Introduction-to-Programming-Using-Java-249.html | 1,500,979,774,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425144.89/warc/CC-MAIN-20170725102459-20170725122459-00230.warc.gz | 353,345,507 | 4,017 | Java Reference
In-Depth Information
When the loop condition evaluates to false , i=5 . So, when the loop terminates,
the invariant says:
the squares of 2..5-1 have been printed
which means:
the squares of 2..4 have been printed
which is exactly the final assertion!
3. How does it make progress? We have to be sure that the loop terminates,
which means that at some point the loop condition will become false . In this
case, i starts out at 2 and ends at 5 . The statement i= i + 1 ; in the repetend
ensures that each iteration makes progress toward the loop condition becoming
false .
4. How does it fix the invariant? The invariant must be true each time the loop
condition is evaluated. So, we have to ensure that each execution of the repetend
fixes the variables to keep it true . In this case, before the repetend is executed,
we know the following from the loop condition and the loop invariant:
i!=5 and the squares of 2..i-1 have been printed
Therefore, the next value to print is i*i . The repetend prints this value and then
increases i , after which the invariant again true .
The above analysis talked about a particular loop. We now discuss the same
questions with regard to a general loop of this form:
initialization
// invariant: explains what is true whenever the condition is evaluated
while ( condition ) {
repetend
}
// postcondition: an assertion of what is supposed to be true at the end
Again, here are the questions we ask when we are understanding a given
loop or writing our own loop:
1. How does it start? In other words, what initialization of variables will make
the invariant true initially? Here are two general strategies for this: try to make a
range empty and set some variable to zero.
2. When does it stop? When it stops, the invariant is true and the loop condition
is false. From these two facts, we should be able to conclude that the postcondi-
tion is true.
3. How does it make progress? What statement in the repetend ensures that
after a number of iterations the loop condition will eventually become false?
4. How does it fix the loop invariant? We have to make sure that the following
precondition-statement-postcondition triple holds:
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IB 2011 PAPER 2 At the IB exam, you must answer all questions from section A and one from section B. You will have 75 minutes. 1
# IB 2011 paper2
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IB 2011 PAPER 2At the IB exam, you must answer all questions from section A and one from section B.You will have 75 minutes.
1
(a) State one piece of evidence that shows that “D” is not proportional to “n” (b) On the graph above, draw the line of best-fit for the data points. (c) Theory suggest that D2 = k x n. A graph of “D2”againts “n” is shown below. Error bars are shown for the first and last data points only. (i) Using the graph on page 1, calculate the percentage uncertainties in D2, of the ring n=7. (ii) Base on the graph below, state one piece of evidence that supports the relationship D2 = k x n. (iii) Use the graph below to determine the value of the constant “k” as well as the uncertainty. (iv) State the unit for the constant “k”
A2. This question is about motion in a magnetic field.
An electron that has been accelerated from rest by a potential difference of 250V, enters a region of magnetic field of strength 0.12 T that is directed into the plane of the page.
(a) The electron’s path while in the region of magnetic field is a quarter circle. Show that the (i) speed of the electron after acceleration is 9.4 x 106 m/sec (ii) radius of the circle is 4.5 x 10-4 m
2
(b) The diagram 0n the right shows the momentum of the electron as it enters and leaves the region of magnetic field. The magnitude of the initial and final momentum is 8.6 x 10-24 N sec. (i) On the diagram, draw an arrow to indicate the vector representing the change in momentum of the electron. (ii) Show that the magnitude of the change in the momentum of the electron is 1.2 x 10-23 N sec. (iii) The time the electron spend in the region of magnetic field is 7.4 x 10-11 s. Estimate the magnitude of the average force on the electron.
A3. A ball of mass 0.25 kg is attached to a string and is made to rotate with constant speed “v” along a horizontal circle of radius r = 0.33 m. The string is attached to the ceiling and makes an angle of 300 with the vertical.
(a) (i) On the diagram above, draw and label arrows to represent the forces on the ball in the position shown. (ii) State and explain whether the ball is in equilibrium.
(b) Determine the speed of rotation of the ball.
SECTION B
This section consist of three questions: B1, B2 and B3. Answer one question.
B1 . This question is in two parts. Part 1 is about a nuclear reactor. Part 2 is about a simple harmonic oscillations.
Part 1. Nuclear reactor
(a) Uranium is used as a fuel in a small research nuclear reactor. The concentration of Uranium U in a sample of uranium must be increased before it can be used as a fuel.
State the name of the process by which the concentration of U is increased.
3
P initial
P final
235 92
235 92
(b) The reactor produces 24 MW of power/ The efficiency of the reactor is 32%. In the fission of one uranium 235 nucleus 3.2 x 10-11 J of energy is released. Determine the mass of uranium 235 that undergoes fission in one year in this reactor.
(c) Explain what would happen if the moderator of this reactor were to be removed.
(d) During its normal operation, the following set of reactions take place in the reactor.
(i) State the name of the process represented by the reaction (II) (ii) Comment on the international implications of the product of these reactions. B1 - Part 2 Simple harmonic oscillations
A longitudinal wave travels through a medium from left to right. Graph 1 shows the variation with time “t” of the displacement “x” of a particle “P” in the medium.
(a) For particle “P”, (i) state how graph 1 shows that its oscillation are not damped. (ii) calculate the magnitude of the maximum acceleration (iii) calculate itrs speed at t = 0.12 (iv) state its direction of motion at t= 0.12 s
(b) Graph 2 shows the variation with position “d” of the displacement “x” of a particle in the medium at a particular instant of time.
4
Determine for the longitudinal wave, using graph 1 and graph 2, (i) the frequency (ii) the speed
(c) The diagram shows the equilibrium positions of six particles in the medium (i) On the diagram above, draw crosses to indicate the position of these six particles at the instant of time when the displacement is given by graph 2. (ii) On the diagram above, label with the letter “C” a particle that is at the center of a compression.
B2: Thos question is in two parts. Part1 is about mechanics and thermal physics. Part 2 is about nuclear physics.
Part 1 Mechanics and thermal physics
The graph shows the variation with time “t” of the speed “v” of a ball of mass 0.50 kg. that has been released from rest above the Earth’s surface.
5
The force of air resistance is not negligible. Assume that the acceleration of the free fall is g = 9.81 m/sec2.
(a) State, without any calculations, how the graph could be used to determine the distance fallen.
(b) (i) Draw and label arrows to represent the forces on the ball at t=2 sec. (ii) Use the graph to show that the acceleration of the ball at 2.0 s is approximately 4 m/sec2
(iii) Calculate the magnitude of the force of air resistance on the ball at 2.0 s. (iv) State and explain whether the air resistance on the ball at t=5.0 s is smaller than, equal to or grater than the air resistance at t=2.0 s
(c) After 10 sec the ball has fallen 190m (i) Show hat the sum of the potential and kinetic energies of the ball has decreased by 780J (ii) The specific heat capacity of the ball is 480 J / (kg K). Estimate the increase in the temperature of the ball. (iii) State the assumption made in the estimate in (C)(ii)
Part 2. Nuclear physics
(a) (i) Define binding energy of a nucleus. (ii) The mass of a nucleus of plutonium Pu is 238.990396 u. Deduce that the binding energy per nucleon for plutonium is 7.6 MeV.
(b) The graph shows the variation with nucleon number “A” of the binding energy per nucleon.
Plutonium ( Pu) undergoes nuclear fission according to the reaction given below
(b) (i) Calculate the number “x” of neutrons produced. (ii) Use the graph to estimate the energy released in the reaction.
(C) Stable nuclei with a mass number greater that about 20, contains more neutrons than protons. By reference to the properties of the nuclear force and of the electrostatic force, suggest an explanation for this observation.
6
239 94
239 94
B3 This question is in two parts. Part 1 is about electrical circuits. Part 2 is about the energy balance of the Earth .
Part 1 Electrical circuits (a) Define (i) electromotive force (emf) of a battery. (ii) electrical resistance of a conductor. (b) A battery of emf “ε” and negligible internal resistance is connected in series to two resistors. The current in the circuit is “I”.
(i) State the equation giving the total power delivered by the battery. (ii) The potential difference across resistor “R1” is “V1” and that across resistor “R2” is “V2”. Using the law of conservation of energy deduce the equation below. ε = V1 + V2
(C) The graph shows the I-V characteristics of two conductors, “X” and “Y”
On the axis below, sketch graphs to show the variation with potential difference “V” of the resistance of conductor “X” (label this graph “X”) and conductor “Y” (label this graph “Y”). You do not need to put any number on the vertical axis.
7
(d) The conductors in (C) are connected in series to a battery of emf ε and negligible internal resistance.
The power dissipated in each of the two resistors is the same. Using the graph given in (C), (i) determine the emf of the battery (ii) calculate the total power dissipated in the circuit.
Part 2. Energy balance of the Earth
(a) The intensity of the Sun radiation at the position of the Earth is approximately 1400 W/m2
Suggest why the average power received per unit of area of the Earth is 350 W/m2
(b) The diagram shows a simplified model of the energy balance of the Earth’s surface. The diagram shows radiation entering or leaving the Earth’s surface only.
The average equilibrium temperature of the earth’s surface is “TE” and that of the atmosphere is TA = 242K
(i) Using the data from the diagram, state the emisivity of the atmosphere. (ii) Show that the intensity of the radiation by the atmosphere toward the Earth’s surface is 136 W/m2
(iii) By reference to the energy balance of the Earth’s surface calculate “TE”
(c) (i) Outline a mechanism by which part of the radiation radiated by the Earth’s surface is absorbed by greenhouse gases in the atmosphere. (ii) Suggest by the incoming solar radiation is not affected by the mechanism you outlined in (C)(i). (iii) Carbon dioxide (CO2) is a greenhouse gas. State one source and one sink (object that remove CO2) of this gas
8
(d) The conductors in (C) are connected in series to a battery of emf ε and negligible internal resistance.
The power dissipated in each of the two resistors is the same. Using the graph given in (C), (i) determine the emf of the battery (ii) calculate the total power dissipated in the circuit.
Part 2. Energy balance of the Earth
(a) The intensity of the Sun radiation at the position of the Earth is approximately 1400 W/m2
Suggest why the average power received per unit of area of the Earth is 350 W/m2
(b) The diagram shows a simplified model of the energy balance of the Earth’s surface. The diagram shows radiation entering or leaving the Earth’s surface only.
The average equilibrium temperature of the earth’s surface is “TE” and that of the atmosphere is TA = 242K
(i) Using the data from the diagram, state the emisivity of the atmosphere. (ii) Show that the intensity of the radiation by the atmosphere toward the Earth’s surface is 136 W/m2
(iii) By reference to the energy balance of the Earth’s surface calculate “TE”
(c) (i) Outline a mechanism by which part of the radiation radiated by the Earth’s surface is absorbed by greenhouse gases in the atmosphere. (ii) Suggest by the incoming solar radiation is not affected by the mechanism you outlined in (C)(i). (iii) Carbon dioxide (CO2) is a greenhouse gas. State one source and one sink (object that remove CO2) of this gas
8
(d) The conductors in (C) are connected in series to a battery of emf ε and negligible internal resistance.
The power dissipated in each of the two resistors is the same. Using the graph given in (C), (i) determine the emf of the battery (ii) calculate the total power dissipated in the circuit.
Part 2. Energy balance of the Earth
(a) The intensity of the Sun radiation at the position of the Earth is approximately 1400 W/m2
Suggest why the average power received per unit of area of the Earth is 350 W/m2
(b) The diagram shows a simplified model of the energy balance of the Earth’s surface. The diagram shows radiation entering or leaving the Earth’s surface only.
The average equilibrium temperature of the earth’s surface is “TE” and that of the atmosphere is TA = 242K
(i) Using the data from the diagram, state the emisivity of the atmosphere. (ii) Show that the intensity of the radiation by the atmosphere toward the Earth’s surface is 136 W/m2
(iii) By reference to the energy balance of the Earth’s surface calculate “TE”
(c) (i) Outline a mechanism by which part of the radiation radiated by the Earth’s surface is absorbed by greenhouse gases in the atmosphere. (ii) Suggest by the incoming solar radiation is not affected by the mechanism you outlined in (C)(i). (iii) Carbon dioxide (CO2) is a greenhouse gas. State one source and one sink (object that remove CO2) of this gas
8
(d) The conductors in (C) are connected in series to a battery of emf ε and negligible internal resistance.
The power dissipated in each of the two resistors is the same. Using the graph given in (C), (i) determine the emf of the battery (ii) calculate the total power dissipated in the circuit.
Part 2. Energy balance of the Earth
(a) The intensity of the Sun radiation at the position of the Earth is approximately 1400 W/m2
Suggest why the average power received per unit of area of the Earth is 350 W/m2
(b) The diagram shows a simplified model of the energy balance of the Earth’s surface. The diagram shows radiation entering or leaving the Earth’s surface only.
The average equilibrium temperature of the earth’s surface is “TE” and that of the atmosphere is TA = 242K
(i) Using the data from the diagram, state the emisivity of the atmosphere. (ii) Show that the intensity of the radiation by the atmosphere toward the Earth’s surface is 136 W/m2
(iii) By reference to the energy balance of the Earth’s surface calculate “TE”
(c) (i) Outline a mechanism by which part of the radiation radiated by the Earth’s surface is absorbed by greenhouse gases in the atmosphere. (ii) Suggest by the incoming solar radiation is not affected by the mechanism you outlined in (C)(i). (iii) Carbon dioxide (CO2) is a greenhouse gas. State one source and one sink (object that remove CO2) of this gas
8 | 3,186 | 13,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-18 | latest | en | 0.90724 |
https://historicalsocietymaleny.com/sightseeing/how-do-you-calculate-monthly-rent-in-australia.html | 1,675,221,986,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00782.warc.gz | 292,423,428 | 18,911 | # How do you calculate monthly rent in Australia?
Contents
In line with Consumer Affairs Victoria, monthly rent is calculated as follows: The weekly rental amount is divided by 7 to determine the daily rental rate, then multiplied by 365 (days per year) to determine the yearly rate and finally divided by 12 to determine the monthly rental amount.
## How do you calculate monthly rent?
How is monthly rent calculated?
1. Step 1: Weekly Rent ÷ 7 = Daily Rent amount.
2. Step 2: Daily Rent x 365 = Yearly Rent amount.
3. Step 3: Yearly Rent ÷ 12 = Monthly rent amount.
## How do you calculate monthly rent to annual salary?
Gross annual salary x 30% = maximum monthly rental income.
## How is rent calculated in Australia?
In line with Consumer Affairs Victoria, monthly rent is calculated as follows: The weekly rental amount is divided by 7 to determine the daily rental rate, then multiplied by 365 (days per year) to determine the yearly rate and finally divided by 12 to determine the monthly rental amount.
## How is actual rent calculated?
Actual rent – It is the actual rent received/receivable by the owner by renting out the property. Expected rent – Higher value between municipal value and fair rent subjected to a maximum of Standard rent is expected rent.
INTERESTING: Is December 24 holiday in Australia?
## What percentage of monthly income should rent be?
When determining how much you should spend on rent, consider your monthly income and expenses. You should spend 30% of your monthly income on rent at maximum, and should consider all the factors involved in your budget, including additional rental costs like renter’s insurance or your initial security deposit.
## How do I calculate 3 times the rent?
If the monthly rent of an apartment is \$2,000, then 3 times the monthly rent is \$2000 x 3 = \$6000 (monthly income required to keep housing payments less than 1/3 of income)
## How do I calculate 2.5 times my rent?
The Rent Calculator Equation:
Monthly Income / 2.5 = Rent you can afford! It is recommended that your income is 2.5 times your monthly rent amount.
## How is daily rent calculated?
It works like this: take the monthly rent and multiple it by 12 to find the total yearly rent. Then divide the sum by 365 to determine the daily rent. Once you find the daily rent, you multiply it by the number of days the tenant will occupy the unit.
## How do I calculate my 6 week free rent?
If the special is based on a number of weeks, such as “6 weeks free,” you’ll use four easy steps: 1) First, multiply the market rent by the number of months in the lease term. 2) Then, divide that number by the number of weeks in the lease term.
## How is standard rent on a house calculated?
Standard rent is the rent determined under Rent Control Act. The property owner cannot charge a rent higher than the standard rent fixed under Rent Control Act. Net Annual Value (NAV) is the value calculated as Gross Annual Value minus Municipal taxes paid.
INTERESTING: Best answer: Is Australian gold a good tanning lotion?
## How do you calculate notional rental income?
How is notional rent calculated? The notional rent is determined on the basis of the annual value of the house. Annual value is the expected rental value of the house in the next year. The annual value is calculated taking into consideration the fair rent, standard rent and municipal value.
## What is reasonable expected rent?
It is the reasonable expected rent which the property can fetch. … It is the maximum rent which a person can legally recover from his tenant under the Rent Control Act. Standard rent is applicable only in case of properties covered under Rent Control Act. | 801 | 3,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-06 | latest | en | 0.925082 |
http://mathhelpforum.com/algebra/144370-primitive-funktion-print.html | 1,513,629,976,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948623785.97/warc/CC-MAIN-20171218200208-20171218222208-00029.warc.gz | 164,376,609 | 3,055 | Primitive funktion
• May 12th 2010, 09:49 AM
DenOnde
Primitive funktion
Need a little help here... :P
what is the primitive funktion of:
$(0,5x^2)^2$
and what is the way of finding it?
Best regards
• May 12th 2010, 10:17 AM
shenanigans87
Quote:
Originally Posted by DenOnde
Need a little help here... :P
what is the primitive funktion of:
$(0,5x^2)^2$
and what is the way of finding it?
Best regards
I don't understand the equation you wrote down...
The primitive function is the antiderivative.
So if the function were $(5x^2)^2=25x^4$
then the primitive function is $5x^5$ because when you take its derivative, you get $25x^4$
• May 12th 2010, 10:24 AM
DenOnde
aha, it was the outer square of the equation that had me dazzled.
the equation was: $((x^2)/2)^2$
so that correct answer should be:
$0,05x^5$?
Thanx for the quick reply by the way
• May 12th 2010, 10:37 AM
shenanigans87
Quote:
Originally Posted by DenOnde
aha, it was the outer square of the equation that had me dazzled.
the equation was: $((x^2)/2)^2$
so that correct answer should be:
$0,05x^5$?
Thanx for the quick reply by the way
The answer would be $\frac{x^5}{20}+C$
the C is necessary because its some unknown constant.
• May 12th 2010, 10:43 AM
DenOnde
Not in this case ;)
I was looking for the primitiv form in a integral so the C is not necessery this time :)
Thank you for the help!
Really appreciate it | 434 | 1,404 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-51 | longest | en | 0.907711 |
http://spotidoc.com/doc/181743/week5 | 1,603,665,716,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890028.58/warc/CC-MAIN-20201025212948-20201026002948-00623.warc.gz | 94,687,436 | 13,609 | # Week5
```Week5
Thursday, August 23, 2012
7:27 PM
This week we are going to learn how to isolate a variable using the four steps below
1) Distribute( Grouping symbols)
2) Combine Like Terms
a) same side =
b)opposite side = opposite sign
4) Multiplication/Division
New Vocab for the week:
Inverse Property
Identity Property
Property of Equality
One-Step Equation
Two-Step Equation
Multi-Step Equation
Isolate a Variable
Math Equation
1st9weeks Page 1
Mon. Review Redo
Thursday, August 23, 2012
7:27 PM
1st9weeks Page 2
Tue One Step
Thursday, August 23, 2012
7:27 PM
Isolating a variable begins this week....
There's a temptation when you are learning how to isolate a variable to see a few simple problems,
understand it enough to get the correct answer and not really understand the mathematics behind it.
Remember, the power of math is the use of logical arguments. If you understand the concepts
underneath what you are learning you don't have to rely on tricks, and when the problems get more
difficult you can use logic to support the correct procedure.
With that said I had you write down 4 steps to memorize because our ultimate goal is to ISOLATE A
VARIABLE:
1) Distribute( Grouping symbols)
2) Combine Like Terms
a) same side =
b)opposite side = opposite sign
4) Multiplication/Division
Last week we covered how to combine like terms and to distribute. This week we are going to learn the
last two steps. Next week we are going to finish off the brief unit by tying all these steps together.
Before reading any further check out this video to prep your mind on the notes
So remember when I wrote about the distributive property, I talked about logic. Today we are learning
WARNING: Your mind is going to go in reverse.....Say "Mary had a little lamb" ...now say it backwards...is
your brain fuzzy...it should be, if not you may want to get that checked out
Let's take a simple problem
x + 31 = 53
What plus 31 will give you 52? Easy right, 22....think about what your mind had to do to figure that out.
Most of us don't have that memorized, so you probably thought
1) I can subtract 31 from 53 and that will give me the answer
2) or I can count up from 31
1st9weeks Page 3
Both are correct procedures. Today we are going to give you the why's and the how's for any problem .
In the world of mathematics solving for that mysterious number is called
Isolating a Variable...
And notice the difference from last week when we looked at math expressions, now we have an = sign.
This is called a mathematical equation. In mathematics language the equal sign is a very powerful
symbol because it means both sides must be the same...not sort of the same...exactly the same, but
x + 31 = 53 doesn't look like
53 = 53, which is exactly the same
This is where the Inverse Property comes in, the Inverse Property states:
When a number is combined with its inverse, it is equal to its identity.
Identity means itself.
Let's look at Addition and Subtraction 1st
,think of a number.....how about 20
20 stands alone.....what can I add with 20 to get 20
20 + ? = 20
There's only one number, that number is zero....So any number adding with zero gives you that
number...that may seem silly, but that is the logic we are going to use to isolate a variable today.....
So let's take
x+5=9
What can I do to find the identify of X? Well using the logic I just used x + 0 = x, right?
An equation is like a scale, if I take something off of a balanced scaled I must do the same to the other
side in order to balance it again. This is called the Property of Equality. Students get freaked out by this
because they aren't used to changing equations. They are used to solving them. Well in math you have
the power to do ANYTHING to an equation. Later on we are going to use the property of equality to
completely change the look of an equation to get what we want. In order to use the property of
equality you must understand what the inverse property is....
If x + 0 =x, what can I do to the left side of the equation to cancel out the 5? The inverse or the opposite
of a +5 is -5 this creates a 0. But what I do to one side I must do to the other side...it looks like this
x+5=9
-5 = - 5
---------------x+0=4
using the logic we used earlier. X must equal 4....we don't ever write the 0, but it's important that you
see why you are cancelling out numbers.
WARNING: Learn your sign rules for addition and subtraction...Sing the SIGN SONG...in 5,4,3,2,1....
1st9weeks Page 4
x + 4 = -10
What is the inverse of +4, it's - 4, that will create a x + 0 , It looks like this
x + 4 = - 10
- 4 = -4
--------------x + 0 = Uh Oh....do you know your sign rules for addition and subtraction?
Same sign add and keep the sign, even if it's negative......
x + 0 = -14
or x = - 14.....Review question, what coefficient is in front of x? + 1 right....that leads us to our next
property
Inverse of Multiplication/Division
So let's practice some logic again. Think of a number....how did I know you were going to pick 8....
8 * ? = 8 ; 8 times what will give me it's identity? There's only 1 number, and that number is +1
8 * 1 = 8 mathematical fact
the same holds true for division....
8/? = 8 ; only 1 will make this true
8/1=8 mathematical fact
So if I have
4x = 32 the logic is "x multiplied with +1 will give me x, so what's the inverse of multiplying by +4"
Common mistake!! At this point student will say divide by - 4 because they just learned the Addition
Inverse property, let's think about that:
does 4 / - 4 = +1 No, never....
So the inverse of multiplying by +4 is divide by +4, it looks like this
4x = 32
--- ---4
4
1 * x = 8 ; x must equal 8 because the logic of math
Fractions
Every year students have a difficult time with fractions. But this is where I can see the students that
understand the logic and those that are just looking to follow a pattern. UNDERSTAND THE LOGIC
1st9weeks Page 5
understand the logic and those that are just looking to follow a pattern. UNDERSTAND THE LOGIC
There are two ways to look at a fraction, let's take
3 x=9
2
1) You can look at a fraction as x is a number multiplying with 3 and dividing by 2...so it's like two
operations
2) You can look at it like it's just a number multiplying with x, one operation
Every student sees it differently. Kids either fall into these two groups. One way will totally confuse you
and the other way will make so much sense. If both ways confuse you come see me for help
I'll show you the first way.....
1) If x is being multiplied by 3, what's the inverse of multiplying by + 3....divide by + 3 to both
sides....now the next one...if x is being divided by 2 what's the inverse, multiply by 2 to both sides, here's
how it looks
3 x = 9
-------2(3)
(3)
-----------------------(2)(1)x = 3 ( 2)
------2
---------------------(1)x
------ = 6........does x times 1 and then divided by 1 = x
(1)
so x must equal 6 mathematically
Now the second way, which as of recently I've had more success with
2) 3/2 is just a number multiplying with x ….then just divide by 3/2 to both sides....Remember how to
divide by a fraction Copy Change Reciprocal (Flip)...It looks like this
3 x
---2
-----------3
----2
=
9
---------3
---2
-----------------------------------------------------1st9weeks Page 6
-----------------------------------------------------(1)x = 6
x must be 6
Now let's look at how it's going to look on your quiz. Knowing math is nothing unless you can put it in
real life terms:
First define what you don't have, or what you are looking for. In this example we want to know how
many people speak Mandarin, let's call that a variable M
Next define the numbers you do have it looks like this
M = Mandarin total
487= English Speaking
Now let's look at the relationship in the problem: 487 was 512 fewer or less that how many spoke
Mandarin...
I know it makes sense to add 487 and 512. But let's try and set up a one step equation. You could also
say that
M - 512 = 487
Again define what you are looking for and assign a letter. We don't know how much Jason weighs
J= Jason's weight
Now define the numbers
144= how much Ben weighs
Now examine the relationship: Ben weighs 3 times more than Jason
1st9weeks Page 7
So it makes sense that
3J = 144
Algebra takes practice....this is when you must drill through repetition. The homework looks like a long
assignment, but all you are really doing is addition, subtraction, multiplication and division....it should
not take more than 20 mins....
I'm available at 615am for morning help
1st9weeks Page 8
Wed Two Step
Thursday, August 23, 2012
7:27 PM
Now that you know how to isolate a variable when there is one operation, now we are going to look at
Two-Step Equations. This is when there are two math operations with a variable.
In order to understand the mathematics you must review the orders of operations a bit...PEMDAS
In order to evaluate a problem you must do these operations first:
P Parentheses
E Exponents
M/D Multiply/Divide left to right
A/S Add or subtract left to right
Now I'm not sure if I was around when PEMDAS was first started, but there's a couple of mistakes in the
acronym.
First off the P in PEMDAS stands for Parentheses. A parentheses is just a Grouping symbol. But there are
other grouping symbols that lock numbers in a BFF situation. Like
X+1
------5
I don't see a parentheses but the long division bar is also a grouping symbol. It creates a BFF situation in
the numerator...what's not shown is (X+1)...it's assumed you know that those two are locked in
together.
Another grouping symbol is the square root symbol
This also creates a BFF situation also...so if I could change PEMDAS I would change it to GEMDAS
G Grouping symbols
E Exponents
M/D Multiply/Divide left to right
A/S Add or subtract left to right
Let's look at a simple two step equation first:
2x + 1
1st9weeks Page 9
Notice there is no equation symbol yet, because you need to look at GEMDAS.
If I were to evaluate this expression using an input of 3, what would my answer be:
It would be 7 correct, didn't you multiply by 2 and then add 1. You used the order of operations.
WARNING BRAIN FREEZE COMING SOON.....
Well in order to isolate a variable you need to work in reverse of what you did to solve the
problem...OUCH, WHAT, BUT???!! You're kidding right?
I wish I were kidding. Now as we get into this you may get frustrated and start thinking "When am I ever
going to use this" What you are learning is how to think in reverse, and how to follow a
procedure...That style of thinking is extremely important when you are going to get out into the big bad
world of real life problem solving. So will you be isolating a variable everyday of your adult life, probably
not. Will you be solving problems and probably have to reverse engineer some situations, absolutely!
Back to the problem
2x + 1
Let's review the steps to solve it
1) Multiply by 2
Now let's throw an equation into the mix
2x + 1 = 11
Now we need to isolate the variable. But to undo the problem we must
****Do the inverse of every operation in reverse*****
Stay with me, hang with me for a few seconds
Instead of doing the two steps from above, I'm going to cancel out the numbers around x using the
properties we learned yesterday.
So to both sides of the equation I'm going to:
1) Subtract 1
2) Divide by 2
It looks like this
1st9weeks Page 10
It looks like this
In the background, what I didn't show is that you created an equation that looks like this
(1)x + 0 = 5
-----(1)
Doesn't anything times +1 / divided by +1 and then added with 0 get you back to the original #….YES
math proves it correct 100% of the time!!
Here's a problem that trips up students all the time:
x+5
-------- = 6
2
Remember that long division bar is a grouping symbol. One of the tricky parts about math is that there
are different ways to write the same equation. Equations that don't look, but are exactly the same thing.
but it's good to keep in the back of your head that it can be re-written
x
--2
+
5
--- = 6
2
OR it can be written
1st9weeks Page 11
OR it can be written
1(x+5)
---= 6
2
Go ahead and try and plug in numbers into all three equations you will get the same thing. The reason
they can be written that way is because the long division bar is just the distributive property. Notice the
two new ways I just distributed the 1/2 into the parentheses.....An the original equation is just following
the rule on how to add fractions, like denominators and add the numerators.
You are welcome to solve any of these variables on your own using the methods we've already learned,
I'm going to teach you a short cut.
In the original equation the long division bar creates a BFF situation. So we must break that up by
canceling out the 2. What's the inverse of dividing by +2....multiplying by +2...Now we have just a +5,
what's the inverse - 5.....here's how the math works
Remember the underlying math is
(+1)x + 0 = 7
1st9weeks Page 12
Thur Multi-Step
Thursday, August 23, 2012
7:27 PM
Let's review what we've gone over this week
1) Solving a one step equation
2) Solving a two step equation
Today we are going to cover multi-step...We are going to combine what you learned last week with
what we've been covering this week. We have covered most of the 4 steps of isolating a variable. The
highlighted steps are what you've learned:
1) Distribute( Grouping symbols)
2) Combine Like Terms
a) same side =
b)opposite side = opposite sign
4) Multiplication/Division
Notice that the only step we've not learned is how to combine like terms on the opposite side of the =.
Hopefully you've gained some confidence of each of the skills learned. Today we are going to combine
those skills in solving for multi-step equations. A multi-step equation is when a student must isolate a
variable by combining like terms and using the inverses.
Let's take for example:
3( x + 1) - 5( x+ 2) = 7
Now we can focus on the 4 steps I've laid out since last week.
First step: You must get rid of any grouping symbols, for the most part this will mean distribute, but
remember that's not the only grouping symbol
Second step: Combine like terms
Third/Fourth step: Use the inverse property to cancel out numbers and isolate a variable
If your skills are strong and you commit to these steps, this is going to be easy, if not, it's going to
frustrate you a bit. Let's solve:
COPY THIS
1st9weeks Page 13
The numbers or symbols that are in different colors are your actual thoughts and actions. Notice we are
pulling together all the skills you've learned so far. SIGN song, Multiplication rules, Combine like terms,
Inverse and Identity to solve this equation.
Let's look at a word problem:
This is a great question because you've really got to analyze what you are looking for, and you've got to
understand numbers.
First what are we looking for: three consecutive multiples of 5
Let's start with understanding what a consecutive number is. Let's define a number any number.
let's call it n.
How do you define three consecutive numbers in a row. You could say that if the first one is n
n....the next one must be n + 1 and the next one must be n + 2
Let's say I evaluate by plugging in 1, wouldn't that give me 1,2,3...three consecutive numbers...But
there's something different about these three numbers they must be multiples of 5. I could say the first
one is
1st9weeks Page 14
one is
5n the next one is 5(n+1) and the last is 5(n+2)
Now if I'm adding these three terms it will look like this
5n + 5(n+1) + 5(n+2) and since they must add up to 90 I can drop an =
5n + 5(n+1) + 5(n+2) = 90
Now that we've figured out the underlying math, let's go ahead and go through the 4 steps
1) Distributive
2) Combine Like TERMS
4) Inverse Multiply/Divide
DONE
Now remember what the question was asking....what are the 3 consecutive multiples of 5...It's NOT
5,6,7 because the first term is 5n...so the three consecutive multiples are 25,30,35....do they add up to
90, yes they do.
I know at first this looks like it's much work, but keeping it neat and organized will save you much time
and wasted energy in the future. As long as you follow the 4 steps of isolating a variable, and
understand the mathematics behind it, you won't have any problems, and it will be pretty systematic
1st9weeks Page 15
Fri Quiz
Thursday, August 23, 2012
7:27 PM
This quiz will not have many review questions from functions.
It will be about what was learned this week and last week
Review
Simplifying
Combine Like Terms
Distributing
One Step
Two Step
Multi Step---Extra Credit since we just learned that
You will be quizzed on setting up an equation, writing it out and solving for a variable. The activities we
did in class
1st9weeks Page 16
``` | 4,345 | 16,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-45 | latest | en | 0.935085 |
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## Easy Worksheets For KindergartenEasy Worksheets For Kindergarten
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## Printable Materials For KindergartenPrintable Materials For Kindergarten
Published at Wednesday, March 25th 2020, 12:33:26 PM by Pensee Perrier. Worksheet. Have the students calculate their score then determine who stacked the most candies and who had the highest score—due to the scoring, this may not be the same person. Discuss any unusual approaches students used, if any.
## Interactive Games For KindergartenInteractive Games For Kindergarten
Published at Thursday, April 02nd 2020, 00:41:47 AM by Carressa Moulin. Worksheet. These addition word problems worksheets will produce 1 digit problems with three addends, with ten problems per worksheet. These word problems worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade.
Published at Saturday, January 04th 2020, 18:19:25 PM. Worksheet By Vafara Allard. Worksheets typically have a ”right answer.” Jamaica is expected to circle the rhyming words or match the pictures of things that start with the letter ”G.” She may learn quickly that putting down a wrong answer is emotionally costly. Worksheet activities may make her feel ignorant and incompetent, so that she learns to stop taking risks by guessing.
## Smart Board Activities For KindergartenSmart Board Activities For Kindergarten
Published at Friday, January 03rd 2020, 17:09:43 PM. Worksheet By Gallia Meyer. These fractions worksheets are great for practicing how to subtract measurement you would find on a tape measure. The problems will use 1/2’s, 1/4’s, 1/8’s. 1/16’s and there is an option to select 1/32’s and 1/64’s. These worksheets will generate 10 tape measurement fraction subtraction problems per worksheet.
### Preschool And Kindergarten WorksheetsPreschool And Kindergarten Worksheets
Published at Saturday, December 28th 2019, 04:50:45 AM. Worksheet By Genevieve Gilbert. Counting money is one of the most practical early math skills. Our grade 2 counting money worksheets help kids learn to recognize common coins and bills and to count money. U.S. and Canadian currencies are used. 2nd grade counting money worksheet
#### English Activity Sheets For KindergartenEnglish Activity Sheets For Kindergarten
Published at Friday, December 27th 2019, 23:43:49 PM. 1st grade By Karoly Guichard. Here is a graphic preview for all of the addition worksheets. These dynamically created addition worksheets allow you to select different variables to customize for your needs. The addition worksheets are randomly created and will never repeat so you have an endless supply of quality addition worksheets to use in the classroom or at home. Our addition worksheets are free to download, easy to use, and very flexible.
##### Classroom Decoration For KindergartenClassroom Decoration For Kindergarten
Published at Thursday, December 26th 2019, 04:06:46 AM. Worksheet By Berneen Dumas. These division worksheets are a good introduction for algebra concepts. You may select various types of characters to replace the missing numbers of the division problems on the division worksheets. The format of the division worksheets are horizontal and the answers range from 0 to 99. These division worksheets can be configured to layout the division problems using the division sign or a slash (/) format. You may select between 12 and 30 problems for these division worksheets.
###### Kindergarten Math Concepts WorksheetsKindergarten Math Concepts Worksheets
Published at Tuesday, December 24th 2019, 06:53:38 AM. kindergarten By Fantina Marty. This section contains all of the graphic previews for the Linear Functions Worksheets. We currently have worksheets covering review of linear equations, graphing absolute value functions, graphing linear inequalities. These Linear Functions Worksheets are a good resource for students in the 9th Grade through the 12th Grade.
## Free Printable Sheets For KindergartenFree Printable Sheets For Kindergarten
Published at Sunday, December 22nd 2019, 19:20:44 PM. Worksheet By Celestiel Guillou. These subtraction worksheets are good for introducing algebra concepts. You may select various types of characters to replace the missing numbers on these subtraction worksheets. The formats of the subtraction worksheets are horizontal and the numbers range from 0 to 99. You may select up to 30 subtraction problems for these worksheets.
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https://models.cellml.org/e/173/n62.cellml/@@cellml_codegen/MATLAB | 1,669,815,500,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00397.warc.gz | 437,458,583 | 6,938 | # Generated Code
The following is matlab code generated by the CellML API from this CellML file. (Back to language selection)
The raw code is available.
```function [VOI, STATES, ALGEBRAIC, CONSTANTS] = mainFunction()
% This is the "main function". In Matlab, things work best if you rename this function to match the filename.
[VOI, STATES, ALGEBRAIC, CONSTANTS] = solveModel();
end
function [algebraicVariableCount] = getAlgebraicVariableCount()
% Used later when setting a global variable with the number of algebraic variables.
% Note: This is not the "main method".
algebraicVariableCount =12;
end
% There are a total of 4 entries in each of the rate and state variable arrays.
% There are a total of 8 entries in the constant variable array.
%
function [VOI, STATES, ALGEBRAIC, CONSTANTS] = solveModel()
% Create ALGEBRAIC of correct size
global algebraicVariableCount; algebraicVariableCount = getAlgebraicVariableCount();
% Initialise constants and state variables
[INIT_STATES, CONSTANTS] = initConsts;
% Set timespan to solve over
tspan = [0, 10];
% Set numerical accuracy options for ODE solver
options = odeset('RelTol', 1e-06, 'AbsTol', 1e-06, 'MaxStep', 1);
% Solve model with ODE solver
[VOI, STATES] = ode15s(@(VOI, STATES)computeRates(VOI, STATES, CONSTANTS), tspan, INIT_STATES, options);
% Compute algebraic variables
[RATES, ALGEBRAIC] = computeRates(VOI, STATES, CONSTANTS);
ALGEBRAIC = computeAlgebraic(ALGEBRAIC, CONSTANTS, STATES, VOI);
% Plot state variables against variable of integration
[LEGEND_STATES, LEGEND_ALGEBRAIC, LEGEND_VOI, LEGEND_CONSTANTS] = createLegends();
figure();
plot(VOI, STATES);
xlabel(LEGEND_VOI);
l = legend(LEGEND_STATES);
set(l,'Interpreter','none');
end
function [LEGEND_STATES, LEGEND_ALGEBRAIC, LEGEND_VOI, LEGEND_CONSTANTS] = createLegends()
LEGEND_STATES = ''; LEGEND_ALGEBRAIC = ''; LEGEND_VOI = ''; LEGEND_CONSTANTS = '';
LEGEND_VOI = strpad('time in component time (ms)');
LEGEND_CONSTANTS(:,1) = strpad('IStim in component stimulus_protocol (uA_per_mmsq)');
LEGEND_STATES(:,1) = strpad('V in component membrane (mV)');
LEGEND_CONSTANTS(:,3) = strpad('Cm in component membrane (uF_per_mm2)');
LEGEND_ALGEBRAIC(:,1) = strpad('i_Na in component sodium_channel (uA_per_mm2)');
LEGEND_ALGEBRAIC(:,11) = strpad('i_K in component potassium_channel (uA_per_mm2)');
LEGEND_ALGEBRAIC(:,12) = strpad('i_Leak in component leakage_current (uA_per_mm2)');
LEGEND_CONSTANTS(:,8) = strpad('IStimC in component membrane (uA_per_mm2)');
LEGEND_CONSTANTS(:,4) = strpad('g_Na_max in component sodium_channel (mS_per_mm2)');
LEGEND_ALGEBRAIC(:,2) = strpad('g_Na in component sodium_channel (mS_per_mm2)');
LEGEND_CONSTANTS(:,5) = strpad('E_Na in component sodium_channel (mV)');
LEGEND_STATES(:,2) = strpad('m in component sodium_channel_m_gate (dimensionless)');
LEGEND_STATES(:,3) = strpad('h in component sodium_channel_h_gate (dimensionless)');
LEGEND_ALGEBRAIC(:,3) = strpad('alpha_m in component sodium_channel_m_gate (per_ms)');
LEGEND_ALGEBRAIC(:,7) = strpad('beta_m in component sodium_channel_m_gate (per_ms)');
LEGEND_ALGEBRAIC(:,4) = strpad('alpha_h in component sodium_channel_h_gate (per_ms)');
LEGEND_ALGEBRAIC(:,8) = strpad('beta_h in component sodium_channel_h_gate (per_ms)');
LEGEND_ALGEBRAIC(:,6) = strpad('g_K1 in component potassium_channel (mS_per_mm2)');
LEGEND_ALGEBRAIC(:,10) = strpad('g_K2 in component potassium_channel (mS_per_mm2)');
LEGEND_STATES(:,4) = strpad('n in component potassium_channel_n_gate (dimensionless)');
LEGEND_ALGEBRAIC(:,5) = strpad('alpha_n in component potassium_channel_n_gate (per_ms)');
LEGEND_ALGEBRAIC(:,9) = strpad('beta_n in component potassium_channel_n_gate (per_ms)');
LEGEND_CONSTANTS(:,6) = strpad('g_L in component leakage_current (mS_per_mm2)');
LEGEND_CONSTANTS(:,7) = strpad('E_L in component leakage_current (mV)');
LEGEND_RATES(:,1) = strpad('d/dt V in component membrane (mV)');
LEGEND_RATES(:,2) = strpad('d/dt m in component sodium_channel_m_gate (dimensionless)');
LEGEND_RATES(:,3) = strpad('d/dt h in component sodium_channel_h_gate (dimensionless)');
LEGEND_RATES(:,4) = strpad('d/dt n in component potassium_channel_n_gate (dimensionless)');
LEGEND_STATES = LEGEND_STATES';
LEGEND_ALGEBRAIC = LEGEND_ALGEBRAIC';
LEGEND_RATES = LEGEND_RATES';
LEGEND_CONSTANTS = LEGEND_CONSTANTS';
end
function [STATES, CONSTANTS] = initConsts()
VOI = 0; CONSTANTS = []; STATES = []; ALGEBRAIC = [];
CONSTANTS(:,1) = 0.0;
CONSTANTS(:,2) = 0.0;
STATES(:,1) = -73.8;
CONSTANTS(:,3) = 0.12;
CONSTANTS(:,4) = 4.0;
CONSTANTS(:,5) = 40.0;
STATES(:,2) = 0.05;
STATES(:,3) = 0.785;
STATES(:,4) = 0.0935;
CONSTANTS(:,6) = 0.0;
CONSTANTS(:,7) = -60.0;
CONSTANTS(:,8) = CONSTANTS(:,1);
if (isempty(STATES)), warning('Initial values for states not set');, end
end
function [RATES, ALGEBRAIC] = computeRates(VOI, STATES, CONSTANTS)
global algebraicVariableCount;
statesSize = size(STATES);
statesColumnCount = statesSize(2);
if ( statesColumnCount == 1)
STATES = STATES';
ALGEBRAIC = zeros(1, algebraicVariableCount);
utilOnes = 1;
else
statesRowCount = statesSize(1);
ALGEBRAIC = zeros(statesRowCount, algebraicVariableCount);
RATES = zeros(statesRowCount, statesColumnCount);
utilOnes = ones(statesRowCount, 1);
end
ALGEBRAIC(:,3) = ( 0.100000.*( - STATES(:,1) - 48.0000))./(exp(( - STATES(:,1) - 48.0000)./15.0000) - 1.00000);
ALGEBRAIC(:,7) = ( 0.120000.*(STATES(:,1)+8.00000))./(exp((STATES(:,1)+8.00000)./5.00000) - 1.00000);
RATES(:,2) = ALGEBRAIC(:,3).*(1.00000 - STATES(:,2)) - ALGEBRAIC(:,7).*STATES(:,2);
ALGEBRAIC(:,4) = 0.170000.*exp(( - STATES(:,1) - 90.0000)./20.0000);
ALGEBRAIC(:,8) = 1.00000./(1.00000+exp(( - STATES(:,1) - 42.0000)./10.0000));
RATES(:,3) = ALGEBRAIC(:,4).*(1.00000 - STATES(:,3)) - ALGEBRAIC(:,8).*STATES(:,3);
ALGEBRAIC(:,5) = ( 0.000100000.*( - STATES(:,1) - 50.0000))./(exp(( - STATES(:,1) - 50.0000)./10.0000) - 1.00000);
ALGEBRAIC(:,9) = 0.00200000.*exp(( - STATES(:,1) - 90.0000)./80.0000);
RATES(:,4) = ALGEBRAIC(:,5).*(1.00000 - STATES(:,4)) - ALGEBRAIC(:,9).*STATES(:,4);
ALGEBRAIC(:,1) = ( CONSTANTS(:,4).*power(STATES(:,2), 3.00000).*STATES(:,3)+0.00140000).*(STATES(:,1) - CONSTANTS(:,5));
ALGEBRAIC(:,6) = 0.0120000.*exp(( - STATES(:,1) - 90.0000)./50.0000)+ 0.000150000.*exp((STATES(:,1)+90.0000)./60.0000);
ALGEBRAIC(:,10) = 0.0120000.*power(STATES(:,4), 4.00000);
ALGEBRAIC(:,11) = (ALGEBRAIC(:,6)+ALGEBRAIC(:,10)+CONSTANTS(:,2)).*(STATES(:,1)+100.000);
ALGEBRAIC(:,12) = CONSTANTS(:,6).*(STATES(:,1) - CONSTANTS(:,7));
RATES(:,1) = (CONSTANTS(:,1) - (ALGEBRAIC(:,1)+ALGEBRAIC(:,11)+ALGEBRAIC(:,12)))./CONSTANTS(:,3);
RATES = RATES';
end
% Calculate algebraic variables
function ALGEBRAIC = computeAlgebraic(ALGEBRAIC, CONSTANTS, STATES, VOI)
statesSize = size(STATES);
statesColumnCount = statesSize(2);
if ( statesColumnCount == 1)
STATES = STATES';
utilOnes = 1;
else
statesRowCount = statesSize(1);
utilOnes = ones(statesRowCount, 1);
end
ALGEBRAIC(:,3) = ( 0.100000.*( - STATES(:,1) - 48.0000))./(exp(( - STATES(:,1) - 48.0000)./15.0000) - 1.00000);
ALGEBRAIC(:,7) = ( 0.120000.*(STATES(:,1)+8.00000))./(exp((STATES(:,1)+8.00000)./5.00000) - 1.00000);
ALGEBRAIC(:,4) = 0.170000.*exp(( - STATES(:,1) - 90.0000)./20.0000);
ALGEBRAIC(:,8) = 1.00000./(1.00000+exp(( - STATES(:,1) - 42.0000)./10.0000));
ALGEBRAIC(:,5) = ( 0.000100000.*( - STATES(:,1) - 50.0000))./(exp(( - STATES(:,1) - 50.0000)./10.0000) - 1.00000);
ALGEBRAIC(:,9) = 0.00200000.*exp(( - STATES(:,1) - 90.0000)./80.0000);
ALGEBRAIC(:,1) = ( CONSTANTS(:,4).*power(STATES(:,2), 3.00000).*STATES(:,3)+0.00140000).*(STATES(:,1) - CONSTANTS(:,5));
ALGEBRAIC(:,6) = 0.0120000.*exp(( - STATES(:,1) - 90.0000)./50.0000)+ 0.000150000.*exp((STATES(:,1)+90.0000)./60.0000);
ALGEBRAIC(:,10) = 0.0120000.*power(STATES(:,4), 4.00000);
ALGEBRAIC(:,11) = (ALGEBRAIC(:,6)+ALGEBRAIC(:,10)+CONSTANTS(:,2)).*(STATES(:,1)+100.000);
ALGEBRAIC(:,12) = CONSTANTS(:,6).*(STATES(:,1) - CONSTANTS(:,7));
ALGEBRAIC(:,2) = power(STATES(:,2), 3.00000).*STATES(:,3).*CONSTANTS(:,4);
end
% Pad out or shorten strings to a set length
req_length = 160;
insize = size(strin,2);
if insize > req_length
strout = strin(1:req_length);
else
strout = [strin, blanks(req_length - insize)];
end
end
```
Source
Derived from workspace EMBC 2013 Tutorial - Noble 1962 reproducibility at changeset e2e13897b3f6.
Collaboration
To begin collaborating on this work, please use your git client and issue this command: | 2,950 | 8,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | longest | en | 0.442263 |
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Consecutive Odd and Prime (Posted on 2006-11-08)
How many sets of 3 or more consecutive positive odd numbers greater than 1 exist such that all of the numbers are prime?
Submitted by Richard Rating: 2.2500 (4 votes) Solution: (Hide) {3,5,7} is the unique set of 3 or more consecutive odd numbers that are primes. Suppose x,y,z are 3 consecutive odd numbers in order and suppose that x is a prime that exceeds 3. Then x equals either 1 or 2 mod 3. If x equals 1 mod 3, then y is a multiple of 3. If x equals 2 mod 3, then z is a multiple of 3.
Subject Author Date Puzzle Solution K Sengupta 2008-11-09 05:33:03 re: Trivial Richard 2006-11-08 23:21:22 Trivial e.g. 2006-11-08 19:08:10 a solution Dennis 2006-11-08 09:25:03
Search: Search body:
Forums (45) | 269 | 863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-09 | latest | en | 0.841403 |
https://uicheritagegarden.org/manual-pdf/12156-fluid-mechanics-white-8th-edition-solution-manual-pdf-437-198.php | 1,642,455,911,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00392.warc.gz | 647,650,503 | 9,296 | # Fluid Mechanics White 8th Edition Solution Manual Pdf
File Name: fluid mechanics white 8th edition solution manual .zip
Size: 1170Kb
Published: 10.04.2021
## College Physics Ppt
The wide variety of topics gives instructors many options for their course and is a useful resource to students long after graduation. Students can progress from general ones to those involving design multiple steps and computer usage. Connect is the only integrated learning system that empowers students by continuously adapting to deliver precisely what they need when they need it how they need it so that class time is more effective. Free Sample: Click Here. Fluid Mechanics in SI Units. No account? Already have an account?
## Fluid Mechanics Pdf
No more searching online and not getting what u desire. Here on stuvera. So to get information on the book fluid mechanics 8th solution manual visit stuvera. The wide variety of topics gives instructors many options for their course and is a useful resource to students long after graduation. Students can progress from general ones to those involving design, multiple steps and computer usage.
Use these values to estimate, crudely, say, within a factor of 2, the number of molecules of air in the entire atmosphere of the earth. Solution: Assume zero shear. Due to element weight, the pressure along the lower and right sides must vary linearly as shown, to a higher value at point C. Vertical forces are presumably in balance with element weight included. But horizontal forces are out of balance, with the unbalanced force being to the left, due to the shaded excess-pressure triangle on the right side BC. Thus hydrostatic pressures cannot keep the element in balance, and shear and flow result.
There is indeed a PDF book site where you can get themechanics of materials 10th edition solutions pdf and all you've got to do is visit. Stress, strain, deformation deflection, torsion, flexure, shear diagram, and moment diagram are some of the topics covered by this subject. An optional chapter discusses failure and modern fracture theory, including stress intensity factors and crack growth. Ch Waves Typed Notes. Excellent service when it comes to textbook solutions.
48 Solutions Manual • Fluid Mechanics, Fifth Edition Weisbach (–) was born near Annaberg, Germany, the 8th of nine children of working-class.
## Fluidos- Frank M. White- Fluid Mechanics- Solutions
Sign In. White Mechanics Of Fluids Potter Wiggert Solution Manual Solutions manual mechanics of fluids 4th edition potter , solutions manual mechanics of fluids. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. Solution manual fundamentals of fluid mechanics Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website.
YOU are the protagonist of your own life. Fluid mechanics is the study of the effects of forces and energy on liquids and gases. Cimbala Author. This online declaration solution manual of problems in fluid mechanics can be one of the options to accompany you once having new time. This powerful problem-solver gives you 2, problems in fluid mechanics and hydraulics, fully solved step-by-step!
College Physics — Raymond A. Serway, Chris Vuille — 8th Edition. Introduction to Heat Transfer — Frank P.
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College Physics — Raymond A. Serway, Chris Vuille — 8th Edition. Introduction to Heat Transfer — Frank P. Incropera — 6th Edition. Nixon, Alberto S.
In this project we try to give an insight into some of the interesting applications that exist. Project Brooklyn 4K. This post give all project free pdf with detail explanation. Now a days the principles of fluid mechanics find wide applications in many situations directly or indirectly. If the packing is too tight, it will impair the movement and possibly damage the stem. Fluid mechanics is also a rich source of fundamental research challenges in the physical or biological sciences, mathematics and scientific computation.
Go to collections. My collections. Find the products you're tracking here.
Go to collections. My collections. Find the products you're tracking here.
В воздухе пахло жженой пластмассой. Вообще говоря, это была не комната, а рушащееся убежище: шторы горели, плексигласовые стены плавились. И тогда она вспомнила .
Вы рассказываете ей только то, что считаете нужным. Знает ли она, что именно вы собираетесь сделать с Цифровой крепостью.
И уже мгновение спустя ее осенило. Ее глаза расширились. Стратмор кивнул: - Танкадо хотел от него избавиться.
Огонь приближался к вершине. ТРАНСТЕКСТ стонал, его корпус готов был вот-вот рухнуть. Голос Дэвида точно вел ее, управляя ее действиями.
Танкадо зашифровал Цифровую крепость, и только ему известен ключ, способный ее открыть.
Это его прерогатива. Я плачу вам за то, чтобы вы следили за отчетностью и обслуживали сотрудников, а не шпионили за моим заместителем. Если бы не он, мы бы до сих пор взламывали шифры с помощью карандаша и бумаги. А теперь уходите! - Он повернулся к Бринкерхоффу, с побледневшим лицом стоявшему возле двери.
Затем он быстро побежит в заднюю часть собора, словно бы за помощью, и в возникшей неразберихе исчезнет прежде, чем люди поймут, что произошло. Пять человек. Четверо. Всего трое. Халохот стиснул револьвер в руке, не вынимая из кармана.
0 Response | 1,407 | 6,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-05 | latest | en | 0.919509 |
https://brainly.com/question/131554 | 1,485,272,740,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00019-ip-10-171-10-70.ec2.internal.warc.gz | 795,361,994 | 9,328 | # Simple Interest:54000 x 0.038 x (3/12) = I don't know what I'm doing wrong
2
by rvs95dance
Can you give the problem and the variables given please?
Principle: 54,000 Rate of Interest 3 (7/8) % Time 3 months Simple Interest?
is the rate 3 \\and// seven eights or 3 \\times// seven eights
The first one
Okay, I know what I did wrong. Thanks
2014-09-26T22:35:59-04:00
The simple interest rate equation is I = Prt
I = interest
P = principle ($54,000) r = rate (3.875%) t = time (3 months, so .25 years) I = Prt I = (54,000)*(.03875)*(.25) I =$523.125
This depends on whether the rate the asker gave was a monthly rate or a yearly rate. If it was monthly, my answer is correct. If it was yearly, your answer is correct.
it says (3/12) originally, so would we assume months or years?
Oh. True. You are correct.
2014-09-26T22:37:47-04:00
I = Prt
I = 54000*.03875*3
54000 is the principal
.00875 is the same thing as 3+(7/8) %, because to make a decimal a percentage you multiply by 100
3 is the time
I = 6277.5
That's the amount that has been made off of interest alone. If you're looking for the total amount of money, it's 54000+6277.5=60277.5 | 376 | 1,143 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-04 | latest | en | 0.942548 |
https://www.writingessays.org/%E7%AE%97%E6%B3%95%E4%BD%9C%E4%B8%9A%E4%BB%A3%E5%86%99/ | 1,695,505,876,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00882.warc.gz | 1,212,498,631 | 19,495 | Search the whole station
# 算法作业代写 Algorithm代写 CS代写 cs算法课业代写
638
## CSCI-570 Homework 4
### 2 Escape From the Building
In this problem, we need to decide whether there is a feasible plan for all the persons in a building to escape when they meet some emergency issues. More specifically, a building is described as an n by n grid and the position of p persons are represented as the integer points (x1, y1), . . . ,(xp, yp) in the building. Note that to ensure safety, we don’t allow any intersection between the paths of any two person. Therefore, your task is to decide whether there exist p vertex-disjoint paths from their starting points to any p different points on the boundary of the grid. Give an algorithm polynomial in n and prove the correctness of it.
### 3 Install Software to Your New Computer算法作业代写
Suppose that you have just bought a new computer and you want to install software on that. Specifically, two companies, which you can think of like Microsoft and Apple, are trying to sell their own copy of n different products, like Operation System. Spread Sheet, Web Browser. For each product i, i ∈ {1, 2, . . . , n}, we have
the price pi0 that Microsoft charges and the price p’i 0 that Apple charges.
the quality qi 0 of Microsoft version and the quality qi 0 of Apple version.
#### For example,
Apple may provide a better Web Browser Safari, but Microsoft a better Word Processor. You want to assemble your favorite computer by installing exactly one copy of each of the n products, e.g. you want to buy one operating system, one Web Browser, one Word Processor, etc. However, you don’t want to spend too much money on that. Therefore, your goal is to maximize the quality minus total price.
#### However, 算法作业代写
As you may know, the products of different companies may not be compatible. More concretely, for each product pair (i, j), we will suffer a penalty τij ≥ 0 if we install product i of Microsoft and product j of Apple. Note that τij may not be equal to τji just because Apple’s Safari does not work well on Microsoft Windows doesn’t mean that Microsoft’s Edge does not work well in Mac-OS. We assume that products are always compatible internally, which means that there is no penalty for installing two products from the same company. All pairwise penalties will be subtracted from the total quality of the system.
Your task is then to give a polynomial-time algorithm for computing which product i to purchase from which of the two companies (Apple and Microsoft) for all i ∈ {1, 2, . . . , n}, to maximize the total system quality (including the penalties) minus the total price. Prove the correctness of your algorithm. (Hint: You may model this problem as a min-cut problem by constructing your graph appropriately.)
### 4 Jumping Frogs
Somewhere near the Algorithmville, a number of frogs are standing on a number of lotus leaves. As they are social animals (and yes, they are never infected by coronavirus!), the frogs would like to gather together, all on the same lotus leaf. The frogs do not want to get wet, so they have to use their limited jump distance d to get together by jumping from piece to piece. However, these lotus leaves just started to grow, they will get damaged further by the force needed to jump to another leaf. Fortunately, the frogs are real experts on leaves, and know exactly how many times a frog can jump off each leaf before it sinks and become unavailable. Landing on leaves does not damage it. You have to help the frogs find a leaf where they can meet.
In this question, we will get the position of N lotus leaves. For each i ∈ [N], we know its position (xi , yi), the number of frogs ni on that leaf and the number of jumps mi before it sinks. The distance between two leaves (xi , yi) and (xj , yj ) is defined as |xi − xj | + |yi − yj |. Design an polynomial algorithm to determine whether each lotus leaf can hold all frogs for a party. The output is an array with length N, containing yes/no solution.
### 5 Preparing for the Exams算法作业代写
My friend Leo wants to have a emergency plan for his final exams on University of Southern Algorithmville. He has N subjects to prepare for, and for each subject, his score is determined only by the time he spend on learning. It’s not surprising that Leo found out he actually spent zero time on preparing before.
At least he knows when he can start learning all of these subjects. For each subject i there is a start time si when he can get all materials ready to 6start learning. And there is also a ending time ei for each subject, when his learning materials expire and he can’t learn anymore. We know that si and ei are integers, and Leo can only dedicate to a single subject within each time phase.
In the University of Southern Algorithmville (USA), a student’s total grade is the minimum grade among all subjects. Leo wants you to help him find out the best outcome. Given N subjects and their time intervals (si , ei), design an algorithm to find out the maximum time possible for the least prepared subject. (Hint: It’s not enough to use the network flow algorithm alone to determine the answer.)
### 6 Help Kumiko!算法作业代写
Kumiko is a member of a high school music group and she is in charge of counting the monthly community fee for each members. This work should be easy but the members are trying to make Kumiko’s life harder. They never pay the fee in an integer amount! The following form is what Kumiko has recorded.
each row and column. Give an polynomial time algorithm to obtain such rounding and show the correctness of it. (Hint: (1). You can check if this kind of rounding is possible by checking whether some flow is feasible. Then show that this flflow always exists. (2). You can first consider the case when Mij ∈ [0, 1] and then generalize it.)
### 7 Edges that Increase Max-Flow算法作业代写
Given a graph G = (V, E), the source-sink pair (s, t) and capacity of edges {ce ≥ 0 | e ∈ E}, design a polynomial-time algorithm to find a set of edges S, such that for every edge e ∈ S, increasing ce will lead to an increase of max-flow value between s and t. Prove the correctness of your algorithm.
The next:
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1 | 1,840 | 7,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | longest | en | 0.912009 |
https://scribesoftimbuktu.com/find-the-distance-between-two-points-38-15/ | 1,669,923,729,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710869.86/warc/CC-MAIN-20221201185801-20221201215801-00324.warc.gz | 542,421,926 | 14,636 | # Find the Distance Between Two Points (-3,8) , (1,5)
(-3,8)(1,5)
Use the distance formula to determine the distance between the two points.
Distance=(x2-x1)2+(y2-y1)2
Substitute the actual values of the points into the distance formula.
(1-(-3))2+(5-8)2
Simplify.
Multiply -1 by -3.
(1+3)2+(5-8)2
42+(5-8)2
Raise 4 to the power of 2.
16+(5-8)2
Subtract 8 from 5.
16+(-3)2
Raise -3 to the power of 2.
16+9 | 161 | 406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-49 | latest | en | 0.640237 |
http://www.rasch.org/rmt/rmt193h.htm | 1,464,541,759,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049281869.96/warc/CC-MAIN-20160524002121-00110-ip-10-185-217-139.ec2.internal.warc.gz | 767,733,224 | 7,311 | # Rasch Dichotomous Model vs. One-parameter Logistic Model (1PL 1-PL)
For most practical purposes these models are the same, despite their conceptual differences.
Aspect Rasch Dichotomous Model Item Response Theory:
One-Parameter Logistic Model
Abbreviation Rasch 1-PL IRT, also 1PL
For practical purposes When each individual in the person sample is parameterized for item estimation, it is Rasch. When the person sample is parameterized by a mean and standard deviation for item estimation, it is 1PL IRT.
Motivation Prescriptive: Distribution-free person ability estimates and distribution-free item difficulty estimates on an additive latent variable Descriptive: Computationally simpler approximation to the Normal Ogive Model of L.L. Thurstone, D.N. Lawley, F.M. Lord
Persons, objects, subjects, cases, etc. Person n of ability Bn, or
Person ν (Greek nu) of ability βn in logits
Normally-distributed person sample of ability distribution θ, conceptualized as N(0,1), in probits; persons are incidental parameters
Items, agents, prompts, probes, multiple-choice questions, etc.; items are structural parameters Item i of difficulty Di, or
Item ι (Greek iota) of difficulty δι in logits
Item i of difficulty bi (the "one parameter") in probits
Nature of binary data1 = "success" - presence of property
0 = "failure" - absence of property
1 = "success" - presence of property
0 = "failure" - absence of property
Probability of binary data Pni = probability that person n is observed to have the requisite property, "succeeds", when encountering item i Pi(θ) = overall probability of "success" by person distribution θ on item i
Formulation: exponential form
e = 2.71828
Formulation: logit-linear form
loge = natural logarithm
Local origin of scale: zero of parameter estimatesAverage item difficulty, or difficulty of specified item. (Criterion-referenced) Average person ability. (Norm-referenced)
Item discriminationItem characteristic curves (ICCs) modeled to be parallel with a slope of 1 (the natural logistic ogive)ICCs modeled to be parallel with a slope of 1.7 (approximating the slope of the cumulative normal ogive)
Missing data allowedYes, depending on estimation methodYes, depending on estimation method
Fixed (anchored) parameter values for persons and itemsYes, depending on softwareItems: depending on software. Persons: only for distributional form.
Fit evaluationFit of the data to the model
Local, one parameter at a time
Fit of the model to the data
Global, accept or reject the model
Data-model mismatchDefective data do not support parameter separability in an additive framework. Consider editing the data.Defective model does not adequately describe the data. Consider adding discrimination (2-PL), lower asymptote (guessability, 3-PL) parameters.
Differential item functioning (DIF) detectionYes, in secondary analysisYes, in secondary analysis
First conspicuous appearanceRasch, Georg. (1960) Probabilistic models for some intelligence and attainment tests. Copenhagen: Danish Institute for Educational Research.Birnbaum, Allan. (1968). Some latent trait models. In F.M. Lord & M.R. Novick, (Eds.), Statistical theories of mental test scores. Reading, MA: Addison-Wesley.
First conspicuous advocateBenjamin D. Wright, University of ChicagoFrederic M. Lord, Educational Testing Service
Widely-authoritative currently-active proponentDavid Andrich, Univ. of Western Australia, Perth, AustraliaRonald Hambleton, University of Massachusetts
Introductory textbookApplying The Rasch Model.T.G. Bond and C.M. FoxFundamentals of Item Response Theory. R.K. Hambleton, H. Swaminathan, and H.J. Rogers.
Widely used softwareWinsteps, RUMM, ConQuestLogist, BILOG
Minimum reasonable sample size30200 (Downing 2003)
See also: Andrich, D. (2004) Controversy and the Rasch model: A characteristic of incompatible paradigms? Medical Care, 42, 7-16. Reprinted in E.V. Smith & R.M. Smith, Introduction to Rasch Measurement: Theory, Models and Applications. JAM Press, Minnesota. Ch. 7 pp 143-166.
Downing S.M. (2003) Item response theory: applications of modern test theory for assessments in medical education. Medical Education, 37:739-745.
Linacre J.M. (2005). Rasch dichotomous model vs. One-parameter Logistic Model. Rasch Measurement Transactions, 19:3, 1032
Rasch Publications
Rasch Measurement Transactions (free, online) Rasch Measurement research papers (free, online) Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch Applying the Rasch Model 3rd. Ed., Bond & Fox Best Test Design, Wright & Stone
Rating Scale Analysis, Wright & Masters Introduction to Rasch Measurement, E. Smith & R. Smith Introduction to Many-Facet Rasch Measurement, Thomas Eckes Invariant Measurement: Using Rasch Models in the Social, Behavioral, and Health Sciences, George Engelhard, Jr. Statistical Analyses for Language Testers, Rita Green
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Journal of Applied Measurement Rasch models for measurement, David Andrich Constructing Measures, Mark Wilson Rasch Analysis in the Human Sciences, Boone, Stave, Yale
in Spanish: Análisis de Rasch para todos, Agustín Tristán Mediciones, Posicionamientos y Diagnósticos Competitivos, Juan Ramón Oreja Rodríguez
Forum Rasch Measurement Forum to discuss any Rasch-related topic
Go to Top of Page
Go to index of all Rasch Measurement Transactions
AERA members: Join the Rasch Measurement SIG and receive the printed version of RMT
Some back issues of RMT are available as bound volumes
Subscribe to Journal of Applied Measurement
Go to Institute for Objective Measurement Home Page. The Rasch Measurement SIG (AERA) thanks the Institute for Objective Measurement for inviting the publication of Rasch Measurement Transactions on the Institute's website, www.rasch.org.
Coming Rasch-related Events
May 27 - June 24, 2016, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 6, 2016, Monday Symposium on Rasch Analysis, Nottingham, UK (D. Andrich), Announcement
June 16-19, 2016, Thur.-Sat. In-person workshop: Introduction to Rasch measurement analysis in the healthcare sciences and education (in English), Barcelona, Spain (L. González de Paz, W. Boone, Winsteps), Announcement
July 1 - July 29, 2016, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
July 6-8, 2016, Wed.-Fri. In-person workshop: IRT and CAT using Concerto, Cambridge, UK, www.psychometrics.cam.ac.uk/
July 30-31, 2016, Sat.-Sun. PROMS 2016 Pre-Conference Workshop, Xi'an, China, confchina.com
Aug. 1-3, 2016, Mon.-Wed. PROMS 2016 Conference, Xi'an, China, confchina.com
Aug. 1 - Nov. 25, 2016, Mon.-Fri. On-line course: Introduction to Rasch Measurement Theory EDU5638 (D. Andrich, RUMM2030), www.education.uwa.edu.au
Aug. 3-5, 2016, Wed.-Fri. IMEKO 2016 TC1-TC7-TC13 Joint Symposium, Berkeley, CA, imeko-tc7-berkeley-2016.org
Aug. 12 - Sept. 9, 2016, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Sept. 2 - Oct. 14, 2016, Fri.-Fri. On-line workshop: Rasch Applications, Part 1: How to Construct a Rasch Scale (W.P. Fisher, Jr.), www.statistics.com
Sept. 28-30, 2016, Wed.-Fri. In-person workshop: Introductory Rasch (M. Horton, RUMM), Leeds, UK, www.leeds.ac.uk/medicine/rehabmed/psychometric
Oct. 3-5, 2016, Mon.-Wed. In-person workshop: Intermediate Rasch (M. Horton. Tennant, RUMM), Leeds, UK, www.leeds.ac.uk/medicine/rehabmed/psychometric
Oct. 6-7, 2016, Thur.-Fri. In-person workshop: Advanced Rasch (M. Horton, RUMM), Leeds, UK, www.leeds.ac.uk/medicine/rehabmed/psychometric
Oct. 14 - Nov. 11, 2016, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Nov. 11, 2016, Fri. In-person workshop: 11th International Workshop on Rasch Models in Business Administration, Tenerife, Spain, www.ull.es
Dec. 7-9, 2016, Wed.-Fri. In-person workshop: Introductory Rasch (M. Horton, RUMM), Leeds, UK, www.leeds.ac.uk/medicine/rehabmed/psychometric
Jan. 6 - Feb. 3, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
May 26 - June 23, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 30 - July 29, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
Aug. 11 - Sept. 8, 2017, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Aug. 18-21, 2017, Fri.-Mon. IACAT 2017: International Association for Computerized Adaptive Testing, Niigata, Japan, iacat.org
Oct. 13 - Nov. 10, 2017, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Jan. 5 - Feb. 2, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Jan. 10-16, 2018, Wed.-Tues. In-person workshop: Advanced Course in Rasch Measurement Theory and the application of RUMM2030, Perth, Australia (D. Andrich), Announcement
Jan. 17-19, 2018, Wed.-Fri. Rasch Conference: Matilda Bay Club, Perth, Australia, Website
May 25 - June 22, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 29 - July 27, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
Aug. 10 - Sept. 7, 2018, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Oct. 12 - Nov. 9, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com | 2,615 | 9,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-22 | longest | en | 0.822626 |
https://ece.uwaterloo.ca/~dwharder/Matlab/others.html | 1,550,695,678,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247496080.57/warc/CC-MAIN-20190220190527-20190220212527-00176.warc.gz | 561,119,639 | 1,186 | ## Other Miscellaneous Information
When you're getting bored of programming, you can always ask why:
```>> why
A very rich programmer wanted it that way.
```
By placing a semicolon at the end of a line, you suppress the output. For example,
```>> diag([1 2 3 4])
ans =
1 0 0 0
0 2 0 0
0 0 3 0
0 0 0 4
>> diag([2 4 6 8]);
```
If you do not assign the last result to anything, it is automatically assigned to the variable ans:
```>> x = 3 + eye(2)
x =
4 3
3 4
4 - 2*eye(2)
>> ans =
2 4
4 2
>> ans + x
ans =
6 7
7 6
``` | 209 | 608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-09 | latest | en | 0.776758 |
https://piping-designer.com/index.php/disciplines/mechanical/stationary-equipment/88-tank/3529-storage-tank-glossry | 1,721,871,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00652.warc.gz | 385,269,975 | 15,255 | # Storage Tank Glossary
on . Posted in Storage Tank
### A
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Absolute Pressure - A pressure at absolute zero can only exist in a total vacuum and any pressure above this is called absolute pressure.
• Acid Gas (Sour Gas) - Any significant amount of hydrogen sulfide (H2S) and carbon dioxide (CO2) contains a natural gas that can cause corrosion.
• Anchor Bolt -
• Anode -
• API Std 650 Tank Calculator - The API Std 650 standard published by the American Petroleum Institute (API) is designed to provide the petroleum industry with welded steel tanks for use in the storage of petroleum products and other liquid products commonly handled and stored by the various branches of the petroleum industry.
• API Tank Size - Based on 8.337 lbs/gallon and is based on stated capacity, not total volume.
• Atmospheric Pressure - The pressure exerted upon the earth's surface by the air because of the gravitational attraction of the earth.
• Atmospheric Storage Tank - A storage tank which has been designed to operate at pressures from atmospheric through 0.5 psig.
• Autoignition Temperature - The temperature to which a flammable mixture of vapor and air in the explosive range must be heated for ignition to occur spontaneously without external source of ignition.
• Automatic Sample - A sample taken by automatic means.
### B
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Baffle - Reduces disturbance and surges from the inlet fluids in horizontal or vertical. The most used type is a perforated plate in the shape of a segmented circle.
• Base Plate - A plate attached to the bottom of the support that sits on the foundation.
• Barlow's Formula - The point just before or at when a pipe or tubing fails under pressure.
• Blind Flange - A piping component for covering or closing the end of a pipe, valve, vessel or tank.
• Bottom Sample - A spot sample taken from the material at the bottom of the tank.
• Breathing Loss - Loss associated with thermal expansion and contraction of the vapor space, resulting from the daily temperature cycle or any such temperature cycle that can be induced by weather conditions such as rainstorms.
### C
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Catalyst - A material that aids or promotes a chemical reaction between other substances but does not react itself.
• Cavitation - The creation and collapse of bubbles in a liquid.
• Closed Gauging System - A method of obtaining measurements of the tank contents without opening the tank. This may be accomplished by using automatic tank gauges or by taking measurements through a pressure/vapor lock standpipe. This type of gauging is done extensively on vessels with inert gas systems.
• Closed System Gauging Unit - Closed gauging system measurement equipment specially designed to be used with a specific type of standpipe/vapor lock. The unit may have a single purpose such as taking temperature, ullages, watercuts or samples, or may be a combined unit capable of performing all necessary measurement functions.
• Coalescer - A device used to cause the separation and removal of one liquid from another such as water from a petroleum liquid.
• Corrosion - The thinning of a pipe wall that is typically caused by a chemical reaction from a corroding fluid or agent and is limited almost exclusively to metal products.
• Corrosion Allowance - The amount of material in a pipe or vessel that is available for corrosion without affecting the pressure containing integrity.
• Corrosion Inhibitor - A substance that slows down the chemical reaction rate of corrosion on metal that is exposed to the environment.
• Critical Zone - A term used to define the liquid level in a floating roof type storage tank from the point where floating of the roof begins to the point where the roof is fully floating. Sometimes known as the “inaccurate zone” or “partially floating region.”
• Crude - A mixture of naturally occuring liquid hydrocarbons before refining.
• Crude Oil - See crude
### D
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Data Plate - On all vessels. Gives the manufacturer, year manufactured, pressure, temperature, and any other indormation needed about the vessel.
• Davit Arm -
• Design Pressure - The pressure the vessel is designed for.
• Design Temperature - The temperature the vessel is designed for.
• Distillate - Products of distillation formed by condensing vapors.
### E
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Enthalpy - Measures the sum of internal energy changes in heat under constant pressure of the system.
• Explosive Limits - These are the limits of the explosive (flammable) range, the range between the minimum and maximum concentrations of a flammable vapor in air, which form explosive mixtures. These conditions may exist in the vapor space of ordinary fixed roof tanks.
### F
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Flange - A bolted connection where two pieces of pipe, equipment, fittings or valves are connected together to form a piping system.
• Flocculation - The process in which small pieces in a solution form clumps of fine particles which can rise or fall to the bottom of the tank.
• Flow-proportional Sample - A sample taken by an automatic sampler from a pipeline at a rate that is proportional to the liquid flow rate.
• Flush Nozzle - A rectangular tank nozzle which enters the side of the tank and has the underside flush with the tank bottom.
• Flush-type Cleanout -
### G
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Gate Valve - One of the most frequently used valve in piping systems and is classified as either "rising-stem" or "nonrising-stem" valves.
• Gas Boot Vapor -
• Gauge Hatch -
• Globe Valve - A type of valve used for regulating fluid flow, both on/off and throttling, it is a control valve.
• Gross Tankage Volume - The total geometric tank volume below the shell height.
• Grounding Lug -
• Gusset Plate - A thick steel plate used to strengthen joints on new and retrofitted structures.
### H
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Head - The end closure of the vessel.
• Heat - A form of energy that causes physical change in what is being heated.
• Heat Transfer - The exertion of power that is created by heat, or the increase in temperature.
• Heating Coils - See Tank Coils
• Heavy Crude - Crude oil with a high proportion of heavy hydrocarbons and metallic content.
• Holiday - A discontinuity in painted or coated surfaces.
### J
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Jacket Vessel - An external cavity around the vessel, controlling the temperature of the contents of the vessel through a cooling or heating fluid.
### L
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Latent Heat - The energy absorbed or released by a substance during a constant temperature or phase change from a solid to liquid, liquid to gas or vise versa.
• Level Interface -
• Level Switch - Provides a discrete input to the PLC. The level switch is typically used as a method of control in tanks and vessels.
• Light Crude - Crude oil with a light proportion of light hydrocarbons fractions and low metallic compounds.
• Liquid - A specific volume and can hold any shape it is contained within.
• Liquefied Natural Gas - Natural gas cooled to a liquid state.
• Lower Sample - A spot sample obtained at the midpoint of the lower third of the tank contents.
### M
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Manway - An opening on a tank or vessel designed as an entryway for personal access and equipment.
• Mass Flow Rate - The average velocity of a mass that passes by a point.
• Maximum Operating Pressure - The maximum pressure permissible at the top of the vessel during normal operation.
• Maximum Operating Pressure - The maximum temperature permissible that provides sufficient flexability during normal operation.
• Middle Sample - A spot sample obtained at the midpoint of the middle of the tank contents.
### N
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Nameplate - On all vessels. Gives the manufacturer, year manufactured, pressure, temperature, and any other indormation needed about the vessel.
• Net Tank Volume - The total usable volume inside a tank. Net volume equals gross volume less the volume equivalent attributable to tank heel.
• Pour Point - The lowest temperature at which oil will pour or flow when it is chilled under prescribed ASTM test condition.
• Nozzle - An opening in a tank or vessel for a flanged connection.
### P
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Petroleum - See crude
• PH - How acidic or alkakine water is.
• Pour Point - The lowest temperature at which oil will pour or flow when it is chilled under prescribed ASTM test conditions.
• Pressure - It is the force exerted perpendicular to the surface of an object and is expressed as force per unit area.
• Pressure Differential - The pressure difference between two points of a system.
• Pressure Loss - The difference in upstream and downstream pressure.
• Pressure Safety Valve - Used to protect vessels and tanks from overpressure. It is designed to protect critical system components such as pressure vessels, tanks and flange ratings.
• Programmable Logic Controller - The types of signals are analog output (AO), analog input (AI), discrete output (DO), and discrete Input (DI).
### R
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Raw Water - Untreated water.
• Reinforcing Pad - A plate formed to the shape of the tank or vessel around a nozzle for extra strength. Also known as repad or weld pad.
• Rundown Line - The pipeline from a process unit to a tank, through which the unit's production flows.
• Running Sample - A sample obtained by submerging an unstoppered beaker or bottle from the surface of the liquid to a point as near as possible to the shore tank draw off point or about one foot above the level of the free water in a ship tank, and then raising it without letting it rest, at a rate so that it will be about 75% full as it emerges from the liquid.
### S
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Saddle - Used to support horizontal and vertical designs in locations to support the concentrated loads at on each end of the vessel.
• Sampling - The process of obtaining a sample of the material in the tank, container or pipeline to use for testing or other purposes. This can be achieved by automatic or manual means.
• Sample Box - A sample box is a series of pipe, valves and fittings located on a pressure vessel or tank that is used to manually determine the height of the liquid or where the interface between oil and water resides.
• Sand Pan -
• Saturated Steam - The point (temperature and pressure) when steam is in contact with the liquid water (boiling) it came from.
• Sensible Heat - The heat added to a substance which increases its temperature but not the phase is called sensible heat.
• Shell - The cylinder portion of the vessel.
• Shell Nozzle -
• Shell Platform -
• Slip-on Flange - Designed to slip over the outside of pipe, long-tangent elbows, reducers, and swages.
• Sour Gas - See acid gas
• Specific Gravity - The density or ratio of any substance to another substance.
• Spot Sample - A sample taken at a specific “spot” within a tank using a stoppered bottle or beaker and lowering it to the level of desired sample then opening it and allowing it to remain at that level until full. A thief or a zone sampler may also be used to obtain spot samples.
• Stack Gas - Anything that comes out of a burner stack in gaseous form.
• Storage Tank Capacity Calculator - This calculates the working volume of a tank. The working volume is the maximum usable capacity of a storage tank.
• Stress - The force per unit area of cross-section.
• Stub End - Allows a flange to swivel in order to mate with another flanges bolt holes that may not align perfectly.
• Sweet Crude - Crude oil with a sulfur content less than 0.5% by weight, no sulfur smell.
### T
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Tail Gas - The lightest hydrocarbon gas released from a refining process.
• Tank Blanketing - The injection of gas into a storage tanks vapor space. The purpose of the blanket is to maintain a layer of gas above the liquid to prevent the entrance of air into the tank.
• Tank Blanketing (Tank Padding) - , also known as tank padding, is the injection of a gas into the open space of a liquid storage tank.
• Tank Coils - Placed in tanks and are designed to transfer heat from steam vapor or condensate to the tank.
• Tank Foundation - The purpose of a tank foundation is to provide a level surface for the tank and to equally transmit the load of the tank and its contents to the soil.
• Tank Vent - It relieves over pressure and under pressure scenarios caused by the process.
• Tap Sample - A sample taken from a valve or connection on a tank or pipeline.
• Temperature - The amount of heat or cold, but it is neither heat or cold.
• Tension Strength - The capacity of a material to resist a force tending to stretch it.
• Thermal Expansion - The increase in length, area or volume due to the increase (in some cased decrease) in temperature.
• Thermowell - Used in temperature measurement and provide isolation from the temperature sensor and the process fluid.
• Time-proportional Sample - A sample taken from a pipeline at regular intervals during a batch transfer period.
• Truck Connection -
### U
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Upper Sample - A spot sample obtained at the midpoint of the upper of the tank contents.
### V
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Vacuum Breaker - A vent on top of the vessel that allows air to be pulled into the tank.
• Vacuum Tank - Uses high-pressure hoses to clean out liquids and sludges without any damage to the object being cleaned.
• Vapor Recovery - A vapor recovery system is a series of components that work together to keep fugitive emissions from escaping into the surrounding environment.
• Venting Roof Nozzle -
• Vessel Internals - Vessels have internals to help them effectively process the fluids go through it.
• Viscosity - The measure of the internal friction/resistance to the flow of a liquid.
• Vortex Breaker - When the flow exits a nozzle it will create a vortex. A cross of plates on the discharge nozzle will prevent the discharge from swirling.
### W
• A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
• Water - Can exist in three of the four phases of matter: gas, liquid, or solid.
• Water Hammer - A valve is suddenly opened or closed.
• Water Outlet -
• Weep - A term applied to a minute leak.
• Welding - The fabrication process that fuses like materials togeather by heating them to a suitable temperatures, this can be acomplished by brazing, soldering or welding.
• Wet Gas - A gas containing a relatively high portion of hydrocarbons that are recoverable as liquids.
• Wier Box -
• Working Pressure - The normal pressure that a system operates at.
### Z
Tags: Storage Tank Glossary | 4,259 | 16,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.817951 |
http://mathhelpforum.com/calculus/206955-urgent-need-help-differential-compound-interest-rate-problem-print.html | 1,508,294,854,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822668.19/warc/CC-MAIN-20171018013719-20171018033719-00001.warc.gz | 210,985,850 | 2,941 | # Urgent! Need help with Differential Compound Interest Rate Problem!
• Nov 7th 2012, 06:22 AM
jojoyce
Urgent! Need help with Differential Compound Interest Rate Problem!
Find the amount in a savings account after one year if the initial balance in the account was \$1000, the interest is paid continuously into the account at a nominal rate of 10% per annum(compounded continuously), and the account is being continuously depleted(by taxes say) at the rate of y^2 per million dollars per year. The balance in the account after t years is given by y=y(t). How large can the account grow? How long will it take the account to grow to half of this maximum balance?
I've worked out that dy/dt=0.1y-y^2/10^6
y=y^2/20-y^3/(3*10^6)
I dont know what to do from here since I'm not really sure how to deal with the ys in the function equation above.
ANY HELP will be greatly appreciated!!!!!!!!!!
$\displaystyle \frac{dy}{dt} = 0.1y-10^{-6}y^2 \implies \frac{dy}{ 0.1y-10^{-6}y^2} = dt$ | 272 | 981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-43 | longest | en | 0.935326 |
http://slideplayer.com/slide/2531786/ | 1,506,313,208,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690318.83/warc/CC-MAIN-20170925040422-20170925060422-00036.warc.gz | 304,156,504 | 30,391 | The World We Live In… What is the Universe? How old is it?
Presentation on theme: "The World We Live In… What is the Universe? How old is it?"— Presentation transcript:
The World We Live In… What is the Universe? How old is it?
How big is it? What is it made of? What laws of nature govern it? What was happening to it in the past? What will happen to it in the future? Assembled by Sergei Zverev
What is the Universe? The universe is defined as everything that physically exists: the entirety of space and time, all forms of matter, energy, and the physical laws and constants that govern them.
How old is it? Our Universe (which means the space, time, matter and energy) was born in a Big Bang 13.7 billion (13,700,000,000) years ago from a singularity, a point in space-time in which gravitational forces cause matter to have an infinite density and zero volume. Before the Big Bang time and space did not exist. Right after the Big Bang all matter and energy were concentrated in an extremely small volume of space. The Universe was expanding since the Big Bang, and the rate of expanding as well as it’s properties were dramatically changing.
Demonstration which shows that our Universe is not thought to be a set of objects moving in a space away from a certain point at which the explosion occurred Our Universe is three-dimensional but you can picture a universe that consists only of the surface of a balloon (get a balloon, prepare it in advance: inflate it and draw the galaxies on the surface; make sure you have plenty of galaxies; let the air out). You inflate the balloon just a little bit and note the galaxies on the surface of the balloon. Then start inflating the balloon more and more. You will see that, while the balloon is inflating, the galaxies are moving farther away from each other and, if you measure the distance between the galaxies before and after the inflation of the balloon, you will see that more distant galaxies will appear to move apart faster.
In this model it is easy to see that every galaxy will observe the same effect, and no one galaxy is in a special location. Then you can ask yourself if you can find the center of expansion on the surface of the balloon, and you will not find it. This will help you make a conclusion that there is no location on the surface of the balloon that can be identified as the "center" of the universe or a point of the Big Bang “explosion.”
How big is it? Our Universe is approximately 96 billion light years across (96,000,000,000 light years). One light year = distance traveled by light in free space in one year = × 1015 meters, which is approximately 1013 kilometers. The Universe is approximately 9.6 × kilometers wide or 6 × miles wide: 600,000,000,000,000,000,000,000 miles
FROM MACRO to MEGA to MICRO COSMOS
ZOOM FROM MACRO to MEGA to MICRO COSMOS
This is a trip at high speed, jumping distances by a factor of 10.
We will start with 100 (equivalent to 1 meter), and will increase the size of observable area by a factor of 10, or 101 (10 meters), 102 (10x10 = 100 meters), 103 (10x10x10 = 1,000 meters), 104 (10x10x10x10 = 10,000 meters), so on, until we get close to the limit of our imagination in the direction of the mega cosmos. Later we will return, a little faster, back to the point where we started and continue our trip in the opposite direction reducing observable distances by factors of 10 into the micro cosmos. We will see how much the human race still needs to learn...
100 1 meter Distance to a bunch of leaves in the garden
101 10 meters Starting our trip upwards We can see the foliage
102 100 meters At this distance we can see the limits of the forest and the edifications
103 1 km We will pass from meters to kilometers...
Now it is possible to jump with a parachute
104 10 km The city could be observed but we really can’t see the houses
105 100 km At this height, the state of Florida is just coming into view...
106 1,000 km Typical sight from a satellite
107 10,000 km The northern hemisphere of Earth, and part of South America
108 100,000 km The Earth starts looking small...
109 1 million km The Earth and the Moon’s orbit in white...
1010 10 million km Part of the Earth’s Orbit in blue
1011 100 million km Orbits of Venus and Earth...
1012 1 billion km Orbits of Mercury, Venus, Earth, Mars and Jupiter
1013 10 billion km At this height of our trip, we could observe the Solar System and the orbits of the planets
1014 100 billion km The Solar System starts looking small...
1015 1 trillion km The Sun now is a small star in the middle of thousands of stars...
1016 1 light-year At one light-year the little Sun star is very small
1017 10 light-years Here we will see just stars in the infinity...
a light-year is the distance that light travels in a vacuum in one year at a speed of 3.0 x 108 m/s or 190,000 mi/s
1018 100 light-years A lot of stars and Nebulae...
1019 1,000 light-years At this distance we are travelling in the Milky Way, our galaxy
1020 10,000 light-years We continue our travel inside the Milky Way
1021 100,000 light-years We begin reaching the periphery of the Milky Way
1022 1 million light-years At this tremendous distance we can see the entire Milky Way and other galaxies as well
million light-years From this distance, all the galaxies look small with immense empty spaces in between The same laws are ruling all bodies of the Universe We could continue traveling upwards (up to 100 billion light years) with our imagination, but now let’s return home
1022
1021
1020
1019
1018
1017
1016
1015
1014
1013
1012
1011
1010
109
108
107
106
105
104
103
102 In this trip “upwards” we went to the power of 23 of 10
101 Now we are going to dig inside of matter in a reverse trip...
100 1 meter 10 centimeters or 4 inches
We arrive at our starting point. We could touch it with our hands...
10-1 10 centimeters Getting closer at 10 cm ... We can delineate the leaves
10-2 1 centimeter At this distance it is possible to observe the structure of the leaf
10-3 1 millimeter The cellular structures begin to show...
10-4 100 microns The cells can be delineated You can see the union between them
10-5 10 microns Starting our trip inside the cell...
10-6 1 micron The nucleus of the cell is visible.
10-7 100 nanometers Again we changed the measuring unit to adapt to the miniscule size You can see the chromosomes
10-8 10 nanometers In this micro universe the DNA chain is visible
10-9 1 nanometer ...the chromosome blocks can be observed.
10-10 1 Angstrom It appears like clouds of electrons... These are carbon atoms, one of the organic components Is there some resemblance of the micro cosmos with the mega cosmos? cosmos ...
10-11 10 picometers In this miniature world we can observe the electron cloud
10-12 1 picometer You can see a small nucleus in the middle of the atom and an empty space between the nucleus and the electron orbits
10-13 100 femtometers At this incredible and minuscule size we could observe the nucleus of the atom
10-14 10 femtometers Now we can observe the nucleus of the carbon atom: 6 protons and 6 neutrons
10-15 1 femtometer Here we are in the field of the scientific imagination, face to face with a proton
10-16 100 attometers Examine the proton in the nucleus
There is nowhere further to go... We are at the limits of current scientific knowledge
What is the contents of the Universe?
There are approximately 100 billion galaxies in the universe, including approximately 3,000 visible galaxies. The diameter of a typical galaxy is 30,000 light-years, it contains between billion stars, and the typical distance between two neighboring galaxies is 3 million light-years. In the center of many galaxies there are compact objects of very large mass, black holes. Gravitational attraction of a black hole is so powerful that nothing, not even electromagnetic radiation (for example visible light), can escape its pull. It makes the hole's interior invisible, and indistinguishable from the black space around it.
What is the contents of the Universe?
What is the contents of the Universe?
What is the contents of the Universe?
What is the contents of the Universe?
What laws of nature govern it?
Laws of nature are observable and measurable. Physical laws are scientific generalizations based on measurements and observations of physical behavior of our environment. They describe observable laws, and they are typically conclusions based on repeated scientific experiments and observations, over many years, and which have become accepted within the scientific community. The production of descriptions in the form of such laws is a fundamental objective of science. Examples of laws of nature include Classical (Newtonian) Mechanics, Quantum Mechanics, Einstein's Theory of Relativity and Standard Model of elementary particles.
What was happening to the Universe after the Big Bang?
A period of "inflation" produced a burst of exponential growth in the universe. For the next several billion years, the expansion of the universe gradually slowed down as the matter in the universe pulled on itself via gravity. More recently, the expansion has begun to speed up again as the repulsive effects of dark energy have come to dominate the expansion of the universe. A representation of the evolution of the Universe over 13.7 billion years.
What was happening to the Universe after the Big Bang?
Observations suggest that the universe as we know it began around 13.7 billion years ago. Since then, the evolution of the universe has passed through three phases: 1. The very early universe was the split second in which the universe was so hot that particles had energies higher than those currently accessible in particle accelerators on Earth (particles had no mass, only one force existed). 2. Following this period, in the early universe, the evolution of the universe proceeded according to known high energy physics. This is when the particle acquired mass, one force split in four, first protons, electrons and neutrons formed, then nuclei and finally atoms. 3. With the formation of neutral hydrogen, the cosmic microwave background was emitted. Then matter started to aggregate into the first stars and ultimately galaxies, quasars, and clusters of galaxies formed.
What was happening to the Universe
after the Big Bang? The Planck epoch - up to 10–43 seconds after the Big Bang: only one fundamental force exists. The grand unification epoch - 10– –36 seconds, gravity separates from the fundamental force. The electroweak epoch - between 10–36 seconds and 10–12 seconds: strong force separates from the electroweak force. The inflationary epoch - between 10–36 seconds and 10–32 seconds: inflation (rapid exponential expansion) occurs. In the end the universe is filled with a quark-gluon plasma. The quark epoch - between 10–12 seconds and 10–6 seconds: particles acquire a mass, fundamental interactions of gravitation, electromagnetism, the strong interaction and the weak interaction have taken their present forms. The hadron epoch - between 10–6 seconds and 1 second: The quark-gluon plasma cools, protons and neutrons and other hadrons form, neutrinos decouple and begin traveling freely through space. The lepton epoch - between 1 second and 3 minutes: The majority of hadrons and anti-hadrons annihilate each other, leptons and anti-leptons dominating the mass of the universe, then most leptons and anti-leptons annihilate leaving a small residue of leptons. The photon epoch - between 3 minutes and 380,000 years: universe is dominated by photons. These photons are still interacting frequently with charged protons, electrons and later with the nuclei, and continue to do so for the next 300,000 years. After 17 minutes the nuclear fusion stops. In the end of the epoch most of the atoms in the universe become neutral, and the photons can now travel freely: the universe has become transparent for radiation. Dark ages - 380,000 – 500 million years: the universe is filled with neutral hydrogen and helium and is relatively opaque at certain wavelengths, and does not emit radiation. Structure formation: after 150 million years: first stars and quasars formed, large volumes of matter form galaxies. Reionization million to 1 billion years: intense quasar radiation re-ionizes the surrounding universe. From this point on, most of the universe is composed of plasma and is transparent. 5.4 billion years after the Big Bang – the Milky Way formed. 8.5 billion years after the Big Bang – the Solar System formed. 13.7 billion years after the Big Bang - you were born. Expansion of the Universe is accelerating.
What was happening to the Universe after the Big Bang?
What will happen to the Universe in the future?
What will happen to the Universe in the future?
What will happen to the Universe in the future?
Big freeze: 1014 years (100,000,000,000,000 = 100,000 billion = 100 trillion) and beyond This scenario is generally considered to be the most likely, as it occurs if the universe continues expanding as it has been. 1. Over a time scale on the order of 100 trillion years or less, existing stars burn out, new stars cease to be created, and the universe goes dark; 2. Over a much longer time scale in the eras following this, the galaxy evaporates as the stellar remnants comprising it escape into space, and black holes evaporate via Hawking radiation; 3. In some grand unified theories, proton decay will convert the remaining interstellar gas and stellar remnants into leptons (such as positrons and electrons) and photons. Some positrons and electrons will then recombine into photons. In this case, the universe has reached a maximum-entropy state consisting of a bath of particles and low-energy radiation at thermodynamic equilibrium. 4. No changes after that.
What will happen to the Universe in the future?
Big rip: 200+ billion years This scenario is possible only if the energy density of dark energy actually increases without limit over time. Such dark energy is called phantom energy and is unlike any known kind of energy. In this case, the expansion rate of the universe will increase without limit. Gravitationally bound systems, such as clusters of galaxies, galaxies, and ultimately the solar system will be torn apart. Eventually the expansion will be so rapid as to overcome the electromagnetic forces holding molecules and atoms together. Finally even atomic nuclei will be torn apart and the universe as we know it will end in an unusual kind of gravitational singularity. In other words, the universe will expand so much that the electromagnetic force holding things together will fall to this expansion, making things fall apart.
Trivia Questions 1. What is the Universe?
Space, time, matter, energy and laws of nature 2. How old is our Universe? 13.7 billion (13,700,000,000) years old 3. How big is our Universe? 96 billion light years wide 4. What is it made of? Atoms and radiation, dark matter and dark energy 5. What is the largest object in our Universe? Super clusters of galaxies up to 500 million light years (or 5x1024 meters) in size and containing up to 10,000 galaxies 6. What is the smallest object in our Universe? A quark and an electron – both are less than meters in size 5. What was happening to the Universe after the Big Bang? It was expanding from a tiny dot, it is expanding now, and will keep expanding in the future
And now Note that going downwards we could only go to the power of minus 16 of 10 to reach the known limits of matter... and upwards we went to the power of 23 of 10 but we could have continued our trip up to the size of our universe – 1027 meters (power of 27 of 10) then What is behind those limits? Are there any limits? What initiated the Big Bang? What are the dark matter and dark energy? Why zillions of elementary particles of the same kind (for example electrons) in the universe are absolutely identical? Are we alone in the universe or there is life in other solar systems and galaxies? How long can the mankind survive on Earth? – etc., etc., etc. Do you want to find answers to these questions?
I hope that one day you will say YES!
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# Grade 5 NEW Ontario Math Curriculum - Number Sense & Place Value Digital Slides
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Are you planning for the NEW Grade 5 Ontario Math Curriculum? This fully editable Google Slide resource from My Call to Teach is what you need!
This resource is for the NUMBER strand and covers all specific expectations for “B1. Number Sense” in regards to whole numbers and decimals. In this resource, students will experience math through engaging real-life questions, storytelling through math, and interactive slides.
For teachers who are given Nelson© Math by your board, this resource loosely follows the order of the Place Value chapter (Chapter 2). This allows you to assign additional questions from this text if needed.
What’s Included:
• 66 UNIQUE and EDITABLE Google Slides in total
• Slides are editable for you to increase student engagement (e.g. you can add your Bitmoji, change the name in the word problem example, add pictures, etc.)
• Interactive student slides – value-added questions to assess student understanding. Students can hand in Google Classroom when completed.
• Each file starts off with an Opening Engagement Question touching on real-life, practical situations told in a storytelling format.
• Specific expectations that are covered
Titles of sections included:
1. Reading and Writing Whole Numbers
2. Comparing and Ordering Whole Numbers
3. Reading, Writing, and Rounding Decimals
4. Comparing and Ordering Whole Numbers
Specific Expectations covered:
B1.1 read, represent, compose, and decompose whole numbers up to and including 100 000, using appropriate tools and strategies, and describe various ways they are used in everyday life
B1.2 compare and order whole numbers up to and including 100 000, in various contexts
B1.5 read, represent, compare, and order decimal numbers up to hundredths, in various contexts
B1.6 round decimal numbers to the nearest tenth, in various contexts
*B1.3, B1.4, and B1.7 are found in the Fractions, Ratios, Percents Unit
Are you looking for extra, supplement practice that aligns with this digital unit? Check out the Number Sense & Place Value Worksheet Package.
Are you looking for SUMMATIVE assessments for the New Ontario Curriculum? Check out this Number Sense & Place Value Assessment Package.
Are you looking for a COMPLETE SOLUTION to implement the entire curriculum? This includes Digital Slides, Worksheets, and Assessments that covers all curriculum expectations. Check out this COMPLETE SOLUTION BUNDLE.
Do you want to see how I use all these resources in my own class? Click HERE.
See MY STORE for all the other units! Follow my store to get all the updates as I am carefully creating each unit to ensure quality resources.
Please note that I take great time in creating these resources. If you have any concerns I would greatly appreciate if you reach out to me through email as I strive to provide value-added resources for all my customers. You can email me at: margaret@mycalltoteach.com
Finally, follow me on Instagram, as this is where I post what I'm working on and what I launch - and I always launch with a flash sale or freebie!
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Remember to leave a review! I truly value your feedback! | 733 | 3,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.834477 |
https://www.quizzes.cc/metric/percentof.php?percent=0.7&of=3610 | 1,582,670,912,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00136.warc.gz | 869,439,076 | 3,088 | #### What is 0.7 percent of 3,610?
How much is 0.7 percent of 3610? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 0.7% of 3610 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 0.7% of 3,610 = 25.27
Calculate another percentage below. Type into inputs
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percent of
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Calculating zero point seven of three thousand, six hundred and ten How to calculate 0.7% of 3610? Simply divide the percent by 100 and multiply by the number. For example, 0.7 /100 x 3610 = 25.27 or 0.007 x 3610 = 25.27
#### How much is 0.7 percent of the following numbers?
0.7 percent of 3610.01 = 2527.007 0.7 percent of 3610.02 = 2527.014 0.7 percent of 3610.03 = 2527.021 0.7 percent of 3610.04 = 2527.028 0.7 percent of 3610.05 = 2527.035 0.7 percent of 3610.06 = 2527.042 0.7 percent of 3610.07 = 2527.049 0.7 percent of 3610.08 = 2527.056 0.7 percent of 3610.09 = 2527.063 0.7 percent of 3610.1 = 2527.07 0.7 percent of 3610.11 = 2527.077 0.7 percent of 3610.12 = 2527.084 0.7 percent of 3610.13 = 2527.091 0.7 percent of 3610.14 = 2527.098 0.7 percent of 3610.15 = 2527.105 0.7 percent of 3610.16 = 2527.112 0.7 percent of 3610.17 = 2527.119 0.7 percent of 3610.18 = 2527.126 0.7 percent of 3610.19 = 2527.133 0.7 percent of 3610.2 = 2527.14 0.7 percent of 3610.21 = 2527.147 0.7 percent of 3610.22 = 2527.154 0.7 percent of 3610.23 = 2527.161 0.7 percent of 3610.24 = 2527.168 0.7 percent of 3610.25 = 2527.175
0.7 percent of 3610.26 = 2527.182 0.7 percent of 3610.27 = 2527.189 0.7 percent of 3610.28 = 2527.196 0.7 percent of 3610.29 = 2527.203 0.7 percent of 3610.3 = 2527.21 0.7 percent of 3610.31 = 2527.217 0.7 percent of 3610.32 = 2527.224 0.7 percent of 3610.33 = 2527.231 0.7 percent of 3610.34 = 2527.238 0.7 percent of 3610.35 = 2527.245 0.7 percent of 3610.36 = 2527.252 0.7 percent of 3610.37 = 2527.259 0.7 percent of 3610.38 = 2527.266 0.7 percent of 3610.39 = 2527.273 0.7 percent of 3610.4 = 2527.28 0.7 percent of 3610.41 = 2527.287 0.7 percent of 3610.42 = 2527.294 0.7 percent of 3610.43 = 2527.301 0.7 percent of 3610.44 = 2527.308 0.7 percent of 3610.45 = 2527.315 0.7 percent of 3610.46 = 2527.322 0.7 percent of 3610.47 = 2527.329 0.7 percent of 3610.48 = 2527.336 0.7 percent of 3610.49 = 2527.343 0.7 percent of 3610.5 = 2527.35
0.7 percent of 3610.51 = 2527.357 0.7 percent of 3610.52 = 2527.364 0.7 percent of 3610.53 = 2527.371 0.7 percent of 3610.54 = 2527.378 0.7 percent of 3610.55 = 2527.385 0.7 percent of 3610.56 = 2527.392 0.7 percent of 3610.57 = 2527.399 0.7 percent of 3610.58 = 2527.406 0.7 percent of 3610.59 = 2527.413 0.7 percent of 3610.6 = 2527.42 0.7 percent of 3610.61 = 2527.427 0.7 percent of 3610.62 = 2527.434 0.7 percent of 3610.63 = 2527.441 0.7 percent of 3610.64 = 2527.448 0.7 percent of 3610.65 = 2527.455 0.7 percent of 3610.66 = 2527.462 0.7 percent of 3610.67 = 2527.469 0.7 percent of 3610.68 = 2527.476 0.7 percent of 3610.69 = 2527.483 0.7 percent of 3610.7 = 2527.49 0.7 percent of 3610.71 = 2527.497 0.7 percent of 3610.72 = 2527.504 0.7 percent of 3610.73 = 2527.511 0.7 percent of 3610.74 = 2527.518 0.7 percent of 3610.75 = 2527.525
0.7 percent of 3610.76 = 2527.532 0.7 percent of 3610.77 = 2527.539 0.7 percent of 3610.78 = 2527.546 0.7 percent of 3610.79 = 2527.553 0.7 percent of 3610.8 = 2527.56 0.7 percent of 3610.81 = 2527.567 0.7 percent of 3610.82 = 2527.574 0.7 percent of 3610.83 = 2527.581 0.7 percent of 3610.84 = 2527.588 0.7 percent of 3610.85 = 2527.595 0.7 percent of 3610.86 = 2527.602 0.7 percent of 3610.87 = 2527.609 0.7 percent of 3610.88 = 2527.616 0.7 percent of 3610.89 = 2527.623 0.7 percent of 3610.9 = 2527.63 0.7 percent of 3610.91 = 2527.637 0.7 percent of 3610.92 = 2527.644 0.7 percent of 3610.93 = 2527.651 0.7 percent of 3610.94 = 2527.658 0.7 percent of 3610.95 = 2527.665 0.7 percent of 3610.96 = 2527.672 0.7 percent of 3610.97 = 2527.679 0.7 percent of 3610.98 = 2527.686 0.7 percent of 3610.99 = 2527.693 0.7 percent of 3611 = 2527.7 | 1,918 | 4,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-10 | latest | en | 0.774604 |
https://www.jiskha.com/display.cgi?id=1374283871 | 1,511,212,804,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00603.warc.gz | 817,696,662 | 3,821 | # math
posted by .
use mathematical induction to prove -1/2^n = 1/2^n - 1
• math -
Must be a typo, since if n=3,
-1/2^3 = -1/8
1/2^3 - 1 = -7/8
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https://open.kattis.com/contests/h4g3ch/problems/honeyheist | 1,708,622,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00459.warc.gz | 455,166,934 | 8,168 | Hide
# Problem CHoney Heist
0x67 is a scout ant searching for food and discovers a beehive nearby. As it approaches the honeycomb, 0x67 can sense an area inside packed with dried honey that can be easily carried back to the nest and stored for winter. However, it must burrow through the honeycomb to reach the cell containing the sweet loot. If 0x67 can create a passage to the honey to help the other ants find it, it will do so before returning to the nest.
The cells of the honeycomb are numbered in row major order, so cell IDs can be assigned as shown below:
When 0x67 discovers the opening to the honeycomb, it enters the cell. Some ants are stronger than others, depending on their age, so 0x67 can only chew through at most $N$ cells before its jaw wears out and must return to the nest to recuperate. The honeycomb is hexagonal, and each edge length is $R$ cells. 0x67 enters through a hole at location $A$ and must get to the honey at location $B$ by chewing a path through no more than $N$ adjacent cells. Because ants can be competitive, 0x67 wants to reach the honey by chewing through the fewest possible cells. 0x67 can also sense some of the cells are hardened with wax and impossible to penetrate, so it will have to chew around those to reach the cell at location $B$.
Scout ants have rudimentary computational skills, and before 0x67 begins to chew, it will work out where it needs to go, and compute $K$, the least number of cells it needs to chew through to get from $A$ to $B$, where $B$ is the $K$th cell. If $K > N$, 0x67 will not be strong enough to make the tunnel. When 0x67 returns to the nest, it will communicate to its nestmates how many cells it chewed through to get to $B$, or will report that it could not get to the honey.
## Input
The input contains two lines. The first line contains five blank separated integers: $R~ N~ A~ B~ X$
$R$: the length (number of cells) of each edge of the grid, where $2 \leq R \leq 20$. The total number of cells in the grid can be determined by taking a difference of cubes, $R^3 - (R-1)^3$.
$N$: the maximum number of cells 0x67 can chew through, where $1 \leq N < R^3 - (R-1)^3$.
$A$: the starting cell ID, This cell is located on one of the grid edges: The cell has fewer than six neighbors.
$B$: the cell ID of the cell containing the honey, where $1 \le B \le R^3 - (R-1)^3$.
$X$: the number of wax-hardened cells, where $0 \leq X < ( R^3 - (R-1)^3 ) - 1$.
The second line contains $X$ integers separated by spaces, where each integer is the ID of a wax-hardened cell.
The ID’s, $A$, $B$, and all the ID’s on the second line, are distinct positive integers less than or equal to $R^3 - (R-1)^3$.
## Output
A single integer $K$ if 0x67 reached the honey at cell $B$, where $B$ is the $K$th cell, otherwise the string No if it was impossible to reach the honey by chewing through $N$ cells or less.
Sample Input 1 Sample Output 1
6 6 1 45 11
15 16 17 19 26 27 52 53 58 65 74
6
Sample Input 2 Sample Output 2
6 3 1 45 11
15 16 17 19 26 27 52 53 58 65 74
No
Hide | 871 | 3,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.943208 |
https://think.design/services/data-visualization-data-design/cartogram/ | 1,718,594,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00131.warc.gz | 507,591,910 | 15,514 | # Cartogram
A cartogram is called a thematic map in which a mapping variable – such as travel time, population, or GNP – is substituted for land area or distance. As the space or geometry of the map is used to convey the information of an alternative variable, the map is usually distorted and sometimes to a large extent. Cartograms have been used for a long time, primarily to display emphasis and minimize the natural distortions caused by our perception of real geography.
### Quick details
What: Discover Rank
Why: Revisualize geographical entities as per prominence
## History of Cartogram
The French engineer Charles Minard is largely credited as the first to use the term cartogram (Friis, 1974). Minard was a pioneer of statistical graphs and charts in the mid-1800s. While maps and atlases had begun to use cartograms of some types, perhaps the earliest example of its use in a general English-speaking publication was in The Washington Post in 1929. Joseph R. Grundy used a cartogram to illustrate his belief that State-based voting powers were unfair as they all had markedly different populations. However, one of the first cartographers to generate cartograms with the aid of computer visualization was Waldo Tobler of UC Santa Barbara in the 1960s. Cartograms can also be constructed manually, either by hand or in a computer-assisted environment. Now a number of software packages generate cartograms.
1921 cartogram of USA based on electrical energy sold for light and power
## When to Use a Cartogram
### 1Represent a population through area cartograms
Use a cartogram when you need to represent an area in the context of the value a variable associated with it holds, making it useful as an isodemographic map. Cartograms are particularly useful as population cartograms, which can illustrate the relative sizes of the populations of the countries of the world by scaling the area of each country in proportion to its population. In such scenarios, the shape and relative location of each country is retained to as large an extent as possible. Cartograms can traditionally show a single data variable at a time but by modifying the color or by using textures or glyphs additional variables could also be incorporated.
Cartogram showing the distribution of the global population. Each of the 15,266 pixels represents the home country of 500,000 people – cartogram by Max Roser for Our World in Data
Source
### 2When data representation can do without preservation of shape and topology
One can also use cartograms when shape and topology preservation might be a lower priority than representing values to enable comparisons. In such cases, cartograms like Dorling replaces actual shapes with circles scaled according to the mapped features, where circles are distributed to resemble the original topology. Demers cartogram is a variation of Dorling cartogram, but it uses rectangles instead of circles and also attempts to retain visual cues at the expense of minimum distance. Schematic maps based on quadtrees can be seen as non-shape-preserving cartograms when some degree of neighborhood preservation might be needed.
Carbon atlas of the world using a Dorling Cartogram
Source
### 3Map out distances while visualizing proximities
Use cartograms to visualize the proximities between points, such as travel time. The distance cartogram is appropriate visualization alters distances between a user-specified origin and the other locations in a map with respect to travel time. Using a distance cartogram one can weigh the relative travel time costs between the origin and potential destinations at a glance because travel times are projected in a linearly interpolated time-space from the origin.
A distance cartogram distorting distances to show actual travel times
Source
## Types of Cartograms
### 1. Contiguous Cartograms
These charts distort the geometry of geographical regions in proportion to the data value associated with that region, such that larger the value, the more distorted and enlarged the region’s area is.
### 2. Non-Contiguous Cartograms
Invented by Judy M. Olson, they preserve the shape or outline of geographical regions and rescale each region from its center-point, in proportion with the data values assigned to them.
### 3. Dorling Maps
Here geographical regions are converted into circles and are organized and positioned in a way that loosely resembles the original topology. The area of the circles are proportional to the values they represent.
### 4. Demers Cartograms
Similar to a Dorling Map, it visualizes data but uses squares instead of circles, reducing the gaps between each geographical region.
### 5. Mosaic Cartograms
Unlike a Demers Cartogram which uses squares of varying sizes, this Cartogram variation keeps the size of the squares uniform.
### 6. Cartogram Hexmaps
Similar to mosaic cartograms these charts use hexagons instead of squares. Tilegrams is a tool which can be used to generate these charts.
### 7. Distance Cartograms
This Cartogram distorts a map (either geographical or an abstract travel network map) to show the travel times from a particular position on that map.
## When Not to Use a Cartogram
### 1When required to make statistically accurate trade-off decisions
Do not use cartograms when balancing statistical accuracy, geographical accuracy and topological accuracy are of importance to you. Maps depend on variation in scale to represent a large area and a cartogram is not a true representation of the real-world area and may give incomplete information. One of the disadvantages of the cartogram is that it inevitably changes the visual representation of geography.
### 2When interpretation needs to be simple and free of bias
Cartography has allowed maps to show only specific features that are important to users at a specific time and not all the features that are available in the area. Cartograms mostly use symbols to represent objects on land and these require interpretation. Also, since maps are drawn and designed by different entities who may have certain biases towards certain areas, hence full information regarding the area might not be represented. | 1,227 | 6,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.94656 |
https://list.ly/list/2xKT-curated-lesson-resources-logarithms | 1,606,689,439,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00577.warc.gz | 379,080,332 | 13,817 | Updated by Caleb Burnham on Apr 21, 2019
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10 items 1 followers 0 votes 2 views
Curated Lesson Resources - Logarithms
1
Rationale: I chose this article because it has good explanations including color-coded equations. It also includes practice problems so students can chaeck their understanding
2
Khan Academy Intro to logarithms Video
Description: This video explains what logarithms are and gives a few examples of finding logarithms.
Rationale: I chose this as an alternative for visual learners to use to learn the material.
3
Wolfram|Alpha: Making the world’s knowledge computable
Description: WoframAlpha is a computational search engine (fancy calculator) that students can use to find solutions to logarithms tied to their graphs.
Rationale: I chose this tool because it is a powerful; aid in doing homework and it will provide explanations of its solutions.
4
Desmos
Description: Online graphing tool
Rationale: I included this tool because students can use it to easily create a graph using just an equation
5
Graphing Calculator - GeoGebra
Description: Interactive, free online graphing calculator from GeoGebra.
Rationale: GeoGebra has a few more advanced features that Desmos does not, so I included it as an alternative if students want to get into some of the more complex manipulations of their graphs.
6
Photomath
Description: Photomath is a free mobile app that acts as an advanced graphing calculator that can scan equations from a page, even handwritten text!
Rationale: I included this resource because I know Siri can solve any high school math problem, so if my students are going to be using their phones to solve math problems I want them to have a tool that is easy to use and will give them the steps to solve the problems so they can understand what is going on.
7
Introduction to Logarithms - Math Is Fun
Description: An explanation of logarithms using bright colors and numerous examples.
Rationale: I chose this resource so that students can see more examples of logarithms if they need to and so they can get additional explanation if needed.
8
Logarithm Laws
Description: An image with all the logarithm laws on it.
Rationale: I chose this resource so that my students can save this resource to their devices and use it for quick reference when working on an assignment.
9
Uses of Logarithms
Description: Explains what logarithms are used for and has some interactive elements to illustrate them.
Rationale: I chose this resource so that my students can understand why they are learning to use logarithms.
10 | 554 | 2,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-50 | latest | en | 0.925844 |
http://solve-variable.com/math-variable/cramer%E2%80%99s-rule/math-tutorialalgebra-downloads.html | 1,519,027,857,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812556.20/warc/CC-MAIN-20180219072328-20180219092328-00608.warc.gz | 305,770,190 | 11,818 | Free Algebra Tutorials!
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# Homology (mathematics)
In mathematics, the term homology[a], originally introduced in algebraic topology, has three primary, closely-related usages. The most direct usage of the term is to take the homology of a chain complex, resulting in a sequence of abelian groups called homology groups. This operation, in turn, allows one to associate various named homologies or homology theories to various other types of mathematical objects. Lastly, since there are many homology theories for topological spaces that produce the same answer, one also often speaks of the homology of a topological space. (This latter notion of homology admits more intuitive descriptions for 1- or 2-dimensional topological spaces, and is sometimes referenced in popular mathematics.) There is also a related notion of the cohomology of a cochain complex, giving rise to various cohomology theories, in addition to the notion of the cohomology of a topological space.
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## Homology of Chain Complexes
To take the homology of a chain complex, one starts with a chain complex, which is a sequence ${\displaystyle (C_{\bullet },d_{\bullet })}$ of abelian groups ${\displaystyle C_{n}}$ (whose elements are called chains) and group homomorphisms ${\displaystyle d_{n}}$ (called boundary maps) such that the composition of any two consecutive maps is zero:
${\displaystyle C_{\bullet }:\cdots \longrightarrow C_{n+1}{\stackrel {d_{n+1}}{\longrightarrow }}C_{n}{\stackrel {d_{n}}{\longrightarrow }}C_{n-1}{\stackrel {d_{n-1}}{\longrightarrow }}\cdots ,\quad d_{n}\circ d_{n+1}=0.}$
The ${\displaystyle n}$th homology group ${\displaystyle H_{n}}$ of this chain complex is then the quotient group ${\displaystyle H_{n}=Z_{n}/B_{n}}$ of cycles modulo boundaries, where the ${\displaystyle n}$th group of cycles ${\displaystyle Z_{n}}$ is given by the kernel subgroup${\displaystyle Z_{n}:=\ker d_{n}:=\{c\in C_{n}\,|\;d_{n}(c)=0\}}$, and the ${\displaystyle n}$th group of boundaries ${\displaystyle B_{n}}$ is given by the image subgroup${\displaystyle B_{n}:=\mathrm {im} \,d_{n+1}:=\{d_{n+1}(c)\,|\;c\in C_{n+1}\}}$. One can optionally endow chain complexes with additional structure, for example by additionally taking the groups ${\displaystyle C_{n}}$ to be modules over a coefficient ring ${\displaystyle R}$, and taking the boundary maps ${\displaystyle d_{n}}$ to be ${\displaystyle R}$-module homomorphisms, resulting in homology groups ${\displaystyle H_{n}}$ that are also quotient modules. Tools from homological algebra can be used to relate homology groups of different chain complexes.
## Homology Theories
To associate a homology theory to other types of mathematical objects, one first gives a prescription for associating chain complexes to that object, and then takes the homology of such a chain complex. For the homology theory to be valid, all such chain complexes associated the same mathematical object must have the same homology. The resulting homology theory is often named according to the type of chain complex prescribed. For example, singular homology, Morse homology, Khovanov homology, and Hochschild homology are respectively obtained from singular chain complexes, Morse complexes, Khovanov complexes, and Hochschild complexes. In other cases, such as for group homology, there are multiple common methods to compute the same homology groups.
In the language of category theory, a homology theory is a type of functor from the category of the mathematical object being studied to the category of abelian groups and group homomorphisms, or more generally to the category corresponding to the associated chain complexes. One can also formulate homology theories as derived functors on appropriate abelian categories, measuring the failure of an appropriate functor to be exact. One can describe this latter construction explicitly in terms of resolutions, or more abstractly from the perspective of derived categories or model categories.
Regardless of how they are formulated, homology theories help provide information about the structure of the mathematical objects to which they are associated, and can sometimes help distinguish different objects.
## Homology of a Topological Space
Perhaps the most familiar usage of the term homology is for the homology of a topological space. For sufficiently nice topological spaces and compatible choices of coefficient rings, any homology theory satisfying the Eilenberg-Steenrod axioms yields the same homology groups as the singular homology (see below) of that topological space, with the consequence that one often simply refers to the "homology" of that space, instead of specifying which homology theory was used to compute the homology groups in question.
For 1-dimensional topological spaces, probably the simplest homology theory to use is graph homology, which could be regarded as a 1-dimensional special case of simplicial homology, the latter of which involves a decomposition of the topological space into simplices. (Simplices are a generalization of triangles to arbitrary dimension; for example, an edge in a graph is homeomorphic to a one-dimensional simplex, and a triangle-based pyramid is a 3-simplex.) Simplicial homology can in turn be generalized to singular homology, which allows more general maps of simplices into the topological space. Replacing simplices with disks of various dimensions results in a related construction called cellular homology.
There are also other ways of computing these homology groups, for example via Morse homology, or by taking the output of the Universal Coefficient Theorem when applied to a cohomology theory such as Čech cohomology or (in the case of real coefficients) De Rham cohomology.
### Inspirations for homology (informal discussion)
One of the ideas that led to the development of homology was the observation that certain low-dimensional shapes can be topologically distinguished by examining their "holes." For instance, a figure-eight shape has more holes than a circle ${\displaystyle S^{1}}$, and a 2-torus ${\displaystyle T^{2}}$ (a 2-dimensional surface shaped like an inner tube) has different holes from a 2-sphere ${\displaystyle S^{2}}$ (a 2-dimensional surface shaped like a basketball).
Studying topological features such as these led to the notion of the cycles that represent homology classes (the elements of homology groups). For example, the two embedded circles in a figure-eight shape provide examples of one-dimensional cycles, or 1-cycles, and the 2-torus ${\displaystyle T^{2}}$ and 2-sphere ${\displaystyle S^{2}}$ represent 2-cycles. Cycles form a group under the operation of formal addition, which refers to adding cycles symbolically rather than combining them geometrically. Any formal sum of cycles is again called a cycle.
### Cycles and Boundaries (informal discussion)
Explicit constructions of homology groups are somewhat technical. As mentioned above, an explicit realization of the homology groups ${\displaystyle H_{n}(X)}$ of a topological space ${\displaystyle X}$ is defined in terms of the cycles and boundaries of a chain complex ${\displaystyle (C_{\bullet },d_{\bullet })}$ associated to ${\displaystyle X}$, where the type of chain complex depends on the choice of homology theory in use. These cycles and boundaries are elements of abelian groups, and are defined in terms of the boundary homomorphisms ${\displaystyle d_{n}:C_{n}\to C_{n-1}}$of the chain complex, where each ${\displaystyle C_{n}}$ is an abelian group, and the ${\displaystyle d_{n}}$ are group homomorphisms that satisfy ${\displaystyle d_{n-1}\circ d_{n}=0}$ for all ${\displaystyle n}$.
Since such constructions are somewhat technical, informal discussions of homology sometimes focus instead on topological notions that parallel some of the group-theoretic aspects of cycles and boundaries.
For example, in the context of chain complexes, a boundary is any element of the image${\displaystyle B_{n}:=\mathrm {im} \,d_{n+1}:=\{d_{n+1}(c)\,|\;c\in C_{n+1}\}}$of the boundary homomorphism ${\displaystyle d_{n}:C_{n}\to C_{n-1}}$, for some ${\displaystyle n}$. In topology, the boundary of a space is technically obtained by taking the space's closure minus its interior, but it is also a notion familiar from examples, e.g., the boundary of the unit disk is the unit circle, or more topologically, the boundary of ${\displaystyle D^{2}}$ is ${\displaystyle S^{1}}$.
Topologically, the boundary of the closed interval ${\displaystyle [0,1]}$ is given by the disjoint union ${\displaystyle \{0\}\,\amalg \,\{1\}}$, and with respect to suitable orientation conventions, the oriented boundary of ${\displaystyle [0,1]}$ is given by the union of a positively-oriented ${\displaystyle \{1\}}$ with a negatively oriented ${\displaystyle \{0\}.}$ The simplicial chain complex analog of this statement is that${\displaystyle d_{1}([0,1])=\{1\}-\{0\}}$. (Since ${\displaystyle d_{1}}$ is a homomorphism, this implies ${\displaystyle d_{1}(k\cdot [0,1])=k\cdot \{1\}-k\cdot \{0\}}$ for any integer ${\displaystyle k}$.)
In the context of chain complexes, a cycle is any element of the kernel${\displaystyle Z_{n}:=\ker d_{n}:=\{c\in C_{n}\,|\;d_{n}(c)=0\}}$, for some ${\displaystyle n}$. In other words, ${\displaystyle c\in C_{n}}$ is a cycle if and only if ${\displaystyle d_{n}(c)=0}$. The closest topological analog of this idea would be a shape that has "no boundary," in the sense that its boundary is the empty set. For example, since ${\displaystyle S^{1},S^{2}}$, and ${\displaystyle T^{2}}$ have no boundary, one can associate cycles to each of these spaces. However, the chain complex notion of cycles (elements whose boundary is a "zero chain") is more general than the topological notion of a shape with no boundary.
It is this topological notion of no boundary that people generally have in mind when they claim that cycles can intuitively be thought of as detecting holes. The idea is that for no-boundary shapes like ${\displaystyle S^{1}}$, ${\displaystyle S^{2}}$, and ${\displaystyle T^{2}}$, it is possible in each case to glue on a larger shape for which the original shape is the boundary. For instance, starting with a circle ${\displaystyle S^{1}}$, one could glue a 2-dimensional disk ${\displaystyle D^{2}}$ to that ${\displaystyle S^{1}}$ such that the ${\displaystyle S^{1}}$ is the boundary of that ${\displaystyle D^{2}}$. Similarly, given a two-sphere ${\displaystyle S^{2}}$, one can glue a ball ${\displaystyle B^{3}}$ to that ${\displaystyle S^{2}}$ such that the ${\displaystyle S^{2}}$ is the boundary of that ${\displaystyle B^{3}}$. This phenomenon is sometimes described as saying that ${\displaystyle S^{2}}$ has a ${\displaystyle B^{3}}$-shaped "hole" or that it could be "filled in" with a ${\displaystyle B^{3}}$.
More generally, any shape with no boundary can be "filled in" with a cone, since if a given space ${\displaystyle Y}$ has no boundary, one can glue ${\displaystyle Y}$ to the cone on ${\displaystyle Y}$, and then ${\displaystyle Y}$ will be the boundary of that cone. (For example, a cone on ${\displaystyle S^{1}}$ is homeomorphic to a disk ${\displaystyle D^{2}}$ whose boundary is that ${\displaystyle S^{1}}$.) However, it is sometimes desirable to restrict to nicer spaces such as manifolds, and not every cone is homeomorphic to a manifold. Embedded representatives of 1-cycles, 3-cycles, and oriented 2-cycles all admit manifold-shaped holes, but for example the real projective plane ${\displaystyle \mathbb {RP} ^{2}}$ and complex projective plane ${\displaystyle \mathbb {CP} ^{2}}$ have nontrivial cobordism classes and therefore cannot be "filled in" with manifolds.
On the other hand, the boundaries discussed in the homology of a topological space ${\displaystyle X}$ are different from the boundaries of "filled in" holes, because the homology of a topological space ${\displaystyle X}$ has to do with the original space ${\displaystyle X}$, and not with new shapes built from gluing extra pieces onto ${\displaystyle X}$. For example, any embedded circle ${\displaystyle C}$ in ${\displaystyle S^{2}}$ already bounds some embedded disk ${\displaystyle D}$ in ${\displaystyle S^{2}}$, so such ${\displaystyle C}$ gives rise to a boundary class in the homology of ${\displaystyle S^{2}}$. By contrast, no embedding of ${\displaystyle S^{1}}$ into one of the 2 lobes of the figure-eight shape ${\displaystyle X_{8}}$ gives a boundary, despite the fact that it is possible to glue a disk onto a figure-eight lobe.
### Homology groups
Given a sufficiently-nice topological space ${\displaystyle X}$, a choice of appropriate homology theory, and a chain complex ${\displaystyle (C_{\bullet },d_{\bullet })}$ associated to ${\displaystyle X}$ that is compatible with that homology theory, the ${\displaystyle n}$th homology group ${\displaystyle H_{n}(X)}$ is then given by the quotient group ${\displaystyle H_{n}(X)=Z_{n}/B_{n}}$of ${\displaystyle n}$-cycles (${\displaystyle n}$-dimensional cycles) modulo ${\displaystyle n}$-dimensional boundaries. In other words, the elements of ${\displaystyle H_{n}(X)}$, called homology classes, are equivalence classes whose representatives are ${\displaystyle n}$-cycles, and any two cycles are regarded as equal in ${\displaystyle H_{n}(X)}$ if and only if they differ by the addition of a boundary. This also implies that the "zero" element of ${\displaystyle H_{n}(X)}$ is given by the group of ${\displaystyle n}$-dimensional boundaries, which also includes formal sums of such boundaries.
## Informal examples
The homology of a topological space X is a set of topological invariants of X represented by its homology groups ${\displaystyle H_{0}(X),H_{1}(X),H_{2}(X),\ldots }$ where the ${\displaystyle k^{\rm {th}}}$ homology group ${\displaystyle H_{k}(X)}$ describes, informally, the number of holes in X with a k-dimensional boundary. A 0-dimensional-boundary hole is simply a gap between two components. Consequently, ${\displaystyle H_{0}(X)}$ describes the path-connected components of X.[1]
The circle or 1-sphere ${\displaystyle S^{1}}$
The 2-sphere ${\displaystyle S^{2}}$ is the outer shell, not the interior, of a ball
A one-dimensional sphere ${\displaystyle S^{1}}$ is a circle. It has a single connected component and a one-dimensional-boundary hole, but no higher-dimensional holes. The corresponding homology groups are given as ${\displaystyle H_{k}\left(S^{1}\right)={\begin{cases}\mathbb {Z} &k=0,1\\\{0\}&{\text{otherwise}}\end{cases}}}$ where ${\displaystyle \mathbb {Z} }$ is the group of integers and ${\displaystyle \{0\}}$ is the trivial group. The group ${\displaystyle H_{1}\left(S^{1}\right)=\mathbb {Z} }$ represents a finitely-generated abelian group, with a single generator representing the one-dimensional hole contained in a circle.[2]
A two-dimensional sphere ${\displaystyle S^{2}}$ has a single connected component, no one-dimensional-boundary holes, a two-dimensional-boundary hole, and no higher-dimensional holes. The corresponding homology groups are[2][3] ${\displaystyle H_{k}\left(S^{2}\right)={\begin{cases}\mathbb {Z} &k=0,2\\\{0\}&{\text{otherwise}}\end{cases}}}$
In general for an n-dimensional sphere ${\displaystyle S^{n},}$the homology groups are ${\displaystyle H_{k}\left(S^{n}\right)={\begin{cases}\mathbb {Z} &k=0,n\\\{0\}&{\text{otherwise}}\end{cases}}}$
The solid disc or 2-ball ${\displaystyle B^{2}}$
The torus ${\displaystyle T=S^{1}\times S^{1}}$
A two-dimensional ball ${\displaystyle B^{2}}$ is a solid disc. It has a single path-connected component, but in contrast to the circle, has no higher-dimensional holes. The corresponding homology groups are all trivial except for ${\displaystyle H_{0}\left(B^{2}\right)=\mathbb {Z} }$. In general, for an n-dimensional ball ${\displaystyle B^{n},}$[2]
${\displaystyle H_{k}\left(B^{n}\right)={\begin{cases}\mathbb {Z} &k=0\\\{0\}&{\text{otherwise}}\end{cases}}}$
The torus is defined as a product of two circles ${\displaystyle T^{2}=S^{1}\times S^{1}}$. The torus has a single path-connected component, two independent one-dimensional holes (indicated by circles in red and blue) and one two-dimensional hole as the interior of the torus. The corresponding homology groups are[4] ${\displaystyle H_{k}(T^{2})={\begin{cases}\mathbb {Z} &k=0,2\\\mathbb {Z} \times \mathbb {Z} &k=1\\\{0\}&{\text{otherwise}}\end{cases}}}$
If n products of a topological space X is written as ${\displaystyle X^{n}}$, then in general, for an n-dimensional torus ${\displaystyle T^{n}=(S^{1})^{n}}$,
${\displaystyle H_{k}(T^{n})={\begin{cases}\mathbb {Z} ^{\binom {n}{k}}&0\leq k\leq n\\\{0\}&{\text{otherwise}}\end{cases}}}$
(see Torus#n-dimensional torus and Betti number#More examples for more details).
The two independent 1-dimensional holes form independent generators in a finitely-generated abelian group, expressed as the product group ${\displaystyle \mathbb {Z} \times \mathbb {Z} .}$
For the projective plane P, a simple computation shows (where ${\displaystyle \mathbb {Z} _{2}}$ is the cyclic group of order 2):[5] ${\displaystyle H_{k}(P)={\begin{cases}\mathbb {Z} &k=0\\\mathbb {Z} _{2}&k=1\\\{0\}&{\text{otherwise}}\end{cases}}}$
${\displaystyle H_{0}(P)=\mathbb {Z} }$ corresponds, as in the previous examples, to the fact that there is a single connected component. ${\displaystyle H_{1}(P)=\mathbb {Z} _{2}}$ is a new phenomenon: intuitively, it corresponds to the fact that there is a single non-contractible "loop", but if we do the loop twice, it becomes contractible to zero. This phenomenon is called torsion.
## Construction of homology groups
The following text describes a general algorithm for constructing the homology groups. It may be easier for the reader to look at some simple examples first: graph homology and simplicial homology.
The general construction begins with an object such as a topological space X, on which one first defines a chain complex C(X) encoding information about X. A chain complex is a sequence of abelian groups or modules ${\displaystyle C_{0},C_{1},C_{2},\ldots }$. connected by homomorphisms ${\displaystyle \partial _{n}:C_{n}\to C_{n-1},}$ which are called boundary operators.[4] That is,
${\displaystyle \dotsb {\overset {\partial _{n+1}}{\longrightarrow \,}}C_{n}{\overset {\partial _{n}}{\longrightarrow \,}}C_{n-1}{\overset {\partial _{n-1}}{\longrightarrow \,}}\dotsb {\overset {\partial _{2}}{\longrightarrow \,}}C_{1}{\overset {\partial _{1}}{\longrightarrow \,}}C_{0}{\overset {\partial _{0}}{\longrightarrow \,}}0}$
where 0 denotes the trivial group and ${\displaystyle C_{i}\equiv 0}$ for i < 0. It is also required that the composition of any two consecutive boundary operators be trivial. That is, for all n,
${\displaystyle \partial _{n}\circ \partial _{n+1}=0_{n+1,n-1},}$
i.e., the constant map sending every element of ${\displaystyle C_{n+1}}$ to the group identity in ${\displaystyle C_{n-1}.}$
The statement that the boundary of a boundary is trivial is equivalent to the statement that ${\displaystyle \mathrm {im} (\partial _{n+1})\subseteq \ker(\partial _{n})}$, where ${\displaystyle \mathrm {im} (\partial _{n+1})}$ denotes the image of the boundary operator and ${\displaystyle \ker(\partial _{n})}$ its kernel. Elements of ${\displaystyle B_{n}(X)=\mathrm {im} (\partial _{n+1})}$ are called boundaries and elements of ${\displaystyle Z_{n}(X)=\ker(\partial _{n})}$ are called cycles.
Since each chain group Cn is abelian all its subgroups are normal. Then because ${\displaystyle \ker(\partial _{n})}$ is a subgroup of Cn, ${\displaystyle \ker(\partial _{n})}$ is abelian, and since ${\displaystyle \mathrm {im} (\partial _{n+1})\subseteq \ker(\partial _{n})}$ therefore ${\displaystyle \mathrm {im} (\partial _{n+1})}$ is a normal subgroup of ${\displaystyle \ker(\partial _{n})}$. Then one can create the quotient group
${\displaystyle H_{n}(X):=\ker(\partial _{n})/\mathrm {im} (\partial _{n+1})=Z_{n}(X)/B_{n}(X),}$
called the nth homology group of X. The elements of Hn(X) are called homology classes. Each homology class is an equivalence class over cycles and two cycles in the same homology class are said to be homologous.[6]
A chain complex is said to be exact if the image of the (n+1)th map is always equal to the kernel of the nth map. The homology groups of X therefore measure "how far" the chain complex associated to X is from being exact.[7]
The reduced homology groups of a chain complex C(X) are defined as homologies of the augmented chain complex[8]
${\displaystyle \dotsb {\overset {\partial _{n+1}}{\longrightarrow \,}}C_{n}{\overset {\partial _{n}}{\longrightarrow \,}}C_{n-1}{\overset {\partial _{n-1}}{\longrightarrow \,}}\dotsb {\overset {\partial _{2}}{\longrightarrow \,}}C_{1}{\overset {\partial _{1}}{\longrightarrow \,}}C_{0}{\overset {\epsilon }{\longrightarrow \,}}\mathbb {Z} {\longrightarrow \,}0}$
where the boundary operator ${\displaystyle \epsilon }$ is
${\displaystyle \epsilon \left(\sum _{i}n_{i}\sigma _{i}\right)=\sum _{i}n_{i}}$
for a combination ${\displaystyle \sum n_{i}\sigma _{i},}$ of points ${\displaystyle \sigma _{i},}$ which are the fixed generators of C0. The reduced homology groups ${\displaystyle {\tilde {H}}_{i}(X)}$ coincide with ${\displaystyle H_{i}(X)}$ for ${\displaystyle i\neq 0.}$ The extra ${\displaystyle \mathbb {Z} }$ in the chain complex represents the unique map ${\displaystyle [\emptyset ]\longrightarrow X}$ from the empty simplex to X.
Computing the cycle ${\displaystyle Z_{n}(X)}$ and boundary ${\displaystyle B_{n}(X)}$ groups is usually rather difficult since they have a very large number of generators. On the other hand, there are tools which make the task easier.
The simplicial homology groups Hn(X) of a simplicial complex X are defined using the simplicial chain complex C(X), with Cn(X) the free abelian group generated by the n-simplices of X. See simplicial homology for details.
The singular homology groups Hn(X) are defined for any topological space X, and agree with the simplicial homology groups for a simplicial complex.
Cohomology groups are formally similar to homology groups: one starts with a cochain complex, which is the same as a chain complex but whose arrows, now denoted ${\displaystyle d_{n},}$ point in the direction of increasing n rather than decreasing n; then the groups ${\displaystyle \ker \left(d^{n}\right)=Z^{n}(X)}$ of cocycles and ${\displaystyle \mathrm {im} \left(d^{n-1}\right)=B^{n}(X)}$ of coboundaries follow from the same description. The nth cohomology group of X is then the quotient group
${\displaystyle H^{n}(X)=Z^{n}(X)/B^{n}(X),}$
in analogy with the nth homology group.
## Homology vs. homotopy
The nth homotopy group ${\displaystyle \pi _{n}(X)}$ of a topological space ${\displaystyle X}$ is the group of homotopy classes of basepoint-preserving maps from the ${\displaystyle n}$-sphere ${\displaystyle S^{n}}$ to ${\displaystyle X}$, under the group operation of concatenation. The most fundamental homotopy group is the fundamental group ${\displaystyle \pi _{1}(X)}$. For connected ${\displaystyle X}$, the Hurewicz theorem describes a homomorphism ${\displaystyle h_{*}:\pi _{n}(X)\to H_{n}(X)}$ called the Hurewicz homomorphism. For ${\displaystyle n>1}$, this homomorphism can be complicated, but when ${\displaystyle n=1}$, the Hurewicz homomorphism coincides with abelianization. That is, ${\displaystyle h_{*}:\pi _{1}(X)\to H_{1}(X)}$ is surjective and its kernel is the commutator subgroup of ${\displaystyle \pi _{1}(X)}$, with the consequence that ${\displaystyle H_{1}(X)}$ is isomorphic to the abelianization of ${\displaystyle \pi _{1}(X)}$. Higher homotopy groups are sometimes difficult to compute. For instance, the homotopy groups of spheres are poorly understood and are not known in general, in contrast to the straightforward description given above for the homology groups.
For an ${\displaystyle n=1}$ example, suppose ${\displaystyle X}$ is the figure eight. As usual, its first homotopy group, or fundamental group, ${\displaystyle \pi _{1}(X)}$ is the group of homotopy classes of directed loops starting and ending at a predetermined point (e.g. its center). It is isomorphic to the free group of rank 2, ${\displaystyle \pi _{1}(X)\cong \mathbb {Z} *\mathbb {Z} }$, which is not commutative: looping around the lefthand cycle and then around the righthand cycle is different from looping around the righthand cycle and then looping around the lefthand cycle. By contrast, the figure eight's first homology group ${\displaystyle H_{1}(X)\cong \mathbb {Z} \times \mathbb {Z} }$ is abelian. To express this explicitly in terms of homology classes of cycles, one could take the homology class ${\displaystyle l}$ of the lefthand cycle and the homology class ${\displaystyle r}$ of the righthand cycle as basis elements of ${\displaystyle H_{1}(X)}$, allowing us to write ${\displaystyle H_{1}(X)=\{a_{l}l+a_{r}r\,|\;a_{l},a_{r}\in \mathbb {Z} \}}$.
## Types of homology
The different types of homology theory arise from functors mapping from various categories of mathematical objects to the category of chain complexes. In each case the composition of the functor from objects to chain complexes and the functor from chain complexes to homology groups defines the overall homology functor for the theory.[9]
### Simplicial homology
The motivating example comes from algebraic topology: the simplicial homology of a simplicial complex X. Here the chain group Cn is the free abelian group or module whose generators are the n-dimensional oriented simplexes of X. The orientation is captured by ordering the complex's vertices and expressing an oriented simplex ${\displaystyle \sigma }$ as an n-tuple ${\displaystyle (\sigma [0],\sigma [1],\dots ,\sigma [n])}$ of its vertices listed in increasing order (i.e. ${\displaystyle \sigma [0]<\sigma [1]<\cdots <\sigma [n]}$ in the complex's vertex ordering, where ${\displaystyle \sigma [i]}$ is the ${\displaystyle i}$th vertex appearing in the tuple). The mapping ${\displaystyle \partial _{n}}$ from Cn to Cn−1 is called the boundary mapping and sends the simplex
${\displaystyle \sigma =(\sigma [0],\sigma [1],\dots ,\sigma [n])}$
to the formal sum
${\displaystyle \partial _{n}(\sigma )=\sum _{i=0}^{n}(-1)^{i}\left(\sigma [0],\dots ,\sigma [i-1],\sigma [i+1],\dots ,\sigma [n]\right),}$
which is considered 0 if ${\displaystyle n=0.}$ This behavior on the generators induces a homomorphism on all of Cn as follows. Given an element ${\displaystyle c\in C_{n}}$, write it as the sum of generators ${\textstyle c=\sum _{\sigma _{i}\in X_{n}}m_{i}\sigma _{i},}$ where ${\displaystyle X_{n}}$ is the set of n-simplexes in X and the mi are coefficients from the ring Cn is defined over (usually integers, unless otherwise specified). Then define
${\displaystyle \partial _{n}(c)=\sum _{\sigma _{i}\in X_{n}}m_{i}\partial _{n}(\sigma _{i}).}$
The dimension of the n-th homology of X turns out to be the number of "holes" in X at dimension n. It may be computed by putting matrix representations of these boundary mappings in Smith normal form.
### Singular homology
Using simplicial homology example as a model, one can define a singular homology for any topological space X. A chain complex for X is defined by taking Cn to be the free abelian group (or free module) whose generators are all continuous maps from n-dimensional simplices into X. The homomorphisms ∂n arise from the boundary maps of simplices.
### Group homology
In abstract algebra, one uses homology to define derived functors, for example the Tor functors. Here one starts with some covariant additive functor F and some module X. The chain complex for X is defined as follows: first find a free module ${\displaystyle F_{1}}$ and a surjective homomorphism ${\displaystyle p_{1}:F_{1}\to X.}$ Then one finds a free module ${\displaystyle F_{2}}$ and a surjective homomorphism ${\displaystyle p_{2}:F_{2}\to \ker \left(p_{1}\right).}$ Continuing in this fashion, a sequence of free modules ${\displaystyle F_{n}}$ and homomorphisms ${\displaystyle p_{n}}$ can be defined. By applying the functor F to this sequence, one obtains a chain complex; the homology ${\displaystyle H_{n}}$ of this complex depends only on F and X and is, by definition, the n-th derived functor of F, applied to X.
A common use of group (co)homology ${\displaystyle H^{2}(G,M)}$is to classify the possible extension groups E which contain a given G-module M as a normal subgroup and have a given quotient group G, so that ${\displaystyle G=E/M.}$
## Homology functors
Chain complexes form a category: A morphism from the chain complex (${\displaystyle d_{n}:A_{n}\to A_{n-1}}$) to the chain complex (${\displaystyle e_{n}:B_{n}\to B_{n-1}}$) is a sequence of homomorphisms ${\displaystyle f_{n}:A_{n}\to B_{n}}$ such that ${\displaystyle f_{n-1}\circ d_{n}=e_{n}\circ f_{n}}$ for all n. The n-th homology Hn can be viewed as a covariant functor from the category of chain complexes to the category of abelian groups (or modules).
If the chain complex depends on the object X in a covariant manner (meaning that any morphism ${\displaystyle X\to Y}$ induces a morphism from the chain complex of X to the chain complex of Y), then the Hn are covariant functors from the category that X belongs to into the category of abelian groups (or modules).
The only difference between homology and cohomology is that in cohomology the chain complexes depend in a contravariant manner on X, and that therefore the homology groups (which are called cohomology groups in this context and denoted by Hn) form contravariant functors from the category that X belongs to into the category of abelian groups or modules.
## Properties
If (${\displaystyle d_{n}:A_{n}\to A_{n-1}}$) is a chain complex such that all but finitely many An are zero, and the others are finitely generated abelian groups (or finite-dimensional vector spaces), then we can define the Euler characteristic
${\displaystyle \chi =\sum (-1)^{n}\,\mathrm {rank} (A_{n})}$
(using the rank in the case of abelian groups and the Hamel dimension in the case of vector spaces). It turns out that the Euler characteristic can also be computed on the level of homology:
${\displaystyle \chi =\sum (-1)^{n}\,\mathrm {rank} (H_{n})}$
and, especially in algebraic topology, this provides two ways to compute the important invariant ${\displaystyle \chi }$ for the object X which gave rise to the chain complex.
Every short exact sequence
${\displaystyle 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0}$
of chain complexes gives rise to a long exact sequence of homology groups
${\displaystyle \cdots \to H_{n}(A)\to H_{n}(B)\to H_{n}(C)\to H_{n-1}(A)\to H_{n-1}(B)\to H_{n-1}(C)\to H_{n-2}(A)\to \cdots }$
All maps in this long exact sequence are induced by the maps between the chain complexes, except for the maps ${\displaystyle H_{n}(C)\to H_{n-1}(A)}$ The latter are called connecting homomorphisms and are provided by the zig-zag lemma. This lemma can be applied to homology in numerous ways that aid in calculating homology groups, such as the theories of relative homology and Mayer-Vietoris sequences.
## Applications
### Application in pure mathematics
Notable theorems proved using homology include the following:
• The Brouwer fixed point theorem: If f is any continuous map from the ball Bn to itself, then there is a fixed point ${\displaystyle a\in B^{n}}$ with ${\displaystyle f(a)=a.}$
• Invariance of domain: If U is an open subset of ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle f:U\to \mathbb {R} ^{n}}$ is an injective continuous map, then ${\displaystyle V=f(U)}$ is open and f is a homeomorphism between U and V.
• The Hairy ball theorem: any continuous vector field on the 2-sphere (or more generally, the 2k-sphere for any ${\displaystyle k\geq 1}$) vanishes at some point.
• The Borsuk–Ulam theorem: any continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point. (Two points on a sphere are called antipodal if they are in exactly opposite directions from the sphere's center.)
• Invariance of dimension: if non-empty open subsets ${\displaystyle U\subseteq \mathbb {R} ^{m}}$ and ${\displaystyle V\subseteq \mathbb {R} ^{n}}$ are homeomorphic, then ${\displaystyle m=n.}$[10]
### Application in science and engineering
In topological data analysis, data sets are regarded as a point cloud sampling of a manifold or algebraic variety embedded in Euclidean space. By linking nearest neighbor points in the cloud into a triangulation, a simplicial approximation of the manifold is created and its simplicial homology may be calculated. Finding techniques to robustly calculate homology using various triangulation strategies over multiple length scales is the topic of persistent homology.[11]
In sensor networks, sensors may communicate information via an ad-hoc network that dynamically changes in time. To understand the global context of this set of local measurements and communication paths, it is useful to compute the homology of the network topology to evaluate, for instance, holes in coverage.[12]
In dynamical systems theory in physics, Poincaré was one of the first to consider the interplay between the invariant manifold of a dynamical system and its topological invariants. Morse theory relates the dynamics of a gradient flow on a manifold to, for example, its homology. Floer homology extended this to infinite-dimensional manifolds. The KAM theorem established that periodic orbits can follow complex trajectories; in particular, they may form braids that can be investigated using Floer homology.[13]
In one class of finite element methods, boundary-value problems for differential equations involving the Hodge-Laplace operator may need to be solved on topologically nontrivial domains, for example, in electromagnetic simulations. In these simulations, solution is aided by fixing the cohomology class of the solution based on the chosen boundary conditions and the homology of the domain. FEM domains can be triangulated, from which the simplicial homology can be calculated.[14][15]
## Software
Various software packages have been developed for the purposes of computing homology groups of finite cell complexes. Linbox is a C++ library for performing fast matrix operations, including Smith normal form; it interfaces with both Gap and Maple. Chomp, CAPD::Redhom Archived 2013-07-15 at the Wayback Machine and Perseus are also written in C++. All three implement pre-processing algorithms based on simple-homotopy equivalence and discrete Morse theory to perform homology-preserving reductions of the input cell complexes before resorting to matrix algebra. Kenzo is written in Lisp, and in addition to homology it may also be used to generate presentations of homotopy groups of finite simplicial complexes. Gmsh includes a homology solver for finite element meshes, which can generate Cohomology bases directly usable by finite element software.[14]
## Some non-homology-based discussions of surfaces
### Origins
Homology theory can be said to start with the Euler polyhedron formula, or Euler characteristic.[16] This was followed by Riemann's definition of genus and n-fold connectedness numerical invariants in 1857 and Betti's proof in 1871 of the independence of "homology numbers" from the choice of basis.[17]
### Surfaces
Cycles on a 2-sphere ${\displaystyle S^{2}}$
Cycles on a torus ${\displaystyle T^{2}}$
Cycles on a Klein
bottle ${\displaystyle K^{2}}$
Cycles on a hemispherical projective plane ${\displaystyle P^{2}}$
On the ordinary sphere ${\displaystyle S^{2}}$, the curve b in the diagram can be shrunk to the pole, and even the equatorial great circle a can be shrunk in the same way. The Jordan curve theorem shows that any closed curve such as c can be similarly shrunk to a point. This implies that ${\displaystyle S^{2}}$ has trivial fundamental group, so as a consequence, it also has trivial first homology group.
The torus ${\displaystyle T^{2}}$ has closed curves which cannot be continuously deformed into each other, for example in the diagram none of the cycles a, b or c can be deformed into one another. In particular, cycles a and b cannot be shrunk to a point whereas cycle c can.
If the torus surface is cut along both a and b, it can be opened out and flattened into a rectangle or, more conveniently, a square. One opposite pair of sides represents the cut along a, and the other opposite pair represents the cut along b.
The edges of the square may then be glued back together in different ways. The square can be twisted to allow edges to meet in the opposite direction, as shown by the arrows in the diagram. The various ways of gluing the sides yield just four topologically distinct surfaces:
${\displaystyle K^{2}}$ is the Klein bottle, which is a torus with a twist in it (In the square diagram, the twist can be seen as the reversal of the bottom arrow). It is a theorem that the re-glued surface must self-intersect (when immersed in Euclidean 3-space). Like the torus, cycles a and b cannot be shrunk while c can be. But unlike the torus, following b forwards right round and back reverses left and right, because b happens to cross over the twist given to one join. If an equidistant cut on one side of b is made, it returns on the other side and goes round the surface a second time before returning to its starting point, cutting out a twisted Möbius strip. Because local left and right can be arbitrarily re-oriented in this way, the surface as a whole is said to be non-orientable.
The projective plane ${\displaystyle P^{2}}$ has both joins twisted. The uncut form, generally represented as the Boy surface, is visually complex, so a hemispherical embedding is shown in the diagram, in which antipodal points around the rim such as A and A′ are identified as the same point. Again, a is non-shrinkable while c is. If b were only wound once, it would also be non-shrinkable and reverse left and right. However it is wound a second time, which swaps right and left back again; it can be shrunk to a point and is homologous to c.
Cycles can be joined or added together, as a and b on the torus were when it was cut open and flattened down. In the Klein bottle diagram, a goes round one way and −a goes round the opposite way. If a is thought of as a cut, then −a can be thought of as a gluing operation. Making a cut and then re-gluing it does not change the surface, so a + (−a) = 0.
But now consider two a-cycles. Since the Klein bottle is nonorientable, you can transport one of them all the way round the bottle (along the b-cycle), and it will come back as −a. This is because the Klein bottle is made from a cylinder, whose a-cycle ends are glued together with opposite orientations. Hence 2a = a + a = a + (−a) = 0. This phenomenon is called torsion. Similarly, in the projective plane, following the unshrinkable cycle b round twice remarkably creates a trivial cycle which can be shrunk to a point; that is, b + b = 0. Because b must be followed around twice to achieve a zero cycle, the surface is said to have a torsion coefficient of 2. However, following a b-cycle around twice in the Klein bottle gives simply b + b = 2b, since this cycle lives in a torsion-free homology class. This corresponds to the fact that in the fundamental polygon of the Klein bottle, only one pair of sides is glued with a twist, whereas in the projective plane both sides are twisted.
A square is a contractible topological space, which implies that it has trivial homology. Consequently, additional cuts disconnect it. The square is not the only shape in the plane that can be glued into a surface. Gluing opposite sides of an octagon, for example, produces a surface with two holes. In fact, all closed surfaces can be produced by gluing the sides of some polygon and all even-sided polygons (2n-gons) can be glued to make different manifolds. Conversely, a closed surface with n non-zero classes can be cut into a 2n-gon. Variations are also possible, for example a hexagon may also be glued to form a torus.[18]
The first recognisable theory of homology was published by Henri Poincaré in his seminal paper "Analysis situs", J. Ecole polytech. (2) 1. 1–121 (1895). The paper introduced homology classes and relations. The possible configurations of orientable cycles are classified by the Betti numbers of the manifold (Betti numbers are a refinement of the Euler characteristic). Classifying the non-orientable cycles requires additional information about torsion coefficients.[19]
The complete classification of 1- and 2-manifolds is given in the table.
Topological characteristics of closed 1- and 2-manifolds[20]
Manifold Euler no.,
χ
Orientability Betti numbers Torsion coefficient
(1-dimensional)
Symbol[18] Name b0 b1 b2
${\displaystyle S^{1}}$ Circle (1-manifold) 0 Orientable 1 1
${\displaystyle S^{2}}$ Sphere 2 Orientable 1 0 1 None
${\displaystyle T^{2}}$ Torus 0 Orientable 1 2 1 None
${\displaystyle P^{2}}$ Projective plane 1 Non-orientable 1 0 0 2
${\displaystyle K^{2}}$ Klein bottle 0 Non-orientable 1 1 0 2
2-holed torus −2 Orientable 1 4 1 None
g-holed torus (g is the genus) 2 − 2g Orientable 1 2g 1 None
Sphere with c cross-caps 2 − c Non-orientable 1 c − 1 0 2
2-Manifold with g holes and c cross-caps (c > 0) 2 (2g + c) Non-orientable 1 (2g + c) 1 0 2
Notes
1. For a non-orientable surface, a hole is equivalent to two cross-caps.
2. Any closed 2-manifold can be realised as the connected sum of g tori and c projective planes, where the 2-sphere ${\displaystyle S^{2}}$ is regarded as the empty connected sum. Homology is preserved by the operation of connected sum.
In a search for increased rigour, Poincaré went on to develop the simplicial homology of a triangulated manifold and to create what is now called a simplicial chain complex.[21][22] Chain complexes (since greatly generalized) form the basis for most modern treatments of homology.
Emmy Noether and, independently, Leopold Vietoris and Walther Mayer further developed the theory of algebraic homology groups in the period 1925–28.[23][24][25] The new combinatorial topology formally treated topological classes as abelian groups. Homology groups are finitely generated abelian groups, and homology classes are elements of these groups. The Betti numbers of the manifold are the rank of the free part of the homology group, and in the special case of surfaces, the torsion part of the homology group only occurs for non-orientable cycles.
The subsequent spread of homology groups brought a change of terminology and viewpoint from "combinatorial topology" to "algebraic topology".[26] Algebraic homology remains the primary method of classifying manifolds.[27]
## Notes
1. ^ in part from Greek ὁμός homos "identical"
## References
1. ^ Spanier 1966, p. 155
2. ^ a b c Gowers, Barrow-Green & Leader 2010, pp. 390–391
3. ^ Wildberger, Norman J. (2012). "More homology computations". YouTube. Archived from the original on 2021-12-11.
4. ^ a b Hatcher 2002, p. 106
5. ^ Wildberger, Norman J. (2012). "Delta complexes, Betti numbers and torsion". YouTube. Archived from the original on 2021-12-11.
6. ^ Hatcher 2002, pp. 105–106
7. ^ Hatcher 2002, p. 113
8. ^ Hatcher 2002, p. 110
9. ^ Spanier 1966, p. 156
10. ^ Hatcher 2002, p. 126.
11. ^ "CompTop overview". Archived from the original on 22 June 2007. Retrieved 16 March 2014.
12. ^ "Robert Ghrist: applied topology". Retrieved 16 March 2014.
13. ^ van den Berg, J.B.; Ghrist, R.; Vandervorst, R.C.; Wójcik, W. (2015). "Braid Floer homology" (PDF). Journal of Differential Equations. 259 (5): 1663–1721. Bibcode:2015JDE...259.1663V. doi:10.1016/j.jde.2015.03.022. S2CID 16865053.
14. ^ a b Pellikka, M; S. Suuriniemi; L. Kettunen; C. Geuzaine (2013). "Homology and Cohomology Computation in Finite Element Modeling" (PDF). SIAM J. Sci. Comput. 35 (5): B1195–B1214. Bibcode:2013SJSC...35B1195P. CiteSeerX 10.1.1.716.3210. doi:10.1137/130906556.
15. ^ Arnold, Douglas N.; Richard S. Falk; Ragnar Winther (16 May 2006). "Finite element exterior calculus, homological techniques, and applications". Acta Numerica. 15: 1–155. Bibcode:2006AcNum..15....1A. doi:10.1017/S0962492906210018. S2CID 122763537.
16. ^ Stillwell 1993, p. 170
17. ^ Weibel 1999, pp. 2–3 (in PDF)
18. ^ a b Weeks, Jeffrey R. (2001). The Shape of Space. CRC Press. ISBN 978-0-203-91266-9.
19. ^ Richeson 2008, p. 254
20. ^ Richeson 2008
21. ^ Richeson 2008, p. 258
22. ^ Weibel 1999, p. 4
23. ^ Hilton 1988, p. 284
24. ^ For example L'émergence de la notion de groupe d'homologie, Nicolas Basbois (PDF), in French, note 41, explicitly names Noether as inventing the homology group.
25. ^ Hirzebruch, Friedrich, Emmy Noether and Topology in Teicher 1999, pp. 61–63.
26. ^ Bourbaki and Algebraic Topology by John McCleary (PDF) Archived 2008-07-23 at the Wayback Machine gives documentation (translated into English from French originals).
27. ^ Richeson 2008, p. 264
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation. | 12,514 | 46,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 249, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-38 | latest | en | 0.896409 |
https://iresearchnet.org/geogrpahy-final-project/ | 1,709,009,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474670.19/warc/CC-MAIN-20240227021813-20240227051813-00280.warc.gz | 319,273,363 | 16,526 | ## Geogrpahy Final Project
• Post category:Geology
# Geogrpahy Final Project
Geology Homework Help
Natural Hazards & Disasters
NAME:
Project, Part 1: Risk Assessment
*** NOTE: YOU MUST READ ALL DIRECTIONS TO SUCCESSFULLY COMPLETE THE PROJECT ***
I. Introduction
For your final project, imagine you are an emergency manager, property insurance agent, or simply a concerned property owner. In this role, you are to use a series of maps and natural hazard data to evaluate the risk to a building structure of your choice in the state of California (Note: the building cannot be associated with a UC or any homes, apartments, or dorms near a UC. If you have questions about this requirement, see the instructor or TA before starting the project).
As a responsible assessor, you need to be aware of the exposure risk to the building. For each hazard, you will rate the potential risk in two dimensions:
(1) Probability – The probability that a hazardous event “may” occur, can range anywhere from 0 to 1 (or, changing 0 and 1 to percentages, you can think of these values as 0% to 100%). Note, the probability can never be 0 (there is always some chance that a hazard may occur) or 1 (cannot be absolutely certain a hazard will occur).
· You will assess the probability that a hazard will affect your building based on the proximity (distance) of previous, historic hazards that have occurred near your building
· First, you will examine the mapped data and determine the probability: Low (not likely to affect your building)Medium, or High (most likely to affect your building).
· Then, you will assign a numeric value to the probability, such as Low: 0.1 – 0.3, Medium: 0.4 – 0.6, High: 0.7 – 0.9.
(2) Severity of Impact – A risk, by its very nature, always has a negative impact. However, the size of the impact varies in terms of cost and impact on health, human life, or some other critical factor. The severity of impact ranges in value from 1 (negligible) to 10 (critical).
· You will assess the severity of the hazard to your building based on the extent (danger level) of previous, historic hazards that have occurred.
· First, you will examine the mapped data and determine if the impact severity to your building is Low (negligible impact, little damage)Medium, or High (significant impact, major damage).
· Then, you will assign a numeric value to the severity, such as Low: 1 – 3, Medium: 4 – 6, High: 7 – 10.
After assessing the probability and severity of the hazard to the building, you are to create a chart representing the probability and severity of natural hazards affecting your building.
II. Plate Boundaries, Volcanoes, & Earthquakes
Use the Project, Part 1 WebGIS to answer the questions below.
1. Turn off all layers except the plate boundaries, volcanoes, volcanic features, faults, and earthquakes layers. Type your building’s address into the webmap search bar to zoom in to your building. Leave the webmap’s location pop-up window open to “point” to the location of your building, then zoom out until the nearest plate boundary, volcano(es) and/or volcanic feature(s), fault(s), and earthquake(s) are within the map extent. Print (or take a screengrab of) the map as a landscaped image and copy/insert the map image into the space below.
2. What tectonic plate is your building on (textbook Fig. 2.14)?
3. Measure the distance from your building to the nearest plate boundary:
(don’t forget units!)
4. What type of plate boundary is this?
5. How many volcanoes or volcanic features are within 25 mi of your building?
6. If you answered ‘0’ to the previous question, answer this question, otherwise, skip this question. How many volcanoes or volcanic features are within 50 mi of your building?
7. If you answered ‘0’ to the previous question, answer this question; otherwise, skip this question. How many volcanoes or volcanic features are within 100 mi of your building?
8. What type(s) of volcanoes are near your building?
(read information in data pop-up window, look at Description, FeatureTyp, Volcan_Type; if there are no volcanoes within 100 mi of your building, answer for the closest volcano.)
9. The volcano(es) are FORMCHECKBOX Active / FORMCHECKBOX Dormant / FORMCHECKBOX Extinct / FORMCHECKBOX Unknown
(read information in data pop-up window, look at Last Erupt or Notes field)
10. Complete the risk assessment table for your building with regards to volcanoes and volcanic features.
· If you have volcanoes or other volcanic features near your building (<25 mi) the probability that the hazard will affect your building is most likely high; conversely, if volcanoes are far away (> 50 mi) then the probability is low. Investigate the data and choose a numeric probability value to low (0.1 – 0.3), medium (0.4 – 0.6), and high (0.7 – 0.9) (see directions on first page and table footnotes).
· NOTE: you must choose a single number within the range; for example, if you decide the probability of the hazard affecting your building is low, then think about the proximity and number of occurrences of that hazard and then choose a value between 0.1 – 0.3, such as 0.2. In this example, you would then put 0.2 in the ‘Probability’ column of the table.
· Investigate the data (e.g., type of plate boundary, type of volcano and associated eruptive features) and consider the severity, or potential damage, from the volcano(es) and volcanic features. Decide whether the impact to your building would be low, medium, or high. Then, assign a numeric value based on your low (1-3), medium (4-6), high (7-10) description.
· Again, you must choose a single number within the range. For example, if you think there would not be much damage, then choose a value within the low range, such as 1. Then put that number in the ‘Severity’ column of the table.
· Lastly, calculate the risk by multiplying the probability value x the severity value. Following the previous examples, 0.2 x 1 = 0.2. Then you would enter 0.2 in the ‘Risk’ column of the table.
Hazard1 Probability2 Severity3 Risk4
1 – List the hazard you are assessing (e.g., volcano, earthquake, tornado, etc.)
2 – Evaluate probability of hazard affecting your building on a scale from 0.1 – 0.9. For example, a low probability of the hazard affecting your building might be assigned a value between 0.1 and 0.3.
– Evaluate the impact severity of the hazard on a scale from 1 – 10. For example, a value between 1 and 3 might be Low (negligible, little damage) impact.
4 – Calculate the risk, Risk = Probability Value x Severity Value
11. Measure the distance from your building to the nearest fault:
(don’t forget units!)
12. What type of fault is it (normal, reverse, strike-slip)?
(read information in data pop-up window, look at SlipSense field)
13. How many earthquakes have occurred within 25 mi of your building?
14. If you answered ‘0’ to the previous question, answer this question, otherwise, skip this question. How many earthquakes have occurred within 50 mi of your building?
15. If you answered ‘0’ to the previous question, answer this question; otherwise, skip this question. How many earthquakes have occurred within 100 mi of your building?
16. Of the earthquakes near your building, what is the largest magnitude?
(read information in data pop-up window, look at magnitude field)
17. Complete the risk assessment table for your building with regards to earthquakes.
· If you have mapped faults and/or earthquakes near your building (<25 mi) the probability that the hazard will affect your building is most likely high; conversely, if faults and earthquakes are mostly far away (> 50 mi) then the probability is low. Investigate the data and assign a numeric probability value, such as low (0.1 – 0.3), medium (0.4 – 0.6), or high (0.7 – 0.9) (see directions on first page and first table footnotes).
· Investigate the data (e.g., type of plate boundary, type of faults and magnitude of earthquakes) and consider the severity, or potential damage, from earthquakes. Decide whether the impact to your building would be low, medium, or high. Then, assign a numeric value based on your low (1-3), medium (4-6), or high (7-10) description.
Hazard1 Probability2 Severity3 Risk4
III. Severe Storms: Hail and Tornadoes
Use the Project, Part 1 WebGIS to answer the questions below.
18. Turn off all layers except the tornado and hail layers. Type your building’s address into the webmap search bar to zoom in to your building. Leave the webmap’s location pop-up window open to “point” to the location of your building, then zoom out until the nearest hail storm(s) and tornado(es) are within the map extent. Print (or take a screengrab of) the map as a landscaped image and copy/insert the map image into the space below.
19. How many hail storms have occurred within 25 mi of your building?
20. If you answered ‘0’ to the previous question, answer this question, otherwise, skip this question. How many hail storms have occurred within 50 mi of your building?
21. If you answered ‘0’ to the previous question, answer this question; otherwise, skip this question. How many hail storms have occurred within 100 mi of your building?
22. Of the hail storms near your building, what is the largest diameter of hail?
(data’s size field, don’t forget units!)
23. Complete the risk assessment table for your building with regards to hail storms.
· If historic hail storms have occurred near your building (<25 mi) the probability that the hazard will affect your building is most likely high; conversely, if hail storms are mostly far away (> 50 mi) then the probability is low. Investigate the data and assign a numeric probability value, such as low (0.1 – 0.3), medium (0.4 – 0.6), or high (0.7 – 0.9) (see directions on first page and first table footnotes).
· Investigate the data (e.g., size of hail, frequency of hail storms) and consider the severity, or potential damage, from hail storms. Decide whether the impact to your building would be low, medium, or high. Then, assign a numeric value based on your low (1-3), medium (4-6), or high (7-10) description.
Hazard1 Probability2 Severity3 Risk4
24. How many tornadoes have occurred within 25 mi of your building?
25. If you answered ‘0’ to the previous question, answer this question, otherwise, skip this question. How many tornadoes have occurred within 50 mi of your building?
26. If you answered ‘0’ to the previous question, answer this question; otherwise, skip this question. How many tornadoes have occurred within 100 mi of your building?
27. Of these tornadoes, what is the highest (E)F scale?
(data’s Fscale field)
28. What is the wind speed of that category (textbook Table 10.2)?
(don’t forget units!)
29. What is the damage associated with that category (textbook Table 10.2)?
30. Complete the risk assessment table for your building with regards to tornadoes.
· If tornadoes have occurred near your building (<25 mi) the probability that the hazard will affect your building is most likely high; conversely, if tornadoes have been mostly far away (> 50 mi) then the probability is low. Investigate the data and assign a numeric probability value, such as low (0.1 – 0.3), medium (0.4 – 0.6), or high (0.7 – 0.9) (see directions on first page and first table footnotes).
· Investigate the data (e.g., tornado scale and associated damage) and consider the severity, or potential damage, from tornadoes. Decide whether the impact to your building would be low, medium, or high. Then, assign a numeric value based on your low (1-3), medium (4-6), or high (7-10) description.
Hazard1 Probability2 Severity3 Risk4
IV. Wind and Fire
Use the Project, Part 1 WebGIS to answer the questions below.
31. Turn off all layers except the strong winds and fires layers. Type your building’s address into the webmap search bar to zoom in to your building. Leave the webmap’s location pop-up window open to “point” to the location of your building, then zoom out until the nearest wind swath(s) and wildfire(s) are within the map extent. Print (or take a screengrab of) the map as a landscaped image and copy/insert the map image into the space below
32. How many strong wind gusts have occurred within 25 mi of your building?
33. If you answered ‘0’ to the previous question, answer this question, otherwise, skip this question. How many strong wind gusts have occurred within 50 mi of your building?
34. If you answered ‘0’ to the previous question, answer this question; otherwise, skip this question. How many strong wind gusts have occurred within 100 mi of your building?
35. What is the strength of the strongest wind gust?
(data’s magnitude field, don’t forget units!)
36. Complete the risk assessment table for your building with regards to strong wind gusts.
· If strong winds have occurred near your building (<25 mi) the probability that the hazard will affect your building is most likely high; conversely, if strong winds occur mostly far away (> 50 mi) then the probability is low. Investigate the data and assign a numeric probability value, such as low (0.1 – 0.3), medium (0.4 – 0.6), or high (0.7 – 0.9) (see directions on first page and first table footnotes).
· Investigate the data (e.g., strength of wind gusts) and consider the severity, or potential damage, from strong winds. Decide whether the impact to your building would be low, medium, or high. Then, assign a numeric value based on your low (1-3), medium (4-6), or high (7-10) description.
Hazard1 Probability2 Severity3 Risk4
37. How many fires have occurred within 25 mi of your building?
38. If you answered ‘0’ to the previous question, answer this question, otherwise, skip this question. How many fires have occurred within 50 mi of your building?
39. If you answered ‘0’ to the previous question, answer this question; otherwise, skip this question. How many fires have occurred within 100 mi of your building?
40. Complete the risk assessment table for your building with regards to historic fires.
· If fires have occurred near your building (<25 mi) the probability that the hazard will affect your building is most likely high; conversely, if fires have occurred mostly far away (> 50 mi) then the probability is most likely low. Investigate the data and assign a numeric probability value, such as low (0.1 – 0.3), medium (0.4 – 0.6), or high (0.7 – 0.9) (see directions on first page and first table footnotes).
· Investigate the data (e.g., size of fires, cause of fires, topography, locations of fuel sources near your building) and consider the severity, or potential damage, from fires. Decide whether the impact to your building would be low, medium, or high. Then, assign a numeric value based on your low (1-3), medium (4-6), or high (7-10) description.
Hazard1 Probability2 Severity3 Risk4
V. Plot
Complete the chart of the two measures your assessed in this exercise: probability and severity. The probability that a hazard will occur near your building is represented on y-axis of the chart and the severity of impact on the x-axis. As an emergency manager, insurance assessor, or home owner, this chart will give you a quick, clear view of the priority that you need to give the evaluated hazards.
41. For each hazard you evaluated, draw a point in the correct location representing the probability (y-axis) and severity (x-axis) you determined. For example, a probability of 0.2 and a severity of 1 will plot at y=0.2 and x=1 in the lower left portion of the graph. Create a legend on the right side of the plot explaining which points belong to which hazard. Hint: One way to make a legend might be to have each point (each hazard) be a different color.
Remove the empty plot and copy/insert your completed plot in the space below
42. For each hazard you evaluated, draw a bar in the graph below representing the risk value you calculated. For example, for a volcanic risk of 0.2 draw a bar above the “Volcanoes” label that reaches up to 0.2 on the y-axis.
Remove the empty plot and copy/insert your completed plot in the space below.
43. Consider the graphs, which hazard is most dangerous to your building?
44. What resources you will allocate or renovations would you make to manage the risk associated with the most dangerous hazard you identified in the previous question. You will be graded on the quality of your response and the clarity with which you communicate your thoughts. | 3,932 | 16,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-10 | latest | en | 0.885648 |
https://www.xpmath.com/forums/showthread.php?s=920a607009bf7ca82a6caa0b7e56f425&t=4335&page=20 | 1,656,983,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00075.warc.gz | 1,125,355,811 | 13,189 | counting game - Page 20 - XP Math - Forums
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12-23-2011 #193 JosePereira11 Join Date: May 2011 Posts: 124 183 __________________ You need only 2 objectives in life: when you look at the past and you say that you coudn't have done any better than that is the 1º one. The 2º one is to do better than that Then, you must learn 2 rules in your life: Rule number one : respect yourself and the others Rule number two: first learn rule number one
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12-23-2011 #195 JosePereira11 Join Date: May 2011 Posts: 124 185 __________________ You need only 2 objectives in life: when you look at the past and you say that you coudn't have done any better than that is the 1º one. The 2º one is to do better than that Then, you must learn 2 rules in your life: Rule number one : respect yourself and the others Rule number two: first learn rule number one
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Contact Us - XP Math - Forums - Archive - Privacy Statement - Top | 629 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.79386 |
https://www.thedataschool.co.uk/lorna-eden/ds2tipweek-lods-of-fun-with-custom-dates/ | 1,643,334,927,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305341.76/warc/CC-MAIN-20220128013529-20220128043529-00602.warc.gz | 1,038,411,895 | 110,687 | ## #DS2TipWeek - LODs of Fun with Custom Dates
by Lorna Brown
Today’s tip involves two handy little tricks I learnt whilst teaching Level of Detail (LOD’s) to DS2.
So the scenario is- you want to see the average sales by month and year but we also want to define the average sales specifically per year, which then can enable you to see which monthly average sales are above or below the yearly average sales. Sounds complicated right? Let me break it down in simple steps.
Let’s create the first view, drag AVG(sales) to columns, discrete year(order date) to rows and discrete month(order date) also to rows. So this view shows the average sales per month per year.
Now if we want to find just the year’s average then we need an LOD calc. However we cannot use Year(Order Date) in an LOD calc. So we have to create a custom date. Go to order date, click Jeremy (Triangle), create, custom date, change detail to years. This now creates a dimension that can be used in a calculated field.
Now lets create an LOD calc that just focuses on the average yearly sales. This equation is fixing the average sales based on the year custom date.
If we put this new calculated field onto detail, and create a reference line per pane you can see the average sales per year for each year.
Just to make the chart more visual, we can create a final calculated field that shows the months that are above or below the average yearly sales.
Then drag this new calculated field onto colour and you can see the months that were above or below the average sales for that year
I tried to use a real world situation. If you have any problems or need a different scenario to work from please let me know!
Lorna Brown
Fri 19 Feb 2016
Tue 16 Feb 2016 | 384 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.920162 |
https://cs.nyu.edu/pipermail/fom/2007-October/012152.html | 1,708,742,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00731.warc.gz | 189,855,701 | 2,323 | # [FOM] Does 2^{\aleph_0} = 2^{\aleph_1}?
James Hirschorn James.Hirschorn at univie.ac.at
Sun Oct 28 16:20:56 EDT 2007
```On Friday 26 October 2007 13:01, joeshipman at aol.com wrote:
> You asked about the result that RVM implies 2^{\aleph_0} =
> 2^{\aleph_1}. I don't know who first proved this, but the stronger
> result that RVM implies no cardinal smaller than the continuum has a
> power set larger than the continuum was proved by Prikry (Bulletin of
> the AMS 81 (1975), 907-909). You can find a short proof online in
> section 5E of Fremlin's monograph "Real-Valued Measurable Cardinals",
>
> http://www.essex.ac.uk/maths/staff/fremlin/rvmc/index.htm
Thanks for the helpful reference.
I think there was some confusion about the statement RVM. Please correct me if
I am wrong, but I believe that by RVM you mean that "2^{aleph-0} is a
real-valued measurable cardinal". There is apparently some precedence for
this, but I'm not sure when or by whom the acronym RVM was first used.
By RVM I meant "Lebesgue measure can be extended to all subsets of the real
line", or equivalently "there exists an atomlessly measurable cardinal". This
was the definition of RVM I used in my recent paper:
http://arxiv.org/abs/math/0604085v3
I was unaware of the precedent for RVM when I wrote it (even though I had
heard of the acronym), and the latter RVM seemed to be the best candidate for
an axiomatization of random forcing.
While Prikry's result entails that the former RVM implies 2^{aleph-0} =
2^{aleph-1}, clearly the latter RVM does not: Starting with GCH and a
measurable cardinal kappa, and adding kappa^{+aleph-(omega-1)} (i.e. the
aleph-(omega-1)'th successor of kappa) many random reals gives a
counterexample.
James Hirschorn
``` | 505 | 1,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.925509 |
https://metinmediamath.wordpress.com/tag/zipf/ | 1,638,838,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00223.warc.gz | 453,881,405 | 17,914 | # Distribution of E-Book Sales on Amazon
For e-books on Amazon the relationship between the daily sales rate s and the rank r is approximately given by:
s = 100,000 / r
Such an inverse proportional relationship between a ranked quantity and the rank is called a Zipf distribution. So a book on rank r = 10,000 can be expected to sell s = 100,000 / 10,000 = 10 copies per day. As of November 2013, there are about 2.4 million e-books available on Amazon’s US store (talk about a tough competition). In this post we’ll answer two questions. The first one is: how many e-books are sold on Amazon each day? To answer that, we need to add the daily sales rate from r = 1 to r = 2,400,000.
s = 100,000 · ( 1/1 + 1/2 + … + 1/2,400,000 )
We can evaluate that using the approximation formula for harmonic sums:
1/1 + 1/2 + 1/3 + … + 1/r ≈ ln(r) + 0.58
Thus we get:
s ≈ 100,000 · ( ln(2,400,000) + 0.58 ) ≈ 1.5 million
That’s a lot of e-books! And a lot of saved trees for that matter. The second question: What percentage of the e-book sales come from the top 100 books? Have a guess before reading on. Let’s calculate the total daily sales for the top 100 e-books:
s ≈ 100,000 · ( ln(100) + 0.58 ) ≈ 0.5 million
So the top 100 e-books already make up one-third of all sales while the other 2,399,900 e-books have to share the remaining two-thirds. The cake is very unevenly distributed.
This was a slightly altered excerpt from More Great Formulas Explained, available on Amazon for Kindle. For more posts on the ebook market go to my E-Book Market and Sales Analysis Pool. | 446 | 1,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-49 | latest | en | 0.913114 |
https://www.hpmuseum.org/forum/showthread.php?mode=threaded&tid=1739&pid=15556 | 1,624,430,051,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488534413.81/warc/CC-MAIN-20210623042426-20210623072426-00223.warc.gz | 711,944,970 | 5,675 | Fractional Part - A Difficult Integral
06-29-2014, 11:36 AM
Post: #1
Gerald H Senior Member Posts: 1,457 Joined: May 2014
Fractional Part - A Difficult Integral
Here some results for integrating FP from zero to 5.8:
Maple V Release 3.0
evalf(int(frac(x), x=0..5.8)) 2.820001887
Maxima 5.23.2
8,3171,0]
PARI 2.4.2
intnum(x=0,5.8,frac(x)) 2.770657207613055496607250297
WolframAlpha 2011-5-28
integrate(frac(x),x,0,5.8) 2.82
CASIO fx-9860GII (& fx-5800P)
∫_0^5.8▒〖Frac X〗 dx 2.72
HP 42S
Gauss-Lagrange 16 point:
Divisions: 1 2.32
2 2.71441362769
3 2.95110930932
4 2.84122268726
5 2.72066778956
6 2.8405122545
HP 50G
Inbuilt integration programme: ∫_0^5.8▒〖FP(X)〗 dX
FIX 6 2.820116 in 1,232.15 seconds
Gauss-Lobatto 4 point formula with 7 and 13 point Kronrod extensions:
FIX 2 2.80
3 2.819
4 2.8199
5 2.82000
6 2.820000 in 12.34 seconds
SHARP EL-9650
∫_0^5.8▒〖fpart X〗 dxX 2.419454779
TI-84 Plus 2.53MP
fnInt(fPart(X),X,0,5.8) 2.720014333
TI-86 (tol = 1E-5)
fnInt(fPart x,x,0,5.8) 2.71992436738
TI-89 (& voyage 200 & 92(plus))
∫_0^5.8▒〖fPart (x)〗 dx 2.72000000036
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Messages In This Thread Fractional Part - A Difficult Integral - Gerald H - 06-29-2014 11:36 AM RE: Fractional Part - A Difficult Integral - Massimo Gnerucci - 06-29-2014, 11:55 AM RE: Fractional Part - A Difficult Integral - pito - 06-29-2014, 05:18 PM RE: Fractional Part - A Difficult Integral - kakima - 06-29-2014, 06:17 PM RE: Fractional Part - A Difficult Integral - Gerald H - 06-29-2014, 07:57 PM RE: Fractional Part - A Difficult Integral - Thomas Klemm - 06-29-2014, 06:47 PM RE: Fractional Part - A Difficult Integral - pito - 06-29-2014, 09:40 PM RE: Fractional Part - A Difficult Integral - ttw - 06-30-2014, 01:17 PM
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MTH302 Assignmnet#1 Fall 2010 Solution & Discussion Due Date:28-10-2010
Assignment # 1
MTH302 (Fall 2010)
Due Date:28-10-2010
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files sent by students in assignment 1 are found to be copied and so
awarded zero. You are therefore reminding here again.
• Above all instructions are for all assignments so may not be mentioned in future.
• There are 7 questions in the assignment but only one question will be graded. However we are not mentioning that which question will be graded so you
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• Solve the assignment on MS word document and upload your word (.doc) files only. Do not solve the assignment on MS excel. If we get any assignment on MS excel or any format
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Question 1:
If the basic salary of an employee is Rs. 37000 and allowances are Rs. 22,000.
What is the taxable income of employee?
Question 2:
If the salary and allowances of an employee is as follows:
Basic salary = Rs. 12,000
House Rent Allowance = 45%
Conveyance Allowance = 5%
Utilities Allowance = 7%
Group Insurance/Medical = 0%
Misc. Social Charges = 5.8%
There are 12 casual, 24 earned and no sick leaves per year where as normal working days per month are 26. Find the Gross remuneration of the employee.
Question 3:
A trade discount series of 20, 10, and 5 is offered on an item which has a list price of Rs. 9100. Find the amount of discount and the net price.
Question 4:
If your goal is to have an amount of 325000 in seven years and you can get interest rate of 9% per annum compounded annually. How much would you need to invest now?
Question 5:
If you start saving Rs. 28,000 at the end of each six month, and you get interest rate 11% (per annum) compounded semi-annually, how much will you have accumulated at the end of 20 years?
Question 6:
The price of an item decreased from Rs. 856 to Rs.752. What is the percentage change in price of item?
Question # 7
Suppose you can afford to pay 9,000 per month on a loan. How much can you borrow if the loan is for a period of 15 years and the interest rate is 5%
per annum compounded monthly?
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MTH302 Assignment#1 Solution Fall 2010
See the attached file pls
don't just copy & paste otherwise u will be get zero marks..
thanks
Attachments:
Thanku Sir. Ya won;t copy just getting an idea.
Regards.
Ans no.1
basic salary=37000
allownces=22000
%allownces =22000/37000*100 =59.45946
=59.5% - 50% =9.45946
= 9.45946*37000 / 100 =350000.02/100=3500
=3500
hence RS 3500 of allownces are taxable income ..
chkk dis1......
Q # 1 ka Answer 3,500 nahi..........
3,500 Plus 37,000 = 40,500 Total Taxable Income hai........
plz Q.2 ka soluion bta dyn...........
1
2
3
4
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HI I'm unable to get this CR
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the argument is that because we know what temp/pressure it takes to create a fullereness in the lab, and that fullereness is found naturally, we can conclude that similar temp/pressure must have happened at some point on earth to form fullereness.
option D is the only one that weakens this because, if the fullereness found naturally is a of a different structural makeup, it means that we no longer know the temperature/pressure it takes to form natural fullereness, and thus can't use it as a test case for geologists.
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# 3.3 Model Training with Stochastic Gradient Descent (Part 1-4)
What we covered in this video lecture
This lecture introduced the training algorithm behind logistic regression: stochastic gradient descent. This is the same training algorithm we use for training deep neural networks.
Stochastic gradient descent is based on calculus: we compute the loss function’s derivatives (or gradients) with respect to the model weights. Why? The loss measures “how wrong” the predictions are. And the gradient tells us how we have to change the weights to minimize (improve) the loss.
The loss is correlated to the accuracy, but sadly, we cannot optimize the accuracy directly using stochastic gradient descent. That’s because accuracy is not a smooth function.
Computing the loss gradients is based on the chain rule from calculus, and if you are not familiar with it, it may look daunting at first. But do not worry. We will introduce PyTorch functions that can handle the differentiation (that is, the calculation of the gradients) automatically for us. This is known as automatic differentiation or autograd.
The following lecture introduces PyTorch functionality that calculates the gradients automatically for us. However, if you are new to calculus or need a refresher and you want to learn more (not required for this course), I have written a concise calculus primer that you might find helpful: Calculus and Differentiation Primer.
Moreover, if you are interested in an alternative introduction to stochastic gradient descent, you may find my article Single-Layer Neural Networks and Gradient Descent helpful.
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